MAAGardner cover v7POD_Layout 1 11/19/12 4:10 PM Page 1
Martin Gardner in the TwentyFirst Century
Martin Gardner in the TwentyFirst Century
Michael Henle and Brian Hopkins, Editors
Michael Henle and Brian Hopkins, Editors
Martin Gardner in the TwentyFirst Century
Martin Gardner enormously expanded the field of recreational mathematics with the Mathematical Games columns he wrote for Scientific American for over 25 years and the more than 70 books he published. He also had a long relationship with the Mathematical Association of America, publishing articles in the MAA journals right up to his death in 2010. This book collects articles Gardner wrote for the MAA in the twentyfirst century, together with other articles the MAA published from 1999 to 2012 that spring from and comment on his work.
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Michael Henle and Brian Hopkins, Editors
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Martin Gardner in the TwentyFirst Century
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c 2012 by the Mathematical Association of America, Inc. Library of Congress Catalog Card Number 2012954077 Print edition ISBN: 9780883859131 Electronic edition ISBN: 9781614448013 Printed in the United States of America Current Printing (last digit): 10 9 8 7 6 5 4 3 2 1
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Martin Gardner in the TwentyFirst Century
edited by Michael Henle Oberlin College and Brian Hopkins Saint Peter’s University
Published and Distributed by The Mathematical Association of America
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Preface Martin Gardner and the MAA share a long history. In 1958, around the time he started his famous “Mathematical Games” column for Scientific American, he submitted the first of many problems to The American Mathematical Monthly. In 1982, as his column wound down, Gardner’s first MAA article was published in The Two Year College Mathematics Journal. He wrote for MAA journals the rest of his life, particularly The College Mathematics Journal and Math Horizons. Gardner contributed to the latter almost annually from its founding in 1993 until 2005. Gardner’s prodigious writing activity continued right until his death in 2010. Articles, stories, problems, solutions, Quickies, and other kinds of contributions continued to flow. His last mathematical article to appear in an MAA journal, “Ltromino Tiling of Mutilated Chessboards,” was the centerpiece of a special puzzle issue of The College Mathematics Journal in 2009, and it is included here. Early in 2010, Math Horizons editors Steve Abbott and Bruce Torrence were surprised to receive a typescript manuscript. Gardner used a typewriter his whole life, never email. The submission was accompanied by a note, “Is this short story something you can use? I wrote the math column in Scientific American for 25 years. If my piece is not right for Math Horizons, there is no need to send it back. All best, Martin.” There was not enough time for the editors to thank Martin for his submission [6]. Fittingly, this story, “Superstrings and Thelma,” is the last piece in this collection. Apart from his own work, Martin Gardner, by enormously expanding the field of recreational mathematics, opened up vast mathematical tracts for exploration by others. This was quite deliberate. In an interview in The TwoYear College Mathematics Journal [1], Gardner said, “I’m defining [recreational mathematics] in the very broad sense to include anything that has a spirit of play about it.” Gardner, of course, had a refined and very welldeveloped sense of play, one quality that made his pieces so enjoyable to read. In almost everything he wrote, Gardner posed problems to challenge his readers, and they responded. He maintained an extensive correspondence with mathematicians, both professional and amateur. Their work fueled his own pieces, but then his correspondents turned around and wrote their own articles. One consequence was the expansion of recreational mathematics into a major research area (also helped by the development of the computer and the corresponding expanded interest in discrete mathematics) that is such a feature of the current mathematical landscape. The CMJ devoted the January 2012 issue to papers on topics that Gardner introduced to the mathematical public. There were so many articles to include that half of the March 2012 issue continued the theme. v
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Preface
Another consequence of the flowering of recreational mathematics is this volume. We have collected MAA journal articles starting from 1999 on topics that Gardner developed. Some are written by Gardner, but most are by others. The tribute articles from the January and March 2012 CMJ issues are all here, but they constitute less than half of this collection. All the MAA journals are represented, Mathematics Magazine and the Monthly, as well as CMJ and Math Horizons. The limitation to pieces published roughly in the twentyfirst century is a practical one. Even so, some puzzle collections, longer articles, and pieces less directly linked to Martin Gardner have been omitted. The 41 pieces collected here are grouped around common themes, such as geometry, number theory and graph theory, and cards and probability. Flexagons, the topic of Gardner’s first Scientific American column, are seen to be associated with Catalan numbers and together merit their own section. Geometric tiling and various “magic” number puzzles are all about “Making Things Fit,” and there are enough other puzzles and games to fill another section. Gardner’s interests ranged far beyond mathematics. A fan of magicians and magic tricks from childhood (“I waste a lot of time on it” [2]), he wrote several books on magic. He annotated Lewis Carroll’s Alice books and other classics, and produced two novels of his own. Other topics included philosophy, religion, literature and pseudoscience, leading to some 70 books. The last section of this volume highlights some of these other facets of Martin Gardner’s wideranging interests. It includes two short stories by Gardner and several other pieces that demonstrate his support for amateur mathematicians, his love of play (about an April Fool’s joke he played on his Scientific American readers), and his interest in debunking false science. Also included is Gardner’s review of a 2004 novel in which an important character seems to be based on him. Our hope is that this volume will play a role in perpetuating the memory of Martin Gardner, modest celebrity, largerthanlife character, selfconfessed amateur as a mathematician, popularizer of recreational mathematics in the broadest sense, prolific and brilliant writer. Given the lasting impression he made on several generations of mathematics enthusiasts of all backgrounds, we are confident that the MAA and others will be publishing articles inspired by Gardner’s work for a long time. Gardner, like the readers of this book, loved mathematics. We close this preface with Gardner’s own words on his background, the community, and why he enjoyed the field so much (from [5], [4], and [3], respectively). I took no math in college. I’m like a person who loves music but can’t hum a tune or play an instrument. My understanding of math does not go beyond a minimal understanding of calculus. I hasten to add that I consider this one reason for the success of my Scientific American column. I had to work hard to understand whatever I wrote about, and this made it much easier for me to write in a way that readers could understand. When I first started the column, I was not in touch with any mathematicians, and gradually mathematicians who were creative in the field found out about the column and began corresponding with me. So my most interesting columns were columns based on material that I got from them, so I owe them a big debt of gratitude.
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Preface
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[I enjoy mathematics] because it has a strange kind of unearthly beauty. There is a strong feeling of pleasure, hard to describe, in thinking through an elegant proof, and even greater pleasure in discovering a proof not previously known.
Aclnowledgments The editors are grateful for the encouragement and hard work of the MAA publications staff: Director of Publications Ivars Peterson, who conceived the book, Production Manager Carol Baxter, who designed it, and especially Electronic Publication Manager Beverly Joy Ruedi, who set the book in type.
Bibliography [1] Anthony Barcellos and Martin Gardner, A Conversation with Martin Gardner, TwoYear College Math. J. 10 (1979) 233–244. [2] Don Albers and Martin Gardner, On the Way to “Mathematical Games”: Part I of an Interview with Martin Gardner, College Math. J. 36, (2005) 178–190. [3] Don Albers and Martin Gardner, “Mathematical Games” and Beyond: Part II of an Interview with Martin Gardner, College Math. J. 36 (2005) 301–314. [4] Colm Mulcahy and Martin Gardner, An Interview with Martin Gardner, Card Colm, October 2006, available at www.maa.org/columns/colm/cardcolm200610.html [5] Michael Henle and Martin Gardner, Interview with Martin Gardner, College Math. J. 40 (2009) 158–161. [6] Bruce Torrence and Stephen Abbott, To Our Readers, Math Horizons 18 (2010) 2–4.
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Contents Preface
v
I Geometry
1
1 The Asymmetric Propeller Martin Gardner
3
Gardner, paying tribute to dentist and geometer Leon Bankoff, discusses some of his unpublished results and concludes with an open question.
2 The Asymmetric Propeller Revisited Gillian Saenz, Christopher Jackson, and Ryan Crumley
7
Three University of Texas students use dynamic geometry software to confirm Bankoff’s results and resolve Gardner’s question.
3 Bracing Regular Polygons As We Race into the Future Greg W. Frederickson
11
A problem Gardner published in 1963 continues to spur generalizations and improved solutions around the world.
4 A Platonic Sextet for Strings Karl Schaffer
19
The professor and dance company codirector details string polyhedra constructions for ten participants.
5 Prince Rupert’s Rectangles Richard P. Jerrard and John E. Wetzel
25
A 17th century puzzle that Gardner posed in higher dimensions is here solved in the case of threedimensional boxes.
II
Number Theory and Graph Theory
6 Transcendentals and Early Birds Martin Gardner
35 37
Gardner moves from Liouville to an “innocent but totally useless amusement” that nonetheless captured the attention of Solomon Golomb.
7 Squaring, Cubing, and Cube Rooting Arthur T. Benjamin
39
The professor and “mathemagician,” inspired as a high school student by Gardner’s tricks for mental calculations, extends some of them here.
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Contents
8 Carryless Arithmetic Mod 10 David Applegate, Marc LeBrun, and N. J. A. Sloane
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Inspired by the carryless arithmetic of the game Nim, this trio of authors explores the number theory of a South Pacific island.
9 Mad Tea Party Cyclic Partitions Robert Bekes, Jean Pedersen, and Bin Sha
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Another playful trio analyzes cyclic arrangements that build from integer partitions in a Lewis Carroll setting.
10 The Continuing Saga of Snarks sarahmarie belcastro
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A type of graph, given a fanciful name by Gardner from Lewis Carroll, was the subject of a Branko Gr¨unbaum conjecture for 39 years.
11 The MapColoring Game Tomasz Bartnicki, Jaroslaw Grytczuk, H. A. Kierstead, and Xuding Zhu
73
Daltonism and halfdollar coins are used in this exploration of a Steven Brams game theory approach to the Four Color Theorem.
III
Flexagons and Catalan Numbers
12 It’s Okay to Be Square If You’re a Flexagon Ethan J. Berkove and Jeffrey P. Dumont
85 87
This article details the 1939 origin of flexagons at Princeton University and focuses on the neglected tetraflexagons.
13 The Vflex, Triangle Orientation, and Catalan Numbers in Hexaflexagons Ionut E. Iacob, T. Bruce McLean, and Hua Wang
103
This trio of Georgia Southern University authors examines a onceillegal variety of flex and makes a connection between “pat classes” and Catalan numbers.
14 From Hexaflexagons to Edge Flexagons to Point Flexagons Les Pook
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An engineer and author of two books on flexagons considers the more general edge flexagons and recently discovered point flexagons.
15 Flexagons Lead to a Catalan Number Identity David Callan
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Examining the descent permutation statistic on flexagon pats leads the author to full binary trees and a combinatorial proof.
16 Convergence of a Catalan Series Thomas Koshy and Z. Gao
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Calculus is brought to bear on the infinite sum of Catalan number reciprocals and related series; and the golden ratio make appearances.
IV
Making Things Fit
17 LTromino Tiling of Mutilated Chessboards Martin Gardner
125 127
In his last MAA mathematics article, Gardner moves from classic chessboard domino tiling problems to new results.
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Contents 18 Polyomino Dissections Tiina Hohn and Andy Liu
xi 135
The authors introduce a new technique for solving dissection problems, often presented in the context of quilts, leaving several puzzles for the reader.
19 Squaring the Plane Frederick V. Henle and James M. Henle
143
A father and son team resolve Golomb’s “heterogenous tiling conjecture” and discuss another dozen open questions.
20 Magic Knight’s Tours John Beasley
153
The author surveys results combining a knight’s tour on the chessboard with magic squares, including a computeraided solution to a Gardner question.
21 Some New Results on Magic Hexagrams Martin Gardner
159
Here Gardner focuses on three types of puzzles about placing numbers on sixpointed stars, mentioning a “rare mistake” of the British puzzlist Henry Dudeney.
22 Finding All Solutions to the Magic Hexagram Alexander Karabegov and Jason Holland
167
The authors relate magic hexagrams to magic edge labelings of cubes, using card shuffling to enumerate distinct solutions.
23 Triangular Numbers, Gaussian Integers, and KenKen John J. Watkins
173
Miyamoto’s contemporary puzzle is expanded to complex numbers where a different unique factorization adds to the challenge.
V
Further Puzzles and Games
24 Cups and Downs Ian Stewart
179 181
One of Gardner’s mathematical successors at Scientific American uses graph theory and linear algebra on two related puzzles.
25 30 Years of Bulgarian Solitaire Brian Hopkins
187
Some recent math history explains this oddlynamed puzzle on integer partitions, visualized with state diagrams and generalized to a new twoplayer game.
26 Congo Bongo HsinPo Wang
195
A high school student uses state diagrams and Dennis Shasha’s detectives to open a tricky treasure chest.
27 Sam Loyd’s Courier Problem with Diophantus, Pythagoras, and Martin Gardner Owen O’Shea
201
A Classroom Capsule extends Gardner’s solution of related Sam Loyd puzzles to other army formations.
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Contents
28 Retrolife and The Pawns Neighbors Yossi Elran
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An inverse version of Conway’s game Life, famously popularized by Gardner, is examined using chessboards.
29 R ATWYT Aviezri Fraenkel
213
The combinatorial game theorist uses the Calkin Wilf tree to devise a rational number version of Wythoff’s Nim.
VI Cards and Probability
219
30 Modeling Mathematics with Playing Cards Martin Gardner
221
In addition to probability applications, Gardner uses a deck of cards for a discrete version of a fluid mixing puzzle and mentions a correction to W. W. Rouse Ball.
31 The Probability an Amazing Card Trick Is Dull Christopher N. Swanson
227
Rook polynomials and the principle of inclusionexclusion help determine the likelihood that the author’s students were underwhelmed.
32 The Monty Hall Problem, Reconsidered Stephen Lucas, Jason Rosenhouse, and Andrew Schepler
231
These authors remind us of Gardner’s early role in this infamous problem that still “arouses the passions” and examine variations.
33 The Secretary Problem from the Applicant’s Point of View Darren Glass
243
Changing perspective, the author reconsiders a classic strategy in order to help job seekers choose the best interview slot.
34 Lake Wobegon Dice Jorge Moraleda and David G. Stork
249
Lake Wobegon Dice, named after Garrison Keillor’s Minnesota town, have the property that each is “better than the set average.”
35 Martin Gardner’s Mistake Tanya Khovanova
257
Another controversial problem about probability and information is carefully discussed, putting Gardner in the company of Dudeney and Ball.
VII
Other Aspects of Martin Gardner
36 Against the Odds Martin Gardner
263 265
In this short story, a principal recognizes the potential in a student whose unconventional thinking irritates his teacher.
37 A Modular Miracle John Stillwell
271
Gardner used an obscure result of Hermite and the limitations of 1970’s calculators for an April Fool’s Day prank.
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38 The Golden Ratio—A Contrary Viewpoint Clement Falbo
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Building on a Gardner article in The Skeptical Inquirer, the author argues that “is not entirely astonishing.”
39 Review of The Mysterious Mr. Ammann by Marjorie Senechal Philip Straffin
285
This Media Highlight discusses an example of Gardner’s support of an amateur mathematician who independently discovered Penrose tiles.
40 Review of PopCo by Scarlett Thomas Martin Gardner
287
This popular 2004 novel includes a character based on Gardner, so he was a natural choice to review the book.
41 Superstrings and Thelma Martin Gardner
289
Gardner’s last MAA submission, a short story about a physics graduate student and a waitress who quips, “How are strings?”
Index
293
About the Editors
297
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I Geometry
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1 The Asymmetric Propeller Martin Gardner The late Leon Bankoff (he died in 1997) was a Beverly Hills, California, dentist who also was a world expert on plane geometry. (For G. L. Alexanderson’s interview with Bankoff, see [1].) We became good friends. In 1979 he told me about a series of fascinating discoveries he had made about what he called the asymmetric propellor theorem. He intended to discuss them in an article, but never got around to it. This is a summary of what he told me. The original propeller theorem goes back to at least the early 1930’s and is of unknown origin. It concerns three congruent equilateral triangles with corners meeting at a point as shown shaded in Figure 1.1. The triangles, which resemble the blades of a propeller, need not form a symmetrical pattern, but may be in any position. They may touch one another or even overlap. Lines BC , DE, and FA are drawn to form a hexagon inscribed in a circle. The midpoints of the three lines mark the vertices of an equilateral triangle. E
D
F
C
B
A Figure 1.1.
A proof of the theorem, using complex numbers, appeared in [2] as the answer to Problem B1 in the annual William Lowell Putnam Competition. H. S. M. Coxeter sent the proof to Bankoff on a Christmas card, asking him if he could provide a Euclidean proof of the theorem. Reprinted from The College Mathematics Journal, Vol. 30, No. 1 (Jan. 1999), pp. 18–22. This article received the George P´olya Award in 2000.
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Part I. Geometry
Bankoff had no difficulty finding such a proof. In a paper titled “The Asymmetric Propeller” [3], Bankoff, Paul Erd˝os, and Murray Klamkin made the first generalization of the theorem. They showed that the three equilateral triangles need not be congruent. They can be of any size, as shown in Figure 1.2, and the theorem still holds. Two proofs are given, one a simple Euclidean proof, the other with complex numbers. As before, and in all subsequent extensions, the triangles may touch one another or even overlap.
Figure 1.2.
Figure 1.3.
Later, Bankoff made three further generalizations. As far as I know they have not been published. Second generalization: The propeller triangles need not meet in a point. They may meet at the corners of any equilateral triangle, as shown in Figure 1.3. Third generalization: The propeller triangles need not be equilateral! They need only be similar triangles of any sizes that meet at a point. The midpoints of the three added lines will then form a triangle similar to each of the propellers, as shown in Figure 1.4.
Figure 1.4.
Fourth generalization: The similar triangles need not meet at a point! If the propellers meet at the corners of a fourth triangle of any size, provided it is similar to each propeller, the midpoints of the added lines will form a triangle similar to each propeller. Vertices of the interior triangle must touch corresponding corners of the propellers. Here is how Bankoff proved his final generalization on a sheet that he typed in 1973. It makes use of Figure 1.5.
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1. The Asymmetric Propeller
5 J
Z H A Q
R B
S C
E
Y
P G
D X F Figure 1.5.
The propellers shown are right triangles, although they can be any type of triangle. Perhaps the proof that follows can be simplified. If ABC , AHJ , DBE, and F C G are similar triangles, all labelled in the same sense and situated so that corresponding angles meet at the vertices of triangle ABC , then X; Y; Z, the midpoints of DF , GH and JE, are vertices of a triangle similar to the other four. Proof. Let †BCA D †GCF D ˛; †DBE D †ABC D ˇ; †JAH D †CAB D ; and let P; Q; R; S denote the midpoints of the segments DC , AC , AE and CH respectively. We proceed stepwise to show that triangles PQR, PSZ and finally XY Z are similar to triangle ABC . If triangle ABD is pivoted about B so that AB falls along BC and DB along EB, it is seen by the relation AB=BC D DB=BE and by the equality of angles ABD and CBE that triangles ABD and CBE are similar and that EC=AD D BC=AB, with †EC; AD D †BC; BA D ˇ. Since RQ is parallel to and equal to half EC while QP is parallel to and equal to half AD, we extend the previous relation to read RQ=QP D EC=AD D BC=BA, with †RQ; QP D †EC; AD D †BC; BA D ˇ. It follows that triangles PQR and ABC are similar. In like manner, because of the relationship of AJ and AH to RZ and QS as well as to RP and QP in both relative length and in direction, we find triangles ZRP and SQP similar. Then ZP =SP D ZR=QS D AJ=AH D AC=AB, with the angles between the segments in the numerator and in the denominator all equal to . As a result, triangles PSZ and ABC are similar.
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Part I. Geometry
Continuing as before, we find triangles ZPX and ZS Y similar since PX=S Y D CF=CB D CA=C G and †PX; S Y D †CF; C G D †CA; CB D ˛. Noting that in the similar triangles ZPX and ZS Y we have ZX=Z Y D ZP =ZS D CA=CB and †ZX; Z Y D †CA; CB D ˛, we conclude that triangles XY Z and ABC are similar. And now a question for interested readers to explore. Do the propellers have to be triangles? It occurred to me that if squares are substituted for triangles, as in Figure 1.6, that equilateral triangle still shows up.
Figure 1.6.
I have written this piece as a tribute to one of the most remarkable mathematicians I have been privileged to know.
Bibliography [1] G. L. Alexanderson, A conversation with Leon Bankoff, College Math. J. 23 (1992) 98–117. [2] J. H. McKay, The William Lowell Putnam Mathematical Competition, Amer. Math. Monthly 75 (1968) 732–739. [3] Leon Bankoff, Paul Erd˝os, and Murray Klamkin, The asymmetric propeller, Math. Mag. 46 (1973) 270–272.
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2 The Asymmetric Propeller Revisited Gillian Saenz, Christopher Jackson, and Ryan Crumley In [1] Martin Gardner proved an asymmetric propeller theorem that was originally proposed by Leon Bankoff. He showed that by connecting one vertex from each of three equilateral triangles inscribed in a circle to the center of the circle, one can form a fourth equilateral triangle. This fourth triangle is formed by connecting the remaining vertices with segments as seen in Figure 2.1. Then the midpoints of these segments are connected to form a triangle. This works regardless of the arrangement of the original triangles. We confirmed Gardner’s findings and proceeded to consider his final question: when working with squares, does the same phenomenon occur? We show that the answer is no. Here we give counterexamples. H
J
I
Figure 2.1.
Using Geometer’s Sketchpad, we attempted to show that the propeller theorem also worked for squares and possibly other polygons. In [1, Fig. 6] a midpoint of a side of each of three squares occurred at the vertices of an equilateral triangle. We had the squares meet at a single point by shrinking the equilateral triangle to a point. If the conjecture is false for this case, then it will not hold true if the point is expanded to a triangle. By experimenting with Geometer’s Sketchpad we found that even when using squares inscribed in a circle Reprinted from The College Mathematics Journal, Vol. 31, No. 5 (Nov. 2000), pp. 347–349.
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Part I. Geometry
J I
K
Figure 2.2.
the propeller theorem was not true, see Figure 2.2. The triangle shown in this figure was not equilateral not did its angles remain constant when the squares were moved. Seeing that the square conjecture would not work, we set up the following proof. We start with three congruent squares inscribed in a circle. The squares meet at the midpoints of one of their edges. The vertices of the squares are then connected as in Figure 2.3. The midpoints of these new segments are then connected as before to form a triangle. Using the distance formula, it is evident that the edges of the triangle are not congruent, because p 3 ¤ 34=2. (1, 2)
(–1, 2)
(2, 1)
(2, –1)
(1, –2)
(–1, –2) Figure 2.3.
We also investigated whether the midpoint connection was essential when working with equilateral triangles. We used the simplest case possible, three congruent equilateral triangles inscribed in a circle meeting at a point. Rather than constructing the midpoints of the connecting segments, we constructed a point one quarter along each segment. These points were connected to form a triangle. Figure 2.4 shows that the triangle formed is not an equilateral triangle. The theorem seems to be true only when the triangle formed is constructed using the midpoints of the connected vertices. Attempts using other polygons were unsuc
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2. The Asymmetric Propeller Revisited
9 K
L
M Figure 2.4.
cessful. Based on our results, a propeller theorem for squares and other polygons seems unlikely.
Bibliography [1] Martin Gardner, The asymmetric propeller, College Math. J., 30 (Jan. 1999) 18–22; Chapter 1 in this volume.
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3 Bracing Regular Polygons As We Race into the Future Greg N. Frederickson If today’s mathandscienceoriented kids could see the resources available to likeminded youths of the 1960s and 70s, they might pity their counterparts back then: no laptop computers, no highspeed internet, no online bookstores. Yet such pity would be wasted on that earlier generation, who could well respond, “Ah, but we had Martin Gardner.” For a quarter century, Gardner wrote Mathematical Games, a wonderful column in Scientific American [3]. Each month he treated his readers to a tour of some mathematical recreation that he made tantalizing for people who, like himself, had no advanced training in mathematics. He was at the center of an amazing human network, referred to as “Martin Gardner’s mathematical grapevine” by Doris Schattschneider [7]. Long before our current electronic networks, Gardner gathered juicy mathematical tidbits from his farflung contacts, turned them into mouthwatering morsels, and encouraged readers to send back their own variations. How many thousands leaped into mathematical, scientific, and engineering fields because of the enthusiasm nurtured by this singular man? Let’s reach back in time to sample a problem that Gardner introduced and explore the magic that he created, magic which would expand as we raced towards our technological future.
Bracing a square In his November 1963 column, Gardner dished up a tasty problem devised by Raphael Robinson, a mathematician at the University of California, Berkeley, who is best known for determining the minimum number of pieces for the BanachTarski paradox. Gardner wrote: Imagine that you have before you an unlimited supply of rods all of the same length. They can be connected only at the ends. A triangle formed by joining three rods will be rigid but a fourrod square will not: it is easily distorted into other shapes without bending or breaking a rod or detaching the ends. The simplest way to brace the square so that it cannot be deformed is to attach eight more rods to form the rigid octahedron. Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 51–57.
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Part I. Geometry
Suppose, however, you are confined to the plane. Is there a way to add rods to the square, joining them only at the ends, so that the square is made absolutely rigid? All rods must, of course, lie perfectly flat in the plane. They may not go over or under one another or be bent or broken in any way. The answer is: Yes, the square can be made rigid. But what is the smallest number of rods required? Readers are asked to try to solve the problem and, if they succeed, to save their solution or send it to this department. Next month we shall publish the minimum answer Robinson has so far been able to achieve. The next month, Gardner displayed two solutions by Robinson, each with 31 rods in addition to the p four rods of the square. Robinson used equilateral triangles to form lengths of 2, 3, and 13, producing a right triangle whose right angle braces the square. Clever? Yes! And yet, instead of administering the coup de grace to this intriguing little problem, the highpowered Berkeley mathematician had merely set in motion a long sequence of improvements as the problem began to bounce around in Gardner’s network.
Superior solutions from readers Two months later, in the February 1964 column, Gardner reported that 44 readers had surprised him with a 25rod solution, bettering each of Robinson’s solutions by 6 rods! Moreover, Gardner was stunned that seven readers had designed an even niftier 23rod solution! E
E
B
C
B
A
D Figure 3.1. (a) 23rod solution for bracing a square;
C A
D (b) Underlying geometry.
The 23rod solution appears in Figure p 3.1a. Two pairs of equilateral triangles guarantee that the distances BE and CE are each 3. Two trusses, each consisting of three equilateral triangles, guarantee that the rods form straight lines between B and D, and between C and D. These straight lines force points A, B, and C to be collinear. That the distances BE and CE are equal then forces reflection symmetry along the axis containing A and D. p Thus AB and AC to both be 2. We †BAE and †CAE are both right angles, forcing distances p can then conclude that the distance AD is also 2, forcing the rhombus containing A and D to actually be a square. Figure 3.1b depicts this underlying structure. Typically, Gardner collected his columns from Scientific American into books that he published every several years. When doing this, he added new material that had come to light once the columns had appeared. Including this column in his Sixth Book of Mathematical Games [4], Gardner noted on page 55 about the square bracing problem:
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The problem can be extended to other regular polygons. The hexagon is simply solved with internal braces (how many?), but the pentagon is a tough one. T. H. O’Beirne managed to rigidify a regular pentagon with 64 additional rods, but it is not known if this number is minimal.
O’Beirne’s solutions for various polygons In 1999 I was selecting material to include in Hinged Dissections [1]. The 23rod solution for bracing a square, along with O’Beirne’s 64rod solution for bracing a regular pentagon, seemed perfect for a short segment in the book. There was just one hitch—I couldn’t find O’Beirne’s braced pentagon anywhere. In desperation I contacted Gardner. As it turned out, neither he nor O’Beirne had ever published the pentagon bracing. Gardner generously sent me all of O’Beirne’s correspondence on this problem. One aerogramme from O’Beirne showed three progressively better solutions for the pentagon, each added as he excitedly discovered a further improvement. During one remarkable week in December 1963, O’Beirne had found solutions not only for the pentagon but also for three other regular polygons as well! Almost certainly prompted by Robinson’s square bracings in Gardner’s December column, O’Beirne had found ways to brace the pentagon (64 rods), the octagon (105 rods), the decagon (183 rods), and the dodecagon (45 rods). I included the pentagon in [1]. Armed with an M.A. in math and physics from the University of Glasgow, Thomas H. O’Beirne (1915–1982) had worked in turn as a government scientist, a chief mathematician for an engineering firm, and a lecturer in the university’s computing department. In 1961– 62, he wrote a weekly series for the New Scientist, which was collected into a book, Puzzles and Paradoxes. Lying dormant for more than thirty years, O’Beirne’s nifty constructions might well have been forgotten. But that would have belied the residual strength of Gardner’s network and the growing strength of modern networks.
An improved octagon bracing After publishing my book with its discussion of the braced square and braced pentagon, I continued to think about bracing problems. In 2003, I discovered an improvement to O’Beirne’s solution for the octagon, using only 51 additional rods, as shown in Figure 3.2a. The solution incorporates several ideas. First, the rods in the interior of the octagon form rhombuses, forcing opposite pairs of sides of the octagon to be parallel. For every pair of opposite sides, not only are those two rods parallel, but also a pair of bracing rods in the interior. Second, some rods outside of the octagon force other rods to be parallel to rods on the boundary of the octagon. Thus a truss consisting of five equilateral triangles forces three rods to be on the line LM. Also, a truss of three equilateral triangles forces two rods to be on the line CN. Third, other rods outside the octagon enforce certain distances between points on the inside of the octagon. The group of four equilateral triangles between M and C force a
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F
L
M E
G
K
J
L
F
D G I
H A
E
J
C
B
H
D
K
C
I A
B
N
Figure 3.2. (a) 51rod solution for an octagon;
(b) Underlying geometry.
p distance of p3 between these vertices. Since rods ME and CI are parallel, distance EI must also be 3. p Similarly, a group of four equilateral triangles between H and N force a distance of 3. Rods HJ and BN transmit this restriction to points J and B. A final group of four p equilateral triangles forces a p distance of 3 between H and L. Two sets of parallel rods force distance BK to also be 3. Yet it is not so clear that the octagon is then rigid. We can verify this claim by modeling the braced structure (Figure 3.2b) with equations that we solve using Mathematica. First, fix the location of an edge, say AB, by xA D 0, yA D 0, and xB D 1, and yB D 0. Next, for each of the three rods along the boundary from B to E, such as CD, force its length to be 1 by .xC xD /2 C .yC yD /2 D 1. Then, for each of the six rhombuses inside the octagon, such as ABCI, force an opposite pair of rods to be parallel by xB xC D xA xI and yB yC D yA pyI . Do the same for the (degenerate) “flat rhombus” LFEF. And to enforce a distance of 3 between each of the three pairs of two vertices, such as H and L, introduce .xH xL /2 C .yH yL /2 D 3. This gives us 24 equations determining the 24 total coordinates of the twelve points. However there may be more than one solution to this system of equations, because we have not specified, for example, whether rod IC is above or below rod AB. So introduce more constraints to force a specific topology as in Figure 3.2b. These constraints would be yC > yB , yD > yB , xC > xB , and xD > xB . The resulting set of restrictions leaves us with a unique solution that then guarantees rigidity.
Theobald’s solution for bracing a pentagon In 2004, after studying O’Beirne’s braced pentagon (in [1]), Gavin Theobald improved it, reducing the number of additional rods from 64 to 56, as in Figure 3.3a. Theobald adapted key ideas of O’Beirne and also introduced new ones. Let’s verify carefully that the pentagon is braced. O’Beirne had first set out to produce an angle of 108ı , the interior angle of the regular pentagon. He had known that a diagonal in a regular pentagon divides the pentagon into
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a trapezoid and an isosceles triangle with this angle. p Furthermore, he had recognized that if the two equal sides of the triangle are of length 3, then the length of the third side is p p 3.1 C 5/=2. Trusses consisting of three equilateral triangles will create 4ABC with sides p 1, 2, and 2, as in Figure 3.3a. By the Pythagorean Theorem, such a triangle has height . 15/=2. Adding thepheight of an equilateral triangle (top of Figure 3.3b) gives the desired p distance of 3.1 C 5/=2. A
A
2
D B
2
B
1
G
C
C
E F
2 D
L M I
H
B
K
2 H
C N J
Figure 3.3. (a) Theobald’s braced pentagon;
I
108 ◦ J (b) Underlying geometry.
While O’Beirne had then gone on to generate the angle of 108ı and transfer it elsewhere in the structure, Theobald chose instead to transfer the distance itself. Theobald noted that he already had a suitable angle, namely †BCD, so that the desired length would be between points B and D (See the middle of Figure 3.3b). Theobald then transferred the desired length by constructing parallelogram BDHI as p follows. He made a truss of four equilateral triangles to force distance BI to be 7, p and then made a “zigzag truss” with five equilateral triangles to force distance DH to be 7. He forced BE to be parallel to DG by creating a sequence of three parallelograms between them. Then BI must be parallel to DH . Since BDHI is its opposite sides pa parallelogram, p must be the same length, and thus distance HI is also 3.1 C 5/=2. Theobald positioned the vertices of an isosceles triangle HIJ with HI as the base. p Then two equilateral triangles force distance IJ to be 3 and four equilateral triangles p force distance HJ to be 3. This gives the isosceles triangle with †IJH D 108ı . The two equilateral triangles touching at J supply two of the sides of the regular pentagon. Once O’Beirne had positioned one angle of 108ı in the regular pentagon, he had positioned the opposite side by using a sequence of an equilateral triangle, two rhombuses, and another equilateral triangle. Theobald used a similar idea to position side LM of his pentagon with just one rhombus. Theobald’s construction forces the side LM of the pentagon to be at the correct angle to make the pentagon be regular: First, rods EL and FM form a rhombus, so that rod LM
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is parallel to rod BC . Rod BC is at an angle of 30ı with the line BD. Then rod LM is at an angle of 30ı with line HI . Pentagon edges JK and JN are rotated 30ı counterclockwise from the lines IJ and HJ , which means that diagonal KN is parallel to LM . Thus LM is at the correct angle. Even knowing O’Beirne’s approach, Theobald still needed creativity and persistence to reduce the number of rods by eight. Gavin Theobald was born in London, England, in 1961. Graduating with a B.Sc. in Computer Studies at Loughborough University in 1983, he worked as a computer programmer on lowlevel highperformance graphics rendering engines. Theobald is an expert in geometric dissections [8], an area into which he was drawn after reading one of Gardner’s books.
Khodulev’s solutions when rods cross Even more progress was made by Andrei Khodulev, who had learned of the problem from the Russian translation of Gardner’s book [5]. Noting that one can use fewer rods if they are allowed to cross, Khodulev discarded the restriction on crossing. Then, for example, a regular octagon can be rigidified with only 23 additional rods: eight rods to create rhomp buses within the octagon, and three sets of five rods each to produce three distances of 3. Published in the bulletin of the Russian puzzle club Diogen (Diogenes) [6], Khodulev’s results are reproduced on Erich Friedman’s webpage [2]. Andrei Khodulev (1953–1999) earned a Ph.D. in computer science in 1984 from Keldysh Institute of Applied Mathematics, in Moscow, where he was a senior research scientist. His research areas were in graphics, imaging, approximation theory, and stochastic simulation.
Gardner’s enduring challenge Thanks to Martin Gardner’s gift for connecting people together, a problem that originated with Raphael Robinson (1911–1995) spread to Gardner’s wide readership, resulting in improvements from dozens of his correspondents. Thomas H. O’Beirne then generalized the problem and produced solutions for four other polygons, one of which Gardner mentioned in his subsequent book. That led the current author to include O’Beirne’s solution in a book, which in turn attracted the attention of Gavin Theobald, who found an improvement for the pentagon, as I found an improvement for the octagon. Separately, Andrei Khodulev made progress by allowing rods to cross. So don’t be surprised when further delectable developments are announced over our networks—just brace yourself! Acknowledgment I thank Gavin Theobald for permission to include his pentagon bracing, Serhiy Grabarchuk for copies of Sharada, Svitlana Mayboroda for translating Russian, and the referees for good suggestions.
Bibliography [1] G. N. Frederickson, Hinged Dissections: Swinging and Twisting, Cambridge Univ. Press, New York, 2002.
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[2] E. Friedman, Problem of the month (January 2000); available at www.stetson.edu/˜ efriedma/mathmagic/0100.html, 2000. [3] M. Gardner, Mathematical games, Scientific American, Nov., 1963, p. 144; Dec., 1963, pp. 144, 146; Feb., 1964, p. 128. [4] ——, Martin Gardner’s Sixth Book of Mathematical Games from Scientific American, Charles Scribner’s Sons, New York, 1971. [5] ——, Mathematical Leisure (Matematiqeskie Dosugi), Mir (Mir), Moscow, 1972. [6] A. Khodulev, A task about a rigid polygon (zadaqa o estkom mnogougolbnike), Sharada (Xarada) 7, no. 61 (1998) 2–8. [7] D. Schattschneider, In praise of amateurs, in The Mathematical Gardner, D. A. Klarner, ed., Prindle, Weber & Schmidt, Boston, 1981, 140–166. [8] G. Theobald, Geometric dissections; available at home.btconnect.com/GavinTheobald/Index.html.
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4 A Platonic Sextet for Strings Karl Schaffer Our dance company, the Dr. Schaffer and Mr. Stern Dance Ensemble, began performing mathematical dance shows in 1990. Scott Kim joined us in 1993, and in 1995 Scott, another Bay area dancer Barbara Susco, and I developed a schoolage show entitled “Through the Loop, In Search of the Perfect Square,” which included the performance of many string figures. Martin Gardner wrote several columns on polyhedra and one on traditional string figures, in addition to his many references to magic tricks and puzzles with string [5, Ch. 1], [6, Ch. 17], [7, Ch. 19], [8, Ch. 10], [6, pp. 74–75]. Traditional string figures, made on the hands with a simple loop of string, are found worldwide, particularly associated with ancient cultures [7]. In 1994 our dance company began experimenting with oversize loops carried on stage by dancers, and manipulated to form a variety of string figures. String figure constructions and the associated mathematics make a pleasurable classroom activity that introduces graph theory and polyhedra to geometry, liberal arts, and discrete math students, or as a handson activity for workshops and math festivals. In the process of creating one of our shows, “Through the Loop,” which includes over 20 string figures, we began altering wellknown figures, using large loops held by several dancers, or using a sequence of finger manipulations different from the customary methods. Traditional string figures often display shapes or actions reminiscent of the natural world, as indicated by such titles as: Bear Climbing a Tree, the Yam Thief, and Mosquito. Yet these figures also involve complex mathematical patterns in both their construction and final designs. We began experimenting with polyhedral designs which we thought might be more recognizable by contemporary audiences as being mathematical. Scott Kim and the author came up with several string constructions of the tetrahedron, octahedron, and cube [8, 9], though Scott’s are particularly beautiful for their symmetry and simplicity. All these constructions keep the string loops taut, and are simplified by involving symmetrical transfers by participants. Several of our mathematical designs, including that of the octahedron and its collapsed form as a six pointed star, were used in one of the dances in the Loop show; other string figures in that show were traditional ones as well as a series of star polygons [7]. Subsequently the author found several ways to build all five Platonic solids sequentially, including a method employing identical loops of three colors which creates interesting symmetric color patterns. This six loop version Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 6469.
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was demonstrated for the first time at a Math and the Arts conference at UC Berkeley in 1998 and at the Gathering for Gardner 3 in Atlanta in 1998. It has since has been handed around at conferences and workshops, in the manner of folk mathematics familiar to fans of Martin Gardner. It is published here for the first time. These multiple loop constructions are inspired by contemporary modular origami which uses identical copies of modular units to construct complex polyhedral forms. Though some observers insist on calling this string sequence a dance, these more complex string polyhedra are too stationary in space and require too many careful manipulations by the hands of participants crowding around the figures to be usable as a part of actual staged dances.
Euler circuits An Euler circuit is a sequence of adjacent edges in a graph that includes each edge once. The name refers to Leonhard Euler’s 1736 solution of the Seven Bridges of K¨onigsberg Problem, though such figures also appear in sand drawings found in West Africa and elsewhere [1, pp. 30–43]. Traditional string figures, which invariably use simple unknotted loops of string, accomplish the same thing. Think of the points of support, where hands or fingers keep the string in tension, as the vertices, and the strands of string as the edges of a polyhedral graph. For our polyhedra we use the thumb and first finger of one hand for each vertex, so that a polyhedron with 20 vertices, such as the dodecahedron, requires ten people. As Euler found, a connected graph can be decomposed into a circuit (or collection of circuits) only if all vertex degrees are even. Therefore some edges of our polyhedra need to be doubled and made with at least two strands to even out the vertex degrees. For example, the four vertices of the tetrahedron are each of degree three, and so we need to double at least two opposite edges in order to Eulerize its graph and so allow it to be made with a circuits (i.e., loops of string). Four of the Platonic solids have odd degree vertices, and require edge doublings; the octahedron has degree four vertices, and so can be made from loops without edgedoubling. By utilizing the symmetries of the polyhedra, we allow all the loops to be of the same size, and this simplifies the constructions and facilitates the directions, as multiple participants do the same string manipulations. After choosing edges to double, we essentially decompose the graph of each polyhedron into cycles, but the struggle to insure that the strings do not end in tangles nor the participants interfere with each other makes the search for these decompositions more challenging. The most pleasing figures seem to allow the least slack in the strings and make use of simple and surprising transformations.
Transformations The overall manner in which we handle the string can be modeled by several standard graph theory operations. An edge subdivision replaces an edge uv having vertices u and v with a new vertex w and two new edges uw and vw. This is accomplished with string when one grasps and pulls on the middle of a string edge. In order to manipulate the string smoothly, each hand involved holds it by making a loop with the thumb and index finger,
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so that the string can slide easily “through” the vertex. This is why we require a whole hand for a single vertex. We also use the reverse operation, smoothing a vertex, releasing a vertex w while simultaneously pulling outwards at all other vertices to take up the slack, and replacing two edges uw and wv by single edge uv. Pulling two or more vertices together and holding them in one hand is vertex amalgamation. We retain the separate identification of the amalgamated vertices, since we sometimes need to reverse the operation and release some of the vertices from the amalgamation. Vertex separation is the release of a string from an amalgamated vertex. We may also perform an edge contraction by sliding two vertices joined by an edge together to amalgamate their endpoint vertices, while removing the edge between them by pulling outward on the other vertices, as with vertex smoothing. In all cases the participants need to adjust their hands slightly to make the polyhedra more regular; letting the strings slide through the loop made by the thumb and forefinger facilitates this, while grabbing the strings makes it more difficult.
Sixloop platonic solids Figure 4.1 shows the sixstring transformations of the Platonic solids. All six strings should be same length, and it is helpful to use two loops of each of three colors. Each vertex is labeled Ri or Li , indicating that it is held by person i with the Right or Left hand. Figure 4.1a shows the opening pattern, seen from above, made with the two loops of the same color held by opposite participants. It is important to keep the doubled loops from crisscrossing, as otherwise the later figures will contain extra edges. One of the pairs of loops must sit inside the other two pairs (that held by R1 and R3 ), and one pair must be external to both the others (that held by R5 and R6 ), so this construction is not as symmetrical as other of the polyhedral string figures; a more symmetric alternative is to link the strings like the Borromean rings, but this leads to tangles! Figure 4.1a shows that first R5 and R6 subdivide and hand their loops off, forming the square R1 R2 R3 R4 in Figure 4.1b. R5 and R6 then grasp new vertices as shown in Figure 4.1b, forming squares R4 R5 R2 R6 , and R1 R5 R3 R6 , and thus the octahedron shown in Figure 4.1c. The six participants still use only one hand apiece, though actually both hands are needed to perform the transition from Figure 4.1a to Figure 4.1b. The octahedron is usually held with one vertex high and the opposite near the floor to make the instructions easier to follow, though one person must reach high and another low in order to grasp the vertices. The octahedron thus lines up an axis of fourfold rotational symmetry with the vertical; we might also sit the octahedron on one of its triangles to accentuate the 3fold symmetries. In order to transform the octahedron to the icosahedron, we must double the number of vertices, and transform the 12 edges of the octahedron to the 30 edges of the icosahedron. This is accomplished by adding one vertex and 3 edges for each of the octahedron’s 8 vertices as follows (see the arrows in Figure 4.1c for the R1 , R2 , and R6 vertices.) Note that all edges in the octahedron are doubled, but only six nonadjacent edges in the icosahedron are doubled. The two vertices from the horizontal R1 R2 R3 R4 loop held at the R1 vertex are separated, one sliding upwards, the other downwards along the R1 R5 R3 R6 loop (in the
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Part I. Geometry R5 R4
R4
R5
R1
R4
R5
R3
R3
R1
R3 R1
R6 R2 R6
(a)
R2
(b)
L5 R5
R1
L1
R2
L4
R4
R9 L2 R8
L1
L10 R4
L3 L9
L6 (d) R7
R2
R1
R6
R10
R7
R5
R3
R2
L2
R6 (c)
L5
R10
L4
R6
R3 L7
L3 L8
L6 (e) R7
R9 R10
L10
R9
R8
L9
L8
(f)
L7
L10
(g)
L8
Figure 4.1. Six loop Platonic solids.
plane of the drawing), to form the new R1 and L1 vertices. Each participant will execute a similar maneuver. The exact same operation as at R1 in Figure 4.1c is performed at the R3 vertex. At R2 and R4 the vertical R4 R5 R2 R6 loop’s amalgamated vertices are separated horizontally to form new R2 ; L2 ; R4, and L4 vertices. At R5 and R6 the R1 R6 R3 R5 loop’s vertices are separated horizontally to form new R5 ; L5 ; R6 , and L6 vertices. All this produces the icosahedron in Figure 4.1d. When performing these operations make sure that the pairs of same colored loops do not cross. Note that the remaining 6 doubled edges are parallel in pairs, forming a perfect matching of the vertices of the icosahedron, and hence a minimal Eulerization. At this point notice that there are 8 triangles, 4 high and 4 low, which have edges of 3 colors, none doubled. To transform the icosahedron, with 12 vertices and 30 edges, to the dodecahedron, with 20 vertices and 30 edges, we need to add 8 vertices, and somehow change 20 triangles to 12 pentagons. To accomplish this, each edge of those eight, 3colored triangles will be subdivided, and the three new vertices on each such face amal
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gamated as one vertex. We call this operation on all the edges of a single face squeezing the face. This is illustrated in Figure 4.1d by arrows (just for the R2 R3 ; L5 and R2 L3 ; R6 triangles), resulting in the dodecahedron with all edges doubled (some doubled edges not shown), in Figure 4.1e. Removing these 8 triangles results in a polyhedron with 20 8 D 12 pentagons. This operation requires 4 new participants, who each will most conveniently squeeze two triangles above and below each other, with say, right hands high, left hands low. Two of the same colored loops are shown as part of sets of doubled edges in Figure 4.1e. These sets of edges form what are known as the Petrie polygons of the dodecahedron, skew polygons no three of whose successive edges are part of the same face. The step from Figure 4.1e to 4.1f is simple: participants 1 through 6 release their vertices, while the new participants (7 through 10) hang on to theirs. Their 8 vertices (white circles in Figure 4.1e) are the vertices of a cube in Figure 4.1f, which magically appears with all its edges doubled. Strings of like colors appear on opposite faces. The transition to the tetrahedron (Figure 4.1g) is equally simple: two of the remaining participants opposite each other on the upper face of the cube, say 7 and 9, hold on to their high vertices with their right hands and release their left hand vertices, while 8 and 10 hold on to their left hand vertices and release their right. In the tetrahedron formed each pair of string loops of the same color are nonincident and skew perpendicular. To complete the cycle back to the octahedron, each of these four participants uses their free hand to squeeze the edges of the three colors of one triangle back to a new vertex, reforming the cube; this operation is indicated by the arrows on one of the faces in Figure 4.1g. Participants 1 through 6 now return to form the rhombic dodecahedron by squeezing one face per participant to a new vertex, taking care to use edges of two colors only per vertex, and such that the same two colors are used for the vertices on opposite faces of the cube. Also, one of the pairs of same color loops must be external to the others, and one pair internal, or the octahedron will become tangled. Participants 1 through 6 now hang on to their vertices, and 7 through 10 release theirs, forming the octahedron again.
Conclusion Many, many other polyhedral string figures similar to these are possible. For example, the author has found fairly direct transformations of the six loop polyhedra shown to the Archimedean polyhedra, as well as a series of constructions of the Platonic solids using either one or four loops. See also [8, pp. 85–97], [9], [3, pp. 280–287], and [7]. These constructions make excellent classroom investigations of Euler circuits and polyhedra. Start simply. For example, a fivepointed star is easy to draw but making one with a loop of string requires 3 people cooperating carefully. Have students find their own constructions of a tetrahedron and octahedron; there are many. If the students find a nice construction have them practice several times so the actions are quick and fluid. Directions and lore on traditional string figures can be found at [7]. Many variations on the figures mentioned in this article are possible.
Note on making loops for use in these constructions We use loops which can be stretched to “arm span,” length or about 5 to 6 feet. We either use cotton string, taping the ends together with cloth or duct tape, or nylon string with the
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ends melted together over a candle flame. Use string that is at least 1=8th to 1=4 inch in diameter. For the Platonic solids, we use 1=2 inch diameter flexible cotton rope, purchased in 3 colors. One can also dye cotton rope a variety of colors with fabric dye. Make sure the rope or string is manufactured with a soft weave, rather than using brittle or inflexible string or rope. The string polyhedra constructed in this paper use multiple loops of the same size. To store the loops without getting them all tangled, double each one over several times and tie it in a knot.
Videos George Csicsery has produced several videos of sequences of string polyhedra, including a threepolyhedra sequence similar to, but shorter than, the one described here, and involving only four participants. This video and other videos relevant to string polyhedra and to Martin Gardner are available on YouTube at http://www.youtube.com/user/g4gman.
Bibliography [1] M. Ascher, Ethnomathematics, a Multicultural View of Mathematical Ideas, Brooks/Cole Publishing Company, Pacific Grove CA, 1991. [2] M. Gardner, The Second Book of Mathematical Diversions, Simon and Schuster, New York, 1961. [3] ——, The Unexpected Hanging and Other Mathematical Diversions, Simon and Schuster, New York, 1968. [4] ——, Martin Gardner’s Sixth Book of Mathematical Diversions from “Scientific American,” Simon and Schuster, New York, 1971. [5] ——, Wheels, Life, and other Mathematical Amusement, W.H. Freeman and Company, New York, 1983. [6] ——, The Last Recreations Hydras, Eggs, and Other Mathematical Mystifications, SpringerVerlag, New York, 1997. [7] K. Schaffer, Star String Polygons, Gathering for Gardner 9, 2010 (to appear). [8] K. Schaffer, E. Stern, and S. Kim, Math Dance with Dr. Schaffer and Mr. Stern, MoveSpeakSpin; available at www.movespeakspin.org, 2001. [9] ——, math dance activities; available at www.mathdance.org. [10] M. Sherman, ed., International String Figure Association, string figure references and instructions; available at www.isfa.org. [11] I. Stewart, Cows in the Maze and Other Mathematical Explorations, Oxford Univ. Press, Oxford, 2010.
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5 Prince Rupert’s Rectangles Richard P. Jerrard and John E. Wetzel Introduction More than three hundred years ago, according to the contemporaneous John Wallis [11, pp. 470–471], Prince Rupert (1619–1682) won a wager that a hole can be cut in one of two equal cubes large enough to permit the second cube to pass through. Nearly a century later the Dutch scientist Pieter Nieuwland (1764–1794) p showed that the largest cube that can be so passed through a cube of side one has side 3 2=4 1:061. An accessible discussion with enlightening anaglyphs appears in Ehrenfeucht [8]. In 1950, D. J. E. Schreck [10] gave an interesting, historically based survey of Prince Rupert’s problem and Nieuwland’s extension. Schreck includes a photograph of a model showing the cube in transit. Nieuwland’s “passage” problem of finding the largest cube that can pass through a unit cube is equivalent to finding the largest square that fits in the unit cube, because once the largest square is located, the hole through the cube having that largest square as its cross section clearly provides the desired passage. In higher dimensions one might seek the side of the largest mdimensional cube that fits in an ndimensional cube of side one. Not much is known about this question, which apparently was first raised, at least in the case of a cube in a hypercube, by Martin Gardner (see Gardner [2, pp. 172–173] or Guy and Nowakowski [7, pp. 967–68]). The literature on Prince Rupert’s and Nieuwland’s problems is extensive and we cannot claim to have examined it all, but nowhere have we found any mention of the analogous questions for rectangles and “boxes” (i.e., rectangular parallelepipeds): Question 1. Find the largest rectangle R with given aspect ratio that fits in the unit cube, i.e., for given with 0 1, find the largest L so that an L L rectangle fits in the unit cube. This question is an instance of a general unsolved fitting problem for boxes in higher dimensions: in d dimensional Euclidean space Ed , find necessary and sufficient conditions Reprinted from The American Mathematical Monthly, Vol. 111, No. 1 (Jan. 2004), pp. 22–31. of the Rhine and Duke of Bavaria, son of Frederick V, the Winter King, Elector Palatine, and king of Bohemia, and Elizabeth, daughter of James I of England. It is convenient to regard a line segment as a rectangle with one side 0, and a rectangle as a box with one edge 0. Count Palatine
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on the k C d edges for a kdimensional box .k d / with edges p1 , p2 ; : : : ; pk to fit in a d dimensional box with edges s1 , s2 ; : : : ; sd . For k D 1 the matter is trivial for each q d : a line segment of length p fits in the d dimensional box precisely when p
s12 C s22 C C sd2 . For k D d D 2 the question of when one rectangle fits in another was asked in 1956 by L. R. Ford [4], and a necessary and sufficient condition was soon given by W. B. Carver [2] (see also Wetzel [12]). The problem was posed for k D d D 3 by F. M. Garnett in 1923, and an incomplete answer was supplied in 1925 by Carver [1].
Question 2. Find the largest box of given shape that can be passed through a suitably perforated unit cube, i.e., for given and with 0 1, find the largest D so that a D D D box can pass through a suitable hole in the unit cube. This â€œpassageâ€? question bears the same relation to the â€œfittingâ€? question 1 as Rupertâ€™s does to Nieuwlandâ€™s: once the largest rectangle with aspect ratio D = in the unit cube is known, then the hole through the cube having that rectangular cross section provides the desired passage; and no smaller hole with similar cross section suffices. As an example, suppose that D 0:7 and p D 0:2.p The largest D D D box that can pass through the unit cube has D D 2. 143 8/=9 2:02885 (see Theorem 1 and its corollary), so that D 1:42020 and D 0:40577. Figure 5.1a shows the largest rectangle with aspect ratio D 2=7 that fits in the unit cube and the cube perforated to accommodate the box. Figure 5.1b shows the box in transit through the perforated cube.
(a) The perforated unit cube.
(b) The box in transit.
Figure 5.1. A box through the unit cube.
The Fitting Lemma Let C be the unit cube, and let @C denote its boundary, viz., the union of its six closed faces. Once the aspect ratio is fixed in Ĺ’0; 1Â?, there is a largest real number Lmax so that
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the Lmax Lmax rectangle Rmax fits in C. The first thing we need to ask is precisely how this largest rectangle Rmax is situated in C. The answer is given by the “fitting lemma,” Lemma 4, whose proof is the objective of this section. Note first that, because of the central symmetry, any rectangle R that fits in C must also fit in C with its center at the center of symmetry of the cube. Lemma 1. If a rectangle R D ABPQ (Figure 5.2) fits in the unit cube C in any way whatsoever, then it also fits in C with its center at the center of C. B
P
A
Q
Figure 5.2. A rectangle with aspect ratio 0.4.
Proof. Suppose that the rectangle R D ABPQ fits in C and let A0 , B 0 , P 0 , Q0 be the points symmetric to A, B, P , Q in the center O of C (Figure 5.3). Then the rectangle R0 D A0 B 0 P 0 Q0 is congruent to R and forms with R a parallelepiped that fits in C with center at the center O of C. The medial section of this parallelepiped midway between R and R0 is a rectangle that is congruent to R and has its center at O.
R A
A' O R'
Figure 5.3. Centering.
As a consequence of this lemma, we may always assume that a rectangle under consideration is centrally placed in C. We call such a rectangle centered. A corner of a centered rectangle lies on an edge or face of C precisely when the opposite corner of the rectangle lies on the opposite edge or face of C. Corners at vertices If one corner of a centered rectangle R in C lies at a vertex of C, then the opposite corner of R lies at the opposite vertex of C, and the two remaining corners of R are at opposite vertices of C. There are only two possibilities: p p D 0 andpR is a diagonal line segment of C regarded as a 3 0 rectangle, or D 1= 2 and R is a 2 1 diagonal cross section of C. In the following analysis, we usually suppose that the corners of R are not at the vertices of C.
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Centered maximal rectangles How does a centered maximal rectangle R fit in C? We show next that its corners must lie on the boundary @C of C. Lemma 2. The corners of a centered maximal rectangle R lie on the boundary @C of C. Proof. Our principal tool is the observation that a rectangle that fits in C with all four of its corners inside C is not maximal. In particular, if a centered rectangle has two adjacent corners in the interior of C, then it is not maximal. It follows that we need examine only those centered rectangles having at least one corner on a face (but not at a vertex) of C. Suppose that one corner A of a centered rectangle R D ABPQ lies inside C and an adjacent corner B lies on a closed face f of C but not at a vertex (Figure 5.4). If B lies in the interior of a face f , then an appropriate small rotation of R in its circumcircle (Figure 5.4a) moves B inside C as well; and consequently R is not maximal. If B lies on an edge e of f (Figure 5.4b), then a suitable small rotation of R about the line ` normal to f through O leaves A inside C and moves B off e into f , the case just considered; so again R is not maximal. It follows as claimed that all four corners of a maximal centered rectangle must lie on faces of C. ` f e
B
`
B
A
O
O f A
(a) Corner B on an open face.
(b) Corner B on an edge.
Figure 5.4. Corner A inside the cube.
But more is true: the corners of a maximal centered rectangle must be on the edges of C. Lemma 3. The corners of a centered maximal rectangle R lie on edges of C. Proof. We show first that at least one of the corners of a centered maximal rectangle R must lie on an edge of C. If adjacent corners A and B of R both lie on the same (open) face of C (Figure 5.5a), then the other two corners P and Q lie on the opposite face of C, and a suitable small rotation about the axis ` through O parallel to AB moves R completely inside C, contrary to Lemma 2. If adjacent corners A and B of R lie on opposite (open) faces of C, then the adjacent corners A and Q lie on the same (open) face of C, which as we have just seen is a contradiction. If adjacent corners A and B of R lie on adjacent (open) faces of C (Figure 5.5b), a suitable small rotation of R about the diagonal AP moves B inside C, again contrary to Lemma 2. Consequently, at least one corner of R must lie on an edge of C; of course, the opposite corner of R then lies on the opposite edge of C.
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29 `
A A
B
`
O
B
P
(a) Corners A and B on the same face.
(b) Corners A and B on adjacent faces.
Figure 5.5. Corners A and B on open faces.
Suppose next that a corner A of the centered maximal rectangle R lies on an edge e of C. If a corner B adjacent to A lies on an (open) face f of C adjacent to e (Figure 5.6a), then a suitable small rotation of R about the line ` through O perpendicular to f moves both A and B into the (open) face f , contrary to the first part of this proof. If a corner B adjacent to A lies on an (open) face f that does not meet e, then the opposite corner Q is adjacent to A and lies on the face f 0 opposite the face f , which is adjacent to e. Again it follows that R is not maximal. Finally, if a corner A of R lies on an edge e and an adjacent corner B lies on a face f that meets e only at a vertex of C (Figure 5.6b), the corner P opposite A lies on the edge opposite e, and a suitable small rotation about the diagonal AP of R moves both B and Q into the interior of C, once more contrary to Lemma 2. A
e
B O
f `
f B e
A
P
(a) Corner B on an open face adjacent to e.
(b) Corner B on an open face normal to e.
Figure 5.6. Corner A on an edge.
It follows, as claimed, that all four corners of a centered maximal rectangle lie on edges of C. Lemma 4 (Fitting Lemma). A centered maximal rectangle R whose corners are not at the vertices of C fits in C in exactly one of the following two ways: a. two adjacent corners of R lie on the same edge of C (Figure 5.8); or b. two adjacent corners of R lie on adjacent edges of C (Figure 5.9).
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Figure 5.7. Rectangle is not maximal.
Proof. The four corners of R lie on a sphere centered at the center O of C, and each lies at the same distance from the nearest vertex of C. If (a) and (b) are both false, then the corners must also lie on four parallel edges of C, one on each. It follows that there are two adjacent corners of R (labeled A and B in Figure 5.7) whose edge is parallel to an edge of C. But then a small rotation about an axis through O parallel to AB would move all four vertices off the edges of C into the (open) faces, contrary to Lemma 3.
The Maximal Rectangles According to Lemma 4, there are only two possibilities to examine. Situation (a). Suppose that two adjacent corners of the maximal centered rectangle R lie on the edge e of C (Figure 5.8). Write a for the distance p from A to the nearest vertex V of C. Since AB 1 AQ, we see that L D p AQ D 2 and L D AB D 1 2a. Since 0 a 1=2, we find that when 0 1= 2 p (5.1) L D 2:
V A Q O B P
Figure 5.8. Situation (a).
Figure 5.9. Situation (b).
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Situation (b). Suppose that two adjacent corners A and B of a centered maximal rectangle R of aspect ratio are on the adjacent open edges eA and eB that share a vertex V (Figure 5.9). Write a for the distance VA and b for the distance VB. Then p p AB D a2 C b 2; BP D 1 C .1 a/2 C .1 b/2 : (5.2)
Since OA D OB, it is easy to see that b must be either a or 1 a. Further, AB could be the smaller dimension L or the larger dimension L of R. Thus situation (b) splits into four apparent subcases, one of which proves to be impossible. Case b1 : b D ap and AB D L. Setting a D b and D 1 in (5.2), we find that a D 3=4 so that L D 3 2=4, which is Nieuwland’s result. More generally, setting a D b, eliminating a, and writing f1 ./ D L, we learn that when 0 1 LD p 3
3
2
and
C
3 aDp p 2 3
p D f1 ./ 2
2 C
p : 2
(5.3)
(5.4)
Case b2 : b D 1 a and AB D L. Substituting in (5.2), eliminating a, requiring p p that 0 a 1, and writing f2 ./ D L, we see that for satisfying 1= 3 1= 2 LD p 1
1
2
D f2 ./:
r
32 1 : 1 2
In this case, 1 1 aD ˙ 2 2
(5.5)
Case b3 : b D a and AB D L. Eliminating a in (5.2) leads to the equation p .1 2 /L2 2 2L C 3 D 0: p Writing L D f3 ./, we discover that when 1= 3 1 3 LD p p D f3 ./: 2 C 32 1 (The second root is too large when < 1.) Then p p 2 2 32 1 aD ; 1 2
(5.6)
(5.7) p and to ensure that a 1 we require additionally that p 1= 2. In other words, the appropriate domain for f3 ./ in (5.6) is described by 1= 2 1: The apparent possibility with b D 1 a and AB D L leads to an equation for L having no real solutions, so this possibility can not be realized.
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Part I. Geometry L max 3
2 f1
f2
f3
3 2 4 1 l 0
l1
l2
1
Figure 5.10. The longer side Lmax in terms of the aspect ration .
Conclusions All that remains is to collect our results. We find the following answer to Question 1 of the introduction (see Figure 5.10). p p Theorem 1. Let 1 D 1 2=2 0:29289 and 2 D 1 1 D 2=2 0:70711. For each with 0 1, the longer side Lmax ./ of the largest rectangle Rmax with aspect ratio that fits in the cube of side 1 is given as follows: 8 3 ˆ ˆ p if 0 1 ; p ˆ ˆ 2 ˆ ˆ <p 3 C 2 Lmax ./ D (5.8) 2 if 1 2 ; ˆ ˆ ˆ ˆ 3 ˆ ˆ p if 2 1: :p 2 C 32 1 Proof. When 0 1 the longer p side Lmax ./ is the larger of the number f1p ./, where f1 is the function of case b , and 2 (from (5.1)). If , L ./ D 2, 1 1 2 max p p p because f1 ./ 2 on the entire interval Œ1 ; 2 and f2 ./ 2 when 3=3 2 . Similarly Lmax ./ D f3 ./ for satisfying 2 1.
Figure 5.11 shows the positions of the maximal rectangles. When 0 1 , there are twelve maximal rectangles, all centered and located as shown in Figure 5.11a. Their longer dimension L and the distance p a D VA are given by formulas (5.3) and (5.4), respectively. If 1 2 , then L D 2, and there are six centered maximal rectangles and infinitely many that are not centered, all located on the diagonal planes of C as shown in Figure 5.8. If 2 1 there are again twelve maximal rectangles, all centered and located as shown in Figure 5.11b, and their longer dimension L and the distance a D VA are given by the formulas (5.6) and (5.7). Theorem 1 gives a necessary and sufficient condition for a rectangle to fit in a unit cube: an a b rectangle with a b fits into the unit cube C if and only if a Lmax .b=a/, where Lmax ./ is given by (5.8).
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A
(a) The case 0 1 .
(b) The case 2 1.
Figure 5.11. Maximal rectangles.
The answer to Question 2 of the introduction follows immediately from this theorem. Corollary. Given and with 0 1, a D D D box can pass through a suitable hole in a unit cube C if and only if 1 D Lmax : Proof. The largest L .=/L rectangle that fits in C has L D Lmax .=/, so the box can pass through the cube precisely when D L.
Final Remarks Greg Huber, a physicist at the University of Massachusetts, and Terry J. Ligocki of Lawrence Berkeley National Laboratory have generated data for the diagonal dmax ./ of Rmax ./ by modeling the problem with an iterative numerical procedure known as Boltzmann simulated annealing. Professor Huber emailed us as follows, â€œThe basic idea comes from statistical mechanics. The size of the embedded rectangle is taken to be an â€œenergy,â€? and configurations (orientations of the rectangle) are perturbed and accepted stochastically with respect to a continuously changing â€œtemperature,â€? according to an annealing schedule. Asymptotic convergence to a global extremum is guaranteed under a few strong conditions on the acceptance function and scheduling, but in practice [we] chose heuristic functions.â€? In March 2002 he forwarded to us two hundred computergenerated values p of dmax ./ for in Ĺ’0; 1Â?. The graph of the corresponding values of Lmax ./ D dmax = 1 C 2 matches closely the function described in Theorem 1 and sketched in bold in Figure 5.10. In response to a problem proposed by Mauldon [9], Chapman [3] provided an algebraic answer for the question of finding the largest square that fits in the unit cube, thereby supplying a solution to this wellknown problem of intuitive geometry that is â€œnot overly dependent on geometric intuition,â€? as the Monthlyâ€™s 1995 Problems Editors remarked. An algebraic argument independent of the fitting lemma can probably be given for Theorem 1 along the same lines. Finally, it is a pleasure to acknowledge the assistance of Jonathan P. Green of the UIUC Department of Germanic Languages and Literatures with John Wallisâ€™s Latin.
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Bibliography [1] W. B. Carver, Solution to Problem 3036, Amer. Math. Monthly 32 (1925) 47–49. [2] ——, Solution to Problem E1225, Amer. Math. Monthly 64 (1957) 114–116. [3] R. J. Chapman, Solution to Problem 10251, Amer. Math. Monthly 102 (1995) 465–467. [4] L. R. Ford, Problem E1225, Amer. Math. Monthly 63 (1956) 421. [5] M. Gardner, The Colossal Book of Mathematics, W. W. Norton, New York, 2001. [6] F. M. Garnett, Problem 3036, Amer. Math. Monthly 30 (1923) 337. [7] R. K. Guy and R. J. Nowakowski, Monthly unsolved problems, 1969–1997, Amer. Math. Monthly 104 (1997) 967–973. [8] A. Ehrenfeucht, The Cube Made Interesting, Pergamon Press, Oxford, 1964. [9] J. G. Mauldon, Problem 10251, Amer. Math. Monthly 99 (1992) 782. [10] D. J. E. Schreck, Prince Rupert’s problem and its extension by Pieter Nieuwland, Scripta Math. 16 (1950) 73–80, 261–267. [11] J. Wallis, De Algebra Tractatus, 1685, in Opera Mathematica, vol. 2, Oxoniae: E Theatro Sheldoniano, 1693, pp. 470–471. [12] J. E. Wetzel, Rectangles in rectangles, Math. Mag. 73 (2000) 204–211.
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II Number Theory and Graph Theory
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6 Transcendentals and Early Birds Martin Gardner It’s hard to believe that it was not until 1844 that transcendental numbers were known to exist! But first, a few definitions. A rational number is one that can written as a=b, where a and b are integers. In decimal form, rational numbers either terminate (1=4 D :25) or they have a pattern of consecutive digits that repeat endlessly (1=7 D :142857 142857 142857 : : :). An irrational number is one that cannot be expressed as a=b where a and b are integers. In decimal form, it never ends, and it has no pattern of consecutive digits that keep repeating. An algebraic number is the root of a polynomial equation with rational coefficients. It can be rational or irrational. Thus 2 is algebraic because it is the root of such equations as 10x D 20. The square root of 2 is algebraic because it is the root of x 2 D 2. Transcendental numbers, such as and e, are irrational numbers that are not the roots of algebraic equations with rational coefficients. In 1844, the French mathematical Joseph Liouville (18091882) first proved the existence of transcendentals by actually constructing an infinite number of them. Liouville numbers, as they are known, can be called artificial because Liouville did not find them anywhere in mathematics. He simply made them up out of whole cloth. The simplest Liouville number is the binary transcendental shown below: :110001000000000000000001 : : : The ones are at positions given by consecutive factorials. The first 1 is at position 1Š D 1, the second one is at position 2Š D 2, the third is at position 3Š D 6, the fourth is at position 4Š D 24, and so on. It’s easy to construct artificial numbers that are obviously irrational but not transcendental (like the square root of two). It is not so easy to construct such numbers that are transcendental. Consider :101001000100001 : : : Liouville showed that any number of this form, where the ones can be replaced by any constant digit from 2 through 9, is transcendental. It was not long until , e, and scores of other famous irrationals were shown to be transcendental. In spite of the great recent progress in proving numbers transcendental, Reprinted from Math Horizons, Vol. 13, No. 2 (Nov. 2005), pp. 5, 34.
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many deep unsolved questions remain. For example, e has been proved transcendental, but no one yet knows if e is transcendental or even if C e or e or e e or are irrational! All are believed to be transcendental, but until some genius proves otherwise, it is possible that the sum of and e is a rational number! Other than Liouville numbers, the most famous of all artificial transcendentals is made simply by putting a decimal point in front of the counting numbers, like so: :123456789101112131415 : : : This is called Mahler’s number, after Kurt Mahler, a mathematician who first proved that this counting number is transcendental is all bases. You will find a neat proof for base 10 in Ivan Niven’sNiven, Ivan marvelous little book Irrational Numbers, recently republished by the Mathematical Association of America. A number is normal if every digit, every doublet, every triplet, or any specified pattern of digits appear the number of times you would expect if the number were generated by a randomizer. So far, , e, and the irrational roots of all algebraic numbers have passed all statistical tests for normalcy, but none has yet been proved normal. In 1934, it was shown ab is transcendental if a is an algebraic number not 0 or 1, and p b is any irrational number. Thus 2 and 10 2 are transcendental. Now for a startling coincidence. Note the last five digits of Mahler’s number. They are to four decimals! (I’m indebted to Jaime Poniachik, an Argentine puzzle maker, for telling me about this.) Every positive integer obviously appears somewhere in Mahler’s number, but many numbers show up far ahead of where they normally would occur as a counting number. Pi turns up so early because 3 is the last digit of 13, followed by 14 and 15. Poniachik points out that to 5 decimals appears as early as 141593141594. Consider 666. Of course it shows up in the counting sequence, but you can spot it early in the sequence 656667. Much innocent but totally useless amusement can be had by searching for the earliest appearance of such familiar numbers as integral powers of roots, the last four digits of your phone number, your house number, and so on. Better still, see if you can formulate a procedure for determining the earliest appearance of any given integer in the counting number. If a number appears ahead of its position as a counting number, let’s call it an early bird. The famous year 1492, for another example, is an early bird (491492). Poniachik points out that 8192, the thirteenth power of 2, is an early bird (181920). Like , the transcendental number e, to four decimals (2.7182) is also an early bird (27182719), though not as early as . What is the smallest early bird prime? When will the next coming year be an early bird? Mathematician Solomon W. Golomb, in a personal letter, mentioned that his home zip code 91011 (91011) is a very early bird. The early bird numbers, Golomb wrote, are a subset of the counting numbers, and therefore countably infinite. He is convinced that “almost all” integers are early birds. There are no onedigit early birds, but half (45 out of 90) twodigit numbers are early birds, with rapidly increasing members of the species as the number of their digits increases.
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7 Squaring, Cubing, and Cube Rooting Arthur T. Benjamin I still recall the thrill and simultaneous disappointment I felt when I first read Mathematical Carnival [4] by Martin Gardner. I was thrilled because, as my high school teacher had told me, mathematics was presented there in a playful way that I had never seen before. I was disappointed because Gardner quoted a formula that I thought I had “invented” a few years earlier. I have always had a passion for mental calculation, and the formula (7.1) below appears in Gardner’s chapter on “Lightning Calculators.” It was used by the mathematician A. C. Aitken to square large numbers mentally.
Squaring Aitken took advantage of the following algebraic identity. A2 D .A
d /.A C d / C d 2 :
(7.1)
Naturally, this formula works for any value of d , but we should choose d to be the distance to a number close to A that is easy to multiply. Examples To square the number 23, we let d D 3 to get 232 D 20 26 C 32 D 520 C 9 D 529: To square 48, let d D 2 to get 482 D 50 46 C 22 D 2300 C 4 D 2304: With just a little practice, it’s possible to square any twodigit number in a matter of seconds. Once you have mastered those, you can quickly square threedigit numbers by rounding up and down to the nearest hundred. Examples 2232 D 200 246 C 232 D 49;200 C 529 D 49;729
9522 D 1000 904 C 482 D 904;000 C 2;304 D 906;304: Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 58–63.
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To do mental calculations of this size, one needs to be quick at multiplying 2digit and 3digit numbers by 1digit numbers, generating the answer from left to right. More details and examples are given in Secrets of Mental Math [1], which Martin encouraged me to write. As I prepared a DVD version of this book [2] recently, I learned some new methods for cubing and cuberooting. Although much of that material did not end up in the DVD, I thought it would be of interest to readers of the College Mathematics Journal.
Cubing To cube a twodigit number in your head quickly, it is worth learning how to multiply two twodigit numbers that are near each other. I call it the “close together method” and it is based on the formula .z C a/.z C b/ D z.z C a C b/ C ab;
(7.2)
where z is typically a number that ends in zero. Example For the problem 23 26, z D 20, a D 3, b D 6, leads to 23 26 D 20 29 C 3 6 D 580 C 18 D 598: Notice that on both sides of (7.2), the twodigit numbers being multiplied have the same sum (23 C 26 D 49 D 20 C 29). Example Thus, to do a problem like 88 86, since the numbers sum to 174 D 90 C 84, we can begin with the product 90 84: 88 86 D 90 84 C . 2/ . 4/ D 7560 C 8 D 7568: To cube a twodigit number, we can exploit the algebra A3 D .A
d /A.A C d / C d 2 A:
Example Thus to cube 23, we can do 233 D 20 23 26 C 32 23 D 20 598 C 9 23 D 11;960 C 207 D 12;167; where we used the closetogether method to do 23 26. But there is a faster way. Since we know that the problem will begin by doing 20 29, let’s build that into the algebra. Writing the problem a different way, to cube z C d , we do .z C d /3 D zŒz.z C 3d / C 3d 2 C d 3 :
(7.3)
Example To cube 23, z D 20 and d D 3 leads to the easier calculation
233 D 20 Œ20 29 C 27 C 33 D 20 607 C 33 D 12;140 C 27 D 12;167:
Equation (7.3) is an instance of Horner’s method for polynomial evaluation. Notice that when cubing a twodigit number, d will always be one of the numbers ˙1; ˙2; ˙3; ˙4; ˙5 and so 3d 2 will always be 3; 12; 27; 48 or 75, which reduces the mental effort. Also notice that in the first twodigit multiplication, the number z C 3d will always end with a digit that is three times the original last digit.
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41
Example For 883 , by tripling the last digit, we know that we will multiply our rounded number (90) by the number closest to 88 that ends in 4, namely 84. Thus, 883 D 90 Œ90 84 C 12 C . 2/3 D 90 7572
23 D 681;480
8 D 681;472:
As a side note, the last part of the calculation is not as hard as it looks. By splitting 7572 D 7500 C 72, 9 7572 D .9 7500/ C .9 72/ D 67;500 C 648 D 68;148: To complete the calculation, simply multiply this number by 10, and subtract 8 at the same time. The calculation of twodigit cubes this way is surprisingly fast. With practice, the squares and cubes of twodigit numbers below 50 will be so quick that you will be able to cube threedigit numbers as well. Example Exploiting the previous calculations: 3233 D 300 Œ300 369 C 3.232 / C 233 D 300 Œ110;700 C 1587 C 233
D 3 112;287 100 C 233 D 33;686;100 C 12;167 D 33;698;267:
Another calculating tip. After the second multiplication, you can almost always say the millions part of the answer (here, 33 million) and hold the hundreds digit on your fingers. Here, just raise 1 finger to “hold onto” the hundreds digit so you only have to remember 686 while you calculate the cube of 23. By the way, if you just want a good cubing approximation, simply compute z.z.z C 3d //. For example, 3233 369 300 300 D 33;210;000: This will come in handy in the calculation of cube roots.
Cube rooting One of the easiest feats of lightning calculation is determining the cube root of a perfect cube when the cube root is two digits. The following description comes from Martin Gardner’s classic book, Mathematics, Magic, and Mystery [5]. The cube root demonstration begins by asking members of the audience to select any number from 1 through 100, cube it, then call out the result. The performer instantly gives the cube root of each number called. To do the trick it is necessary first to memorize the cubes of the numbers from 1 through 10. An inspection of this table reveals that each cube ends in a different digit. The digit corresponds to the cube root in all cases except 2 and 3 and 7 and
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Part II. Number Theory and Graph Theory Table 7.1. Table of cubes. 1
2
3
4
5
6
7
8
9
10
1
8
27
64
125
216
343
512
729
1000
8. In these four cases the final digit of the cube is the difference between the cube root and 10. To see how this information is used by the lightning calculator, let us suppose that a spectator calls out the cube 250047. The last number is a 7 which tells the performer immediately that the last number of the cube root must be 3. The first number of the cube root is determined as follows. Discard the last three figures of the cube (regardless of the size of the number) and consider the remaining figures—in this example they are 250. In the above table, 250 lies between the cubes of 6 and 7. The lower of the two figures—in this case 6 —will be the first figure of the cube root. The correct answer, therefore, is 63. One more example will make this clear. If the number called out is 19,683, the last digit, 3, indicates that the last digit of the cube root is 7. Discarding the final three digits leaves 19, which falls between the cubes of 2 and 3. Two is the lower number, therefore we arrive at a final cube root of 27. I learned this trick in a book on mental magic [3] as a high school student, and wondered how it might be extended to larger problems, where the cube root was a threedigit number. I didn’t pursue that question seriously at the time, since the cube of a threedigit number can be as large as nine digits, and most calculators back then only went up to eight. Now that calculators with greater capacity are quite common, I have learned two quick ways to do this problem, using casting out nines and casting out elevens.
Casting out nines Ask someone to cube a threedigit number, and ask how many digits are in the answer. (It should be seven, eight, or nine digits.) Ask the volunteer to recite (or write down) the answer, and you can pretty easily determine the cube root in your head. The first digit and last digit of the cube root are determined exactly as before. The last digit of the cube root is uniquely determined by the last digit of the cube. (In fact, the last digit of the cube root is the last digit of the cube of the last digit of the cube!) For the first digit, we look at the magnitude of the millions. Example Let’s find the cube root of the perfect cube 377,933,067. The first digit of the cube root must be 7 (since 377 lies between 73 D 343 and 83 D 512) and the last digit of the cube root must be 3 (since the cube ends in 7). Hence the answer must be of the form 7‹3. How do we determine the middle digit? The method I use is simply to add the digits of the cube, mod 9. Here the digits sum to 45, which is a multiple of 9. This can only happen if the cube root is a multiple of three. In other words, the cube root must be 723 or 753 or 783. Since 377 is much closer to 343 than 512, we would go with a cube root of 723. For extra verification, we could estimate 7203 700 700 760 D 372;400;000.
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This same approach can be used when the cube is not a multiple of nine, since there are only two other possible outcomes, as seen in Table 7.2. If the cube reduces to 1 (mod 9), then the cube root reduces to 1 (mod 3). If the cube reduces to 8 (mod 9), then the cube root reduces to 2 (mod 3). Table 7.2. Table of cubes mod 9. n
0
1
2
3
4
5
6
7
8
n3
0
1
8
0
1
8
0
1
8
Example If the 3digit cube results in 19,248,832, then we know that the cube root is of the form 2‹8. To determine the middle digit, notice that the cube has digit sum 37, which reduces to 1 mod 9. Hence the original number is congruent to 1 mod 3, so the cube root must be either 208 or 238 or 268 or 298. Judging from the proximity of 19 to its surrounding cubes, 8 and 27, the answer 268 seems most likely. Since 2703 300 300 210 D 18;900;000, we have more confidence in our answer of 268.
Casting out elevens With a little extra mental effort and memory, we can determine the middle digit without any guesswork, because the cubes of the numbers 0 through 10 are all distinct mod 11, as shown in Table 7.3. Table 7.3. Table of cubes mod 11. n
0
1
2
3
4
5
6
7
8
9
10
n3
0
1
8
5
9
4
7
2
6
3
10
Example revisited For the cube root of 19,248,832, we know that the answer will begin with 2 and end in 8 as in the last example. By alternately adding and subtracting its digits from right to left, we see that it reduces to 2 3 C 8 8 C 4 2 C 9 1 D 9 .mod 11/: Hence the cube root must reduce to 4 (mod 11). Since the answer is of the form 2?8, then 2 C 8 ‹ D 4 (mod 11) results in an answer of 268, as well. Note that if the cube has 7 or 9 digits, then you can alternately add and subtract the digits in their natural order. If the cube has 8 digits, you can start with 11 then alternately subtract and add the subsequent digits. Example To find the number with cube 111,980,168, we see immediately that the answer will be of the form 4‹2. Casting out 11s from left to right, we compute 1 1 C 1 9 C 8 0 C 1 6 C 8 D 3. Here, the cube root must reduce to 9 (mod 11). Thus 4 C 2 ‹ D 9 .mod 11/ tells us that the missing digit is 8. Hence the cube root is 482. Between the two approaches, I prefer casting out nines over casting out elevens, since it is easier to add the digits of the cube as they are called out. I recommend that the reader play with a few examples using both methods, choose one to master, and cast out the other.
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Acknowledgment The author wishes to thank the referee for many helpful suggestions, and especially Martin Gardner for being such an important factor in his life.
Bibliography [1] A. Benjamin and M. Shermer, Secrets of Mental Math: The Mathemagician’s Guide to Lightning Calculation and Amazing Math Tricks, Random House, New York, 2006. [2] A. T. Benjamin, The Secrets of Mental Math (DVD Course), The Teaching Company, Chantilly, VA, 2011. [3] T. Corinda, 13 Steps to Mentalism, 4th ed., Tannen Magic Publishing, New York, 1968. [4] M. Gardner, Mathematical Carnival: A New Roundup of Tantalizers and Puzzles from Scientific American, Random House, New York, 1965. [5] ——, Mathematics, Magic, and Mystery, Dover, New York, 1956.
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8 Carryless Arithmetic Mod 10 David Applegate, Marc LeBrun, and N. J. A. Sloane Nim Forms of Nim have been played since antiquity and a complete theory was published as early as 1902 (see [3]). Martin Gardner described the game in one of his earliest columns [7] and returned to it many times over the years ([8]–[16]). Central to the analysis of Nim is Nimaddition. The Nimsum is calculated by writing the terms in base 2 and adding the columns mod 2, with no carries. A Nim position is a winning position if and only if the Nimsum of the sizes of the heaps is zero [2], [7]. Is there is a generalization of Nim in which the analysis uses the baseb representations of the sizes of the heaps, for b > 2, in which a position is a win if and only if the modb sums of the columns is identically zero? One such game, Rimb (an abbreviation of RestrictedNim) exists, although it is complicated and not well known. It was introduced in an unpublished paper [6] in 1980 and is hinted at in [5]. Despite his interest in Nim, Martin Gardner never mentions Rimb , nor does it appear in Winning Ways [2], which extensively analyzes Nim variants. In the present paper we focus on b D 10, and consider, not Rim10 itself, but the arithmetic that arises if calculations, addition and multiplication, are performed mod 10, with no carries. Along the way we encounter several new and interesting number sequences, which would have appealed to Martin Gardner, always a fan of integer sequences.
The Carryless Islands The fabled, carefree residents of the Carryless Islands in the remote South Pacific have very few possessions, which is just as well, since their arithmetic is illsuited to accurate bookkeeping. When they add or multiply numbers, they follow rules similar to ours, except that there are no carries into other digit positions. Sociologists explain this by noting that the Carryless Islands were originally penal colonies, and, as penal institutions are generally known to have excellent dental care, the islanders were, happily, generally free of carries. We will use C and for their operations , and C and for the standard operations used Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 43–50. We prefer not to use outlandish symbols such as  and , since C and are perfectly reasonable operations,
although to our eyes they have rather strange properties. As Marcia Ascher remarks, writing about mathematics in indigenous cultures, “in many cases these cultures and their ideas were unknown beyond their own boundaries, or misunderstood when first encountered by outsiders” [1].
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by the rest of the world. Addition and multiplication of singledigit numbers are performed by “reduction mod 10.” Carry digits are simply ignored, so 9C4 D 3, 5C5 D 0, 9C4 D 6, 54 D 0, and so on. Adding or multiplying larger numbers also follows the familiar procedures, but again with the proviso that there are no carries. For example, adding 785 and 376 produces 51, and the product of 643 and 59 is 417 (see Figure 8.1). (a)
7 8 5 C3 7 6 0 5 1
(b)
6 4 3
5 9 4 6 7
0 0 5 0 4 1 7 Figure 8.1. (a) Carryless addition. (b) Carryless multiplication.
What does elementary number theory look like on these islands? Let’s start with the carryless squares nn. For n D 0; 1; 2; 3 we get 0, 1, 4, 9, Then for n > 3 we have 44 D 6, 55 D 5, 66 D 6, 77 D 9, 88 D 4, 99 D 1, 1010 D 100; : : : ; giving the sequence 0; 1; 4; 9; 6; 5; 6; 9; 4; 1; 100; 121; 144; 169; 186; 105; 126; 149; 164; : : :: It turns out that this is entry A059729 in the OEIS [17], contributed by Henry Bottomley on February 20, 2001, although without any reference to earlier work on these numbers. Bottomley also contributed sequence A059692, giving the carryless multiplication table, and several other sequences related to carryless products. Likewise the sequence of values of nCn, 0; 2; 4; 6; 8; 0; 2; 4; 6; 8; 20; 22; 24; 26; 28; 20; 22; 24; 26; 28; 40; 42; : : : ; is entry A004520, submitted to the OEIS by one of the present authors around 1996, again without references. (If these numbers are sorted and duplicates removed, we get the carryless “evenish” numbers, that is, numbers all of whose digits are even, A014263.) Carryless arithmetic must surely have been studied before now, but the absence of references in [17] suggests that it is not mentioned in any of the standard texts on number theory.
The carryless primes If we require that a prime is a number whose only factorization is 1 times itself, we are out of luck, since every carryless number is divisible by 9, and there would be no primes at all. (For 91 D 9, 92 D 8, 93 D 7; : : : ; 99 D 1. So if we construct a number by replacing all the 1’s in by 9’s, all the 2’s by 8’s, : : : then D 9, and would not be a prime.) There are primes, when defined in the right way. Since 11 D 1, 37 D 1 and 99 D 1, all of 1; 3; 7 and 9 divide 1 and so divide any number. We call 1; 3; 7 and 9 units, the usual name for integers that divide 1. Units should not be counted as factors when considering if
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a number is prime (just as factors of 1 are ignored in ordinary arithmetic: 7 D . 1/. 7/ doesn’t count as a factorization when considering if 7 is a prime). So we define a carryless prime to be a nonunit whose only factorizations are of the form D u where u is a unit. Computer experiments suggest that the first few primes are 21; 23; 25; 27; 29; 41; 43; 45; 47; 49; 51; 52; 53; 54; 56; 57; 58; 59; 61; 63; : : : ;
(8.1)
but there are surprising omissions in this list, resulting from some strange factorizations: 2 D 251, 10 D 5665, 11 D 5161. It is hard to be sure at this stage that the above list is correct, since there exist factorizations where one of the numbers is much larger than the number being factored, such as 2 D 45005505553. One property that makes carryless arithmetic interesting is the presence of zerodivisors: the product of two numbers can be zero without either of them being zero: 25 D 0, 62855 D 0. Perhaps 21 is the product of two really huge numbers? Nonetheless, the list is correct, as we will see (it is now sequence A169887 in [17]).
Algebra to the rescue The secret to understanding carryless arithmetic is to introduce a little algebra. Let R10 denote the ring of integers mod 10, and R10 ŒX the ring of polynomials in X with coefficients in R10 . Then we can represent carryless numbers by elements of R10 ŒX: 21 corresponds to 2X C 1, 109 to X 2 C 9, and so on. Carryless addition and multiplication are simply addition and multiplication in R10 ŒX: our first example, 785C376 D 51 corresponds to .7X 2 C 8X C 5/ C .3X 2 C 7X C 6/ D 5X C 1; where the polynomials are added or multiplied in the usual way, and the coefficients then reduced mod 10. Conversely, any element of R10 ŒX represents a unique carryless number (just set X D 10 in the polynomial). In fact arithmetic in R10 ŒX is clearly exactly the same as the arithmetic of carryless numbers. This could be used as a formal definition of carryless arithmetic mod 10. It also shows that this arithmetic is commutative, associative and distributive. Since R10 ŒX is a ring, we can not only add and multiply, we can also subtract, something the Carryless Islanders never considered. The negatives of the elements of R10 are 1 D 9, 2 D 8; : : : ; 9 D 1, and similarly for the elements of R10 ŒX. So the negative of a carryless number is its “10’s complement,” obtained by replacing each nonzero digit d by 10 d , for example 702 D 308. To subtract A from B, we add A to B: 650 702 D 650C308 D 958. This is equivalent to doing elementary school subtraction where we can “borrow” but don’t have to pay back! The units in R10 ŒX, that is, the elements that divide 1, are the constants 1; 3; 7; 9, and the carryless primes that we defined are the irreducible elements in R10 ŒX, that is, nonunits f10 .X/ 2 R10 ŒX whose only factorizations are of the form f10 .X/ D ug10 .X/, where u is a unit and g10 .X/ 2 R10 ŒX. The units can also be written as 1; 1; 3 and 3,
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which more closely relates them to the units 1 and 1 in ordinary arithmetic (3 and 3 act in some ways like the imaginary units i and i , squaring to 1, for example). The key to further progress is to notice that R10 is the direct sum of the ring R2 of integers mod 2 and the ring R5 of integers mod 5. Given r10 2 R10 , we read it mod 2 and mod 5 to obtain a pair Œr2; r5 with r2 2 R2 , r5 2 R5 . The elements 0; 1; : : : ; 9 2 R10 (or equivalently the carryless digits 0; 1; : : : ; 9) and their corresponding pairs Œr2 ; r5 are given by the following table. The Chinese Remainder Theorem guarantees that this is a onetoone correspondence. 0 1 2 3 4 5 6 7 8 9 [0,0] [1,1] [0,2] [1,3] [0,4] [1,0] [0,1] [1,2] [0,3] [1,4]
(8.2)
As a check, we note that f1g is the (singleton) set of units in R2 , while f1; 2; 3; 4g is the set of units in R5 , so the pairs Œ1; 1, Œ1; 2, Œ1; 3 and Œ1; 4 correspondingly produce the units 1, 7, 3 and 9 of R10 . Similarly, polynomials f10 .X/ 2 R10 ŒX correspond to pairs of polynomials Œf2 .X/; f5 .X/, obtained by reading f10 .X/ respectively mod 2 and mod 5. Conversely, given any such pair of polynomials Œf2 .X/; f5 .X/, there is a unique f10 .X/ 2 R10 ŒX that corresponds to them, which can be found using (8.2). We indicate this by writing f10 .X/ $ Œf2 .X/; f5 .X/. If also g10 .X/ $ Œg2 .X/; g5 .X/, then and
f10 .X/ C g10 .X/ $ Œf2 .X/ C g2 .X/; f5 .X/ C g5 .X/ f10 .X/g10 .X/ $ Œf2 .X/g2 .X/; f5 .X/g5 .X/: We are now in a position to answer many questions about carryless arithmetic.
The carryless primes, again What are the irreducible elements f10 .X/ 2 R10 ŒX? If f10 .X/ $ Œf2 .X/; f5 .X/ is irreducible then certainly f2 and f5 must be either units or irreducible, for if f2 D g2 h2 then we have the factorization Œf2 ; f5 D Œg2 ; f5 Œh2; 1. Also Œf2 ; f5 D Œf2 ; 1Œ1; f5, so one of f2 , f5 must be irreducible and the other must be a unit. So the irreducible elements in R10 ŒX are of the form Œf2 .X/; u, where f2 .X/ is an irreducible polynomial mod 2 of degree 1 and u 2 f1; 2; 3; 4g, together with elements of the form Œ1; f5.X/, where f5 .X/ is an irreducible polynomial mod 5 of degree 1. The irreducible polynomials mod 2 are X, X C 1, X 2 C X C 1; : : : ; and the irreducible polynomials mod 5 are uX, uX C v; : : : ; where u; v 2 f1; 2; 3; 4g (see entries A058943, A058945 in [17]). The first few irreducible elements in R10 ŒX are therefore ŒX; 1, ŒX; 2, ŒX; 3, ŒX; 4, ŒX C 1; 1, ŒX C 1; 2; : : : ; and Œ1; X, Œ1; 2X, Œ1; 3X, Œ1; 4X, Œ1; X C 1, Œ1; 2X C 1; : : : : The corresponding carryless primes, according to (8.2), are 56; 52; 58; 54; 51; 57; : : :; and 65; 25; 85; 45; 61; 21; : : : : And so we can verify that the list in (8.1) is correct. We will call a number with at least two digits in which all digits except the rightmost are even but the rightmost is odd an etype number (A143712), and a number with at least two digits in which all digits except the rightmost are 0 or 5 and the rightmost is
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neither 0 nor 5 an ftype number (A144162). Similarly, we call the primes corresponding to the irreducible elements Œ1; f5 .X/ etype primes, and the primes corresponding to the irreducible elements Œf2 .X/; u ftype primes. We also see that our earlier concern about the primality of 21 was groundless. It is impossible for the length (in decimal digits) of a nonzero carryless product to be less than the length of both of the factors. This follows from the fact that if `.n/ is the number of decimal digits in the number n > 0 corresponding to a pair Œf2 .X/; f5 .X/, then `.n/ D 1 C maxfdeg f2 ; deg f5 g. So if mn > 0, `.mn/ minf`.m/; `.n/g. Also, since we know how many irreducible polynomials mod 2 and mod 5 there are of given degree (see A001037, A001692 in [17]), we can write down a formula for the number of kdigit carryless primes, something that we cannot do for ordinary primes, namely 4 k
1
X
d divides k 1
k
1 d
.2d C 5d /;
for k 2, where is the M¨obius function (A008683). There are 28 primes with two digits (the twenty listed in (8.1), together with 65; 67; 69; 81; 83; 85; 87; 89), 44 with three digits, : : : (A169962). For large k the number is about 4 5k 1 =.k 1/, whereas the number of ordinary primes with exactly k digits is much larger, about 910k 1 =.k log 10/, so carryless primes are much rarer than ordinary primes. Incidentally, the prime ideals in R10 ŒX, as distinct from the irreducible elements, all have a single generator, which is one of Œ0; 1; Œ1; 0; Œ1; 1; Œf2.X/; 1; Œ1; f5.X/, where f2 .X/, f5 .X/ are irreducible (cf. [18, Chap. III, Thm. 30]).
The carryless squares, again Squaring a mod 2 polynomial is easy: f2 .X/2 D f2 .X 2 /. So if n corresponds to the pair Œf2 .X/; f5 .X/, n2 corresponds to Œf2 .X 2 /; f5 .X/2 D Œf2 .X 2 /; 0 C Œ0; f5.X/2 . This gives a twostep recipe for producing all carryless squares. First find, using (8.2), the carryless number m corresponding to Œ0; f5 .X/2 , where f5 .X/ is any polynomial mod 5. The effect of adding a nonzero Œf2 .X 2 /; 0 changes some subset of the digits in positions 0; 2; 4; : : : of m by the addition of 5 mod 10. For example, if f5 .X/ D X C 2, f5 .X/2 D X 2 C 4X C 4, and by (8.2) Œ0; f5.X/2 corresponds to the carryless square m D 644. We now add 5 mod 10 to any subset of the digits in positions 0; 2; 4; 6; : : : of m (considering m extended by prefixing it with any number of zeros), obtaining infinitely many squares 644; 649; 144; 149; : : : ; 50644; 5050649; : : :: This also leads to a formula for the number of kdigit carryless squares. For even k the number is 0, and for odd k it is 1 9 10.k 2
1/=2
C 2.k
3/=2
(zero is excluded from the count). There are five squares of length 1 (namely 1; 4; 5; 6 and 9), 46 of length 3, : : : (see A059729, A169889, A169963). For large odd k there are about twice as many kdigit carryless squares as ordinary squares.
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Divisors and factorizations What about the factorization of numbers into the product of carryless primes? Unfortunately, the existence of zerodivisors complicates matters, and it turns out that there is no natural way to define, for example, an analog of the usual sumofdivisors function .n/. In our analysis we define several classes of carryless numbers: U WD f1; 3; 7; 9g, the units, E WD f0; 2; 4; 6; 8; 20; 22; : : :g, the â€œevenishâ€? numbers, in which all digits are even (A014263), F WD f0; 5; 50; 55; : : :g, the â€œfiveishâ€? numbers, in which all digits are 0 or 5 (A169964), Z WD E [ F D f0; 2; 4; 5; 6; 8; 20; 22; : : :g, the zerodivisors (A169884), N WD f1; 3; 7; 9; 10; 11; 12; 13; : : :g, the positive numbers not in Z (A169968).
Suppose d is a carryless divisor of n, that is, there is a number q such that d q D n. What can be said about the possible choices for q? One can showâ€”we omit the straightforward proofsâ€”that if d 2 N then there is a unique q, if d 2 E then d q 0 D n if and only if q 0 D q C v for some v 2 F , if d 2 F then d q 0 D n if and only if q 0 D q C e for some e 2 E. The same distinctions are needed to describe factorizations into primes. If n 2 N then n has a unique factorization as a carryless product of primes, up to multiplication by units. For example, we already saw 10 D 5665. But we also have 10 D .356/.765/ D 5825 D .956/.965/ D 5445 D 95225, etc., illustrating the nonuniqueness. Also 11 D 5161; 101 D 212951, 1234 D 232323515152. It follows that any nonunit in N can be written both as ef and e 0 Cf 0 , where e and e 0 are etype numbers and f and f 0 are ftype numbers. For example, 12 D 8152 D 61C51. If n 2 E then n has a unique factorization as 2 times a product of etype primes, up to multiplication by units (in this case, every ftype prime divides n). For example, 20 D 265, 22 D 261, 2468 D 2696969. If n 2 F then n has a unique factorization as 5 times a product of ftype primes, up to multiplication by units (in this case, every etype prime divides n). For example, 50 D 552, 505 D 55151. Here are the analogous statements about divisors: if n 2 N , n has only finitely many divisors. If d divides n and u 2 U, then d u divides n. The divisors may be grouped into equivalence classes d U. Since the sum of the elements of U is zero, so is the sum of the divisors of n. if n 2 E, d divides n, u 2 U and v 2 F , then d u C v divides n. So n has infinitely many divisors, belonging to equivalence classes d U C F . if n 2 F , d divides n, u 2 U and e 2 E, then d u C e divides n. So n has infinitely many divisors, belonging to equivalence classes d U C E. Any attempt to define a sumofdivisors function must specify how to choose representatives from the equivalence classes. There seems to be no natural way to do this. One possibility would be to choose the smallest decimal number in each class, but this seems unsatisfactory (since it depends on the ordering of decimal numbers, another concept the islanders seem not to be familiar with).
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Further number theory In summary, we can help the Carryless Islanders by defining subtraction, prime numbers, and factorization into primes. But further concepts such as the number of divisors, the sum of divisors and perfect numbers seem to lie beyond these Islands. However, many other carryless analogs are welldefined, including including triangular numbers (A169890), cubes (A169885), partitions (A169973), greatest common divisors and least common multiples, and so on. Some seem exotic, while other familiar sequences simply become periodic. For example, the analog of the Fibonacci numbers coincides with the sequence of Fibonacci numbers read mod 10, A003893, which becomes periodic with period 60 (the periodicity of the Fibonacci numbers to any modulus being a wellstudied subject, see sequence A001175). Similarly, the analogue of the powers of 2 (A000689) becomes periodic with period 4. We might also generalize beyond simple squares, cubes, etc., and investigate the properties of polynomials or power series based on carryless operations—How do these factor? What are their fixed points?—and so on. Taking a different tack, carryless mod 10 partitions are enumerated in A169973, which may be derived as the coefficients of z n in the formal expansion of the analog of the classic Q k partition generating function 1 kD1 .1 C z /, wherein powers of z are multiplied together by combining their exponents with carryless mod 10 addition instead of the ordinary sum.
Afterword There’s a great deal yet to be explored in these Carryless Islands! Watch for our next paper on another carryless arithmetic, in which operations on single digits are defined by a ˚ b D maxfa; bg, a ˝ b D minfa; bg. We call this “dismal arithmetic.” When the Handbook of Integer Sequences was published 39 years ago, Martin Gardner was kind enough to write in his Mathematical Games column of July 1974 that “every recreational mathematician should buy a copy forthwith.” That book contained 2372 sequences: today its successor, the OnLine Encyclopedia of Integer Sequences (or OEIS) [17], contains nearly 200,000 sequences. We were about to write to Martin about carryless arithmetic when we heard the sad news of his death. This article, the first of a series on various kinds of carryless arithmetic, is offered in his honor.
Bibliography [1] Marcia Ascher, Mathematics Elsewhere: An Exploration of Ideas Across Cultures, Princeton Univ. Press, Princeton, NJ, 2002. [2] E. R. Berlekamp, J. H. Conway, and R. K. Guy, Winning Ways for Your Mathematical Plays, A K Peters, Wellesley MA, 2nd ed., 4 vols, 2004. [3] C. L. Bouton, Nim, a game with a complete mathematical theory, Ann. Math. 3 (1902) 35–39; available at dx.doi.org/10.2307/1967631 [4] J. H. Conway, On Numbers and Games, Academic Press, NY, 1976. [5] T. S. Ferguson, Some chip transfer games, Theoret. Comput. Sci. 191 (1998) 157–171; available at dx.doi.org/10.1016/S03043975(97)001357 [6] J. A. Flanigan, NIM, TRIM and RIM, unpublished document, Mathematics Department, University of California at Los Angeles, 1980; available at citeseer.ist.psu.edu/viewdoc/summary?doi=10.1.1.74.955.
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[7] Martin Gardner, Nim and Tac Tix, The Scientific American Book of Mathematical Puzzles and Diversions, Simon & Schuster NY, 1959. [8] ——, Jam, Hot and Other Games, Mathematical Carnival, Vintage Books NY, 1977. [9] ——, Nim and Hackenbush, Wheels, Life and Other Mathematical Amusements, W. H. Freeman NY, 1983. [10] ——, Sim, Chomp and Race Track, Knotted Doughnuts and Other Mathematical Entertainments, W. H. Freeman NY, 1986. [11] ——, Dodgem and Other Simple Games, Time Travel and Other Mathematical Bewilderments, W. H. Freeman NY, 1988. [12] ——, Wythoff’s Nim, Penrose Tiles to Trapdoor Ciphers : : : and the Return of Dr. Matrix, W. H. Freeman NY, 1989. [13] ——, Matches, Mathematical Circus, Mathematical Association of America, Washington DC, revised ed., 1992. [14] ——, The Rotating Table and Other Problems, Fractal Music, Hypercards and More : : : ; W. H. Freeman NY, 1992. [15] ——, Lavinia Seeks a Rule and Other Problems, The Last Recreations: Hydras, Eggs and Other Mathematical Mystifications, Springer NY, 1997. [16] ——, Surreal Numbers, The Colossal Book of Mathematics, W. W. Norton NY, 2001. [17] The OEIS Foundation Inc., The OnLine Encyclopedia of Integer Sequences (2011); available at oeis.org. [18] O. Zariski and P. Samuel, Commutative Algebra, Van Nostrand NY, vol. I, 1958.
The Mad Tea Party Ride
Alice and friends begin a mathematical adventure (see next page).
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9 Mad Tea Party Cyclic Partitions Robert Bekes, Jean Pedersen, and Bin Shao Alice, the March Hare, the Hatter, and the Dormouse, were standing by the Mad Tea Party (MTP) ride (see opposite page). “We’ll start with the same number of teacups as people,” said the Hatter, bossy as usual. “The teacups are arranged in a circle, and each person sits in his or her own teacup.” “Won’t that be lonely?” objected Alice. “Don’t worry,” replied the Hatter, “At the end of this ride, everyone stands up, one of the teacups is removed, then everyone finds a new place to sit. Every teacup has to be occupied by at least one person, so on the second ride there will be one teacup with two people in it. After that ride, another cup will disappear and eventually, when there is just one teacup, no one will be lonely.” The rest said in unison, “You must be mad. Everyone will be squished and it will be quite unbearable.” “Not my problem,” the Hatter said, and walked away. “Wait,” said Alice, who was a keen amateur mathematician, “If there are n people, how many ways are there to fit them into k cups arranged in a circle, assuming, naturally, all people are alike?” She quickly wrote out the possibilities for n D 5 and k 5 in Table 9.1, where the actual partitions are in the second row and the number of partitions are in the third row (consistent with the results in [7]). Alice explained to the group that the sum of the numbers in the third row (which is 7) is the answer to her question. Then she noticed that product of the numbers in the third row (which is 4) tells us the number of different paths through the various partitions available at each stage. (This paper obtains both results, but concentrates on the nature and number of the actual partitions.) Table 9.1. k D Number of cups
5
4
3
2
1
The listing of partitions for k cups
11111
1112
1 1 3 or 1 2 2
1 4 or 2 3
5
The number of paritions for k cups
1
1
2
2
1
But finding them became much tougher with more teacups. “This is hard,” the March Hare said, “Martin Gardner wrote several times about partitions [5]–[7]. The scariest part, Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 25–36.
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for us, is in [7] where Gardner wrote: Although there are many recursive procedures for counting unordered partitions, using at each step the number of known partitions for all smaller numbers, an exact asymptotic formula was not obtained until recent times. The big breakthrough was made by the British mathematician G. H. Hardy, working with his Indian friend Srinivasa Ramanujan. Their not quite exact formula was perfected by Hans A. Rademacher in 1937. The HardyRamanujanRademacher formula is a horribly shaggy infinite series that involves (among other things) pi, square roots, complex roots, and derivatives of hyperbolic functions! So, how can we find the answer to an even more difficult problem?” Now Alice had read the paperfolding results of Hilton and Pedersen ([4, pp. 129– 136], [5, pp. 107–109], and [6]) and thought to apply their ideas. A recursive algorithm would do. Though it wouldn’t give a closed form, it would be, conceptually, far simpler than the HardyRamanujanRademacher formula. “Sometimes it’s easier to solve a more difficult problem and use that solution to answer the easier problem,” Alice reminded the others, “That’s what we should do here. The first thing we need to do is give our cyclic arrangements a name and decide on an appropriate notation. Let’s call them MTP partitions. Our problem is, how many MTP partitions are there for any positive integer n?” For notation we will write, as an example, .2; 1; 3/ to mean there are 2, 1, and 3 people in three cups arranged clockwise in a circle beginning with a teacup containing 2 people. Sometimes, when no confusion occurs, instead of .2; 1; 3/ we will simply write 2 1 3. The March Hare commented that the partition .1; 3; 2/ must be the same as .1; 2; 3/, because if you turned the MTP ride upside down they would be the same. “No way.” the Hatter responded contemptuously, “Disney won’t turn the ride upside down. And if you can’t do that, then these two partitions are different. People in teacups on the left and right side of the teacup with 2 people in it are clearly different in these 2 cases.” Notice that in Table 9.1 the partition 1 1 1 2 would be the same as 1 1 2 1, 1 2 1 1, or 2 1 1 1, because they are cyclic permutations of each other. The modest Dormouse said, “All I know how to do is subtract odd numbers from odd numbers, divide even numbers by 2, and add numbers that sum to less than 100.” “That’s great!” said Alice, ”you have just the mathematical skills we need to tackle this problem.” And she began to explain: First, I’ll teach you, with some examples, how to construct coaches, and how they constitute symbols. We always start with an odd number b and use all the odd numbers ai , where ai < 2b . You’re going to wonder what on earth these symbols are about— but I can’t take time to discuss that; you simply must consult [4]–[6] if you want to know more. Second, I’ll describe how to construct the coaches in symbols in purely mathematical language. After more examples I’ll state two theorems about the coaches and symbols, give a reconstruction algorithm, and observe some persistent patterns among coaches in two special sequences. All this information gives us a nonrecursive algorithm that answers our main question: How many MTP partitions are there for any positive integer n?
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Constructing coaches and symbols We are going to show how to write a coach that involves an initial odd number b and odd ai , such that ai < 2b . As an example, I’ll begin with b D 35. The possible values for ai are 1; 3; 5; 7; 9; 11; 13; 15 and 17. We begin with the smallest value, which we call a1 and write the following ˇ ˇa1 D 1 b D 35 ˇˇ :
Now subtract 1 from 35 and divide the answer (34) as many times as possible by 2 (in this case, once), obtaining remainder 17. The number of factors of 2 that were divided out (1) is written at the bottom of the first column. Then 17 is written at the top of next column. This gives ˇ ˇa1 D 1 a2 D 17 b D 35 ˇˇ : k1 D 1
We repeat this process (subtracting ai from b and dividing by factors of 2 until we get an odd number) with a2 ; a3 ; : : : until the next value of ai is equal to the original a1 D 1 (in this case), as must happen (see [4] or [5]). In the last step we subtract 3 from 35 and divide the answer (32) as many times as possible by 2 (5 times), obtaining 1. We write 5 under the 3 and stop, since 1 is our original number a1 for this coach. This completed verbose coach is: ˇ ˇ ˇa D 1 a2 D 17 a3 D 9 a4 D 13 a5 D 11 a6 D 3ˇ ˇ: b D 35 ˇˇ 1 k1 D 1 k2 D 1 k3 D 1 k4 D 1 k5 D 3 k5 D 5ˇ Eliminating the labels ai ; ki ; we have ˇ ˇ1 b D 35 ˇˇ 1
In sum, the arithmetic for this coach is 35 35 35 35 35 35
1 D 34 D 21 17; 1
17 D 18 D 2 9; 1
9 D 26 D 2 13; 1
13 D 22 D 2 11; 3
11 D 24 D 2 3; 5
3 D 32 D 2 1;
ˇ 17 9 13 11 3ˇˇ : 1 1 1 3 5ˇ
k1 D 1; a2 D 17I
k2 D 1; a3 D 9I
k3 D 1; a4 D 13I
k4 D 1; a5 D 11I
k5 D 3; a6 D 3I
k6 D 5; a7 D 1 D a1 .StopŠ/:
Now not all the odd numbers ai < 35 2 have appeared in the top row of this coach. So we begin again, using the smallest unused ai (so we begin with a1 D 5), and construct the next coach of the symbol obtaining: ˇ ˇ ˇ5 15ˇ ˇ ˇ: b D 35 ˇ 1 2ˇ The arithmetic for this coach is: 35 35
5 D 30 D 21 15; 2
15 D 20 D 2 5;
k1 D 1; a2 D 15I
k2 D 2; a3 D 5 D a1 .StopŠ/:
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But 7 still hasn’t appeared in the top row of a coach, so we begin again, with a1 D 7, and construct the next coach of the symbol obtaining: ˇ ˇ ˇ7ˇ b D 35 ˇˇ ˇˇ : 2 The arithmetic for this coach is:
7 D 28 D 22 7; k1 D 2; a2 D 7 D a1 .StopŠ/:
35
Now that all possible ai s have appeared in the top row of some coach we have all the possible coaches. We combine them to obtain the complete symbol for b D 35: ˇ ˇ ˇ ˇ ˇ1 17 9 13 11 3 ˇ5 15ˇ 7ˇ ˇ ˇ ˇ ˇ: b D 35 ˇ 1 1 1 1 3 5 ˇ1 2 ˇ 2ˇ
In order to make our symbols unique we will always require that each coach have the smallest available ai in the position of a1 . In general, each coach of a symbol is written: ˇ ˇ ˇa a2 ar ˇ ˇ; b ˇˇ 1 (?) k1 k2 kr ˇ
where b and ai are odd, with each ai < b2 , and b
ai D 2ki ai C1 ; i D 1; 2; : : : ; r; ar C1 D a1 :
(??)
Note that ki is maximal (that is, all powers of 2 have been factored from b ai /, since ai C1 must be odd. In [4]–[6] it is shown that, given any two odd numbers a and b, with a < b2 , there is always a completely determined, unique symbol consisting of coaches as determined by (?) and .??); sometimes a symbol may consist of only one coach. If for any i , gcd.b; ai / D 1, we say that the coach (?) is reduced, and if there are no repeats among all the ai ’s we say that the coach (?) is contracted. At this stage, we do not assume that our coaches are reduced, but we do assume they are contracted. The Dormouse said, “Aren’t all coaches contracted because of the way we constructed them?” “Yes,” said Alice, “but later we’ll have good reason to construct noncontracted coaches.”
Alice produces examples and theorems If a coach in a symbol is contracted, but not reduced (allowing gcd.b; ai / ¤ 1) we call that coach and the symbol that contains it hybrid. If all the coaches in a symbol are contracted and reduced we call it pure. Now let’s look at some examples (including b D 35) of pure and hybrid symbols. “Look for patterns,” Alice instructed the Dormouse and the March Hare. “See if you can anticipate the theorems. There is a special reason for singly or doublyunderlining some coaches and boxing some entries in Figure 9.1.” Can you see why?
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9. Mad Tea Party Cyclic Partitions Ë‡ Ë‡1 b D 23Ë‡Ë‡ 1 Ë‡ Ë‡1 b D 35Ë‡Ë‡ 1 Ë‡ Ë‡1 b D 41Ë‡Ë‡ 3 Ë‡ Ë‡1 b D 57Ë‡Ë‡ 3 Ë‡ Ë‡1 b D 63Ë‡Ë‡ 1
Ë‡ 11 3 5 9 7Ë‡Ë‡ 2 2 1 1 4Ë‡
Ë‡ 17 9 13 11 3Ë‡Ë‡ 5 1 1 1 3 5Ë‡ 1
Ë‡ Ë‡9 Ë‡ Ë‡1
Ë‡ Ë‡ 15Ë‡Ë‡ 7 Ë‡Ë‡ 2Ë‡ 2 Ë‡
Ë‡ Ë‡ 5 9Ë‡Ë‡3 19 11 15 13 7 17Ë‡Ë‡ 2 5Ë‡1 1 1 1 2 1 3 Ë‡ Ë‡ Ë‡ Ë‡ 7 25Ë‡Ë‡ 3 27 15 21 9Ë‡Ë‡5 13 11 23 17Ë‡Ë‡ 19 1 5Ë‡ 1 1 1 2 4Ë‡ 2 2 1 1 3 Ë‡ 1 Ë‡ 31Ë‡Ë‡ 3 5Ë‡ 2
Ë‡ Ë‡ 15Ë‡Ë‡5 29 17 23Ë‡Ë‡ 7 4 Ë‡1 1 1 3 Ë‡ 3
Ë‡ Ë‡ Ë‡ Ë‡1Ë‡3 31 17Ë‡ 5 Ë‡ Ë‡ Ë‡ b D 65Ë‡ Ë‡ 6 1 1 4Ë‡ 2 Ë‡ Ë‡1 b D 91Ë‡Ë‡ 1
57
Ë‡ 15 25Ë‡Ë‡7 1 3 Ë‡1
Ë‡ Ë‡9 Ë‡ Ë‡1
Ë‡ Ë‡ Ë‡ Ë‡
Ë‡ Ë‡ 27Ë‡Ë‡11 13 25 19Ë‡Ë‡ 21 2Ë‡2 1 1 2Ë‡ 1
Ë‡ Ë‡ 29 9Ë‡Ë‡11 27 19 23 21Ë‡Ë‡ 13 2 3Ë‡ 1 1 1 1 2 Ë‡ 2
Ë‡ Ë‡ 45 23 17 37 27Ë‡Ë‡3 11 5 43Ë‡Ë‡ 7 1 2 1 1 6 Ë‡3 4 1 4 Ë‡ 2
Ë‡ 41 25 33 29 31 15 19Ë‡Ë‡ 13 1 1 1 1 2 2 3Ë‡ 1
Ë‡ 39Ë‡Ë‡ 2Ë‡
Ë‡ 21 35Ë‡Ë‡ 1 3Ë‡
Ë‡ Ë‡ Ë‡ Ë‡
Ë‡ Ë‡ Ë‡ Ë‡
Figure 9.1. Examples of symbols.
Did you notice that: Where the symbol has no underlined coaches, b is prime?
The underlined coaches are not reduced, furthermore there is a common divisor (other than 1) between all of the numbers in the top row (the ai ) and b? Each of the boxed numbers divides into b and all the ai for that coach?
In the doublyunderlined hybrid coaches the sum of the bottom row is a proper divisor of the sum of the bottom row for the pure coaches?
If we eliminate the underlined coaches in Figure 9.1, those remaining constitute pure symbols (gcd.b; ai / D 1). For a given b what do you notice about the sum of the ki s in the coaches of the pure symbols? For a given b what is r (the number of columns) in each pure coach? These are crucial questions to attempt to answer before going on. Your observations may help you to understand the following (proved in [4]â€“[6]): The QuasiOrder Theorem. Suppose that a coach of a pure symbol has the form (?), P and formula (??) holds. Let riD1 ki D k. Then k is the smallest positive integer such that 2k Ë™1 .mod b/. In fact, 2k . 1/r .mod b/. In other words the sum of the numbers, ki , in the bottom row of any coach of a pure symbol is the smallest number, k, such that b exactly divides 2k . 1/r . For the pure symbols in Figure 9.1 the QuasiOrder Theorem results are in Table 9.2. Notice that the conclusion of the QuasiOrder Theorem can be obtained from any coach of the pure symbol since k is the same and r is always either even or odd.
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Part II. Number Theory and Graph Theory Table 9.2. b
23
35
41
57
63
65
91
k
11
12
10
9
6
6
12
6
6
3 or 7
3 or 5
2 or 4
1; 3; or 5
4; 6; or 8
4047
4096
1025
513
63
65
4095
r 2k
.1/r
Here is another pertinent HiltonPedersen theorem [4] connected with the QuasiOrder Theorem. The Coach Theorem. Let be the Euler phi function. In a complete, pure symbol, with Pr i D1 ki D k, in .?/, and c the number of coaches, we have (b/ D 2kc. Recall that if p is a prime then .p n / D .p 1/p n 1 , and if p1 , p2 are distinct prime n n n n numbers then (p1 1 p2 2 / D .p1 1 /.p2 2 ). The data confirming that .b/ D 2kc for the pure symbols in Figure 9.1 is in Table 9.3. Table 9.3. b
23
35
41
57
63
65
91
k
11
12
10
9
6
6
12
c
1
1
2
2
3
4
3
.b/
22
24
40
36
36
48
72
The Dormouse complained bitterly that Alice had said they knew enough arithmetic to work this problem and now she was describing unbelievable things—things about divisibility, primes, and other stuff. Alice said that she had once been told by the White Queen that, “Why, sometimes I’ve believed as many as six impossible things before breakfast.” Nevertheless Alice told them that if they didn’t want to apply themselves by thinking about extraordinary things they should take a walk and return later. The Dormouse left and Alice began to explain to the March Hare the Reconstruction Algorithm. Given .k1 , k2 , : : : ; kr /, one can recover b, and a1 , a2 , : : : ; ar , to obtain the coach .?/, which may be pure or hybrid. The proof for pure coaches is in [4]. Here we present an example of a particular, but not special, case that should enable you to reproduce the process for yourself in other instances. Suppose you are given k1 D 2; k2 D 1; k3 D 3; and you are asked to produce a coach of the form ˇ ˇ ˇa1 a2 a3 ˇ ˇ ˇ: bˇ 2 1 3ˇ By .??/ we have the three simultaneous linear equations in a1 ; a2 ; a3 : b
a1 D 22 a2 ;
b
a2 D 21 a3 ;
b
a3 D 23 a1 :
Alice solved this system of linear equations using Cramer’s rule, evaluating the determinants by a condensation method invented by Charles Dodgson (Lewis Carroll) ([8];
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[9, pp. 118–119]). She told the March Hare to pay particular attention to the form of the solution compared with the numbers in the original problem: a1 26 D b a2 26 D b a3 26 D b
3
2
1
26 26 26 26 26 26
3 1
2
1
C 26 . 1/3 3 C 26 . 1/3 2 C 26 . 1/3
3 1 2
22 C 20 5 D ; so a1 D 5; b D 65I 6 2 C1 65 24 21 C 20 15 D D ; so a2 D 15; b D 65I 6 2 C1 65 25 23 C 20 25 D D ; so a3 D 25; b D 65: 6 2 C1 65 D
2 3 1
1 2 3
23
The March Hare checked this result by constructing the coach of the hybrid symbol for b D 65, starting with a1 D 5, getting the following coach, which satisfies the original data: ˇ ˇ ˇ5 15 25ˇ ˇ: b D 65 ˇˇ 2 1 3ˇ
But, as Alice noticed immediately, all the b’s are the same and you really didn’t need to solve for a2 , or a3 , because once you had the solution for a1 and b you could then construct the symbol to get the remaining values for ai . “To obtain a coach for a pure symbol with the given values of ki ;” Alice explained, “all 1 15 3 25 5 5 D 13 ; 65 D 13 , and 65 D 13 . Then, you need to do is reduce each of the fractions 65 from the first reduced fraction you see that you can begin with b D 13 and a1 D 1 and construct the following coach, which also satisfies the original data: ˇ ˇ ˇ1 3 5ˇ ˇ :” b D 13 ˇˇ 2 1 3ˇ
The Dormouse and the Hatter returned at this point. Alice convinced them to construct systematically some (possibly) hybrid symbols involving b D 2n 1, and b D 2n C 1.
Persistent coach patterns The Dormouse set to work, constructing the symbols in Figure 9.2, with only doublyunderlined markings, while Alice, the Hatter, and the March Hare looked for patterns. As soon as the last symbol was written down the March Hare said, “For all n 3 it looks as if when b D 2n 1 and a1 D 1, we have a persistent coach pattern 2n
ˇ ˇ ˇ1 2n 1 1ˇ ˇ :” 1 ˇˇ 1 n 1 ˇ
Alice proved it using .??/. Not to be outdone, the Hatter said, “For n 3, except for n D 4, every coach with b D 2n 1 that begins with a1 D 3 always has the form 2
n
ˇ ˇ ˇ3 2n 2 1ˇ ˇ ˇ :” 1ˇ 2 n 2 ˇ
(~n )
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Part II. Number Theory and Graph Theory 22 23 24
2
5
26
ˇ ˇ ˇ1ˇ 1 D 3ˇˇ ˇˇ 1 ˇ ˇ1 1 D 7ˇˇ 1 ˇ ˇ1 1 D 15ˇˇ 1
ˇ 3ˇˇ 2ˇ
ˇ ˇ ˇ 7ˇˇ3ˇˇ5ˇˇ 3ˇ2ˇ1ˇ
ˇ ˇ ˇ ˇ ˇ1 15ˇ3 7ˇ5 13 9 11ˇ ˇ ˇ ˇ ˇ 1 D 31ˇ 1 4 ˇ2 3ˇ1 1 1 2 ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ1 31ˇ3 15ˇ5 29 17 23ˇ7ˇ9 27ˇ11 13 25 19ˇ21ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ 1 D 63ˇ 1 5 ˇ2 4 ˇ1 1 1 3 ˇ3ˇ1 2 ˇ 2 1 1 2 ˇ 1 ˇ Figure 9.2. The Dormouse’s symbols.
The cautious Dormouse began to check the Hatter’s work. When n D 3, in (~n ) he got the coach ˇ ˇ ˇ3 1ˇ ˇ; 23 1 D 7 ˇˇ 2 1ˇ which is the same as
ˇ ˇ ˇ 1 3ˇ ˇ ˇ 7ˇ 1 2ˇ
because it is cyclic. But when he substituted n D 4 in (~n ) he got a coach no one had ever seen before (since they had always stopped when a repeat occurred). It looked like this: ˇ ˇ ˇ3 3ˇ 4 ˇ ˇ: (~4 ) 2 1 D 15 ˇ 2 2ˇ At the same time the Hatter came up with another coach, valid for all n n D 6 using b D 2n 1 and a1 D 9 in (??), ˇ ˇ9 2n 1 22 1 2n 2 2n 3 C 1 2n 1 2n 3 C 2n 4 2n 1 ˇˇ 1 2 1 n 4 By substituting n D 6 into (n ), the group obtained: ˇ ˇ ˇ9 27 9 27ˇ 6 ˇ ˇ: 2 1ˇ 1 2 1 2ˇ
5, except for
ˇ 1ˇˇ ˇ:
(n )
(6 )
Both (~4 ) and (6 ) caused great puzzlement. They could see that these coaches were not contracted, but they really didn’t want to reject coaches that fit the persistent coach patterns (~n / and (n /. “Maybe these curious coaches give us a clue that we shouldn’t always stop at the first repeat in a contracted symbol,” Alice suggested, “Maybe we should go on until the bottom row adds up to n.” The group studied (~n ), (~4 ), (n ), and (6 ) (as the reader should also), along with other coaches with a fixed a1 where b is always 2n 1, or 2n C 1. They discovered that
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in each sequence of coaches, every time a new number ai enters the top row of a coach it represents the beginning of yet another persistent coach pattern, where if ai is the first element in the top row, the ki values all remain the same, except for kr (the last entry in the bottom row) which will always increase by 1 every time n increases by 1. To see another example of this look at the form of the coaches where b D 2n C 1, and a1 D 3, for n D 4, 5, and 6. It turns out that if a coach, with fixed a1 where b is always 2n 1 or 2n C1, is such that the k value for that coach is less than n then the process of duplicating the columns in that coach until the resultant value of k in that coach equals n, will always make that coach appear as it would in some persistent coach sequence without being contracted. Now the Dormouse asked Alice when they were going to get to the main problem. Alice said “That’s next! We have everything we need to answer the question now.”
The big payoff—A nonrecursive algorithm for finding the number of MTP partitions Let’s begin with n D 5. Because we know the MTP partitions in this case (Table 9.1) it will give us a plausibility check. Alice instructed them to look at the following two, (possibly) hybrid, symbols: ˇ ˇ ˇ ˇ ˇ1 15ˇ3 7ˇ5 13 9 11ˇ b D 25 1 D 31ˇ ˇ ˇ ˇ; 1 4 2 31 1 1 2 ˇ ˇ ˇ ˇ ˇ ˇ1ˇ 3 15 9ˇ5 7 13ˇ 11 ˇ b D 25 C 1 D 33ˇ ˇ ˇ ˇ ˇ: 5 1 1 32 1 2 1
Because of the form of b in these two cases, we know in advance (by the QuasiOrder Theorem) that if the symbol is reduced and contracted the value of k for each coach is 5. In one coach the value of k is less than 5. However, in that deficient case it is possible to repeat columns until the sum of entries in the bottom row sums to 5. For our purposes, we only need to know the duplicate entries in the bottom row. Thus, in bottom of the fourth coach for b D 33 we repeat (1) to get 1 1 1 1 1. Having done this, you can simply record the numbers that appear in the bottom row of the various coaches to obtain all the MTP partitions of 5. The partitions contributed by b D 31 are all the even MTP partitions of 5. The values contributed by b D 33 are all the odd MTP partitions of 5. In Table 9.4 we record the MTP partitions by reading across the bottom row of the coaches, repeating the doublyunderlined entries where appropriate. Table 9.4. MTP partition for n D 5. Even MTP partitions from the coaches of b D 31
1 4, 2 3, 1 1 1 2
Odd MTP partitions from the coaches of b D 33
5, 1 1 3, 2 1 2, 1 1 1 1 1
There were, in fact, exactly the same partitions that they had found earlier, even though some were in a different cyclic order. In this case if each partition is written in ascending order there are no repeats so these yield the 7 unordered partitions for n D 5. Then the Dormouse asked, “How can we be sure we have all the MTP partitions?”
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Alice explained, “This is easy! If you think one is missing, just tell me which one it is. By the reconstruction algorithm, I can construct the coach having that partition. But, we know such a coach must already exist because in [4]–[6] it was shown that (??) defines a unique permutation of the ai s and our symbol contains all of the ai s. Finally, it turns out that, if the coach connected with a set of the ai s involves a repeat of the columns then this coach must fit one of the persistent coach patterns. This means we must have all the MTP partitions!” She then summarized their algorithm for finding MTP partitions, for a given n: 1. Construct the (possibly) hybrid symbol for b D 2n
1 and for b D 2n C 1.
2. In any hybrid coaches for either symbol repeat the bottom row, if necessary, so that the k value for that coach is n. 3. From the bottom row of the coaches record the MTP partitions.
Example with n D 8 That’s it! The bottom row of the coaches in the symbols for b D 2n 1, or b D 2n C 1, are the even or odd, MTP partitions, respectively. And, you can always write each of the MTP partitions in descending order, strike out the duplicates, and obtain the unordered partitions of n. This algorithm will work for any n—and a computer can be programmed to produce the symbols. The March Hare, the Hatter, and the Dormouse were really into the process now. They decided to try it for n D 8. Figures 9.3 and 9.4 give the (possibly hybrid) symbols for b D 28 1 D 255 and b D 28 C 1 D 257, as calculated by the Dormouse. ˇ ˇ ˇ ˇ ˇ ˇ ˇ1 127ˇ 3 63ˇ 5 125 65 95ˇ7 31ˇ9 123 33 111ˇ ˇ ˇ ˇ ˇ ˇ ˇ 255ˇ 6ˇ 1 1 1 5 ˇ3 5 ˇ1 2 1 4 ˇ 1 7 ˇ2 ˇ ˇ ˇ ˇ ˇ ˇ ˇ11 61 97 79ˇ13 121 67 47ˇ 15 ˇ 17 119ˇ19 59 49 103ˇ ˇ ˇ ˇ ˇ ˇ ˇ ˇ2 1 1 4ˇ1 1 2 4ˇ 4 ˇ 1 3 ˇ2 2 1 3 ˇ ˇ ˇ ˇ ˇ21 117 69 93 81 87ˇ23 29 113 71ˇ25 ˇ ˇ ˇ ˇ1 1 1 1 1 3ˇ3 1 1 3ˇ1 ˇ ˇ ˇ ˇ27 57 99 39ˇ37 109 73 91 41 107ˇ43 ˇ ˇ ˇ ˇ2 1 2 3ˇ1 1 1 2 1 2 ˇ2 ˇ ˇ ˇ ˇ ˇ45 105 75ˇ 51 ˇ 85 ˇ ˇ ˇ ˇ ˇ ˇ1 1 2ˇ 2 ˇ 1 ˇ
ˇ 115 35 55ˇˇ 2 2 3ˇ
ˇ 53 101 77 89 83ˇˇ 1 1 1 1 2ˇ
Figure 9.3.
In Table 9.5 they recorded the MTP partitions from the bottom row of the coaches, repeating the doublyunderlined coaches as appropriate. Ordering the numbers in each of these 35 MTP partitions and striking out the duplicates they verified that the number of ordinary, unordered partitions for n D 8 is 22.
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9. Mad Tea Party Cyclic Partitions ˇ ˇ ˇ1ˇ3 257ˇˇ ˇˇ 81 ˇ ˇ13 ˇ ˇ2 ˇ ˇ23 ˇ ˇ1 ˇ ˇ37 ˇ ˇ2 ˇ ˇ45 ˇ ˇ2
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ˇ ˇ ˇ ˇ ˇ 127 65ˇˇ5 63 97ˇˇ7 125 33ˇˇ9 31 113ˇˇ11 123 67 95 81ˇˇ 1 6 ˇ2 1 5 ˇ1 2 5 ˇ3 1 4 ˇ1 1 1 1 4ˇ ˇ ˇ ˇ ˇ 61 49ˇˇ15 121 17ˇˇ19 119 69 47 105ˇˇ21 59 99 79 89ˇˇ 2 4ˇ1 3 4ˇ1 1 2 1 3 ˇ2 1 1 1 3ˇ ˇ ˇ ˇ 117 35 111 73ˇˇ25 29 57ˇˇ27 115 71 93 41ˇˇ 2 1 1 3ˇ3 2 3ˇ1 1 1 2 3ˇ ˇ ˇ 55 101 39 109ˇˇ43 107 75 91 83 87 85ˇˇ 1 2 1 2 ˇ1 1 1 1 1 1 2ˇ ˇ 53 51 103 77ˇˇ 2 1 1 2ˇ Figure 9.4.
As we leave our friends the Dormouse went off to buy a small Mickey Mouse hand calculator so that he could more easily do the MTP calculations for n D 9. Table 9.5. MTP partitions for n D 8. Even MTP partitions from the coaches of Figure 3
1 7, 2 6, 1 1 1 5, 3 5, 1 2 1 4, 2 2 1 4, 1 1 2 4, 4 4, 1 3 1 3, 2 2 1 3 1 1 1 1 1 3, 3 3 1 3, 1 2 2 3, 2 1 2 3, 1 1 1 2 1 2, 2 1 1 1 1 2, 1 1 2 1 1 2, 2 2 2 2, 1 1 1 1 1 1 1 1
Odd MTP partitions from the coaches of Figure 4
8, 1 1 6, 2 1 5, 1 2 5, 3 1 4, 1 1 1 1 4, 2 2 4, 1 3 4, 1 1 2 1 3 2 1 1 1 3, 1 2 1 1 3, 3 2 3, 1 1 1 2 3, 2 1 2 1 2, 1 1 1 1 1 1 2, 2 2 1 1 2
Acknowledgment The authors are grateful to Don Albers, Gerald L. Alexanderson, Victor Garcia, Jennifer Hooper, Geoffrey Shephard, and Robin Wilson for reading earlier versions of this manuscript and offering valuable suggestions for improving the text; and we also thank Veronica Garcia for producing the cartoon. The authors accept full responsibility for any typos that may have escaped our scrutiny— but we blame any arithmetical errors on the Dormouse who tended to nod off doing the longer calculations.
Bibliography [1] M. Gardner, Fractal Music, Hypercards, and More Mathematical Recreations from Scientific American Magazine, W. H. Freeman, 1992, 24–38. [2] ——, Mathematical Carnival, Mathematical Association of America, 1989, 137. [3] ——, The Last Recreations: Hydras, Eggs, and Other Mathematical Mystifications, Copernicus Series, Springer, NY, 1997, 37–42. [4] P. Hilton, D. Holton, and J. Pedersen, Mathematical Reflections—In Room with Many Mirrors, Second printing, Undergraduate Texts in Mathematics, Springer, NY, 1998. [5] P. Hilton and J. Pedersen, A Mathematical Tapestry: Demonstrating the Beautiful Unity of Mathematics, Cambridge Univ. Press, 2010.
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[6] P. Hilton, J. Pedersen, and B. Walden, A property of complete symbols: An ongoing saga connecting geometry and number theory, Homage to a Pied Puzzler, A K Peters, 2009, 47–50. [7] A. Knopfmacher and N. Robbins, Some properties of cyclic compositions, Fibonacci Quart. 48 (2010) 249–255. [8] A. Rice and E. Torrence, “Shutting up like a Telescope,” Lewis Carroll’s “curious” condensation method for evaluating determinants, College Math. J. 38 (2007) 85–95. [9] R. Wilson, Lewis Carroll in Numberland: His Fantastical Mathematical Logical Life, W. W. Norton, NY, 2010.
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10 The Continuing Saga of Snarks sarahmarie belcastro Way back in 1852, Francis Guthrie conjectured that every map drawn on the plane can be colored so that regions sharing a border have different colors—and only four colors are necessary. This became known as the Four Color Conjecture. In 1880, Tait [16] proved that the Four Color Conjecture is equivalent to a problem of edge coloring graphs. This is where our story begins, because the study of snarks grew from exactly this edgecoloring problem. To avoid confusion, we note that there is no relationship between the English word ‘snark’ (or its derivatives snarky, snarkiness, etc.) and the mathematical term; in fact, the origin of the latter is a pivotal point in our story. Recall that a graph is a collection of nodes, called vertices, and a collection of connections between these nodes, called edges. You may also remember that a graph is planar if it can be drawn in the plane with no edgecrossings, at which point the regions of the plane demarcated by the graph are called faces. You are less likely to know that a bridgeless graph has no edge whose deletion disconnects the graph, that a 3regular graph has three edges meeting at every vertex, and that a proper edge coloring assigns a color to every edge of a graph such that no two edges meeting at a vertex have the same color. A graph to which all three of these terms apply is shown in Figure 10.1.
Figure 10.1. A bridgeless, 3regular, properly 3edge colored planar graph, but not a snark.
In this language, Tait showed that the Four Color Conjecture holds if and only if every bridgeless, 3regular, planar graph can be properly edge colored using only three colors. That is, instead of coloring regions of maps, one can approach the Four Color Conjecture by coloring edges of bridgeless, 3regular, planar graphs. As Martin Gardner pointed out in his 1976 introduction to this subject [9], it follows that the search for a counterexample Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 82–87.
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to the Four Color Conjecture is really a search for a bridgeless, 3regular, planar graph that is not 3edge colorable. In Gardnerâ€™s language, this was a search for a nontrivial (no bridges or cycles of lengths 4 or less), uncolorable (using only three colors), trivalent (another word for 3regular) planar graph. Because â€œNontrivial Uncolorable Trivalentâ€? is a mouthful, Gardner sought a shorter name. He considered â€˜NUTâ€™, decided it wasnâ€™t serious enough, and then proposed â€˜Snarkâ€™ from the Lewis Carroll poem The Hunting of the Snarkâ€”which does not seem, at least to this author, to be a more serious name. (Carroll invented the name Snark, but his poetic use was not related to the already existing English noun, or, it seems, to the verb and adjective that came later [15].) The name stuck, and has invigorated all subsequent discourse on the topic.
Classical snarkiness We now reveal the full technical definition. Definition. A snark is a bridgeless, 3regular graph that is not edge colorable using only three colors, has smallest cycle length (girth) at least 5, and
is cyclically 4edge connected, meaning that at least 4 edges must be removed in order to separate the graph into two components that each contain a cycle.
Figure 10.2. The Petersen graph, BlanuLsa1, and BlanuLsa2.
The three smallest snarks are shown in Figure 10.2. In 1976, Gardner gave several examples both of individual snarks and of infinite families of snarks [9]. Much of this column was a translation into lay language of an excellent paper by Isaacs [11], published the year before and still cited in modern research. As with many of his columns, Gardnerâ€™s writing on snarks was exciting partly because he gave puzzlelike exercises, some of which led toward proofs! Martin Gardnerâ€™s showcasing (and naming) of snarks is only one part of their story. Earlier, in 1968, Branko GrÂ¨unbaum [10] posed a generalization of the Four Color Conjecture that can be phrased in terms of snarks. To understand it, we need a few more definitions. A planar graph should really be drawn on a sphere to avoid the unbounded face present in a planar drawing. Focusing on the faces of a graph takes us beyond ordinary graph theory
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into topological graph theory. The topology referred to here is the topology of surfaces such as the sphere, torus, and M¨obius band. (The lovely classification theorem for surfaces is presented superbly in [8].)
(a)
(b)
Figure 10.3. The complete graph K5 is nonplanar (a) but does embed on the torus (b).
In Figure 10.3b we show a small nonplanar graph embedded (drawn without edges crossing) on the torus, as well as a way of representing the torus that allows us to avoid the use of dotted lines for unseen parts of edges. The arrows indicate gluing; when an edge of a graph goes off the boundary of the rectangle, it returns on the matching arrowed boundary segment. It is quite an enjoyable exercise to embed the Petersen graph on a torus. We only consider cellular embeddings, in which every face is a topological disk. (In other words, faces cannot have punctures or doughnut holes.) A graph is polyhedrally embeddable on a surface if it can be drawn so that no face uses an edge twice and no two faces share more than one edge. Figure 10.3b depicts a cellular embedding, but because some edges do not separate distinct faces, the embedding is not polyhedral. Gr¨unbaum conjectured that every 3regular graph polyhedrally embeddable on an orientable surface (a sphere, torus, or nholed torus) is threeedge colorable. Phrased in terms of snarks, this says that no snark has a polyhedral embedding on any orientable surface. When Martin Gardner published [9], there had been no progress on Gr¨unbaum’s Conjecture, but a few months later, Appel and Haken turned the Four Color Conjecture into the Four Color Theorem [2], [3]. They did not use edge coloring in their proof (for an excellent exposition of the saga, see [20]), and so by Tait’s theorem we obtain, as a corollary of the Four Color Theorem, that every bridgeless 3regular planar graph is 3edge colorable; equivalently, no snark embeds on a sphere. It’s not just the connection to the Four Color Theorem that makes snarks interesting. Vizing’s theorem (proved only a few years before Gr¨unbaum made his conjecture) implies that every 3regular graph needs at most 4 colors to properly color its edges. Of course, at least 3 colors are needed, so there is not much wiggle room in the land of edge coloring. Snarks are simultaneously barely not 3edge colorable, and as difficult to edgecolor as possible.
Modern snarkiness For a long time, the main research goal was to find snarks, as very few were known. In 1950 there were only four known snarks, but, during the 1970s, five infinite families of snarks and about five individual snarks (not in any families) were found [7]; some are pictured
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Figure 10.4. Six sample snarks from the seventies (drawn by Mathematica).
in Figure 10.4. Twenty years later, the computer age made it possible to generate snarks, so in the 1990s, most work on snarks took a structural approach. These results are highly technical; thus, we give only some flavor of the ideas. First, we consider ways to break down snarks. If one can remove some set of k edges from a snark so that the remaining graph is a disconnected collection of subgraphs of snarks, then the original snark is considered reducible, and if this is not possible, the snark is irreducible. Many snarks are reducible in this sense, and irreducible snarks have the property that removal of two vertices leaves a 3edge colorable graph. By definition, a snark cannot have a Hamilton circuit (a circuit containing all vertices), as otherwise it would be 3edge colorable (2edge color the circuit and give the remaining edges a third color). However, many snarks almost have Hamilton circuits in the sense that they have circuits including all but one vertex. And, some snarks have the stronger property that leaving out any one vertex, the remainder of the graph forms a circuit. All such snarks are irreducible [6]. Tutte conjectured in 1966 that every snark contains a subgraph that is a Petersen graph but subdivided with extra vertices [18]. This can be rephrased by saying that every snark has a Petersen minor. Robertson, Sanders, Seymour, and Thomas announced a proof in 1999, much of which is yet to be published [17, p.14]. Because the Petersen graph occupies such a revered position in graph theory, the fact that every snark has a Petersen minor is in some sense the ultimate structural result. In contrast to the structural approach is the topological approach. From this perspective, we are interested in knowing on which surface(s) a given snark embeds, and with what type of embedding. This essentially amounts to investigating Gr¨unbaum’s Conjecture. Aside from the resolution of the Four Color Conjecture in 1976—proved using a structural approach!—no serious progress was made on Gr¨unbaum’s Conjecture for more than 30 years after it was posed. The graph theory community was divided on whether the conjecture was true, with some leading figures taking opposite sides of the issue at a 2003 conference.
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Very little was known about embedding properties of snarks until the 2000s, though thousands of snarks were tested by computer and proved not to be polyhedrally embeddable on any orientable surface [14], [4] (and therefore none provided a counterexample to Gr¨unbaum’s Conjecture). It was only known that the Petersen graph and one of the two Blanuˇsa snarks embed on the torus. In 2004, several infinite families of snarks were exhibited that embed on the torus [5], [19] and on the 2holed torus [5] (none polyhedrally). Because they are pretty, two seed snarks for these families are shown in Figure 10.5. 1
7
5
9
8
2
3
4
17
11
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1
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Figure 10.5. The smallest snarks in two distinct infinite families of toroidal snarks.
That same year, it was proved that for each number of M¨obiusbandlike twists g 1, there exists a snark with a polyhedral embedding on the corresponding nonorientable surface [14]. (Whereas orientable surfaces are classified by the number of torus holes they have, nonorientable surfaces are classified by the number of M¨obiusbandlike twists they have.) Both of these results provided evidence in favor of Gr¨unbaum’s Conjecture. At a conference late in 2006, Gr¨unbaum himself was asked whether he still thought his eponymous conjecture was true. He demurred, but noted that saying it should be true caused more people to work on it! Then, in 2007 everything blew up. Counterexamples to Gr¨unbaum’s Conjecture were found on the 9holed torus, and then shortly thereafter, on the 5holed torus [13] using snarks exhibited in [5]. (In fact, infinitely many counterexamples were found! The powerful superposition technique of [12] did the trick.) At the same time, Gr¨unbaum’s Conjecture was affirmed for most threeregular graphs embedded on the torus [1]. While this deflates an inspiring balloon, a smaller (and perhaps more manageable?) balloon still exists: What is true about Gr¨unbaum’s Conjecture restricted to nholed tori for
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1 n 4? While perhaps all of mathematical research is about refining the boundaries between different categories of objects, it somehow seems that this is especially true for snarks. It often feels that doing research on snarks is like honing the edge of a knife used to separate two crumbly loaves of bread. Much, much more can (and should!) be said about snarks. The interested reader is encouraged to consult the references and contact the author for further information!
Bibliography [1] M. O. Albertson, H. Alpert, sm. belcastro, and R. Haas, Gr¨unbaum colorings of toroidal triangulations, J. Graph Theory 63 (2010), 68–81; available at dx.doi.org/10.1002/ jgt.20406 [2] K. Appel, W. Haken, and J. Koch, Every planar map is four colorable, I: Discharging, Illinois J. Math. 21 (1977) 429–490. [3] K. Appel and W. Haken, Every planar map is fourcolorable, II: Reducibility, Illinois J. Math. 21 (1977) 491–567. [4] D. Archdeacon, Problems in Topological Graph Theory: ThreeEdgeColoring Orientable Triangulations (1995); available at www.emba.uvm.edu/˜ archdeac/problems/ grunbaum.htm. [5] s.m. belcastro and J. Kaminski, Families of dotproduct snarks on orientable surfaces of low genus, Graphs Combin. 23 (2007), 229–240; available at dx.doi.org/10.1007/ s0037300707299. [6] A. Cavicchioli, T. E. Murgolo, B. Ruini, and F. Spaggiari, Special classes of snarks, Acta Appl. Math. 76 (2003), 57–88; available at dx.doi.org/10.1023/A:1022864000162. [7] A. G. Chetwynd and R. J. Wilson, Snarks and supersnarks, in The Theory and Applications of Graphs (Kalamazoo, Mich., 1980), Wiley, New York, 1981, 215–241. [8] G. K. Francis and J. R. Weeks, Conway’s ZIP proof. Amer. Math. Monthly 106 (1999), 393–399; available at dx.doi.org/10.2307/2589143; also available at new.math.uiuc.edu/zipproof/. [9] M. Gardner, Mathematical games, Scientific American, April 1976, 126–130. [10] B. Gr¨unbaum, Conjecture 6, in Recent Progress in Combinatorics, W.T. Tutte, ed., Academic Press, New York, 1969, 343. [11] R. Isaacs, Infinite families of nontrivial trivalent graphs which are not Tait colorable, Amer. Math. Monthly 82 (1975), 221–239; available at dx.doi.org/10.2307/2319844. [12] M. Kochol, Snarks without small cycles, J. Combin. Theory Ser. B 67 (1996), 34–47; available at dx.doi.org/10.1006/jctb.1996.0032. [13] ——, Polyhedral embeddings of snarks in orientable surfaces, Proc. of the Amer. Math. Soc., 137 (2009), 1613–1619; available at dx.doi.org/10.1090/S0002993908096986. [14] B. Mohar and A. Vodopivec, On polyhedral embeddings of cubic graphs, Combin. Prob. Comput. 15 (2006), 877–893; available at dx.doi.org/10.1017/S0963548306007607. [15] snark, v. and n. The Oxford English Dictionary, 2nd ed., Oxford Univ. Press, 1989. [16] P. G. Tait, Note on a theorem in geometry of position. Trans. Roy. Soc. Edinburgh 29 (1880), 657–660. [17] R. Thomas, Recent excluded minor theorems for graphs, Surveys in combinatorics, 1999 (Canterbury), London Math. Soc. Lecture Note Ser., 267, Cambridge Univ. Press, Cambridge, 1999, 201–222; available at people.math.gatech.edu/˜ thomas/PAP/bcc.pdf.
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[18] W. T. Tutte, On the algebraic theory of graph colorings, J. Combin. Theory Ser. B 1 (1966), 15–50; available at dx.doi.org/10.1016/S00219800(66)800042. [19] A. Vodopivec, On embeddings of snarks in the torus, Discrete Math. 308 (2008), 1847–1849; available at dx.doi.org/10.1016/j.disc.2006.09.051. [20] R. Wilson, Four Colors Suffice: How the Map Problem Was Solved, Princeton Univ. Press, 2003.
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11 The MapColoring Game Tomasz Bartnicki, Jarosław Grytczuk, H. A. Kierstead, and Xuding Zhu Introduction Suppose that Alice wants to color a planar map using four colors in a proper way, that is, so that any two adjacent regions get different colors. Despite the fact that she knows for certain that it is eventually possible, she may fail in her first attempts. Indeed, there are usually many proper partial colorings not extendable to proper colorings of the whole map. Thus, if she is unlucky, she may accidentally create such a bad partial coloring. Now suppose that Alice asks Bob to help her in this task. They color the regions of a map alternately, with Alice going first. Bob agrees to cooperate by respecting the rule of a proper coloring. However, for some reason he does not want the job to be completed—his secret aim is to achieve a bad partial coloring. (For instance, he may wish to start the coloring procedure over and over again just to stay longer in Alice’s company.) Is it possible for Alice to complete the coloring somehow, in spite of Bob’s insidious plan? If not, then how many additional colors are needed to guarantee that the map can be successfully colored, no matter how clever Bob is? This mapcoloring game was invented about twentyfive years ago by Steven J. Brams with the hope of finding a gametheoretic proof of the Four Color Theorem, avoiding perhaps the use of computers. Though this approach has not been successful, at least we are left with a new, intriguing mapcoloring problem: What is the fewest number of colors allowing a guaranteed win for Alice in the mapcoloring game in the plane? Brams’s game was published by Martin Gardner in his Mathematical Games column in Scientific American in 1981. Surprisingly, it remained unnoticed by the graphtheoretic community until ten years later, when it was reinvented by Hans L. Bodlaender [1] in the wider context of general graphs. In this version Alice and Bob play as before by coloring properly the vertices of a graph G. The game chromatic number g .G/ of G is the smallest number of colors for which Alice has a winning strategy. As every map is representable by a graph whose edges correspond to adjacent regions of the map, Brams’s question is equivalent to determining the game chromatic number of planar graphs. Since then the problem has been analyzed in serious combinatorial journals and gained the attention of Reprinted from The American Mathematical Monthly, Vol. 114, No. 9 (Nov. 2007), pp. 793–803.
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several experts. In this article we present what is known and what remains unknown about this fascinating game.
Four colors do not suffice One can see quickly that there are maps demanding more colors in the game than normally. Perhaps the simplest example of this phenomenon is due to Robert High. Consider the map with six regions corresponding to the faces of a cube. Clearly, the map can be 3colored, but Bob can win even if there are four colors available. Suppose that Alice uses color 1 in her first move. Then Bob answers with color 2 on the opposite face. Thus colors 1 and 2 cannot be used any further, and Alice has to play color 3 (see Figure 11.1). Then Bob answers with color 4 on the opposite face and wins the game. Bob 2
1 Alice
Bob
1
1 Alice
Figure 11.1. First two moves on the cube and the dodecahedron.
Earlier, Lloyd Shapley gave another elegant example based on a regular dodecahedron, showing that even five colors do not suffice. Here Bob’s strategy is similar, but this time he replies with the same color on the face opposite to the face that Alice has just colored (see Figure 11.1). In this way, Alice must use a new color in each of her moves, which leads to a win for Bob after the fifth round. When Steven Brams and Martin Gardner started to think that perhaps six colors would be the maximum number that Bob could force, Robert High found a map for which Alice needed seven colors to win. Indeed, there is a modification of the dodecahedron map for which a win by Alice requires eight colors. Unfortunately, it is not possible to verify this without a tedious casebycase analysis. A natural suspicion arose that perhaps there was no finite bound at all, a suspicion that persisted for thirteen years.
Possible strategies for Alice Let us consider the problems posed for Alice in designing an effective strategy for the mapcoloring game. Whenever Bob colors a vertex v, he poses a threat to each of its uncolored neighbors. Alice can counter this threat at one of these neighbors w by immediately coloring it, but this in turn creates new threats for the neighbors of w. Moreover, the remaining uncolored neighbors of v are still open to attack, and Bob can strengthen his position on subsequent moves by coloring their uncolored neighbors. Thus Alice should design a strategy that prioritizes responses to threats to uncolored vertices so that she can keep the total threat under control. She must also limit the damage done when she colors a vertex.
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The following idea allows Alice to avoid creating new points of vulnerability when she colors a vertex. Let .G/ denote the usual chromatic number of a graph G, that is, the least number of colors in a proper coloring of the vertices of G. Suppose that f W V .G/ ! fc1; : : : ; c t g is an optimal coloring, with t D .G/, and that the game is played on a set of kt colors of the form fcji W i 2 Œk; j 2 Œtg, where we use the standard abbreviation Œn for f1; 2; : : : ; ng. We refer to cji as a shade of cj . We say that a color is correct for a vertex v if it is a shade of f .v/; otherwise it is incorrect. Alice should always try to color a vertex v with a correct color. If she is successful, then her early moves will never conflict with her later moves. This is possible as long as she can play in such a way that at any time any uncolored vertex is adjacent to fewer than k vertices incorrectly colored by Bob. This leaves Alice with the problem of limiting the number of neighbors colored incorrectly by Bob that an uncolored vertex can have. In ordinary (noncompetitive) graphcoloring there is a wellknown strategy for managing such threats. The coloring number col.G/ of a graph G is the smallest integer k such that every subgraph of G has a vertex with degree less than k. For example, it is a simple consequence of Euler’s formula that the coloring number of an arbitrary planar graph is at most six. The vertices of a graph with coloring number at most k can be ordered as v1 ; : : : ; vn so that each vertex vi has fewer than k neighbors that precede it in the ordering: we simply construct the ordering from right to left, always choosing a vertex whose degree among the unchosen vertices is less than k. It is also clear that the existence of such an ordering implies that col.G/ k. Now, if Alice colors the vertices from left to right, she will need at most k colors, since each uncolored vertex will have fewer than k colored neighbors. In other words, .G/ col.G/ for all graphs G. Suppose that Alice attempts to modify this strategy for the mapcoloring game. She might naturally try to color the leftmost (in the ordering) neighbor of the last vertex colored by Bob. This vertex will have fewer than k backward (left) neighbors colored by Bob, but the number of its forward (right) neighbors colored by Bob is unbounded. This was the situation with the problem in the early nineties when help came from an unexpected source.
Ramsey numbers and arrangeability A simple graph G consists of a vertex set V .G/ and an edge set E.G/ where each edge is an unordered pair of distinct vertices. A complete graph or clique is a simple graph in which every pair of vertices forms an edge. A clique on t vertices is denoted by K t . According to Ramsey’s famous theorem, any graph G has a Ramsey number r .G/ defined as the smallest integer t such that for any redblue coloring of the edges of K t there is a monochromatic copy of G in K t . For a class C of graphs define rC .n/ by rC .n/ D maxfr .G/ W G 2 C; jV .G/j D ng. Burr and Erd˝os conjectured that rC .n/ is linear in n for any class of graphs satisfying jE.G/j D O.jV .G/j/. This conjecture remains open in general, although some special cases have been proved. Chv´atal, R¨odl, Szemer´edi, and Trotter [5] verified the conjecture for the special case of graphs with bounded degree by using Szemer´edi’s regularity lemma. The proof involved constructing an embedding of G into a monochromatic part of Kn one vertex at a time in an order v1 ; : : : ; vn . It was
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crucial to their argument that each vertex have a bounded number of backward and forward neighbors. If the argument had required only that the backward degree be bounded, they would have established the whole conjecture. Notice that planar graphs satisfy jE.G/j D O.jV .G/j/ but may have arbitrarily large degree. Chen and Schelp made the brilliant observation that something less was required: the number of vertices vh with h < i that are adjacent to vi or to a forward neighbor vj of vi should be bounded (see Figure 11.2). This idea, in one form or another, has turned out to be absolutely crucial for the theory of coloring games.
v1
vh
vi
vj
vn
Figure 11.2. The number col2 .G/ bounds the number of predecessors of vi that are adjacent to vi (grey nodes) or to a forward neighbor of vi (black nodes).
Assume that the vertices of a graph G are ordered as v1 ; v2 ; : : : ; vn . A vertex vh is called a loose backward neighbor of vi if h < i and either (a) vh is adjacent to vi or (b) vh is adjacent to vj and vj is adjacent to vi for some j with j > i (see Figure 11.2). We define the 2coloring number col2 .G/ of a graph G to be the least integer k such that there is an ordering of V .G/ for which each vertex has fewer than k loose backward neighbors. This is an inessential variant of the parameter that Chen and Schelp called arrangeability in [3], where it is proved that the conjecture of Burr and Erd˝os is true for the family of graphs with bounded 2coloring numbers. They then proved the BurrErd˝os conjecture for planar graphs by showing that the 2coloring number of any planar graph is at most 761. Kierstead and Trotter [14] realized that this was the missing tool for bounding the game chromatic number g .G/ of planar graphs G. They proved the following theorem: Theorem 1. Every graph G satisfies g .G/ .G/.1 C col2 .G//. Proof. Let .G/ D t and col2 .G/ D k. To prove the assertion we describe a strategy for Alice using at most t.kC1/ colors. Alice should begin by fixing an optimal coloring f of G using the colors c1; : : : ; c t and an ordering v1 ; : : : ; vn of G that satisfies col2 .G/ D k. She prioritizes responses to threats according to this ordering. Whenever Bob colors a vertex v, Alice colors the leastindexed uncolored backward neighbor u of v with a correct color; if v has no such backward neighbor then Alice correctly colors the leastindexed uncolored vertex. To be certain that u has a correct color available (i.e., a color not used by any of its colored neighbors), it suffices to show that u has at most k neighbors that are colored by Bob. Assume that Bob has colored s loose backward neighbors of u and that Alice has colored s 0 loose backward neighbors of u. In accordance with the stated strategy, each time Bob colors a forward neighbor z of u, Alice colors the leastindexed uncolored backward neighbor u0 of z. Since u is an uncolored backward neighbor of z, u0 precedes u, so u0 is a loose backward neighbor of u. As Alice has colored s 0 loose backward neighbors of u, we conclude that Bob has colored at most s 0 forward neighbors of u. Therefore the number of neighbors of u colored by Bob is at most s C s 0 .< k/. Note that this counts only the
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number of colored neighbors of u at the end of Alice’s previous move. After that move, Bob may have colored another neighbor of u, meaning that u now has at most k neighbors colored by Bob. Thus u has an available correct color, and Alice can win the game with t.k C 1/ colors. This reasoning implies that g .G/ 3044 for a planar graph G. Kierstead and Trotter obtained much tighter bounds on the game chromatic number of planar graphs. They demonstrated that col2 .G/ 10 holds for each planar graph G (which gives g .G/ 44) and provided an example of a planar graph G with col2 .G/ D 8. They further improved their bound on the game chromatic number of planar graphs to g .G/ 33 by using a more carefully designed parameter than 2coloring number.
Acyclic coloring A cycle on vertices v1 ; : : : ; vn is a graph whose edges are v1 v2 ; v2v3 ; : : : ; vn v1 . A forest is a graph without cycles. A coloring of the vertices of a graph G is acyclic if it is a proper coloring such that no cycle of G is 2colored. In other words, a subgraph of G whose vertex set is a union of any two color classes is a forest. The minimum number of colors needed is the acyclic chromatic number of G, denoted by a.G/. This notion was introduced by Gr¨unbaum, who conjectured that a.G/ 5 holds for any planar graph G. The conjecture was turned into a theorem by Borodin in 1979 [2]. Just as the chromatic number of a graph G is bounded by its coloring number, its acyclic chromatic number is bounded by its 2coloring number. Let v1 ; : : : ; vn be an ordering of the vertices of G that realizes its 2coloring number. Color vertices recursively so that no vertex is colored the same color as a loose backward neighbor. This yields a proper coloring using at most col2 .G/ colors. We still must check that each cycle C in G receives at least three colors. The largestindexed vertex vj in C has two backward neighbors vh and vi with h < i < j . Since vh is a loose backward neighbor of vi , they have different colors. Since vj is adjacent to both vi and vh , it has a third color. Twenty years after Gr¨unbaum introduced acyclic coloring, Dinski and Zhu [6] applied the acyclic chromatic number to the Brams game for arbitrary graphs. They proved the following theorem, which implies that thirty colors suffice for Alice to win on any planar map: Theorem 2. Every graph G satisfies g .G/ a.G/.a.G/ C 1/. Proof. The argument is similar to the proof of Theorem 1. Suppose that G has an acyclic coloring using the color set fc1; : : : ; c t g, and let V1 ; : : : ; V t be the corresponding color classes of vertices. When i ¤ j , the subgraph Fij of G consisting of all edges with ends in Vi [ Vj is a forest. Fix an orientation of each Fij so that every vertex has outdegree at most one. Since every edge of G appears in exactly one of the forests Fij , this provides an orientation of G such that every vertex has at most one outneighbor in each color class. Now a vertex v in Vi is endangered if it is as yet uncolored but there is an inneighbor u of v colored with a shade of ci (see Figure 11.3). Since Alice always colors with a correct color, an endangered vertex can be created only by Bob. Clearly, in a single move Bob can create at most one endangered vertex v. Alice should color this vertex immediately after it
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Vi
v
u Figure 11.3. An endangered vertex v will be colored by Alice in her next move.
is created by Bob. This is always possible, for there are at most t neighbors of v colored with shades of ci at this moment (at most t 1 outneighbors and only one inneighbor u that has just been colored by Bob).
Daltonism may help Both strategies for Alice considered earlier were local. Alice responded to any threat posed when Bob colored a vertex by coloring one of its neighbors. As any Go player knows, global strategies can be more powerful. Our next aim is to develop a global strategy for Alice. Curiously, further progress in bounding the game chromatic number was achieved by making the game more difficult for Alice. Suppose Alice suffers from full daltonism (colorblindness): she just cannot distinguish between colors. Because of her disease she proposes to modify the rules of the coloring game so that she can still have some fun playing it. Rather than coloring vertices, the players create an ordering <g of vertices by taking turns choosing the next vertex in the ordering. For a fixed positive integer k Alice’s goal is to have <g demonstrate that col.G/ k. Bob, wishing to remain in her company, tries to thwart her. For the purposes of actual play it is convenient for the players to indicate that they have chosen a vertex by coloring it gray. Bob wins if at some time some uncolored vertex has k gray neighbors, and Alice wins if this never occurs. We denote the least number k allowing success for Alice by colg .G/ and call it the game coloring number of G. Clearly, if Alice can win this ordering game for some integer k, she can also win the coloring game with k colors (even as a completely colorblind person!). In other words, g .G/ colg .G/ is true for every graph G. Alice is now forced to use a strategy that does not rely on shades of correct colors to control threats created by her moves. Suppose that she again selects a specific ordering v1 ; : : : ; vn (not to be confused with <g ) of the vertex set for the purpose of prioritizing responses to threats. Suppose also that no vertex has too many backward neighbors in this order. When Bob chooses a vertex v (colors it gray), Alice faces conflicting requirements. On the one hand, she should choose a backward neighbor x of v, addressing the threat to x posed by v. But this in turn produces a threat from x to its backward neighbors. Alternatively, she could put off choosing x and instead choose a backward neighbor y of x so that y would already be protected when later she finally did choose x. Or should she, perhaps, choose a backward neighbor of y? Here we propose a compromise plan: the first time Alice contemplates choosing x she should move on to considering y, but the second time she considers choosing x she should actually do so.
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Here are the details of Alice’s activation strategy. Imagine that the vertices of G are made of light bulbs that shine a warning when activated. Let N C .v/ and N .v/ denote the sets of backward and forward neighbors of v, respectively. On her first move Alice activates vertex v1 and colors it. The basic step of the strategy is to activate a vertex (leaving a warning that it is in danger) and then jump to its leastindexed uncolored backward neighbor. Suppose that Bob has just colored a vertex v. Then Alice starts by activating v (provided it has not been activated hitherto) and jumps to the uncolored vertex x of smallest index in N C .v/. (See Figure 11.4.) If x is already active, then it is endangered, so Alice stops and colors it. Otherwise she repeats the activation step for x, that is, she activates x and jumps to the leastindexed uncolored vertex y in N C .x/. This process goes on until she stops at some vertex u, either because u is active or because there are no uncolored vertices in N C .u/. In each case she activates and colors u. If it happens that N C .v/ (see Figure 11.4, second line) already contains no uncolored vertices, then she picks the uncolored vertex of smallest index, activates it (provided it has not yet been activated), and colors it. Notice that after Alice’s move all gray vertices are active, whereas after Bob’s move there can be only one gray inactive vertex.
u
y
x
v
u
y
x
v
Figure 11.4. Basic step of the activation procedure.
For example, let’s see how this strategy works for trees (see Figure 11.5). Any tree T can be ordered so that each vertex has at most one backward neighbor. Suppose that Bob has colored a vertex v, and let x be its unique backward neighbor. Then Alice should activate all vertices on the unique backward path starting from v and ending at some active vertex u. Notice that Alice can jump to the same vertex only twice (the first time she activates it, the second time she colors it). It follows that if x stayed uncolored after her move, it could have no more than two active neighbors (one backward and one forward). Hence, after Bob’s move there are at most three colored vertices around x, which gives u
u
u
x
v
x
x
v
v
Figure 11.5. Activation on a tree.
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colg .G/ 4. This result was proved by Faigle, Kern, Kierstead, and Trotter in [8]. The following result reveals the information provided by the activation strategy for general graphs: Theorem 3. Every graph G satisfies colg .G/ 3 col2 .G/
1.
Proof. Fix an order of the vertices v1 ; : : : ; vn realizing col2 .G/ D k, and assume that Alice applies the activation strategy with respect to this order. For each vertex vj let S.vj / be the set of loose backward neighbors of vj . By the definition of col2 .G/, the size of S.vj / is at most k 1. We count the number of active forward neighbors that vj could have before it gets colored. From each active vertex in N .vj / Alice jumps to the leastindexed uncolored backward neighbor in S.vj / or possibly to vj . If she jumps to vj , then she immediately colors it or immediately jumps to another vertex in S.vj /. Since she can jump only twice to a vertex, there can be at most 2.k 1/ active vertices in N .vj / after Alice’s move. Hence, there are at most 3.k 1/ active vertices around vj after Alice’s move. Counting Bob’s last move, there are at most 3k 2 colored vertices around vj when it gets colored.
v1 v2 v3
v Figure 11.6. Construction of a 3tree.
There is a nice class of graphs for which the bound in Theorem 3 is optimal (as long as we assume that Alice cannot distinguish between colors). Fix a number k and start with a clique on k vertices v1 ; : : : ; vk . Then in each subsequent step add a new vertex v and join it to k vertices of any clique in the graph constructed so far. In this recursive manner we produce graphs called ktrees (see Figure 11.6). The resulting ordering of the vertices shows that col2 .G/ D k C 1, which gives colg .G/ 3k C 2 for any ktree G. It can be proved that this bound is optimal for each k (> 1). For planar graphs the theorem gives g .G/ 29, which is not bad but is still not optimal.
Six dollar order We now demonstrate that the estimate g .G/ 18 holds for every planar graph G, which is almost the best upper bound discovered so far. This time we take more care in ordering the vertices v1 ; : : : ; vn so as to make the activation strategy more efficient. We define this
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order inductively (in reverse order), starting with any vertex vn of degree at most five. Suppose that we have already picked vertices vn ; vn 1 ; : : : ; vi C1 and are looking for a candidate for vi . Partition the vertex set V into two parts, C D fvn ; vn 1 ; : : : ; vi C1 g and U D V n C , and construct a new graph H as follows: 1. Delete each edge between pairs of vertices in C .
2. Delete each vertex v in C with at most three neighbors in U . 3. For each deleted vertex v in C add edges between its neighbors in U so that these neighbors form a clique. It can be verified that the new graph H is planar: the new edges in H can be drawn near deleted edges incident to deleted vertices of C . By Euler’s formula H satisfies the inequality jE.H /j 3jV .H /j 6. Now suppose that each edge of H is assigned $2 (in halfdollar coins). These coins are distributed to the vertices of H as follows: if an edge e links two vertices in U , then e gives $1 to each of the two vertices; if e links a vertex x in C and a vertex y in U , then e gives $1.50 to x and $0.50 to y (see Figure 11.7). As the total amount of dollars is equal to 2jEj (< 6jV .H /j), there is a vertex v that receives less than $6. This will be the i th vertex vi in our order. Notice that vi cannot belong to C , for each remaining vertex in C is incident to at least four edges of H , hence receives at least $6.
U
$1
C $1
vi Z
$0.5
$1.5
D
N(vi)
Figure 11.7. Ordering the set of vertices of G by finding a poor vertex in a graph H .
Now we are ready to prove the result stated at the beginning of this section: Theorem 4. Every planar graph G satisfies colg .G/ 18. Proof. Let v1 ; : : : ; vn be the ordering of the vertices of G obtained in the manner just described. It suffices to show that if Alice applies the activation strategy, then at the end of each of her turns there will be at most sixteen active neighbors around any uncolored vertex. Let v D vi be any uncolored vertex, let H , C , and U be as indicated in the discussion preceding Theorem 4 (notice that all of them depend on vi ), let D be the set of those forward neighbors of v (in G) that were deleted while constructing H , and let Z be the set of those vertices of U that are adjacent to v in H . Since we didn’t delete any edges between vertices of U , we see that N C .v/ is contained in Z. Also Z includes all vertices, other than v, to which Alice can jump after activating any vertex in the set D (see Figure 11.7). Hence, arguing as in the previous proof, there can be at most 3jZj C jN .v/ n Dj
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active vertices around v at the end of any move by Alice. Since v received strictly less than $6 in the ordering procedure, 1 jZj C 0:5 jN .v/ n Dj < 6: This implies that 3jZj C jN .v/ n Dj 16, which completes the proof.
Variants of the game Currently, the best upper bounds for a planar graph G are g .G/ colg .G/ 17, as proved in [20]. The proof is based on a modification of the activation strategy. At present, the best lower bounds for planar graphs are 8 g .G/ and 11 colg .G/. There is evidence that the upper bound on the game coloring number colg .G/ may be sharp. However, we cannot even anticipate what the correct answer is for the game chromatic number g .G/ of planar graphs. This situation probably exists because of the rather strange general behavior of the parameter g .G/. For instance, g .G/ is already unbounded for graphs satisfying .G/ D 2 and is also not monotone with respect to subgraphs. Indeed, let U D fu1 ; : : : ; un g and V D fv1 ; : : : ; vn g be disjoint sets of vertices, and let Kn;n be the graph on these vertices obtained by joining each vertex in U with all vertices in V . Let M be a perfect matching in Kn;n (i.e., a set of n edges no two of which are incident with the same vertex; for instance, M D fu1 v1 ; : : : ; un vn g). Clearly we have g .Kn;n / D 3, but g .Kn;n M / D n. Bob’s strategy is to copy Alice’s moves on the other ends of the missing matching (see Figure 11.8). Alice
Bob
Figure 11.8. Copy cat strategy allows Bob to win on Kn;n
M if fewer than n colors are available.
More surprisingly, it is not clear what influence increasing the number of colors has on the coloring game. Suppose that Alice wins with k colors on a graph G. Then it would seem trivial that she should also win on G with k C 1 colors. But can you prove it? As yet, nobody has been able to do so! In view of such pathologies it is natural to look for more predictable variants of the coloring game. For instance, in a modification considered in [3] Bob is not allowed to use a new color until it is absolutely necessary, that is, unless all uncolored vertices are surrounded by all previously used colors. Chen, Schelp, and Shreve prove there that g .T / 3 holds for any tree T , where g .G/ is the related game chromatic number. However, no better bounds for arbitrary planar graphs are provided. In this setting it is clear that increasing the number of colors cannot help Bob. However, we are still left with the anomaly that g .Kn;n M / D n.
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There are many natural variations of the coloring and ordering games. One variation is to allow Alice more moves so as to limit Bob’s ability to interfere [12], [13], [17]. Other variants consider oriented coloring [15], [16], [18] and d relaxed coloring [4], [7]. It is remarkable that the activation strategy is still effective in these diverse settings. Acknowledgment We would like to thank Steven Brams for useful materials and comments on the origin of the mapcoloring game. The research of the fourth author is supported in part by the National Science Council under grant NSC952115M110013MY3.
Bibliography [1] H. L. Bodlaender, On the complexity of some coloring games, Internat. J. Found. Comput. Sci. 2 (1991) 133–147. [2] O. V. Borodin, On acyclic colorings of planar graphs, Discrete Math. 25 (1979) 211–236. [3] G. Chen, R. H. Schelp, and W. E. Shreve, A new game chromatic number, European J. Combin. 18 (1997) 1–9. [4] C. Chou, W. Wang, and X. Zhu, Relaxed game chromatic number of graphs, Discrete Math. 262 (2003) 89–98. [5] V. Chv´atal, V. R¨odl, E. Szemer´edi, and W. T. Trotter, The Ramsey number of a graph with a bounded maximum degree, J. Combin. Theory Ser. B 34 (1983) 239–243. [6] T. Dinski and X. Zhu, Game chromatic number of graphs, Discrete Math. 196 (1999) 109–115. [7] C. Dunn and H. Kierstead, A simple competitive graph coloring algorithm III, J. Combin. Theory Ser. B 92 (2004) 137–150. [8] U. Faigle, U. Kern, H. A. Kierstead, and W. T. Trotter, On the game chromatic number of some classes of graphs, Ars Combin. 35 (1993) 143–150. [9] M. Gardner, Mathematical games, Scientific American (April, 1981) 23. [10] D. Guan and X. Zhu, The game chromatic number of outerplanar graphs, J. Graph Theory 30 (1999) 67–70. [11] H. A. Kierstead, A simple competitive graph coloring algorithm, J. Combin. Theory Ser. B 78 (2000) 57–68. [12] ——, Asymmetric graph coloring games, J. Graph Theory 48 (2005) 169–185. [13] ——, Weak acyclic coloring and asymmetric graph coloring games, Discrete Math. 306 (2006) 673–677. [14] H. A. Kierstead and W. T. Trotter, Planar graph coloring with an uncooperative partner, J. Graph Theory 18 (1994) 569–584. [15] ——, Competitive colorings of oriented graphs, Electron. J. Combin. 8 (2001) #12, 15 pp. [16] H. A. Kierstead and Z. Tuza, Marking games and the oriented game chromatic number of partial ktrees, Graphs Combin. 8 (2003) 121–129. [17] H. A. Kierstead and D Yang, Very asymmetric marking games, Order 22 (2005) 93–107. [18] J. Neˇsetˇril and E. Sopena, On the oriented game chromatic number, Electron. J. Combin. 8 (2001) #14, 13 pp. [19] X. Zhu, The game coloring number of planar graphs, J. Combin. Theory Ser. B 75 (1999) 245– 258. [20] ——, Refined activation strategy for the marking game, J. Combin. Theory Ser. B (2007, in press).
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12 It’s Okay to Be Square If You’re a Flexagon Ethan J. Berkove and Jeffrey P. Dumont It has been said that a mathematician can be content with only paper and pencil. In fact, there are times when one doesn’t even need the pencil. From a simple strip of paper it is possible to make a surprisingly interesting geometric object, a flexagon. The flexagon can credit its creation to the difference in size between Englishruled paper and American binders. The father of the flexagon, Arthur Stone, was an English graduate student studying at Princeton University in 1939. To accommodate his smaller binder, Stone removed strips of paper from his notebook sheet. Not being wasteful, he creased these lengths of paper into strips of equilateral triangles, folded them in a certain way, and taped their ends. Stone noticed that it was possible to flex the resulting figure so that different faces were brought into view—and the flexagon was born [1]. Stone and his colleagues, Richard Feynman, Bryant Tucker, and John Tukey, spent considerable time cataloging flexagons but never published their work. Like many geometric objects, flexagons can be appreciated on many levels of mathematical sophistication (the first author remembers folding flexagons in elementary school). With so adaptable a form, it is not surprising that flexagons have been studied from points of view that vary from art to algebra. Our interest in flexagons was sparked by a question posed in a paper by Hilton, Pedersen, and Walser [9]. They studied one of the hexaflexagons, sonamed because the finished model has the shape of a hexagon. They calculated the group of motions for a certain hexaflexagon, then inquired about other members of the hexaflexagon family. We have determined that the trihexaflexagon is exceptional, as it is the only member of the hexaflexagon family whose collection of motions forms a group. Working to generalize this result, we shifted our attention to tetraflexagons, which are constructed from strips of squares folded into a 2 2 square final form. We discovered that tetraflexagons are, if anything, more complicated and interesting than their hexagonal cousins. These results convinced us that tetraflexagons, often only mentioned in passing in the literature, deserve to be brought into the limelight. In this paper, we will summarize the results of our investigations. Although most of the material on hexaflexagons is known, the material on tetraflexagons includes new results and open questions. It is our intent to Reprinted from Mathematics Magazine, Vol. 77, No. 5 (Dec. 2004), pp. 335348.
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give interested readers enough material to start their own explorations of these fascinating objects.
The hexaflexagon family d c
f
tab
b a
h g
e
i
Figure 12.1. Trihexaflexagon net.
It is easy to fold a flexagon, and we highly recommend making one of your own as this experience will be helpful in following the results in this section (and it’s fun). Construct a strip of nine equilateral triangles and a tab as in Figure 12.1; this strip is called the net of the flexagon. Each triangle in the strip, and in general each polygon in a flexagon net, is called a leaf of the flexagon. You may want to label both sides of each leaf and precrease all edges in both directions. Hold the leaf marked a in your hand, fold leaf c over leaf b, f over e, and i over h. Finish the flexagon by gluing or taping the tab onto leaf a. The final model should look like the flexagon depicted in Figure 12.2. Clockwise from the top, one can read off the leaves f; d; c; a; i; g. We call .f; d; c; a; i; g/ a face of the flexagon.
f g
d
i
c a
Figure 12.2. The .f; d; c; a; i; g/ face of the trihexaflexagon.
To flex your new creation, bring the three corners at the dashed lines down together so they meet. The hexagon will form a Y, at which point it will be possible to open the configuration at the middle. (This is the only possible way to perform a flexdown for this flexagon. There is also an inverse operation, a flexup.) The result is a different face .f; e; c; b; i; h/ of the flexagon. The flex can be repeated to get a third face .g; e; d; b; a; h/, and one more flex returns the flexagon to .g; f; d; c; a; i /, the original face rotated clockwise through an angle of =3. Since it has three distinct faces, this flexagon is known as the trihexaflexagon—it is the simplest member of the hexaflexagon family. The three faces can be seen more easily if they are marked somehow: Wheeler [15] shows how to color the net so each flex brings out a new color, and Hilton et al. [9] give a way of marking the net so flexes bring out happy and sad pirate faces. We would like a way to keep track of all the faces of a flexagon while we flex, which we can do using a graph: vertices represent the faces of the flexagon, and an edge joins two vertices if there is a flex that takes one face of the flexagon to the other. On occasion
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fdcaig
gedbah
fecbih
Figure 12.3. The structure diagram for the trihexaflexagon: vertices denote faces, edges denote flexes between faces.
we will use a directed graph, where an arrow points towards the face that is the result of a flexdown. We choose to ignore the orientation of a face in the graph as this has a tendency to make the graph overly complicated. The completed graph is called a structure diagram. The cycle in Figure 12.3 is the structure diagram for the trihexaflexagon. It shows the three distinct faces, as well as their relationship via flexing. The flexdown we described is a motion of the flexagon, a transformation that takes one hexagonal face of the flexagon to another hexagonal face. We require that our flexagons have no faces containing loose flaps that can be unfolded or moved so the hexagonal shape is lost. This becomes a significant issue as the number of triangles in the net increases. Indeed, in larger nets it is increasingly likely that a random folding of the net yields a face containing a loose flap, which in turn causes the entire flexagon to fall apart into a M¨obius band with multiple twists. Therefore, we only consider flexagons that are folded in such a way that every flex is a motion. To determine the structure of the set of motions, let f denote a flexdown (hence f 1 denotes a flexup), and denote by a flip along the xaxis of the flexagon. Then f 3 D 2 D id and it is easy to confirm that f D f 1 . The group with this presentation is well known as S3 , the symmetric group on 3 letters, which has 6 elements. This analysis ignores the fact that f 3 is not strictly the identity flex, but is instead a rotation through =3 degrees. The complete set of motions of the trihexaflexagon, including rotations, is analyzed in [9]; the only difference from the argument above is that f 18 D id , and the resulting group is D18 , the dihedral group with 36 elements.
Motions of the hexahexaflexagon We perform a similar analysis for the member of the hexaflexagon family with six faces. This hexahexaflexagon can be constructed from the net of 18 triangular leaves with a tab as in Figure 12.4. To create the flexagon, label all leaves front and back and precrease all edges as before. Fold leaf a under the rest of the strip. Then fold the edge between leaves c b
c
d
e
f
g
j
h i
l k
p
n m
o
r q
tab
a
Figure 12.4. The hexahexaflexagon net.
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and d so that a and d are adjacent. Next fold the edge between e and f so that c and f are adjacent. Continue rolling the strip in this manner until o and r are adjacent. The finished roll should look like the initial strip for the trihexaflexagon with leaf b on the far left. Fold this like the trihexaflexagon, then tape the tab to a to complete the model. In contrast to the trihexaflexagon, from the initial face of the hexahexaflexagon either alternating set of corners flexes down. We can distinguish the two flexes by looking at what they do to the number of leaves in a triangular segment of the hexagon. Following Oakley and Wisner [11], we call the entire triangular segment a pat. In our newly folded hexahexaflexagon, pats alternately contain 2 and 4 leaves, or .2; 4/ for short. One of the flexes, which we call f , preserves thicknesses so is patpreserving. The other flex, g, is patchanging, from .2; 4/ to .1; 5/ and vice versa. The pat thicknesses are invariant under , a flip along the xaxis. Starting with the initial face of the hexahexaflexagon, one finds that f 3 rotates the hexagon clockwise through an angle of =3, so f 18 D id . In addition, as f D f 1 , f and generate a copy of D18 in the collection of hexahexaflexagon motions. On the other hand, g leads to a face where the only possible flex is the patpreserving f , which leads to a face where the only possible flex is the patchanging g. The threeflex combination gfg rotates the hexagon clockwise through an angle of =3, and g satisfies g D g 1 . Therefore, there is at least one other copy of D18 in the motions of the hexahexaflexagon. We turn this information into the structure diagram in Figure 12.5. Solid arcs between faces denote patpreserving flexes, while dotted arcs denote patchanging flexes. There is a primary cycle that can be traversed using patpreserving flexes, and three subsidiary cycles that can be entered at faces where two flexes are possible. From the cycle structure we learn an important fact: the collection of motions of the hexahexaflexagon must have at least two generators, f and g. However, there are faces where only one of f and g can be applied, so it is not always possible to apply f or g twice in a row. In fact, g3 does not even make sense. We now have an answer to the question posed in [9]: the hexahexaflexagon’s motions do not form a group! This conclusion surprised us, as the collection of motions for every other geometric object we know of has a group structure. Furthermore, all hexaflexagons except the trihexaflexagon share this characteristic, as they contain patchanging flexes. Patchanging flexes
f g
f
Figure 12.5. Structure diagrams for the hexahexaflexagon: dotted edges are patchanging flexes, solid edges are patpreserving flexes.
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occur at faces with two possible flexes, and the only hexaflexagon structure diagram without intersecting cycles is the trihexaflexagon’s. Structure diagrams are necessarily finite, and results in Wheeler [15] imply that cycles cannot form a closed link. As a result, in any hexaflexagon there are only so many times the patchanging flex can be applied. Thus, for every hexaflexagon with a patchanging flex g there is a k such that gk is undefined, which implies that the collection of motions for that hexaflexagon cannot form a group. The astute reader might have noted that the structure diagram in Figure 12.5 seems to have nine faces, not six, as the hexahexaflexagon’s name suggests. This discrepancy can be explained by carefully studying the hexahexaflexagon’s faces. Upon closer inspection, three identical sets of triangles occur in two separate faces, but in different orders. If a face depends on both the triangles and their order, then indeed the hexahexaflexagon has nine faces. If order is disregarded, however, there are only six faces. The latter is the standard accounting, hence the name (although Oakley and Wisner [11] distinguish between the six physical faces and the nine mathematical faces). We can construct other members of the hexaflexagon family by increasing the number of triangles in the initial net. For example, starting with a straight strip of 36 leaves, the strip can be rolled, then rerolled to yield the net in Figure 12.1, then folded as in the trihexaflexagon case to yield the dodecahexaflexagon. One can fold other members of the family by starting with nets that are not straight. The article by Wheeler [15] and Pook’s book [14] contain some nice directions for folding the tetrahexa and pentahexa cases. There is also a HexaFind program [3], which generates all nets for hexaflexagons with any given number of faces. We have only discussed a bit of what is known about hexaflexagons. In [11], Oakley and Wisner introduce the concept of the pat, then use it to count the total number of hexaflexagons with a particular number of faces. Madachy [10], O’Reilly [12], and Wheeler [15] describe connections between hexaflexagons and their structure diagrams. McIntosh [7], Madachy [10], and Pook [14] provide lengthy bibliographies to other work on hexaflexagons. Gilpin [8], Hilton et al. [9], and Pedersen [5] study the motions of the trihexaflexagon. Furthermore, the group of motions of the trihexaflexagon is identified in [8] and [9].
The distant cousins: tetraflexagons We were introduced to tetraflexagons in a Martin Gardner Mathematical Games column in Scientific American [6]. We immediately noticed some differences; the direct analog of a strip of triangles, a straight strip of squares, makes for a poor flexagon; when you fold square over square you end up with a roll that does not flex at all (and has a trivial structure diagram). Therefore, we allow the nets to have rightangled turns. These turns occur at what we call corner squares, those attached to their neighbors on adjacent edges rather than opposite ones. Because so much is known about hexaflexagons, we felt that the tetraflexagon family would be readily analyzed. Our actual experience mirrored Stone’s, as reported by Gardner [6]: “Stone and his friends spent considerable time folding and analyzing these foursided forms, but did not succeed in developing a comprehensive theory that would cover all their discordant variations.” However, we have established some notation and proved some results; our investigations have convinced us that tetraflexagons are as fascinating as, and more subtle than, their hexaflexagon relatives.
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a
l
c
b
k
d
j
e g
h
i
f
Figure 12.6. A “4 4 ring” net.
Here is one way to fold a tetraflexagon using a 4 4 net with a cut adjacent to a corner square. Label a net as in Figure 12.6 and lay it flat. Mark the cut edges of a and l so you can tape them together after you have folded the tetraflexagon. 1. Fold leaf c over leaf d . Leaves a and b will remain to the left of c, but will flip over. 2. Fold c and d over e; the back of leaf c will touch the front of leaf e, and again, a and b are left free. 3. Fold f over g; move the flap with a and b so it points to the right. 4. Fold f and g over h so f and h are touching. Then slip the flap with a and b under j . 5. Fold j over i (without moving a and b), and flip the partially folded tetraflexagon upside down. Position the tetraflexagon so k and l are at the top of the model and a and b are upside down and facing to the left. 6. Fold i and j over k so i and k are touching. 7. Slide the flap with a and b under l , then fold a over l . Finally, tape a and l together on the right, along the edges that were originally cut. The finished tetraflexagon will have 90degree rotational symmetry, and should look like Figure 12.8a. Now that you have a tetraflexagon in front of you, let’s introduce some notation (see also Figure 12.7). The tetraflexagon has four pats, which conveniently look like the four quadrants in Cartesian coordinates. Therefore, we will refer to pats by the quadrant they are in: pat I, pat IV, etc. If you look carefully at your tetraflexagon, you’ll notice that adjacent pats are connected by a layer of paper, called a bridge. The two leaves, one in connected sides
loose flap
pocket free sides foldable
edge
Figure 12.7. Tetraflexagon notation.
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each quadrant, that make up this layer are bridge leaves. Leaves b and c are bridge leaves, as are k and l . Bridges will figure prominently in our analysis. Define a pocket as a strict subset of leaves within a pat attached to the rest of the tetraflexagon on two adjacent sides. Without the strictness condition, an entire pat is a pocket, connected to the rest of the flexagon by the two bridges; this is an extreme case we wish to avoid. On the other hand, a pocket may consist of a single leaf, in which case the pocket is a bridge leaf. As an example, squares f and g form a pocket in the tetraflexagon you just folded and this is the only pocket in its pat. The attached sides of the pocket are connected, whereas the other two sides are free. When a free side of a pocket lies along the outside edge of the tetraflexagon (like the outside edge of h and i ), we call the entire tetraflexagon edge a foldable edge. The foldable edge will be folded in half during the flex. There are two conditions that a tetraflexagon must satisfy in order to flex: there must be a pocket with a component bridge leaf, and the bridge in the pocket must be at right angles to the foldable edge. (Since the bridge goes from a pocket to an adjacent pat it cannot cross the pocket’s free side.) To perform a flex, orient the tetraflexagon so the foldable edge is forward and the pocket lies on top of the tetraflexagon, as in Figure 12.8a. Fold the tetraflexagon in half perpendicular to the foldable edge so that the pocket remains on the outside. Put your thumbs into the pocket and its kittycorner companion (Figure 12.8b), then pull outwards in the direction of the arrows. The pocket layers will separate from the rest of the pat, rotating outwards 180 degrees but staying in the same quadrant. The rest of the pat layers will maintain their orientation but move to the top of the adjacent pat. Simultaneously, the layers in the adjacent pats will rotate outwards 180 degrees. When you flatten the tetraflexagon, you will see a new face as in Figures 12.8c and 12.8d.
a
b
c
d
Figure 12.8. Flexing a tetraflexagon.
We call the flex with the pockets up and the foldable edge forward a bookdown flex. It is only one of four possibilities, depending on the position of the pocket and foldable edge. We therefore distinguish between four types of flexes: flexes up or down along the yaxis (bookup and bookdown) and flexes up or down along the xaxis (laptopup and laptopdown). From Figure 12.8, we see that an up flex is the inverse of a down flex and vice versa. As before, the top layer of leaf of the tetraflexagon, like .a; d; g; j /, is called a face, and a flex is a motion if it takes one tetraflexagon face to another. Before we go further into our exploration, we introduce some assumptions about the tetraflexagons we consider to aid in our analysis. 1. Net assumption. All tetraflexagon nets have exactly four corner squares. The two vertical strips are each composed of m leaves, and the two horizontal strips of n leaves. All nets are folded into fourpat (2 2) tetraflexagons. 2. Winding assumption. There is only one bridge between adjacent pats of the tetraflexagon. In other words, as the net is folded, the strip of paper enters and exits each
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Part III. Flexagons and Catalan Numbers quadrant exactly once. This assumption is easily justified; we will see in the Flexing Lemma below that subsequent flexes maintain all four bridges as single layers. 3. Rigidity assumption. No tetraflexagon face contains a loose flap. A loose flap is a collection of two or more leaves connected to the rest of the tetraflexagon on only one side (see Figure 12.7). Since a loose flap determines one of two possible tetraflexagon faces depending on its position, movement of a loose flap can be considered a motion. The rigidity assumption ensures that the only tetraflexagon motions are flexes. 4. Symmetry assumption. Folded tetraflexagons have 180degree rotational symmetry (in contrast to the hexaflexagons’ 120degree symmetry).
Figure 12.9. Rings, bolts, and snakes.
The net assumption implies that nets are one of three shapes, depending on parity. All nets are shown in Figure 12.9: when m and n are both even, the only possible net is a ring; when both m and n are odd, the only possible net is a lightning bolt; and when exactly one of m and n are odd, the only possible net is a snake. We remark that analogs of the other three assumptions hold for most hexaflexagons.
Basic analysis: folding 101 and structure diagrams with dead ends Flexing for tetraflexagons is a bit more complicated than for hexaflexagons. We carefully describe what happens during a flex in the following Flexing Lemma. The following hold for tetraflexagons under our assumptions: 1. Corner squares remain in their pats during flexes. 2. A pocket that has a component bridge leaf is either a single layer (that is, it is the bridge leaf), or consists of every leaf in the pat except for either the top or bottom one. In the latter case, the leaf not belonging to the pocket is also a bridge leaf. 3. A flex maintains all four bridges as single layers. Proof. To simplify the arguments, we will assume for this proof that the pocket is always on top of pat IV with the foldable edge forward, as in Figure 12.8a. The other cases will follow by rotation and mirror images. Moreover, by the symmetry assumption we only need focus on pats III and IV. We start with the first claim. When we perform the flex shown in Figure 12.8, all leaves in pat III remain in pat III, which means that the only corner square that can change pats
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lies in pat IV. However, if the bookdown flex moves this corner square to pat III, then the flex is also forced to move the strip of leaves connecting the corner squares in pats I and IV. This is impossible as the corner square in pat I is fixed by the symmetry assumption. For the second claim, assume that the pocket with a bridge leaf consists of more than one layer. We claim that the corner square in pat IV must be a leaf of the pocket. By way of contradiction, assume the corner square is below the pocket. Notice that the corner squares in pats I and IV are connected by a vertical strip. As this strip is woven from the corner square in pat IV to the corner square in pat I, it must at some point become the unique bridge to pat I (by the winding assumption). One of the bridge leaves is a leaf of the pocket, so the vertical strip of squares must also cross from the bottom of pat IV to the pocket. Because the vertical strip is woven back and forth, it can only cross to the pocket on the front or back edge of pat IV. The vertical strip cannot cross the gap at the front edge of pat IV because that edge is free by assumption. The vertical strip cannot cross the gap at the back edge of the pocket either, as that would lock the pocket to the rest of pat IV, making a flex impossible. We conclude that the corner square must be a leaf of the pocket. An analogous argument shows that the vertical strip cannot cross from the corner square to the bottom layer either. By the rigidity assumption, this bottom layer must therefore consist of a single leaf. Now if the bridge leaf to pat III were any layer but the bottom, it would force the left side of the pocket in pat IV to be connected, rather than free. This is impossible as all pockets have two free sides. For the final claim, the winding assumption implies that, when first folded, the tetraflexagon has single layer bridges between pats. Consider the effect of a bookdown flex. This flex flips the bridge leaves that lie in the “pages” of the book upside down, so those bridges remain single layers. For the two bridges that cross the spine of the book, focus on pats III and IV, and assume first that the pocket is a single layer; then the bookdown flex leaves a single layer in pat IV, which must be the bridge leaf. Otherwise, the pocket has more than one leaf and the bottom leaf of pat IV is a single layer. In this case, the completed flex moves the single layer from pat IV to pat III, again maintaining the bridge as a single layer. The Flexing Lemma gives us insight about what happens to a tetraflexagon during a single flex. What about the larger picture? What can we say about the set of all the motions of the tetraflexagon we folded? Again, we answer these questions with a structure diagram, but now use the orientation of the graph’s edges to distinguish among the four possible types of flexes. Denote a book flex by a horizontal line segment in the structure diagram, a laptop flex by a vertical one, and let arrows point to the faces that are the result of a flex down. With a little flexing and experimentation, we find that the structure diagram for the tetraflexagon we folded is an L shape, with five vertices (corresponding to faces), and four edges, two in each part of the L (corresponding to flexes). On all edges, the arrows point away from the corner of the L. This structure diagram, which appears in the bottom left of Figure 12.11, has a feature we have not seen before: there are vertices incident to a single edge. We call the faces associated to these vertices dead ends, because once these faces are reached via a flex, the only motion that can be applied is the flex’s inverse. For the hexaflexagon cases, a flex of finite order appeared in the structure diagram as a cycle. The structure diagram we just constructed has no cycles, hence no elements of finite
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order. In addition, any flex can be applied at most twice, at which point some other flex must be applied. In other words, there is really no group structure at all in the motions of the tetraflexagon we constructed. An immediate question is whether there is a tetraflexagon whose collection of motions forms a group. There is, and it can be folded from the 3 3 lightning bolt net shown in Figure 12.10. Start with a copy of this net—you may want to specially mark the shaded sides of a and h as shown in the diagram, as those are the edges that get taped together at the end. Fold leaf b over c, d (and a, b, and c) over e, and g over f . Make sure h lies on top of a, then tape a and h together on the right, along their shaded sides. Your finished tetraflexagon should have 180degree rotational symmetry, as per the symmetry assumption. From this initial face, the only possible down flex is a laptopdown. If this is followed with bookdown, laptopdown, and bookdown flexes (again, the only possible down flexes) you get back to the initial face. The resulting structure diagram is a square cycle, corresponding to the symmetry group Z=4Z, the cyclic group with 4 elements. We believe that this is the only tetraflexagon whose collection of motions forms a group (although a rigorous proof eludes us). g
h
f c
d
e
b a Figure 12.10. A 3 3 bolt net.
You may have noticed that although we call the net in Figure 12.10 a 3 3 lightning bolt, there are only two leaves, g and h, in one row. Actually, that row does contain a third leaf, a, which becomes part of the row after we finish taping. In general, when we refer to a net as m by n, we include the corner squares in the count. For the cases of the bolt and snake nets, this means that either one row or one column will be a leaf short when the net is first constructed. By reducing larger nets to the 3 3 case, one can construct many tetraflexagons whose structure diagrams contain cycles. For example, starting with the net in Figure 12.6, folding b over c and i over h turns the 4 4 ring net into a 4 3 snake net. Another pair of folds results in a 3 3 lightning bolt net, which can then be folded as above. More generally, one can turn an m n net into either an m 1 n net or an m n 1 net by making appropriate folds. Here, appropriate means that the pair of folds results in a net that satisfies the symmetry assumption. Also, a given sequence of edge choices must result in a tetraflexagon that satisfies our four assumptions. Once the net is a 3 3 lightning bolt, it is folded to guarantee at least one cycle in the structure diagram. This process is reminiscent of how a hexaflexagon is folded from a strip containing 9.2n / triangles by rolling it n times to form the trihexaflexagon net with 9 triangles. A difference, though, is that there are many ways of choosing edges in pairs to reduce an m n net to the 3 3 net, so a generic tetraflexagon net can yield a large number of tetraflexagons with distinct structure diagrams.
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Figure 12.11. Tetraflexagon structure diagrams folded from a 4 4 ring.
As an example, through exhaustive folding, both figurative and literal, we have identified in Figure 12.11 most (all?) of the possible structure diagrams that result from a 4 4 ring. Some of the diagrams are very basic, but notice that three of them contain cycles, all but one contain dead ends, and two contain both. This variety in structure diagrams from the same net demonstrates how difficult it is to classify even simple tetraflexagons. Despite their “discordant variations,” the tetraflexagons we investigated have some features in common. In a number of the examples we folded, we came across faces where all four flexes were possible. We call such faces crossroads. For the reader who wants to see a crossroad in action, start with the net in Figure 12.6, labeled front and back with each letter. Fold b under c, c over d , e (and a–d ) under f , g (and the rest of the net) over f , i over h, j over i , and l over k. Lift a over l , then tape a and l together on the right (outside) edge. The final tetraflexagon is shown in Figure 12.12d. (What is the resulting structure diagram?) Crossroads are among the most interesting features in a structure diagram, and we wondered how many crossroads a structure diagram could contain. We noticed that after we flexed a crossroad, we never saw another crossroad, which led us to prove the Crossroad Theorem. Crossroad faces are never adjacent in tetraflexagons built from ring, lightning bolt, or snake nets. Proof. Let’s analyze the shape of a tetraflexagon at a crossroad face. In order to perform all four types of flexes, every edge of the tetraflexagon must be a foldable edge and there must be two pockets per edge to allow both types of flexes. It is easy to show that the pockets cannot be on the same side of the foldable edge, as in Figure 12.12a. If the pockets were on the same side, the remaining leaves of the two pats would be the bridge between the pats, by the second claim of the Flexing Lemma. By the symmetry assumption, the same would be true for the other two pats of the tetraflexagon, and the entire tetraflexagon would simply fall apart. We conclude that one pat along the foldable edge must have a pocket on the top of the pat to allow a flex down, while the adjacent pat along the edge must have a pocket on the bottom of the pat to allow a flex up. There are essentially two cases to consider, shown in Figures 12.12b and 12.12c (Figure 12.12d is Figure 12.12c’s mirror image). We can show that Figure 12.12b is impossible by a careful analysis of pat IV. Since one of the two pockets in pat IV contains all the layers but one, assume, without loss of generality, that it is the pocket associated to the foldable edge on the right. Then the bridge to pat III crosses the edge marked by the arrow, and the component bridge leaf in pat IV is one of
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a
b
c
d
Figure 12.12. Two impossible configurations and two crossroads.
the leaves in the pocket by the second claim of the flexing lemma. However, this implies that the edge marked by the arrow is a connected edge, which makes a bookdown flex impossible. Therefore, the only possible crossroad faces have configurations like Figures 12.12c and its mirror image 12.12d. We next determine when a flex from a crossroad yields another crossroad. By symmetry arguments, it is sufficient to consider Figure 12.12c. We claim that is impossible to perform a bookdown flex from Figure 12.12c and get the crossroad in Figure 12.12d. This time, consider pat III. Note that by the second claim of the Flexing Lemma one of the two pockets in pat III must consist of every layer in the pat except one; assume (by symmetry again) that it is the bottom pocket. After a bookdown flex is applied, the original layers in pat III are flipped upside down, and are covered by a single layer from pat IV. This single layer must be part of the new bridge to pat IV. Therefore, it is not possible for the right edge of pat III to be free, as in Figure 12.12d. Thus, if two crossroads are connected by a flex, they must be identical. We rule out this possibility too. Consider the enlargement of pat IV of Figure 12.12c, shown in Figure 12.13. In order for a bookdown flex to yield the same configuration, pat IV must contain a pocket as part of the bridge. But the bridge must be a single layer, by the second claim of the Flexing Lemma.
Figure 12.13. Closeup of pat IV.
The Crossroad Theorem tells us that cycles cannot be too dense in a structure diagram. For example, we have a corollary to the theorem, named after the symbol that appears on all Purina pet foods. Purina Corollary. Under our assumptions, there is no “Purina Tetraflexagon” with the structure diagram shown in Figure 12.14.
Surgery: adding new parts to an old flexagon So far, we have a very general idea of what can and cannot be part of a structure diagram. On the other hand, it would be desirable to add a given component to a structure diagram,
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Figure 12.14. Purina structure diagram.
allowing us to tailormake tetraflexagons with interesting properties. There is a technique we learned from Harold McIntosh’s online notes [7] that allows us to do exactly that at an appropriate spot in a structure diagram. We call this technique surgery. One performs surgery on a tetraflexagon by symmetrically grafting two strips of squares to the model to add the new feature. An instance of surgery is shown in Figure 12.15. Start with a tetraflexagon at a face where one of the pats is a single leaf, that is, a corner square. Place this tetraflexagon so the single leaves are in pats II and IV with the foldable edge forward, as in the top left picture in Figure 12.15. By considering mirror images, if necessary, we may assume that the pocket in pat III is on the bottom of the pat. Tape a 2 1 strip to the right edge at pat IV and make a cut in the tetraflexagon along the positive xaxis as in the top right picture. Fold the strip on top of pat IV, then fold the strip up and in half as in the bottom left picture. Finally, tape the top edge of the strip (in pat IV) to the cut edge in pat I as in the bottom right picture. Perform the same procedure symmetrically on pat II; when you are done, the front and back foldable edges will have pockets on both top and bottom. The resulting tetraflexagon’s structure diagram will be the same as before, but with a cycle tacked on at the appropriate face. Following this technique, it is straightforward, at least in theory, to build a tetraflexagon whose structure diagram consists of an arbitrarily long row of fourcycles connected one to the other at opposite corners. (In light of the Crossroad Theorem, in some sense this is
Figure 12.15. How to add a cycle to a tetraflexagon.
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Figure 12.16. A challenging structure diagram.
as crowded as collections of cycles can become.) In practice, the tetraflexagons quickly become too thick to flex easily as the size of the net increases. We challenge the reader to use surgery to construct a tetraflexagon with the structure diagram shown in Figure 12.16. One can also use surgery to add dead ends to a structure diagram. Start with a tetraflexagon in the same position as before and make the same cut along the positive xaxis, but this time add a 1 n strip horizontally to the right edge of pat IV. Roll the strip counterclockwise until it lies on top of pat IV, then tape the top edge of the middle square of the roll to the cut edge of pat I. Repeat this symmetrically on pat II. This procedure will add n horizontal segments to the structure diagram at the appropriate place. To finish this section we mention that besides appearing in Gardner articles [2] and [6], “square” flexagons are the topic of a short note by Chapman [2] where he uses primary and secondary colors to distinguish tetraflexagon faces. This note also contains directions for constructing tetraflexagons whose structure diagrams are a cycle and two linked cycles. For more information on surgery and tetraflexagons, the reader should consult McIntosh’s notes [7] or Pook’s book [14].
Where do we go from here? At this point, we know something about tetraflexagons—how they flex, how and where cycles can occur, and why some structure diagrams are impossible—but there are many questions we haven’t answered. What are the possible tetraflexagon structure diagrams? How many tetraflexagons, up to rotation and mirror image, can be made from an m n net with four corners? We can get a rough upper bound by recalling that the corner squares in a tetraflexagon stay fixed in their quadrant. Flex a tetraflexagon so that one pat consists of a single leaf and its adjacent pats contain n C m 1 leaves. With the exception of the bottom (or top) leaf, these might be in any order, so there are at most .n C m 2/Š possible tetraflexagons that can be folded from a given m n net. Many of these tetraflexagons will not satisfy our four assumptions; is there a better upper bound? Finally, there is the question that provided our initial motivation to study flexagons: Are there any other tetraflexagons besides the one built from the net in Figure 12.10 whose collection of motions forms a group? We developed most of our tetraflexagon results before looking through McIntosh’s material [7], and you can imagine our surprise when we saw directions for the Purina Tetraflexagon in his notes! McIntosh introduced this creature in a discussion on surgery, and we were no less surprised when we constructed it and confirmed that it worked. The apparent discrepancy between beautiful theory and ugly counterexample was explained
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once we took a scissors to the flexagon and opened it up; in the net, every one of the 24 leaves was a corner square! Clearly, in order to do surgery while maintaining the net assumption, one must be careful that the transplanted leaves do not add any more corner squares. More importantly, this example shows that there are many interesting tetraflexagons that do not satisfy the net assumption. In fact, perhaps our favorite flexagon violates both the net and symmetry assumptions, as it is folded from regular pentagons. Of course, this flexagon cannot be folded flat, but we were impressed that such an object could exist at all. This flexagon, and many others, are described in McIntosh’s notes [7] as well as Pook’s book [14]. We highly recommend these references, as well as an impressive report by Conrad and Hartline [1] (also found in [7]) for the reader interested in deepening his or her background about flexagons and their kin. We have found flexagons to be interesting mathematical objects at many levels. Since they are easily folded, they are a good addition to an introductory mathematics class, giving students an opportunity to look for patterns and explore a topic at their own pace. In a more advanced setting, such as a course in geometry or algebra, flexagons can be used to introduce notions of symmetry and transformation. They are also an excellent topic of study for a senior thesis, as most of the flexagon materials are written at an elementary level and are appropriate for a student’s introduction to the reading of mathematical articles. And, of course, there are many open questions that are easily asked but difficult to answer. Despite their long history, flexagons still give an exciting twist to an otherwise boring strip of paper, and are well worth a little study. Acknowledgments We would like to thank Harold McIntosh for informative correspondence, and Liz McMahon, Kyra Berkove, and the anonymous referees, whose comments greatly improved the exposition of this paper.
Bibliography [1] Anthony Conrad and Daniel Hartline, Flexagons, RIAS Technical Report 6211, Baltimore, MD, 1962. [2] P. B. Chapman, Square flexagons, Math. Gaz. 45 (1961) 192–194. [3] Antonio De Queioroz, The HexaFind Program, www.coe.ufrj.br/˜ acmq/hexaflexagons. [4] Martin Gardner, Hexaflexagons and Other Mathematical Diversions, Univ. of Chicago Press, Chicago, 1988. [5] ——, The 2nd Scientific American Book of Mathematical Puzzles and Diversions, Simon and Schuster, New York, 1961. [6] ——, Mathematical games, tetraflexagons and tetraflexigation, Scientific American, May 1958, 122–126. [7] Harold McIntosh, Flexagons, delta.cs.cinvestav.mx/˜ mcintosh/oldweb/pflexagon.html. [8] Michael Gilpin, Symmetries of the trihexaflexagon, Math. Mag. 49 (1976) 189–192. [9] Peter Hilton, Jean Pedersen, and Hans Walser, The faces of the trihexaflexagon, Math. Mag. 70 (1997) 243–251.
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[10] Joseph Madachy, Mathematics on Vacation, Charles Scribner, New York, 1966. [11] C. O. Oakley and R. J. Wisner, Flexagons, Amer. Math. Monthly 64 (1957) 143–154. [12] Thomas O’Reilly, Classifying and counting hexaflexagrams, J. Recreat. Math. 8 (1975) 182– 187. [13] Jean Pedersen, Sneaking up on a group, Two Year College Math. J. 3 (1972) 9–12. [14] Les Pook, Flexagons Inside Out, Cambridge Univ. Press, Cambridge, 2003. [15] Roger Wheeler, The flexagon family, Math. Gaz. 42 (1958) 1–6.
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13 The Vflex, Triangle Orientation, and Catalan Numbers in Hexaflexagons Ionut E. Iacob, T. Bruce McLean, and Hua Wang Using only the pinch flex, the flex described by Martin Gardner in [4], the six triangles of each hexagonal face of a hexaflexagon stay together. If the faces are colored, the face facing up is always monochrome. To scramble the triangles and mix the colors, you need other flexes. In this paper, we describe the Vflex. With the Vflex, faces become multicolored when flexed. It takes persistence to master, but the Vflex is worth it. A hexahexaflexagon has only 9 mathematical faces under the pinch flex; with the Vflex it has 3,420. We conjecture that many people performed a Vflex accidentally as they read Martin Gardner’s 1956 article and had no expert help. In 1963, Bruce McLean discovered the Vflex and, in 1978, provided a graph of the faces. He presented it to Arthur Stone at an MAA meeting that year and was told “I never allowed my students to do that!” In that paper, McLean approached the Vflex systematically as a new flex. The view that the Vflex is somehow wrong, however, persists to the present. In 2011, a referee of this paper wrote, “I’ve taken the majority view, that it was an illegal move and that my only purpose in understanding it was to fix ‘broken’ flexagons.” The Vflex is now understood as a welldefined operation (see [8] for an overview) that adds considerable complexity to an already interesting object. After the meeting with McLean, Stone expressed pleasure that order still prevailed under the Vflex. We hope the reader will be pleased too.
Vflexing Begin by constructing a hexahexaflexagon (see Figure 12.4 in the previous chapter). For clarity, we prefer to number all triangles in the strip of paper from left to right on the front and back of the paper as in our Figure 13.1, and add arrows pointing to each triangle edge not shared with another triangle. The tab is helpful for construction, but not required. Fold the triangle UP 2 under the 1 and continue to spiral in the same direction to obtain a shorter strip of 9 triangles as in Figure 13.1c. Next, fold triangle UP 5 under the 4, UP 11 under the 10, and UP 17 under the 16. Finish by taping the tab to UP 1. The result is the hexahexaflexagon shown in Figure 13.2a. Note that each triangular region of the hexahexaflexagon, called a pat, has a thickness, i.e., the number of triangles. This number Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 6–10.
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Face UP
1
2
3
17
18
Face DOWN
1
2
17
(a)
18
(b)
4 1
16 5
17 (c)
Figure 13.1. Constructing a hexahexaflexagon.
is the degree of the pat. The strip would continue straight and a hexagon could not be constructed if the degree of any pat was divisible by three, hence pat degrees are always congruent to 1 or 2 (mod 3). Pat 1
UP Pat 2
1u 4u
16u
1
Pat 6 4 3
10u
13u
16
14 13
6 7
Pat 5
Pat 4 (a)
(b) Figure 13.2. Starting the Vflex.
For the Vflex, separate Pat 2 with your left thumb into two halves (see Figure 13.2a). It may help to fold the whole flexagon in half, so that 7, 10 and 13 are on one side. You may also need your right thumb to assist in separating Pat 5. The flexagon should begin to open (Figure 13.2b) and eventually look like Figure 13.3a (as seen from the top). Note that Figure 13.3a is threedimensional: an inverted pyramid with a square base and tetrahedra Pat 1 3 Pat 2 7
6
14
16
Pat 6
15 18
14
3
11
13 Pat 3
6
Pat 5
Pat 4 (a)
(b) Figure 13.3. Completing the Vflex.
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on two opposite faces. Pinch 13 with your right thumb on top and pull it towards you while pushing 6 away from you. Rotating 13 under 11 completes the flex (see Figure 13.3b). (An online video may be helpful; see [9].) Note that a Vflex is not possible if either Pat 2, 5, or 6 contains only 1 triangle. To return to the original, flip the flexagon on the horizontal axis, rotate 60 degrees clockwise, and Vflex again. Enjoy!
Triangle orientation Many flexagon lovers wrote Martin Gardner [5] to point out that numbers alone are not enough to describe a flexagon configuration completely. A triangle can come up face up or face down and with different orientations, that is, the arrow mentioned in the previous section may point in different directions within the face. We define the state of a triangle as the following three pieces of information: (i) face position: up (U) or down (D); (ii) triangle number: even (E) or odd (O); and (iii) triangle orientation (the arrow’s direction): outside the hexagon (O), towards the next pat in the face counterclockwise (N), or towards the previous pat in the face (P). There are thus 12 distinct states. For instance, the state UOO means that the triangle is facing up, is oddnumbered, and the arrow points outwards in the hexaflexagon; the state DOP means a face down triangle, oddnumbered, and with the arrow pointing clockwise. Simple observations lead to the fact that from one triangle to the next, in increasing numerical order, parity changes (evidently), but facing up or down changes only if both triangles are in the same pat. We say that a transition from one triangle to the next one is a folding transition (FT), if both triangles are in the same pat; or a pat transition (PT), if they are in consecutive pats. Then, it is easy to observe that, for instance, from a DON triangle we can only have a folding transition (FT) to the next triangle in the sequence, which therefore must be UEO (U because it is in the same pat; E because it comes after an odd numbered triangle; and O as one can determine by looking at the original strip of paper). We summarize these observations with the labeled, directed graph in Figure 13.4, where each edge is either a FT (folding transition) or a PT (pat transition). The diagram shows clearly which states allow a FT or PT, or both, and that the number of triangles folded in consecutive pats forms a sequence consisting of numbers congruent alternately to 1 and 2 (mod 3). Given any snapshot of triangle numbers in a pat, we can easily determine the complete state (with orientation) of every triangle. For instance, consider the pat consisting of triangles 2, 1, 4, 3 (from top to bottom) in a hexahexaflexagon. Triangle 1 must be at a tip of a PT edge (hence in either DOO or UON states) and must begin a path of three FT edges leading to a state that allows the next PT (since the pat contains 4 triangles). This puts triangle 1 in DOO state. Then, as we move to triangle 2, we must follow a FT edge (it is the same pat), which places triangle 2 in UEP state. We continue with a FT to 3 in DON, and another FT to 4 in UEO.
Catalan numbers in flexagons The enumeration of pat classes (under a natural equivalence relation according to how the degree splits, see [1]) enjoys a beautiful connection to the wellknown Catalan numbers,
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FT DON
PT
UOP FT
FT DOP
PT
FT
DEO
FT
UON
FT
FT
DEN
FT
UOO
FT
PT
UEP
DEP
PT
FT
FT
UEN FT
DOO
Figure 13.4. Triangle state diagram.
which we proceed to describe. A pat of degree n; n > 1, splits naturally where a “thumbhole” is formed, the one place in the physical pat where the thumb can be inserted without encountering a pocket [4]. This divides the pat into two parts whose degrees sum to n. Thus each pat class can be represented as a partition of n into an ordered sum of two positive integers representing the degrees of its two principle parts. Like n, these integers are equivalent either to 1 or to 2 (mod 3). Then each part in this partition is in turn a sum of two such parts and so on, until every part equals 1. For example, a pat class of degree 10 may be represented as follows: 10 D 2 C 8
D .1 C 1/ C .1 C 7/
D .1 C 1/ C .1 C .5 C 2//
D .1 C 1/ C .1 C ..1 C 4/ C .1 C 1///
D .1 C 1/ C .1 C ..1 C .2 C 2// C .1 C 1///
D .1 C 1/ C .1 C ..1 C ..1 C 1/ C .1 C 1/// C .1 C 1///: From a combinatorial point of view, such a representation assigns pairs of parenthesis to the sum of n 1’s keeping the sum within each pair of parenthesis either 1 or 2 modulo 3. The number of pat classes is the number of ways to complete this process. The well known Catalan numbers (which Martin Gardner also popularized [3]) count, among other objects (including planar trees, lattice paths, triangulations of a polygon), exactly this quantity without the (mod 3) constraint. The number of pat classes can therefore be analyzed using the same combinatorial tools used for the Catalan numbers (see [2]). In the recent paper of Anderson et al. [1], a pair of recursive relations for the number of pat classes with a given degree is given, namely bn D
n X i D0
ai an
i
and an D
n X1
bi bn
i 1;
i D0
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where an is the number of pat classes of degree 3n C 1 and bn is the number of pat classes of degree 3n C 2. The reader may have already noticed that when partitioning a number of the form 3n C 1, the two parts must be of the form 3n C 2, and vice versa. This is the reason for this bivariable recursion. The enumeration of various objects related to the flexagons that Martin Gardner introduced to a wide audience in 1952 is now an interesting area of research. Both sequences fan g and fbn g are in the OnLine Encyclopedia [11]. Among other interesting concepts associated with an are the number of 4ary (or quartic) trees with 3n C 1 leaves. It turns out that ! ! 4n 2 4n C 1 1 and bn D : an D 3n C 1 n 3n C 2 n (For more on the combinatorics of flexagons, see [1] and [4].)
Bibliography [1] T. Anderson, T. B. McLean, H. Pajoohesh, C. Smith, The combinatorics of all regular flexagons, European J. Combin. 31 (2010) 72–80; available at dx.doi.org/10.1016/ j.ejc.2009.01.005. [2] R. Brualdi, Introductory Combinatorics, 5th ed., Pearson, 2008. [3] M. Gardner, Catalan Numbers, Time Travel and Other Mathematical Bewilderments, W. H. Freeman, 1987. [4] ——, Hexaflexagons, College Math. J., 43 (2011) 2–5; available at dx.doi.org/ 10.4169/college.math.j.43.1.002. [5] ——, Hexaflexagons and Other Mathematical Diversions, Cambridge Univ. Press, 2008. [6] I. E. Iacob, T. B. McLean, and H. Wang, About general order regular flexagons (submitted). [7] T. B. McLean, VFlexing the hexahexaflexagon, Amer. Math. Monthly 86 (1979) 457–466; available at dx.doi.org/10.2307/2320415. [8] ——, Flexagons; available at math.georgiasouthern.edu/˜ bmclean/flex/index.html. [9] R. Moseley, Flexagon.net; available at www.flexagon.net/index.php?option=com_content&task=view&id=22. [10] C. O. Oakley and R. J. Wisner, Flexagons, Amer. Math. Monthly 64 (1957) 143–154; available at dx.doi.org/10.2307/2310544. [11] N. Sloane, The OnLine Encyclopedia of Integer Sequences; available at oeis.org/.
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14 From Hexaflexagons to Edge Flexagons to Point Flexagons Les Pook Edge flexagons Hexaflexagons, the subject of Martin Gardner’s column [5], were the first family of flexagons to be discovered. They are, however, only one example of an edge flexagon, a folded band of hinged, usually identical, convex polygons (called leaves). The strip of polygons used to construct a flexagon is called a net. In a main position an edge flexagon has the appearance of a ring of hinged polygons (see Figure 14.1). The rings are not necessarily flat and what look like single polygons are folded piles of one or more leaves, called pats. By definition, edge flexagons can be flexed to other main positions to display other pairs of faces, usually in a cyclic order. When flexing, leafbending is sometimes allowed, but in main positions the leaves are flat. The preceding chapter in this part concerns hexaflexagons (Figure 14.1a). Here we describe some nonhexagonal forms and say something about their classification.
(a) Trihexaflexagon.
(b) Square flexagon.
(c) Pentagonal flexagon.
Figure 14.1. Main positions of some edge flexagons.
Workable paper models of flexagons are easy to make; however, the mathematics is complex and just how a flexagon works is not immediately obvious from casual examination of a paper model. Even square flexagons (Figure 14.1b), the second family of flexagons to be discovered and the subject of Chapter 12, are far from simple. Martin Gardner wrote [6], “Stone and his friends spent considerable time folding and analyzing these foursided Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 11–14.
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forms, but did not succeed in developing a comprehensive theory that would cover all their discordant variations,” a significant statement as Stone’s friends included John Tukey and Richard Feynman.
Classification of edge flexagons This is a difficult subject, as Martin Gardner’s remark indicates. Certain edge flexagons, however, can be classified on the basis of their characteristic polygon, which describes how the hinge edges are linked by the leaves and is related to the vertex figures used to characterize polyhedra [1]. In particular, by describing how the leaves are linked, the characteristic polygon contains most of the information needed to construct the flexagon. For those edge flexagons that have a characteristic polygon, classification amounts to classification of all possible types of characteristic polygons. We give two examples (in Figures 14.2 and 14.3) of edge flexagons and their characteristic polygons. To make paper models of any of the flexagons described here, use heavy paper stock and enlarge the nets to about twice the printed size. Transfer the numbers in parentheses from the upper side to the lower side and delete the numbers from the upper side. In Figure 14.2a, we have the net for a square edge flexagon. To assemble this model, fold together pairs of leaves numbered 3 and 4, and join the ends (dashed lines). As assembled the flexagon is in main position (Figure 14.1b). This is flat and consists of alternate pats of single leaves and fan folded piles of three leaves. The flexagon can be flexed around a 4cycle by folding it in two (to an intermediate position) and then unfolding into a new main position, and so on. The operation of this flexagon is similar to the traditional Chinese falling block toy sometimes called Jacob’s Ladder [8]. d
c
a
b
3(2) c
b
d
1(4) a
1(2)
d
b
b
c
d
1(4) a
1(2)
b
a
3(2)
c
3(4) a
d
c
d
3(4) a
b
c
b
d
a
c
c
(a) Net.
(b) Characteristic polygon.
Figure 14.2. A square edge flexagon.
The characteristic polygon is obtained by joining the midpoints of the hinge edges on each leaf in the order they are linked by the leaves. This results in a twicetraced square (Figure 14.2b). Incidentally, joining endtoend two copies of the net shown in Figure 14.2a results in the net for another square edge flexagon, the octopus flexagon. Main positions are not flat, but it can still be flexed around a 4cycle using a pinch flex similar to the flex used for the trihexaflexagon. Other flexes are possible, and its overall behaviour is bewilderingly complex [6]. There are two regular pentagons: the usual convex one and the pentagram. This gives rise to two series of pentagonal flexagons, one of whose net is in Figure 14.3a. To assemble,
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c b
d
d
d
3(2)
e 5(4)
c
1(2) e
d
a
a
c
e a
3(4) a
111
d
b
d
5(1) b
b
c
b
e
a
(b) Characteristic polygon.
(a) Half the net. Figure 14.3. A pentagonal edge flexagon.
first make two copies of Figure 14.3a, turn one over and join them endtoend. Next, fold together pairs of leaves numbered 2, 3, 4 and 5, and join the ends. Assembled, the flexagon is in an intermediate position which looks like a pair of pentagons with a common edge. From here, it can be opened into a type of main position (Figure 14.1c), and then closed into another main position in one continuous aesthetically satisfying movement, and so on around a 5cycle. It can also be opened into another type of main position, but flexing round a 5cycle is then complicated [5]. This is a difficult flexagon to handle.
Point flexagons Point flexagons are the duals of edge flexagons. They are a recent discovery, first described by Scott Sherman in 2007 [5]. Nets are derived by truncating the leaves of the net of an edge flexagon at the vertices, so that vertices become edges and edges become vertices (see Figure 14.4). Edge hinges, therefore, become point hinges, usually modelled by short paper strips. By definition, a point hinge has only two degrees of freedom. The triangles in Figure 14.4 may rotate relative to each other about the yaxis, and the zaxis, but not be twisted relative to the xaxis. y Edge hinge
Point hinge
x
Figure 14.4. Truncating the edgehinged triangles (dashed) creates pointhinged triangles (solid).
Figure 14.5a gives a net for a point flexagon with three faces. To assemble, crease the lines between the leaves to form hinges. Fold pairs of leaves numbered 2 and 3 together and glue the two part paper strips together at AA. Assembled, the visible leaves, above and below, will both be numbered 1. The flexagon can be flexed around a 3cycle by turning over the top leaf so that it becomes the bottom leaf, and so on, rather like the way cards are flipped in a Rolodex rotary file. This is analogous to the 3cycle of the trihexaflexagon described by Martin Gardner [5]. The flex is not completely smooth since the hinge folds must be reversed and they are not parallel to the axis of rotation.
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3(2) A
A
1(2)
3(1)
(a) Net.
(b) Assembled.
Figure 14.5. A triangle point flexagon (dual to the trihexaflexagon).
Conclusion Martin Gardner opened a Pandora’s box when he introduced flexagons to a wide audience. Fifty years later, they are a mature topic with an extensive literature. They continue to fascinate with new results and new flexagons still being discovered and still to be discovered. Acknowledgment Thanks to the anonymous reviewers for helpful comments. The figures are copyrighted by the author.
Bibliography [1] H. S. M. Coxeter, Regular Polytopes, 2nd ed., The Macmillan Company, New York, 1963. [2] M. Gardner. Hexaflexagons, College Math. J. 43 (2012) 2–5; available at dx.doi.org/10.4169/college.math.j.43.1.002. [3] ——, Mathematical games: tetraflexagons and tetraflexigation, Scientific American, May 1958, 122–126; available at dx.doi.org/10.1038/scientificamerican0558122. [4] E. Iacob, T. B. McLean, and H. Wang, The Vflex, College Math. J. 43 (2012) 6–10; Chapter 13 in this volume and available at dx.doi.org/10.4169/college.math.j.43.1.006. [5] L. P. Pook. Serious Fun with Flexagons. A Compendium and Guide. Springer, Dordrecht, 2009. [6] ——, Flexagons; available at www.pook.org.uk; accessed 11 April 2011. [7] S. Sherman. Point Flexagons; available at loki3.com/flex/pointflexagon.html, Accessed 17 April 2007. [8] Wikipedia, Jacob’s Ladder, Wikipedia: the free encyclopedia; available at en.wikipedia.org/wiki/Jacob%27s_ladder_%28toy%29.
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15 Flexagons Lead to a Catalan Number Identity David Callan Introduction Flexagons are remarkable objects obtained by suitably folding and glueing a strip of paper. Several websites give detailed instructions on how to construct various kinds. After construction, the flexagon lies flat with a number of faces, delineated by creases in the paper, visible on top and bottom. The name derives from the fact that they can be pinched, or flexed, in various ways into a 3dimensional shape and then flattened again to reveal different faces. First discovered by Arthur Stone at Princeton in 1939, they were popularized by Martin Gardner [2] in his 1956 debut column in Scientific American and continue to attract mathematical interest [1]. At about the same time as Gardner’s column appeared, Oakley and Wisner [4] considered the variety now called hexahexaflexagons—hexagonal shape when flat, with 6 triangular faces visible. They showed that these flexagons can be constructed recursively using a critical fold that splits a flexagon into two smaller ones. In particular, they are equinu2n 1 merous with Dyck paths, counted by the Catalan numbers Cn D nC1 n [5]. A Dyck path of size n is a lattice path of n upsteps .1; 1/ and n downsteps .1; 1/ that starts at the origin and never goes below ground level. The first downstep that returns the path to the xaxis splits a Dyck path into two smaller ones, in complete analogy to the decomposition of flexagons into smaller ones. Oakley and Wisner used this decomposition to show that hexahexaflexagons can be encoded by permutations called pats. Pats themselves are defined recursively, with permutations represented as lists. A singleton permutation is a pat, and a permutation p of length n 2 is a pat if and only if (i) there is a unique split point that divides p into subpermutations p1 and p2 such that all entries in p1 are greater than all entries in p2 , and (ii) the reverse of each of p1 and p2 is a pat. The only pat on [[2]] is 21; the pats on [[3]] are 231 and 312, and on [[4]] are 2431, 3241, 3412, 4132, 4213. The number of pats on ŒŒn C 1 is Cn . The purpose of this note is to find the distribution of the statistic “number of descents” on pats (a descent in a permutation p is a pair of adjacent entries .pi ; pi C1/ with pi > pi C1 ). To do so, we establish a recurrence that leads to a quartic equation for the generating Reprinted from The American Mathematical Monthly, Vol. 119, No. 5 (May 2012), pp. 415–419.
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function. Lagrange inversion then yields a simple closed form for the number of pats on ŒŒn with k descents, thereby giving a combinatorial interpretation of the identity ! ! n X 1 2n 2k C 1 2k D Cn : (15.1) 2n 2k C 1 k n k kD0
The last section exhibits a bijection from pats to full binary trees showing that descents on pats are distributed as evenlevel vertices on binary trees and concludes with a generalization of (15.1).
Lagrange inversion and the distribution of descents on pats From its definition, a pat p of length n determines two pats p1 ; p2 of lengths i and n i respectively for some i 2 Œ1; n 1. The connecting entries from p1 to p2 contribute a descent to p and the remaining descents of p correspond to the ascents in p1 and p2 . Hence we obtain the recurrence u.n; k/ D
n i 1 X1 X
u.i; j /u.n
i; n
j
k
1/
i D1 j D0
for n 2 and k 0
for the number u.n; k/ of pats on Œn with k descents, with u.1; 0/ D 1. The recurrence leads directly to a functional equation for the generating function F .x; y/ WD P n k n1; k0 u.n; k/x y , F .x; y/ D x C
1 F y
1 2 xy; : y
(15.2)
Iterating (15.2) once yields the algebraic equation F D x C y x C F2
2
;
(15.3)
where F .x; y/ is now abbreviated to F . The Lagrange inversion formula is a powerful technique for solving functional equations such as (15.3). In its basic form, it says that for a function f .x/ analytic about the origin and nonzero at the origin, the coefficient of x n in the (compositional) inverse f ı of f is given by Œx n f ı.x/ D n1 Œx n 1 g.x/n where g is defined by f .x/ D x=g.x/. The standard analytic proof (see [6, Chapter 5.1], for example) is based on the special property of the case n D 1 in the contour integral of z n around the origin; it is the only value of n that yields a nonzero integral. This is a little mysterious, but in the context of formal power series, there is a transparent combinatorial proof involving ordered (plane) trees with weighted edges (see [5, Theorem 5.4.2], for example). To apply the Lagrange inversion formula, one typically formulates an equation of the form y D x=g.x/ and wishes to solve for x in terms of y. With f .x/ WD x=g.x/, we then have x.y/ D f ı.y/ and so the formula applies (with a little care not to get x and y mixed up). Direct application of the formula to solve (15.3) is cumbersome but a trick shown to me by Ira Gessel (see [3]) rapidly solves it. Introduce a new variable z and consider
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F D F .x; y; z/ defined by 2 F D z x C y x C F2 :
(15.4)
Equation (15.4) has the form z D F=.F /where
.F / D x C y x C F 2 and so Lagrange inversion says that
2
;
1 h k 2j i y F .F /2j C1 2j C 1 ! ! 1 2j C 1 2k j CkC1 D x : 2j C 1 k j
i h y k z 2j C1 F .x; y; z/ D
The coefficient of x n y k in F .x; y; 1/ is obtained by setting j D n yielding
k
(15.5) 1 in (15.5),
u.n; k/ D Œx n y k F .x; y/
D Œx n y k F .x; y; 1/ D
2n
2n
1 2k
1
2k k
1
!
! 2k ; n k 1
and hence identity (15.1) after replacing n by n C 1. Here are the first few values of u.n; k/. Table 15.1. Values of u.n; k/, the distribution of descents on pats nnk
0
1
2
3
4
5
6
1
1
0
0
0
0
0
0
2
0
1
0
0
0
0
3
0
2
0
0
0
0
0
4
0
1
4
0
0
0
0
5
0
0
12
2
0
0
0
6
0
0
12
30
0
0
0
7
0
0
4
100
28
0
0
8
0
0
0
140
280
9
0
9
0
0
0
90
980
360
0
10
0
0
0
22
1680
2940
220
Pats as trees A pat on ŒŒn C 1 can be represented, using its successive split points, by a vertexlabeled full binary tree on 2n edges as illustrated in Figure 15.1.
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Part III. Flexagons and Catalan Numbers
5
7
6
98 56
798 4
2
1 3
79856
21
479856
213
479856213 Figure 15.1. Full binary tree for pat 4 7 9 8 5 6 2 1 3, interior vertices at even level are enlarged.
In fact, the labels are unnecessary; they can be uniquely recovered from the underlying tree. So we have a bijection from pats on ŒŒn C 1 to full binary trees on 2n edges. Under this bijection, a descent in the pat shows up as an interior vertex at even level in the tree (where rain water would collect between the two leaves corresponding to the descent). Pruning the leaf edges in a full binary tree to obtain a binary tree, we obtain the following. Theorem. Descents on pats are distributed as evenlevel vertices in binary trees. Similar considerations for ternary and higher order trees yield a generalization of (15.1), ! ! ! n X rn rk C 1 rk 1 rn C 1 1 D rn rk C 1 k n k rn C 1 n kD0
for r 2. Acknowledgments I thank Ira Gessel for suggesting the Lagrange inversion technique used in the second section.
Bibliography [1] T. Anderson, T. B. McLean, H. Pajoohesh, C. Smith, The combinatorics of all regular flexagons, European J. Combin. 31 (2010) 72–80; available at dx.doi.org/10.1016/ j.ejc.2009.01.005. [2] M. Gardner, Hexaflexagons and Other Mathematical Diversions: The First Scientific American Book of Puzzles and Games, Univ. of Chicago Press, 1988. [3] I. Gessel, A combinatorial proof of the multivariable Lagrange inversion formula, J. Combin. Theory Ser. A 45 (1987) 178–195; available at dx.doi.org/10.1016/00973165(87)900136. [4] C. O. Oakley, R. J. Wisner, Flexagons, Amer. Math. Monthly 64 (1957) 143–154; available at dx.doi.org/10.2307/2310544.
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[5] R. P. Stanley, Enumerative Combinatorics, Cambridge Studies in Advanced Mathematics, Vol. 62, Cambridge Univ. Press, Cambridge, 1999. [6] H. S. Wilf, Generating Functionology, Academic Press, New York, 1990; also available at www.math.upenn.edu/˜ wilf/DownldGF.html
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16 Convergence of a Catalan Series Thomas Koshy and Zhenguang Gao The well known Catalan numbers Cn are named after Belgian mathematician Eugene Charles Catalan (1814–1894), who found them in his investigation of wellformed sequences of left and right parentheses. As Martin Gardner (1914–2010) wrote in Scientific American [2], they have the propensity to “pop up in numerous and quite unexpected places.” They occur, for example, in the study of triangulations of convex polygons, planted trivalent binary trees, and the moves of a rook on a chessboard [1, 2, 3, 4, 6]. 2n 1 The Catalan numbers Cn are often defined by the explicit formula Cn D nC1 , n 2n where n 0 [1, 4, 6]. Since .nC1/ j n , it follows that every Catalan number is a positive integer. The first five Catalan numbers are 1, 1, 2, 5, and 14. Catalan numbers can also be C C , where C0 D 1. So lim CnC1 D 4. defined by the recurrence relation CnC1 D 4nC2 nC2 n n n!1 P1 1 Here we study the convergence of the series nD0 Cn and evaluate the sum. Since P C xn lim nC1 D 4, the ratio test implies that the series 1 nD0 Cn converges for jxj < 4. Conn!1 Cn P 1 sequently, the series 1 nD0 Cn converges. We evaluate this infinite sum using generating functions, plus fundamental tools from the differential and integral calculus.
Sum of the series
1 P nD0
1 Cn
To this end, let f .x/ be the generating function of the reciprocals of Catalan numbers, P xn f .x/ D 1 nD0 Cn . We compute the sum in three steps. First, we find an ordinary differential equation satisfied by f .x/, then, after solving the differential equation in the interval .0; 4/, we compute f .1/. We first rewrite f .x/ as 1 X xn f .x/ D 1 C : C nD1 n So f 0 .x/ D
1 X nx n Cn nD1
1
D
1 X nC1 n x : C nD0 nC1
Reprinted from The College Mathematics Journal, Vol. 43, No. 2 (March 2012), pp. 141–146.
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Part III. Flexagons and Catalan Numbers nC2 Cn
D
4nC2 , CnC1
by the recurrence relation, this yields 1 1 X n C 2 n X 4n C 2 n x D x ; Cn CnC1 nD0 nD0
1 1 1 X X X n n xn 4.n C 1/ n x C2 D x C C CnC1 n n nD0 nD0 nD0
2 Œf .x/ x
xf 0 .x/ C 2f .x/ D 4f 0 .x/ x.x
2
1 X xn ; C nD0 nC1
1;
4/f 0 .x/ C .2x C 2/f .x/ D 2:
(16.1)
This is a firstorder differential equation for f .x/ with the initial conditions f .0/ D 1 D f 0 .0/. ˇ3=2 ˇ To facilitate solving (16.1) for x 6D 0, we introduce the function g.x/ D ˇ 4 x x ˇ . 0 .x/ Then gg.x/ D x.4 6x/ . This implies that Œx.x
4/g.x/ 0 D .2x C 2/g.x/:
(16.2)
Multiplying (16.1) by g.x/, we get x.x
4/f 0 .x/g.x/ C .2x C 2/f .x/g.x/ D 2g.x/:
Using (16.2), this can be rewritten as Œx.x
4/f .x/g.x/ 0 D 2g.x/:
But, again using (16.2), fx.x
1g.x/g0 D Œx.x
4/Œf .x/
4/f .x/g.x/ 0
D 2g.x/
Œx.x
4/g.x/ 0
.2x C 2/g.x/
D 2xg.x/;
consequently, x.x
4/Œf .x/
1g.x/ D 2
Z
xg.x/ dx C C1 R 2 xg.x/ dx C1 f .x/ D 1 C ; x.4 x/g.x/
where C1 is a constant. Suppose 0 < x < 4. Then Z
xg.x/ dx D D
Z
Z
x
.4
4
x x
3=2
x/3=2
x 1=2
dx
dx:
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121
Letting x D u2 , this implies that Z Z xg.x/ dx D 2 .4 u2 /3=2 du
1 u D u.4 u2 /3=2 C 3u.4 u2 /1=2 C 12 arcsin C C2 2 2p p 1p x x.4 x/3=2 C 3 x.4 x/1=2 C 12 arcsin C C2 ; D 2 2
where C2 is another constant. Therefore, we have p p p x x.4 x/3=2 C 6 x.4 x/1=2 C 24 arcsin 2 C 2C2 C1 f .x/ D 1 C 3=2 x.4 x/ 4 x x p p p x.4 x/3=2 C 6x.4 x/1=2 C 24 x arcsin 2x C C x ; D1C .4 x/5=2 where C D 2C2
C1 . Since f .0/ D 1 D f 0 .0/, C D 0. Thus
f .x/ D 1 C
x.4
p p 1 X x/3=2 C 6x.4 x/1=2 C 24 x arcsin 2x xn D : 5=2 Cn .4 x/ nD0
When x D 1, this yields 1 X 1 33=2 C 6 31=2 C 24 arcsin 1=2 D 1C C 35=2 nD0 n p 4 3 : D 2C 27
(16.3)
p
P1
converges to the limit 2 C 4273 , which is approximately P 1 2.80613305077. Note that already 22 nD0 Cn 2:80613305077; so the series converges to the limit remarkably fast. Thus, the series
1 nD0 Cn
Additional consequences of the differential equation Letting x D 1 in (16.1), we get 3f 0 .1/ D 4f .1/
2 p ! 4 3 D4 2C 27 p 16 3 0 f .1/ D 2 C : 81
Since f 0 .x/ D
P1
nC1 n nD0 CnC1 x ,
2
it follows that
p 1 X nC1 16 3 D 2C : C 81 nD0 nC1
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Part III. Flexagons and Catalan Numbers
Since the differential equation is infinitely differentiable, it follows from (16.1) that 4/f 00 .x/ C .4x
x.x
2/f 0 .x/ C 2f .x/ D 0:
This yields 3f 00 .1/ D 2f 0 .1/ C 2f .1/, so f 00 .1/ D f 00 .x/ D this implies that
8 3
C
p 56 3 . 243
(16.4)
Since
1 X .n C 1/.n C 2/ n x ; CnC2 nD0
p 1 X 8 56 3 .n C 1/.n C 2/ D C : CnC2 3 243
nD0
Differentiating (16.4) with respect to x, it follows similarly that f 000 .1/ D 2f 0 .1/, so p 1 X .n C 1/.n C 2/.n C 3/ 32 3 D4C : CnC3 81 nD0 Clearly, this technique can be employed to evaluate further sums of the form 1 X .n C 1/.n C 2/ .n C k/ ; CnCk nD0
where k 1.
Sum of the series
1 P nD0
. 1/n Cn
We now turn to to solving (16.1) for 4 < x < 0. We have Z
xg.x/ dx D D
Z
ˇ ˇ ˇ 4 x ˇ3=2 ˇ ˇ xˇ dx x ˇ Z .4 x/3=2 p dx: x
Letting x D u2 , this gives Z Z .u2 C 4/3=2 xg.x/ dx D . 2u/ du u Z D 2 .u2 C 4/3=2 du " # ˇ ˇ p u.u2 C 4/3=2 3 p 2 ˇ ˇ 2 D2 C u u C 4 C 6 ln ˇu C u C 4ˇ C C3 4 2 ˇ ˇ p p 1 ˇ ˇ D u.u2 C 4/3=2 C 3u u2 C 4 C 12 ln ˇu C u2 C 4ˇ C C3 2 p p p 1p D jxj.4 x/3=2 C 3 jxj.4 x/ C 12 ln jxj C j4 xj C C3 ; 2
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123
where C3 is a constant. Consequently, as before, we have p p p p jxj.4 x/3=2 C 6 jxj.4 x/ C 24 ln jxjCC4 j4 xj f .x/ D 1 C ˇ ˇ3=2 x.4 x/ ˇ 4 x x ˇ p p p p xC 4 x jxj.4 x/3=2 C 6 jxj.4 x/ C 24 jxj ln C4 D1 : .4 x/5=2 where C4 is a nonzero constant. Since f .0/ D 1 D f 0 .0/, C4 D 2. Thus p p p p xC 4 x jxj.4 x/3=2 C 6 jxj.4 x/ C 24 jxj ln 2 f .x/ D 1 : .4 x/5=2 P . x/n Since f . x/ generates the alternating series 1 nD0 Cn for 0 < x < 4, setting x D 1 we get 1 X . 1/n D1 Cn nD0
53=2 C 6 51=2 C 24 ln 55=2 p 14 24 5 D ln ; 25 125
(16.5)
p 1C 5 . 2
where is the wellknown golden ratio Formulas (16.3) and (16.5) can be employed to compute the sums of the subseries P1 P1 1 1 nD0 C2nC1 : nD0 C2n and p p 1 X 1 32 2 3 12 5 D C ln ; C2n 25 27 125 nD0 p p 1 X 1 18 2 3 12 5 D C C ln : C 25 27 125 nD0 2nC1 Acknowledgment The authors would like to thank the editor and reviewers for their unwavering enthusiasm, and suggestions for the improvement of the exposition.
Bibliography [1] D. I. A. Cohen, Basic Techniques of Combinatorial Theory, Wiley, New York, 1978. [2] M. Gardner, Catalan Numbers: An Integer Sequence that Materializes in Unexpected Places, Scientific American, 234 (1976), 120–125; available at 10.1038/scientificamerican0676120. [3] H. W. Gould, Bell and Catalan Numbers: Research Bibliography of Two Special Number Sequences, Revised Edition, Combinatorial Research Institute, Morgantown, West Virginia, 1978. [4] T. Koshy, Catalan Numbers with Applications, Oxford Univ. Press, New York, 2009. [5] ——, Fibonacci and Lucas Numbers with Applications, Wiley, New York, 2001. [6] R. P. Stanley, Enumerative Combinatorics, Vol. 1, Wadsworth & Brooks/Cole, Monterey, California, 1986.
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17 LTromino Tiling of Mutilated Chessboards Martin Gardner Suppose a standard chessboard is ‘mutilated’ by the removal of two diagonally opposite corner cells. Can the remaining 62 squares be tiled with 31 dominos? The answer is ‘no’ because the removed squares are the same color. Say the color is white. The remaining 62 squares will have an excess of two black cells. Each domino covers one black and one white cell. After 30 are placed, two black cells will remain uncovered. They cannot be adjacent, therefore they can’t be covered by a domino. This famous puzzle, solved by a simple parity check, is a simple example of a tiling problem on a mutilated chess board. Less well known is the following related problem. Assume the chessboard is mutilated by having two cells removed of opposite color from anywhere on the board. Can the remaining 62 squares always be tiled by dominoes? The answer is yes, and there is a lovely proof by Ralph Gomory [2].
Figure 17.1. Gomory’s proof.
Imagine heavy lines drawn on the chessboard as shown in Figure 17.1. They outline a closed path along which the squares are like beads of alternating color on a necklace. If any two cells of opposite color are taken from the path, it will cut the path into two openended segments, or one segment if the removed cells are adjacent. Each segment will consist of an even number of cells of alternating colors, therefore it can be tiled with Reprinted from The College Mathematics Journal, Vol. 40, No. 3 (May 2009), pp. 162–168.
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dominoes. Gomory’s clever proof is easily generalized to all square boards with an even number of cells. If, instead of dominos, we tile with Ltrominos, also called bent, or V, or right trominos, then all square boards with a number of cells divisible by 3 can be tiled except for the 3 3 board. We will not be concerned with such ‘whole’ boards, but only with mutilated boards with a number of cells that is a multiple of 3 after a single cell has been removed from any spot on the board. We will call such boards deficient. In other words, a board of side n is deficient if n2 1 is a multiple of 3, i.e., n is not a multiple of 3. The sides of such boards form the sequence 2; 4; 5; 7; 8; 10; 11; 13; 14; : : : ./ We will call these numbers the orders of a board and, from now on, the word tromino will mean an Ltromino exclusively. Our basic question is this: What deficient boards with sides in the sequence ./ can be tiled without gaps or overlaps with Ltrominos after a cell has been taken from anywhere on the board? We will take up these boards roughly in numerical order, culminating with a statement of the complete solution.
Powers of 2 Consider the order2 board first. It obviously is tilable with any cell missing (see Figure 17.2, left). Figure 17.2, right, shows how the order4 can be tiled. The 2 2 square takes care of a missing cell in each of its four corners. The rest of the board is tiled by taking advantage of what Solomon Golomb named a reptile—a tile that can form an enlarged replica of itself. The top left 2 2 square rotates to put its missing cell in four places, and the entire order4 square rotates to carry the missing cell to any of its sixteen places.
Figure 17.2. Orders 2 and 4.
In 1953 Golomb, the “father” of polyominoes (he named them and was the first to study them in depth) discovered a beautiful proof by induction that all boards with sides in the doubling sequence 2; 4; 8; 16; : : : could be tiled with trominoes when any cell is missing. The proof was first published in [3]. It is repeated on pages 27–28 of [4]. Numerous mathematicians have since included the proof in their books, often without credit to Golomb. Roger Nelsen, in [6], gives Golomb’s proof with a wordless single diagram. Golomb’s famous proof starts with the 2 2 case shown on the left of Figure 17.3. This square is placed in the corner of the order4 as shown at the center of Figure 17.3. The 4 4 then goes in the corner of an order8 (shown on the right) and a tromino placed at the corner of the shaded order4. We know the dark square can be tiled with any cell missing, and we know the three unshaded quadrants can be tiled with trominoes because each has a
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129
Figure 17.3. Golomb’s induction proof.
missing corner cell. By rotating the board, a missing cell at any spot in the shaded quadrant can be brought to any spot on the order8 board.
Orders 5 and 7 The order5 board is next, as 5 is the next unsolved number in the sequence ./. It has a neat symmetrical tiling when the center cell is gone, as shown in Figure 17.4, left. I have tiled this board with four 2 3 tiles. Each is tilable with two trominoes in two different ways. Using 2 3 tiles is a valuable device for solving tromino problems. When the missing cell is the one shown black in Figure 17.4, center, the cell above it must be covered by a tromino on either side. In each case, shown here with a tromino above and on the right, this produces two cells (numbered 1 and 2) that cannot be covered with a tile. Indeed, the order5 square can be tiled only when the missing cell is one of the nine shown in black in Figure 17.4 right. As a pleasant exercise, see if you can tile the board when the missing cell is at a corner.
1 2
Figure 17.4. The order5 square.
The order 7 board is more difficult to analyze. I was unable to find a single diagram that would prove this board tilable, but Golomb sent me his unpublished way of proving tilability with the aid of three diagrams. His proof goes like this. Figure 17.5 shows three tilings of the order7 board. In each tiling, the 2 2 square obviously can be tiled with a tromino so that the missing cell is at any of the four corners. By rotating the three patterns, the missing cell can be placed at any spot on the board.
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Figure 17.5. Golomb’s proof that order7 is tilable.
Somewhat more difficult is to find tilings that maximize the number of 2 3 tiles. As a challenge, can the reader find a tiling of the 7 7 board using six 2 3 tiles and 4 trominoes (see Figure 17.6)? The solution is unique except for a single reflection (see page 134).
? Figure 17.6. A challenge to the reader.
Note in Figure 17.5, that in each pattern the number of free trominoes—trominoes not in any 2 3 tile—is always even. This is no coincidence. It led me to the following trivial little law. When a board’s order is even, the number of free trominoes in a tiling pattern is odd, and vice versa. When the board’s number is odd, the number of free trominoes must be even. The parity proof is simple. If a board’s order is even, after a cell is removed there will be .n2 1/=3 trominoes in any tiling, an odd number. Each 2 3 tile contains two trominoes, so the total number of trominoes in 2 3 tiles will be even. Subtracting this number from the odd total of trominoes and you get an odd number of trominoes not in any 2 3 tiles. Suppose the board’s order is odd. After a cell is removed there will remain an even number of cells. Subtracting the even number of trominoes in the 2 3 tiles leaves an even number of trominoes not in a 2 3 tile.
Beyond 7 Golomb’s induction proof can be applied to an infinity of other doubling sequences. In particular, now that we have tiled the 7 7 board, we can tile boards of size n n where n is of the form 2k 7. For example, consider the order14 board. Divide it into quadrants with a shaded order7 board in the top left corner, and attach a tromino to its lower right corner
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as before. Because the 7board is tilable, the proof for order14 follows, and of course leads by induction to proofs for orders 28; 56; 112; : : : : A similar proof for the order10 board can’t be obtained by placing an order5 in the corner because order5 is not tilable, but we can handle it in a slightly different way. Put in the top left corner an order8 which we know is tilable. The remaining area forms a path of width 2 along the bottom and right sides of the large square (see Figure 17.7). By rotations and reflections, each missing cell in the order8 can be transferred to any cell on the board. This leads to proofs for orders 20, 40, 80, and so on. A similar proof for order11 has an order7 square in the corner, and a path of width 4 along bottom and side. It leads by induction to solutions for orders 22; 44; 88; : : : : Clearly this technique provides an infinity of doubling sequences for tilable boards. Simply, put in the top left corner of any board a tilable board with a side equal to or smaller than the larger board. If you can tile the path it leaves at the bottom and side, then the board is tilable.
8x8
Figure 17.7. Proof that order10 is tilable.
Boards with sides that are primes are usually the hardest to tile. Order 17 is solved by a corner square of side 13 and a path of width 4. Order 19 is solved by a corner square of order 14, in turn based on order7, and a path of width 5. (See Figure 17.8.)
The complete result By working with these patterns I came close, but not close enough, to finding an induction proof that all deficient squares are tilable except for order5. A proof was finally obtained by I. Ping Chu and Richard Johnsonbaugh [1]. Chu and Johnsonbaugh not only took care of all deficient squares, but also all deficient rectangles! Their induction proof is too technical to repeat here. To summarize, they showed tilability for all m by n rectangles (including squares when m D n) which have a number of cells that is a multiple of 3 after a cell is removed. Such boards are tilable if and only if all of the following are true: 1. m is equal to or greater than 2. 2. n is equal to or greater than m.
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14
Figure 17.8. Order19 is tilable.
3. if m is 2, n must be 2, 4. m is not 5. A 4 7 rectangle is the smallest deficient rectangle, not a square, that is tilable with Ltrominoes. As another exercise, see how long it takes you to tile it when the missing cell is at a corner, and there are two 2 3 tiles. Christopher Jensen, in an unpublished paper, showed that if two cells are taken from a corner of any board, as shown in Figure 17.9, the board obviously cannot be tiled with trominoes. However, if none of these five cases is allowed, a 3m 1 by 3n C 1 board, with any two cells missing can be tiled if an only if n D 1 or m and n are each equal to or greater than 3.
A final word Kate Jones, who founded and runs Kadon Enterprises, a firm that makes and sells handsome mechanical puzzles, games, and other recreational math items, has on the market a game called Vee21 [5]. The Vee is for Vtrominoes, and 21 for the 21 tromino tiles in the set. The trominoes are brightly colored, and there is an order8 board on which to place them. The basic task is to put a monomino (order1 tile) at any spot on the board, then cover the remaining 63 cells with the trominoes, thus solving an order8 board. A 40page brochure comes with the set. It contains a short article on “The Deficient Checkerboard”
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Figure 17.9. Impossible tiling patterns when two cells are missing at a corner.
by Norton Starr, and pictures of rectangular fields that offer other challenges. Our final tiling (see Figure 17.10) is a beautiful, symmetric tiling of the standard chessboard.
Figure 17.10. An order8 tiling with no 2 3 tiles and 5 reptiles.
Bibliography [1] I. P. Chu and R. Johnsonbaugh, Tiling deficient boards with trominoes, Math. Mag. 59 (1986) 34–40. [2] M. Gardner, The Eight Queens and Other Chessboard Diversions, in The Unexpected Hanging and Other Mathematical Diversions, Univ. of Chicago Press, 1991. [3] S. W. Golomb, Checker boards and polyominoes, Amer. Math. Monthly 61 (1954) 675–682. [4] ——, Polyominoes, Scribner, New York, 1965. [5] K. Jones, Vee21; available at www.gamepuzzles.com/polycub2.htm#V21. [6] R. Nelsen, Proofs Without Words II: More Exercises in Visual Thinking, Mathematical Association of America, Washington, 2000.
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Solution to the puzzle on page 130
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18 Polyomino Dissections Tiina Hohn and Andy Liu The damaged patchwork quilt Those of us fortunate to have been born before Martin Gardner’s retirement can relate to the excitement of eagerly awaiting his monthly column Mathematical Games in the magazine Scientific American. Particularly popular were those containing a selection of short puzzles. Here is one from the very last column of this kind, for April, 1981. The patchwork quilt in Figure 18.1 was originally made up of 108 unit squares. Part of the quilt’s center became worn, making it necessary to remove 8 squares as indicated. Cut the quilt along the lines into just two parts that can be sewn together to make a 1010 quilt.
Figure 18.1.
This beautiful problem is rather old, and had appeared earlier in [1] and [10, p. 64–65]. Gardner’s version was later anthologized in [7, pp. 249–250] and [8, p. 168]. We give the readers a chance to solve this problem on their own now, before we return to it later. The father of the field of geometric dissections is generally acknowledged to be Harry Lindgren who wrote the fundamental treatise [12]. The torch has now been passed on to Greg Frederickson, with three outstanding books so far, [2]–[4]. He was motivated by Lindgren’s book, which might not have been written without the influence of Gardner’s November 1961 column, later anthologized in [5, pp. 43–51]. Martin Gardner devoted another column in July 1977 to geometric dissections, titled “Cutting Shapes into N Congruent Parts.” This is later anthologized in [6, pages 165– Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (January 2012), pp. 88–94.
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181]. About half of the problems share a common feature with the Damaged Patchwork Quilt Problem, in that the figures to be dissected, as well as the pieces resulting from the dissections, are polyominoes, plane figures consisting of unit squares joined edge to edge. They have a long history. Solomon Golomb was the first to popularize them, and Gardner wrote several columns about them. One of Gardner’s passions was to introduce puzzles into the classroom. From a theoretical point of view, polyomino dissections are an excellent topic. They require little background, and provide training in geometric visualization. The most important reason of all is that they are fun. However, there are practical difficulties. How we should we present such problems, say the Damaged Patchwork Quilt Problem? Providing lots of scissors and precut rectangle with holes leaves a lot to be desired. Nobody would look forward to the preparation, and mixing kids and scissors is never a good idea. Besides, there will eventually be a ton of litter on the floor. Practical objections aside, there is a more important shortcoming of this model. It does not help anyone find a solution to the problem! Cutting at random will not work. In fact, one has to have the solution already in mind before any meaningful cutting can be done.
Preliminary work How do we solve the Damaged Patchwork Quilt problem? Since it appears quite difficult, we “downsize” it. This also yields the benefit of providing subsidiary problems at lower levels. We generalize the 9 12 rectangle with a 1 8 hole in the middle to a .2n 1/ .2n C 2/ rectangle with a 1 .2n 2/ hole, and the 10 10 square to an 2n 2n square. The original problem is the case n D 5. For n D 1, we have a 1 4 rectangle with a 1 0 hole in the middle and a 2 2 square. This is going too far down. The case n D 2 turns out to be quite easy. Since the top row of the rectangle has length 6 while the target square has side 4, we must have a cut as indicated in Figure 18.2. Another cut in the symmetric position yields the desired two parts.

Figure 18.2.
In the case n D 3, we also have to make a cut along the top edge of the 5 8 rectangle at the point 6 units from a corner. However, if we cut through to the hole as we did in the case n D 2, it will not work. After a bit of fiddling around, we may arrive at the solution in Figure 18.3. From these preliminary investigations, it is not unreasonable to assume that the two parts in the solution are congruent to each other and rotationally symmetric. However, we are certainly not confident at this point of solving the original problem.
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Figure 18.3.
Eureka! After much deep thought, we finally came up with a fantastic model. Not only does it solve the practical problems mentioned before, it actually leads us step by step to a solution. We illustrate with the case n D 4. (In [8], there is an erroneous statement that there are no twopiece solutions to this case. That statement is removed in [7].) We draw two playing boards, one the initial 7 10 rectangle with a 1 6 hole in the middle, and the other the target 8 8 square, and provide ourselves with a large supply of bingo chips of two different colors, say yellow and blue. The objective is to fill both boards with chips in such a way that those of the same color form the same shape on both boards. This solves the problem. We begin with a yellow chip at the top left corner of the square board and a blue chip at the bottom right corner. This signifies that these two squares do not belong to the same part, a most reasonable assumption. We define these two corners as the principal corners of the respective parts. We mark them in the same orientation since we anticipate that the two parts are in rotational symmetry. See Figure 18.4a. e ?
u6
?b e e e e e e e e e e e e e
(a) b b b b b b b u
b ee ee b b b b b b u uu uu
u u u u u u u
u u u u u u r
6
(b) e r r r r r r r
e r r r r r r r
b b b b b b b
b b b b b b b r b ee e e r b r b r b r b u uu ur b r r r r r r r
(c)
r r r r r r r
(d) Figure 18.4.
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We now move over to the punctured rectangular board and mark the principal corners. Besides playing the chips at the corners, we observe that in the square board, the yellow chips cannot extend beyond the eighth column from the left while the blue chips cannot extend beyond the eight column from the right. This allows us to place the additional chips on the rectangular board. See Figure 18.4b. Moving back to the square board, we can fill the top row with yellow chips and the bottom row with blue ones, because here, the yellow chips cannot extend beyond the seventh row from the top and the blue chips cannot extend beyond the seventh row from the bottom. See Figure 18.4c. Moving over to the rectangular board, we notice that the two blue chips on the seventh row from the bottom cannot extend at all to the left. This means that we must have six yellow chips on the second row from the top, going from left to right. Similarly, there must be six blue chips on the second row from the bottom, going from right to left. See Figure 18.4d. Continuing with the same strategy, we can complete the solution of the problem as in Figure 18.5. b b b b b b b r
b b b b b b b r
b b b b b b b b b b r r uu r r r r r r ee r r r r r r r r r r r r r r
b b b b b b b
b b b b b b b
(a)
b b b b b b r r b b b b r r r r ee r r r r u ur r b b r r r r r r r r r r r r r r (b)
uu uu ee ee
(c)
(d) Figure 18.5.
Onward This approach may be used to solve related problems. In the above problem, the two parts we end up with are congruent to each other. Of course, they also have to fit together to form a square. Since the punctured rectangle has rotational symmetry, there are many trivial divisions into two congruent parts. However, if a figure does not have any obvious symmetry, the task may be quite challenging. Here is an example from [11, p. 37]. There are two natural choices for the pair of principal corners. Since the figure does not have rotational symmetry, we choose opposite orientations for the corners. In the first attempt as shown in Figure 18.7a, the boundary squares are easily filled out. We observe
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Figure 18.6. Dissect into two congruent pieces.
that in the third row from the bottom, we can have at most two blue chips going from right to left. This means that in the third column from the left, we must have at least four blue chips going from bottom to top. This in turn means that in the second row from the bottom, we must have at least four yellow chips going from left to right. However, this is impossible as the blue chips just placed are in the way. The other choice leads easily to the solution shown in Figure 18.7b. b b b b b b ? b b u b u b u b r r r r r
r r r r r 6 6 r 
(a)
?
(b) Figure 18.7.
Many polyomino dissection problems have appeared in recent years, no doubt influenced by Gardnerâ€™s columns. (Serhiy Grabarchuk dedicated his book [9] to Martin Gardner.) We offer the readers a choice selection here. 1. Solve the â€œDamaged Patchwork Quiltâ€? problem. 2. Dissect each of the following figures into two congruent pieces. The first is taken from [13] and the last two from [6].
3. The patchwork quilt in the diagram below was originally made up of 35 unit squares. Part of the quiltâ€™s became worn, making it necessary to remove 5 of the squares as indicated, Cut the quilt along the lines into just two parts that can be sewn together to make a 5 6 quilt. This is taken from [13].
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4. Dissect each of the following figures into two congruent pieces. The first two are taken from [6] and the last four from [11]. The last one is a composition of Solomon Golomb.
5. Dissect each of the following figures into three congruent pieces. The first one is taken from [13] and the second from [9].
6. Dissect each of the following figures into two parts that can be reassembled to form an 8 8 square. The first one is taken from [13] and the second one from [9].
A final word Any trick used twice becomes a technique. The technique discovered by us and described here is good enough at solving certain polyomino dissection problems that it can be presented to school children. On October 21, 2010, which would have been Martin Gardner’s 96th birthday, The Damaged Patchwork Quilt was one of Gardner’s problems presented at a Math Fair by students of Stratford Junior High School at the Telus World of Science Edmonton. We hope this was something that would please Martin.
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Bibliography [1] H. Dudeney, Puzzles And Curious Problems, Thomas Nelson & Sons, 1931. [2] G. Frederickson, Dissections: Plane & Fancy, Cambridge Univ. Press, Cambridge, 1997. [3] ——, Hinged Dissections: Swinging & Twisting, Cambridge Univ. Press, Cambridge, 2002. [4] ——, PianoHinged Dissections: Time To Fold!, A K Peters, Natick MA, 2006. [5] M. Gardner, The Unexpected Hanging, Univ. of Chicago Press, Chicago, 1991. [6] ——, Penrose Tiles To Trapdoor Ciphers, Mathematical Association of America, Washington, 1997. [7] ——, The Last Recreations, SpringerVerlag, New York, 1997. [8] ——, The Colossal Book Of Short Puzzles and Problems, ed. Dana Richards, W. W. Norton & Co., New York, 2006. [9] S. Grabarchuk, The New Puzzle Classics, Sterling Publishing Co., New York, 2005. [10] B. Kordemsky, The Moscow Puzzles, Dover Publications Inc., New York, 1987. [11] R. Kurchan, Mesmerizing Math Puzzles, Sterling Publishing Company, New York, 2001. [12] H. Lindgren, Geometric Dissections. Dover Publications, New York, 1972. [13] N. Yoshigahara, Geometric dissections, Puzzle World 1 (1992) 48–51.
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19 Squaring the Plane Frederick V. Henle and James M. Henle Introduction This research was inspired by two lovely pieces of mathematics. The first is the discovery by William T. Tutte, A. H. Stone, R. L. Brooks, and C. A. B. Smith of squares with integral sides that can be tiled by smaller squares with integral sides, no two alike. Tutte tells the story in “Squaring the Square,” a beautifully written article that conveys vividly the excitement of mathematical research [9]. It became widelyread in 1958 when it was reprinted in Martin Gardner’s Mathematical Games column in Scientific American. It undoubtedly played a role in inspiring many to become mathematicians. The second piece is the wellknown tiling of the plane by squares whose sides are the Fibonacci numbers (Figure 19.2). The tiling is elegant, but in light of Tutte’s work, possesses a minor flaw—it contains two squares of the same side. The flaw is easily remedied. We can use the squared square in Figure 19.1, together with a square of side 175 (Figure 19.3). This works, but some of the elegance is lost.
39 55 16
9
81
14 20
18 56 38
30 51
31
29 8
64 33
35
43
Figure 19.1. 175 175 perfect square. Reprinted from The American Mathematical Monthly, Vol. 115, No. 1 (Jan. 2008), pp. 3–12. tiling appears in the same volume of Gardner’s columns as “Squaring the Square” [1].
Coincidentally, this
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Part IV. Making Things Fit 5950
34 875 1400
5 8
55
175
525
1 1
3
9625
2
350
3675
21
2275
13
Figure 19.2. Fibonacci tiling.
Figure 19.3. Fibonacci rectangle seeded with Tutte’s perfect square.
In looking for a natural tiling that doesn’t repeat squares, one quickly discovers that the first five squares fit together neatly (Figure 19.4) but it becomes progressively more difficult to add consecutive squares without overlapping or leaving gaps. This suggests the following question: Can the squares with wholenumber sides, one of each size, be fitted together to tile the plane? The answer is that they can. In succeeding sections we will prove this, discuss the history of the problem, and pose a number of questions. 2
3 1 4
5
Figure 19.4. The first five squares.
Squaring the plane We are going to show that the plane admits a tiling that uses exactly one square of sidelength n (n D 1; 2; : : :). Our approach focuses on rectangles and ells. By “ell,” we mean any sixsided figure whose sides are parallel to the coordinate axes.
Figure 19.5. An ell.
The following extends Tutte’s use of “perfect” for rectangles and squares. Definition 1. A figure is perfect if it is composed entirely of squares of different sides. All the remaining figures in this paper will be perfect. The key to our result is Lemma 9, which states that given any perfect ell, it is possible to add squares to it to form a perfect rectangle.
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33
36
5 9
7 28
25
5
16 2
Figure 19.6. A perfect ell.
Figure 19.7. The ell of Figure 19.6 squared up.
When we add squares to a perfect figure, keeping it perfect, we’ll say we are “puffing it up.” When we puff an ell up to form a perfect rectangle, we’ll say we are “squaring up the ell.” We can then “square the plane” as follows: 1. Start with any perfect ell and square it up. 2. Create a new ell by attaching to the rectangle the smallest square not yet used. 3. Square this ell up, making sure that new squares are added in all four directions. 4. Repeat steps 2 and 3 ad infinitum. Definition 2. An ell in standard position is an ell oriented so that the single reflex angle is at the upper right. We refer to the sides by the uppercase letters in Figure 19.8 and their lengths by the corresponding lower case letters.
a
8 ˆ ˆ ˆ ˆ ˆ ˆ < ˆ ˆ ˆ ˆ ˆ ˆ :
‚
b …„ B
d ƒ‚ …„ ƒ
C
A D E
„
F ƒ‚ f
9 > = > ; )
c
e
…
Figure 19.8. An ell in standard position.
When we add squares to a figure, we always do so in such a way that an added square’s sidelength matches that of at least one side of the figure against which it is placed. That means that the only squares we add to an ell in standard position are squares adjacent to sides A, B, C , D, E, or F of the appropriate lengths. We call these operations A, B, C, D, E, and F. So, for example, applying B then A to an ell produces a larger ell (Figure 19.9). We denote sequences of operations by sequences of letters with the understanding that they are applied from left to right (so the combination of operations in Figure 19.9 would be written BA).
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Part IV. Making Things Fit B A
Figure 19.9. An application of BA to the ell of Figure 19.8.
Note that in any ell in standard position, the length of side A is the sum of the lengths of sides C and E, a D c C e; and the length of side F is the sum of the lengths of sides B and D, f D b C d: Thus, to describe the dimensions of such an ell, we need describe only the lengths of sides B, C , D, and E. Definition 3. A 4tuple hb; c; d; ei describes the dimensions of an ell if the sides B, C , D, and E of the ell are b, c, d , and e, respectively. Definition 4. A side of an ell is composite if it is not the side of a single square in the ell. In squaring up an ell, we make particular use of the operations B, F, and ED. For that reason, the following definition is most useful: Definition 5. A perfect ell is regular if each of the moves B, F, and ED either results in a perfect ell in standard position or results in a perfect rectangle. Figure 19.10 shows the smallest regular ell.
10 17 3 1
7 4
Figure 19.10. The smallest regular ell has dimensions h27; 12; 1; 5i.
If an ell with dimensions hb; c; d; ei is regular, then there can be no squares with sides b, f , e, or d C e in the ell. In particular, sides B, F , and E must be composite. In addition, it must be that c d C e (or else move ED would create an ell that is not in standard position). These properties—no squares of sides b, f , e, or d C e, and c d C e—are both necessary and sufficient for regularity. Lemma 5. Every perfect ell P in standard position can be puffed up to form a regular ell without increasing the length of side D.
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Proof. Let the dimensions of P be hb; c; d; ei. In every perfect ell in standard position, either A or F is the longest side, and whichever side is longest is necessarily composite. We may assume that a f , since if not, then side A is the longest side and composite, so A can be performed, and in the new ell the length of side A is less than the length of side F (Figure 19.11).
A
Figure 19.11. A makes a f .
Since a f , side F is the longest side and is composite, so F can be performed. This allows us to assume further that c > d . If not, we can perform FABA, and in the new ell, the length of side C is greater than the length of side D (Figure 19.12). Note that FABA does indeed puff up the ell—the first square, added by F, is larger than any in the ell, and each new square is larger than the one before. Note also that the length of side D is unaffected.
B A
A F
Figure 19.12. The operation FABA.
With our assumptions now (f a, c > d ), we perform FABA (Figure 19.13). Let hb 0 ; c 0 ; d; e 0 i be the dimensions of the ell formed after FABA (d , since FABA doesn’t affect side D). It should be clear that performing either F or B to this new ell would result in a perfect ell in standard position. To check move ED, note first that move E adds a square of side e 0 D b C d C e. This is larger than b C d but smaller than b C c C d C e, hence a square not previously used. Applying D after E adds a square of side d C e 0 D b C 2d C e. This is larger than the square just added, but again less than b C c C d C e, since c > d . The result then, is a perfect ell. To show that the ell is either a rectangle or in standard position, we need only the fact that c 0 d C e 0 . But c 0 D 2b C 2c C d C e and e 0 D b C d C e, and using c > d , we have c 0 D b C 2c C e 0 > d C e 0 . Key to proving our main lemma (Lemma 9) is an analysis of c e modulo d . For a regular ell c d C e. Consequently, we can express c as kd C e C i where k 1 and 0 i < d.
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Part IV. Making Things Fit b0 = 5b + 3c +3d +3e …„
‚
b+ c + d + e
ƒ
d 9 > > > > > > > > > > =
c 0 = 2 b + 2c + d + e > > > > > > b > ‚ …„ ƒ > > > n ; c+e 9 „ ƒ‚ … > = e0 = b + d + e b+ d > ;
8 ˆ ˆ < ˆ ˆ :
Figure 19.13. New dimensions after FABA.
Lemma 6. If a regular ell has dimensions hb; c; d; ei such that c D kd C e C i , with k 2 and 0 i < d , then the sequence of moves BFA produces a regular ell with dimensions hb 0 ; c 0 ; d; e 0 i, where c 0 D .k 1/d C e 0 C i . Proof. Performing BFA (Figure 19.14) produces an ell with dimensions hb 0 ; c 0 ; d; e 0 i D h3b C c C d C e; b C c; d; b C d C ei: Computing, we obtain c 0 D b C c D b C kd C e C i D .k
1/d C e 0 C i:
1/d C b C d C e C i D .k
To see that the new ell is regular, first note that a move of B or F would result in a perfect ell in standard position.
2b + c + d + e
8 ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ < ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ :
‚
A
c+e
b …„ B
ƒd
(
o
c e
F
„
Figure 19.14. The operation BFA.
ƒ‚ b+ d
…
To see that ED would add only new squares, observe that BFA added squares of sides f D bCd , b, and a0 D 2bCc Cd Ce. Move ED would add squares of sides e 0 D bCd Ce and d C e 0 D b C 2d C e. On the other hand, we have 2b C c C d C e > b C 2d C e > b C d C e > b C d > b; the first inequality following from the fact that c > d . All that remains to show that the ell is regular is that c 0 d C e 0 . This follows from 0 c D .k 1/d C e 0 C i , where k 2.
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Lemma 7. If a regular ell with dimensions hb; c; d; ei is such that c D kd C e for some positive integer k, then the ell can be squared up. Proof. We can apply Lemma 6 until we have a regular ell whose dimensions hb 0 ; c 0 ; d; e 0 i satisfy c 0 D d C e 0 . This ell can then be squared up with ED. Lemma 8. Every regular ell can be squared up. Proof. Given a regular ell with dimensions hb; c; d; ei, we have c D kd C e C i with k 1 and 0 i < d . We do induction on d to show that the ell can be squared up. If d D 1, then i D 0 and we are done by Lemma 7. Now assume that the result holds for any ell with dimensions hb ; c ; d ; e i such that d < d . We apply Lemma 6 iteratively until we have an ell with dimensions hb 0 ; c 0 ; d; e 0 i with c 0 D d C e 0 C i . Then ED produces an ell with dimensions hb 0 ; i; d C e 0 ; d C 2e 0i (Figure 19.15). By flipping this over we obtain in standard position an ell with dimensions h2e 0 C d; e 0 C d; i; b 0 i (Figure 19.16).
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Figure 19.15. The operation ED.
Figure 19.16. A flip after ED.
By Lemma 5 we can puff this up to a regular ell without increasing the length of side D (which is i ). Since 0 i < d , by the inductive hypothesis this flipped ell can be squared up. Clearly, the unflipped ell can also be squared up, completing the proof of the lemma. Our main lemma follows immediately: Lemma 9. Every perfect ell can be squared up. Proof. Given an ell, we puff it up to a regular ell (Lemma 5), then square it up (Lemma 8).
In other words, every perfect ell is a part of a perfect rectangle. We have now completed the spadework necessary to answer the tiling question raised in the introduction: Theorem. It is possible to tile the plane with nonoverlapping squares using exactly one square of each integral side.
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Proof. It is easy to see how to carry out the plan described earlier. We start with any perfect ell and grow it by adding squares as described in the algorithm whose steps we presented in Section 2. Note that we don’t move squares around; once added to the figure they are fixed. In particular, we never actually flip the figure as in the proof of Lemma 9—the flipping described there is simply to demonstrate that we can square up a particular ell. Note also that we are guaranteed to fill the plane since we are careful to grow the figure in all four directions in each cycle of the algorithm. Finally, for any n the square of side n will be added to the figure by or before the nth time we perform step 2 of the algorithm. Since we perform step 2 infinitely many times, we are guaranteed to incorporate in the tiling a square of each integral side.
Reflections In preparing this paper, we learned some of the history of the problem. The question was first posed by Solomon Golomb in a 1975 article in the Journal of Recreational Mathematics. He called it the “heterogeneous tiling conjecture” and challenged readers to prove or disprove it. Martin Gardner wrote about the question four years later in his column in Scientific American, later anthologized in [2]. He described one approach to a solution, a fairly orderly tiling of roughly threequarters of the plane with squares and reported that Verner Hoggatt Jr., then editor of The Fibonacci Quarterly, had shown that no square in the tiling appeared more than once. Gr¨unbaum and Shepard wrote about the problem in their 1987 book Tilings and Patterns [4]. They described there a second way in which a squared square S can generate a tiling of the plane (in addition to the method indicated in the present paper): Take a second copy of S and expand it to a square S1 such that the smallest square in S1 is the size of the original square S and fit S into that square. Take another copy of S and expand it to S2 so that its smallest square is the size of S1 , and so on. Gr¨unbaum and Shepard record the observation of Carl Pomerance that in every tiling of the plane by unique squares known at that time, the sides of the squares grow exponentially. In 1997, Karl Scherer [6] succeeded in tiling the plane using multiple copies of squares of all integral sides. The number t.n/ of squares of side n is finite but not bounded. He describes his tiling as “sizealternating,” in that no two squares of the same side share any portion of an edge (though they may share a corner).
Questions Tiling is an enormous field. This theorem might be said to reside in the subfield that concerns infinite tilings of the plane that use exactly one specimen each of a welldefined collection of similar figures. Much work has been done here and many questions remain: Efficiency The algorithm presented in this paper is extravagant in that the ratio of the largest square used so far to the smallest square not yet used rapidly diverges. The proce
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dure for squaring up, for example, when applied to the smallest possible ell, a 2 2 square next to a 1 1 square, ends in a rectangle with dimensions 1106481365205154721693 2659648557852203795117. The smallest square not used at this point is 4 4. The squaringup procedure can certainly be improved. By way of illustration, take the ell in Figure 19.6. This can be squared up to a 69 61 rectangle with the sequence of operations FEFEABC (Figure 19.7). Our procedure, however, doesn’t square the ell up until a rectangle is reached with dimensions approximately .5:01014272 /.5:81014272 /. Can our squaringup procedure be improved in some welldefined way? Does an algorithm exist for tiling the plane that methodically expands a connected island of squares in such a way that the ratio of the largest square used to the smallest not yet used is bounded by a polynomial? Simple tilings A perfect figure is simple if it contains no perfect subrectangle. Our tiling is far from simple. Is there a simple tiling of the plane using one specimen of each integral side? The halfplane and quarter plane Can the halfplane be squared? Can the quarterplane be squared? We are especially interested in this question because if it is possible to tile a quarter plane four times using, altogether, every integral square just once, then it’s possible to tile the plane using all the integral squares plus one square of any given side (say ).
¼
Figure 19.17. .
Rational squares The algorithm of our theorem can easily be used to tile the plane using one square of every rational side. As with integral squares, we don’t know if a similar procedure works for the half or quarterplane. Positive and negative squares We can tile the plane with squares whose sides are natural numbers and with squares whose sides are rationals. What about squares whose sides are (positive and negative) integers? We interpret the effect of placing a small negative square on top of a large positive one as removing a part of the large square. Once again, our algorithm works easily for this. Just as placing a positive square next to a rectangle creates an ell, so does placing a negative square on a corner of a rectangle. With care, no point of the plane will be touching more than three squares (one negative, two positive). Odd squares Can the plane be squared with all the odd squares? This seems unlikely to us. Can the plane be squared with some of the odd squares? In general, what welldefined subsets of the natural numbers can be used to tile the plane?
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Coloring Neither our tiling nor the Fibonacci tiling can be 3colored. Is there a 3colorable tiling of the plane using exactly one square of each integral side? Is there a simple algorithm for 4coloring the tiling described in this paper? Space
Can space be cubed?
Triangles Scherer proved that the plane cannot be tiled with equilateral triangles of different sizes if one triangle is smallest [7]. He has found a way of tiling the plane, however, with different sizes of iscoceles right triangles and with enlargements of certain other triangles [8]. Pomerance proved that it is possible to tile the plane with one rational triangle of each congruence class such that any two neighboring triangles share either an entire side or just a vertex [5]. Left open is the question: Can the plane be tiled with all rational equilateral triangles so that no triangle has an infinite number of neighbors? Acknowledgments We would like to thank Joe O’Rourke for his encouragement, suggestions, and enthusiasm. We are also grateful for the improvements to the paper suggested by the anonymous referees.
Bibliography [1] M. Gardner, The Second Scientific American Book of Mathematical Puzzles & Diversions: A New Selection, Simon and Schuster, New York, 1961. [2] ——, Fractal Music, Hypercards and More . . . , W. H. Freeman, New York, 1992. [3] S. W. Golomb, The heterogeneous tiling conjecture, J. Recreat. Math. 8 (1975) 138–139. [4] B. Gr¨unbaum and G. C. Shephard, Tilings and Patterns, W. H. Freeman, New York, 1987. [5] C. Pomerance, On a tiling problem of R. R. Eggleton, Discrete Math. 18 (1977) 63–70. [6] K. Scherer, New Mosaics, privately printed, 1997. [7] ——, The impossibility of a tessellation of the plane into equilateral triangles whose sidelengths are mutually different, one of them being minimal, Elem. Math. 38 (1984) 1–4. [8] ——, Nutts and Other Crackers, privately printed, 1994. [9] W. Tutte, Squaring the square, Canad. J. Math. 2 (1950) 197–209.
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20 Magic Knight’s Tours John D. Beasley In Mathematical Magic Show [4], Martin Gardner looked at the classic problem of the knight’s tour on a chess board, paying particular attention to tours in which the numbers in each row and column added to 260. At that time, the question of whether the numbers in each long diagonal could also be made to add to 260 was still open. Thanks to advances in computer power since Martin wrote, this question has now been decided, and perhaps readers will be interested in the complete statement that it is now possible to make.
A brief history of magic knight’s tours In knight’s tour literature, the term magic is used for all tours in which the rows and columns add to the same number. Any diagonal or other properties are a bonus. Figure 20.1a is an example. This is a closed tour (the starting and finishing squares are a knight’s move apart), and it has twofold rotational symmetry (numbers which are diametrically opposite differ by 32, so if we draw the lines joining each number to the next, and 64 to 1, we get a pattern which repeats itself on rotation through 180 degrees). No knight’s tour on an 8 8 board can be laterally or diagonally symmetric, nor can it have more than twofold rotational symmetry, so Figure 20.1a has as much symmetry as we can hope for. We may notice that the numbers along a row or column are alternately odd and even. 2
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(a) Wenzelides, 1849
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Figure 20.1. Two magic knight’s tours, with starting and finishing squares in bold. Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (January 2012), pp. 72–75.
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The first magic knight’s tour to appear in print was published by William Beverley in 1848, but this was an open tour (there was no knight’s move from the final square back to the first square). Figure 20.1a, which was found by Karl Wenzelides and appeared in Schachzeitung in 1849, was the first closed magic knight’s tour [12, 18]. Several more closed magic tours appeared during the next few years, including Figure 20.1b, by C. F. Jaenisch, also rotationally symmetric, which appeared in The Chess Monthly in 1859 [5] and was described by its discoverer as “La solution la plus parfaite du probl`eme du cavalier” ([5] is presented bilingually, but the French was the author’s original). Although in neither of these tours do the long diagonals add to 260, the two long diagonals together do add to 520, and Figure 20.1b has a further property at which we look in a moment. By 1940, 126 magic tours had been discovered, 59 of them closed. They were listed in 1951 in an unpublished monograph The Magic Knight’s Tours, a Mathematical Recreation by H. J. R. Murray [9, 16], but there was no convenient list in print until George Jelliss gave one in Chessics in 1986 [7, 8]. He accompanied it with a note that with modern computer methods it should be possible to ascertain whether all such tours had been discovered, and T. W. Marlow immediately showed that they hadn’t by discovering five more, which he published in 1987 and 1988 [14, 15]. T. S. Roberts found two more in January 2003 [10]. Still, none had proved to be diagonally magic, and a distributed computing exercise was set up by Hugues Mackay, JeanCharles Meyrignac, and Guenter Stertenbrink later in 2003 to put the matter finally to rest. This exercise rediscovered all the magic tours previously known and added seven new ones, three open and four closed, giving a grand total of 140 (63 closed), none diagonally magic [11, 17]. These conclusions were subsequently verified by Yann Denef using a program written independently [17]. The complete catalogue is on at least two web sites [11, 17] and in Variant Chess [1, 2]. Each tour can be presented in any of eight orientations and can be numbered from either end, so these 140 tours give rise to 2240 different arithmetical matrices.
Job done? Not quite. Only the 8 8 case was settled. We can ignore boards of odd side, because the row sums of a tour on such a board will be alternately odd and even, and so cannot all be the same. We can also ignore boards of side 4n C 2, since George Jelliss has produced an elegant and ingenious proof implying that no magic tour on such a board exists [11]. Proofs in this field tend to be either elementary or nonexistent, but this is an exception. Theorem (G. P. Jelliss). A magic knight’s tour on a board of side 4n C 2 is impossible. Lemma. In a magic knight’s tour, the number of entries of the form 4k C 2 or 4k C 3 in any row or column, counted together, must be even. Proof of the lemma. Let h be half the length of the side. The row sum is now 4h3 C h. If we express the numbers in the row in binary, we see that h of them, being odd, have a nonzero units digit, and adding these digits gives h. Remove these units digits from the entries, and the numbers represented by the remaining digits must sum to 4h3 . This is a multiple of four and the units digits in the contributing numbers have been removed, so the number of nonzero twos digits in them must be even. But each entry of the
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form 4k C 2 or 4k C 3 contributes one such digit and no entry of the form 4k or 4k C 1 contributes any, so the result follows. A similar argument applies to the columns. Proof of the theorem. Label the squares of the chessboard thus: A C A C
B A B A B ... D C D C D ... B A B A B ... D C D C D ... ..............................
We reflect the tour, if necessary, so that the number in the top left corner is odd; then all the A and D squares will contain odd numbers, and all the B and C squares even numbers. Now consider the distribution of the entries of the form 4k C 3. There are 4n2 C 4n C 1 of these, which is an odd number, and they must all be in A or D squares, so either the A squares contain an odd number and the D squares an even number, or vice versa. Similarly, either the B squares contain an odd number of entries of the form 4k C 2 and the C squares an even number, or vice versa. Now suppose first that the A squares contain an odd number of 4k C 3 entries and the B squares an odd number of 4k C 2 entries; then the C squares will contain an even number of 4k C 2 entries, and the total number of 4k C 3 and 4k C 2 entries on A and C squares, counted together, will be odd. But these squares appear only in alternate columns, and by the lemma each individual column must contain an even number of such entries. This is a contradiction, so the situation is impossible. A similar argument applies whichever of the A or D squares contain the odd number of 4k C 3 entries and whichever of the B or C squares the odd number of 4k C 2 entries. Thus there can be no magic knight’s tour on a board of side 4n C 2. So only boards of side 4n need be considered. T. H. Willcocks had discovered an open magic tour on a 12 12 board in which one diagonal added to the magic constant, and in April 2003 Awani Kumar found four such tours with both diagonals magic [10, 17]. As far as I know, the existence of closed magic tours on this board with both diagonals also magic remains an open question, but two such tours on a 16 16 board were discovered by H. E. de Vasa and published in 1962 [3, 11], and others have been discovered since [13, 17].
Return to the 8 8 board In the field of ordinary magic squares, where the numbers are not constrained to form a knight’s tour, particular attention is paid to pandiagonal squares, whose numbers add to the magic constant not only along rows, columns, and principal diagonals, but also along broken diagonals (in algebraic chess notation, diagonals such as a4e8/f1h3). Even while the possibility of a magic knight’s tour with each long diagonal also adding to 260 was open, it was realised that no pandiagonal tour could exist, because the numbers on a diagonal are either all odd or all even, and the 32 odd numbers from 1 to 64 don’t add to 4260; they add to 4 256, and the 32 even numbers to 4 264. So while preparing the catalogue of magic tours for Variant Chess, I looked to see if any had a property which I called
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quasimagic, where the odd diagonals, principal and broken, each add to 256, and the even diagonals add to 264. Given the inherent disparity between odd and even diagonals, this seemed to be as close to a truly pandiagonal magic knight’s tour as we could hope for. It turned out that none had this property in full, but Figure 20.1b is halfway there. The principal odd diagonal adds to 256, as do each of the three parallel odd broken diagonals, and the principal even diagonal and the three parallel even broken diagonals each add to 264. The same is true of two further tours which Jaenisch published in 1862 [6], but those tours lack the rotational symmetry of Figure 20.1b. Because of this combination of properties, therefore, Figure 20.1b is indeed the “most perfect” solution to the problem of the knight’s tour on the 8 8 board. Now comes a curious question: Was Jaenisch aware of this property of the broken diagonals in Figure 20.1b, which to my mind adds greatly to the elegance of this tour? I can find no mention in any of the sources I have examined from the nineteenth to the twentyfirst century. The property will have become obvious as soon as it occurred to somebody to look for it, and I wrote in 2008 that I could not believe it had remained unspotted until then. Even so, nobody has yet drawn my attention to an earlier report of its existence. Perhaps a reader will be able to enlighten me. Acknowledgment I have been merely the reporter, and in addition to the workers named above I am grateful to the Bodleian Library for access to The Magic Knight’s Tours, a Mathematical Recreation and to other Murray papers, to Cambridge University Library ´ for access to Trait´e des Applications de l’Analyse Math´ematique au Jeu des Echecs, and to George Jelliss for photocopies of the relevant pages from Schachzeitung and The Chess Monthly (made for him by Marian Stere and Ken Whyld). I am also grateful to George Jelliss for constructive comments on my original draft.
Bibliography [1] J. D. Beasley, Another look at 8 8 magic knight’s tours, Variant Chess 57 (2008) 50–53. [2] ——, Credit where credit is due, Variant Chess 58 (2008) 72.
[3] G. D’Hooghe, Les Secrets du Cavalier, Brepols, Bruxelles, 1962. [4] M. Gardner, Knights of the square table, in Mathematical Magic Show, George Allen and Unwin, London, 1977, 188–202, 283. [5] C. F. Jaenisch, De la solution la plus parfaite du probl`eme du cavalier, The Chess Monthly (1859) 110–115, 146–151, 176–179. ´ [6] ——, Trait´e des Applications de l’Analyse Math´ematique au Jeu des Echecs, vol. 2, St Petersburg, 1862. [7] G. P. Jelliss, Catalogue of 8 8 magic knight’s tours, Chessics 26 (1986) 122–128. [8] ——, Notes on Chessics 26, Chessics 29/30 (1987) 163.
[9] ——, H. J. R. Murray’s history of magic knight’s tours, The Games and Puzzles Journal 14 (1996) 238–244, 15 (1997) 266–267. [10] ——, Recent advances in magic knight’s tours, Variant Chess 43 (2003) 40–41. [11] ——, Knight’s tour notes (2001–2005); available at www.mayhematics.com. [12] D. E. Knuth, letter to G. P. Jelliss, quoted in The Games and Puzzles Journal 14 (1996) 243.
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[13] A. Kumar, Studies in magic tours of knight on 16 16 board, J. Recreat. Math. 34 (2005–2006) 275–285. [14] T. W. Marlow, Magic knight tours, The Games and Puzzles Journal 1 (1987) 11. [15] ——, Magic knight tours, The Problemist 12 (1988) 379. [16] H. J. R. Murray, The Magic Knight’s Tours, a Mathematical Recreation, manuscript in the Bodleian Library, shelfmark MS Eng d.2370. [17] G. Stertenbrink, Computing magic knight’s tours; available at magictour.free.fr. [18] K. Wenzelides (as “...l .........s in P.......”), Bemerkungen u¨ ber den R¨osselsprung, Schachzeitung 4 (1849) 41–97.
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21 Some New Results on Magic Hexagrams Martin Gardner Combinatorial problems involving magic squares, stars, and other geometrical structures often can be solved by brute force computer programs that simply explore all possible permutations of numbers. When the number of permutations is too large for a feasible running time, an algorithm can frequently be reduced to manageable time by finding ingenious shortcuts. Such planning makes computer solving less trivial and much more interesting. A superb example of such planning was described in a littleknown short article in the Mathematical Gazette (vol. 75, June 1991, pp. 140–142). The authors, Brian Bolt, Roger Eggleton, and Joe Gilks, posed for the first time a problem based on the pattern shown in Figure 21.1. It is the traditional hexagram or Star of David with its inner hexagon divided into six equilateral triangles. Can numbers 1 through 12 be placed inside the twelve triangles so that each of the six rows, indicated by the arrow, have the same sum? The authors point out that there are 12Š ways of arranging the numbers. When the pattern’s six rotations, and six reflections, are excluded, the number reduces to 11Š or 39,916,800. Can this number be further reduced?
Figure 21.1.
Reprinted from The College Mathematics Journal, Vol. 31, No. 4 (September 2000), pp. 274–280.
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It can. Each of the six rows must have the same sum, and this “magic constant” must be either odd or even. It is not hard to discover that, by neglecting rotations, reflections, and complements, there are just three ways that odd and even numbers from 1 to 12 can be distributed on the hexagram (see Figure 21.2). 0
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A complement, I should explain, is obtained by subtracting each number of a magic square or star from the pattern’s largest number plus 1. For example, consider the Chinese lo shun or 3 3 magic square in Figure 21.3. If each number is replaced by the remainder when it is taken from 10, we obtain the complement (Figure 21.4).
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Note that the complement in this case is the same as before except for a rotation and reflection. In the case of magic hexagrams, complements are always different, but are considered trivial variations. The complements of the three patterns shown in Figure 21.2 are obtained by replacing each 1 with 0 and each 0 with 1. Each 1 stands for an odd number, each 0 for an even number. By neglecting complements, as well as rotations and reflections, the three oddeven patterns provide a way of establishing upper and lower bounds for the star’s magic constant. In pattern A all the even numbers go in outside triangles. They add to 42, but because each number appears in two rows, we double 42 to get 84 as the contribution the even numbers make to the total of the sums of all six rows. The odd numbers in pattern A go on interior triangles. They add to 36, but because each number is in three rows we triple 36 to get 108 as the contribution the odd numbers make to the total of all six rows. The sum of all the rows sums is 84 C 108 D 192. There are six rows, so to find the magic constant for this pattern we divide 192 by 6 to get 32. This is the magic constant
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for pattern A on the assumption that the numbers can be arranged to make the hexagram magic. As it turns out, pattern A has no solution. We turn now to patterns B and C. Each has four odd numbers and two even numbers on the interior triangles. To make the magic constant as low as possible, we place on these inside triangles, where each number appears in three rows, the four lowest odd numbers .1; 3; 5; 7/ and the two lowest even numbers .2; 4/. The sum of the inside numbers is 22. Tripling it gives 66. On the outside triangles go odd numbers 9; 11 and even numbers 6; 8; 10; 12. They add to 56. Each number appears in two rows so we double 56 to get 112. Adding 66 to 112 gives 178. We know that this total must be a multiple of 6 because six rows must have the same sum. So we raise 178 by the smallest amount to arrive at a multiple of 6, namely 180. Dividing 180 by 6 yields 30. This is the lower bound for the magic constants of patterns B and C. To obtain the upper bound we put the largest two even numbers .10; 12/ and the four largest odd numbers .5; 7; 9; 11/ on the interior triangles. They add to 54 and three times 54 is 162. The outside numbers .1; 2; 3; 4; 6; 8/ add to 24 and twice 24 is 48. The sum of 162 and 48 is 210. Dividing 210 by 6 gives 35 as the upper bound for the magic constants of patterns B and C. We now have shown that the magic constant, if such a magic hexagram does indeed exist, must be 30; 31; 32; 33; 34, or 35. Knowing this, and knowing how odd and even numbers must be distributed to make the hexagrams magic, allowed the three authors to reduce their computer program to a manageable length. To the authors’ vast surprise, their computer search produced just one solution! It is shown on the left of Figure 21.5. Its complement (each number taken from 13) is on the right. “When we discovered the beautiful solution,” the authors write, “we rushed from our baths into the street shouting ‘Eureka!’” The two patterns make excellent puzzles. Place numbers 1 through 12 on the hexagram so each row adds to 33, or so each row adds to 32. In both cases the solution is unique. Note that in the pattern on the left each pair of opposite corners add to 12, and on the right, they total 14.
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A much older magic hexagram problem, now thoroughly explored, consists of placing numbers 1 through 12 on the vertices of the traditional Star of David shown in Figure 21.6. Because each number appears in two rows, the sum of all the row sums is twice the sum of numbers 1 through 12, or 782 D 156. There are six rows, so 156=6 D 26, the hexagram’s magic constant.
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Figure 21.6.
The British puzzlist Henry Ernest Dudeney, in Modern Puzzles (1926) claimed there are 37 fundamental solutions, or 74 if complements are included. This was one of Dudeney’s rare mistakes. He missed three basic patterns. Von J. Christian Thiel, in the German periodical Archimedes (vol. 5, September 1963, pp. 65–72) displays 15 basic solutions. By applying well known transformations in which numbers are interchanged a certain way, Thiel raises the total of patterns, not counting rotations and reflections, to 40. These, together with their complements, make 80 solutions. Dudeney, in Modern Puzzles, and later in A PuzzleMine, identifies six solutions which have the additional feature that the points of the star also add to 26. They are shown in Figure 21.7. Each has a complement in which the interior numbers add to 26. Note that on each hexagram each large triangle has a sum of 13, so together they add to 26, and the sum of any small triangle is the same as the sum of the small triangle opposite.
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6
3
10 2
4
8
8 3
12
3
6
2
9
1
5
8 1
6
10
9
12
3
1
Figure 21.7.
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29 53
31
137 59
61
149
67
71 73
43
41
157
179
193 173
47
37
191
163
167
139
181
151 Figure 21.8.
Akira Hirayama and Gakuho Abe, in Researches in Magic Squares (Osaka, 1983)—the text is in Japanese—give a slightly different way of arriving at 80 fundamental solutions. The book also includes Abe’s amazing constructions of two magic hexagons made with consecutive primes! They are shown in Figure 21.8. There is now a sizeable literature on stars with more than six points. The fivepointed star, or pentagram, is easily shown to have no solution for numbers 1 through 10. You’ll find one such proof in the chapter on magic stars in my Mathematical Carnival. In the same chapter I explain how the numbers in a solution for the traditional hexagram can be transferred to the twelve edges of the cube so that four numbers around each face total 26. Because the cube’s “dual” is the octahedron (faces and vertices exchanged), the same numbers can be placed on the octahedron’s edges so that the four edges around each vertex add to 26. Figure 21.9 shows how a hexagram solution is transferred to a magic cube and to a magic octahedron. 10 7 7
4
9
11
12
4
9
8 2
3 1 11
7
6
2
12
6
4
5
10
5
8 1
6
9
5
10 2
8 1
3 12
11
3 Figure 21.9.
A third type of hexagram is formed by adding three diagonals to the traditional structure as shown in Figure 21.10. The question arises: Can numbers 1 through 19 be placed on the vertices of this pattern so eveiy line of five has the same sum? Harold Reiter, at the University of North Carolina, Charlotte, posed this as an unsolved problem in his article “A Magic Pentagram,” in Mathematics Teacher (March 1983, pp. 174–177). By trial and error I discovered the solution shown; it has 1; 2; 3; 4; 5; 6 on the outside points, and a magic constant of 46. Its complement (numbers taken from 20) puts 14; 15; 16; 17; 18; 19 on the outside points, and raises the magic constant to 54.
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Part IV. Making Things Fit 1
4
12
9
15
10
8
13
18
16
11 5
2
19
7
17
14
3
6 Figure 21.10.
Finding all solutions to this problem by a brute force computer algorithm is out of the question because it would require examining 19Š=12 possible permutations, Reiter and his associate David Ritchie, in “A Complete Solution To the Magic Hexagram Problem,” in The College Mathematics Journal (vol. 21, September 1989, pp. 307–316), describe the clever shortcuts they used to reduce their algorithm to an examination of a mere 18,264,704 patterns. In less than five minutes their Pascal program found 2,190 basic solutions, not counting rotations, reflections, and complements, or 4,380 solutions if complements are included. The magic constants range from 46 through 54. The hexagram pattern in Reiter’s problems is an interesting one. It solves an old treeplant puzzle of placing 19 trees to form six straight rows with four trees in each row. I used this pattern for a checkerlike game I named Solomon. It can be purchased from Kadon Enterprises, 1227 Lorene Drive, Pasadena, Maryland 21122. Frank Bernhart wrote to point out that there are many symmetric patterns with vertices and lines that are combinatorially equivalent to Reiter’s hexagram, and therefore have the same set of solutions. Three are shown in Figure 21.11.
Figure 21.11.
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The traditional hexagram obviously contains eight different triangles. The modified hexagram of the kind shown in Figure 21.1 has 20 triangles. (Many old puzzle books give this as a problem.) How many different triangles can you find in Reiter’s hexagram? Counting them all is not so easy. I leave this as a question to be answered in the next issue (see next page).
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In The College Mathematics Journal (Vol. 31, No. 5, page 146), the number of triangles in Reiter’s hexagram was given as 50. However, the current editors have found 56 distinct triangles. 1
4
12
9
15
10
8
13
18
16
11 5
2
19
7
17
14
3
6
The triangles are grouped by number of regions below with the count and a representative for each of seven types of congruent triangles. 1 region: 12 triangles congruent to 1; 9; 12, 2 regions: 6 triangles congruent to 1; 12; 19, 2 regions: 12 triangles congruent to 1; 10; 13, 4 regions: 6 triangles congruent to 1; 5; 13, 5 regions: 6 triangles congruent to 1; 6; 18, 6 regions: 12 triangles congruent to 1; 5; 17, and 12 regions: 2 triangles congruent to 1; 3; 5.
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22 Finding All Solutions to the Magic Hexagram Alexander Karabegov and Jason Holland Problem 394 in Henry Dudeney’s 536 Puzzles and Curious problems [1] poses this puzzle: Put the numbers from 1 to 12 in the circles on the left in Figure 22.1 so that the sum of the four numbers on each line is 26. One solution is shown on the right. 1
A
K
L
3
D
J
I
C
B
H
F
G
8
11
7
12
E
2
4
5
9
10
6
Figure 22.1. The hexagram along with one solution.
The solution pictured in Figure 22.1 is listed as solution number 16 in Table 22.1 at the end. Table 22.1 contains 20 solutions to the puzzle. The letters in Figure 22.1 on the left are used as positions for the solutions in Table 22.1. There are many puzzles referred to as magic stars (see, for example, [2, Ch. 5]). While such puzzles are usually fun, they often require the “brute force” method (also called the method of exhaustion) for their solution. In this brief article, we present a systematic approach to finding all solutions of Dudeney’s puzzle. The key to this approach is something known as an outshuffle in cards. It is known that if two solutions are considered to be the same when one can be obtained from the other by rotation or reflection, then there are 80 different solutions to this puzzle. (Curiously, Dudeney erred in his count, finding only 74, see [1, p. 350]). Martin Gardner [2, chapter 21 in this collection] called this puzzle the Magic Hexagram and pointed out Reprinted from The College Mathematics Journal, Vol. 39, No. 2 (Mar. 2008), pp. 102–106.
167
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Part IV. Making Things Fit Table 22.1. The twenty nonequivalent solutions to the Magic Hexahedron. A
B
C
D
E
F
G
H
I
J
K
L
1
1
7
2
8
10
4
12
3
9
5
6
11
2
1
10
2
8
7
4
12
6
9
5
3
11
3
1
5
7
9
11
4
6
8
3
10
2
12
4
1
8
4
6
11
7
9
5
3
10
2
12
5
1
5
7
11
9
6
2
8
3
12
4
10
6
1
10
8
4
11
5
9
3
7
12
2
6
7
1
7
3
12
6
2
9
8
10
4
5
11
8
1
5
3
12
8
2
9
6
10
4
7
11
9
1
4
2
10
11
5
9
3
7
6
8
12
10
1
9
3
11
5
4
8
10
7
6
2
12
11
1
9
2
6
10
7
11
4
5
8
3
12
12
1
6
3
11
8
2
10
7
9
4
5
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13
1
4
7
11
10
2
6
9
5
8
3
12
14
1
7
2
10
8
3
11
6
9
4
5
12
15
1
9
2
10
6
11
3
4
5
12
7
8
16
1
8
4
7
10
9
6
5
2
12
3
11
17
1
2
5
12
11
3
6
4
8
7
9
10
18
1
9
5
12
4
3
6
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2
10
19
1
7
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6
5
2
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4
12
20
1
6
5
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9
7
4
8
2
11
3
12
that there are indeed 80 different solutions. He also observed that it is equivalent to placing the numbers from 1 to 12 on the edges of a cube so that the sum of the four labels on every face is 26. This identification is illustrated in Figure 22.2. There seems little choice but to call Gardner’s version the Magic Hexahedron (unless it be called the Magic Cube). The two versions are not entirely equivalent however—the 1 9 3
8
11
7
4
6 4 5 10
7
12
3
8 2
5
9
10
2 1
12 11
6 Figure 22.2. The identification between the hexagram and the cube.
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4 7
6
5
9
9
10
2
3 8
7
12
5
10 11 1
12
11
3
4
8
2 6 Figure 22.3. Solution generated from Figure 22.2 by a reflection.
cube has more symmetries than the hexagram (or, equivalently, the hexagon). While the symmetry group of a hexagon has order 12, that of the cube has order 48. For each solution to the Magic Hexahedron, there are four Magic Hexagram solutions. We use Figure 22.2 to illustrate the latter point. If one reflects the cube across the vertical plane through the midpoints of the edges labeled 7, 6, 12, and 1, one gets the solution shown on the left in Figure 22.3. The corresponding Magic Hexagram is shown on the right in Figure 22.3. Clearly, this is different from the solution of the Magic Hexagram in Figure 22.2, that is, no symmetry of one of these hexagrams can produce the other. (For an excellent discussion of the full symmetry group of the cube, see [3, pp. 112–113].) We now turn to determining that there are twenty different Magic Hexahedron solutions. To facilitate our enumeration, we consider two ways of classifying a face in a solution. First, a face is paritybalanced if it has two odd labels and two even ones. Note that the only nonparity balanced faces have all labels odd or all even. A solution is then called paritybalanced if all of its faces have this property. Observe that in a solution that is not paritybalanced, one face must have all labels odd, one must have all labels even, and the other four labels must alternate in parity around the cube. Since the only possibilities for a face that is not paritybalanced are f1; 5; 9; 11g, f3; 5; 7; 11g, f2; 4; 8; 12g, and f2; 6; 8; 10g, it is straightforward (and not difficult) to find the ten solutions that are not paritybalanced. These are the first ten solutions in Table 22.1. We now turn to our second way of classifying a solution. We call a face magnitudebalanced if two of its labels are 6 or less and the other two are 7 or more, and a solution is magnitudebalanced if all of its faces are. Note that of the solutions 1–10 in Table 22.1, only the first two do not have this property. Given a magnitudebalanced face, its four labels can be written as i , j , k C 6, and l C 6, where each of i , j , k, and l is at most 6. Observe that since i C j C k C l D 14, the set f2i 1; 2j 1; 2k; 2l g then is a paritybalanced set with sum 26. Hence, the mapping M with ( 2n 1 if 1 n 6, M.n/ D 2.n 6/ if 7 n 12,
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transforms a magnitudebalanced solution to one that is paritybalanced. If we apply the mapping M to the eight solutions 3–10 in Table 1 (perhaps twice), we get the ten solutions 11–20 in Table 22.1. This process is illustrated in Table 22.2 where arrows to the right indicate that the mapping M has been applied. Table 22.2. Solution numbers from Table 22.1 along with what happens when M is applied. 3
!
11
!
12
4
!
13
!
14
5
!
15
6
!
16
7
!
17
8
!
18
9
!
19
10
!
20
All that remains is to show that there are no other solutions to the Magic Hexahedron puzzle. For this, we use the inverse P of the mapping M :
P .n/ D
8 < 1 .n C 1/ if n is odd, 2 :1
2n
C6
if n is even.
This is a permutation of f1; 2; : : : ; 12g that takes a paritybalanced solution to one that is magnitudebalanced. In a deck of 12 cards, this permutation is known as an outshuffle [4]. We now show that every paritybalanced solution eventually leads through repeated application of P to a nonparity solution. Suppose not, and let S be such a solution. Note that our outshuffle P leaves 1 and 12 fixed and permutes the other ten numbers in the cycle .2; 7; 4; 8; 10; 11; 6; 9; 5; 3/. One of the faces in the solution S contains 12 and not 1. Sooner or later, if we apply P repeatedly, the labels on this face include 12, 11, and two other numbers, neither of which can be 1. Since their sum is 26, this is impossible. Hence, every paritybalanced solution eventually leads to a nonparitybalanced solution. Therefore, every paritybalanced solution arises from a nonparitybalanced one by applying the mapping M , either once or twice. Thus, Table 22.1 gives all solutions to the Magic Hexahedron puzzle. Theorem. The Magic Hexahedron puzzle has precisely twenty nonequivalent solutions. Corollary. The Magic Hexagram puzzle has precisely eighty nonequivalent solutions. Acknowledgments The authors wish to thank John Barton who wrote a program that verified all 960 solutions to the Magic Hexagram and thus gave us plenty of data to work with. We also wish to thank Amy Niemeyer who pointed out the term outshuffle, Stephanie Wong who brought this wonderful puzzle to our attention, and Lowell Beineke for suggesting the name Magic Hexahedron.
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Bibliography [1] H. E. Dudeney, 536 Puzzles & Curious Problems, Charles Scribner’s Sons, 1967. [2] M. Gardner, Mathematical Carnival, Mathematical Association of America, 1989. [3] F. Goodman, Algebra Abstract and Concrete, Prentice Hall, 1998. [4] E. Weisstein, OutShuffle Mathworld webpage, mathworld.wolfram.com/OutShuffle.html.
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23 Triangular Numbers, Gaussian Integers, and KenKen John J. Watkins One of the first of Martin Gardner’s Mathematical Games columns I ever read was “Euler’s Spoilers,” in November 1959 [1], and after all these years I think it is still my favorite (maybe this is just because I love the way he rhymed ‘Euler’ and ‘spoiler’). It dealt with what we now call Latin squares, n n arrays using n symbols such that each symbol appears exactly once in each row and column. Latin squares are extremely useful in the design of statistical experiments, but they are perhaps better known now for their appearance in recreational puzzles such as sudoku where each row and each column of a 9 9 array contains each of the integers 1; 2; : : : ; 9 exactly once. In 2004, a Japanese mathematics teacher, Tetsuya Miyamoto, invented a sudokulike puzzle called KenKen (loosely translated, this means “wisdom squared”). KenKen quickly became so popular that it is now a standard feature in newspapers around the world, and there are two very good web sites ([2], [3]) that offer new puzzles daily online. As in sudoku, the goal in KenKen is to fill an n n grid with the numbers 1 through n so as not to repeat a number in any row or column. Despite the fact that it uses numbers, sudoku is not an arithmetical puzzle; any nine symbols could be used in place of the integers 1–9. Solving a KenKen puzzle, on the other hand, depends heavily upon several important ideas about numbers.
Triangular numbers In Figure 23.1, we see a typical KenKen where each heavily outlined ‘cage’ indicates what must happen inside that particular cage. For example, inside the cage labeled ‘24’ the product of the three numbers must be 24; hence, the three numbers in this cage are forced to be 4, 3, and 2. The cage labeled ‘5’ must contain the three numbers 5, 1, and 1 (because 5 is prime); moreover, we can immediately place these three numbers in this cage in their correct positions since the two 1s must appear in different rows and different columns. Since the only ways to partition the integer 10 into a sum of three distinct integers Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 37–42.
173
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Part IV. Making Things Fit 5×
24×
10+
2
1–
4–
12×
3
2÷
1–
10+
(a) Figure 23.1. A typical 5 5 KenKen.
1; 2; 3; 4, and 5 are as 10 D 5 C 4 C 1 or as 10 D 5 C 3 C 2, we see that we have two choices for the numbers to put in each of the cages labeled ‘10C’. It almost goes without saying that prime factorization and partitions of integers are important in solving KenKen puzzles. What is quite surprising is that another important notion in number theory, the ancient Greek concept of triangular numbers, can also often be used to solve a KenKen because in any solution to a KenKen the sum of the numbers in any row or column of the grid is 1 C 2 C C n; that is, the sum is the nth triangular number. In Figure 23.1, the sum of the numbers in any row or column must be 15 (that is, the triangular number 15 D 1 C 2 C 3 C 4 C 5). For example, since in the bottom row the sum in the first cage is 10, the sum in the second cage must be 5. Hence, that cage contains 2 and 3. But, there is already a 3 in the fourth column, so we can immediately place the 2 and 3 in their correct positions in the bottom row. At this point we can use a similar argument for the lefthand column to conclude that the sum of the two numbers in the upper and lower lefthand corners must also be 5. Thus, these numbers are either 2 and 3, or 1 and 4. But, they can’t be 2 and 3 because we have already placed both 2 and 3 in the bottom row. Therefore, we know these two numbers must be 1 and 4, and we even know that the 4 goes in the upper corner since the cage labeled ‘24’ contains 4, 3, and 2. It is now a straightforward matter to finish solving this puzzle. (Solutions for the ordinary KenKen puzzles in this article are on p. 178.) In Figure 23.2, we see a 4 4 KenKen puzzle that normally would require a lot of trial and error to solve. For example, it is clear that the ‘3 ’ cage contains a 1 and 4 and the cage below contains a 2 and 3, but a good deal of experimenting would need to take place to determine the exact placement of these four numbers. Fortunately, we can use triangular numbers to avoid a tedious trial and error process. It is clear that the two numbers a and b in the ‘2’ cage must be 1 and 2 in some order. Since the sum of the numbers in the two bottom rows is an even number .20 D 10 C 10/, we immediately conclude that b is also even (because 9 is odd and the sum of the two numbers in the ‘1 ’ cage must also be odd). Therefore, b D 2, and it is now easy to solve
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23. Triangular Numbers, Gaussian Integers, and KenKen 3–
1–
175
12×
2÷
a 9+
1–
b 4
(b) Figure 23.2. Using triangular numbers to find b.
this puzzle without any trial and error (the first column can now be completed and the ‘12’ cage cannot contain a 1). Note that we could have added all the known quantities in the bottom two rows and then subtracted the resulting sum, 18, from 20 to find b, but that takes more work, and often, as in this case, a simple parity argument provides the needed information. In Figure 23.3, we have a rather difficult 6 6 KenKen that can also be solved with the aid of triangular numbers. We focus first on the ‘4’ cage in the bottom two rows. The question is whether the 4 factors as 4 1 1 or as 2 2 1. Note that, in either case, since the 120 in the ‘120’ cage in the two bottom rows must factor as 6 5 4, we can conclude that the ‘2’ cage in the bottom row does not contain both 2 and 4 (because in the first case there are already two 4s in the bottom two rows, and in the second case there are already 12×
11 +
12+
1–
120×
4×
a b
2
3÷
3–
2÷
3÷
3
120×
10+
7+
c 2÷
d (c)
Figure 23.3. Using triangular numbers to find a; b and c.
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two 2s in the bottom two rows). Hence, this ‘2’ cage contains either 1 and 2, or 3 and 6, and either way the sum of the two numbers in this cage is odd. Now, 3 is also odd, as is the sum of the numbers in the ‘120’ cage (6 C 5 C 4 D 15); however, the sum in the ‘10C’ cage is even, as is the entire sum of the bottom two rows (being twice the triangular number 21). Therefore, we immediately see that a C b C c must be odd. Hence, the numbers in the ‘4’ cage are 2, 2, and 1. With this information we can quickly fill in the bottom two rows (using the fact that the ‘11C’ cage in the top row must contain a 5 and 6), and easily complete the rest of the puzzle without ever having to resort to trial and error.
More tricks The puzzle in Figure 23.3 can also be used to illustrate a very sneaky technique I learned from Barry Cipra in Atlanta at the eighth Gathering for Gardner conference. He used his idea to solve sudoku puzzles, but the same idea can also be helpful for KenKen. So, let’s start from scratch on the puzzle in Figure 23.3, and try to determine the value of d based on an argument involving the assumption that there is a unique solution to this problem. It is usually implicit, though rarely stated, in puzzles such as KenKen and sudoku, that there is a unique solution to the puzzle, and this fact can sometimes be invoked as an aid in finding that solution. Again, we consider the situation in the ‘120’ cage at the bottom of the puzzle. This time we look at the ‘11C’ cage in the top row directly above this cage. Since the ‘11C’ cage must contain 5 and 6, if we assume that there is a unique solution to this puzzle then there cannot also be a 5 and 6 in the two squares in the bottom row in the ‘120’ cage (otherwise, simply switching the 5s and 6s in the top and bottom rows yields two distinct solutions). And, of course, 5 and 6 can’t both appear in the 4th column in the ‘120’ cage since either a 5 or a 6 appears in that same column in the ‘11C’ cage in the top row. The only remaining option in the ‘120’ cage then is if d D 4. Factorials, the multiplicative analog of triangular numbers, can also be used in solving KenKen puzzles. For example, in a 6 6 puzzle, if there is a straight ‘120’ cage with four squares in a row, then the product of the two extra squares in this row is 6 (because 6Š 120 D 6). So, if this ‘120’ cage is next to an angled ‘12’ cage with two squares in the same row and one square in the row above, then the single square in the row above must contain a 2.
Gaussian puzzles One of the most enjoyable features of KenKen puzzles, especially larger ones, is the way in which the various possibilities that arise from the prime factorization of an integer come into play when dealing with irregularly shaped cages. Another number system with unique factorization into primes is the Gaussian integers, all complex numbers of the form a C bi where a and b are integers. What is fun about the Gaussian integers is that some familiar numbers such as 2 and 5 are no longer prime since they can be factored: 2 D .1 C i /.1 i / and 5 D .1 C 2i /.1 2i /. We conclude this tribute to Martin Gardner with several KenKen puzzles for the reader to solve featuring the Gaussian integers (solutions are on p. 186). Here are a few hints to get you started. Consider the ‘4’ cage in puzzle (d). Since 4 D 2 2 and we are not
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23. Triangular Numbers, Gaussian Integers, and KenKen 4×
3+
i–
1+ i –
177 8×
i–
4×
3+
1– i –
i–
2×
2×
i–
(d)
(e)
Figure 23.4. Solve using the four numbers 1, 1 C i , 1
i , and 2.
allowed to repeat a 2 in a column, there is only one useful way to factor 4 as a product of three distinct numbers: 4 D 2.1 C i /.1 i /. Thus, we know which three numbers belong in this cage (though not their order). This tells us immediately that a 1 goes in the lower lefthand corner (and then a 2 next to it in the ‘2’ cage). There are two ‘3C’ cages in this puzzle, one with three squares and one with two squares; it is easy to see that the only possibility is for the cage with three squares to contain 1; 1 C i , and 1 i , and for the cage with two squares to contain 1 and 2. What about the ‘i ’ cage in this puzzle? Clearly 2 cannot appear here since 2 minus any of the other three numbers leaves a 1 behind. Also, this cage cannot contain both 1 C i and 1 i since .1 C i / .1 i / D 2i . So, the only options for this cage are for it to contain 1 and 1 C i (since .1 C i / 1 D i ), or for it to contain 1 and 1 i (since 1 .1 i / D i ). 4+
5×
3+
2×
i÷
1 +2 i–
i–
3i –
2–2 i ×
1+ 2i –
2+ 2i × 4×
5×
3+
(f) Figure 23.5. Solve using the six numbers 1, 1 C i , 1
i , 1 C 2i , 1
2i , and 2.
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Similarly, the ‘1 C i ’ cage must contain a 2 (since, otherwise, the 1s completely cancel during subtraction). Thus, the only possibility is for this cage to contain 2 and 1 i (since 2 .1 i / D 1 C i ). The Gaussian integers, as lattice points in the complex plane, are also vectors. For example, 1 C i can be thought of as the vector from the origin .0; 0/ to .1; 1/. Geometrically, the difference ˛ ˇ between any two complex numbers ˛ and ˇ is just the vector from point ˇ to point ˛. Thus, in puzzle (d), it is easy to spot which pairs of numbers can go in the ‘i ’ cage because the i vector can only be drawn from .1; 0/ to .1; 1/, or from .1; 1/ to .1; 0/; similarly, we can determine the pair of numbers that must go in the ‘1 C i ’ cage merely by observing that we can only draw a 1 C i vector from .1; 1/ to .2; 0/.
Bibliography [1] M. Gardner, Euler’s spoilers: the discovery of an order10 GraecoLatin square, New Mathematical Diversions, Mathematical Association of America, Washington DC, 1995, 162–172. [2] The official KENKEN web site with puzzles posted daily; available at www.kenken.com. [3] The New York Times puzzle web site with puzzles posted daily; available at www.nytimes.com/kenken.
Solutions to the ordinary KenKen puzzles (a)–(c) on pages 174–175 (a)
(c)
4 3 2 5 1
3 2 5 1 4
2 4 1 3 5
1 5 3 4 2
5 1 4 2 3
3 5 6 4 1 2
4 1 5 6 2 3
1 4 2 5 3 6
5 3 1 2 6 4
6 2 3 1 4 5
(b)
1 4 3 2
3 1 2 4
4 2 1 3
2 3 4 1
2 6 4 3 5 1
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24 Cups and Downs Ian Stewart Martin Gardner was fond of mathematicallybased magic tricks using simple apparatus —in particular, pennies. Chapter 2 of his Mathematical Carnival [2, pages 12–26] is entirely about penny puzzles. About five years later, Gardner and Karl Fulves invented a delightfully simple trick with three pennies (see Demaine [1]). Its explanation involves an important mathematical idea called a state diagram, useful in many combinatorial problems. A related trick, usually described using cups instead of coins, has the same state diagram with ‘heads’ and ‘tails’ replaced by ‘up’ and ‘down.’ This article explores the relations between the two tricks, develops the deeper mathematics that they have in common, and discusses a generalization of the cups trick. (For other developments from Gardner’s penny puzzles, see [1].)
Gardner’s threepenny trick The trick is performed by a blindfolded magician. A volunteer places three pennies in a row, and chooses at will whether each coin shows heads or tails. However, both heads and tails must appear, otherwise the trick ends before it begins. The magician announces that even though she cannot see the coins, she will give instructions to turn coins over so that all three coins show the same face, heads or tails. The instructions are: 1. Flip the lefthand coin. 2. Flip the middle coin. 3. Flip the lefthand coin. After steps 1 and 2 the magician asks whether all three coins show the same face, and if the answer is ‘yes’, the trick stops, otherwise the magician requests the third flip. Figure 24.1 shows an example. Although it is plausible that enough flips will eventually get all coins the same way up, it is a little surprising that at most three flips are needed. In this case it is straightforward to list all possible starting configurations, and check that three moves always suffice. However, this is not a very satisfactory explanation. All beReprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 15–19.
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(a) Typical start. . .
(b) Flip the left. . .
(c) Flip the middle. . .
(d) Flip the left. Figure 24.1.
comes crystalclear, if we draw the state diagram, which shows all possible configurations, together with the transitions that occur when the instructions are obeyed. The diagram itself naturally forms a cube, Figure 24.2a, in which each coin corresponds to an axis in 3dimensional space, and H, T (respectively) are the coordinates 0, 1. The effects of moves 1 and 2 are shown in the figure, and it can easily be checked that starting anywhere except HHH and TTT, a sequence of at most three moves always leads to HHH or TTT. HTT
flip left
flip middle
HTH flip middle
flip left
flip middle
TTH
TTS
STS
flip middle
TST
SST
THT
HHT
TTT
STT
TTT
flip left
HHH
flip left
SSS
THH
TSS
Figure 24.2. (a) State diagram for the coin trick. (b) State diagram for the cup trick.
Three cups The same state diagram occurs in the analysis of an old trick, which starts by placing three cups upright on a table or a bar [[[ ; then turning the middle one over:
[\[
:
You announce that you will turn all three of them upside down in exactly three moves, where a move inverts exactly two cups. Like this:
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183 [\[ \[[ [[\ \\\
:
It can be done in one move, but asking for three is a classic piece of misdirection, setting up the second part of the trick. Which is to casually turn the middle cup upright: \[\ ;
and invite someone to repeat the trick. But now, whatever they do, they can’t get all three cups upside down. What they haven’t spotted is that you’ve altered the initial position, and the parity of the upright cups has changed from even to odd. The original start position has even parity, and so does the required end position. But the new start position has odd parity. Since inverting two cups preserves parity, the task becomes impossible. There are several different ways to think about this puzzle mathematically. One is to draw its state diagram, which shows all possible states of the three cups and uses connecting lines to show what happens if you turn cups upside down one at a time. Figure 24.2b shows this. Of course, you have to invert them two at a time, so you redraw the diagram to show which states are connected by two 1cup moves, obtaining Figure 24.3. This falls into two disconnected pieces: one contains the first starting state and the required finishing one, and the other contains the second starting state. Obviously you can’t change components by going along the lines. TTS
STT
SSS
SST
TST
TTT
TSS
STS
Figure 24.3. State diagram for 2cup moves. (Each edge represents two or three distinct combinations of 1cup moves.)
The problem doesn’t depend on the order in which the cups are arranged, just how many are up (and hence how many are down). So you can ‘factor out the symmetry’ by considering only the number of ‘up’ cups, getting Figure 24.4a for 1cup moves and Figure 24.4b for 2cup moves. Now the obstacle is even more obvious.
3
2
1
0
3
1
2
0
Figure 24.4. State diagrams identifying symmetrically related states. (a) 1cup moves. (b) 2cup moves.
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You can also represent parity by colouring odd numbers grey and even ones white. The result, for Figure 24.4a, is Figure 24.5. Now any twostep move preserves colour (that is, parity) so again there can’t be a solution.
3
1
2
0
Figure 24.5. Parity diagram for 1cup moves.
Matrix solution and generalization If you prefer algebra to graph theory, you can use the transition matrix. For 1cup moves this is a 4 4 matrix A whose rows and columns correspond to the four states in Figure 24.4, with entry aij being 1 if there is a move from state j to state i , and 0 if not. Clearly 2
0 61 AD6 40 0
1 0 1 0
0 1 0 1
3 0 07 7: 15 0
The transition matrix for 2cup moves is just the square of A, and 2
1 6 0 A2 D 6 41 0
0 2 0 1
1 0 2 0
3 0 17 7: 05 1
The entry 2 shows not just that there is a connection, but that there are two ways to realise it, but this is information we don’t really need. So let’s introduce a new symbol, 1, subject to the rules 1 C 1 D 1; 1 1 D 1. Then we can use 1 to represent the existence of a connection, and 0 when there isn’t one. Now 2 3 1 0 1 0 6 0 1 0 17 7 A2 D 6 41 0 1 0 5: 0 1 0 1 Sequences of 2cup moves now correspond to powers of A2 . But you can use the properties of 1 to show that .A2 /2 D A2 , so such sequences can’t achieve anything new. Since the start and end states are not connected, the problem is impossible. A few years ago ManKeung Siu, a mathematician at Hong Kong University, wondered what happens if you allow more than three cups to start with, and each move inverts a fixed number of cups. In particular, when can you start with all cups up and end with all cups down, and what is the shortest sequence of moves that achieves this? Together, we discovered that this problem has a surprisingly complicated answer. If there are n cups, and any m of them can be inverted at each move, then this minimal number is shown in Table 24.1. Here dxe is the ceiling function, the smallest integer x.
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185 Table 24.1.
n even, m even 2m n dn=me
n odd, m odd 2d.n
m/=2me C 1
n odd, m even n even, m odd no solution
2m > n 1 if m D n; 3 if m < n 1 if m D n; 3 if m < n no solution
2dn=2me 2dn=2.n
m/e
The proof uses the transition matrix and is found in [3]. Exploring some of the simpler cases is highly recommended: you will quickly appreciate that the general pattern is not at all obvious.
Bibliography [1] Erik D. Demaine. Puzzles and tricks from Martin Gardner inspire math and science, American Scientist 98 (2010) 452–456; available at www.americanscientist.org/ issues/pub/2010/6/recreationalcomputing. [2] M. Gardner. Penny puzzles, in Mathematical Carnival, Knopf, New York, 1975. [3] I. Stewart and M.K. Siu, How to invert n cups m at a time? Math. Today (SouthendonSea) 46 (2010) 34–38.
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Solutions to the Gaussian integer KenKen puzzles (d)–(f) on page 177 (d)
(e)
(f)
1Ci 2 1 i 1
1 1 i 1Ci 2
2 1Ci 1 1 i
i 1 2 1Ci
1Ci 1 2 1 i
2 1Ci 1 i 1
1
1 1 i 1Ci 2
2 1 C 2i 1 2i 1 i 1 1Ci
1 2i 1 i 1 1 C 2i 1Ci 2
i 2 1 1Ci
1
1 C 2i 1 1Ci 1 2i 2 1 i
1 1Ci 1 i 2 1 C 2i 1 2i
1
i
2 1 C 2i 1Ci 1 2i 1
1Ci 1 2i 2 1 1 i 1 C 2i
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25 30 Years of Bulgarian Solitaire Brian Hopkins “Oh, you’re a mathematician! Let me show you something interesting.” But I’m trying to work on my talk, I thought. As the train sped along, the man sitting across from me looked eager. OK, let’s get this over with. “Here are fifteen playing cards. Arrange them into piles; as many piles as you like, each with as many cards as you like.” I made five piles with heights 3, 1, 4, 1, 6. The reference went unnoticed. “Now take one card from each pile to make a new pile.” The operation left me with piles of 3 1 D 2 cards, 4 1 D 3 cards, 6 1 D 5 cards, two empty piles from 1 1 D 0, and a new pile of 5 cards. I realized that the order of the piles does not matter, so for consistency I put them in nonincreasing order, 5; 5; 3; 2. “Now do it again and again. I know what will happen!” He looked away. Did he already think through the iterations? How long will this go? I was curious now. Here is the sequence of pile sizes: .6; 4; 3; 1; 1/
! !
.5; 5; 3; 2/ .5; 4; 2; 2; 2/
! !
.4; 4; 4; 2; 1/ .5; 4; 3; 1; 1; 1/
! !
.5; 3; 3; 3; 1/ .6; 4; 3; 2/
Oh no, that’s almost where I started. When will this end? But then, suddenly, it did end: .6; 4; 3; 2/
!
.5; 4; 3; 2; 1/
!
.5; 4; 3; 2; 1/ again.
“Hmm,” I said. “You ended with one pile of 5 cards, one of 4, one of 3, one of 2, and one of 1, didn’t you?” He looked at the cards. “Yes! That’s always what happens. Try again!” I started with a single pile of 15 cards. It took more moves, but I did indeed end up with the 5; 4; 3; 2; 1 pattern. Then I started with three piles of 5 cards. It took even more moves, but ended at the same fixed point. Three examples is not a proof, but the claim was now reasonable. Why would it always go there? How long could it take to reach this fixed point? What happens with other numbers of cards? “This is interesting,” I admitted.
Reprinted from The College Mathematics Journal, Vol. 43, No. 2 (Mar. 2012), pp. 135–140.
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This puzzle was popularized by Martin Gardner in 1983 [9] with the unusual name Bulgarian solitaire. In this article, we examine the earlier history of the puzzle including its name, summarize subsequent research, and consider a new twoplayer variation.
Before Gardner Around 1980, Konstantin Oskolkov of the Steklov Mathematical Institute in Moscow traveled by train to give a talk in Leningrad (now Saint Petersburg). A man on the train told him of the problem, although the details of the dialog above are fictional. Oskolkov shared this with his colleagues at the institute; reportedly when one number theorist heard about it “his face took a Satanic expression, he ran to his office, closed the door and did not come out until he solved the problem.” The puzzle reached Victor Gutenmacher, editor of the journal Kvant. The problem appeared there in 1980, in the context of stacks of books. Andrei Toom, a research scientist at Moscow State University, published a solution in 1981 [19]. The material was also included in a 1981 book on mathematics olympiads [20] whose authors include Gutenmacher and Toom. Later in 1980, Anatolii Alexeevich Karatsuba traveled from Steklov to the Institute of Mathematics of the Bulgarian Academy of Sciences in Sofia. After a lecture on approximation theory, he shared Oskolkov’s story and the puzzle. Again, it captured the attention of many mathematicians, and Milko Petkov published it in the “competition problems” section of a high school mathematics journal in 1980, in the context of heaps of balls. No student solved the puzzle, so the solution of Petkov’s institute colleague Borislav Bojanov was published in 1981 [4]. Gert Almkvist of Lund University was visiting Sofia while Karatsuba was there. When Almkvist returned to Sweden, he shared the puzzle with colleagues, including Henrik Eriksson of the KTH Royal Institute of Technology, who published a solution also in 1981 [7] with the name “Bulgarisk patiens,” in the context of piles of cards. Jørgen Brandt, finishing his masters at Aarhus University, somehow learned of the puzzle as well (without the name). “The problem appeared so pure, that I did not think much about where it came from,” he explained recently. His thorough solution, in the setting of integer partitions, was submitted to Proceedings of the American Mathematical Society in 1981 and published in 1982 [5]. Meanwhile, Eriksson traveled from Stockholm to California, where he called the puzzle Bulgarian solitaire. He recently explained, “The silly name is my invention, silly because it is neither Bulgarian nor a solitaire.” Donald Knuth started a fall 1982 programming and problemsolving seminar with Bulgarian solitaire [12]. Ron Graham passed it on to Gardner, who used Bulgarian solitaire as the culminating example of “tasks you cannot help finishing no matter how hard you try to block finishing them” [9]. The silly name has stuck.
Gardner and beyond A partition of n is a collection of positive integers D .1 ; : : : ; t / whose sum is n; as the order does not matter, we index the parts in nonincreasing order, i.e., 1 2
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t . Bulgarian solitaire can be considered as an operation on partitions, B./ D .t; 1 1; : : : ; t 1/ where there may be zeros to remove and the parts may need to be reordered. This can also be visualized through a graphic representation of partitions. The Ferrers diagram of a partition has a column of 1 dots followed by a column of 2 dots, etc. Figure 25.1 shows the Bulgarian solitaire operation on .6; 4; 3; 1; 1/, choosing to remove the bottom dot of each column, that is, the bottom row. 6
4
3
! 1
1
5
5
3
2
Figure 25.1. Example of the Bulgarian solitaire operation.
The number of partitions of n grows very quickly as n increases, but it is helpful to look at the effect of B on all partitions of n for small n. Figure 25.2 shows all partitions of 6 under the operation. 2211
 411
 33
 23
 313
 42
 321
6  6  51 16 214 Figure 25.2. Bulgarian solitaire on all partitions of 6; exponents denote repetition. Like the 15 card puzzle of the introduction, Bulgarian solitaire on partitions of 6 leads to a fixed point, the partition 3 D .3; 2; 1/. Both 6 and 15 are triangular numbers, i.e., can be written as Tk D 1 C C k D k.k C 1/=2. The effect of Bulgarian solitaire can be very different on other values of n. Exercise. Work out Bulgarian solitaire on the 22 partitions of 8. (For the solution, see [13, Figure 4].) The three 1981 solutions in European journals [4, 7, 19] use very similar ideas to show that, for n D Tk , there is indeed just the single fixed point k D .k; k 1; : : : ; 2; 1/. Each author argues that a ball / book / card cannot move to a longer diagonal in the configuration, and will eventually move into the shortest diagonal possible, filling any gaps and leaving a triangular shape. Brandt [5] and Toom [19] showed that for other values of n, the â€œalmost triangularâ€? partitions form cycles. For example, the partitions .4; 2; 2/ and .3; 3; 1; 1/ form a 2cycle in one connected component of the partitions of 8 from the exercise; see Figure 25.3. From this characterization, Brandt used PÂ´olya enumeration theory to determine the number of connected components. For instance, the partitions are in a single component exactly when n is within one of a triangular number. See also [11], which revisits and generalizes Toomâ€™s solution, and [1], which elaborates on Brandtâ€™s work. The partitions of Figure 25.3 illustrate another relationship. Reading the row lengths of .4; 2; 2/ rather than the column heights gives .3; 3; 1; 1/. Considered another way, reflect
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Part V. Further Puzzles and Games 4
2
2
!
3
3
1
1
Figure 25.3. A twocycle among the partitions of 8.
ing the Ferrers diagram for .3; 3; 1; 1/ across the diagonal line y D x gives .4; 2; 2/. These partitions are said to be conjugates. Gardnerâ€™s article [9] focuses on the fixed point result for triangular numbers, with a discussion of using 45 cards, and the analog of Figure 25.2 for partitions of 10. He mentions two other conjectures, which were both proven within a few years. First, for n D Tk , data suggest that the greatest possible distance to the fixed point is linear in n, specifically k.k 1/. Igusa [15] shows that the partition k D .k 1; k 1; k 2; k 3; : : : ; 3; 2; 1; 1/ is at distance k.k 1/ from k and that this distance is maximal; see also [3] and [8]. Second, look carefully at the sequence of partitions from 3 D .2; 2; 1; 1/ to 3 D .3; 2; 1/ in Figure 25.2: until the fixed point, the sequence consists of nested conjugate pairs, shown in Figure 25.4. Bentz [3] proves that this occurs along the â€œmain trunkâ€? of the diagram for all n D Tk , which distinguishes k among the partitions at maximal length from the fixed point (for k 4, there are several partitions at that distance). See also [2]. 2211
 411
 33
 23
 313
 42
 321
Figure 25.4. Conjugate pairs on the â€œmain trunkâ€? of Figure 25.2.
Gardner also mentioned â€œEdenâ€? partitions that have no predecessor under the operation, such as 2 D .2; 2; 1; 1/, .2; 1; 1; 1; 1/, and .1; 1; 1; 1; 1; 1/ in Figure 25.2. Recently, Hopkins and Sellers [14] determined a formula for the number of these partitions for general n that resembles a recurrence result of Euler. There remain many open questions about Bulgarian solitaire, several of which are surveyed in Hopkins and Jones [13]. Griggs and Ho [10] conjectured maximal lengths to a cycle partition for n Â¤ Tk . How many partitions are at given length from a cycle partition or fixed point? What is the relation between B and conjugation outside the â€œmain trunkâ€?? When there are multiple components, is there an easy way to tell if two partitions are in the same component? What can be said about component sizes? There are many questions about asymptotic and random behavior; some are addressed by Popov [18]. Bulgarian solitaire continues to arise in the teaching literature, such as Nicholson [16] and DorÂ´ee [6]. The operation has been rediscovered at least once, see [17]. Several variants have been considered in the literature, but there is not space here to discuss them.
A new game Figure 25.1 shows the effect of changing the bottom row of the Ferrers diagram of .6; 4; 3; 1; 1/ to a column. Suppose you could change any row to a column. Figure 25.5
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shows the effect of changing the length 2 row to a column. Under this operation, .6; 4; 3; 1; 1/ could be followed by one of three partitions: .5; 5; 3; 2/ by selecting the bottom row, .5; 3; 3; 2; 1; 1/ selecting the length 2 or a length 3 row, or .5; 4; 3; 1; 1; 1/ selecting a length 1 row. 6
4
3
5
! 1
1
3
3
2
1
1
Figure 25.5. Changing the length 2 row of .6; 4; 3; 1; 1/ to a column gives .5; 3; 3; 2; 1; 1/.
For a 2player game, we start with the singlepart partition .n/. In turn, each player chooses a single row to change into a column. The loser is the first player who creates a partition that has occurred before in play. The first player can only chose a row of length 1, giving the partition .n 1; 1/. The second player then can choose a row of length 1 or 2, making the partition .n 2; 1; 1/ or .n 2; 2/, respectively (for sufficiently large n). All possible moves for partitions of 6 are shown in Figure 25.6.  16


6
411
51
? 42
MB B ? B ?? B 3 31 321 33 B B B ? ? B ?? 3 4 2 21 2211 Figure 25.6. All possible moves in the twoplayer game on partitions of 6. Double arrows indicate where selecting either of two different row lengths produces the same partition.
Here are some open questions about this game. As a finite impartial twoplayer game of perfect information with no draws, either the first or second player has a winning strategy. Which player? What is the strategy? Do the answers depend on n? Notice that there is a Hamiltonian cycle in the directed graph of Figure 25.6. That is, there is a sequence of play where every partition arises before there is repeti
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Part V. Further Puzzles and Games tion. I conjecture that this occurs for all n. If so, is there a nice way to describe the corresponding sequence of moves?
Conclusion A mathematical recreation that reportedly arose from a conversation in a train has gone on to inspire several lines of research and student activities. This is due largely to Martin Gardner’s attention to it in his mathematics column from Scientific American. Bulgarian solitaire is just one example of the impact his work has had on all of us who enjoy mathematics. Acknowledgment The history of Bulgarian solitaire comes from personal communication with Borislav Bojanov, Jørgen Brandt, Vesselin Drensky, Victor Gutenmacher, Pencho Petrushev, Andrei Toom, and especially Henrik Eriksson. Recollections of events thirty years ago can be imperfect; hopefully the account provided here approaches what occurred. Aleksandar Nikolov kindly provided translations from Bulgarian. Suggestions from two anonymous referees improved the article. I would like to thank Eriksson, Suzanne Dor´ee, and Mizan Khan for their interest and help.
Bibliography [1] E. Akin and M. Davis, Bulgarian solitaire, Amer. Math. Monthly 92 (1985) 237–250; available at dx.doi.org/10.2307/2323643. [2] T. Bending, Bulgarian Solitaire, Eureka 50 (1990) 12–20. [3] H.J. Bentz, Proof of the Bulgarian solitaire conjectures, Ars Combin. 23 (1987) 151–170. [4] B. Bojanov, Problem solution 4, Obuchenieto po matematika (Mathematics Education) 24 (5) (1981) 59–60. [5] J. Brandt, Cycles of partitions, Proc. Amer. Math. Soc. 85 (1982) 483–486; available at dx.doi.org/10.1090/S00029939198206561295. [6] S. Dor´ee, Bulgarian Solitaire, Resources for Teaching Discrete Mathematics, B. Hopkins, ed., Mathematical Association of America, Washington DC, 2009, 83–92. [7] H. Eriksson, Bulgarisk patiens, Elementa 64 (4) (1981) 186–188. [8] G. Etienne, Tableaux de Young et solitaire bulgare. J. Combin. Theory Ser. A 58 (1991) 181– 197; available at dx.doi.org/10.1016/00973165(91)90059P. [9] M. Gardner, Mathematical Games: Tasks you cannot help finishing no matter how hard you try to block finishing them, Scientific American 249 (1983) 12–21; available at dx.doi.org/10.1038/scientificamerican088312. Also available as Bulgarian solitaire & other seemingly endless tasks, The Last Recreations, SpringerVerlag NY, 2007, 27– 43. [10] J. Griggs and C.C. Ho, The cycling of partitions and compositions under repeated shifts, Adv. in Appl. Math. 21 (1998) 205–227; available at dx.doi.org/10.1006/aama.1998.0597. [11] T. Hart, G. Khan, and M. Khan, Revisiting Toom’s proof of Bulgarian solitaire, Ann. Sci. Math. Qu., to appear. [12] J. Hobby and D. Knuth, A Programming and ProblemSolving Seminar, Stanford University Department of Computer Science report, 1983.
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[13] B. Hopkins and M. Jones, ShiftInduced Dynamical Systems on Partitions and Compositions, Electr. J. Comb. 13 (2006) R80 19 pp. [14] B. Hopkins and J. Sellers, Exact Enumeration of Garden of Eden Partitions, Integers 7 (2) (2007) A19. [15] K. Igusa, Solution of the Bulgarian solitaire conjecture, Math. Mag. 58 (1985) 259–271. [16] A. Nicholson, Bulgarian Solitaire, The Mathematics Teacher 83 (1996) 84–86. ´ Thierry, Dynamics of the Picking transformation on integer partitions, [17] H. D. Phan and E. Discrete Math. Theor. Comput. Sci. AB(DMCS) (2003) 43–56. [18] S. Popov, Random Bulgarian Solitaire, Random Struct. Alg. 27 (2005) 310–330; available at dx.doi.org/10.1002/rsa.20076. [19] A. Toom, Problem solution M655, Kvant (Quantum) 1981 (7) 28–30. [20] N. Vasilyev, V. Gutenmacher, J. Rabbot, and A. Toom, Zaochnye matematicheskie olimpiady (Mathematical olympiads by correspondence), Nauka, Moscow, 1981.
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26 Congo Bongo HsinPo Wang An expedition into Congo uncovered a treasure chest in the shape of a regular octagon. At each corner was a bongo drum. A scroll attached to the chest, written in French, explained that there was a genie inside each bongo drum. A genie is either standing upright or doing a handstand. One may strike a number of bongo drums at the same time. When a bongo drum is struck, the genie inside will change its posture from right side up to upside down, or vice versa. The treasure chest will open if and only if all genies are right side up, or all are upside down. However, each time some bongo drums are hit the treasure chest will spin rapidly on its vertical axis. As the bongo drums are all identical in appearance, after the rotation it is impossible to tell which of them had just been hit. Unfortunately, the scroll did not record the exact procedure by which the treasure chest might be opened. However, it mentioned that such a procedure had been documented. This document was considered so valuable that it was put inside the treasure chest for safekeeping. It was a lot safer than was originally thought. The sponsor of the expedition was definitely not pleased with the current state of affairs, so she hired a team of mathematicians to try to open it. The task was seemingly hopeless, and many gave up, until Dr. Jacob Ecco arrived with his sidekick, Professor Justin Scarlet. “My dear professor,” said Dr. Ecco, “we must approach this challenge systematically. What is the most basic principle in problem solving?” “Downsizing,” replied the professor, who was well versed with the methods of Dr. Ecco. “Suppose there is only one bongo drum. Then the treasure chest will open automatically.” “What if there are two bongo drums?” asked Dr. Ecco. “Well, if the treasure chest is not already open, hitting either of the bongo drums will do it.” “Excellent! As usual, my dear friend, you have provided me with the inspiration that leads to the solution.” “How?” cried Professor Scarlet. “I have made but the most trivial observation.” “Nevertheless, a solution to the case with two bongo drums leads to a solution to the case with four bongo drums, which in turn leads to a solution to the case with eight bongo drums,” Dr. Ecco proclaimed. “I can smell that you are going to use induction, but l still do not see the connection from one case to the next,” replied Professor Scarlet. Reprinted from Math Horizons, Vol. 18, No. 1 (Sept. 2010), pp. 18–21.
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“Be patient. Obviously, we will be treating pairs of bongo drums as a single bongo drum. How can we identify pairs of bongo drums? Adjacency is unsatisfactory because after the treasure chest spins around, it is impossible to tell whether a bongo drum is to be paired with its neighbor to the left or its neighbor to the right.” “The obvious solution is to identify opposite pairs of bongo drums,” pronounced Professor Scarlet. “What insight!” said Dr. Ecco with a slight smile. “This is fine if the genies in each opposite pair of bongo drums are both right side up or both upside down. But there is no guarantee that they start off like that,” his companion pointed out. “True, but there is a way to make them end up that way. Let 0 or 1 indicate whether a genie is right side up or upside down, and consider a foursided chest. If we are really lucky, the initial state may already be (0,0,0,0) or (1,1,1,1).” “You may be good enough to be that lucky, but that never works with me,” complained Professor Scarlet. “I am not counting on luck. I am just singling out the most favorable scenario. We may regard (0,0,0,0) and (1,1,1,1) as the same state. Call it state 0. It is also referred to as an absorbing state, in that once we enter it we do not leave (for the problem is then solved). How many other states are there?” asked Dr. Ecco. Professor Scarlet paused. “Well, we have three other states, namely, (0,1,1,1), (0,0,1,1), and (0,0,0,1).” “You are correct as far as the number of other states is concerned, but wrong about their composition,” Dr. Ecco said. “By symmetry, we may consider (0,1,1,1) and (0,0,0,1) as the same state. We call it state 3. Your (0,0,1,1) is state 2, which is not the same as (0,1,0,1). This I call state 1.” “Yes!” exclaimed Professor Scarlet excitedly. “In state 1, the genies in each opposite pair of bongo drums are both right side up or both upside down. In the case with two bongo drums, we hit one of them. So here we hit an opposite pair of bongo drums, and the treasure chest will open.” “We will call hitting an opposite pair of bongo drums operation A, and as you keenly observed, performing this operation from state 1 will unlock the chest. What happens when I perform operation A if we are in state 2 or state 3?” “Let me see,” Professor Scarlet said. “Ah, we will remain in the same state as before.” “How can I move from state 2 to state 0 or 1?” Dr. Ecco asked. “Well, we may hit an adjacent pair of bongo drums. I suppose this will be operation B. If we hit both zeros, or both ones, we will be in state 0 immediately. If we hit one 0 and one 1, we will be in state 1.” “Indeed, you can check that state 3 is left unchanged by operation B, just as it was unchanged by operation A,” Dr. Ecco agreed. “The rest is easy now. We perform operation C by striking just one bongo drum. This will change state 3 into state 0, state 1, or state 2,” Professor Scarlet said triumphantly. “Not so fast,” Dr. Ecco cauntioned. “After operation B, we may be in state 1, and if we rush into operation C now, state 1 will become state 3, and we will be going in circles.” “I was hasty,” admitted Professor Scarlet. “We must perform operation A once more to clear state 1 before taking care of state 3. After operation C, we will clear the lower states with the sequence ABA again. So the overall procedure for the foursided chest is ABACABA.”
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“Excellent. Here, l have drawn a state transition diagram for you,” Dr. Ecco said. State 0
State 1
State 2
B
A
0 0
0 0 1
1
1
0
1 0
1
0 0
0
0 A
State 3
1 1
1 1
1
0
1
1 0
C
A/B
Here is the same drawing, with an emphasis on the structure of the transition operations: State 0 C
A
C
State 1
A
State 3
B C
B
B
State 2
A
“That explains everything clearly!” said the professor. “Now I am confident we are onto something that will crack the eightsided treasure chest. Let’s see. We wish to treat each pair of opposing bongo drums as a single entity. As before, the simplest states are those in which the genies in each opposite pair of bongo drums are both right side up or both upside down. Here is my drawing of this part of the state transition diagram. It is really the same as yours, except that operation A means hitting every other pair of opposing bongo drums; operation B means hitting any two adjacent pairs of opposing bongo drums; operation C means hitting any pair of opposing bongo drums. By performing the sequence ABACABA when starting from any such simple state, the treasure chest will open.” “Excellent, my dear professor,” exclaimed Dr. Ecco. “These states together form an expanded absorbing state in the overall diagram below. It is the box marked 4, where the box marked m contains all states with m matching opposite pairs, for 0 D m D 4.”
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B
0
0 0
0 0 0
0
1
1
1 1
0
1
1
0 0 1
A 1 0 1
1 1
0
A 0
1 1
1 1 0
0
1
0
0 1
0
1 0 0
0
1
1
1 0
0 1 1
C
1
“This diagram looks suspiciously like the ones before, except that states 4 and 0 are no longer equivalent. Why do you have a state 2 and a state 20 ? And what are the operations D, E, F, and G?” 4
D
G
D
0 G E E
E
F
1 or 3
F G 2
G
F
F
D 2¢ E
D
“The states with 2 matching pairs are classified according to whether those matching pairs are alternating or adjacent,” Dr. Ecco said. “The former are grouped under state 2 while the latter are grouped under state 20 . Operation D hits any four adjacent bongo drums. Operation E hits any two bongo drums separated by one other drum. Operation F hits any two adjacent bongo drums. Operation G hits any one bongo drum.” “Let me see. If we denote the sequence ABACABA by X, then the sequence that will open the treasure chest is XDXEXDXFXDXEXDXGXDXEXDXFXDXEXDX,” summed up Professor Scarlet. “We keep repeating X so that whenever we enter state 4, we will not return to another state. Whatever the initial state of the treasure chest may be, it will open by the end of this sequence.” The sponsor of the expedition was delighted with their analysis. She hired a team of Congo natives to strike the bongo drums in the prescribed fashion. Lo and behold, the treasure chest opened right before her eyes. Unable to contain her excitement, she reached inside, but found only another scroll. It was written in Kituba, but the most prominent line read something like XDXEXDXFXDXEXDXGXDXEXDXFXDXEXDX.
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Later, back in his New York apartment, Dr. Ecco said to Professor Scarlet, “That was hilarious, wasn’t it? By the way, have you thought of the obvious question?” “Yes, l have,” replied the professor. “For what number of bongo drums can such a treasure chest be opened? From our analysis, the general procedure when the number is a power of two is clear. We treat each opposing pair as a single entity, thereby reducing to the preceding case. Then we progressively move all states into the expanded absorbing state. But what happens if the number of drums is not a power of two?” “Then unlocking the chest is not always possible,” Dr. Ecco replied. “For in this case the number of bongo drums has an odd prime factor p. Suppose the pagan god Hemba is having fun with us. He will choose p evenly spaced bongo drums and make sure that the genies inside are not all right side up and not all upside down. Now ignore all other bongo drums except these p. The two types of bongo drums are not equal in number since p is odd. In order for us to succeed, we must strike precisely the bongo drums of one type. But Hemba will spin the treasure chest so that the bongo drums we plan to hit include at least one from the opposite group. This way, he can keep us from opening the treasure chest forever.” See problem 249 in the Playground below for more fun with these treasure chests.
Further Reading This problem was posed in the Senior ALevel paper in the 2009 Fall Round of the International Mathematics Tournament of the Towns. It is related to a problem posed in Martin Gardner’s famous “Mathematical Games” column in Scientific American (February 1979), which is reproduced in his anthology Fractal Music, Hypercards and More Mathematical Recreations (W.H. Freeman, 1992). Gardner’s version of the problem is also treated in three other sources: Ted Lewis and Steve Willard, The rotating table, Math. Mag., 53 (1980): 174–179. William Laaser and Lyle Ramshaw, Probing the rotating table, in The Mathematical Gardner, edited by David Klarner, (Wadsworth, 1981) 288–307. Albert Stanger, Variations on the rotating table problem, J. Recreat. Math., 19 (1987): 307–308 and 20 (1988): 312–314. Jacob Ecco and Justin Scarlet are fictitional characters created by Dennis Shasha of the Courant Institute, New York. He is a leading creator of mathematical puzzles. Shasha has written a number of puzzle books featuring these clever detectives. A good place to start is his book The Puzzling Adventures of Doctor Ecco (Dover, 1998).
Playground Problem 249 The end of the article explains why no sequence of drum strikes is guaranteed to open an ndrum chest unless n is a power of 2. However, for other n it may be possible to have a pretty good chance of opening the chest during a certain sequence of drum strikes, assuming that the spinning between strikes is truly random. This leads us to Problem 249, Congo Bongo not so Wrongo: What is the smallest number of drum strikes needed on a threedrum treasure chest to give you at least a 90 percent chance of opening it? (As a separate further challenge, which may require computer assistance, try to answer the same question for a fivedrum chest.)
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Problem 249 Solution The Math Club from the University of Minnesota Duluth submitted a correct solution that 6 drum strikes are required; there were two incorrect submissions. First, the club members realized that there are only two possible states for the chest: either all three of the genies have the same orientation (state A, which causes the chest to open) or just two of the genies have the same orientation (state B). Then, they realized that striking either one or two drums will cause a chest in state B to move to state A with probability 1/3 and stay in state B with probability 2/3. Assuming that the chest is initially in state B (otherwise it would open immediately!), the probability that the chest opens in k or fewer strikes is 1=3 C .2=3/.1=3/ C .2=3/2 .1=3/ C C .2=3/k
1
.1=3/ D 1
.2=3/k
which first becomes greater than 0.9 when k D 6.
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27 Sam Loyd’s Courier Problem with Diophantus, Pythagoras, and Martin Gardner Owen O’Shea In his classic collection Cyclopedia of Puzzles, published in 1914, Sam Loyd has two versions of the Courier Problem ([2, p. 315]): For the reason that many communications are being received relating to a very ancient problem, the authorship of which has been incorrectly accredited to me, occasion is taken to present the original version which has led to considerable discussion. It has been reproduced, in many forms, generally accompanied by an absurd statement regarding the impossibility of solving it, which produced letters of inquiry as well as correct answers from some, who, under the misapprehension of having mastered a hitherto unsolved problem, desire to have the same published. It is a simple and pretty problem which yields readily to ordinary methods, and can be solved by experimental analysis upon the plan generally adopted by puzzlists. The trouble is that the terms of the problem are seldom given correctly and are not generally understood, for which reason,. . . , we will first look at the ancient version which appears in the oldest mathematical works: A courier starting from the rear of a moving army, fifty miles long, dashes forward and delivers a dispatch to the front and returns to his position in the rear, during the exact time it required the entire army to advance just fifty miles. How far did the courier have to travel in delivering the dispatch, and returning to his previous position in the rear of the army? A better puzzle is created by the following extension of the theme given as problem No. 2: If a square army, fifty miles long by fifty miles wide, advances fifty miles while a courier makes the complete circuit of the army and returns to the starting point in the rear, how far does the courier have to travel? Reprinted from The College Mathematics Journal, Vol. 39, No. 5 (Nov. 2008), pp. 387–391.
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The problem appeared in print long before Loyd published it. The first known appearance was in A Companion to the Gentleman’s Diary of 1798, and a year later it was published in The Gentleman’s Mathematics Companion. It also appeared much later in William R. Ransom’s One Hundred Mathematical Curiosities in 1955 [3]. Loyd describes the puzzle as an ancient problem and gives the answers to both parts. However, he does not indicate how to tackle the problems. In his book More Mathematical Puzzles of Sam Loyd [1, p. 103], Martin Gardner discusses both problems and their solutions. The answer to the first question is that the courier p travels 1 C 2 times the length of the army, or about 120.71 miles for a fiftymilelong army. Here is Gardner’s solution: For convenience, assume that the length of the army is 1 unit, and also assume that the time it takes the army to march its length is 1. Thus its speed is also 1. Let x denote the total distance traveled by the courier, so that is also his speed. On his forward trip, the courier’s speed relative to the army is thus x 1, and on the return trip it is x C 1. Since each trip is a distance of 1 relative to the moving army and the courier’s total journey is completed in unit time, it follows that 1 x
1
C
1 D 1; xC1
p or equivalently, x 2x 1 D 0. Consequently, x D 1 C 2. This means that in Loyd’s p problem the courier travels 50 1 C 2 miles, or, as asserted earlier, about 120.71 miles. Gardner’s solution to the second question is similar. As before, let the length of the army and the time it takes to travel its length be 1 unit, so its speed is also 1. Additionally, we let x be the total distance traveled by the courier and also his speed. On the forward trip, the courier’s speed relative to the army is x 1, on the return trip, x C 1, and on each p of the diagonal portions, x 2 1. Since each trip has length 1 relative to the army and the total journey is completed in unit time, it follows that 2
1 x
1
C
1 2 Cp xC1 x2
1
D 1:
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27. Sam Loyd’s Courier Problem with Diophantus, Pythagoras, and Martin Gardner 203 This time a fourthdegree equation results: x 4 4x 3 2x 2 C 4x C 5 D 0. The only solution that fits the conditions of the problem is approximately 4.18. Thus, in traversing the entire army, the courier now travels approximately 4.18 times as far as the army, or about 209 miles. In the spirit of classroom exercises, it would be nice to have versions of the courier problems with integer solutions; that is, Diophantine problems. In exploring this possibility, we discovered that certain Pythagorean triples, such as .3; 4; 5/, .5; 12; 13/, and .7; 24; 25/, that is, those in which the two largest numbers differ by 1, can be used to generate such problems.
Army in single file Assume that an army 60 miles long is marching at a constant rate. A courier, also going at a constant rate, rides from the rear of the army up to the front, delivers a message, and returns, arriving at the rear just as the army has gone 45 miles. How far did the courier go? Following Gardner, we again let the speed of the army (but not its length) be 1, and we let x be the speed of the courier. As before, the speed of the courier relative to the army is x 1 on the outward journey and x C 1 on the return journey, so the total time that 60 . This must equal 45, the time it takes the army to go the given he travels is x601 C xC1 distance. Gardner’s equation is therefore 60 x
1
C
60 D 45: xC1
This reduces to 3x 2 8x 3 D 0, for which the only realistic solution is x D 3. Thus, the courier travels three times as fast as the army, and therefore rides 135 miles. We thus have a Diophantine version of the courier problem. Consider now a general singlefile problem, in which the army is ` miles long and travels d miles during the time that the courier makes the trip. Gardner’s equation becomes ` x Setting ˛ D
` d
1
C
` D d: xC1
, we find that this reduces to x2
2˛x
1 D 0;
p which has x D ˛ C ˛ 2 C 1 as its only positive solution. This is where Pythagorean triples come to the fore. If .a; b; b C 1/ is such a triple and ˛ D ab , then x D a is a solution to Gardner’s equation. Our example used the triple .3; 4; 5/ with distances multiplied by 15. Since in these triples b D 21 .a2 1/ and c D 12 .a2 C 1/, we can pose a singlefile courier problem with an integer solution for any odd positive integer a and any constant k: Problem 1 (single file army) An army 21 .a2 1/k miles long travels a distance of ak miles during the time that the courier makes his journey. How far does the courier travel?
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Army in square formation We can do something similar for the second of Loyd’s problems. Here is an example (also based on the .3; 4; 5/ triple): An army in the formation of a square 36 miles on a side is advancing at a constant rate. A courier, also moving at a steady rate, travels all the way around the army just as it advances 256 miles. How far does the courier go? Again we take the speed of the army to be 1 and that of the courier x, so Gardner’s equation becomes 36 36 2 36 C Cp D 256: x 1 xC1 x2 1 In polynomial form, this is 81x 4
576x 3
162x 3 C 576x C 1105 D 0;
but all we care about is that x D 45 is a root (this is easy to check in the original equation, not so easy to discover). The same Pythagorean triples .a; b; c/ as before (with a an odd integer, b D 21 .a2 1/, and c D 12 .a2 C 1/) all generate squarearmy courier problems with Diophantine solutions: Problem 2 (square army) An army b 2 miles on a side advances 2a .b C c/ miles during the time that a courier completes his circuit. How far does the courier go? Gardner’s equation for this army can be written as x 1 2 2b Cp D 2a .b C c/ ; x2 1 x2 1
which has x D ac as a solution. That is, the courier travels ac times as fast as the army and hence goes 2c.b C c/ miles. Note that the roles of a and b can be interchanged here; that is, the army could be b 2 on a side and travel 2b.a C c/ miles while the courier completes his circuit.
Army in rectangular formation Of course, armies do not always march in either singlefile or square formation; sometimes they are proper rectangles. Suppose that an army is ` miles long and w miles wide and that it goes a distance d in the time that the courier transverses it, going at a speed x times that of the army. Gardner’s equation for this situation is `
` 2w Cp D d: x 1 xC1 x2 1 For example, assume that the army is 40 miles long and 30 miles wide, and that it travels 120 miles while the courier completes his circuit. Then the courier rides 5=3 as fast as the army travels, and hence goes 200 miles. Thus, Pythagorean triples of the given type can be used in this equation to generate Diophantine solutions for a rectangular army in two ways, with .a; b; c/ again a Pythagorean triple with c b D 1. C
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27. Sam Loyd’s Courier Problem with Diophantus, Pythagoras, and Martin Gardner 205 Problem 3 (rectangular army) An army’s length is a multiple of 21 b 2 and its width a multiple of 12 b. It advances 2a .b C c/ miles during the time that courier completes his circuit. How far does the courier travel? Acknowledgment The author thanks the referees and the editors for helpful suggestions.
Bibliography [1] Martin Gardner, More Mathematical Puzzles of Sam Loyd, Dover, 2007. [2] Sam Loyd, Cyclopedia of Puzzles, Lamb Publishing, 1914. www.mathpuzzle.com/loyd/ [3] William R. Ransom, One Hundred Mathematical Curiosities, J. Weston Walch, 1955.
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28 Retrolife and The Pawns Neighbors Yossi Elran Martin Gardner created no less than a revolution in the popularization of mathematics through his many books, and, in particular, his Scientific American columns. Some of his most important columns, which spurred not only popular interest but also a wealth of innovative research and even practical applications, were his columns describing John Conway’s “Game of Life” [5, 6]. Gardner himself said [7], Probably my most famous column was the one in which I introduced Conway’s game of Life. Conway had no idea, when he showed it to me, that it was going to take off the way it did. He came out on a visit, and he asked me if I had a Go board. I did have one, and we played Life on the Go board. He had about 50 other things to talk about besides that. I thought that Life was wonderful—a fascinating computer game. When I did the first column on Life, it really took off. There was even an article in Time magazine about it. “It really took off” is an understatement.
The Game of Life The “Game of Life” was invented in 1970 by British mathematician John Conway. It is best described as the archetype of cellular automata. The player (there is only one) places an initial distribution of checkers on an infinite checkerboard; one checker per square. The squares are called cells and the initial distribution is called a population. This terminology reflects one of the original objects of the game—to simulate biological evolution. Cells are live if they contain a checker and dead if empty. The population evolves in time in discrete time steps known as generations, according to a simple set of rules, applied simultaneously. The rules determine the fate of each cell according to the number of live cells in its eight neighboring cells. If a cell is empty and exactly three of its eight neighboring cells are alive, the cell itself will be born—a checker is placed there. If a cell is alive and two or three of its eight neighboring cells are also alive, nothing changes—the cell survives. A live cell dies (i.e., the checker is removed from the board) in two situations. If a cell is alive and four or more of its eight neighboring cells are also alive, the cell dies (from overcrowding). A live cell also dies from isolation, if it has less than two live neighbors. Figure 28.1 shows three consecutive generations of Life. Reprinted from The College Mathematics Journal, Vol. 43, No. 2 (Mar. 2012), pp. 147–151.
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Figure 28.1. Three generations of the Game of Life.
Although these rules are simple and straightforward, iterating Life for many generations can result in a surprising variety of intriguing outcomes. A population can die out. Another possibility is that the population evolves into a stable, constant form which doesn’t change anymore. A 2by2 block of live cells is an example of such a form. Populations may also cycle, regenerating themselves after a certain time. Such populations come in forms known as oscillators and gliders. Three isolated live cells in a row are an example of a simple oscillator. Populations can also grow indefinitely. In 1970, a team from the Artificial Intelligence Project at M.I.T. won a 50 dollar prize from Conway for proving that an initial finite population can grow without limit. Such populations are generated by shapes variously known as glider guns, agars and puffertrains. Martin Gardner wrote: “You will find the population constantly undergoing unusual, sometimes beautiful and always unexpected change.” Stephen Silver’s lexicon [10] lists over 700 terms applied to various forms of Life. Much work on Life was done as a result of Martin Gardner’s original publication in 1970, in particular by Bill Gosper from the original M.I.T. group, Robert Wainwright (who managed a newsletter called “Lifeline”) and other “Lifeenthusiasts” (as Gardner calls them). The three articles that Martin Gardner wrote in Scientific American and thereafter reprinted in Wheels, Life and Other Mathematical Amusements [6] have spurred so much research that it is impossible to summarize it all without doing injustice to some people. Life is considered the prototypical computer simulation. Among other things, it is a universal Turing machine [3]. Paul Rendell proved the universality of Life by showing a direct simulation of counter machines [9], and Dean Hickerson created an initial population that, when evolved, generates the Prime Numbers [8]. Variations on Life [2] have also been developed. Some interesting games, where only the rules are changed, are Highlife, which has the same rules as Life and in addition six live neighbors also cause a birth, and Seeds where a birth is caused when a cell has exactly 2 live neighbors and in all other cases the cell dies. The twodimensional Moore neighborhood grid can also be changed, resulting in games played on triangular and hexagonal boards. Andrew Adamatzky recently commemorated Life’s 40th birthday by publishing a unique collection of works that provide a comprehensive and uptodate study of Life, with contributions from many renowned mathematicians and computer scientists [1].
Retrolife An important aspect of Life is that applying the rules to any given generation results in a unique outcome. The reverse, however, is not true. In fact, there are an infinite number of possible population distributions (hereafter referred to as states) in the preceding gen
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209
eration. Finding a particular state from the variety of possible states is the essence of the Retrolife puzzles [4]. In its original form, the task in Retrolife is to find predecessors of a given initial state that comply with some extra constraints. Many constraints can be chosen, each generating a slightly different puzzle. One might, for instance, demand that all the cells in the initial state survive, that is, cells live in the puzzle (the initial state in Life) are also live in the solution (the previous generation). Another constraint is to demand the opposite, that all the live cells in the initial state are born. Figure 28.2 shows three Retrolife puzzles and their solution using this last constraint.
Figure 28.2. Three Retrolife puzzles. Top row: puzzles. Bottom row: possible solutions.
To explain the rules of Retrolife, one first needs to explain Life and then suggest the reverse problem. Needless to say, this is rather cumbersome. Instead, we therefore suggest the following rephrasing of Retrolife.
The Pawns Neighbors A player is given an infinite chessboard, any number of pawns, and some tokens. Each square on the board has eight neighboring squares (in the directions: north, south, west, east and on the diagonals). The pawns are placed in some pattern on the board. The challenge: surround each pawn (and no other square on the board) with exactly three tokens so that no token itself has either two or three tokens as neighbors. A token may have less than two tokens as neighbors, or more than three. Note also that no empty square in the board is allowed to be surrounded with three tokens; however it may be surrounded by more or less. Figure 28.3 shows one setup and a possible solution. The Pawns Neighbors is really Retrolife in disguise, however it can be described in a straightforward manner and solved without any prior knowledge of Life or Retrolife.
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Figure 28.3. A Pawns Neighbors puzzle and a solution.
Variations on The Pawns Neighbors The Pawns Neighbors is isomorphic to Retrolife with the constraint that all the cells in the initial state are born. We can change the rules to create an isomorphic Retrolife problem with the constraint that all the cells in the initial state survive. In this case, we have to surround each pawn with tokens so that the pawns have either two or three neighbors, either pawns or tokens, but no token has two or three neighbors (again, neighbors can be either pawns or tokens). No empty square on the board is allowed to have three neighbors (pawns or tokens). Many other variations are possible. The computational complexity of Pawn Neighbor games is an open question. We leave the reader the following Pawns Neighbors problem to solve. Surround each pawn (and no other square on the board) with exactly three tokens. No token can have either 2 or 3 neighbors. The solution is on p. 211.
Figure 28.4. A Pawns Neighbors puzzle.
Martin Gardner—in memoriam Martin Gardner was particularly fond of Life and related problems. Shortly before he died, he wrote to me enthusiastically about Retrolife and offered to add a piece about them in one of his upcoming books. It’s a pity I can’t share with him this article. I’m sure he would have enjoyed it. I miss Martin a lot, and for all the many books he wrote, I still yearn for more.
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211
Bibliography [1] A. Adamatzky (ed.), Game of Life Cellular Automata, SpringerVerlag, London, 2002. [2] B. Bays, The Game of Life in Nonsquare Environments, in Game of Life Cellular Automata, A. Adamatzky (ed.), SpringerVerlag, London, 2002, 319–330. [3] E. R. Berlekamp, J. H. Conway and R. K. Guy III, Winning Ways (for your Mathematical Plays), Volume 2, A. K. Peters, Wellesley, MA, 2003. [4] Y. Elran, Retrolife, in Homage to a Pied Puzzler, E. Demaine, M. Demaine and T. Rodgers (ed.), A. K. Peters, Wellesley, MA, 2008, 129–136. [5] M. Gardner, Scientific American 223 (1970) 120–123. [6] ——, Wheels, Life and Other Mathematical Amusements, W. H. Freeman, New York, 1983. [7] ——, “Mathematical Games” and Beyond: Part II of an Interview of Martin Gardner, College Math J. 36 (2005) 301–314. [8] C. Hickerson, Dean Hickerson’s Game of Life Page; available at radicaleye.com/DRH/ life.html [9] P. Rendell, Turing Universality of the Game of Life, in Collision Based Computing, A. Adamatzky (ed.), SpringerVerlag, London, 2002, 513–540. [10] S. Silver, The Life Lexicon; available at www.argentum.freeserve.co.uk/lex.htm
Solution to the problem on p. 210.
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29 R ATWYT Aviezri S. Fraenkel In memory of Martin Gardner, who was and remains enchantingly influential and inspiring.
W YTHOFF In 1907, the Dutch mathematician, Willem Abraham Wythoff [13] invented this game, later vividly explained by Martin Gardner in [7]. W YTHOFF is played on a pair of nonnegative integers, .M; N /. A move consists of either (i) subtracting any positive integer from precisely one of M or N such that the result remains nonnegative, or (ii) subtracting the same positive integer from both M and N such that the results remain nonnegative. The first player unable to move loses. Given the position .3; 3/, say, the next player wins in a single move: .3; 3/ ! .0; 0/. The position .3; 3/ is called an N position, because the Next player wins. If M D N D 0, the next player loses, and the previous player, the one who moved to .0; 0/, wins. Thus .0; 0/ is a P position, because the Previous player wins. If M > 0, it is easy to see that .0; M / and .M; M / are N positions, since the next player can win in one move. On the other hand, .1; 2/ is a P position because all its followers—positions reached in one move—are N positions. The first few P positions are listed in Table 29.1. Note that every N position has at least one P follower, but all followers of a P position are N positions. From an N position, in order to win, a player must move to a P position. Further, the P  and N positions partition the set of all game positions: every game position is either a P position or an N position but never both. Table 29.1. The first few P positions .An ; Bn / for Wythoff’s game. n
0 1 2 3
4
5
6
7
8
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
An 0 1 3 4
6
8
9 11 12 14 16 17 19 21 22 24 25 27 29 30 32 33 35 37 38 40 42 43
Bn 0 2 5 7 10 13 15 18 20 23 26 28 31 34 36 39 41 44 47 49 52 54 57 60 62 65 68 70
The sequences, An and Bn in Table 29.1, each strictly increasing, have remarkable properties. Note that Bn D An C n for all n. But how is An computed? A study of the table reveals that for n 1, An is the smallest positive integer not yet appearing in the sequences. Thus, the next entries in Table 29.1 are A28 D 45, B28 D 73. It follows that the Reprinted from The College Mathematics Journal, Vol. 43, No. 2 (Mar. 2012), pp. 160–164.
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sequences are complementary: every positive integer appears precisely once in these two, recursively defined sequences. Can a takeaway game, such as W YTHOFF, be played on the rational numbers, rather than the integers? We present such a game here.
A rational takeaway game Given a rational number p=q in lowest terms, a step is defined by p q p ! ; q q if p=q 1, otherwise
p p ! : q q p
The game R ATWYT is played on a pair of reduced rational numbers .p1 =q1 , p2 =q2 /. A move consists of either (i) doing any positive number of steps to precisely one of the rationals, or (ii) doing the same number of steps to both. The first player unable to play (because both numerators are 0) loses. For example, from .3=5; 2=3/, suppose that Alice moves to .3=2; 2=3/. Then Bob can do three steps to both rationals: .3=2; 2=3/ ! .1=2; 2=1/ ! .1=1; 1=1/ ! .0=1; 0=1/, and thereby win. Could Alice have made a better initial move? Can Alice win? Suppose she does two steps to each of the initial rationals: .3=5; 2=3/ ! .3=2; 2=1/ ! .1=2; 1=1/. If Bob then moves to .1=1; 1=1/, Alice can move to .0=1; 0=1/ D .0; 0/, winning. We leave it to the reader to verify that in the three remaining possible moves for Bob, Alice can also win in one move. Thus .3=5; 2=3/ is an N position in R ATWYT. Is there a nice winning strategy for R ATWYT? If so, what is it? Continuing with our example, we expand 3=5 into a continued fraction: 3=5 D 0 C
1 1C
1
;
1 1C 2
or 3=5 D Œ0; 1; 1; 2, to use the short notation for continued fractions. Similarly, 2=3 D Œ0; 1; 2. The sum of the partial quotients: 0 C 1 C 1 C 2 D 4 is called the integer induced by the corresponding rational. We claim that playing R ATWYT on .3=5; 2=3/ is equivalent to playing W YTHOFF on their induced integers, .3; 4/! Supposing this established, then, because .3; 4/ is not in Table 1, it is not a P position, rather an N position, in W YTHOFF. Hence .3=5; 2=3/ is an N position in R ATWYT. So Alice can, in fact, win. Let us reexamine her initial move: .3=5; 2=3/ ! .3=2; 2=3/. Now 3=2 D Œ1; 2 3 (where denotes the inducing process), so .3=2; 2=3/ .3; 3/. Alas, .3; 3/ is another N position in W YTHOFF, so Alice missed her opportunity to move to a P position! Had she made the twostep move .3=5; 2=3/ ! .1=2; 1=1/ D .Œ0; 2; Œ1/ .1; 2/, she could have won, since .1; 2/ is a P position in W YTHOFF.
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The CWtree To substantiate our claim, that playing R ATWYT on .p1 =q1 ; p2 =q2 / is equivalent to playing W YTHOFF on their induced integers, we resort to a method of counting the rationals, described in the short, elegant, and influential paper of Calkin and Wilf [4]. The Calkin Wilf tree, CWtree for short, is a binary tree whose nodes are all the nonnegative rational numbers without repetition! The root is the fraction 0=1. The rest of the tree is described inductively by the rule that each vertex i=j has at most two children: a right child .i C j /=j ; and, if i > 0, a left child i=.i C j /. The first five levels (levels 0 4) are drawn in Figure 29.1. Note that at each step in R ATWYT we move towards the root of the tree by one level, so we are guaranteed to eventually reach 1=1 on level 1, and then the root on level 0.
0/1 1/1 1/2
2/1
1/3 1/4
3/2 4/3
3/5
2/3 5/2
2/5
3/1 5/3
3/4
4/1
Figure 29.1. The first five levels of the CWtree of reduced fractions.
Here are some key properties of the CWtree. 1. The numerator and denominator at each vertex are relatively prime: proved by induction on the level of the tree. 2. The induced integer of every rational on level k of the tree is k. Indeed, an element r has the right child r C 1. Incrementing by 1 means incrementing the first term of the continued fraction, hence incrementing the induced integer. The left child is 1=.1 C 1=r /. Since taking the reciprocal of a continued fraction amounts to either prefixing or removing a leading 0, neither of which changes the induced integer [8], this also increments the induced integer by one. (See Figure 29.2.) 3. Every rational number p=q in reduced form occurs precisely once in the CWtree, namely on level k, where k is the integer induced by p=q. There is a unique path from p=q to the root whose length is precisely k—the same as for every element on level k of the tree, in particular of the rational k=1 D k. Hence playing on the rational p=q is equivalent to playing on its induced integer k, as we set out to show.
More games In [13], Wythoff also discovered that the sequences An , Bn can be described p explicitly: An D bnc, Bn D bn 2 c, where bxc is the floor function; and D .1 C 5/=2 is the
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[0] [1] [0,2] [0,3] [0,4]
[2] [1,2]
[1,3]
[0,1,1,2]
[0,1,2] [2,2]
[0,2,2]
[1,1,2]
[3] [0,1,3]
[4]
Figure 29.2. The first five levels of the CWtree of continued fractions.
golden ratio. This observation leads to a winning strategy for Wythoff that is polynomial time in the succinct input size log.xy/ of any given game position .x; y/. The preceding section explained how playing R ATWYT on the rationals is equivalent to playing on their induced integers. The same holds for any takeaway game. Playing on the integers means restricting travel along the right edge of the CWtree. Playing on the integer k is equivalent to playing on any of its 2k 1 rational siblings on level k of the tree .k 1/. Here are a few sample takeaway games, played on reduced rationals. Keep in mind that the number of steps performed must always preserve nonnegativity, and that the first player unable to move loses. Game I. Let t be a fixed positive integer. A position is a pair .p1 =q1 , p2 =q2/ of rationals. There are two types of moves: (i) do any positive number of steps on precisely one of the rationals, or (ii) do k > 0 steps on one of the rationals and ` > 0 on the other, such that jk `j < t. This game, if played on the integers induced by p1 =q1, p2 =q2, is a generalization of W YTHOFF (the case t D 1 is W YTHOFF). The winning strategies given in [5] apply directly to every rational number with the same induced integers. P positions of generalized W YTHOFF partition the integers, so Game I partitions the rationals. Are there any meaningful partitions of the rationals that transcend tree level? Game II. Let t be fixed positive integer. A position is a single rational, p=q. A move consists of performing up to t steps. The P positions are all the rationals on nonnegative integer multiples of level .t C 1/. Game III. A position consists of m rationals .p1 =q1 ; : : : ; pm =qm /. A move consists of selecting any one of the rationals and performing a positive number of steps. This game amounts to playing Nim on the integers induced by those rationals. Though playing on the integer k is equivalent to playing on any of its 2k 1 rational siblings on level k of the tree .k 1/, there are games on the rationals with no obvious integer parallels. For example, one may restrict every move to a single tree direction: either move up right, or move up left, both for impartial (Wythofflike) and partizan (chesslike) games.
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Our game steps amount to single steps in the “slow Euclidean Algorithm.” The game E UCLID corresponds to playing along the Euclidean Algorithm at a pace chosen by the player. There is a large bibliography on E UCLID including [3], [11], [6]. For related games, see [2] and [10]. Like the CWtree, the older SternBrocot tree [12], [1] also contains all the reduced rationals. The former has the advantage that the transition from any level to its neighbor is simple; the latter that it is a search tree. Both trees are intimately related to the famous Stern diatomic sequence available at http://oeis.org/A002487. A construction of the rationals using recurrence is given in [9]. Final remark In a different context, an anonymous referee recently asked whether there are any “bridges” between combinatorial game theory (CGT) and classical game theory. In the latter, of course, the notion of “rationality” plays a prominent role (in the form of the rational players). Now we see that rationality also plays a role in CGT.
Bibliography [1] A. Brocot, Calcul des rouages par approximation, nouvelle m`ethode, Revue Chronom`etrique 6 (1860) 186–194. [2] G. Cairns and N. B. Ho, Min, a combinatorial game having a connection with prime numbers, Integers 10 G03 (2010) 765–770; available at dx.doi.org/10.1515/ INTEG.2010.103. [3] G. Cairns, N. B. Ho, and T. Lengyel, The SpragueGrundy function of the real game Euclid, Discrete Math. 311 (2011) 457–462; available at dx.doi.org/10.1016/ j.disc.2010.12.011. [4] N. Calkin and H. S. Wilf, Recounting the rationals, Amer. Math. Monthly 107 (2000) 360–363; available at dx.doi.org/10.2307/2589182. [5] A. S. Fraenkel, How to beat your Wythoff games’ opponent on three fronts, Amer. Math. Monthly 89 (1982) 353–361; available at dx.doi.org/10.2307/2321643. [6] A. S. Fraenkel, Euclid and Wythoff games, Discrete Math. 304 (2005) 65–68; available at dx.doi.org/10.1016/j.disc.2005.06.022. [7] M. Gardner, Wythoff’s Nim, Ch. 8, pp. 103–118 in: Penrose Tiles to Trapdoor Ciphers, Revised Edition, Mathematical Association of America, 1997. [8] J. Gibbons, D. Lester and R. Bird, Functional pearl: Enumerating the rationals, J. Funct. Programming 16 (3) (2006) 281–291; available at dx.doi.org/10.1017/ S0956796806005880. [9] S. P. Glasby, Enumerating the rationals from left to right, Amer. Math. Monthly 118 (2011) 830–835; available at dx.doi.org/10.4169/amer.math.monthly.118.09.830. [10] S. Hofmann, G. Schuster, and J. Stending, Euclid, Calkin & Wilf—playing with rationals, Elem. Math V. 63 (2008) 109–117; available at dx.doi.org/10.4171/EM/95. [11] G. Nivasch, The SpragueGrundy function of the game Euclid, Discrete Math. 306 (2006) 2798–2800; available at dx.doi.org/10.1016/j.disc.2006.04.020. ¨ [12] M. A. Stern, Uber eine zahlentheoretische Funktion, J. Reine Angew. Math. 55 (1858) 193–220; available at dx.doi.org/10.1515/ crll.1858.55.193. [13] W. A. Wythoff, A modification of the game of Nim, Nieuw Arch. Wisk. 7 (1907) 199–202.
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VI Cards and Probability
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30 Modeling Mathematics with Playing Cards Martin Gardner Because playing cards have values 1 through 13 (jacks 11, queens 12, kings 13), come in two colors, four suits, and have fronts and backs, they provide wonderfully convenient models for hundreds of unusual mathematical problems involving number theory and combinatorics. What follows is a choice selection of little known examples. One of the most surprising of card theorems is known as the Gilbreath principle after magician Norman Gilbreath who first discovered it. Arrange a deck so the colors alternate. Cut it so the bottom cards of each half are different colors, and then riffle shuffle the halves together. Take cards from the top in pairs. Amazingly, every pair will consist of a red and black card! Here is a simple proof by induction that this must happen. Assume that the first card to fail on the table during the shuffle is black. If the next card to fail is the card directly above it in the same half, that card will be red. This places on the table a redblack pair. If the next card after the first one comes from the other half, it too will be red and will put a redblack pair on the table. In either case, after two cards have dropped, the bottom cards of each half will be of different colors, so the situation is exactly the same as before, and the same argument applied for the rest of the cards. No matter how careful or careless the shuffle, it will pile redblack pairs on the table. Gilbreath’s principle generalizes. Arrange the deck so the suits are in an order, say spades, hearts, clubs, diamonds, that repeats throughout the pack. Deal as many cards as you like to form a pile. This of course reverses the order of the suits. When the pile is about the same size as the remaining portion of the deck, riffle shuffle the two portions together. If you now take cards in quadruplets from the top of the shuffled pack, you will find that each set of four contains all four suits. The ultimate generalization is to shuffle together two decks, one with its cards in the reverse order of the other deck. After the shuffle, divide the 104card pack exactly in half. Each half will be a complete deck of 52 different cards! What mathematician David Gale has called the “nonmessingup theorem” is another whimsical result. From a shuffled deck, deal the cards face up to form a rectangle of any proportion. In each row, rearrange the cards so their values do not decrease from left to right. In other words, each card has a value higher than the one on its left, or two cards Reprinted from The College Mathematics Journal, Vol. 31, No. 3 (May 2000), pp. 173–177.
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of the same value are side by side. After ordering the rows, do the same thing with the columns. This of course drastically alters the order of cards in the rows. After rearranging the columns, you may be amazed to find that the rows are still ordered! The theorem is at least a hundred years old. You will find it proved as the answer to a problem in the American Mathematical Monthly (70:2, 1963, 21213), and in a monograph by Gale and Richard Karp, published in 1971 by the operations research center of the engineering school of the University of California, Berkeley. Donald Knuth discusses the theorem in the third volume of The Art of Computer Programming in connection with a method of sorting called “shellsort.” In my The Last Recreations, Chapter 11, I describe a clever card trick based on the theorem. Is it possible to arrange a deck so that if you spell the name of each card by moving a card from top to bottom for each letter, then turning over the card at the end of the spell and discarding it, it will always be the card you spelled? For example, can you so arrange the cards that you can first spell all the spades, taking them in order from ace through king, then do the same thing with the hearts, clubs, and diamonds? You might imagine it would take a long time to find out how to arrange the deck, assuming it is possible to do so, in a way that permits the spelling of all 52 cards. Actually, finding the order is absurdly easy. First arrange the deck from top down in the order that is the reverse of your spelling sequence. Take the King of Diamonds from the top of the deck, then take the queen, place it on top of the king, and spell “Queen of Diamonds” by moving a card at each letter from bottom to top. In brief, you are reversing the spelling procedure. Continue in this way until the new deck is formed. You are now all set to spell every card in the predetermined order. Of course you can do the same thing with smaller packets, such as the thirteen spades, or with cards bearing pictures, say of animals whose names you spell. Remember the old brainteaser about two glasses, one filled with water, the other with wine? You take a drop of water, put it into the wine, stir, then take a drop of the mixture, move it back to the water, and stir. Is there now more or less wine in the water than water in the wine? The answer is that the two quantities are exactly equal. The simplest proof is to realize that, after the transfers, the amounts of liquid in each glass remain the same. So the quantity missing from the water is replaced by wine, and amount of wine missing from the other glass is replaced by the same amount of water. This is easily modeled with cards. Divide the deck into two halves, one of all red cards, the other of all black. Randomly remover n red cards, insert them anywhere in the black half, and shuffle. Now randomly remove n cards from the half you just shuffled, put them back among the reds, and shuffle. Inspection will show that the number of black cards in the red half exactly equals the number of red cards in the black half. It doesn’t matter in the least if the red and black portions are not equal at the start. Closely related to this demonstration is the following trick. Cut a deck exactly in half, turn over either half and shuffle the two parts together. Cut the mixedup deck in half again, and turn over either half. You’ll find that the number of facedown cards in either half exactly equals the number of facedown cards in the other half. The same is true, of course, for the faceup cards. Do you see why this is the case? The trick is baffling to spectators if they don’t know that the deck is initially divided exactly in half, and if you secretly turn over one half as you spread its cards on the table.
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Playing cards provide a wealth of counterintuitive probability questions. The notorious Monty Hall problem can be modeled with cards. There are three cards face down on the table and you are told that one card only is an ace. Put a finger on a card. Clearly the chance you have selected the ace is 1=3. A friend now secretly peeks at all three cards and turns face up a card that is not the ace. Two cards remain face down, one of which you know is the ace. What now is the probability your finger is on the ace? Many persons think the probability has risen from 1=3 to 1=2. A little reflection should convince you that it remains 1=3 because your friend can always turn a nonace. Now switch your finger from the card it is on to the other card. The probability you have now chosen the ace jumps form 1=3 to 2=3. This is obvious from the fact that the card you first selected has the probability of 1=3 being the ace. Because the ace must be one of the two facedown cards, the two probabilities must add to 1 or certainty. A similar seeming paradox also involves three facedown cards dealt from a shuffled deck. A friend looks at their faces and turns over two that are the same color. What’s the probability that the remaining facedown card is the same color as the two faceup cards? You might think it is 1=2. Actually it is 1=4. Here’s the proof. The probability that three randomly selected cards are the same color is two out of eight equal possibilities, or 1=4. Subtract 1=4 from 1 (the card must be red or black) and you get 3=4 for the probability that the facedown card differs in color from the two faceup cards. This is the basis for an ancient sucker bet. If you are the operator, you can offer even odds that the card is of opposite color from the two faceup cards, and win the bet three out of four times. Here’s a neat problem involving a parity check. Take three red cards from the deck. Push one of them back into the pack and take out three black cards. Push one of them back into the deck and remove three reds. Continue in this manner. At each step you randomly select a card of either color, return it to the deck and remove three cards of opposite color. Continue as long as you like. When you decide to stop you will be holding a mixture of reds and blacks. Is it possible that the number of black cards you hold will equal the number of reds? Unless you think of a parity check it might take a while to prove that the answer is no. After each step you will always have in your hand an odd number of cards, therefore the two colors can never be equal. Magicians have discovered the following curiosity. Place cards with values ace through nine face down in a row in counting order, ace at the left. Remove a card from either end of the row. Take another card from either end. Finally, take a third card from either end. Add the values of the three cards, then divide by six to obtain a random number n. Count the cards in the row from left to right, and turn over the nth card. It will always be the four! I leave it to readers to figure out why this works and perhaps to generalize it to longer rows of numbers. For example, use twelve cards with values 2; 3; 4; 5; 6; 7; 8; 9; 10; J; Q; K to make the row. Take a card three times from either end, divide their sum by 9, and call the result n. The nth card from the left will always be the five. A classic card task, going back more than two centuries, is to arrange all the aces, kings, queens, and jacks—sixteen cards in all—in a square array so that no two cards of the same value, as well as no two cards of the same suit, are in the same row, column, or diagonal. Counting the number of different solutions is not trivial. W. W. Rouse Ball, in his classic Mathematical Recreations and Essays, said there are 72 fundamental solutions, not counting rotations and reflections. This is a mistake that persisted through the
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book’s eleventh edition, but was dropped from later editions revised by H. S. M. Coxeter. Dame Kathleen Ollerenshaw, a noted British mathematician who was once Lord Mayor of Manchester, found there are twice as many fundamental solutions, 144, making the number of solutions including rotations and reflections 8 144 D 1;152. She recently described a simple procedure for generating all 1;152 patterns in an article written for the blind. (Dame Ollerenshaw, now 87, is slowly losing her vision, and energetically learning how to read Braille.) This is her procedure. Number the sixteen positions of the square from 1 through 16, left to right, top down. Place an arbitrary card, say the Ace of Spades, in position 1, the top left corner. A second ace, say the Ace of Hearts, goes in the second row. It can’t go in the same column or diagonal as the Ace of Spades, so it must go in either space 7 or 8. Place it arbitrarily in space 7. Two aces remain to go in rows 3 and 4. Put the Ace of Diamonds in the third row. It can go only in space 12. The Ace of Clubs is now forced into space 14 of the bottom row. Had the second ace gone in space 8, the last two aces would be forced into spaces 10 and 15. Consider the other three spades. They can’t go in the top row or leftmost column, or in a main diagonal. This forces them into spaces 4, 10, and 15. Arbitrarily place the King of Spades in 4, the Queen of Spades in 10, and the Jack of Spades in 15. The pattern now looks like this: Aª A© Kª Qª A§
A¨ Jª
The remaining nine cards are forced into spaces that complete the following pattern: Aª K© Q¨
J§
Q§
J¨
A© Kª
J©
Qª
K§ A¨
K¨ A§
Jª
Q©
Multiply the number of choices at each step, 16 3 2 2 3 2, and you get the total of 1,152 patterns. For more examples of mathematical theorems, problems, and tricks with playing cards, see my Dover paperback Mathematics, Magic, and Mystery, and Karl Fulves’s SelfWorking Card Tricks, also a Dover softcover, and the following chapters in my collections of Scientific American columns: The Scientific American Book of Mathematical Puzzles and Diversions, Chapter 10; Mathematical Carnival, Chapter 10 and 15; Mathematical Magic Show,
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Chapter 7; Wheels, Life, and Other Mathematical Amusements, Chapter 19; Penrose Tiles to Trapdoor Ciphers, Chapter 19, and The Last Recreations, Chapter 2. Now for two puzzles that can be modeled with cards. Solutions appear on page 226. 1. Arrange nine cards as shown in Figure 30.1. Assume the aces have a value of 1. Each row, each column, and one diagonal has a sum of 6. The task is to alter the positions of three cards so that the matrix is fully magic for all rows, columns, and diagonals. A
2
3
A
2
3
3
A
2
4
5
6
2
3
A
7
8
9
Figure 30.1.
Figure 30.2.
2. Nine cards arranged and shown in Figure 30.2 have the property of minimizing the sum of all absolute differences between each pair of cells that are adjacent vertically and horizontally. Assume that the matrix is toroidal; that is, it wraps around in both directions. The sum of the differences is 48. This was proved minimal by Friend Kirstead, Jr., in the Journal of Recreational Mathematics (18, 1985–86, 301). The challenge is to take nine cards of distinct values (court cards may be used) and form a toroidal square that will maximize the sum of all absolute differences.
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Solutions to the playing card puzzles above 1. Move the bottom two of three cards to the top, or move the leftmost column to the right. Either change produces the desired magic square in Figure 30.3a. 2
3
A
A
K
Z
A
2
3
Q
Y
2
3
A
2
X
3
J
(a)
(b) Figure 30.3.
2. Figure 30.3b shows how nine cards of distinct values can be placed in a toroidal square so as to maximize the sum of absolute differences of adjacent values in rows and columns. Any value 4 through 10 can go in cells, X, Y, and Z to make a total of 120. This was proved maximal by Brian Maxwell, of Middlesex, England, in the Journal of Recreational Mathematics, vol. 18 (1985–86), p. 300.
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31 The Probability an Amazing Card Trick Is Dull Christopher N. Swanson The Ashland University student chapter of the MAA holds biweekly meetings, typically consisting of a short business meeting to plan activities that the group is sponsoring followed by a social time when the group plays mathematical games and munches on brownies. As a new feature in Fall 2002, I told the students that I would perform a new mathematical card trick at each of these meetings. My source for most of these card tricks is the delightful book [2]. One of the finest tricks described there is due to an amateur New York magician named Henry Christ. A spectator shuffles a deck of cards several times, and then the magician deals nine cards face down on a table. The spectator selects one of these cards, looks at it, and stacks the nine dealt cards on the table with the selected card on top. The magician places this stack on the bottom of the deck, and hands the deck to the spectator. The spectator is told to deal the cards out face up in a pile, counting backwards out loud from ten while dealing the cards. If at some point, the denomination of the card matches the number the spectator says, then the spectator repeats this procedure with a second pile, again counting backwards from ten. If 1 is reached without any cards matching, the spectator places the next card face down on top of the pile, and repeats this procedure with the second pile. The spectator does this until four piles are created in this manner, and then adds the numbers appearing on the cards that are face up. The spectator counts through the remaining cards of the deck to find the card in the position represented by this sum, and this card is the one that was originally selected. The reason why this card trick works is that the final card will always be the 44th card in the deck. This can be seen by noting that any unmatched pile will contain 11 cards that are set aside. If a pile has a match with a value i showing, then the match occurred after counting down 11 i cards. This card with value i will contribute i to the sum determining how many cards in the final deck to count down, and thus, a total number of 11 i Ci D 11 cards will be set aside for this pile. After practicing this card trick a few times, I was ready to present it to the students at our meeting. Unfortunately, when I performed the trick at the meeting, a match did not occur in any of the four piles. This is unfortunate for two reasons. First, nothing is done with the pile of cards left in the spectator’s hands, and the magician must tell the Reprinted from The College Mathematics Journal, Vol. 36, No. 3 (May 2005), pp. 209–212.
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spectator to simply turn over the card appearing face down on the final pile. Second, and more serious from the perspective of the magician, the trick seems quite simplistic when no matches occur because it is clear that you set aside 44 cards. Thus, a magician may ask, “What is the probability that no matches occur and my trick is dull?” This question falls within a class of problems known as counting permutations with restricted positions. For a more general discussion of this class of problems, see [1]. For a more advanced treatment of the topic, see [7] or [8], and for additional applications, see [3], [4], [5], [6].
The probability of no matches Our first step in studying this card trick problem is to note that after the nine cards are returned to the bottom of the deck, the order of the cards in the deck is one of the possible 52! permutations. Hence, for our purposes, we just need to focus on the latter part of the card trick. Let D be the number of ways to shuffle the deck of cards such that there is not a 10 in the 1st, 11th, 21st, or 31st positions, there is not a 9 in the 2nd, 12th, 22nd, or 32nd positions, etc. Thus, each card from ace and ten has four positions where it cannot occur. We find D by using the principle of inclusionexclusion: Given a collection of finite sets A1 ; A2 ; : : : ; An , the number of elements in the union of these sets A1 [ A2 [ [ An is X X jA1 [ A2 [ [ An j D jAi j jAi \ Aj j C 1i n
nC1
C . 1/
1i <j n
jA1 \ A2 \ \ An j:
Essentially this means to find the number of elements that appear in at least one of these sets, count the number of elements in each set, then subtract off the number of elements that appear in two sets, then add the number of elements that appear in three sets, and continue to alternate these additions and subtractions up to the number of elements that appear in all of the sets. To apply this principle to our problem, we number the forty cards with restricted positions and let Ai be the set of permutations that have card numbered i in one of its restricted positions. Then the number of permutations without any cards in restricted positions is D D 52Š D 52Š D 52Š C
jA1 [ A2 [ [ A40 j 0 X X @ jAi j 1i 40
40 X
1i <j 40
. 1/k
kD1
X
jAi \ Aj j C
1i1 <i2 <<ik 40
1
jA1 \ A2 \ \ A40 jA
jAi1 \ Ai2 \ \ Aik j:
The inner sum of this formula overcounts the number of ways to place at least k cards in restricted positions. Let rk (sometimes called a rook number due to an interpretation concerning placing nonattacking rooks on a chessboard) be the number of ways of placing k cards in restricted positions, ignoring the placement of any other cards. Then the remaining cards can be placed in .52 k/Š ways, and we see that X jAi1 \ Ai2 \ \ Aik j D rk .52 k/Š: 1i1 <i2 <<ik 40
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Taking r0 D 1 (as is standard), we see that our formula becomes DD
40 X
. 1/k rk .52
k/Š;
(31.1)
kD0
for which we need the numbers rk . To calculate rk , we find the generating function (called the rook polynomial) for this sequence, R.x/ D r0 C r1 x C r2 x 2 C C r40 x 40 . (Note that ri is simply the coefficient of x i .) We do this by first finding the generating polynomial P .x/ for the number of ways for having i cards for a particular denomination in a restricted position, and then observing that R.x/ D .P .x//10 . For ease of description, we focus on placing10’s in restricted positions, but clearly the same holds for each denomination. There are 41 4 D 16 ways to place one 10 since there are 41 ways to choose the 10 and four positions for it. Similarly, for two 10’s, there are 42 ways to choose the cards, and then 4 3 ways to place them in restricted positions (with the suits in alphabetical order say, four choices for the “first” and three for the “second”) for a total of 42 4 3, or 72. In the same way, we find there are 96 ways to place three 10’s and 24 ways to place all four 10’s in restricted positions. Therefore, with there being just one way to have no 10’s in restricted positions, P .x/ D 1 C 16x C 72x 2 C 96x 3 C 24x 4. Now let’s consider the cards of two denominations, say the 10’s and the 9’s. We can find the number of ways of having various combinations of these eight cards in restricted positions (without regard to any others) by squaring P .x/ W 1 C 32x C 400x 2 C 2496x 3 C 8304x 4 C14592x 5 C12672x 6 C4608x 7 C576x 8. For example, consider the term 2496x 3. This says there are 2496 ways of having three 9’s and 10’s in restricted positions, arising from no 9’s and three 10’s (1 96 ways), one 9 and two 10’s (16 72 ways), two 9’s and one 10 (72 16 ways), and three 9’s and no 10’s (96 1 ways). The multiplicative rule is justified since the restricted positions for different denominations are disjoint. From this line of reasoning it follows that the coefficient of x k in .P .x//10 is the number of ways of having k number cards in restricted positions, ignoring the placement of the other cards. Thus, R.x/ D .1 C 16x C 72x 2 C 96x 3 C 24x 4/10 . We used MAPLE to expand R.x/ to find the numbers rk , and then used equation (31.1) to get D D 3286792937574955359428826782591912851876783001405845125895657881600 3:29 1066 :
D Therefore, the probability that no card is in a forbidden position is 52Š 0:0407. We see that although it is unlikely that the magician would not get any cards to match while performing the magic trick, it will happen often enough that the magician better know how to adjust the trick. (An anonymous referee noted that we can also produce the approximate value of 0:0407 using doubleprecision arithmetic.)
Exercises 1. The game Euchre uses only 6 cards .9; 10; Jack; Queen; King; Ace/ from each suit. Suppose you are at a party and you wish to demonstrate the card trick using such
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Part VI. Cards and Probability a deck. You plan to have the spectator create 4 piles while counting backwards Jack109. How many cards should you first deal? What is the probability that no match occurs in the four piles?
2. Consider the general version of Henry Christ’s trick. You have a deck of cards consisting of k suits having f face values and in the trick the spectator is to create p piles while counting backwards from m to 1. What are the possible values for m? How many cards should you first deal? Find an expression for the probability that no match occurs in the p piles. Acknowledgment The author would like to thank the editor and the anonymous referees for improving the clarity of this article. He would also like to thank the students of Ashland University’s student chapter of the MAA for putting up with his bad humor and card tricks during meetings.
Bibliography [1] R. A. Brualdi, Introductory Combinatorics, 3rd ed., PrenticeHall, 1999, Ch. 6. [2] M. Gardner, Mathematics, Magic and Mystery, Dover Publications, 1956. [3] A. W. Joseph, A problem in derangements, J. Inst. Actuaries Students’ Soc. 6 (1946) 14–22. [4] F. F. Knudsen and I. Skau, On the asymptotic solution of a cardmatching problem, Math. Mag., 69 (1996) 190–197. [5] B. H. Margolius, The dinnerdiner matching problem, Math. Mag., 76 (2003) 107–118. [6] S. G. Penrice, Derangements, permanents, and Christmas presents, Amer. Math. Monthly 98 (1991) 617–620. [7] J. Riordan, An Introduction to Combinatorial Analysis, Princeton Univ. Press, 1978. [8] R. P. Stanley, Enumerative Combinatorics: Volume I, Cambridge Univ. Press, 1998.
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32 The Monty Hall Problem, Reconsidered Stephen Lucas, Jason Rosenhouse, and Andrew Schepler In its classical form, the Monty Hall Problem (MHP) is the following: Version 1. (Classic Monty) You are a player on a game show and are shown three identical doors. Behind one is a car, behind the other two are goats. Monty Hall, the host of the show, asks you to choose one of the doors. You do so, but you do not open your chosen door. Monty, who knows where the car is, now opens one of the doors. He chooses his door in accordance with the following rules: 1. Monty always opens a door that conceals a goat. 2. Monty never opens the door you initially chose. 3. If Monty can open more than one door without violating rules one and two, then he chooses his door randomly. After Monty opens his door, he gives you the choice of sticking with your original choice or switching to the other unopened door. What should you do to maximize your chances of winning the car? In the entire annals of mathematics, you would be hardpressed to find a problem that arouses the passions like the MHP. It has a history going back at least to 1959, when Martin Gardner introduced a version of it in Scientific American [4, 5]. When statistician Fred Moseteller included it in his 1965 anthology of probability problems [9], he remarked that it attracted far more mail than any other problem. In his 1968 book Mathematical Ideas in Biology [18], biologist John Maynard Smith wrote, “This should be called the Serbelloni problem since it nearly wrecked a conference on theoretical biology at the villa Serbelloni in the summer of 1966.” In its modern game show format the problem made its first appearance in a 1975 issue of the academic journal The American Statistician [16]. Mathematician Steve Selvin presented it as an interesting classroom exercise on conditional probability. Though he presented the correct solution, (that there is a big advantage to be gained from switching), he found himself strongly challenged by subsequent letters to the editor [17]. Reprinted from Mathematics Magazine, Vol. 82, No. 5 (Dec. 2009), pp. 332–342.
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The problem really came into its own when Parade magazine columnist Marilyn vos Savant responded to a reader’s question regarding it. There followed several rounds of angry correspondence, in which readers challenged vos Savant’s solution. The challengers later had to eat crow when it was shown by a Monte Carlo simulation that vos Savant was correct, but not before the fracas reached the front page of the New York Times [20]. The whole story is recounted in the books by Rosenhouse and vos Savant [13, 15]. In the end, the situation has been best summed up by cognitive scientist Massimo PalmatelliPalmarini who wrote that, “. . . no other statistical puzzle comes so close to fooling all the people all the time. . . The phenomenon is particularly interesting precisely because of its specificity, its reproducibility, and its immunity to higher education” [10].
Why all the confusion? The trouble, you see, is that most people argue like this: “Once Monty opens his door only two doors remain in play. Since these doors are equally likely to be correct, it does not matter whether you switch or stick.” We will refer to this as the fiftyfifty argument. This intuition is supported by a wellknown human proclivity. A negative consequence incurred by inaction hurts less than the same negative consequence incurred through some definite action. In the context of the MHP, people feel worse when they switch and lose than they do after losing by sticking passively with their initial choice. There is a large literature in the psychology and cognitive science journals documenting and explaining the difficulty people have with the MHP. Burns and Wieth [3] summarized the findings of numerous such studies by writing, These previous articles reported 13 studies using standard versions of the MHD, and switch rates ranged from 9% to 23% with a mean of 14.5%. This consistency is remarkable given that these studies range across large differences in the wording of the problem, different methods of presentation, and different languages and cultures. (Note that MHD stands for “Monty Hall Dilemma.”) Gilovich, Medvec, and Chen [6] studied people’s reactions to losing by switching versus their reactions to losing by sticking. They used boxes instead of doors, and crafted an experimental situation in which players would lose regardless of their decision to switch or stick. Their findings? Because action tends to depart from the norm more than inaction, the individual is likely to feel more personally responsible for an unfortunate action. Thus, subjects who switched boxes in our experiment were more likely to experience a sense of “I brought this on myself,” or “This need not have happened,” than subjects who decided to keep their initial box. It would seem the defenders of sticking can point both to a plausible mathematical argument and to certain fine points of human psychology. How can the forces for switching fight back?
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Focus on Monty, not the doors There are a variety of elementary methods for solving the MHP. Working out a tree diagram for the problem, as in Figure 32.1, establishes that switching wins with probability 23 , while sticking wins with probability 13 . Consequently, we double our chances of winning by switching.
1 3
1 3
1 3 Car behind Door 2
Car behind Door 1
Car behind Door 3
1 2
1
1
Monty opens Door 2
Monty opens Door 3
Monty opens Door 3
Monty opens Door 2
Switching loses Probability = 16
Switching loses Probability = 16
Switching wins Probability = 13
Switching wins Probability = 13
1 2
Figure 32.1. Probability tree for the classical MHP, when the player initially chooses door one.
Monte Carlo simulations are also effective for establishing the correct answer. The Monty Hall scenario is readily simulated on a computer. The large advantage to be gained from switching quickly becomes apparent by playing the game multiple times. Such methods, however, do little to clarify why the fiftyfifty argument is incorrect. Practical results obtained from a simulation can show you that something is wrong with your intuition, but they will not make the correct answer seem natural. The trouble lies in the difficulty people have in recognizing what is and is not important in reasoning about conditional probability. The mantra about focus goes a long way towards pointing people in the right direction. When Monty opens door X, there is a tendency to think, “I have learned that door X conceals a goat, but I have learned nothing of relevance about the other two doors.” This is what we mean by “focusing on the doors.” The proper approach involves focusing on Monty, specifically on the precise manner in which he chooses his door to open. We should think, “Monty, who makes his decisions according to strict rules, chose to open door X. Why this door as opposed to one of the others?” Let us assume the player initially chose door one and Monty then opened door two. According to the rules, we can be certain that one of the following two scenarios has played out:
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1. The car is behind door one. Monty chose door two at random from among doors two and three. 2. The car is behind door three. Since the player initially chose door one, Monty was now forced to open door two. The second of these scenarios is more likely than the first. Since the car is behind the first door onethird of the time, and since Monty then opens door two in onehalf of those cases, we see that scenario one occurs onesixth of the time. Scenario two, on the other hand, happens whenever the car is behind door three (and the player has chosen door one). That happens onethird of the time. Scenario two is twice as likely as scenario one. Thus, we should think, “I have just witnessed an event that is twice as likely to occur when the car is behind door three than it is when the car is behind door one. Consequently, the car is more likely to be behind door three, and I am more likely to win the car by switching.”
An exotic selection procedure The general principle here is that anything affecting Monty’s decisionmaking process is relevant to updating our probabilities after Monty opens his door. To further illuminate this point, let us consider an altered version of the problem: Version 2. (HighNumbered Monty) As before, we have three identical doors concealing one car and two goats. The player chooses a door that remains unopened. Monty now opens a door he knows to conceal a goat. This time, however, we stipulate that Monty always opens the highestnumbered door available to him (keeping in mind that Monty will never open the door the player chose). Will the player gain any advantage by switching doors? For reasons of concreteness, we will assume once more that the player initially chooses door one. Any time door one conceals a goat, Monty has no choice regarding which door to open. He can not open door one (since the player chose that door), and he can not open the door that conceals the car. This leaves only one door available to him. The interesting cases occur when door one conceals the car. Unlike Classic Monty, who now chooses randomly, HighNumbered Monty will always open door three when he can. It follows that if we see him open door two instead we know for certain that the car is behind door three. And if HighNumbered Monty opens door three? Since Monty is certain to open door three whenever the car is behind door one or door two, we now have no basis for deciding between them. It really is a fiftyfifty decision in this case. Take this as a cautionary tale. Whether we are playing Classic Monty or HighNumbered Monty, it is certain that Monty will open a goatconcealing door. In the former case the probability that our initial choice concealed the car did not change while in the latter case it did. This shows that any proposed solution to the MHP failing to pay close attention to Monty’s selection procedure is incomplete.
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Monty meets Bayes The main point thus far is that the probability that door X conceals the car, given that Monty has shown us the goat behind door Y , depends on a detailed consideration of Monty’s selection procedure. More precisely, it depends on the probability that Monty will open door Y under the assumption that door X conceals the car. The precise manner in which these probabilities are related is given by Bayes’ theorem. We denote by Ci the event that the car is behind door i , and by Mj the event that Monty opens door j to reveal a goat. Also assume the player initially chooses door one, and Monty then opens door two. The probability that the player’s door conceals the car, given Monty’s display, can be found using Bayes’ theorem: P .C1 jM2 / D
P .C1 /P .M2 jC1 / : P .M2 /
Expanding the bottom of this fraction via the law of total probability leads to P .C1 jM2 / D
P .C1 /P .M2 jC1 / : P .C1 /P .M2 jC1 / C P .C2 /P .M2 jC2 / C P .C3 /P .M2 jC3 /
In both of our versions of the MHP we have P .M2 jC2 / D 0, since it is given that Monty will never open the door concealing the car. Also, since we are given that the doors are identical, we have 1 P .C1 / D P .C2 / D P .C3 / D : 3 Making these substitutions leads to P .C1 jM2 / D
P .M2 jC1 / : P .M2 jC1 / C P .M2 jC3 /
In both versions of the game we have P .M2 jC3 / D 1. That is, when the player chooses door one and the car is behind door three, Monty is certain to open door two. In Classic Monty we have P .M2 jC1 / D 21 , since Monty chooses at random when the car is behind the door initially chosen by the player. In HighNumbered Monty we have P .M2 jC1 / D 0, since Monty is required by his rules to open door three. Plugging everything into Bayes’ Theorem shows that for Classic Monty we now have P .C1 jM2 / D
1 2 1 2
C1
D
1 3
[Classic],
while for HighNumbered Monty we have P .C1 jM2 / D
1 2
0C1
D
1 2
[HighNumbered].
These are precisely the answers we obtained in the previous section. Let us go one more round:
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Version 3. (Random Monty) As always, assume that the player has initially chosen door one and Monty subsequently opened door two to reveal a goat. This time, however, you know that Monty chose his door randomly, subject only to the restriction that he not open the door the player chose. Should we switch? The novelty here lies in the nonzero probability of Monty opening the door concealing the car. Intuitively we would reason as follows: Since Monty opened door two after I selected door one, since door two concealed a goat, and since I know Monty chose randomly between doors two and three, I conclude that one of two scenarios has played out: 1. The car is behind door one, Monty chose door two randomly. 2. The car is behind door three, Monty chose door two randomly. Since the car is equally likely to be behind doors one and three, these scenarios are equally likely to occur. The conclusion is that the remaining doors are equiprobable, and therefore there is no advantage to switching. Our intuition is confirmed via Bayes’ Theorem. We know that Monty will not open door one, and we know that door two conceals a goat. We now have 1 ; 3 1 P .M2 jC1 / D P .M2 jC3 / D ; 2 P .M2 jC2 / D 0:
P .C1 / D P .C2 / D P .C3 / D
Bayes’ Theorem now says P .C1 jM2 / D
1 3
1 2
1 1 3 2 C 13 .0/
C
1 3
1 2
D
1 : 2
The tree diagram in Figure 32.2 might be helpful for visualizing the situation.
Twoplayer Monty Threedoor versions of the MHP can become remarkably complex. The following version comes from a paper by philosopher Peter Baumann [1, 2]. For the remainder of the paper we will refer simply to the probability of door X, thereby avoiding the more cumbersome expression, “The probability that door X conceals the car.” Version 4. (TwoPlayer Monty) We begin with three identical doors concealing two goats and one car. There are two players in the game. Each player chooses one of the doors, but does not open it. Each player knows there is another person in the game, but neither knows which door the other player selected. Monty now opens a door according to the following procedure. 1. If both players selected the same door, then everything proceeds as in the classical game. Monty opens a goatconcealing door, choosing randomly if he has a choice.
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1 3
Monty Opens Door 2
1 3
1 3
Car behind Door 3
Car behind Door 2
Car behind Door 1 1 2
237
1 2
1 2 Monty Opens Door 3
1 2
Switching loses Switching loses Probability = 16 Probability = 16 Monty Opens Door 2 Game ends
Monty Opens Door 2
1 2
Switching wins Probability = 61
1 2 Monty Opens Door 3 Game ends
Monty Opens Door 3 Switching wins Probability = 16
Figure 32.2. Probability tree for Random Monty, when the player initially chooses door one.
2. If the players selected different doors, then Monty opens the one remaining door, regardless of what is behind it. We assume that both players select their initial doors randomly. If you are one of the players and you have just seen Monty open a goatconcealing door, should you switch? This will be a fine test of our newfound intuition. For concreteness, suppose that Player A initially chose door one, and Monty has now opened the goatconcealing door two. What do Monty’s actions tell us about Player B’s choice? Initially we consider it equally likely that Player B chose door one, door two or door three. After seeing Monty open door two we reason that one of three scenarios has played out: 1. Player B chose door three. In this case Monty was forced to open door two, which conceals a goat with probability 23 . 2. Player B chose door one and door one conceals the car. In this case Monty opens door two with probability 12 . Since door one conceals the car with probability 13 , this scenario occurs with probability 16 . 3. Player B chose door one and door three conceals the car (which happens with probability 13 ). In this case Monty is again forced to open door two. Combining items two and three above shows that Player B chose door one with probability 1 C 61 D 12 . Item one shows that Player B chose door three with probability 23 . We conclude 3 that the event in which Monty opens the goatconcealing door two after Player A chooses door one is 43 more likely to occur when Player B has chosen door three than when he has chosen door one.
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It is a consequence of Bayes’ theorem that the probabilities we now assign to “Player B chose door three,” and “Player B chose door one” must preserve this 4 W 3 ratio. (A proof of this assertion can be found in the paper by Rosenthal [14].) Consequently, we assign probabilities of 74 and 37 respectively. To continue the analysis, note that from Player A’s perspective there are now four possibilities. Player B could have chosen door one or door three, and the car could be behind either of those doors. Let us denote these possibilities via ordered pairs of the form (Player B’s Door, Location of the Car). Thus, the four remaining possibilities are .3; 1/; .3; 3/; .1; 1/; .1; 3/: Consider the first two pairs. If Player B chose door three, then Monty was forced to open door two. Consequently, we learn nothing regarding the probability of doors one and three. Since these two scenarios collectively have a probability of 74 , and since they are equally likely, we now assign the following probabilities: P .3; 1/ D P .3; 3/ D
2 : 7
The remaining two pairs, however, are not equiprobable. Suppose that Player B chose door one, just as Player A did. If the car is behind door one, then Monty chose door two randomly, which happens with probability 12 . If the car is behind door three, then Monty was forced to choose door two. It follows that it is twice as likely that the car is behind door three than that it is behind door two. Since these scenarios have a collective probability of 3 7 , we assign the following probabilities: P .1; 1/ D
1 7
and P .1; 3/ D
2 : 7
Of the four scenarios, the two in which Player A wins by switching are .1; 3/ and .3; 3/. Since both have probability 72 , this gives a total probability of winning by switching of 47 . That is our solution. The really amusing part is that both players will go through this analysis, and both will decide to switch doors. In those scenarios in which the players chose different doors, this implies that someone is definitely making the wrong decision. Such are the cruelties of probability. Twoplayer Monty has also been discussed by Baumann [1], Levy [7], Rosenhouse [13], and Sprenger [19].
Many doors Ready for the final exam? Version 5. (Progressive Monty) This time there are n identical doors, concealing one car and n 1 goats. The player chooses a door, but does not open it. Monty now opens a
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goatconcealing door, choosing randomly from among his options. The player is now given the choice of sticking or switching. The player makes his choice, but again does not open his chosen door. Monty opens another goatconcealing door. The player is again given the opportunity to stick or switch. This continues until Monty has opened n 2 doors. The player makes his final selection, and wins whatever is behind his door. What strategy will maximize his chances of winning the car? To help us get our bearings, let us try a case study. Suppose we begin with five doors. At any stage of the game we represent the probabilities of the doors, based on all available knowledge, via an ordered 5tuple, which we call the probability vector. As the game begins, we have probability vector 1 1 1 1 1 ; ; ; ; : 5 5 5 5 5 As always, let us assume the player chooses door one and Monty now opens door two. Our past experience suggests that the probability of our door does not change, and this is confirmed by Bayes’ theorem. In the following calculation, the notation Ci denotes the event where the car is not behind door i . We now compute P .C1 jM2 / D
P .C1 /P .M2 jC1 /
P .C1 /P .M2 jC1 / C P .C1 and C2 /P .M2 jC1 and C2 / 1 1 1 5 4 D 1 1 3 1 D : 5 5 4 C 5 3
Since the other doors are identical and since their probabilities must sum to 54 , we now have probability vector 1 4 4 4 ; 0; ; ; : 5 15 15 15 What if we now switch to door three and then see Monty open door five? Known probabilities are now 4 ; 15 1 P .M5 jC1 / D P .M5 jC4 / D ; 2 1 1 P .C1 / D and P .M5 jC3 / D : 5 3 P .C3 / D P .C4 / D P .C5 / D
If we use the law of total probability to write P .M5 / D P .C1 /P .M5 jC1 / C P .C3 /P .M5 jC3 / C P .C4 /P .M5 jC4 / D
29 ; 90
and plug the results into Bayes’ Theorem, the result is the probability vector 9 8 12 ; 0; ; ; 0 : 29 29 29 The probabilities of all the remaining doors went up.
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What if Monty had opened door one after we switched to door three? The reader can supply the details that lead to the vector 1 3 3 0; 0; ; ; : 4 8 8 4 to 14 . Our chosen door actually Notice that the probability of door three went down, from 15 seems less likely as the result of Monty’s actions. A surprising result! Things get messy indeed in this version. Plainly we need some guidelines to aid our intuition. The first principle is simple. Any time Monty chooses not to open a door different from your present choice, the probability of that door goes up. In our case study, Monty opened door two after we chose door one. The event, “Monty does not open door three,” is more likely to happen when the car is behind door three than when it is elsewhere. Consequently, we will revise upward our probability of door three. The second principle is that if the doors different from your present choice are equiprobable, then the probability of your choice does not change when Monty opens a door. In our case study, after Monty opened door two, we reason that the event, “Monty does not open door one,” has probability one regardless of the location of the car. Consequently, we learn nothing from the occurrence of that event. The calculation in our case study confirms this intuition. Why, though, does it matter that the other doors are equiprobable? The answer is that Monty’s failure to open a door is not the only source of information to which we have access. The probability of the event, “Monty opens door X,” depends in part on the probability of the event, “Door X conceals the car.” Specifically, the more likely a door is to conceal the car, the less likely Monty is to open that door. Once more returning to our case study, we switched to door three at a moment when doors three through five were equiprobable and collectively very likely to conceal the car. By opening door five, Monty eliminated one element of this collection. This revelation does nothing to shake our confidence that the car is more likely to be found among doors three through five than it is to be found among any collection of three doors that includes door one. Consequently, we will revise upward the probability of our chosen door. Why did the probability of door three go down when Monty opened door one? This one is harder to explain, but our calculation suggests the proper way to think about it. If all four doors had been equiprobable at the moment we switched to door three, then we would simply be playing Classic Monty on four doors. In that case, our chosen door would retain its 14 probability after Monty opens a door. The probability vector we computed for our present situation is identical to what we would have obtained were we playing fourdoor Classic Monty. The implication is that by eliminating door one, Monty has essentially erased the prior history of the game. We are now faced with three doors that were equiprobable at the moment we chose among them, and these doors were among an ensemble of four doors, just as in fourdoor Classic Monty. That door one had a different probability from the other doors does not distinguish our situation in a relevant way. This observation leads to our final clue. If we select a door at a moment when precisely k doors remain, the probability of that door can never be smaller than k1 . Even if we have been careless in extracting the maximum amount of information from Monty’s actions, we still know the door was chosen from among k possibilities.
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32. The Monty Hall Problem, Reconsidered
241
As a test of our principles, let us go another round with our case study. We left off with the player having chosen door three and with probability vector, 1 3 3 0; 0; ; ; : 4 8 8 Imagine that we now switch to door four. If Monty now opens door three then only doors four and five remain in play. We would reason that these two doors were equiprobable at the moment we switched to door four, but that door four was selected from among three possibilities. We are, in effect, playing Classic Monty, and we would expect our updated probability vector to be 1 2 : 0; 0; 0; ; 3 3 The calculation is P .C4 /P .M3 jC4 / P .C4 /P .M3 jC4 / C P .C5 /P .M3 jC5 / 3 1 1 8 2 D 3 1 3 D : 3 8 2 C 8 .1/
P .C4 jM3 / D
And if Monty opens door five instead? Our intuition tells us that both doors should see their probabilities go up: door three, because it might have been opened but was not; door five, because it was part of an equiprobable ensemble that has decreased in size. Bayes’ Theorem confirms our intuitions. We compute P .C3 jM5 / D
P .C3 /P .M5 jC3 / D P .C3 /P .M5 jC3 / C P .C4 /P .M5 jC4 /
and obtain probability vector
4 3 0; 0; ; ; 0 : 7 7
1 4 .1/ 1 .1/ C 38 4
1 2
D
4 ; 7
Remarkably, our arguments to this point are already enough to justify the correct solution to Progressive Monty. Consider the strategy in which we switch at the last minute (SLM). That is, we will stick with our initial choice until only two doors remain, and then we will switch. Our initial choice has probability 1n . Since the other doors are equiprobable, this probability will not change so long as we keep it as our choice. At the moment when only two doors remain, the other door will have probability n n 1 . That is the probability that we win with SLM. We also know that there will never be a moment in the game when a door has a probability smaller than n1 . Thus, at the moment when only two doors remain it is impossible to produce a door with probability greater than n n 1 . This shows that SLM is optimal. Very nice. A full, rigorous proof that SLM is, in fact, uniquely optimal can be found in the book by Rosenhouse [13]. You might also wonder what can be said about other strategies. For example, what if we are playing with fifty doors and we are absolutely determined to switch exactly seven times during the game? What is our best strategy? A consideration of such questions can be found in the paper by Lucas and Rosenhouse [8].
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Progressive Monty receives further attention in one paper by Paradis, Viader, and Bibiloni [11] and another by Rao and Rao [12]. For a variation on the problem, see the paper by Zorzi in the December 2009 Mathematics Magazine. It would seem that a bit of clear thinking can steer us through even the densest of Montyinspired forests. Once our intuition has been tuned to what is important, it is not so difficult to ferret out the correct answer.
Bibliography [1] Peter Baumann, Three doors, two players, and singlecase probabilities, American Philosophical Quarterly 42(1) (January, 2005) 71–79. [2] ——, Singlecase probabilities and the case of Monty Hall: Levy’s view, Synthese 162(2) (May 2008) 265–273. dx.doi.org/10.1007/s1122900791856 [3] Bruce Burns and Mareike Wieth, The collider principle in causal reasoning: why the Monty Hall Problem is so hard, Journal of Experimental Psychology, General 103(3) (2004) 436–449. [4] Martin Gardner, Problems involving questions of probability and ambiguity, Scientific American 201(4) (April, 1959) 174–182. [5] ——, How three modern mathematicians disproved a celebrated conjecture of Leonhard Euler, Scientific American 201(5) (May, 1959) 188. [6] T. Gilovich, V. H. Medvec, and S. Chen, Commission, omission, and dissonance reduction: coping with regret in the Monty Hall Problem, Personality and Social Psychology Bulletin 21(2) (February 1995) 182–190. dx.doi.org/10.1177/0146167295212008 [7] Ken Levy, Baumann on the Monty Hall Problem and singlecase probabilities, Synthese 158(1) (September 2007) 139–151. dx.doi.org/10.1007/s1122900690655 [8] Stephen K. Lucas and Jason Rosenhouse, Optimal Strategies for the Progressive Monty Hall Problem, Math. Gaz. (to appear). [9] Fred Mosteller, Fifty Challenging Problems in Probability, With Solutions, AddisonWesley Inc., Reading, 1965. [10] Massimo PiattelliPalmarini, Probability blindness, neither rational nor capricious, Bostonia (March/April 1991) 28–35. [11] J. Paradis, P. Viader, and L. Bibiloni, A mathematical excursion: from the threedoor problem to a Cantortype set, Amer. Math. Monthly 106 (1999) 241–251. dx.doi.org/ 10.2307/2589679 [12] V. V. Bapeswara Rao and M. Bhaskara Rao, A threedoor game show and some of its variants, Math. Sci. 17(2) (1992) 89–94. [13] Jason Rosenhouse, The Monty Hall Problem, Oxford University Press, New York, 2009. [14] Jeffrey Rosenthal, Monty Hall, Monty fall, Monty crawl, Math Horizons (October 2008) 5–7. [15] Marilyn vos Savant, The Power of Logical Thinking, St. Martin’s Press, New York, 1996. [16] Steve Selvin, A problem in probability, (Letter to the Editor), Amer. Statist. 29(1) (1975) 67. [17] ——, On the Monty Hall Problem, (Letter to the Editor), Amer. Statist. 9(3) (1975) 134. [18] John Maynard Smith, Mathematical Ideas in Biology, Cambridge University Press, London, 1968. [19] Jan Sprenger, Probability, rational singlecase decisions, and the Monty Hall Problem, Synthese (to appear). [20] John Tierney, Behind Monty Hall’s doors, puzzle, debate, and answer? The New York Times, July 21, 1991, 1A.
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33 The Secretary Problem from the Applicant’s Point of View Darren Glass Searching for a job is always stressful and, with unemployment rates at their highest levels in years, never more so than now. Applicants can and should use every advantage at their disposal to obtain a job which is rewarding, financially and otherwise. While this author believes a math major gives applicants many advantages as they search for their dream job, one often overlooked is the ability to strategize and schedule their interviews to maximize the chance of landing that job. The secretary problem helps an employer find the optimal candidate for a job out of a large pool of applicants. The set up is as follows: only one person can be hired, and, for any pair of applicants, the employer has a strict preference for one of them that they can discern after seeing both. However, after each interview the employer must either accept or reject the candidate. If the candidate is the final person, the interviewer simply must accept them, as rejected candidates cannot be recalled. In the classical formulation of the problem, the goal is to select the best applicant overall. What strategy can the employer use to maximize the probability of hiring the best overall applicant? It is clear that, other than the final candidate, you only hire an applicant if they are the best applicant you have seen to that point. Otherwise you are certainly not hiring the best person. It follows that the best strategy is to reject an initial number of candidates and then hire the first candidate who is better than everyone seen so far. With a little work, one can show that if you know there will be a total of n candidates, where n is large, then you should initially reject k ne candidates. For a proof, see [3]. A more recent article discussing this problem and its implications for students’ dating lives can be found in [6]. While there is some uncertainty regarding the origin of this problem, its first published appearance was in Martin Gardner’s Mathematical Games column in February of 1960 (reprinted in [2]). A detailed history of the secretary problem and other classical stopping problems is in [1]. The secretary problem is an outstanding example of how a seed planted by one of Gardner’s columns has grown and flourished. Over the intervening halfcentury, people have studied many variants. For example, how does the answer change if the interviewer is allowed to recall the last few applicants [7]? Or what if there is a full committee of Reprinted from The College Mathematics Journal, Vol. 43, No. 1 (Jan. 2012), pp. 76–81.
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interviewers rather than a single interviewer [4]? Or how should one deal with the costs of interviewing and pressures to hire an early applicant to reduce costs [5]? As of this writing, there are over 130 papers in MathSciNet referring to the Secretary Problem and its extensions. Almost all published variants look at the problem from the employerâ€™s perspective. At a recent senior thesis presentation on one such variant at Gettysburg College, another senior, clearly concerned about his own job prospects, asked â€œWhat does this mean for applicants? If I know that an employer is going to behave optimally, when should I schedule my interview?â€? This is the question we examine here.
When rank is unknown First we consider the situation of an applicant who has no idea how strong they are compared to the rest of the pool. Throughout this note, we assume that there are n applicants for a single job, and that the employerâ€™s strategy is initially to reject the first k and then hire the first subsequent candidate who is better than all candidates already seen, if such a candidate exists, or the final candidate otherwise. We call this the optimal strategy. We assume that all nĹ possible ordered rankings of the n candidates are equally likely and examine the probability that each position in the interview order is the one where the chosen applicant is found. Let us define the random variable X to be the interview position of the applicant who is finally chosen. Using the optimal strategy, the employer rejects the first k applicants, and therefore P .X D i / D 0 if 1 i k. In order for the .k C 1/st candidate to be chosen, they must be better than all the candidates in the rejected group. This occurs exactly when the .k C 1/st candidate is the best among the first k C 1 candidates, which occurs with 1 . More generally, for k C 1 i n 1, the i th candidate will be chosen probability kC1 if and only if: The i th candidate is the best of the first i candidates.
The best of the first i 1 candidates is in the initial rejected group of k candidates. The second condition ensures that we get to the i th candidate without choosing someone else; the first condition ensures that we then choose the i th candidate. The probability of these two conditions simultaneously holding is 1i i k 1 . Finally, there are two separate situations in which the last candidate is selected: either the best candidate overall was in the initial group of rejected candidates, in which case the final candidate is chosen as a last resort, or the best overall candidate is the last one interviewed and the second best candidate was in the initial group of rejectees. The first scenario occurs with probability kn ; the second with probability n1 n k 1 . Adding these two cases together, P .X D n/ D kn C n.nk 1/ D n k 1 . In summary, we have the following result. Theorem 1. The probability that the i th candidate is chosen is 8 Ë† if i k, Ë† <0 k P .i / D i.i 1/ if k < i < n, Ë† Ë† : k if i D n. n 1
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33. The Secretary Problem from the Applicant’s Point of View
k i.i
245
The goal of the applicant is to obtain interview slot i which maximizes P .i /. Note that is a decreasing function in i . Therefore the candidate should target the .k C 1/st slot 1/
1 or the final slot, depending on whether kC1 or nk 1 is larger. An applicant will prefer the 1 . This is the case when final slot if n k 1 > kC1 p 1 C 4n 3 k> ; (33.1) 2
agreeing with the intuition that rejecting a larger number of candidates makes it more likely that the best candidate is in the rejected group and that the employer accepts the final candidate. If the employer uses the optimal strategy with k D ne , then the applicant must check whether (33.1) holds, which happens when k 2 C .1 e/k C 1 > 0. A simple calculation shows this is always the case, and therefore the candidate should choose to be interviewed in the final slot if they have no information about their relative standing with the other candidates. Of course, ne is never an integer, so in reality the employer chooses either k D b ne c or d ne e. In the latter case, the candidate will still wish to choose the final interview p
slot, because if ne > 1C 24n 3 then certainly d ne e is as well. If the employer rounds down then one can check that the candidate will be better off choosing last as long as k > 2:5. In particular, being interviewed in the final position is optimal if n 10. One can manually check that, if the interviewer is going to reject k D b ne c applicants, one will be no worse off being interviewed last except when n D 4; 5, or 8. Thus we have proved Theorem 2. If the number of applicants to a position is at least nine and an employer uses the optimal strategy then the probability that they hire the final person interviewed is higher than any other person.
When rank is known On the whole, in the absence of information about their standing, a candidate should choose to be interviewed last. This is also true if you know you are the worst candidate, as you will never be better than all the candidates in the rejected group, so you can only be chosen if you are the last resort. On the other hand, if you know you are the strongest candidate, then you prefer the .k C 1/st slot over all others. You are then guaranteed to be chosen: you are certainly better than anyone in the rejected group, and in any later slot there is a chance that someone else is picked before you are interviewed. In general, then, it seems that stronger applicants want to be interviewed early and weaker applicants later. In this section, we prove that this is so by considering the point of view of a candidate who knows they are the j th best in a pool of n candidates. Assume first that our candidate interviews last. As in the previous section, if j ¤ 1 then the only way that they can be chosen is if the best candidate is in the rejected group, which happens with probability n k 1 , since there are n 1 slots remaining for the best applicant to be in, all equally likely. On the other hand, if j D 1 then the only way to be chosen from the last slot is if the second best candidate is rejected, which also happens with probability n k 1 . In other words, the probability that the candidate in the final slot is chosen is nk 1 independent of how strong they are.
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As before, if a candidate is interviewed in slot i , k < i < n, they will be chosen if and only if they are the best candidate seen so far and the second best candidate to that point is rejected. Assume that i > 1, which will is the case as long as k > 0, i.e., at least one candidate is rejected. The probability that the j th best candidate overall is the best candidate seen so far, when they are interviewed in the i th slot, is the probability that all i 1 candidates seen previously come from the n j candidates who are worse than them. This happens with probability ni 1j = ni 11 . Moreover, the probability that the best candidate seen before the i th slot was in the rejected group is i k 1 . Putting this together, we get the following result: Theorem 3. Assume the j th best candidate is interviewed in the i th slot. The probability of getting chosen is: 8 0 if i k, ˆ ˆ < n j .i 1/ k Pj .i / D if k < i < n, .n 1/ i 1 ˆ ˆ : ki 1 if i D n. n 1 Again, Pj .i / is a decreasing function in the range k < i < n. Explicitly, we note that if i and i C 1 are in this range, then n j n 1 Pj .i / k i i 1 i D n 1 n j Pj .i C 1/ i 1 k i 1 i
.n i /Š .n i j /Š i D .n i 1/Š .n i j C 1/Š i 1 n i i D > 1; n i j C1i 1
and therefore Pj .i / > Pj .i C 1/, as desired. This means that for any fixed value of j , an applicant should either choose to be interviewed in the .k C 1/st slot or in the nth slot. In order to see which of these options gives the higher probability, we must compare Pj .k C 1/ D n k j = n k 1 with Pj .n/ D n k 1 . After some algebraic manipulation, we find that Pj .k C 1/ > Pj .n/ if and only if n k j > nk 21 . We wish to consider this inequality separately for different values of j . If j D 1, then we compare n k 1 to nk 21 . Since n k 1 D n k 2 C nk 21 , the best candidate will prefer to be interviewed in the .k C 1/st slot, as we argued earlier. If j D 2, then we compare n k 2 and nk 21 . The former is larger exactly when k < n 1 . If the employer chooses k n=e, then this holds for large values of k, implying that 2 the second best candidate should also choose the .k C 1/st slot. If j D 3, then we compare nk 3 and nk 21 . Looking at the quotient of these terms and expanding algebraically, we see that the former is larger exactly if k 2 C .5 p
2
3n/k C .n2
3n C 2/ > 0;
p
which occurs if k < 3n 5 5n2 18nC17 . For large n, this is true if k < 3 2 5 n :38n; which is (barely) guaranteed if the employer chooses k D n=e :367n. More specifically, one can work out that the thirdbest candidate should choose the .k C 1/st slot for all cases except for 33 specific values of n, the largest of which is n D 98.
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33. The Secretary Problem from the Applicant’s Point of View For j 4, comparing n k j and former term is greater exactly when
n
k k
n 2 k 1
1
247
is difficult. In particular, one can show that the
jY2 `D1
n
k n
` `
1 1
> 1:
Assuming that n D ek and letting n be large, we see that this holds if and only if .e 1/j 1 1, which is false for j 4. In particular, for sufficiently large n one prefers to ej 2 be interviewed last if ones ranking is fourth or worse. We summarize the preceding results: Theorem 4. If you are the j th best applicant then you want to be interviewed in the .k C 1/st slot for j 3, and in the final slot if j 4 and n is sufficiently large. We also note that for any n and j 3 and k > n 2 j we will have that kn 12 > nk j1 n j . If the employer sets k D d ne e then k will be greater than n 2 j if jn > e e 2 :26. In k particular, this says that regardless of n, if a candidate suspects they are not in roughly the top quarter of candidates, then they would prefer to be interviewed in the last slot.
Final thoughts Unless you are a truly extraordinary applicant, if your prospective employer uses the classic strategy, then you should try to be interviewed last. However, attempting to game the system is likely to backfire as there is no margin of error. Being the final interviewee gives one the highest probability of being hired, but being the secondtolast has one of the lowest! Students going on the job market should put their energy into improving their resum´es rather than strategizing interview timing. We note that a student can do both by generalizing our work to some of the variants of the secretary problem mentioned earlier—and publishing their work in future issues of the College Mathematics Journal. Acknowledgment The author thanks the two students who inspired this question, Brian Lemak and Paul Smith, as well as his colleagues in the math department at Gettysburg College and the anonymous referees.
Bibliography [1] T. S. Ferguson, Who solved the secretary problem? Statist. Sci. 4 (1989) 282–296; available at dx.doi.org/10.1214/ss/1177012493 [2] M. Gardner, New mathematical diversions, Mathematical Association of America, Washington, DC, revised edition, 1995. [3] J. P. Gilbert and F. Mosteller, Recognizing the maximum of a sequence, J. Amer. Statist. Assoc. 61 (1966) 35–73; available at 10.2307/2283044 [4] H. Glickman, A bestchoice problem with multiple selectors, J. Appl. Probab. 37 (2000) 718– 735; available at 10.1239/jap/1014842831 [5] Z. Govindarajulu, The secretary problem: optimal selection with interview cost, in Proceeding of the Symposium on Statistics and Related Topics, Carleton Univ., Ottawa, 1975, 19.
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[6] K. Merow, The view from here: Finding your match mathematically, Math Horizons 17 (2009) 18–20; available at 10.4169/194762109X468346 [7] O. K. Zakusilo, Optimal choice of the best object with possible returning to previously observed, Theory Stoch. Process. 10 (2004) 142–149.
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34 Lake Wobegon Dice Jorge Moraleda and David G. Stork Nontransitive dice, discovered by Bradley Efron, but introduced to the public by Martin Gardner [2], are sets of three dice which if rolled in pairs, die A most frequently beats die B, B beats C , and C beats A. This is a statistical analogue of the game Rock, paper, scissors (also called rochambeau, roshambo, and jankenpon), in which each element or action beats another and is beaten by yet another. In such a set there is no best or dominant element. Of course the scalar mean values of the dice cannot exhibit nontransitivity; it is only in the statistical case of rolls of different pairs of dice where nontransitivity can arise. For an example of nontransitive dice, see Figure 34.1.
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