Chalkdust, Issue 11

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In this issue... 4

In conversation with Trachette Jackson Ellen Jolley and Eleanor Doman talk to the professor of mathematical oncology

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How to cheat at cards Kevin Houston has a problem

16

Polynomials’ order Yuliya Nestarova walks into a bar

29

Unsupervised joke generation I like my towns like I like my Alex: Bolton

46

Kepler’s barrels Sam Hartburn orders wine by the barrel

49

Weirdonacci numbers To get the next article, add Rob J Low to Theirry Platini

61

Polytopia: adopt a polyhedron Mara Kortenkamp, Anna Maria Hartkopf and Erin Henning’s home for peculiar polytopes

49

61

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22

What can you do with this space? Nobody can draw a space filling curve by hand, but that doesn’t stop Andrew Stacey

3 8 20 28 34

Page 3 model

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On the cover: Apollonian packing by David Sheard

44 48 55

Puzzles

56 58 60 65

Crossnumber

66 68

Letters

What’s hot and what’s not Dear Dirichlet A mathematical song Comic: The inverse homotopy by Tom Hockenhull

Joke generator The big argument: is zero a natural number? Reviews What’s the shape of you? How to make... ... Ecki the polytope Top ten: pictures of scorpions spring 2020

chalkdust Welcome to Chalkdust issue 11. Whether you like it or not, we are going to start with an acrostic:

The team Carmen Cabrera Arnau Charlotte Connolly Petru Constantinescu ThĂšy DĆ°ĆĄng “TDâ€? Ă?ạng Eleanor Doman Sean Jamshidi Ellen Jolley Nikoleta Kalaydzhieva Jean LagacĂŠ Johnny Nicholson Sam Porritt Tom Rivlin Matthew Scroggs Belgin SeymenoÄ&#x;lu David Sheard Jakob Stein Adam Townsend Cartoonist Tom Hockenhull

d c a b l n e

chalkdustmagazine.com contact@chalkdustmagazine.com @chalkdustmag chalkdustmag chalkdustmag @chalkdustmag@mathstodon.xyz Chalkdust Magazine, Department of Mathematics, UCL, Gower Street, London WC1E 6BT, UK.

Space filling curves, How to make Ecki, Apollonian packing, Polynomial orders, Ed Sphere-an, Some polygons. Speaking of �-gons, the most recent year of the dragon ended in 2013; and 2013 pops up in a few articles about recent mathematical breakthroughs. First, in Yuliya Nestarova’s article about polynomials entering the origin, then exiting in a different order (pages 16– 19): it wasn’t known until 2013 which orders were possible and which were impossible. If you think our jokes are bad, then you’ll appreciate the second appearance of 2013 in Alex Bolton’s article about computer generated jokes (pages 29–33). Next time you’re at a party, crack some of these jokes to impress your friends. If this doesn’t impress them, why not try a card trick (pages 9–15), a mathematical song (page 28), or a poetry recital (page 46). And iff all else fails, then give up on your friends, sit in the corner, and enjoy reading a magazine for the mathematically curious. The Chalkdust team

Acknowledgements We would like to thank: all our authors for writing wonderful content; our sponsors for allowing us to continue making the magazine; Helen Wilson, Helen Higgins, Luciano Rila and everyone else at UCL’s Department of Mathematics; Luciano Rila, the LMS, and all the speakers for their help with the Black Role Models in STEM event; everyone at Achieve Fulfilment for their help with distribution; Matt Parker, Rob Eastaway, and Sarah Cooper for inviting us to Maths Fest, and Ben Baker for helping with our Maths Fest stall; Isaac Newton for discovering gravity; and Sandra Bullock for starring in Gravity. ISSN 2059-3805 (Print). ISSN 2059-3813 (Online). Published by Chalkdust Magazine, Dept of Mathematics, UCL, Gower Street, London WC1E 6BT, UK. Š Copyright for articles is retained by the original authors, and all other content is copyright Chalkdust Magazine 2020. All rights reserved. If you wish to reproduce any content, please contact us at Chalkdust Magazine, Dept of Mathematics, UCL, Gower Street, London WC1E 6BT, UK or email contact@chalkdustmagazine.com

chalkdustmagazine.com

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Eleanor Doman

You there! Yes—you cooking the spaghetti! How do you tell when it’s cooked? No, don’t throw it against the wall... maths is here for you. Cooking pasta increases the amount of water it contains until the pasta is fully hydrated, at which point we say it’s cooked. But this means that the pasta is no longer a rigid material—it’s exible. Nathaniel Goldberg and Oliver O’Reilly, in their 2020 paper, model the process of spaghetti cooking in a saucepan:

Measure the arclength, đ?‘ , of your spaghetto from đ?‘ = đ?&#x;˘ to đ?‘ = đ??ż and call the angle to the horizontal at any point đ?œƒ(đ?‘ ): đ?‘Ś

đ?‘Ľ

đ?‘ =0

đ?œƒ

đ?’?

đ?’Ž

đ?‘ =đ??ż

Flexible rod theory tells you that the moment along the spaghetto, đ?’Ž, depends on the angle, đ?œƒ, and the weight force, đ?’?. For a bent spaghetto, you can write the curvature, đ?œ…, in terms of the intrinsic curvature, đ?œ…đ?&#x;˘ , and the moment, divided by the rigidity, đ??¸đ??ź: đ?œ•đ?œƒ đ?‘š = đ?œ… = đ?œ…đ?&#x;˘ + . đ?œ•đ?‘ đ??¸đ??ź

In the pan!

But the intrinsic curvature changes as water is absorbed. Goldberg & O’Reilly use a model from plant stem growth for the rate of intrinsic curvature change, making it dependent on the amount of time the spaghetto has been in the water for: đ?œ•đ?œ…đ?&#x;˘ đ?œ•đ?‘Ą

= đ?›ź(đ?‘Ą)(đ?œ… − đ?œ…đ?&#x;˘ )

where đ?›ź(đ?‘Ą) = đ?›źâˆž

đ?&#x;Ł − e−đ?‘Ą/đ?œ? . đ?&#x;Ł + e−(đ?‘Ąâˆ’đ?‘Ąđ?&#x;˘ )/đ?œ?

Want the spaghetti al dente? Better get working out the parameters đ?›źâˆž and đ?œ?...

Goldberg NN & O’Reilly OM (2020). Mechanics-based model for the cooking-induced deformation of spaghetti, Physical Review E 101 013001

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I n c o n versat io n w it h . . .

Trachette Jackson Ellen Jolley and Eleanor Doman

O

ncology, the study of cancer, is just one of many specialisms which increasingly employs the predictive power of applied mathematics. This issue we chat with Trachette Jackson, professor of mathematics at the University of Michigan, to learn about the surprising effectiveness of mathematics in the treatment of cancer, as well as to hear about her own journey into mathematical oncology.

Modelling in medicine Trachette starts by bringing us up to date on how mathematics has been used in cancer treatment. “Mathematical approaches have been applied to just about every aspect of tumour growth, starting decades ago.” One aspect is cancer therapeutics: “We write down equations that describe the mechanism of action of new drugs and how the tumour responds, then make predictions about how best to deliver those drugs.” Another aspect is more fundamentally biological, such as how cells are transformed: “You can find mathematical equations describing the probabilities of acquiring Cancer cells mutations and under what circumstances a tumour forms, as well as what the composition of that tumour will be and how many cells in that tumour will have chalkdustmagazine.com

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chalkdust these different mutations.” This sort of modelling allows us to diagnose or assess the risk of cancer developing, as well as treat it. Trachette’s recently published paper, on the modelling of a drug targeting a chemical needed by cancerous tumours to grow, demonstrates the value of mathematical modelling. “We built a mathematical model that could look at the crosstalk between tumour cells and endothelial cells that secrete the chemical, then used it to predict responses of the tumour to this new treatment, along with traditional chemotherapy.” The model predicted something unexpected—administering Wikimedia Commons user Wapcaplet, CC BY-SA 3.0 the two drugs at the same time, as suggested by experimenMichigan talists, “would give you the worst results possible. Instead, the model predicted that fortnightly injections of chemotherapy along with continuous weekly administration of low amounts of targeted drugs would lead to synergistic responses.” Could these same achievements have been made without the models? “I think the answer is probably yes. But it would be very expensive, and it would take a lot longer. What does math do? It speeds it up, and it saves the lives of millions of mice these drugs are tested on.” We were happy to hear, then, that mathematics is fundamentally better than biology and will soon be replacing it entirely. But Trachette kept us grounded by reminding us that it goes both ways. In fact, it is through her collaboration with experimentalists that she decides where to focus her attention. “They’re the ones who know what the big questions in cancer biology are, and which problems they would love answers to; those they think mathematical modelling might help them address.” We couldn’t hold in the burning question any longer—what kind of maths are we talking about here? “I mainly use differential equations, both ordinary and partial, and sometimes delay differential equations.” In a delay differential equation, the derivative at a certain time is given in terms of the value of the independent variable at previous times. “I also use hybrid models, which are methods that combine continuous differential equations with a discrete approach, to answer questions that look at the interplay between something that diffuses like a chemical, the biochemistry, with the biomechanics, like a cell actually moving on tissue structures.” So, it all boils down to solving differential equations. Should be easy enough? “I’m talking about systems of non-linear differential equations, so we’re past the point of pen and paper analysis. We have to go to the computer and run simulations.” But that doesn’t stop Trachette from having fun. “Sometimes, we can do some model reduction to get the system in a form where we can say something about it analytically, but often the systems I work with require solving numerically.” In the future, she expects to see big improvements in how patients’ data is used in mathematical modelling, in what is called ‘personalised medicine’. “When a patient is diagnosed with cancer, they’re put through a battery of tests that gives information about the molecular details of that cancer. We need much more of that data collected, and we need to be able to use all the information clinically available: from a specific person’s genetics, to the way that their body processes these drugs. Instead of having a blanket regime for every type of cancer, we really start looking at what to do for individual patients.” 5

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A leopard can change its spots Trachette credits her parents for fostering her love of learning. “They really stressed the importance of education. They let me know that learning is fun, it’s OK to be a smart girl, and it’s OK to be a girl who is good at math.” She moved around a lot as a child, because her father was in the US air force, but she eventually settled in Arizona where she went to high school and became interested in mathematics. She went to a local university to take classes in the summer, and this is where she met her first maths professor. “He didn’t look anything like I thought a math professor would look. I was shocked that a normal looking person could be a mathematician. He became a very big role model.” This professor helped to point her in the right direction when she went on to study for her degree at this same university. Initially, she wasn’t studying maths. “Believe it or not, I was an engineering major. I quickly learned I didn’t like engineering, so he called me into his office and said you should be a math major.” She took the advice, but at first chose pure maths, studying “nothing applied at all”. It took a second role model to show her what applied mathematics can do and to inspire her to go into mathematical biology. James Murray, a pioneer in the field of mathematical biology, came to her university and gave a talk about how leopards get their spots. “I thought, ‘This does not belong in the math department! How is math going to have anything to do with leopards and spots?’ But I went to the talk, and… my mind was blown.”

Spencer Wright, CC BY 2.0

A leopard’s spots

James Murray later became her PhD advisor. At the time, she didn’t understand all the mathematics in his talk, but she took away what turned out to be a much more important message. “I understood that math can play a role in science, in biology, in medical science, and developmental biology, and I had never seen that before. That was when I knew I wanted to be an applied mathematician, because I saw real benefit of math in those areas. I wanted to be like this person I saw giving this talk.”

Dreaming of the future Our conversation turns onto the topic of diversity in academia. Trachette hopes that today’s young women can be inspired the way she was to consider mathematics. She says she is a “big fan” of recent initiatives to get more women interested in mathematics and has a lot of praise for the Association of Women in Mathematics in the US and their networking schemes, saying they are “just amazing”. She says they have communities where “women can find like-minded women in their own areas and form networks”. She is also very impressed by their mentoring programmes, which enable women to “find women at higher career stages, so they can ask those difficult questions through all the critical transitions in your career. Sometimes it feels a little isolating to be a woman in math. And this is a way to get, even at undergraduate level, connected to women.” Trachette is very passionate about increasing opportunities in mathematics for people of colour. “Being an African-American mathematician at a research institution is very rare, and I want to chalkdustmagazine.com

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chalkdust change that demographic.” This is why she “started a programme to help underrepresented minorities progress through the pipeline. I think it’s very important to support the movement of diverse people through the PhD and beyond, so that we actually change the face of the professoriate.” Trachette takes lots of opportunities to be a visible role model in her career, for example by appearing in Power in Numbers by Talitha Williams, which tells the stories of over 30 women’s journeys through maths. You can find Chalkdust’s review on our blog. She sees two approaches to improving diversity in universities. First, there is the “top-down” approach, involving national level organisations. “I know these big societies have diversity, equity, and inclusion committees, and are aware of the problem. The mathematical community often looks to these societies, so it trickles down.” But change can also be from the bottom-up, and it is important that more effort is made to make these kinds of improvements. “There needs to be proponents of diversity in each department to Power in Numbers by Taltry to locally start changing the culture to be more welcoming and itha Williams inclusive, and to take seriously the issue of recruiting a diverse graduate student body and faculty. I think the mathematical community really needs a deep dive into where we are and how we got here and what we can do change it.” She seems positive about the future, and her closing message about the next generation of female mathematicians is very inspiring. “I hope they will blossom and find networks of other people to work with and find mentors at all levels. I hope that they find people to give them good advice, and to advocate for them, so they can eventually be agents of change.” Ellen Jolley Ellen is a PhD student at UCL studying fluid mechanics. She specialises in the flow around droplets and ice particles.

Eleanor Doman Eleanor is a maths PhD student at UCL. She works on biomechanical models of nerves and is interested in the applications of mathematical modelling to biological and medical problems.

a @eleanordoman Six months ago, we didn’t know... ...that the Collatz conjecture (see page 64) is true if and only if a group constructed by Ian Leary is trivial? How hard can it be to check that a group is trivial?!

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Maths is a ďŹ ckle world. Stay Ă la mode with our guide to the latest trends.

& WHAT’S

WHAT’S

HOT NOT

Agree? Disagree? a @chalkdustmag b chalkdustmag l chalkdustmag f chalkdustmag

Epidemiological models Keeping us safe, informing policy, saving lives. Maths is rarely this visibly important.

HOT

HOT

Polytopes

All the hot maths celebs are joining in with #PolytopeSelďŹ e. Adopt your own on pages 61–64.

Tea

HOT

No-one wants to move to Albuquerque.

NOT HOT

Indoor maths

Read articles (see pages 1–68), solve the crossnumber (see page 56), or write something for issue 12.

See page 31.

Coee See page 31.

Isotopes

Outdoor maths

HOT NOT

đ?œˆ (nu)

No more calculating the height of the church spire (or tower) using trigonometry.

NOT

One of the least useful Greek letters.

Fake đ?œˆs

Tensors f

For example: đ?‘Ł , đ?‘˘, đ?œ?, , or đ?‘Ś .

HOT

HOT

A useful way to represent multidimensional data.

Ten saws Once you’ve got three saws, you probably don’t need any more.

NOT More free fashion advice online at d chalkdustmagazine.com

Pictures Background: Flickr user Harmon, CC BY-SA 2.0. Graph: Johannes Kalliauer, CC BY-SA 4.0.

chalkdustmagazine.com

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chalkdust

How to cheat at cards Kevin Houston

A

s a child I didn’t want to be a mathematician—I didn’t even know that was a job. Instead, for a short time, my dream was to be a crime-fighting card cheat. My interest arose from the TV series Sword of Justice where the main character obtained some help from card cheats in his quest to take down the despicable crime syndicate that had framed him and sent him to prison. (The opening credits on YouTube will attest to the programme’s brilliance.) Despite the incompatibility of combining crime-fighting and card cheating I told my family that that was to be my chosen career. Unfortunately, I wasn’t very good at it, usually announcing my intentions to cheat before offering to play a quick game with someone. Nonetheless, to this day my family refuse to play cards with me—even Uno. Along the way, I did learn a thing or two though. So allow me to explain to you an impressive trick that all the best card cheats know and that has some interesting mathematics hidden behind it.

The perfect shuffle Every card cheat needs to be able to give the appearance of shuffling a deck of cards so that it remains in order. However, there is an entirely legitimate shuffle that allows us to cheat: the perfect shuffle. It mixes the cards but in a perfectly known and determined way. A standard deck has 52 cards. Split them into two equal piles and then interweave the piles so that 9

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chalkdust the new order of cards alternates between the two piles. The effect on the deck is shown on the right, where the cards are labelled đ?&#x;˘ to đ?&#x;§đ?&#x;Ł for reasons which will become clear later.

0 1 2 3

Instructions on how to do the shuffle are given below. Admittedly, it is not easy and plenty of practice is required but is satisfying once learned and has an impressive ‘wow!’ reaction when performed.

49 50 51

â‹Ž â‹Ž

0 1 2 3

26 27 28 29 49 50 51

â‹Ž â‹Ž

23 24 25

A perfect (out) shuffle

Eight perfect shues

What can we do with the perfect shuffle when we have learned it? First we have an interesting— and hopefully surprising—fact: Perfect shuffle a deck eight times. It is now back in the order you started with! With this you can perform a surprising trick. Put the deck in new deck order. If you perfect shuffle it five times, then the deck looks shuffled. There are patterns if one looks closely but you can show it to someone and with a straight face claim it is mixed up. Do the perfect shuffle three times for them—so now you have done it eight times in total—and to your spectator’s bafflement the deck is suddenly in order! How can we use mathematics to prove this surprising fact? We start by numbering our cards. As above, from top to bottom, we shall number them đ?&#x;˘ to đ?&#x;§đ?&#x;Ł, rather than the usual đ?&#x;Ł to đ?&#x;§đ?&#x;¤.

Next we use modular arithmetic. Arithmetic modulo đ?‘› is where we Modular arithmetic is consider all numbers that have the same remainder after division by also used by clocks: they đ?‘› to be equivalent numbers. So đ?&#x;§ = đ?&#x;Łđ?&#x;Š mod đ?&#x;Ś means that đ?&#x;§ and đ?&#x;Łđ?&#x;Š count the time mod 12. are equivalent modulo đ?&#x;Ś as they both have remainder đ?&#x;Ł after division by đ?&#x;Ś. Sometimes people write đ?&#x;§ ≥ đ?&#x;Łđ?&#x;Š mod đ?&#x;Ś rather than using the equals sign.

With our đ?&#x;˘ to đ?&#x;§đ?&#x;Ł numbering the formula for finding where a card goes to after a perfect shuffle is relatively easy: (Except for card 51) the card at position đ?‘Ľ goes to position đ?&#x;¤đ?‘Ľ mod đ?&#x;§đ?&#x;Ł.

For example, card đ?&#x;Š goes to đ?&#x;¤ Ă— đ?&#x;Š mod đ?&#x;§đ?&#x;Ł = đ?&#x;Łđ?&#x;Ś mod đ?&#x;§đ?&#x;Ł, ie position đ?&#x;Łđ?&#x;Ś. Note that this is the đ?&#x;Łđ?&#x;§th physical card in the deck as we started counting at đ?&#x;˘. chalkdustmagazine.com

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chalkdust Similarly card đ?&#x;Ľđ?&#x;§ goes to

đ?&#x;¤ Ă— đ?&#x;Ľđ?&#x;§ mod đ?&#x;§đ?&#x;Ł = đ?&#x;Šđ?&#x;˘ mod đ?&#x;§đ?&#x;Ł

= đ?&#x;§đ?&#x;Ł + đ?&#x;Łđ?&#x;Ť mod đ?&#x;§đ?&#x;Ł = đ?&#x;Łđ?&#x;Ť.

That is, card đ?&#x;Ľđ?&#x;§ goes to position đ?&#x;Łđ?&#x;Ť.

Note that the card đ?&#x;Š gets lower in the deck and card đ?&#x;Ľđ?&#x;§ gets higher. So immediately we see that a perfect shuffle is beginning to mix the cards. The formula has the exception of card 51. It is easy to see in the picture on the previous page that this card does not move and so we don’t have to worry about it in the formula—we will ignore it from now on.

But why does this đ?‘Ľ ↌ đ?&#x;¤đ?‘Ľ mod đ?&#x;§đ?&#x;Ł result hold? Consider first the top half of the deck, ie all the cards in positions đ?&#x;˘ to đ?&#x;¤đ?&#x;§. During the shuffle, the two cards in a pair of consecutive cards get a card between them. So the new positions of the top half cards will be multiplied by đ?&#x;¤. Since we can have at most đ?&#x;¤ Ă— đ?&#x;¤đ?&#x;§ = đ?&#x;§đ?&#x;˘, which is less than đ?&#x;§đ?&#x;Ł, we get đ?‘Ľ ↌ đ?&#x;¤đ?‘Ľ mod đ?&#x;§đ?&#x;Ł since đ?&#x;¤đ?‘Ľ is in this case actually equal to đ?&#x;¤đ?‘Ľ mod đ?&#x;§đ?&#x;Ł (rather than just congruent). For the bottom half cards we see that their positions should also be multiplied by đ?&#x;¤ since they get cards between them. However, we note that card đ?&#x;¤đ?&#x;¨ (the top card of the bottom half of the deck) needs to go to position đ?&#x;Ł (the second physical card position!). One way to do this is to take mod 51 (since đ?&#x;¤ Ă— đ?&#x;¤đ?&#x;¨ mod đ?&#x;§đ?&#x;Ł = đ?&#x;Ł). It is easy to check that we can use this for all the other cards.

Now let’s see what happens when we do eight perfect shuffles on a particular card, say card đ?&#x;Łđ?&#x;¤. We just repeat the process that the card in position đ?‘Ľ will go to position đ?&#x;¤đ?‘Ľ mod đ?&#x;§đ?&#x;Ł: once twice three times four times five times six times seven times eight times

đ?&#x;Łđ?&#x;¤ đ?&#x;¤đ?&#x;Ś đ?&#x;Śđ?&#x;Ş đ?&#x;Śđ?&#x;§ đ?&#x;Ľđ?&#x;Ť đ?&#x;¤đ?&#x;Š đ?&#x;Ľ đ?&#x;¨

→ → → → → → → →

đ?&#x;¤đ?&#x;Ś đ?&#x;Śđ?&#x;Ş đ?&#x;Ťđ?&#x;¨ đ?&#x;Ťđ?&#x;˘ đ?&#x;Šđ?&#x;Ş đ?&#x;§đ?&#x;Ś đ?&#x;¨ đ?&#x;Łđ?&#x;¤

mod đ?&#x;§đ?&#x;Ł mod đ?&#x;§đ?&#x;Ł mod đ?&#x;§đ?&#x;Ł mod đ?&#x;§đ?&#x;Ł mod đ?&#x;§đ?&#x;Ł mod đ?&#x;§đ?&#x;Ł mod đ?&#x;§đ?&#x;Ł mod đ?&#x;§đ?&#x;Ł

= = = =

đ?&#x;Śđ?&#x;§ đ?&#x;Ľđ?&#x;Ť đ?&#x;¤đ?&#x;Š đ?&#x;Ľ

mod đ?&#x;§đ?&#x;Ł mod đ?&#x;§đ?&#x;Ł mod đ?&#x;§đ?&#x;Ł mod đ?&#x;§đ?&#x;Ł

= đ?&#x;Łđ?&#x;¤ mod đ?&#x;§đ?&#x;Ł.

Of course, that doesn’t prove that this happens for the whole deck. I may have been cheating by picking one that worked—and you should never trust a card cheat! So let’s do the calculation for a general đ?‘Ľ . If we repeat the đ?&#x;¤đ?‘Ľ mod đ?&#x;§đ?&#x;Ł operation we get đ?&#x;¤(đ?&#x;¤đ?‘Ľ) mod đ?&#x;§đ?&#x;Ł. Continuing in this way for the full đ?&#x;Ş times we see card đ?‘Ľ goes to position đ?&#x;¤đ?&#x;Ş đ?‘Ľ mod đ?&#x;§đ?&#x;Ł. Simplifying mod đ?&#x;§đ?&#x;Ł we get đ?&#x;¤đ?&#x;Ş đ?‘Ľ mod đ?&#x;§đ?&#x;Ł = đ?&#x;¤đ?&#x;§đ?&#x;¨đ?‘Ľ mod đ?&#x;§đ?&#x;Ł

= ((đ?&#x;§ Ă— đ?&#x;§đ?&#x;Ł)đ?‘Ľ + đ?‘Ľ) mod đ?&#x;§đ?&#x;Ł = (đ?&#x;˘ + đ?‘Ľ) mod đ?&#x;§đ?&#x;Ł

= đ?‘Ľ.

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chalkdust Hence, we have proved that eight shuffles move the card in position đ?‘Ľ back to position đ?‘Ľ .

If we have a different number of cards in our deck, then the number of perfect shuffles needed to return the deck to its original order varies. For example, a deck with đ?&#x;§đ?&#x;˘ cards needs đ?&#x;¤đ?&#x;Ł perfect shuffles. The number of shuffles required for the deck is called the order. The following table shows the order of the perfect shuffle for a selection of decks where đ?‘ denotes the number of cards: đ?‘ order

4 2

6 4

8 3

10 6

12 10

14 12

16 4

18 8

‌ ‌

50 21

52 8

54 52

The word order arises due to a connection with group theory and the order of elements of a certain group. The details of this and many more hardcore mathematical facts can be found in Magic Tricks, Card Shuffling and Dynamic Computer Memories by S Brent Morris: one of its appendices is a list of all orders for đ?‘ up to 200.

Binary fun with in and out perfect shues Up to this point I have talked of a perfect shuffle, ie singular. In fact, there are two types: • A perfect out shuffle: the top card remains the top card. • A perfect in shuffle: the top card becomes the second card. The perfect out shuffle is just what have been calling the perfect shuffle. The in shuffle is the new one. Its effect on deck order can be seen below—the top card ends up ‘in’ the deck. By combining out and in shuffles we can move the top card to anywhere in the deck. And, as we shall see, from a mathematical perspective we get a surprising appearance of binary.

0 1 2 3

Alex Elmsley (1929–2006) was a mathematician and magician. He studied physics and mathematics at the University of Cambridge before becoming interested in computers. Among magicians he is known for inventing a convincing false count of cards which is now known as the Elmsley Count.

49 50 51

â‹Ž â‹Ž

0 1 2 3

26 27 28 29 49 50 51

â‹Ž â‹Ž

23 24 25

A perfect in shuffle.

Elmsley was very interested in the perfect shuffle (which among magicians is also known as the Faro shuffle) and investigated the mathematics behind it. In particular, he found a method for moving the top card to any position in the deck through a sequence of in and out shuffles. Let us suppose we wish to move the top card to position đ?‘˜ in the deck. (We keep the numbering of the cards from đ?&#x;˘ to đ?&#x;§đ?&#x;Ł so the top card is in position đ?&#x;˘.) To move the card via a sequence of in and out shuffles we proceed as follows: chalkdustmagazine.com

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chalkdust Translate đ?‘˜ to binary. Then do out and in shuffles where out = đ?&#x;˘ and in = đ?&#x;Ł. Note how brilliant it is that ‘o’ looks like đ?&#x;˘ and ‘i’ looks like đ?&#x;Ł.

For example, suppose we want to move the top card to position đ?&#x;Łđ?&#x;Ľ. Well, đ?&#x;Łđ?&#x;Ľ is written as 1101 in binary. So we do ‘i i o i’: in in out in.

in

â‹Ž â‹Ž

out

in

â‹Ž

â‹Ž

in

â‹Ž

â‹Ž

â‹Ž

â‹Ž

â‹Ž

â‹Ž

Moving the top card to position 13.

So why does this work? Imagine we have a card in position đ?‘Ľ in the top half of the deck. (The case of the bottom half is similar and is, of course, left as an exercise.) What does an out shuffle do to the binary representation of the number? Since an out shuffle is đ?‘Ľ ↌ đ?&#x;¤đ?‘Ľ mod đ?&#x;§đ?&#x;Ł it just adds a zero to the representation. So, a card at position đ?&#x;Łđ?&#x;˘đ?&#x;Ł moves to đ?&#x;Łđ?&#x;˘đ?&#x;Łđ?&#x;˘. This is no different to when we tell children that to multiply by đ?&#x;Łđ?&#x;˘ they can just place a đ?&#x;˘ at the end. Now, an in shuffle is given by

đ?‘Ľ ↌ đ?&#x;¤đ?‘Ľ + đ?&#x;Ł mod đ?&#x;§đ?&#x;¤.

That is, multiply by đ?&#x;¤, add đ?&#x;Ł and take mod đ?&#x;§đ?&#x;¤ (note, not đ?&#x;§đ?&#x;Ł this time). This can be checked in the same way as for the out shuffle. For a binary representation this will just add a đ?&#x;Ł to the end. Thus, the card in position đ?&#x;Łđ?&#x;˘đ?&#x;Ł moves to position đ?&#x;Łđ?&#x;˘đ?&#x;Łđ?&#x;Ł.

To put this all together, consider what happens in the above example where we move the top card to position đ?&#x;Łđ?&#x;Ľ. That is, we apply in in out in and track the progress of the initial top card. The first in shuffle moves the top card to position đ?&#x;Ł in the deck. The second in shuffle moves it to đ?&#x;Łđ?&#x;Ł. The out moves it to đ?&#x;Łđ?&#x;Łđ?&#x;˘ and the last in moves it to position đ?&#x;Łđ?&#x;Łđ?&#x;˘đ?&#x;Ł. OK, so this is an unexpected and charming appearance of binary but it only moves one card. That’s useful, but let’s see how we can move more cards and cheat in a real game.

Dealing four aces An impressive bit of cheating is to deal yourself four aces in a game of poker—that’s a hand that is hard to beat. For our version we need a game with three other players. Take out the four aces and place them on top. A card cheat can do this secretly with ease. Now, do two perfect out shuffles. 13

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chalkdust Deal the cards starting with you, ie go round in a circle giving each person a card until everyone has five cards each. You now have a hand with the four aces! The reason that this works is simple. At the start, the pack (starting at the top) looks like AAAAxxxxxxx‌ where A denotes an ace and x denotes an indifferent card (whose value is unimportant). Do the first shuffle and indifferent cards get placed between the aces. So we get out

AxAxAxAxxxxxxx‌ Do the second shuffle and we get a set up ready for a good hand when there are three other players:

out

â‹Ž

â‹Ž â‹Ž

â‹Ž

AxxxAxxxAxxxAxxxxxxx‌

â‹Ž â‹Ž

Two perfect out shuffles give you all the aces (green).

Well, of course if you are going to cheat and win all the time, people are going to notice. To dodge this professional card cheaters would often work in pairs and the winning would alternate between the two cheaters, reducing the chances of discovery. So in this poker game of four players you need an accomplice. Suppose you are sitting in a circle with you dealing first to the person on your left and then going clockwise. Wherever your accomplice sits you will be able to deal them the four aces. As before arrange the four aces on top. Suppose your partner is sitting at position 3. You need to deal them the third card—the card at position 2 in our numbering. Let’s move that top ace to position đ?&#x;¤. In binary the number đ?&#x;¤ is 10 so we do ‘io’ perfect shuffles. Hey presto, the top ace is now in the right place and furthermore so are the other aces as the two shuffles put three cards between each of them. A normal deal now will give your partner-in-crime all four aces. Let’s see that in a diagram: start: in shuffle: out shuffle:

AAAA!

AAAAxxxxxxx‌ xAxAxAxAxxxxx‌ xxAxxxAxxxAxxxAxxxxx‌

Wherever your partner sits you can do the right sequence of in and out shuffles to ensure that they receive the aces! If you wish to know more about card shuffling and mathematics, then the best book to read is (as mentioned already) Magic Tricks, Card Shuffling and Dynamic Computer Memories by S Brent chalkdustmagazine.com

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chalkdust Morris. And remember, card cheating is a crime—so don’t get caught!

How to do the perfect shue The perfect shuffle requires cards in good condition so buy a new deck on which to learn and don’t use it for anything else—cards buckled in card games will not weave very well. There are various methods to achieve a perfect shuffle. This version is done ‘in the hands’ but some can be done on the table and don’t look in any way suspicious.

C Houston

Splitting the deck

Hold the lower half of the deck in your left hand with fingers arranged around as above. The right hand has a similar grip but does not go all the way round and the index finger sometimes contacts the top of the deck, sometimes not. Butt the cards together to square them up. If the cards are not square, then perfect shuffling is impossible. Press the halves together, with C Houston the right index finger touching both top cards so that the halves can Squaring up the cards be put in position. For an out shuffle the right hand part should be slightly higher so that the left hand top card will go under the right hand top card at the end of the shuffle. For an in shuffle the right hand part needs to be lower. Remove the index finger and begin applying pressure at the bottom corner where the two packs meet. Once interweaving has started let the deck ‘do the work’. It took me many weeks to be able to do this consistently so don’t be discouraged if you can’t do it straight away. Like many skills it needs practice!

C Houston

A perfect shuffle

Kevin Houston Kevin is a senior lecturer in the School of Mathematics at the University of Leeds. He is the author of the bestselling textbook How To Think Like A Mathematician and is surprisingly lucky at card games.

d kevinhouston.net a @k_houston_math Did you know... ...that eđ?&#x;¤iĎ€ + đ?&#x;Ł = đ?&#x;¤? 15

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Polynomials’ order Yuliya Nestarova

T

hree polynomials walk into a bar. Well, a đ?‘Ś -axis, actually. In all seriousness, suppose three distinct polynomials approach the axis đ?‘Ľ = đ?&#x;˘ from the left: one polynomial is bound to be on top, one on the bottom...

Are they guaranteed to come out the other side in much the same way? And if not, how can we permute them? And if we may permute them, are all permutations possible? How many are possible? Suddenly, the problem becomes far less a joke and far more a recently-closed problem. It was only in 2013 that Étienne Ghys demonstrated it in its generality: a permutation is realisable by polynomials if and only if it avoids being one of two types. Let’s make our way to the intersection, order up a drink, wiggle some polynomials, and explore the mystery!

The origin Formally, we wish to know if it is possible for three polynomials to cross the đ?‘Ś -axis in one configuration and come out the other side with their order switched. By adjusting our polynomials up and down as necessary, we can restate the problem to be about polynomials crossing through the chalkdustmagazine.com

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chalkdust point (đ?&#x;˘, đ?&#x;˘). Let us label the top polynomial to the left of the đ?‘Ś -axis as ‘1’, the one below it as ‘2’, and the bottom one as ‘3’: in what order do ‘1’, ‘2’, and ‘3’ emerge on the right side of the axis? Using cycle notation, we can keep track of these wiggling polynomials. Dipping into a bit of group theory, a permutation đ?œŽ ∈ đ?‘†đ?‘› is code for a rearrangement of the numbers đ?&#x;Ł, đ?&#x;¤, đ?&#x;Ľ, ..., đ?‘›. The symmetric group, đ?‘†đ?‘› , is a much studied object: we can detect it when looking at the set of all possible permutations, or studying the symmetries of an đ?‘›-simplex, and a whiff of it is present in many more maths problems. To every configuration of polynomials entering the origin from the left as đ?&#x;Ł, đ?&#x;¤, đ?&#x;Ľ and emerging in some new order đ?œŽ(đ?&#x;Ł), đ?œŽ(đ?&#x;¤), đ?œŽ(đ?&#x;Ľ) so that the new position of 1 is đ?œŽ(đ?&#x;Ł) etc, we may attach a permutation (đ?&#x;Ł đ?œŽ(đ?&#x;Ł) đ?œŽ(đ?œŽ(đ?&#x;Ł))). For example, for the polynomials in the picture on the right, đ?&#x;Ł, đ?&#x;¤, đ?&#x;Ľ left of the origin becomes đ?&#x;¤, đ?&#x;Ľ, đ?&#x;Ł on the right as đ?‘Ľ đ?&#x;Ľ − đ?&#x;Ľđ?‘Ľ dips from top place to come out on the bottom on the right, so the permutation here is (đ?&#x;Ł đ?œŽ(đ?&#x;Ł) đ?œŽ(đ?œŽ(đ?&#x;Ł)) = (đ?&#x;Ł đ?&#x;Ľ đ?œŽ(đ?&#x;Ľ)) = (đ?&#x;Ł đ?&#x;Ľ đ?&#x;¤). Are all 6 permutations in the symmetric group đ?‘†đ?&#x;Ľ of 3-object permutations achievable?

1

2

(đ?&#x;Ł đ?&#x;¤ đ?&#x;Ľ)

đ?‘Ľ

−đ?‘Ľ đ?‘Ľ đ?&#x;¤

(đ?&#x;Ł đ?&#x;Ľ đ?&#x;¤)

đ?‘Ľ

đ?&#x;˘

đ?&#x;Ł đ?&#x;Ľ đ?‘Ľ đ?&#x;¤

1 2

3

3

A short search can find all 6 such 3configurations. In fact, we need look no further than cubics to observe all possibilities. Here they are in full:

đ?‘Ľđ?&#x;¤ đ?‘Ľđ?&#x;Ľ

đ?‘Ľđ?&#x;¤

đ?‘Ľ đ?&#x;Ľ − đ?&#x;Ľđ?‘Ľ

These polynomials give the permutation (đ?&#x;Ł đ?&#x;Ľ đ?&#x;¤)

đ?‘Ľ đ?&#x;Ľ −đ?‘Ľ đ?&#x;¤

(đ?&#x;Ł đ?&#x;¤)

đ?‘Ľ

đ?&#x;˘ −đ?‘Ľ

(đ?&#x;Ł đ?&#x;Ľ)

đ?&#x;˘ −đ?‘Ľ đ?&#x;Ľ đ?‘Ľ đ?&#x;¤ (đ?&#x;¤ đ?&#x;Ľ)

Possibilities with four polynomials In 2009, Maxim Konsevitch had shown the following: the permutations (đ?&#x;Ł đ?&#x;Ľ đ?&#x;Ś đ?&#x;¤) and (đ?&#x;Ł đ?&#x;¤ đ?&#x;Ś đ?&#x;Ľ) cannot be realised by polynomials. All other 22 permutations of {đ?&#x;Ł, đ?&#x;¤, đ?&#x;Ľ, đ?&#x;Ś} can be realised, yet these 2 mysterious configurations are not possible to achieve with polynomial functions. To prove the permittable permutations is to write down 22 cases: an engaging activity of which it would be criminal to deprive the reader. To prove the impossibility of our mysterious pair, we will need to 17

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chalkdust pull the concept of a valuation from our algebraic minibar. Suppose đ?‘“ (đ?‘Ľ) = đ?‘Žđ?‘› đ?‘Ľ đ?‘› + ... + đ?‘Žđ?&#x;Ł đ?‘Ľ + đ?‘Žđ?&#x;˘ is a real polynomial: a polynomial with real numbers for coefficients. The valuation of đ?‘“ at đ?‘Ľ = đ?&#x;˘, đ?‘Ł(đ?‘“ ), is the smallest đ?‘– with đ?‘Žđ?‘– ≠đ?&#x;˘. For example, đ?‘Ł(đ?&#x;Şđ?‘Ľ đ?&#x;¤ + đ?&#x;¤đ?&#x;˘đ?‘Ľ) = đ?&#x;Ł

and

đ?‘Ł(đ?‘Ľ đ?&#x;¤đ?&#x;˘ + đ?&#x;¨đ?&#x;˘đ?&#x;˘đ?‘Ľ đ?&#x;Łđ?&#x;˘ − đ?‘Ľ) = đ?&#x;Ł.

Close to the origin, the smallest terms serve the most decisive role. The valuation of a constant is defined to be ∞.

If you have yourself been gathering the 22 polynomial cases, you have perhaps noticed that it is useful to treat the đ?‘Ľ -axis itself as a polynomial, simplifying the problem. (To achieve this, we may pick a polynomial đ?‘“đ?‘˜ to straighten out and replace all polynomials đ?‘“đ?‘– under consideration by đ?‘“đ?‘– − đ?‘“đ?‘˜ . For the purposes of our proof, pick đ?‘“đ?‘˜ to be đ?‘“đ?&#x;Ś , one that’s the bottom of the bunch to the left of the axis.) Imagine the four polynomials ambling from left to right in time, crossing over the origin and switching places in the process in accordance with the target permutation (đ?&#x;Ł đ?&#x;Ľ đ?&#x;Ś đ?&#x;¤). Under our assumption, to the left of the origin and before the mixing, đ?‘“đ?&#x;Ś is at the bottom (our newly minted đ?‘Ľ -axis) and the other 3 polynomials are above. Since đ?‘“đ?&#x;Ś goes to second place under (đ?&#x;Ł đ?&#x;Ľ đ?&#x;Ś đ?&#x;¤), we must have 2 polynomials which were previously above đ?‘“đ?&#x;Ś for đ?‘Ľ < đ?&#x;˘ but now are below for đ?‘Ľ > đ?&#x;˘. Making use of our assumption that đ?‘“đ?&#x;Ś is the đ?‘Ľ -axis, we look for 2 polynomials (our đ?‘“đ?&#x;Ł and đ?‘“đ?&#x;Ľ ) which cross the đ?‘Ľ -axis at the origin, with đ?‘“đ?&#x;Ł , now in 3rd place, on top of the đ?‘“đ?&#x;Ľ , now fourth.

1

2 3

? 2

4

3

Crucially, we may observe that close to the origin, ? the non-constant polynomial with đ?‘Ł(đ?‘“ ) = đ?‘› is ap(đ?&#x;Ł đ?&#x;Ľ đ?&#x;Ś đ?&#x;¤) gives us trouble proximated by đ?‘Žđ?‘› đ?‘Ľ đ?‘› : all other terms go to zero faster than the valuation’s term. But what of our valuation? It is useful to us precisely for the fact that đ?‘“ (đ?‘Ľ) crosses the đ?‘Ľ -axis at the origin if and only if đ?‘Ł(đ?‘“ ) is odd. And by definition it is the odd functions which have rotational symmetry and switch sign either side of the origin. Thus, near the origin, if đ?‘› is odd and đ?‘Žđ?‘› is positive, đ?‘“ (đ?‘Ľ) ≈ đ?‘Žđ?‘› đ?‘Ľ đ?‘› which is {

>đ?&#x;˘ <đ?&#x;˘

for đ?‘Ľ < đ?&#x;˘, for đ?‘Ľ > đ?&#x;˘.

What of it? We now have established đ?‘Ł(đ?‘“đ?&#x;Ľ ) and đ?‘Ł(đ?‘“đ?&#x;Ł ) to be odd and đ?‘Ł(đ?‘“đ?&#x;¤ ) to be even. The parity is different. Let’s look at the order in which these polynomials appear more closely. To the left of the origin,

chalkdustmagazine.com

đ?‘“đ?&#x;Ł (đ?‘Ľ) > đ?‘“đ?&#x;¤ (đ?‘Ľ) > đ?‘“đ?&#x;Ľ (đ?‘Ľ) > đ?&#x;˘

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chalkdust by our choice of indices; to the right, đ?‘“đ?&#x;¤ (đ?‘Ľ) > đ?&#x;˘ > đ?‘“đ?&#x;Ł (đ?‘Ľ) > đ?‘“đ?&#x;Ľ (đ?‘Ľ).

And near the origin, in the good company of fractions, each function looks much like its valuationexponent power of đ?‘Ľ . Inching closer to (đ?&#x;˘, đ?&#x;˘) as necessary, we find đ?‘Ł(đ?‘“đ?&#x;Ł ) ⊞ đ?‘Ł(đ?‘“đ?&#x;¤ ) ⊞ đ?‘Ł(đ?‘“đ?&#x;Ľ ). We had glimpsed earlier that đ?‘“đ?&#x;Ł (đ?‘Ľ) > đ?‘“đ?&#x;Ľ (đ?‘Ľ) to the right of the origin, so đ?‘Ł(đ?‘“đ?&#x;Ľ ) ⊞ đ?‘Ł(đ?‘“đ?&#x;Ł ), as the 3rd function decreases faster that the 1st. There is only one way to interpret đ?‘Ł(đ?‘“đ?&#x;Ł ) ⊞ đ?‘Ł(đ?‘“đ?&#x;¤ ) ⊞ đ?‘Ł(đ?‘“đ?&#x;Ľ ) ⊞ đ?‘Ł(đ?‘“đ?&#x;Ł ): equality all round! Yet in the last paragraph, we noted the parity must be different. This leads us to a contradiction. But must we repeat the same song and dance for the (đ?&#x;Ł đ?&#x;¤ đ?&#x;Ś đ?&#x;Ľ) configuration? No: it is enough to note that this is the opposite cycle. Take the picture we painted for (đ?&#x;Ł đ?&#x;Ľ đ?&#x;Ś đ?&#x;¤), hold it up to the mirror, and see the impossibility of (đ?&#x;Ł đ?&#x;¤ đ?&#x;Ś đ?&#x;Ľ)!

Back for more (polynomials) What wonders lie beyond groups of four polynomials crossing at the origin? In his 2013 paper Intersecting Curves, Étienne Ghys showed that every permutation in đ?‘†đ?‘› is realisable—provided that the permutation did not ask any subset of the four polynomials to alternate in the unachievable (đ?&#x;Ł đ?&#x;Ľ đ?&#x;Ś đ?&#x;¤) and (đ?&#x;Ł đ?&#x;¤ đ?&#x;Ś đ?&#x;Ľ) configurations. Make sure those 2 permutations are not in your entourage and the polynomials can enter from the left and emerge to the right of (0,0) in any order they please! But if you pick four of them and try to force the bottom of the group to be second, and the third to be the bottom, you’ll have an impossibility on your hands. You’ll be the butt of the joke! It is easy to show necessity: erase all other polynomials but any four and we see that the forbidden permutations are as forbidden as ever. Yet is tricky to show this as a sufficient condition. For this proof, we need graph theory: we must walk out of our pub at the origin and into a garden of pruned trees. We must build a tree for every collection of polynomials, starting from a constantterm-is-zero-for-all root, to linear terms and beyond, assigning a level to each degree and a node to each set of polynomials with identical terms of that degree. The leaves of such a tree will stand for the highest degree terms which help set all polynomials in a group apart. But alas, we must prune these trees if we have any hope that a computer may count the great many of them! Finally, by induction, we can describe, from a tree of đ?‘› − đ?&#x;Ł polynomials, how to introduce the final đ?‘“đ?‘› and how we both can and cannot tack on an extra leaf. No (đ?&#x;Ł đ?&#x;Ľ đ?&#x;Ś đ?&#x;¤) and (đ?&#x;Ł đ?&#x;¤ đ?&#x;Ś đ?&#x;Ľ) overlap will form. Our tale ends here for now.

Yuliya Nesterova Yuliya is a master’s student of algebraic geometry at Queen’s University, Canada. She likes to celebrate the informal mathematics that happens in the cracks of academia. She volunteers at MathQuest camp, a mathematics camp for girls, creates Pi Day celebrations, and hosts Ottawa’s MathsJams.

d mathquest.ca a @ottawamathsjam 19

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chalkdust

Moonlighting agony uncle Professor Dirichlet answers your personal problems. Want the prof’s help? Contact c deardirichlet@chalkdustmagazine.com

Dear Dirichlet, You’ll be pleased to hear I’ve jus t finished my PhD in abstract alg ebra, but now I’m stressed about jobs. I need to wri te this grant application for a big postdoc position studying a set combined with two binary operations, but all I want to do is curl up and read my favourite book: JRR Tolkein’s The Two Towers. What should I do?

— Elven Safety, Birmingham

DIRICHLET SAYS: Congraduations on completing your degree! I know you just want to get to the fun bits, but before you can read The Two Towers, you must complete the Fellowship Application of the Ring. May the Force be with you.

■

Dear Dirichlet,

line rse. I’m sure the Waterloo & City My journey to work is getting wo TfL? have any suggestions to send to is down more than it’s up. Do you pfully m of the route which I have hel For your perusal, I attach a diagra mocked up in TikZ.

Bank

Waterloo

nard’s Warrior Square

— Marcus McMacDaniel, St Leo

■

DIRICHLET SAYS: I have performed some analysis on your linear

map. First note that since it is embedded in a finite-dimensional 0 l ). Could you sub(way)space, your commute can be represented by ( -l 0 save yourself time by just leaving the house and turning clockwise 90o instead? I attempted to shine a beam of light through your map to observe its spectrum, σ( ), but only saw a pair of faint green circles. Alas... sigmal failure. I wouldn’t start a war of nerves with TfL exclusively, though: I think you would see much the same for any operator. Bank

Waterloo

chalkdustmagazine.com

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chalkdust

Dear Dirichlet, recently I finished Finnegans Wa ke, and have been inspired to create a similarly infinitely looping stream of con sciousness... but in musical form ! I’m thinking of pressing an LP: one track, one side only, which once you press pla y, would never end! Now I just need to get a few mates together to record it. Wo uld you promote it for us? I ask because

— Rejames! Rejoyce!, Riverrun

DIRICHLET SAYS: An interesting twist! I offer you the best of luck putting together your Möbius band. Although if you tried to play this inevitable trash in my pub (The Boisterous Badger, no entry fee), it’d be your vinyl visit: you’d be Möbius banned.

■

Dear Dirichlet,

Now that this the Winter Olympics in issue 07. I enjoyed your advice regarding er them. Any sugned, I have time to train to ent year’s Games have been postpo gestions for events I could enter?

— Emily Al-Jabr, Tokyo

■ DIRICHLET SAYS: Test your three-dimensional trigonometric skills with the hery-l . Don’t get caught using Cauchy’s residue theorem in the pole vault. Get expanding those brackets of yours by fencing with FOIL! The modern pentagon . W8lifting. l0nis. The fifty-metre 3style. ϱing, χak, ψcling. Scorpion wrestling f.

Dear Dirichlet,

suitable mathsI’m throwing a maths-themed seminar and am looking for some their names with themed biscuits. Perhaps there are some out there which share famous mathematicians? — Chuck O’Lye , Biarritz

■

DIRICHLET SAYS: Can’t think of any. More Dear Dirichlet, including two seasonal specials, online at d chalkdustmagazine.com 21

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What can you do with this space? Andrew Stacey

A

few years back, the students and staff at my school were set an art challenge over the summer break. Those who chose to take part were given a blank postcard and asked the question: What can you do with this space?

I feel that the obvious answer to this question is ‘fill it’. But in mathematics, the combination of the words space and fill links to a very specific concept: space-filling curves. In this article, I’m going to explain what these are, and how this leads to my entry (shown above).

Space-ďŹ lling curves There are many variations on the theme of space-filling curves, but the original and simplest version is a continuous curve that passes through every point in the unit square. Formally, it is a continuous surjective function đ?›ž âˆś [đ?&#x;˘, đ?&#x;Ł] → [đ?&#x;˘, đ?&#x;Ł]đ?&#x;¤ . Note that it only needs to be continuous and doesn’t have to be differentiable, so sharp corners are allowed. The first space-filling curve was designed by Peano in 1890. SurprisThe detail of my postcard ingly, his paper has no pictures. A year later, based on Peano’s work, Hilbert came up with a slightly different construction and included pictures. Hilbert curves seem chalkdustmagazine.com

22

chalkdust to be the more popular of the two, with about 20 times more search hits on the internet. You might conclude that pictures are important in mathematics; I couldn’t possibly comment. The reason why Peano and Hilbert (and others) were interested in these curves was because of Cantor. Not long before Peano published his paper, Cantor had started on his exploration of infinity and established that the unit interval and the unit square had the same quantity of points (this is what caused Cantor’s famous utterance “I see it, but I don’t believe it.”). This meant that there was a function from the unit interval to the unit square that hit every single point. Cantor constructed such a function using continued fractions to represent real numbers, but his function was discontinuous. It was established that there couldn’t be a continuous bijection between the unit interval and unit square, but the question remained as to whether it was possible to create a surjective function that was continuous. That is, a function that visits every point in the unit square but possibly more than once. Peano’s paper answers this with a definitive ‘yes’. Interestingly, Peano notes that his function is not differentiable. It would later be shown that no such differentiable function could exist. My interpretation of what Peano and Hilbert were doing is that they were experimenting with mathematics. Cantor’s ideas were spreading and they wanted to understand them better. To do so, they made examples testing the ideas, pushing them to their limits to see if they would break.

Peano curves Of the several variants out there, I’m going to focus on Peano curves. I’ll explain why once I’ve shown what they are. A space-filling curve is actually very easy to draw (here’s one: ) it is defining them that takes a little more work. The modern way to construct a space-filling curve is as a limit of a family of curves that are more straightforward to define. This is not how Peano originally constructed his curve, which was defined by giving a formula based on a way of writing numbers using a base 3 variant of decimals. It is more akin to how Hilbert constructed his curve. The construction of Peano’s curve that I know best starts by dividing the unit square into 9 smaller squares and joining their centres in a specific order, as I’ve illustrated to the right (top left, in blue).

3

4

9

2

5

8

6

1

7

The first two iterations of the Peano curve

The second iteration splits each of the 9 squares into 9 smaller squares and repeats the pattern in each smaller grid, with some rotations and reflections so that they join up. There are some options as to how to choose the rotations. The

2

3

1

4

The first two iterations of the Hilbert curve

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chalkdust one I’ve used is shown next to the first iteration. This process can be repeated and the Peano curve is what you get by doing this infinitely many times. Formally, it is the limit of this family. The Hilbert curve is constructed in a similar fashion except that the unit square is divided initially into 4 smaller squares, which are then each further divided into 4 and so on. The first two iterations are shown beneath the Peano curves on the previous page (in orange). I suspect that the real reason Hilbert curves are more well-known is because of the superiority of this division-by-two strategy and not due to Hilbert drawing pictures. So to fill the postcard with a space-filling curve, I could either colour in the whole postcard to show the final curve or show one of the stages. Neither of these felt like a good entry for an art project. For one, they were both a bit simplistic, and for two, it wouldn’t be obvious what was happening to someone who didn’t already know. So I wanted to make my design more interesting and somewhat self-explanatory while remaining true to the idea of a space-filling curve.

From space-filling to postcard-filling Ideas rarely come from nothing. I had seen the travelling salesman art of Bob Bosch and so the idea of using a single curve to draw a picture was not far from my mind. Henry Segerman has made some 3D printed shapes which interpolate from one level of a Hilbert curve to another, which introduces the idea of the different levels existing together. I had also seen some animations using different stages of Hilbert curves to indicate levels of darkness in an image, though not as a single curve. Putting these together, I wanted to make an image using different levels of Peano curves to indicate light and dark, while keeping it as a single curve. I’m not the first to have the idea of using space-filling curves in this fashion, but I like exploring ideas and I find that eventually I have to try things out for myself without worrying about whether I’m retreading old ground or breaking new.

Henry Segerman

a @henryseg d segerman.org

Henry Segerman’s 3D printed developing Hilbert curves. You can download the files needed to print your own from d thingiverse.com/thing:1696383

This also seems an appropriate point to explain why I settled on Peano curves rather than Hilbert curves or some other variant. One reason is quite simply that I prefer to use the object less studied. I’d rather adapt something from a well-known case to a less well-known one than simply repeat what has already been done. So the fact that Peano curves are used less than Hilbert curves naturally inclines me towards them. But more importantly, with a Peano curve then the last point on the curve is in the opposite corner to the starting point whereas with a Hilbert curve it is chalkdustmagazine.com

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chalkdust in a neighbouring corner. My eventual construction worked better with opposite corners than neighbouring ones. 3 2 1

4 5 6

9 8 7 Travelling diagonally

The key insight was that the path taken from the centre of one square to the next was relatively unimportant. So long as the route from one centre to the next doesn’t stray outside the two squares themselves, the limit will still be the Peano curve. This is because as the squares get smaller the distance between the original construction and a variation will be no bigger than the width of the squares and this gets vanishingly small. So rather than joining the centres with straight lines, we travel from one corner to the opposite, as in the leftmost picture above. This is a different, but known, construction of Peano’s curve. As the curve now meets itself, it can be somewhat tricky to trace its actual path. To combat that, I’ve squared-off the corners in the second image. An alternative is to tie a little knot at each corner, which is an approach that I’m quite fond of—and as far as I’m aware hasn’t been studied by anyone else—but which ultimately I didn’t use in this particular design. With this basic pattern it is possible to replace an individual square by a copy of itself (suitably scaled) without disturbing the rest of the curve, as shown in the rightmost picture above. This can also be viewed as a type of path replacement where a single diagonal line is replaced by the crinkled line, and then each segment of that is replaced by a copy of itself, and so on. More on this later. Using this I could make a picture: start with a grid of squares of suitable dimensions. Create The picture and its Peanoification a black-and-white picture by colouring in some squares, then convert that to a Peano curve by using the next level for the darker squares. The diagram to the right shows the pixel picture and its conversion to a Peano curve. 25

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chalkdust This made it a more interesting image but it didn’t make it self-explanatory. But once I realised I could replace the paths with anything suitable, the last step was not hard to conceive. Just hard to implement. I turned the curve into a text written as the ultimate in joined-up writing.

Exploring beyond The construction of the Peano curve which works by replacing line segments by more complicated curves is an example of a method which is typically used to produce fractal images. Technically, only the actual Peano curve is a fractal as stopping after a finite number of steps produces a curve that doesn’t have the ‘infinite zoom’ of a genuine fractal. So drawing the fractal that results from the Peano curve results in just a filled square. Not quite as exciting as, say, the Mandelbrot set. But using selective replacements we can produce genuine fractals that have a more aesthetic appearance. Continually replacing the line in the central square, we get a curve that looks a bit like the top left image on the left. The top right image is built similarly but with the top right square being replaced each time. In this image, after traversing the eight segments of one level, you are two-thirds of the way closer to your destination by a direct route, but you still have the same distance to travel if you stick to the path. In the opposite direction, we could apply level deepening only to the squares on the outside edges, leading to the bottom left image, while applying the deepening to non-central squares leads to a melding of Peano curves with the Sierpinski carpet on the bottom right. Another route to explore is to interpolate between the levels. Playing around with different ways to adjust the path gave me the idea of making them flexible, and that led to thinking of the basic path as the end of a family of paths starting from just a diagonal line and slowly crinkling up, as shown below. This interpolation gives a sense of greyscale or can be used to animate between pictures. This can be used to make a more intricate image. Peano fractals

Interpolating the diagonal version of the Peano curve

So by careful choices, it is possible to find a family of curves starting with just a diagonal line and ending with the full Peano curve and which makes a picture of Peano himself somewhere along the way, as you can see opposite. chalkdustmagazine.com

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chalkdust Just as Peano and Hilbert experimented with the ideas of Cantor to produce the original spacefilling curves, I’ve experimented with their ideas to produce my postcard and more. I suppose my real response to the question ‘what can you do with this space?’ is to pose a question of my own: ‘what can you do with this curve?’. So far, the answer seems to be ‘quite a lot’.

Peano in a Peano curve, and detail of his left eye

Further reading There is a lot of literature on space-filling curves, ranging from viewing then as art forms, as I have done here, through hearing them as music, to practical applications in the field of image analysis. I’ve already mentioned Henry Segerman and Bob Bosch; other names I’ve encountered while researching this article are Herman Haverkort, Jeffrey Ventrella, and Doug McKenna. From the historical perspective, the original papers of Cantor, Peano, and Hilbert can be found online (albeit in their original languages!) and there is a chapter on them in the book Curves for the Mathematically Curious. They are a fascinating class of objects to study and after making my postcard I’ve found myself returning to them again and again to play and explore. Somewhat like the curves themselves, the closer I look, the more detail I see. Ultimately, the space that they are filling is that of my imagination. Andrew Stacey Andrew is a maths and computer science teacher at Oxford High School, part of the Girls’ Day School Trust. Prior to that, he was an academic mathematician specialising in differential topology. In his copious spare time he writes programs that make images by combining weird mathematical effects such as Peanoification.

d loopspace.mathforge.org c loopspace@mathforge.org a @mathforge o Andrew Stacey 27

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Ada, Ada (to the tune of Daisy Bell (Bicycle Made for Two) by Henry Dacre)

Ada, Ada, translate this stuff for me, Manabrea wrote it in Italy, In England they are so critical Of my engine Analytical, So please have a bash, Then I might get some cash From the rich aristocracy.

Words by Sue de Pomerai a @suedepom

Babbage, Babbage, the methods are hard to see, I’ve written some explanations in notes labelled A to G, Your Bernoulli sums are wrong, And really far too long (so) I’ve added a table With loops, so it’s able To solve it efficiently.

Charles Babbage (1791–1871) was an English mathematician and inventor. He designed a difference engine—a mechanical calculator—and an analytical engine—a general purpose computer. Ada Lovelace (1815–1852) was an English mathematician and writer. She published the first ever computer algorithm; her algorithm was designed to be run on the analytical engine and would compute Bernoulli numbers. The building of the engines was never completed, but the ideas behind them led to the development of modern computers.

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Automated joke generation Alexander Bolton

Y

ou wouldn’t expect a computer to be able to make up a funny joke—it seems like a uniquely human ability. In order for a computer to understand the jokes that it is making, it would need to have deep semantic understanding of words, which is infeasible. And it can be difficult to explain why a joke is funny. However, it is feasible to use factor graphs to define a relatively simple model for what makes a specific class of jokes funny. A computer can then apply this methodology to make funny jokes, although it will not understand them. Mathematicians can use factor graphs to represent the factorisation of a function. As an example, suppose we have the function đ?‘” â„Ž đ?‘“ (đ?‘‹đ?&#x;Ł , đ?‘‹đ?&#x;¤ , đ?‘‹đ?&#x;Ľ ) = đ?‘‹đ?&#x;Ł (đ?‘‹đ?&#x;Ł + đ?‘‹đ?&#x;¤ + đ?‘‹đ?&#x;Ľ ). Then đ?‘“ can be factorised as

đ?‘“ (đ?‘‹đ?&#x;Ł , đ?‘‹đ?&#x;¤ , đ?‘‹đ?&#x;Ľ ) = đ?‘”(đ?‘‹đ?&#x;Ł ) â„Ž(đ?‘‹đ?&#x;Ł , đ?‘‹đ?&#x;¤ , đ?‘‹đ?&#x;Ľ ),

đ?‘‹đ?&#x;Ł

đ?‘‹đ?&#x;¤

đ?‘‹đ?&#x;Ľ

where đ?‘”(đ?‘‹đ?&#x;Ł ) = đ?‘‹đ?&#x;Ł and â„Ž(đ?‘‹đ?&#x;Ł , đ?‘‹đ?&#x;¤ , đ?‘‹đ?&#x;Ľ ) = đ?‘‹đ?&#x;Ł + đ?‘‹đ?&#x;¤ + đ?‘‹đ?&#x;Ľ . The factor graph for đ?‘“ is above on the right. It shows how đ?‘“ can be factorised, and which variables appear in each factor. To make the graph, we label the factors in square boxes, the variables in circular boxes, and connect a variable to a factor if that variable is a term in the factor. 29

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chalkdust Factor graphs are useful for representing the factorisation of probabilities and they aid in visualising and piecing together the model’s connections. As another example, the combined probability that there will be rain on Monday and Tuesday (đ?‘…đ?‘€ and đ?‘…đ?‘‡ ) can be factorised as đ?‘ƒ(đ?‘…đ?‘€ , đ?‘…đ?‘‡ ) = đ?‘ƒ(đ?‘…đ?‘€ )đ?‘ƒ(đ?‘…đ?‘‡ | đ?‘…đ?‘€ )

đ?‘“đ?&#x;Ł

= đ?‘“đ?&#x;Ł (đ?‘…đ?‘€ )đ?‘“đ?&#x;¤ (đ?‘…đ?‘€ , đ?‘…đ?‘‡ ).

đ?‘…đ?‘€

đ?‘“đ?&#x;¤

��

Factor đ?‘“đ?&#x;Ł determines whether it rains on Monday, and factor đ?‘“đ?&#x;¤ determines if it rains on Tuesday given that it rained on Monday. The factor graph is, once again, above on the right. Unsupervised joke generation from big data, a paper by SaĹĄa Petrović and David Matthews from 2013, is one of the most cited papers on automated joke generation. It uses factor graphs to generate jokes of the form I like my đ?‘‹ like I like my đ?‘Œ : đ?’ .

For example:

I like my secret like I like my subset of â„?đ?&#x;¤ : open.

I like my men like I like my coee: hot.

The authors assumed that the probability that a joke of this form is funny is a function of its inputs, (đ?‘‹ , đ?‘Œ , đ?‘? ). Technically the authors created a model for the probability that you would randomly choose the triple (đ?‘‹ , đ?‘Œ , đ?‘? ) when choosing from a population of funny jokes. But we are interested in choosing triples that are likely to be funny. Fortunately, Bayes’ theorem tells us that the two quantities are proportional to each other. If đ??´ and đ??ľ are random events, Bayes’ theorem says đ?‘ƒ(đ??´ | đ??ľ) âˆ? đ?‘ƒ(đ??ľ | đ??´)đ?‘ƒ(đ??´).

We now let đ??´ be the event that we choose the triple (đ?‘‹ , đ?‘Œ , đ?‘? ), and let đ??ľ be the event that we choose a funny triple; so if a particular triple (đ?‘‹ , đ?‘Œ , đ?‘? ) is more likely to be chosen by the model then its associated joke is more likely to be funny. Petrović and Matthews further assumed that đ?‘ƒ(đ??´ | đ??ľ) can be factorised using the factor graph below, which I will now go through. đ?‘‹

Co-ocđ?‘‹

Rare đ?‘?

Diff Co-ocđ?‘Œ

Sens

đ?‘Œ chalkdustmagazine.com

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Factor 1: Co-ocđ?‘‹ , Co-ocđ?‘Œ

A joke of the form ‘I like my đ?‘‹ like I like my đ?‘Œ : đ?‘? ’ will only make sense if the adjective đ?‘? is often used to describe đ?‘‹ and đ?‘Œ . Without this assumption we would get anti-jokes such as I like my goat like I like my Ferris wheel: intentional.

The factor Co-ocđ?‘‹ is a measure of the co-occurrence. If we are given non-random values đ?‘Ľ, đ?‘§ then we can evaluate: đ?‘“ (đ?‘Ľ, đ?‘§) Co-ocđ?‘‹ (đ?‘Ľ, đ?‘§) = , ∑đ?‘Ľ ′ ,đ?‘§ ′ đ?‘“ (đ?‘Ľ ′ , đ?‘§ ′ )

where đ?‘“ (đ?‘Ľ, đ?‘§) counts the number of times that adjective đ?‘§ is used to describe đ?‘Ľ in a large corpus (a collection of written works). Petrović and Matthews used the Google Ngram data set as their corpus. We divide đ?‘“ (đ?‘Ľ, đ?‘§) by the sum of đ?‘“ (đ?‘Ľ ′ , đ?‘§ ′ ) over all pairs (đ?‘Ľ ′ , đ?‘§ ′ ) in order to scale Co-ocđ?‘‹ (đ?‘Ľ, đ?‘§) to be in the interval [đ?&#x;˘, đ?&#x;Ł], like a probability (it will be used like a probability in Factor 2). Co-ocđ?‘Œ is just the same but with đ?‘Ś replacing đ?‘Ľ .

Factor 2: Di(đ?‘‹, đ?‘Œ)

‘I like my đ?‘‹ like I like my đ?‘Œ : đ?‘? ’ will only work as a joke if đ?‘‹ and đ?‘Œ are quite different nouns. Without this assumption we would get anti-jokes such as: I like my tea like I like my coee: hot.

So the authors wanted to give a higher score to dissimilar nouns: Diff(đ?‘Ľ, đ?‘Ś) =

đ?&#x;Ł , similarity(đ?‘Ľ, đ?‘Ś)

where the similarity function is estimated from the corpus. We set the probability of seeing the (noun, adjective) pair (đ?‘Ľ, đ?‘§) to be đ?‘?(đ?‘Ľ, đ?‘§) = Co-ocđ?‘‹ (đ?‘Ľ, đ?‘§), and then define đ?‘?(đ?‘Ľ) =

đ?‘“ (đ?‘Ľ) , ∑đ?‘Ľ ′ đ?‘“ (đ?‘Ľ ′ )

where đ?‘“ (đ?‘Ľ) counts the number of occurrences of đ?‘Ľ in the corpus. So đ?‘?(đ?‘Ľ) is the probability that we would choose đ?‘Ľ if we chose a random word from the corpus. We also define đ?‘?(đ?‘§ | đ?‘Ľ) =

đ?‘?(đ?‘Ľ, đ?‘§) đ?‘?(đ?‘Ľ)

,

so đ?‘?(đ?‘§ | đ?‘Ľ) is the probability that you have chosen adjective đ?‘§ , given that you chose a random (noun, adjective) pair from the data set and you see that you have chosen noun đ?‘Ľ . We define đ?‘?(đ?‘Ś) and 31

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chalkdust đ?‘?(đ?‘§ | đ?‘Ś) similarly. The authors then chose a similarity function that resembles a correlation function between đ?‘‹ and đ?‘Œ : ∑đ?‘§ đ?‘?(đ?‘§ | đ?‘Ľ)đ?‘?(đ?‘§ | đ?‘Ś) . similarity(đ?‘Ľ, đ?‘Ś) = đ?&#x;¤ đ?&#x;¤ √∑đ?‘§ đ?‘?(đ?‘§ | đ?‘Ľ) Ă— ∑đ?‘§ đ?‘?(đ?‘§ | đ?‘Ś) If đ?‘Ľ = đ?‘Ś then similarity(đ?‘Ľ, đ?‘Ś) = đ?&#x;Ł, and if đ?‘Ľ and đ?‘Ś are commonly described using the same adjectives then the similarity will be close to đ?&#x;Ł. And if đ?‘Ľ and đ?‘Ś are very rarely described using the same adjectives then similarity(đ?‘Ľ, đ?‘Ś) will be close to đ?&#x;˘.

Factors 3 & 4: Rare(đ?‘?) & Sens(đ?‘?)

The third factor encodes that the joke is funnier if đ?‘? is not a common adjective. Without this assumption we would get anti-jokes such as: I like my dogs like I like my quality of life: good.

The third factor is just defined as the reciprocal of the frequency of adjective � : Rare(�) =

đ?&#x;Ł

� (�)

.

The final factor encodes that the joke is funnier if đ?‘? has many different senses. ‘I like my đ?‘‹ like I like my đ?‘Œ : đ?‘? ’ jokes are often funny because one meaning of đ?‘? is used for đ?‘‹ and another is used for đ?‘Œ . Without this assumption we would get anti-jokes such as: I like my rectangles like I like my parallelograms: quadrilateral.

Just because đ?‘? is chosen to have multiple senses doesn’t guarantee that it will be used in different senses for đ?‘‹ and đ?‘Œ : it is just an approximation. The final factor is Sens(đ?‘§) = senses(đ?‘§),

where senses(đ?‘§) is derived from the Wordnet corpus. The probability that the authors chose a particular triple (đ?‘Ľ, đ?‘Ś, đ?‘§) is then given by Co-ocđ?‘‹ (đ?‘Ľ, đ?‘§) Co-ocđ?‘Œ (đ?‘Ś, đ?‘§) Diff(đ?‘Ľ, đ?‘Ś) Rare(đ?‘§) Sens(đ?‘§) , ∑đ?‘Ľ ′ ,đ?‘Ś ′ ,đ?‘§ ′ Co-ocđ?‘‹ (đ?‘Ľ ′ , đ?‘§ ′ ) Co-ocđ?‘Œ (đ?‘Ś ′ , đ?‘§ ′ ) Diff(đ?‘Ľ ′ , đ?‘Ś ′ ) Rare(đ?‘§ ′ ) Sens(đ?‘§ ′ )

where we divide by the sum of all possible triples in order to make sure that the probabilities all sum to đ?&#x;Ł. It was too expensive to compute the sum over all possible triples (đ?‘Ľ, đ?‘Ś, đ?‘§) so the authors restricted themselves to some selected values of đ?‘Ľ , and for each of these they computed the conditional probability of (đ?‘‹ , đ?‘Œ , đ?‘? ) given that đ?‘‹ = đ?‘Ľ , which is computationally feasible. The chalkdustmagazine.com

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chalkdust authors tested their model, comparing the jokes that they generated to a random selection of jokes found on Twitter. A small panel judged that their method can generate funny jokes: 16.3% of the jokes generated by their model were found funny, compared to 33.1% for the Twitter jokes. Here are two of the jokes generated by their model: I like my relationships like I like my source: open.

I like my coee like I like my war: cold.

In summary, factor graphs can be used to represent probability models, and these can also be used to generate jokes! Why not have a go at making your own jokes using these principles—Chalkdust would love to hear from you, especially if you come up with a good mathematical joke. Alexander Bolton Alex did a statistics PhD at Imperial College London, studying change point detection. He now works as a quantitative analyst at G-Research.

a @AlexDBolton Multilingual by Daniel Griller One thousand delegates are attending a conference. Half of them speak English, half speak French, half speak German, and half speak Spanish. At most how many of the delegates speak exactly one of these four languages? You can find more puzzles by Daniel in his (highly recommended) book Elastic Numbers, on his blog d puzzlecritic.wordpress.com, and on Twitter a @puzzlecritic.

My favourite conjecture Conjectures are statements that we think are true, but no-one has been able to prove. If you’re really lucky, your conjecture could become one of the most famous unsolved problems in maths; and it could even become named after you. We’ve spread some of our favourite conjectures throughout this issue.

ErdĹ‘s–Oler conjecture Belgin SeymenoÄ&#x;lu

Stanton McCandlish, CC BY-SA 3.0

Here’s a problem: pack đ?‘› unit circles into the smallest possible (equilateral) triangle. ErdĹ‘s and Oler conjectured that if đ?‘› is a triangular number, then the optimal triangles for đ?‘› and đ?‘› − đ?&#x;Ł circles will have the same length! It’s now known to be true for đ?‘› < đ?&#x;Łđ?&#x;§, but the jury’s still out for other cases. I like this conjecture because it reminds me of snooker balls held in a triangular frame!

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On the cover

Apollonian packing David Sheard

W

hat can you do with this space? So asks Andrew Stacey (page 22). ‘Fill it’ is the prompt reply, but fill it with what? Maybe like Andrew you want to use a single curve, but I want to use circles. If you do this in the way shown above in blue, the result is called an Apollonian packing, a variant of which can be seen on the cover of this issue. Here we shall explore the history of this entrancing object, which spans over 2000 years, and percolates into a surprising variety of mathematical disciplines. Starting in the familiar world of Euclidean geometry, Apollonian packings extend into fractal geometry and measure theory; Möbius transformations and the hyperbolic plane; and then on into the distant reaches of geometric group theory, number theory, orbital mechanics, and even ship navigation. At times we may wonder off into thickets of more obscure mathematics, so those readers who get lost should feel free to skip ahead to the next section.

Apollonius of Perga Apollonius (c 230 BC) was a Hellenistic mathematician, considered one of the greatest after Euclid and Aristotle. Perhaps his most important work was his eight book treaties Κωνικα on conic sections—once lost to European civilisation, but fortuitously preserved by the more enlightened Middle Eastern scholars and later reintroduced by Edmund Halley in 1710. The same unfortunately cannot be said of Έπαφαι (De Tractionibus or Tangencies). Although now lost, we have accounts chalkdustmagazine.com

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chalkdust of the work from other ancient authors, particularly in the writings of Pappus of Alexandria. In it, Apollonius posed and solved the following problem.

Problem Given three geometric objects in the plane (points, lines, and/or circles), find all circles which meet all three simultaneously (ie which pass through any points, and are tangent to any lines or circles). So for example, given three points which don’t lie on the same line, there is exactly one circle which passes through all three. The case which interests us at present is when we are given three circles, each of which is tangent to the other two. In the very special case that all three are tangent at the same point there are infinitely many circles tangent to all three. Usually, however, the circles will be pairwise tangent at three distinct points, in which case there are exactly two other circles tangent to all three simultaneously.

Given three mutually tangent circles (black) there are usually exactly two others (blue) tangent to all three.

This is as far as Apollonius went; the next step would not be taken until 1643, when RenĂŠ Descartes discovered a formula for the size of the two tangent circles, which he wrote in a letter to Princess Elizabeth of the Palatinate. The same formula was later rediscovered by Frederick Soddy and published as a poem in Nature in 1936. The size of a circle is determined by its radius đ?‘&#x; . If đ?‘&#x; is small, the circle will be small, but it will also be very curved. We can define the curvature of the circle to be đ?‘˜ = đ?&#x;Ł/đ?‘&#x; . Descartes showed that if three given circles are mutually tangent at three distinct points, and have curvatures đ?‘˜đ?&#x;Ł , đ?‘˜đ?&#x;¤ , and đ?‘˜đ?&#x;Ľ , then a fourth circle which is tangent to all three has curvature đ?‘˜đ?&#x;Ś satisfying (đ?‘˜đ?&#x;Ł + đ?‘˜đ?&#x;¤ + đ?‘˜đ?&#x;Ľ + đ?‘˜đ?&#x;Ś )đ?&#x;¤ = đ?&#x;¤(đ?‘˜đ?&#x;Łđ?&#x;¤ + đ?‘˜đ?&#x;¤đ?&#x;¤ + đ?‘˜đ?&#x;Ľđ?&#x;¤ + đ?‘˜đ?&#x;Śđ?&#x;¤ )

For technical algebraic reasons, sometimes this equation gives negative values for the curvature đ?‘˜đ?&#x;Ś , which we can interpret as corresponding to a circle with curvature |đ?‘˜| which contains the other circles in its interior. Notice that this equation is quadratic in the variable đ?‘˜đ?&#x;Ś , so there are two solutions; these will correspond to the two possibilities for the fourth circle found by Apollonius.

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(f)

đ?‘˜đ?&#x;¤

đ?‘˜đ?&#x;Śâˆ’

đ?‘˜đ?&#x;Ś+ đ?‘˜đ?&#x;Ł đ?‘˜đ?&#x;Ľ

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Apollonian packings So far we have constructed at most 5 mutually tangent circles. The step to infinity may seem obvious, but took another 63 years and some 1900 years after Apollonius. The earliest description seems to appear in a letter from Leibniz to des Bosses (11 March 1706):

Imagine a circle; in it draw three other circles that are the same size and as large as possible, and in any new circle and in the space between circles again draw the three largest circles of the same size that are possible. Imagine proceeding to infinity in this way… What Leibniz is describing is in fact a nested Apollonian packing, since at each step he fills in every circle as well as the gaps between circles. This early description makes the nested Apollonian packing one of the first fractals, although it wasn’t studied properly until mathematicians like Cantor, Weierstrass, von Koch, and Sierpinski started discovering other fractals in the late nineteenth and early twentieth centuries. This may be because Leibniz was not interested in the mathematical construction, but rather was trying to draw an analogy to argue against the existence in infinitesimals in nature. Henceforth we shall only consider the un-nested Apollonian packing. As a fractal, it has a number of inA finite iteration of a nested Apollonian teresting properties: it is a set of measure 0, which packing similar to the one described by means that if you tried to make it by starting with a Leibniz. disc of metal, and then drilled out infinitely many ever smaller holes (and if you ignore that metal is made out of atoms), then you would finish up with a single piece of metal (you haven’t removed everything), but nevertheless with exactly 0 mass. It has fractal dimension approximately 1.30568, which means that mathematically it lives somewhere between a 1D curve and a 2D area. Finally, if you look at just the portion of an Apollonian packing which lies in the triangular region between three tangent circles, this is homeomorphic to the Sierpinski triangle, which means that one can be bent and stretched to look like the other.

Adapted from Todd Stedl, CC BY-SA 4.0

bend and stretch

Todd Stedl, CC BY-SA 4.0

Beojan Stanislaus, CC BY-SA 3.0

A portion of an Apollonian packing is homeomorphic to the Sierpinski triangle—just squash all of the circles to make them triangular.

There is a curious combinatorial consequence of Descartes’ formula for Apollonian packings. If we chalkdustmagazine.com

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chalkdust start with three mutually tangent circles with curvatures đ?‘˜đ?&#x;Ł , đ?‘˜đ?&#x;¤ , and đ?‘˜đ?&#x;Ľ , we can solve (f) to find that the curvatures đ?‘˜đ?&#x;Ś+ and đ?‘˜đ?&#x;Śâˆ’ of the other two circles are đ?‘˜đ?&#x;ŚÂą = đ?‘˜đ?&#x;Ł + đ?‘˜đ?&#x;¤ + đ?‘˜đ?&#x;Ľ Âą đ?&#x;¤âˆšđ?‘˜đ?&#x;Ł đ?‘˜đ?&#x;¤ + đ?‘˜đ?&#x;¤ đ?‘˜đ?&#x;Ľ + đ?‘˜đ?&#x;Ľ đ?‘˜đ?&#x;Ł

Adapted from Todd Stedl, CC BY-SA 4.0

The integral Apollonian packing starting with curvatures −đ?&#x;Łđ?&#x;˘, 18, 23, and 27.

(ff)

Now suppose we start constructing an Apollonian packing by drawing four mutually tangent circles whose curvatures đ?‘˜đ?&#x;Ł , đ?‘˜đ?&#x;¤ , đ?‘˜đ?&#x;Ľ , and đ?‘˜đ?&#x;Ś+ are all integers. From equation (ff) it follows that đ?&#x;¤âˆšđ?‘˜đ?&#x;Ł đ?‘˜đ?&#x;¤ + đ?‘˜đ?&#x;¤ đ?‘˜đ?&#x;Ľ + đ?‘˜đ?&#x;Ľ đ?‘˜đ?&#x;Ł must be an integer since đ?‘˜đ?&#x;Ś+ is an integer, and so đ?‘˜đ?&#x;Śâˆ’ is also an integer. Now we can build the packing by filling in a fifth circle wherever we see four mutually tangent circles. By the observation above, if the four circles have integer curvatures, the fifth circle will also have integer curvature. Inductively therefore we will end up with an Apollonian packing consisting of infinitely many tangent circles, all of which have integer curvatures.

Hyperbolic geometry If you have some familiarity with non-Euclidean geometry, Apollonian packings may remind you of the PoincarĂŠ model of the hyperbolic plane. The hyperbolic plane â„?đ?&#x;¤ is a 2D surface on which we can do geometry just like we can on the flat Euclidean plane. Whereas a sphere has constant positive curvature (it curves the same way in all directions), and the Euclidean plane has constant zero curvature (it’s flat), â„?đ?&#x;¤ is an infinite surface which has constant negative curvature, which means that at every point it curves in the same way as a Pringle. This negative curvature makes the surface crinkle up on itself more and more as you move out towards infinity, which is inconvenient when we try to work with it. Usually then we represent it on a flat All these scorpions have the same hyperbolic size. surface so we can draw pictures of it in magazines and the like. One way to do this is with the PoincarĂŠ model. This views the hyperbolic plane as a disc. In order to fit the whole infinity of â„?đ?&#x;¤ into a finite disc, we have to shrink distances as we move out towards the edge of the disc. Using this skewed way of measuring distances, the circular edge of the disc is infinitely far away from its centre. We can think of an Apollonian packing as living in the PoincarĂŠ disc, with the outermost circle of the packing as the boundary circle of â„?đ?&#x;¤ . Then the circles in the packing which are not tangent to this boundary are also circles in the strange hyperbolic way of measuring distance, that is, all points are equidistant from some other point in the plane—the circle’s hyperbolic centre. Circles in the packing which are tangent to the boundary are called horocycles (in Greek this literally means border circle), which are circles with infinite radius in the hyperbolic metric. Horocycles have no analogue in the Euclidean plane. 39

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chalkdust Something interesting happens when we see what an Apollonian packing looks like in the upper half-plane (UHP) model for â„?đ?&#x;¤ . This model is similar to the PoincarĂŠ model, but instead of using a disc, we use the half-plane above the đ?‘Ľ -axis {(đ?‘Ľ, đ?‘Ś) ∈ â„?đ?&#x;¤ âˆś đ?‘Ś > đ?&#x;˘}, where the đ?‘Ľ -axis behaves like the boundary circle and should be thought of as at infinity. There is a problem, in that in the PoincarĂŠ disc, the boundary of â„?đ?&#x;¤ was a circle, and so it closed up on itself. In the UHP, the boundary is a line which doesn’t close up on itself, but these are supposed to be models for the same thing. To fix this, we imagine there is a point at infinity ∞ which joins up the two ends of the boundary to form an infinite diameter circle. If we start with any Apollonian packing living in the PoincarĂŠ disc, there is a map from the disc to the UHP preserving hyperbolic distances, under which the outer circle of the packing becomes the đ?‘Ľ -axis (together with the point at infinity), and exactly one of the horocycles (one of the circles tangent to the outer circle in the packing) becomes the horizontal line đ?‘Ś = đ?&#x;Ł. All other circles and horocycles in the packing are sent to circles which are tangent to each other as before, but are now sandwiched between the lines đ?‘Ś = đ?&#x;˘ and đ?‘Ś = đ?&#x;Ł.

If we focus on just those circles which meet the đ?‘Ľ -axis we get what are called Ford circles. Remarkably each of these circles is tangent to the đ?‘Ľ -axis at a rational number đ?‘?/đ?‘ž , and has radius đ?&#x;Ł/đ?&#x;¤đ?‘ž đ?&#x;¤ . Moreover every rational number is the point of tangency of one of the circles (see below). Now some magic happens: suppose the Ford circles at đ?‘Ž/đ?‘? and đ?‘?/đ?‘‘ are tangent to each other, then there is a unique circle sandwiched between these two circles and the đ?‘Ľ -axis. The rational point at which this circle meets the đ?‘Ľ -axis is given by the Farey sum of đ?‘Ž/đ?‘? and đ?‘?/đ?‘‘ đ?‘Ž đ?‘?

⊕

Note that for this to be well-defined, đ?‘Ž/đ?‘? and đ?‘?/đ?‘‘ must be written in their simplest form. This Farey sum, and the associated Farey sequences đ??šđ?‘› you get by looking at all rational numbers between 0 and 1 which can be written as a fraction with denominator at most đ?‘›, turn up in several places across number theory. These include rational approximation of irrational numbers and the Riemann Hypothesis.

MĂśbius transformations

đ?‘?

đ?‘‘

=

đ?&#x;˘ đ?&#x;Ł

đ?‘Ž+đ?‘? đ?‘?+đ?‘‘

đ?&#x;Ł đ?&#x;Ł đ?&#x;§ đ?&#x;Ś

đ?&#x;Ł đ?&#x;Ľ

đ?&#x;¤ đ?&#x;§

đ?&#x;Ł đ?&#x;¤

đ?&#x;Ľ đ?&#x;§

đ?&#x;¤ đ?&#x;Ľ

=

đ?&#x;Ľ đ?&#x;Ś đ?&#x;Ś đ?&#x;§ đ?&#x;Ł đ?&#x;¤ ⊕ đ?&#x;¤ đ?&#x;Ľ

đ?&#x;Ł đ?&#x;Ł

If you haven’t seen hyperbolic geometry before, you may wonder how we can map the PoincarÊ disc model to the UHP model, and in such a way that the strange distance measure in the two models is preserved—for a start one is a finite region while the other is an infinite half-plane. chalkdustmagazine.com

40

chalkdust The answer is to view both models as living inside the complex plane â„‚ (or more accurately the Ě‚ = â„‚ âˆŞ {∞}): the PoincarĂŠ disc is the unit disc {đ?‘§ ∈ â„‚ âˆś |đ?‘§| < đ?&#x;Ł}, and the extended complex plane â„‚ UHP is the region above the real axis {đ?‘§ ∈ â„‚ âˆś Im(đ?‘§) > đ?&#x;˘}. Then a function like đ?‘§ ↌ −i

đ?‘§ + đ?&#x;Ł −iđ?‘§ − i = đ?‘§âˆ’đ?&#x;Ł đ?‘§âˆ’đ?&#x;Ł

(fff)

will do the trick. This function is an example of a MĂśbius transformation, which in general is a complex function of the form đ?‘Žđ?‘§ + đ?‘? đ?‘§â†Ś đ?‘?đ?‘§ + đ?‘‘ were we require đ?‘Žđ?‘‘ − đ?‘?đ?‘? ≠đ?&#x;˘ so that this function is invertible. The function (fff) sends the unit disc to the UHP, but it is not the only MĂśbius transformation which does this. In fact there are infinitely many such functions, all of which preserve the hyperbolic metric. In the previous section I claimed that starting with any Apollonian packing, we could choose one of these MĂśbius transformations such that the image had a very specific form, sandwiched between the lines Im(đ?‘§) = đ?&#x;˘ and Im(đ?‘§) = đ?&#x;Ł.

The upshot of this is that all Apollonian packings are the same in the hyperbolic plane, because they can all If you have seen MÜbius transforbe mapped to the same packing by (invertible) funcmations before, you may wish to tions which preserve hyperbolic distance. Once we have try and prove that the purported started thinking about the Apollonian packing living in mapping exists yourself. (Hint: rethe complex plane, the whole world of complex functions member that MÜbius transformais open to us, and we can start to do crazy things. If tions send circles and lines to circles we don’t restrict ourselves to just MÜbius transformaand lines, and are completely detertions, but see what happens when we apply holomorphic mined by their image on 3 distinct or anti-holomorphic functions to the packing (these are points.) complex functions with a good notion of derivative in the sense of calculus, which in particular have the property that they preserve angles between intersecting curves), we can get some very pretty designs. We need not even require (anti-)holomorphicity. The patterns featured on the front and back covers were drawn in this way.

Beyond the packing Let us return to Apollonius of Perga. Remember that his treaties ÎˆĎ€ÎąĎ†ιΚ, where he stated and solved the problem of finding tangent circles, is lost to history—how then do we know what he proved and how? The answer is that we don’t. The only record we have appears in the writings of Pappus of Alexandria, who lived some 400 years after Apollonius, but who references many of Apollonius’ works, including six which are no longer extant. All he says of Tangencies is the general problem which Apollonius was interested in, and that he solved it by solving many simple special cases and working up from there. The first person to reprove Apollonius’ results in ‘modern’ times was Adriaan van Roomen in 1596. His solution, however, does not use ruler and compass constructions, so cannot have been the one 41

spring 2020

chalkdust Apollonius used. The result was later proved using methods available to Apollonius, and in the way described by Pappus, by van Roomen’s friend François Viète. The method of Viète was later reworked and simplified by several mathematicians, including Isaac Newton in his Principia. Newton related the position of the centre of the fourth circle to its distance from the centres of the three circles to which it is supposed to be tangent. This is called hyperbolic positioning or trilateration. Newton used this viewpoint to describe the orbits of planets in the solar system, but it can also be used to help navigate ships, and to locate the source of a signal based on the different times the signal is received at three different locations. In the first world war this was used to locate artillery based on when shots were heard. This is also how modern GPS works (not by triangulation as is commonly believed).

A ship’s location determined by its distance from three points.

So this 2000-year-old problem in abstract geometry turned out to have extremely useful applications in the real world. The Apollonian packing also shows up in lots of different areas of mathematics. For example, Ford circles inspired the Hardy–Littlewood circle method, an important tool in analytic number theory which was used to solve Waring’s Problem: for an integer đ?‘˜ , can every integer be written as a sum of at most đ?‘› đ?‘˜ th powers for some value of đ?‘›? This is true: for example, every integer is the sum of 4 squares, 9 cubes, 19 fourth powers, and so on. In 2013, Harald Helfgott used the circle method to prove the weak Goldbach conjecture: every odd number greater than 5 is the sum of 3 primes.

The limit set of some Schottky groups are twisted Apollonian packings.

As a final application, I am a geometric group theorist, so let me talk about one place the Apollonian Ě‚ , it turns out, can be thought of as the packing shows up in my field. The extended complex plane â„‚ đ?&#x;Ľ đ?&#x;Ľ boundary of three-dimensional hyperbolic space â„? . â„? can be modelled as the upper half-space Ě‚ identified with the plane {(đ?‘Ľ, đ?‘Ś, đ?‘§) ∈ â„?đ?&#x;Ľ âˆś đ?‘§ = đ?&#x;˘} âˆŞ {∞}. Then {(đ?‘Ľ, đ?‘Ś, đ?‘§) ∈ â„?đ?&#x;Ľ âˆś đ?‘§ > đ?&#x;˘} âˆŞ {∞}, with â„‚ chalkdustmagazine.com

42

chalkdust Ě‚ act on the whole of â„?đ?&#x;Ľ preserving hyperbolic distance. MĂśbius transformations acting on â„‚

If we choose just a few MĂśbius transformations to start with, these generate a group which acts on â„?đ?&#x;Ľ , and in doing so, creates a pattern on the complex plane called the limit set of the group, which is a picture of how the group acts ‘at infinity’. Choosing the right MĂśbius transformations gives a group called a Schottky group, whose limit set is precisely the Apollonian packing. If we perturb the starting MĂśbius transformations just slightly, we get a Schottky group whose limit set is a twisted Apollonian packing. Even though this picture—see opposite—looks like it is still made up more or less of circles, it is in fact one continuous closed curve which is fractal, and does not intersect itself anywhere. If you would like to read more about Schottky groups, limit sets, and twisted Apollonian packings, there are more details in the version of this article on the Chalkdust website, or alternatively you can read the excellent book Indra’s Pearls: the Vision of Felix Klein. David Sheard David is a PhD student at UCL, who spends most of his time thinking about triangles, except on special days when he also thinks about squares. In his free time he likes playing the flute, singing, and dreaming of circles.

d davidsheard.co.uk c david.sheard.17@ucl.ac.uk Orange and green by Catriona Shearer

The orange square has area 30 and covers two-thirds of the equilateral triangle. What’s the area of the green square?

You can find Catriona and loads more of her puzzles on Twitter a @Cshearer41.

My second-favourite conjecture

A limit

Sam Porritt

Consider the sequence đ?‘Ľđ?‘› =

đ?&#x;Ł

, đ?‘›đ?&#x;¤ sin đ?‘›

for đ?‘› = đ?&#x;Ł, đ?&#x;¤, đ?&#x;Ľ, ‌ . It is conjectured that đ?‘Ľđ?‘› → đ?&#x;˘ as đ?‘› → ∞ but no one can prove it. 43

spring 2020

Puzzles

Looking for a fun puzzle but not got time to tackle the crossnumber? You’re on the right page.

Arrange the digits Put the numbers 1 to 9 (using each number exactly once) in the boxes so that the sums are correct. The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, đ?&#x;Ś + đ?&#x;Ľ Ă— đ?&#x;¤ is 14, not 10.

+ − = 4

1 3

2 4

5 6

á á

+ − = 1

Ă— Ă— Ă—

+ − = 2

=đ?&#x;Ľ = đ?&#x;Łđ?&#x;§ = đ?&#x;Łđ?&#x;Ś

Sums and products by Humbug As usual, no numbers begin with 0. The sum of the digits of each across clue is equal to the clue’s number. For example, the sum of the digits of 5A is equal to 5. The product of the digits of each down clue is equal to the clue’s number. For example, the product of the digits of 4D is equal to 4.

≈

≈

>

<

≈

>

44

>

<

Some boxes have ≈ symbols between them: the numbers in these boxes differ by 1.

≈

≈

≈

Some boxes have > symbols between them: the boxes to the larger size contain numbers greater than the boxes to the smaller side.

>

More, less, or nearly equal Put the numbers 1 to 5 in the boxes so that each row and each column contains each digit exactly once.

chalkdustmagazine.com

á

>

chalkdust Not hot

Arrow word Write the answers to the clues in the white boxes in the directions shown by the arrows.

57

opp adj

∑

Prime

Ă—đ?&#x;Ľ

đ?‘Ś = đ?‘Ľđ?&#x;¤

1 Game

Basic truth

đ?›˝

√

đ?&#x;Łđ?&#x;˘đ?&#x;Łđ?&#x;˘

đ?‘Ľđ?&#x;¤ − đ?‘Śđ?&#x;¤ at (đ?&#x;˘, đ?&#x;˘)

_ists

10

đ?&#x;Łđ?&#x;˘đ?&#x;˘

{đ?&#x;˘, đ?&#x;¤, đ?&#x;Ľ} _cahtoa

đ?œ‚

Clever robot

Coke is _

đ?œ™

Current unit

∈

_ term eđ?‘Ľ +e−đ?‘Ľ

Ě‚

adj

đ?&#x;¤

∋

opp

4 bits

_(đ?&#x;¨, đ?&#x;Ş) is 2

đ?‘“ âˆś â„?→â„?

Inverse of eđ?‘Ľ

Donut

đ?œ

18

Ď€đ?‘&#x; đ?&#x;¤ (

đ??´ or đ??ľ, not both

â„? đ?&#x;Śâˆśđ?&#x;Ľ

45

spring 2020

Kepler’s barrels Kepler went out shopping for a barrel full of wine, He had a problem with the price the merchant would assign, The merchant used a measure stick to calculate the price, Into the barrel it did drop, Through a hole at centre top, And down and left until it stopped, He thought it imprecise.

He said there’d be a certain price for a barrel short and fat, And another that was tall and thin would cost the same as that, But while the short fat barrel would hold lots and lots of wine, The tall thin one would hold much less, This observation caused distress, Kepler started to obsess, He didn’t think it fine.

Though Kepler paid the merchant then, he never could forget, That inconsistent pricing, well, it really made him fret, Day after day he pondered as he went about his work, He just had to investigate, The fairness of the merchant’s rate, He could no more procrastinate, Or he would go berserk.

For any given pricing, how much wine could one consume? What measurement of barrel gave the maximum volume? He noted that a cylinder was almost the same shape, As the barrel he’d been using, For his celebratory boozing, And so he began his musing, He could approximate. 46

For cylinders of different heights he worked out the diameter, Assuming that the merchant’s measure was a fixed parameter, Trying different values gave a groundbreaking conclusion, Diameter, times by root 2, Would give the perfect height value, A maxed-out volume would ensue, It was no illusion.

Then Kepler found out something new that really made him cheer, All of Austria’s barrels had this ratio, or near, So the barrel that he’d purchased had a fair price after all, A few might slightly deviate, But Kepler could appreciate, The volume change that this creates, Is imperceptible.

But the story doesn’t end here, because Kepler’s observation, That the points around the max have but a tiny deviation, Inspired Fermat’s theorem about stationary points, It might not seem so obvious, But this led on to calculus, A field meritorious, That never disappoints.

Sam Hartburn Sam is a freelance proofreader and copy-editor and a hobbyist maths geek. She likes to make maths rhyme.

d qbfproofreading.co.uk c sam.hartburn@qbfproofreading.co.uk a @SamHartburn d samhartburn.co.uk 47

Joke generator A focus group

“Where’s your belt?”

“Get up!”

0?

8

purple

represents the

symmetries

An Abelian grape

chalkdustmagazine.com

of a polygon?

of a Platonic solid? 48

Benôit B Mandelbrot?

short for?

Benôit B Mandelbrot

Benôit B Mandelbrot

Lyndon B Johnson

B in

the

Closer to the chicken

the road?

call

funny?

is

and

get if you cross

Schrödinger’s joke

Linear owl-gebra

a bird of prey?

a vector with

do you

not

recursion?

commutes?

another vector?

What

green

use when designing their new

Zero (the vectors were parallel)

method of surveying

did

Ford Focus

Ford

did the famous car company

∞?

say to

car?

Start at the yellow box and follow the arrows to create hilarious maths jokes.

A dihedral grape

A polyhedral grape

short for?

Binomial

chalkdust

Weirdonacci numbers Robert J Low and Thierry Platini

W

e all know how the Fibonacci numbers, đ??šđ?‘› , work: each number in the sequence is the sum of the previous two numbers, and if we make the usual start, we get đ??šđ?&#x;Ł đ??šđ?&#x;¤ đ??šđ?&#x;Ľ đ??šđ?&#x;Ś đ??šđ?&#x;§ đ??šđ?&#x;¨ đ??šđ?&#x;Š â‹Ż đ?&#x;Ł

đ?&#x;Ł

đ?&#x;¤

đ?&#x;Ľ

đ?&#x;§

đ?&#x;Ş

đ?&#x;Łđ?&#x;Ľ â‹Ż

The numbers in this sequence grow larger and larger, and can be expressed with the help of the golden ratio đ?œ™ = (đ?&#x;Ł + √đ?&#x;§)/đ?&#x;¤, by the surprising formula đ??šđ?‘› =

đ?œ™ đ?‘› − (−đ?œ™)−đ?‘›

. √đ?&#x;§ There is a vast amount of literature about this number, đ?œ™ , to which this article will make no contribution whatsoever.

Instead, we’re going to think about the consequences of interpreting the rule slightly differently. Since this interpretation of the rules is a little weird, we’ll call these the Weirdonacci numbers, đ?‘Šđ?‘› . One can imagine that this new sequence comes from someone who has misunderstood the construction of the Fibonacci numbers. Instead of adding up the last two numbers of the sequence, the person would add the last two digits in the sequence. As a consequence, we are still going to get each term by adding together the previous two numbers, but we will think of đ?&#x;Łđ?&#x;Ľ as two numbers (digits), đ?&#x;Ł and đ?&#x;Ľ. The sequence will now grow differently. 49

spring 2020

chalkdust We get:

đ?‘Šđ?&#x;Ł đ?‘Šđ?&#x;¤ đ?‘Šđ?&#x;Ľ đ?‘Šđ?&#x;Ś đ?‘Šđ?&#x;§ đ?‘Šđ?&#x;¨ đ?‘Šđ?&#x;Š đ?‘Šđ?&#x;Ş đ?‘Šđ?&#x;Ť đ?‘Šđ?&#x;Łđ?&#x;˘ đ?‘Šđ?&#x;Łđ?&#x;Ł đ?‘Šđ?&#x;Łđ?&#x;¤ â‹Ż đ?&#x;Ł

đ?&#x;Ł

đ?&#x;¤

đ?&#x;Ľ

đ?&#x;§

đ?&#x;Ş

đ?&#x;Ł

đ?&#x;Ľ

đ?&#x;Ś

Notice that after 10 terms, the sequence starts repeating itself: we can write this as đ?‘Šâ„“ +đ?‘— = đ?‘Šđ?‘— , where â„“ = đ?&#x;Łđ?&#x;˘ is the length of the cycle.

đ?&#x;Š

đ?&#x;Ł

đ?&#x;Ł

â‹Ż

Sadly, now that we’ve seen it repeat, there doesn’t seem to be much more to say about this sequence. But there is. Since the digits are what matters here, it is important to notice that this sequence depends on the base in which the numbers are expressed.

The inevitability of cycles Let’s see what happens with a couple of other choices of base. (đ?‘?) We will write đ?‘Šđ?‘› for the Weirdonacci numbers generated in base đ?‘? . In base eight, we have the following sequence with cycle length â„“ = đ?&#x;Š: (đ?&#x;Ş)

đ?‘Šđ?&#x;Ł

đ?&#x;Ł

(đ?&#x;Ş)

đ?‘Šđ?&#x;¤

đ?&#x;Ł

(đ?&#x;Ş)

đ?‘Šđ?&#x;Ľ

đ?&#x;¤

(đ?&#x;Ş)

đ?‘Šđ?&#x;Ś

đ?&#x;Ľ

(đ?&#x;Ş)

đ?‘Šđ?&#x;§

đ?&#x;§

(đ?&#x;Ş)

đ?‘Šđ?&#x;¨

đ?&#x;Ł

(đ?&#x;Ş)

đ?‘Šđ?&#x;Š

đ?&#x;˘

Cuckoo: Wellcome Trust, CC BY 4.0.

Weirnardo of Pisa

In base nine we have the sequence below with cycle length â„“ = đ?&#x;Łđ?&#x;Ł: (đ?&#x;Ť)

đ?‘Šđ?&#x;Ł

đ?&#x;Ł

(đ?&#x;Ť)

đ?‘Šđ?&#x;¤

đ?&#x;Ł

(đ?&#x;Ť)

đ?‘Šđ?&#x;Ľ

đ?&#x;¤

(đ?&#x;Ť)

đ?‘Šđ?&#x;Ś

đ?&#x;Ľ

(đ?&#x;Ť)

đ?‘Šđ?&#x;§

đ?&#x;§

(đ?&#x;Ť)

đ?‘Šđ?&#x;¨

đ?&#x;Ş

(đ?&#x;Ť)

đ?‘Šđ?&#x;Š

đ?&#x;Ł

(đ?&#x;Ť)

đ?‘Šđ?&#x;Ş

đ?&#x;Ś

(đ?&#x;Ş)

đ?‘Šđ?&#x;Ť

(đ?&#x;Ť)

đ?‘Šđ?&#x;Łđ?&#x;˘

đ?‘Šđ?&#x;Ş

đ?&#x;Ł

đ?‘Šđ?&#x;Ť

đ?&#x;§

(đ?&#x;Ş)

â‹Ż

đ?&#x;Ł

â‹Ż

(đ?&#x;Ť)

đ?‘Šđ?&#x;Łđ?&#x;Ł

đ?&#x;Ł

(đ?&#x;Ť)

đ?&#x;˘

â‹Ż

â‹Ż

which brings us back to đ?&#x;Ł, đ?&#x;Ł, so we see that in each case the sequence jumps back to the start after some number of terms. We might be tempted to conjecture that this always happens, but we should also bear in mind that three cases are a rather slender foundation to build a conjecture on. It’s worth looking at a few other possibilities before committing to trying to prove something. This tempting conjecture would also prove true in base seven (you should use the margins of this page to check this), but in fact fails in base six: (đ?&#x;¨)

đ?‘Šđ?&#x;Ł

đ?&#x;Ł

(đ?&#x;¨)

đ?‘Šđ?&#x;¤

đ?&#x;Ł

(đ?&#x;¨)

đ?‘Šđ?&#x;Ľ

đ?&#x;¤

(đ?&#x;¨)

đ?‘Šđ?&#x;Ś

đ?&#x;Ľ

(đ?&#x;¨)

đ?‘Šđ?&#x;§

đ?&#x;§

(đ?&#x;¨)

đ?‘Šđ?&#x;¨

đ?&#x;Ł

(đ?&#x;¨)

đ?‘Šđ?&#x;Š

đ?&#x;¤

â‹Ż

â‹Ż

Here, it has gone into a cycle, but it didn’t return to the start of the sequence. So in every case we still get a cycle, but now we know that the loop might not go back all the way to the beginning. chalkdustmagazine.com

50

chalkdust This leads to a reasonable conjecture: no matter which base you choose, the Weirdonacci numbers will eventually go into a cycle and repeat forever. It isn’t too hard to see that this must be the case (once you think about it the right way). If we work in base đ?‘? , then there are only đ?‘? distinct numbers that can occur in the sequence. But if there are only đ?‘? distinct numbers, there can’t be more than đ?‘? đ?&#x;¤ pairs of numbers. Then a sequence of more than đ?‘? đ?&#x;¤ terms must have some repeated pairs. Wherever we spot a pair repeating, we have identified a cycle. So not only do we know that the Weirdonacci numbers always eventually repeat, we know that the length of the cycle can’t be any longer than đ?‘? đ?&#x;¤ . (Question: can it be as long as đ?‘? đ?&#x;¤ ?)

We can see from the above example that the cycle doesn’t have to start with the pair of numbers đ?&#x;Ł, đ?&#x;Ł. In fact we can start a Weirdonacci sequence with any pair of numbers we like, and it doesn’t matter what the starting values are, or what base we use, eventually the sequence must enter a cycle. One of the first questions we started thinking about was what the length of a cycle could be.

Cycles can be any length We’re going to look at what lengths cycles can be now. The maths gets a bit more complicated, so don’t feel embarrassed about skipping to the conclusions after the scorpions at the end of this section, and coming back to this later. Fibonacci numbers Let us write F đ?‘› (đ?›ź, đ?›˝) for the đ?‘›th Fibonacci number, with initial values đ?›ź and đ?›˝ . For example, if we start the sequence with đ?&#x;¤, đ?&#x;§ we obtain đ?&#x;¤, đ?&#x;§, đ?&#x;Š, đ?&#x;Łđ?&#x;¤, ‌ and so Fđ?&#x;Ľ (đ?&#x;¤, đ?&#x;§) = đ?&#x;Š and Fđ?&#x;Ś (đ?&#x;¤, đ?&#x;§) = đ?&#x;Łđ?&#x;¤.

We can show (this is left as an exercise for the reader) that these new sequences are related to the usual Fibonacci numbers by the formula Exercise for the reader

Fđ?‘› (đ?›ź, đ?›˝) = đ?›źđ??šđ?‘›âˆ’đ?&#x;¤ + đ?›˝đ??šđ?‘›âˆ’đ?&#x;Ł ,

where the usual Fibonacci sequence had to be a little extended by adding the numbers đ??šâˆ’đ?&#x;Ł = đ?&#x;Ł and đ??šđ?&#x;˘ = đ?&#x;˘. Look at the same example as above (the sequence starting with đ?&#x;¤, đ?&#x;§), this formula correctly tells us that Fđ?&#x;Ś (đ?&#x;¤, đ?&#x;§) = đ?&#x;§đ??šđ?&#x;Ľ + đ?&#x;¤đ??šđ?&#x;¤ = đ?&#x;§ Ă— đ?&#x;¤ + đ?&#x;¤ Ă— đ?&#x;Ł = đ?&#x;Łđ?&#x;¤. We easily verify that Fđ?&#x;Ł (đ?›ź, đ?›˝) = đ?›ź and Fđ?&#x;¤ (đ?›ź, đ?›˝) = đ?›˝ , as well as Fđ?‘› (đ?&#x;Ł, đ?&#x;Ł) = đ??šđ?‘› . 51

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chalkdust Weirdonacci numbers (đ?‘?)

Let us write đ?‘Šđ?‘› (đ?›ź, đ?›˝) for the đ?‘›th Weirdonacci number, generated in base đ?‘? , and with initial num(đ?‘?) bers đ?›ź and đ?›˝ . As long as Fđ?‘› (đ?›ź, đ?›˝) is smaller than the base đ?‘? , the Weirdonacci numbers đ?‘Šđ?‘› (đ?›ź, đ?›˝) are equal to Fđ?‘› (đ?›ź, đ?›˝). We then define as đ?‘˜ the smallest integer such that Fđ?‘˜ (đ?›ź, đ?›˝) < đ?‘? ⊽ Fđ?‘˜+đ?&#x;Ł (đ?›ź, đ?›˝). Therefore for any đ?&#x;Ł ⊽ đ?‘› ⊽ đ?‘˜, (đ?‘?)

đ?‘Šđ?‘› (đ?›ź, đ?›˝) = đ?›źđ??šđ?‘›âˆ’đ?&#x;¤ + đ?›˝đ??šđ?‘›âˆ’đ?&#x;Ł .

When we reach a number greater than the base đ?‘? , the next two Weirdonacci are generated: re(đ?&#x;Łđ?&#x;˘) (đ?&#x;Łđ?&#x;˘) (đ?&#x;Łđ?&#x;˘) member that in base đ?&#x;Łđ?&#x;˘, đ?‘Šđ?&#x;§ = đ?&#x;§, đ?‘Šđ?&#x;¨ = đ?&#x;Ş and since đ?&#x;Ş + đ?&#x;§ = đ?&#x;Łđ?&#x;Ľ, one writes đ?‘Šđ?&#x;Š = đ?&#x;Ł and (đ?&#x;Łđ?&#x;˘) đ?‘Šđ?&#x;Ş = đ?&#x;Ľ. We’ll call this a collapse.

In general, we can write Fđ?‘˜+đ?&#x;Ł (đ?›ź, đ?›˝) = đ?‘žđ?‘? + đ?‘&#x; , so that the next two Weirdonacci numbers are (đ?‘?) (đ?‘?) đ?‘Šđ?‘˜+đ?&#x;Ł (đ?›ź, đ?›˝) = đ?‘ž and đ?‘Šđ?‘˜+đ?&#x;¤ (đ?›ź, đ?›˝) = đ?‘&#x; . In fact, đ?‘ž must be 1, because otherwise at least one of the two previous Weirdonacci numbers would have to be greater than đ?‘? . Consequently, we can write đ?‘&#x; = Fđ?‘˜+đ?&#x;Ł (đ?›ź, đ?›˝) − đ?‘? .

It follows that every loop must at least contain one pair of the form (đ?&#x;Ł, đ?›˝), which it is natural to consider as the start of the cycle. Starting with 1 and đ?›˝ , we can write (đ?‘?)

(đ?‘?)

(đ?‘?)

(đ?‘?)

(đ?‘?)

đ?‘Šđ?&#x;Ł (đ?&#x;Ł, đ?›˝) đ?‘Šđ?&#x;¤ (đ?&#x;Ł, đ?›˝) â‹Ż đ?‘Šđ?‘˜ (đ?&#x;Ł, đ?›˝) đ?‘Šđ?‘˜+đ?&#x;Ł (đ?&#x;Ł, đ?›˝) đ?‘Šđ?‘˜+đ?&#x;¤ (đ?&#x;Ł, đ?›˝) â‹Ż đ?&#x;Ł

where đ?‘&#x; = Fđ?‘˜+đ?&#x;Ł (đ?&#x;Ł, đ?›˝) − đ?‘? .

đ?›˝

(đ?‘?)

â‹Ż Fđ?‘˜ (đ?&#x;Ł, đ?›˝)

đ?&#x;Ł

đ?‘&#x;

â‹Ż

Simple loops

A loop of length â„“ = đ?‘˜ would be formed if đ?‘&#x; = đ?›˝ . We’ll call loops that arise this way simple loops. Of course this is not the only way to generate a loop, as it may take several collapses for the loop to close, but it makes sense to think about the simplest possibility first. The equation đ?›˝ = Fđ?‘˜+đ?&#x;Ł (đ?&#x;Ł, đ?›˝) − đ?‘?

gives us a relation between đ?‘? , đ?›˝ and đ?‘˜ for all simple loops. We can then rewrite this equation as đ?‘? − đ??šđ?‘˜âˆ’đ?&#x;Ł = đ?›˝(đ??šđ?‘˜ − đ?&#x;Ł).

If we pick a desired cycle length đ?‘˜ = â„“ , we can use this formula to find a base and starting numbers to give this cycle length. This means that cycles of every length are possible. There’s lots more we could think about here, including the following challenge. chalkdustmagazine.com

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chalkdust

Challenge 1 (đ?&#x;¨)

The Weirdonacci sequences đ?‘Šđ?‘›

(đ?&#x;Łđ?&#x;˘)

and đ?‘Šđ?‘›

both have a cycle of length 4.

Show that if the base đ?‘? is even, then it is possible to pick two starting numbers that lead to a Weirdonacci sequence with a cycle of length 4. If you’re stuck, there is a hint at the end of the article. Non-simple loops Of course, one could try to find equations describing non-simple loops—those with more than one collapse. For example, we could look for loops like this (we’ve left out the (đ?&#x;Ł, đ?›˝)s in the top row): (đ?‘?)

đ?‘Šđ?&#x;Ł

đ?&#x;Ł

(đ?‘?)

đ?‘Šđ?&#x;¤ đ?›˝

â‹Ż

(đ?‘?)

đ?‘Šđ?‘˜

(đ?‘?)

â‹Ż Fđ?‘˜ (đ?&#x;Ł, đ?›˝)

(đ?‘?)

(đ?‘?)

đ?‘Šđ?‘˜+đ?&#x;Ł đ?‘Šđ?‘˜+đ?&#x;¤ â‹Ż đ?&#x;Ł

These constraints lead to the set of equations

�′

(đ?‘?)

đ?‘Šđ?‘˜+đ?‘ž

(đ?‘?)

â‹Ż Fđ?‘ž (đ?&#x;Ł, đ?›˝ ′ )

(đ?‘?)

(đ?‘?)

đ?‘Šđ?‘˜+đ?‘ž+đ?&#x;Ł đ?‘Šđ?‘˜+đ?‘ž+đ?&#x;¤ â‹Ż đ?&#x;Ł

đ?›˝

â‹Ż

đ?‘? + đ?›˝ ′ = đ??šđ?‘˜âˆ’đ?&#x;Ł + đ?›˝đ??šđ?‘˜ ,

where đ?‘˜ + đ?‘ž is the period of the loop.

đ?‘? + đ?›˝ = đ??šđ?‘žâˆ’đ?&#x;Ł + đ?›˝ ′ đ??šđ?‘ž ,

We have in fact already seen two examples for such loops, in base đ?‘? = đ?&#x;Łđ?&#x;˘ and đ?‘? = đ?&#x;Ť, where we had đ?›˝ = đ?&#x;Ł, đ?›˝ ′ = đ?&#x;Ľ, đ?‘˜ = đ?&#x;¨, đ?‘ž = đ?&#x;Ś; and đ?›˝ = đ?&#x;˘, đ?›˝ ′ = đ?&#x;Ś, đ?‘˜ = đ?&#x;Ş, đ?‘ž = đ?&#x;Ľ (respectively). Again, there are many other things to think about here, including the following challenge.

Challenge 2 Show that it is not possible to have a non-simple loop with đ?‘˜ = đ?‘ž . If you’re stuck, there is a hint at the end of the article.

We’ve found out a bit about which cycles can happen. We know that if we are allowed to choose the base and initial values, we can get any length of cycle that we care to. We have some idea how simple loops—where the value only exceeds the base once—behave, but loops where this happens more than once are much harder to analyse. We’re now going to hand the question‌ Cycles

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‌ over to you We are now in a position to ask a few questions: (đ?‘?)

(đ?‘?)

1. What is the sequence, generated in base đ?‘? , following the initial two numbers đ?‘Šđ?&#x;Ł and đ?‘Šđ?&#x;¤ ?

2. After how many steps does the sequence enter a cycle?

3. What is the length of the cycle reached by the sequence? 4. For a given base đ?‘? , is there one unique cycle? If not, how many cycles are there?

There are lots of other questions you might be asking about these sequences. We leave it to you to think about them using the ideas from the previous section—and, of course, whatever other ideas you have yourself! If you find out something interesting, why not write it up for Chalkdust issue 12? Robert J Low Robert teaches maths at Coventry University.

d robjlow.blogspot.co.uk c mtx014@coventry.ac.uk a @RobJLow Thierry Platini Thierry teaches maths at Coventry University.

c ab3334@coventry.ac.uk My favourite conjecture

An elementary inequality Sam Porritt

đ?‘›

Let đ?œŽ(đ?‘›) = ∑đ?‘‘|đ?‘› đ?‘‘ be the sum of the divisors of đ?‘› and đ??ťđ?‘› = ∑đ?‘Ž=đ?&#x;Ł đ?&#x;Ł/đ?‘Ž. Then for all integers đ?‘› > đ?&#x;Ł, the following inequality holds đ?œŽ(đ?‘›) < đ??ťđ?‘› + exp(đ??ťđ?‘› ) log(đ??ťđ?‘› ).

Inspired by and building on an inequality suggested by Guy Robin in 1984, this inequality was conjectured to hold by Jeffrey Lagarias in 2001. Despite the fact that this inequality looks like it might have come from a maths Olympiad or Putnam Competition paper, remarkably, it is in fact equivalent to the Riemann hypothesis! Challenge 2: Try setting đ?‘˜ = đ?‘ž in the two equations and see what happens. Challenge 1: Try starting your sequence with 1 and đ?‘?/đ?&#x;¤ − đ?&#x;Ł. Challenge hints chalkdustmagazine.com

54

THIS ISSUE‌

IS ZERO A NATURAL NUMBER?

E TH E TH

G I B

E TH

T EN M GU AR

NO

YES No one hates zero!

No, one hates zero!

The natural numbers are often informally called the ‘counting numbers’, and đ?&#x;˘ ďŹ ts as a counting number. Many programming languages, including Python, start counting from đ?&#x;˘, and that seems to be the norm rather than an exception. But people use zero when counting, too–we just take it for granted. What happens when you’re counting your pencils and ďŹ nd you don’t have any? What number do you use to describe this? That’s right—it’s zero!

Well I say it’s about time we stop listening to computer scientists: they may like including đ?&#x;˘, but us mathematicians need to worry about consequences. If we add đ?&#x;˘ to the natural numbers, then we can no longer divide by every natural number, and everything gets very complicated when we try to think about factors: we can divide đ?&#x;˘ by đ?&#x;Š, so is đ?&#x;Š a factor of đ?&#x;˘?! I’m against writing â„•đ?&#x;Ł (it’s ugly) or ℤ⊞đ?&#x;Ł (it’s very ugly): if they want to include đ?&#x;˘, then they should write â„•đ?&#x;˘ , and we can keep our â„• zerofree.

If that’s not convincing enough, note that we use natural numbers to denote the size (or cardinality) of a ďŹ nite set. That includes zero—it’s the size of the empty set!

If we give in and accept that 0 is natural, then what’s next? Will they suggest that −đ?&#x;Ł is natural? Or Ď€?! How irrational!

ARGUES

ARGUES

BELGIN SEYMENOÄžLU

JOHNNY NICHOLSON 55

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#11 Set by Humbug 1 5

2

4

3

7

6

9

8 10 11 14

15

12

13

16

17

18 20

21

19

22

23

24

25 27

26 28

29

30

31

32 33 34

35

36 37

38

Although many of the clues have multiple answers, there is only one solution to the completed crossnumber. As usual, no numbers begin with 0. Use of Python, OEIS, Wikipedia, etc is advised for some of the clues. To enter, send us the sum of the across clues via the form on our website (d chalkdustmagazine.com) by 3 September 2020. Only one entry per person will be accepted. Winners will be notified by email and announced on our blog by 1 October 2020. One randomly-selected correct answer will win a ÂŁ100 Maths Gear goody bag, including non-transitive dice, a Festival of the Spoken Nerd DVD, a dodecaplex puzzle and much, much more. Three randomly-selected runners up will win a Chalkdust T-shirt. Maths Gear is a website that sells nerdy things worldwide. Find out more at d mathsgear.co.uk chalkdustmagazine.com

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chalkdust

Across

Down 1 Equal to 9A.

(2)

5 The product of the digits of this number is (6) prime.

2 1D multiplied by 9A.

(3)

8 A multiple of 10 that is equal to the difference (2) between 29D and 12D.

4 Larger than 28D, but divisible by (6) 28D.

1 38A backwards.

(10)

9 Equal to 1D.

(2)

10 This number is equal to the sum of the cubes (3) of its digits. 13 The difference between 18A and 26A.

(3)

15 This number is a multiple of each of its digits. (7) 17 20A multiplied by a factor of 17A.

(3)

18 Each digit of this number (except the first) is (3) one less than the previous digit.

3 One more than a multiple of 35D. (3)

6 For each đ?‘› from 0 to 9, the digit đ?‘› (9) appears either 0 or đ?‘› times in this number. 7 This number is equal to the sum (5) of the fifth powers of its digits. 11 A non-square multiple of 6 that is (2) equal to 3 more than the sum of the digits of 15A.

(2)

12 A square number whose first digit (2) is equal to the first digit of 15A.

21 The final digit of this number is the same as (2) the final digit of 10A and the first digit of this number is the same as the first digit of 24A.

14 Larger than 16D, but divisible by (4) 16D.

20 The sum of the digits of 6D.

24 The final digit of this number is the same as (2) the final digit of 33A and the first digit of this number is the same as the first digit of 21A. 25 The sum of the digits of 19D.

(2)

26 Each digit of this number (except the first) is (3) one less than the previous digit. 27 25A multiplied by a factor of 27A.

(3)

29 This number is a multiple of each of its digits. (7) 32 The difference between 18A and 26A.

(3)

33 This number is equal to the sum of the cubes (3) of its digits.

16 14D backwards.

(4)

19 For each đ?‘› from 0 to 9, the digit đ?‘› (9) appears either 0 or đ?‘› times in this number. 22 23D backwards.

(4)

23 Larger than 22D, but divisible by (4) 22D. 28 4D backwards.

(6)

29 A square number whose first digit (2) is equal to the first digit of 29A.

(2)

30 A non-square multiple of 6 that is (2) equal to 3 more than the sum of the digits of 29A.

36 A multiple of 10 that is equal to the difference (2) between 29D and 12D.

31 This number is equal to the sum (5) of the fifth powers of its digits.

37 The product of the digits of this number is (6) prime.

34 One more than a multiple of 2D. (3) 35 37D multiplied by 35A.

(3)

38 Larger than 1A, but divisible by 1A.

37 Equal to 35A.

(2)

35 Equal to 37D.

(10) 57

spring 2020

On this page, you can find out what we think of recent books, films, games, and anything else vaguely mathematical. Full reviews of many of the items featured here can be found at d chalkdustmagazine.com

The Royal Institution Christmas lectures Hannah Fry Hannah Fry takes us on a whirlwind tour of real-world maths over three hour-long episodes (available on YouTube: m TheRoyalInstitution). The series covers topics from volcanology to viral videos, and featuring a host of your favourite science communicators getting their hands dirty in illuminating experiments and games. Fry is brilliantly engaging and lively as usual, and also takes the opportunity to address some serious questions such as the ethics of big data. A must-watch for science fans of any age.

ggggg Bach, The Universe, and Everything

The Secret Service A wonderfully bonkers supermarionation series about a vicar who is also a secret agent and defeats his enemies by speaking gobbledygook.

Orchestra of the Age of Enlightenment The sublime union of the music of Bach and a secular sermon on a topic from maths or science—a transcendent experience.

gggii

ggggh Captain Scarlet and the Mysterons

MathsJam Gathering 2019

A highly entertaining and also quite scary supermarionation series. The best (and most mathematical) episode is Big Ben Strikes Again.

Another excellent weekend of 5 minute talks, puzzles, and entertainment. Perhaps the best MathsJam yet.

ggggg

ggggg

Fireball XL5 Torchy the Battery Boy

“I wish I was a spaceman, the fastest guy alive...”

So cursed.

ggggi chalkdustmagazine.com

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Chalkdust Book of the Year 2019 The Maths of Life and Death by Kit Yates This book, as its title suggests, is all about the maths of life and death. It looks at the maths related to law, biology, medicine, the spread of diseases, and many more areas. Kit takes you through these areas with a series of interesting and well-written stories. This book was selected by the editors of Chalkdust to be the Chalkdust Book of the Year 2019, based on our four judging criteria: style, control, damage and aggression.

Chalkdust Readers’ Choice 2019 A Compendium Of Mathematical Methods by Jo Morgan This book contains details of various methods of completing common mathematical tasks, including subtraction, multiplication, simplifying surds, and polynomial division. Each method is demonstrated by example and the reasons why each method works are discussed. This book is an excellent reference for secondary school teachers, as well as an interesting book to browse through for the mathematically curious. This book was voted by our readers to be the Chalkdust Readers’ Choice 2019.

Shortlisted The winners were selected from our shortlist of nine books released in 2019. The seven non-winning books are all also very good. They were:

So You Think You’ve Got Problems? by Alex Bellos; The Hidden Half by Michael Blastland; The Art of Logic by Eugenia Cheng; Maths on the Back of an Envelope by Rob Eastaway; Here Come the Numbers by Kyle D Evans and Hana Ayoob; Humble Pi by Matt Parker; and Geometry Puzzles in Felt Tip by Catriona Shearer. 59

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What is the shape of you?

START Tea or coffee?

COFFEE

YOU ARE A

torus TEA YOU ARE A

DOUGHNUTS

cone What is your favourite food?

SPONGE CAKE

ICE CREAM

BUFFET

YOU ARE A

GINGER SNAPS

Menger sponge

What’s your favourite TV show?

BLIND DATE SURPRISE, SURPRISE! YOU ARE

Ed Spheran Ed Spheran

YOU ARE A

YOU ARE A

Platonic solid

trapezium

GAME OF THRONES

YOU ARE A

pyramid YOU, PIKACHU!

I choose…

I DON’T

YOU ARE

THE CIRCUS

FRIENDS

Cylinder Black BANACH & TARSKI

EGYPT

GREECE

Where are you most likely to run away to?

YOU ARE

Ed Spheran

Are you immature?

NO

I KNOW YOU ARE BUT WHAT AM I? YOU ARE A

Koch snowflake YOU ARE A

Minkowski sausage

chalkdustmagazine.com

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chalkdust

Adopt a polyhedron Adapted from Everdon sign, Wikimedia Commons user Everdon, CC BY-SA 3.0

Mara Kortenkamp, Anna Maria Hartkopf and Erin Henning

W

ould you like to have a mathematical object named after yourself? You no longer need to be Pythagoras to eternalise yourself in the terminology of mathematics.

A polyhedron is not just for Christmas Mathematicians currently know over a hundred trillion∗ different kinds of polyhedron and only a very small percentage of them actually have a name, like cube or pyramid. Not only does the cube possess a name and fame, it is also realised in gigantillions of models all over the world. But not for most polyhedra. These helpless shapes are nameless, unrealised, and lost in abstraction. You can help change that, today.

Who are we? Polytopia is a foster home for all polyhedra with up to nine vertices. We give them a home, regardless of their prominence or whether they’ve al-

This is Ecki. He lives in Polytopia.

∗

114,683,158,574,357 to be precise. This is the number of combinatorial types of polyhedron with up to 18 vertices. No mathematician has been able to give the number for 19 vertices exactly: so far for more than 18 vertices, only upper bounds on the number of combinatorial types are known. Even the computers we have today would take forever to construct and count all types for 19 vertices.

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chalkdust ready got a name. When visiting Polytopia, you’ll be able to see all the polyhedra who have already found loving homes and, more importantly, meet the ones still waiting to be adopted.

But what can you do? You can help us make a change and adopt one of these little geometrical creatures. By giving them a name and building a physical model of them, you can bring them into the real world. Visit us at Polytopia today and be inspired to rescue your new best friend from the realm of abstraction.

About our polyhedra There are several definitions of polyhedra in mathematics. Let us start with talking about polygons, which you are probably already familiar with. A polygon is a two-dimensional shape, consisting of straight line segments, called edges, and points, called vertices, where two edges meet. The simplest example of a polygon is a triangle. One important type of polygon is the regular polygon: those which have edges of the same length, and with all internal angles the same. Examples of regular polygons are the equilateral triangle and the square.

“I named my polyhedron Seems to be Piez and had a lot of fun crafting it.”

chalkdustmagazine.com

“My polyhedron is called Oktavius and makes a lovely decoration for my room.”

62

One of the polyhedra that lives in Polytopia.

From polygons we can naturally build up to polyhedra. A polyhedron is a geometrical object bounded by a finite number of polygons. In other words, a polyhedron is a three-dimensional object, whose faces are polygons, whose edges are the straight line segments where two such faces meet, and whose vertices are points where three or more faces meet. One of the simplest (and most famous) polyhedra is the cube. It is built by glueing squares together and it consists of eight vertices connected by twelve edges and six faces.

chalkdust In Polytopia, we are only interested in convex polyhedra, which means that all inner angles between two edges or two faces are less or equal to 180°. Thus, no indentations, holes or cavities are allowed, and all vertices and edges are ‘pointing outwards’. The superstars of the polyhedral world are the Platonic solids. These five polyhedra, each consisting only of identical regular polygons, are named after the ancient Greek philosopher Plato. Starting with regular triangles, we can use them to build three of the five Platonic solids: the tetrahedron, the octahedron, and the icosahedron. Do you already know the two remaining Platonic solids? (Hint: one of them is the cube.)

Three Platonic solids. Who’s missing?

Now you might ask yourself: how do we differentiate the polyhedra? We at Polytopia distinguish polyhedra by their combinatorial type and not their geometric realisation. The combinatorial type describes the structure of the vertices, edges, and faces of a polyhedron but it is not sensitive to the size or even the shape of a polyhedron. Given two polyhedra, take out the faces as if they were made of glass. “Since I was in need of a new “It was so much fun modelling What remains is the frame of the pillow, I decided to sew my my polyhedron Pullyeder out of polyhedron Flori into one.” chocolate ice cream, especially edges. Now imagine that you can because I ate it afterwards!” mould and transform this frame in any way you like, but you cannot disconnect any of the edges or vertices. If you can transform the frame of one polyhedron into the other without making or breaking any connections, then these two polyhedra are considered combinatorially equivalent. The two pictures on the next page show the cube but in different geometric realisations. The one on the left looks like what you might expect and the one on the right was generated by modifying the size and shape of the usual cube. But notice that the vertices can still be matched to one another, preserving the edges’ connections. Therefore, they are combinatorially equivalent, and are both called Cube. 63

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Cube standing up straight, and Cube slouching.

Unfortunately, even with combinatorial equivalence, there are so many different polyhedra that it is infeasible for mathematicians to name them all. That’s why we here at Polytopia invite you to name as many polyhedra as you can. Every little helps. Mara Kortenkamp and Erin Henning Erin is studying mathematics, and Mara is studying computer science. Both are working as student assistants within the discrete geometry workgroup at Freie Universität Berlin. Anna Maria Hartkopf Anna Maria is a professional maths communicator and is currently writing her PhD about her project, Polytopia.

d polytopia.eu b polytopia.eu l polytopia.eu My favourite conjecture

Collatz conjecture David Sheard

Consider the function:

�↌{

đ?&#x;Ľđ?‘› + đ?&#x;Ł đ?‘›/đ?&#x;¤

if đ?‘› is odd, if đ?‘› is even.

Start say with đ?‘› = đ?&#x;Łđ?&#x;Ľ and repeatedly apply this—what do you notice?

It is conjectured that starting with any positive integer you will eventually hit 1.

My favourite conjecture

Perfect numbers Petru Constantinescu

A perfect number is a positive integer that is equal to the sum of its divisors, excluding the number itself. The first few are 6, 28, 496, and 8128. Euclid in his Elements gave a formula for producing even perfect numbers: đ?&#x;¤đ?‘›âˆ’đ?&#x;Ł (đ?&#x;¤đ?‘› − đ?&#x;Ł) is perfect whenever đ?&#x;¤đ?‘› − đ?&#x;Ł is prime. More than 2000 years later, Euler proved that all even perfect numbers must be of this form. It is still not known whether or not there are infinitely many even perfect numbers.

The situation for odd perfect numbers is even more interesting. It is unknown whether there are any odd perfect numbers. chalkdustmagazine.com

64

Ecki the polytope You will need A copy of Chalkdust issue 11, scissors, glue, pipe cleaners, a pen

Instructions

1

Cut out this net.

Fold along the lines, glue the tabs, and make the 3D solid.

3

2

Draw on Ecki’s eyes and mouth, and use pipe cleaners to make arms, legs, and glasses.

Tube map platonic solids, FrĂśbel stars and slide rules: more How to make at d chalkdustmagazine.com 65

spring 2020

We love it when our readers write to us. Here are some of the best emails, tweets and letters we’ve been sent. Send your comments by email to c contact@chalkdustmagazine.com, on Twitter a @chalkdustmag, or by post to e Chalkdust Magazine, Department of Mathematics, University College London, Gower Street, London WC1E 6BT, UK.

Dear Chalkdust, Do you have any back copies of issues 01–09 left? My husband recently came across a copy of issue 10 and is really enjoying it. I also liked the crochet article as that’s my favourite hobby! I’d love to surprise him with some back issues as a Christmas present. Could you let me know if there’s anything I can do to get hold of some? Emily

Dear Chalkdust,

Thank you for the Christmas card—the whole family were impressed by the Raspberry Pi stamp :)

I thought I had been forsaken by the Chalkdust folks, but I return to the office after the break to find a lovely Christmas card waiting in my pigeonhole. Neat stamp too!

Sam Hartburn, Kent

Peter Rowlett a @PeterRowlett

What a great article from Chalkdust. I happen to have a croch et hook handy! d chalkdustmagazine.com/features/crocheting-fractals Amanda Hogan a @hogesonline

Delighted that The Maths of Life and Death has been chosen as the Chalkdust book of the year for 2019. If you get a chance to read it, I hope you enjoy. Kit Yates a @Kit_Yates_Maths

This series of reviews is in danger of costing me some serious money. Good job I’ve already bought three of the upcoming books.

a @UsrBinPRL

Thanks, Catriona [see page 43]—I’m a fan of yours, too! (Also of Chalkdust, obviously. Everyone loves Chalkdust!) Colin Beveridge a @icecolbeveridge chalkdustmagazine.com

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TOP TEN

This issue features the top ten pictures of scorpions. To vote on the top ten mathsthemed days out, go to d chalkdustmagazine.com

At 10, it’s Wind of Change by The Scorpions.

At 9, it’s scorPioneers by Bloc Party.

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9

At 8, it’s scorpIonisation, Edgard Varèse’s work that features only percussion instruments.

At 7, it’s scorpiOne by U2.

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7

At 6, it’s sCorpus Christi Carol by Jeff Buckley.

At 5, it’s scorpiOnce in a Lifetime by The Talking Heads.

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5

At 4, it’s scorpiOn The Road Again by Willie Nelson.

At 3, it’s scorpiOne of These Days by Camper Van Beethoven.

4

3

At 2, it’s I Write Sins Not Tragedies by Panic! at the Discorpion.

Topping the pops this issue, it’s the album Scorpion by Drake.

2

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