Chalkdust, Issue 14 by Chalkdust

c ha l k dus t Amagaz i nef ort hemat hemat i cal l ycur i ous

1 4/ /

2021

ONTHECOVER:VHAT?VHERE?VENN MAKI NGACHRI STMASLECTURE+DEARDI RI CHLET PAWNS,PUZZLES&PROOFS+PRI ZECROSSNUMBER

IN THIS ISSUE 3

Page 3 model

Dear Dirichlet

The big argument d𝑥/d𝑡 , 𝑥 ′ or 𝑥 ̇ ? Derivative notations fight for your heart

What’s hot and what’s not

Pawns, puzzles, and proofs Dimitrios Roxanas promotes retrograde chess puzzles

How Polish cryptographers first broke the unbreakable cipher Kimi Chen deciphers the 1920s story you haven’t heard

Handbook of equations Tilly Pitt throws us a lifeline

On the cover: Vhat? Vhere? Venn? Venn is a diagram not a diagram? Madeleine Hall explores.

Puzzles

Southgate ⩾ Sven? An update by Paddy Moore

A walk on the random side Tanmay Kulkarni intentionally gets lost on the Tokyo subway

Prize crossnumber

Celebrating simple mathematical models Hollis Williams catches an infectious disease

Cryptic crossword

Top ten: Waves

4 In conversation with Dominique Sleet Ellen Jolley learns how to run a maths outreach programme

10 That’s a Moiré Donovan Young’s favourite interference patterns hit your eye like a big pizza pi

44 An odd card trick Self-working tricks: Michael Wendl keeps his cards close to his chest 1

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chalkdust From Subject

Chalkdust <contact@chalkdustmagazine.com> Re: Issue 14 02:11

Sorry we forgot the attachment. Here it is. The team Jamie Handitye Jeong Gyu Im Ellen Jolley King Ming Lam Sophie Maclean Matthew Scroggs Belgin Seymenoğlu David Sheard Jakob Stein Beatrice Taylor Adam Townsend Cover artwork Beatrice Taylor

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.

The Chalkdust team > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > >

On 19 Oct at 02:07, Chalkdust wrote: To whom it may concern, We attach for your perusal the 14th issue of Chalkdust, a magazine for the mathematically curious. We are sure that you will agree that this edition is a particularly good one. It's lovely that we're currently writing this message as we sit in the same place (±1m), in some cases for the first time. We're sure you'll enjoy the freshness brought by our new faces on the team: more artwork, cleaner design, less mistakes to annoy grammar pedants. As you read the attached issue, you will notice that one of its strongest features is the wide variety of topics. There really is something for everyone to enjoy in this issue, although keen readers like you are likely to enjoy it all. We hope you are well, and look forward to receiving your draft article for issue 15. All the best, The Chalkdust team

Acknowledgments 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; the organisers of TMiP for sending copies of Chalkdust to all attendees, as well as letting us promote the magazine; Clare Wallace for her discussions with Professor Dirichlet; Kit “Kit Yates Maths” Yates for consultation; everyone at Achieve Fulfilment for their help with distribution. ISSN 2059-3805 (Print). ISSN 2059-3813 (Online). Published by Chalkdust Magazine, Dept of Mathematics, UCL, Gower Street, London WC1E 6BT, UK. © Copyright for articles is retained by the original authors, and all other content is copyright Chalkdust Magazine 2021. 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

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Jakob Stein

As the weather turns from grey to cold and grey, it’s time to get out your favourite woolly jumper. Only, it’s a bit longer than you remember. Has someone has been stretching it out behind your back? Instead of jumping to conclusions, let’s try to understand how this wool-d happen. Your jumper is made up of yarn, made from woollen fibres spun together. Let’s assume the fibres are in one of two states: 𝐴, unstretched, or 𝐵, completely stretched out. The jumper as a whole is much harder to stretch with the fibres in state 𝐵 than in state 𝐴, because each individual fibre is already elongated. As we start stretching the jumper, more and more fibres go from state 𝐴 to state 𝐵, and the rate at which it stretches slows down. Let’s model this stretching over time as d𝑙 = 𝛼 − 𝛽𝑙, d𝑡 where 𝑡 is time, 𝑙 is how much the jumper has stretched, and 𝛼, 𝛽 are constants depending on the mechanical properties in states 𝐴 and 𝐵. Assume that the jumper started off unstretched (unless you got fleeced), so that 𝑙(𝟢) = 𝟢. Then this equation has a unique solution, giving an explicit formula for how much your jumper has stretched: 𝑙(𝑡) =

𝟣 𝛼 (𝟣 − ). 𝛽 exp (𝛽𝑡)

On the bright side, even if someone has been wearing it, this shows your jumper won’t go on stretching forever: as 𝑡 gets very large, 𝑙 ≈ 𝛼/𝛽. This approximate solution was reproduced by experimental evidence from the appropriately named Wool Textile Research Laboratory, just in case you thought this whole ‘two-state’ yarn was a bit of a stretch. 3

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IN CONVERSATION WITH

Dominique Sleet Ellen Jolley talks to the science communication expert about museums, masterclasses, Christmas lectures, and representation

here are many qualities attributed to mathematicians that we can be proud of: we’re logical, meticulous, intelligent, even creative. Despite maths being revered for needing all of these excellent attributes, one thing we are perhaps not so renowned for is our communication skills: most of the world still finds maths intimidating and opaque. That’s why Chalkdust sat down with science communication expert Dominique Sleet to learn the secrets that will help us to share the beauty of maths as far and wide as possible.

Science explained Dominique began her science communication life as an explainer at the Science Museum in London. Many London-based Chalkdust readers will be keenly familiar with the Science Museum, but for those who are not, the Science Museum is pretty much exactly what it says on the tin. As well as many more traditional static galleries, it has several interactive galleries aimed at young people, which are like science-themed playgrounds.

Wikimedia Commons user Shadowssettle, CC BY-SA 4.0

s The Science Museum chalkdustmagazine.com

Explainers are tasked with (you guessed it) explaining the science behind what they’re doing. “You get to play with some fun things, like putting flowers into liquid nitrogen and smashing them or blowing up a hydrogen balloon. The whole museum ethos is to build an association between fun and science. If there’s some learning in there then that’s great, but there doesn’t have to be; it’s just about nurturing that relationship.” But there’s plenty to learn if you’re looking. “Some of the science behind the exhibits is beautiful. We had this exhibit where you would look through polarising lenses at a thin layer of ice and you could see all these feathery beautiful patterns with amazing colours. I’m a nerd, I like science.” 4

chalkdust Those of us who grew up to be Chalkdust editors could spend a cheerful afternoon churning out algebra, but for most children the liquid nitrogen has the more evident appeal. “Maths has its challenges. It’s a lot more abstract. With science, as long as you use the right language, you can make almost anything accessible, whereas with maths, you often need to have prior knowledge.” And don’t forget that intimidation. “People have this barrier, it’s almost like a badge, ‘I don’t do maths.’ And it’s really sad. So before you’ve even begun, you have to overcome this preconception of maths being like an alien language, and only for clever people.” People are often more receptive to the content if they don’t know that it’s maths—“Pattern Pod was my favourite gallery, which was for under-eights and all about maths, but we were looking at patterns and didn’t label it as maths.”

The whole museum ethos is to build an association between fun and science. If there’s some learning in there then that’s great, but there doesn’t have to be, it’s just about nurturing that relationship.

Unfortunately, the plausible deniability can’t last forever, and inevitably your audience will notice that they are being subjected to maths: what then? “You need to show why something’s important and make it relevant to everyday life. You can’t get into the depth that you’d need to understand all the maths behind the exhibits, but you can go into some detail about how the maths was used, how it changed the world, and what impact it had on people.” The work does not end there however. How do we get people to turn up for maths in the first place?

A royal invitation Receiving millions of visitors per year, the Science Museum is well-placed to reach out to people who wouldn’t usually be interested. “But even then there are barriers. I remember doing an outreach programme in south-east London and people hadn’t even heard of the Science Museum. That’s why outreach is really important. Going out into local communities and finding the people where they are in their everyday lives.” Dominique’s next job was at the Royal Institution (also in London), where she worked on evs Dominique talking in the Royal Institution erything from their famous annual Christmas lecture theatre lectures to their extensive year-round masterclass programme. So what’s the trick to running a maths masterclass? “Pick a topic that interests you because if you’re passionate about the topic then you’re halfway there. The kids aren’t going to be excited about something if you’re not excited yourself. In the same breath, you need to realise that, while you might find something amazing, other people really don’t. You’ve got to show them why it’s interesting.” 5

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chalkdust

s A braid made during one of Matthew Scroggs’s RI masterclasses

It was nice to hear a shoutout for one of Chalkdust’s own, who apparently is quite the master of masterclasses himself. “This will sound like I’m sucking up, but I remember a session with Matthew Scroggs getting the kids to explore different braiding patterns. There’s actually some really interesting maths going on, because some combinations of braid would work and some of them wouldn’t. But at the end he was saying he doesn’t really understand it, he doesn’t know what makes a good braid and what doesn’t.” This open-ended aspect of maths often isn’t apparent until university, and school often leaves people with the idea that all the maths has been done. “Kids have this idea that maths can only be right or wrong, but in fact there can be lots of exploration. And maths can be really quite creative.”

When Dominique learned that the 2019 Christmas lectures would be focused on maths, and feature veritable maths celebs Hannah Fry and Matt Parker, she jumped at the chance to be involved. “My role was Christmas lectures assistant, a kind of catch-all. On the night itself, I’m the one in the front row, on the laptop with little prompts for Hannah—and at the same time, trying to keep an eye on messages from livestreaming venues, to make sure they’re all happy.” And it was a lot of hard work. “Very high pressure and some of the team would work until two o’clock in morning. It was crazy.” One of her contributions turned out to be very prescient, in a segment designed to show how mass vaccination succeeds. “I suggested that we use surgical masks to indicate that the kids were vaccinated and that they can’t catch the virus. Now looking back on it—oh my God! Told the future!” Of course, communicating maths for TV brings with it some new challenges. “Sometimes there will be conflicting priorities between the production team and the Royal Institution. The production team want everything to look flashy. Whereas obviously the RI still want it to be interesting, but we also want to make sure the integrity of the maths is still there.” Those who watched the Christmas lectures (if you didn’t, you should hang up your maths fanatic hat right now) will recall a specific sequence which involved schoolchildren lining up and then taking a step to either the left or the right based on the result of a coin toss. “What we were trying to show was that probabilities can help you predict outcomes. So we wanted to get this lovely bellshaped curve from all the students moving about, but we didn’t get what we were hoping for. Partly chalkdustmagazine.com

s Dominique with Christmas lecturer Hannah Fry

chalkdust because it’s hard to instruct a large group of people to do exactly what you’re asking, but also simply because probability doesn’t offer any guarantees.” So where did that leave the narrative of the lecture? “The fact that the kids didn’t do what we thought is actually a really interesting point in itself. But from a TV perspective, that’s the opening demonstration and we can’t go off on a tangent. So we ended up having a montage of two different schools in the lecture.”

Widening participation As Dominique moves on to her next job working on the outreach programme at Imperial College London, she finds that the university setting has an increased focus on widening participation—so how do we convince young people from underrepresented groups to consider maths? She says an obvious start is making your event free if possible (since financial barriers often have a big impact), and ensuring diverse role models are present. “Representation does matter—look for different people from different backgrounds, from different areas, as well as different topics.” Marketing is also crucial. “You can have the best outreach in the world, but if nobody knows about it and it’s not reaching the right people, then it’s not doing anything.” But don’t think your job is over once you have them in the room. “The audience should be kept at the forefront of your mind. You need to be thinking about who your activity is for, what you want them to learn and how are you going to make sure they actually understand?” Finally, she encourages everyone to take the necessary time and effort to accommodate accessibility needs. “Putting a bit of effort in to make it as accessible as possible, whether that’s looking at the colour scheme you’re using for colour blindness, not putting too many words on a screen, or having materials available in advance or in large print. All of those accessibility things can feel like extra work but they’re only ever going to improve it. Good for everyone—not just the person you are trying to accommodate.” Having only been at Imperial for three months, she Representation does matter—look is is still reasonably new to the job, but has a lot of for different people from differpositive things to say about what she has seen. Her current project focuses on sixth formers. “I think ent backgrounds, from different the programme itself is really worthwhile, it’s very areas, as well as different topics. intense. There are online courses, with mentoring sessions in small groups throughout the year as well as large welcome and closing events on campus, although as you can imagine these have had to convert to online events in recent times.” Looking to the future, she says: “There is a move to intervene earlier in childrens’ maths education so Imperial, like many other organisations, have a growing number of outreach programmes aimed at younger students.” We wish her all the best in her latest endeavour. Ellen Jolley Ellen is a PhD student at UCL studying fluid mechanics. She specialises in the flow around droplets and ice particles.

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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, I’m not much of a football fan but I enjoyed the Euros over the sum mer. I’d like to learn a bit more about the spo rt and maybe support a club or two. Any suggestions?

— Steve Orgizovic, Unknown, possibl

y kidnapped

■ dirichlet says: I once attended an Imaginary Madrid game, although Ventry City and Coventry City are good and cogood (respectively). Of course, much of the fun is in the quality punditry: I often spend my Saturday evenings watching Gary Linear but if the scores are exponential, I switch over to ln(Shearer) on Logglebox.

Dear Dirichlet,

ry of his birth, our ned two. At the exact anniversa My baby son Norman recently tur ng on? a square into a circle. What’s goi house suddenly changed from — Joan, Norfolk

dirichlet says: I think I can explain this: I think you’ve switched from a lNorm to a 2Norm. Be warned that further expansion of your house may happen on little Norm’s future birthdays. But the shape of your house will remain bounded by the larger square made by the infinite norm, so thankfully no part of it will be going where Norman has gone before. ■

Dear Dirichlet, I’m looking after my sister’s spa niel for a fortnight. As a little sur prise, I’m thinking of renaming it. Got any good dog names?

— Peggy, Mourne Mountains

■

dirichlet says: Rhover. Dogarithm. Xorgi. Barctangent. Scooby Doodecahedron.

chalkdustmagazine.com

chalkdust

Dear Dirichlet,

TV pitches ng to come up with some reality tryi I’m y? tell of lot a tch wa Do you s (Rake Off), learner e: jousting with outdoor broom for the BBC but so far only hav ams of Heathrow c lights (Brake Off), and livestre drivers pulling away from traffi — Square eyes, Salford runway 𝟢𝟫L (Take Off). Help.

dirichlet says: How about... members of the public appear on stage and try to draw straight lines next to circles: Britain’s Got Tangent! Or celebs pair up with professionals to hold hands and make arm waves: Strictly Cos Dancing! Athletes running into an arena with Ulrika Jonsson and pointing in the direction the temperature increases the most: Gradiators! If none of that sticks, you could try generalising The Circle to n dimensions.

■

Dear Dirichlet, Thank you for your advice last issu e on helping me find actors for my play. Alas, the September cabinet reshuffle ren dered your advice useless and the wh ole project was a disaster. But no fear, I hav e a new muse. It’s a musical! It cro sse s urb an, hiphop, R&B sounds with traditional show tunes while telling a sto ry about an 18th century statesman. I’ve manag ed to arrange dates playing jus t opp osite the illustrious Dominion Theatre. How can I guarantee its success?

— Kimberly Donglesworth, Newca

stle

dirichlet says: Take your actors’ kinetic energy, T; add the play’s potential, V, and consider H = T + V. You must ensure H is conserved over time! If so, you can be sure your production is... Hamiltonian!

■

Dear Dirichlet,

r England batsman, I was playing a game of cricket against Trescothick, the forme ly on my previous who claimed he could tell how I was going to bowl based entire What’s his secret? delivery. Also the pitch was around 20 metres longer than usual.

— Rob Eastaway, Test Match Special studio

■

dirichlet says: Alas, you were playing on a Marcus chain (but not furlong!) 9

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chalkdust

chalkdustmagazine.com

chalkdust

◀ That’s a Moiré Donovan Young interferes in wave patterns

Love is like a never ending melody Poets have compared it to a symphony A symphony conducted by the lighting of the moon But our song of love is slightly out of tune... — from the song Desafinado by Antônio Carlos Jobim. English lyrics by Jon Hendricks & Jessie Cavanaugh.

was sitting in the back seat of my parents’ car, stopped at a junction. I could see two cars in front of ours and both had their indicators flashing. The amber lights were out of phase: one turned on just as the other went off, but then, over the course of a few minutes, they teamed up and were flashing in unison. Then, a few minutes later, they returned to being out of phase. That memory has always stayed with me. I was a young child, and it would be years before I studied physics and learned about wave oscillations and beat frequency—the technical term for the phenomenon I had witnessed. The flashing frequencies of the cars’ indicator lights were slightly different, causing them to slowly drift in and out of phase with one another. The beat frequency is the frequency with which this phase oscillates, and as we’ll see below, it is equal to the difference between the two flashing frequencies. My second encounter with beat frequency came when I first picked up the guitar as a teenager and learned how to keep it in tune. The method I learned is familiar to any player. By fretting one string so that it plays the note which the next string plays unfretted and playing the two together, the second string can be tuned. When the second string is slightly out of tune, one can hear a ‘wah-wah-wah-wah’—a periodic variation in sound intensity. One then adjusts the tension in the string until the wah-wahs slow down to a standstill; the string is then in tune. I had no idea at the time that this tuning method had anything to do with my childhood memory of flashing indicator lights, but the frequency of the wah-wahs is the beat frequency: the difference between the vibrational frequencies of the two strings. 11

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chalkdust In order to visualise what’s happening, a basic trigonometric identity is very useful: sin(𝐴 + 𝐵) + sin(𝐴 − 𝐵) ≡ 𝟤 sin 𝐴 cos 𝐵. We can represent the sound wave from the first fretted string as sin 𝜔𝟣 𝑡 and that from the second, open string as sin 𝜔𝟤 𝑡 , where 𝜔𝟤 is (hopefully) nearly equal to 𝜔𝟣 . What arrives at your ear is a sum of these two waves, which looks like:

and is represented mathematically as sin 𝜔𝟣 𝑡 + sin 𝜔𝟤 𝑡 = 𝟤 cos (

𝜔𝟣 − 𝜔 𝟤 𝜔 + 𝜔𝟤 𝑡) sin ( 𝟣 𝑡) . 𝟤 𝟤

Here we have used the identity with 𝐴 = (𝜔𝟣 + 𝜔𝟤 )/𝟤 and 𝐵 = (𝜔𝟣 − 𝜔𝟤 )/𝟤. The cosine function oscillates at the beat frequency and defines the amplitude of the sine function, which oscillates at the mean frequency. If you’re looking for a little intuition as to why this happens, think of it this way. Waves can interfere either constructively, when they are in phase and so work to augment each other, or destructively, when they are out of phase, and as one is decreasing as the other is increasing; in this case they cancel each other out. When the frequency of the waves is slightly different, they slowly drift from being in phase, to out of phase, and back again, just like the indicator lights.

When the shapes hit your eye, but there’s something awry... My next encounter with beat frequency came later in life, long after having studied at university, where I learned some important facts about the mathematics of periodic functions. I began to notice visual beat frequencies all around me. I was driving along the motorway and noticed an interesting pattern in the gantry supporting the overhead signs. If you look closely at the black region just below the signs, a remarkable pattern is visible. The most striking feature for me was the symmetry—the light blobs were arranged in a honeycomb pattern. What was going on? chalkdustmagazine.com

David Dixon, CC BY SA 2.0

chalkdust The black structure is in fact two identical perforated metal sheets, let’s call them screens, one on each side of the scaffolding which runs across the bottom of the signs. The perforations are much too small to see at a distance, but they are small circular holes arranged—you guessed it—in a honeycomb pattern. What we are seeing is an interference effect between the perforation pattern of the front and back screens. The back screen is further away from the observer, and so appears slightly smaller. The spatial frequency of the holes is therefore slightly increased compared to the front screen. The two patterns then beat against each other. The light blobs repeat with the beat frequency—the difference of the apparent spatial frequencies of the tiny invisible holes in the two screens. This is an example of what is called a Moiré pattern—an interference pattern formed by two sets of superimposed lines. A simple mock-up demonstrating the same effect is relatively easy to do yourself:

f thss sse maath ett f ustst m d lk a h C u d lk a Ch

1313 14 14 15 15

1 1

1212

6t)hs) 0 in0(1 in6t(1hs

1111

1010

prooof f ererpro att ShSahtt

0 cm 11 0 cm

Take two rulers with identical measurement scales, and place one a few centimetres above the other. The visual beat frequency phenomena is visible created by the distance tick marks. In the picture, the millimetre and 16th of an inch scales produce patters with different periods— approximately 2 cm and 1¼ in respectively as measured on the upper ruler.

When the screens perforate, and you then integrate... We can model the pattern of perforations on the screen by a periodic transmission function 𝑇 (𝑥, 𝑦). This function takes values between 𝟢 and 𝟣, representing the amount of light which the screen lets through at coordinates (𝑥, 𝑦). So, for example, at a location (𝑥𝟢 , 𝑦𝟢 ) where there is a perforation in the screen we would have 𝑇 (𝑥𝟢 , 𝑦𝟢 ) = 𝟣, whereas at another location (𝑥𝟣 , 𝑦𝟣 ) where the screen has no perforation, 𝑇 (𝑥𝟣 , 𝑦𝟣 ) = 𝟢. When one copy of the screen is placed behind the other, the transmission function of the back screen is relatively compressed due to the effect of perspective. We will take the 𝑥 -𝑦 coordinate system to refer to the image coordinates of the front screen and take the front screen’s transmission function to be 𝑇 (𝑥, 𝑦). The apparent transmission function of the back screen is then 𝑇 (𝑠𝑥, 𝑠𝑦), ie a compressed version of the same transmission function with a compression factor 𝑠 = 𝟣 + 𝜀 , where 𝜀 is very small, applied to the coordinates. The product of the two transmission functions gives the overall transmission function of the screens taken together. Since the eye cannot resolve the perforations in the screens, what is perceived is a shade of grey—an average transmission taken over the smallest resolvable scale. We will take 13

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chalkdust this scale to be ℓ , and also assume that ℓ is significantly larger than the spacing between the perforations. Let 𝐼 (𝑥, 𝑦) be the resulting brightness that we perceive at the point (𝑥, 𝑦). We then have 𝐼 (𝑥, 𝑦) = =

𝟣

ℓ𝟤 𝟣

ℓ𝟤

𝑥+ℓ /𝟤

𝑦+ℓ /𝟤

∫

𝑇 (𝑋 , 𝑌 ) 𝑇 (𝑠𝑋 , 𝑠𝑌 ) d𝑌 d𝑋

∫

𝑇 (𝑋 , 𝑌 ) 𝑇 (𝑋 + 𝜀𝑋 , 𝑌 + 𝜀𝑌 ) d𝑌 d𝑋 .

𝑥−ℓ /𝟤 𝑦−ℓ /𝟤 𝑥+ℓ /𝟤 𝑦+ℓ /𝟤 𝑥−ℓ /𝟤

𝑦−ℓ /𝟤

Now, since 𝜀 is very small, and ℓ is also small in comparison to the scale of the Moiré pattern, we can make the approximations 𝑋 + 𝜀𝑋 ≈ 𝑋 + 𝜀𝑥 and 𝑌 + 𝜀𝑌 ≈ 𝑌 + 𝜀𝑦 . This brings us to 𝐼 (𝑥, 𝑦) ≈

𝟣

ℓ𝟤

𝑥+ℓ /𝟤

∫

𝑥−ℓ /𝟤

𝑦+ℓ /𝟤

∫

𝑦−ℓ /𝟤

𝑇 (𝑋 , 𝑌 ) 𝑇 (𝑋 + 𝜀𝑥, 𝑌 + 𝜀𝑦) d𝑌 d𝑋 .

We can now use the periodicity of the transmission function to shift the integration domain to be centred around the origin, 𝐼 (𝑥, 𝑦) ≈

𝟣

ℓ𝟤

ℓ /𝟤

∫

−ℓ /𝟤

ℓ /𝟤

∫

−ℓ /𝟤

𝑇 (𝑋 , 𝑌 ) 𝑇 (𝑋 + 𝜀𝑥, 𝑌 + 𝜀𝑦) d𝑌 d𝑋 .

This is called the autocorrelation of 𝑇 (𝑥, 𝑦). An autocorrelation is an integral of a function against a shifted version of itself, as a function of that shift, and has applications in other areas of mathematics including signal processing, Fourier analysis and statistics. Notice that the autocorrelation is magnified, as both 𝑥 and 𝑦 appear dressed with a factor of 𝜀 , ie they have been stretched using a scale factor of 𝟣/𝜀 . The image below shows an example using a square mesh pattern:

A perforated screen whose transmission function takes the value 1 in the white regions and 0 in the black regions is shown on the left. The autocorrelation of the transmission function is shown on the right; this is the resulting macroscopic pattern when two such microscopic screens are superposed. chalkdustmagazine.com

chalkdust Again, the intuition follows from the flashing indicator signals, or the guitar strings. Along certain lines of sight, the perforations of the front and back screens are aligned, allowing the light to pass through both to your eye, and producing a bright spot. Along others, you see through the front screen’s perforations only to be blocked by the back screen’s material; the result is a darker spot.

From the big things to small, you’ll see this in them all... The most impressive aspect of these visual beat frequencies is that they act as a kind of microscope: the pattern of the underlying mesh, too small to see, is magnified and revealed due to interference. When one of the screens moves a little (or the observer moves their head a little), the effect is even more pronounced. A small translation of one of the screens by an amount equal to the spacing between the perforations results in a light blob turning into a dark blob—a change which really grabs one’s attention and is easy to spot. It’s very similar to the way interference is used in very large baseline interferometry, a technique in radio astronomy. Signals from many widely separated receiving dishes are combined to form a powerful radio telescope, one with the resolving power of a single dish whose diameter is equal to the largest separation between the individual dishes. In 2019 this technique was used to form the first image of a black hole. When I see these visual beat frequencies, I’m reminded that maths is not only about the big stuff—abstract ideas, abstruse proofs of theorems, cutting-edge science and technology, etc. These things are awe inspirESO/EHT Collaboration, CC BY 4.0 ing, but also sometimes intimidating for people when they start learning the subject. Maths is also in the little things, the patterns that are all around us. Mathematics is a beautiful lens to view the world through and can be a profound way to commune with nature. Keep your eyes open: mathematics is hiding in plain sight.

Donovan Young Donovan is a maths teacher who is transfixed by the science and mathematics of light and vision.

autumn 2021

THIS ISSUE…

DERIVATIVE NOTATION: . IS dy/dx BETTER THAN y′ or x?

G I B

E TH

T EN M GU AR

YES ARGUES ELLEN JOLLEY

ARGUES SOPHIE MACLEAN As mathematicians, we love dealing with fundamental truths and as truths go, ‘humans are lazy’ is about as fundamental as it gets. Mathematicians, as a subset of humans (it’s true—I looked it up) must therefore also be lazy. And we can see empirical evidence of this—I once sat staring at a problem for an hour trying to work out which method of solving it required the least writing. I’m pretty confident the phrase ‘work smart, not hard’ was invented for mathematicians.

Do you take me for an engineer, sir? The only acceptable notation for a derivative is the original notation created by Leibniz: d𝑦/d𝑥 . This is the only notation that demonstrates precisely what the derivative is: the limit of the change in 𝑦 over the change in 𝑥 as the change in 𝑥 tends to zero. I will not stand idly by as the integrity and beauty of the derivative is lost in a sea of dashes and dots. Anyone who insists on such primitive notation clearly could not have ventured far in their study of calculus: any student of multivariable calculus can see plainly that these pathetic diacritics are simply not up to the task. A simple extension to Leibniz’s notation allows me to write with ease any order of partial derivative with respect to any number of variables I choose, meanwhile the dashing-dotting hooligans are left scrambling.

So then why would anyone ever write d𝑦/d𝑥 ?! The effort taken when compared to 𝑥̇ is vast. Furthermore this will occur repeatedly throughout a paper! If you thought writing d𝑦/d𝑥 out by hand is slow, wait until you try typing it in LATEX. I should point out here that I’m not the one typesetting this argument, hence I have no qualms about repeatedly writing d𝑦/d𝑥 . d𝑦/d𝑥 , d𝑦/d𝑥 , d𝑦/𝑚𝑎𝑡ℎ𝑟𝑚{𝑑}𝑥 .

High schoolers can also reap the benefits of Leibniz’s original notation: they may initially be bemused by the concept of a fraction that is not a fraction— but as soon as they study integration by substitution, off and away they go writing d𝑢 = 𝟤𝑥 d𝑥 and so on. What exactly are we to do with 𝑢′ ? Prime-root it? And I suppose 𝑢̇ ’s reciprocal is ụ? chalkdustmagazine.com

E TH E TH

It’s also so much quicker to read 𝑥̇ than d𝑦/d𝑥 . I’m all about the marginal gains, but in this case the gains are on an astronomical scale! And then you get on to the environmental impact. Wasting paper space writing d𝑦/d𝑥 , when 𝑥̇ does exactly the same job, is frankly not justifiable. What would Greta say? 16

WHAT’S

& WHAT’S

Alan Turing

Boulton and Watt

Maths is a fickle world. Stay à la mode with our guide to the latest trends.

HOT NOT What a specimen!

More like Boulton and who?

HOT

NOT Winning an Olympic medal

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

Winning a Fields medal

HOT

NOT

Getting an A* in A-level maths

Winning the US Open

We’re very proud of Emma Raducanu for achieving this

Does anyone even remember who won it this year?

NOT

HOT

c ha l k dus t Chess puzzles Amagaz i nef ort hemat hemat i cal l ycur i ous

1 4/ /

HOT

2021

See pages 18–24

Chess TV shows No, Netflix, I don’t want to rewatch the Queen’s Gambit

NOT

ONTHECOVER:VHA T?VHERE?VENN MAKI NGACHRI STMASLECTURE+D EARDI RI CHLET PAWNS,PUZZLES& PROOFS+PRI ZECROSSNUMBER

Writing for Chalkdust two issues in a row

Writing for Chalkdust three issues in a row

See pages 37–41; and issue 13, pages 22–26

Get to work on that draft, Maddy.

HOT

NOT

More free fashion advice online at d chalkdustmagazine.com

Pictures Background: Flickr user Harmon, CC BY-SA 2.0. £50: Bank of England.

autumn 2021

Pawns, puzzles, and proofs Dimitrios Roxanas tells us why playing chess backwards is the new black (and white)

am sure you will agree that problem solving is an integral part of learning and creating new mathematics, but for many mathematicians recreational problem solving is also a favourite pastime. I am no exception to this; I have a fondness for riddles and brainteasers, an interest which later expanded to include chess problem solving. The amount one could write about chess problems could fill a book (in fact, many books have been written on the topic!). For today though, I want to introduce you to the world of one particular type of chess problem, namely retrograde chalkdustmagazine.com

chalkdust problems. I am going to start by letting you in on a secret: retrograde chess problems are essentially logic puzzles set on a chessboard. And what’s more, they can be approached in the same way as typical maths problems and puzzles. Now I’ve really got your attention, haven’t I?

Back to square one In retrograde problems one is typically given a chess position, possibly with some additional rules imposed by the composer. What do I mean by rules? They could tell you, for example, that the white king has not moved more than twice, or that pieces only moved to squares of the same colour they started. The composer then asks a question, like, what was Black’s last move? The answer needs to be inferred from the position, the additional stipulations, if any, and the rules of the game of chess. In other words, every position has to be legal, which means that it was reached by a sequence of legal moves. We have to accept the conditions imposed in the same way that we accept mathematical axioms. The term retrograde is quite appropriate because to find the answer to the posed question, you always have to investigate the history of the position. That is to say, you must determine which moves were played to reach the given position, even if the question does not explicitly ask you to. I want to emphasise that there should be no guessing in deducing the solutions; they will follow logically and deterministically from the position and the stipulation(s), and without violating the rules of the game. It is therefore no surprise that retrograde analysis is truly mathematical or that one of the champions of the genre was logician Raymond Smullyan (whose retrograde puzzles and books served as my own introduction to retrograde analysis).

Three maths problems in a trenchcoat! In the puzzles that follow, we will see examples of retrograde chess problems, and what’s more, we will approach them as typical mathematical problems. From the positions and the axioms, we will make observations, observe patterns, make claims, formulate conjectures, and prove true statements the same way you would prove a lemma or a theorem. This approach to solving retrograde problems shares a lot in common with the daily work of a researcher in mathematics! Of course, this article is merely an introduction to this exciting area, and there are many interesting ways to see true mathematical thinking. But still, even in this short introduction we will be able to use problem solving approaches, such as exploiting invariants, and use counting and contradiction. I will assume that you are familiar with how chess pieces move, and that you understand the basic rules of the game. Don’t be discouraged if you are not a ‘strong’ chess player: to solve retrograde problems one does not need to know chess well enough to win a game—the only thing that matters is the legality of the moves up to the current position.

Notation: written in black and white Before we move to the problems, a quick word about notation. In chess, there is a standardised way of recording the moves that take place on the chessboard. This is called algebraic notation. The name may sound scary but it’s quite simple really: the chessboard is an 𝟪 × 𝟪 grid, so, very 19

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chalkdust much like a matrix, each square can be uniquely identified by two coordinates. The 𝑥 -coordinate takes values from the letter a to the letter h (from left to right), and the 𝑦 -coordinate ranges from 1 to 8 (from bottom to top). White’s 16 pieces are originally arranged in rows 1 and 2 (called the 1st and 2nd rank), while Black’s occupy rows 7 and 8 (called the 7th and 8th rank). For example, below, the white queen is resting on c3 and the black king on b6. The naming convention for the pieces is:

rZ0ZRZrZ 7 ZpZ0Z0Z0 6 0j0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0ZKZ0Z 3 Z0L0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0

K king K, Q queen Q, B bishop B, R rook R, N knight N,

and p pawns are simply denoted by their position (no letter). To avoid explaining every time whether a piece is white or black, the first move recorded is always White’s and the next one is Black’s. We don’t record the square from which the piece moved unless there is ambiguity. For example, the next move recorded in the game on the left is 1… Rg×e8+

We must specify that it is the rook on g8 that moves to e8, and not the one on a8 which was also an admissible move. The + symbol denotes that the white king is placed in check with this move, and the ellipsis (…) is used to say that we didn’t care about White’s previous move and the first move of interest was Black’s. Finally the × symbol signifies that the rook captures the white piece currently residing at e8.

Hardworking pawns deserve a promotion Now that we know how to read a recorded sequence of chess moves, we are ready to discuss and enjoy retrograde problems! To make the most of these problems, I recommend that you cover the explanations and solutions and first try to solve each problem on your own. Our first problem asks that you look at the position on the right, and determine whether a promotion has taken place. For those not already familiar with the promotion rule, it states that when a pawn reaches the 8th rank, it is replaced by the player’s choice of piece, called a promoted piece. You can’t promote to a pawn or a king (and it has to be a piece of your own colour), but other than that, anything goes! chalkdustmagazine.com

0Z0Z0Z0A Z0Z0Zko0 6 0Z0Z0Z0o 5 Z0Z0ZKZ0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0Z0Z0Z0 8 7

s Problem 1: Has there been a promotion?

chalkdust What’s immediately striking in this problem is the rather awkward placement of the white Bh8. The white bishop has somehow managed to trap itself behind the black g7 pawn. Notice that the g7 pawn has never moved, which makes the position of the bishop a real conundrum. Shouldn’t the bishop pass through g7 to get to h8? Well, there is another possibility here, the bishop on h8 must be a promoted piece! How did that happen? We can’t tell exactly when or in what order, but a white pawn must have reached g6, captured a black piece on h7 and promoted to a bishop. The captured piece could have been a queen, a rook, a bishop, or a knight, but it couldn’t have been a pawn. (I’ll leave it to you to work out why!) Remember: in retrograde problems, the moves don’t have to be good, they just have to be legal. I am reiterating this now because it’s unlikely that promoting a pawn to a trapped bishop was the best move from the viewpoint of the chess player, but it matters not for the chess detective!

Let’s get this parity started In the next problem, we are presented with an almost full board, except the white rook on h1 is missing. We are told that it is now Black’s turn to move. The problem setter here asks us to find a move that Black must have played in this game. While we are not able to completely reconstruct the game prior to this point, we can still deduce (with proof!) a move that was played by Black, by arguing in a logical manner.

TR Dawson The Chess Amateur, Feb 1927

rmblkans opopopop 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 POPOPOPO 1 SNAQJBM0 8 7

s Problem 2: Black to move. Indicate a move Black must have played.

We immediately notice that since every pawn is still sitting at its original square, the only pieces that could have moved in this scenario are the knights and the rooks. What’s more, the rooks can only have moved when their adjacent knight was not on its original square. This is true for both sides. Evidently the white rook was captured by a knight on either h1 or g1. OK, so the rook was captured by a black knight on either g1 or h1 which means that a black knight must have reached one of the squares f3/g3/h3, captured the rook and gone back to its original square. We are narrowing possibilities down but we haven’t yet found a specific move that must have been played.

How to make progress? We certainly don’t want to try to list all possible paths that each of the black knights took, and even if we did, we still couldn’t prove which one of them actually appeared on the board. Nevertheless, looking at possible knight itineraries will prove to be useful. 21

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chalkdust A preliminary observation is that if a knight has returned to its original square after a number of moves by tracing its original path backwards, it must have completed an even number of moves. Right? It jumped 𝑘 times to go from 𝐴 to 𝐵, so it needs to jump 𝑘 times to return from 𝐵 to 𝐴 if it stays on the same path. In total 𝟤𝑘 moves. Hmm, that’s interesting. Parity (whether a number is odd or even), and invariants (things that never change) are, in general, extremely useful problem solving tools. This is not an invariant just yet, but perhaps we can show that a knight always needs an even number of moves to return to its original square. We’ve seen that this is true in the case when the knight retraces its steps, but what if it decides to come back through a different path? For this we will exploit another invariant:

When a knight jumps, it always lands on a square of a different colour.

(This is what I would call a lemma!) Let’s prove this! The knight moves two steps either vertically or horizontally, which by the checkerboard pattern on the chessboard means that before it takes its last step (right/left or up/down) it was on a square with the same colour as the original. Hence it lands on a square of a different colour. QED! In view of this observation, a light-squared knight will visit dark squares on every odd move and light squares on even ones. A similar result follows for a dark-squared one. The final destination is the original square. Crucially, this means that we start and end on the same colour, so the knight took an even number of moves to return. Let’s summarise our findings:

A knight always needs an even number of moves to return to its original square.

That is a universal truth, proved by rules of logic within the accepted set of axioms. If that’s not a theorem, I don’t know what it is. OK! We got ourselves an invariant! How can we use it? Both of the black knights are now resting on their original squares. By our theorem above they must have made an even number of moves each, so an even number of moves for Black in total. Every black move follows a white move, and since White was the last to move (remember, they tell us it’s Black’s move) White has moved an odd number of times! We’ve already established that White’s moves were completed by its knight(s) and the rook on h1. The white knights have moved an even number of times, which means that the rook made an odd number of moves. This implies that the rook was captured on g1! Parity again: this rook can only travel between g1 and h1, it’s on h1 on every even move it makes and on g1 on odd ones. So a black knight captured the rook on g1. chalkdustmagazine.com

chalkdust Almost there! We have now narrowed the possibilities to just two moves, one of which must have been played: ...Nh3×g1 or ...Nf3×g1. If the black knight were to ever get to the f3 square, it would have put the white king in check. Only a capture can make the check go away because clearly the king can’t move. The black knight is obviously not captured, ergo, it never made it to f3. Therefore the move Nh3×g1 must have been played by Black. Defeated. Next!

Whose turn is it anyway? The question for our final problem is one you would expect from mainstream chess puzzles. Mate in one. A quick look at the diagram on the right explains why this is a retrograde problem: White can mate in one with N×f7, but Black can also mate with N×c2. Which one is it? In other words, this is a creative way of asking us to determine which side is to move next. We have acquired enough experience to suspect that this is going to be another counting problem.

J-L Turco Diagrammes 60, 01/1983

rZbjnarM opopopop 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0Z0Z0Z 3 Z0Z0Z0Z0 2 POPOPOPO 1 mRANJBZR 8 7

We can immediately see that no pawn or bishop a b c d e f g h has moved in this game. It’s therefore not hard to deduce that the queens were also captured on s Problem 3: Mate in one. their original squares. So the only pieces that have moved are the knights, at least one rook from either side, and at least one of the kings (the black one has certainly moved). What about the rooks? Each side has made an odd number of moves with their rooks: odd for the wRb1 and the bRg8, and even (possibly 0) for the wRh1 and the bRa8. The black king has made an odd number of moves while his white counterpart has made an even number of moves: this is because the kings here can only move between the d and e squares of their corresponding rows. Let’s summarise our observations so far: excluding knight moves, White has made an odd number of moves (rooks: odd, king: even), while Black has moved an even number of times (king: odd, rook: odd). Let’s now take a closer look at the knights. If the wNd1 originally started at b1, then by the same arguments we used in the previous problem, it must have made an even number of moves, since d1 and b1 are the same colour. OK, but if that was the case, then the wNh8 must have started on g1 and, by the same argument, it has made an even number of moves as well. The other possibility is that the wNd1 started off at g1, in which case it has made an odd number of moves. The wNh8 must also have made an odd number of moves, for the same reason. No matter what, the white knights have in total made an even number of moves. Arguing in exactly the same way we can prove that the black knights, combined, have made an even number of jumps. Put it all together: White has made odd + even = odd number of moves, while Black has made 23

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chalkdust even + even = even number of moves. Since a white move always precedes a black move, and the total number of moves made by both sides is always even after the last black move, it is Black’s move to play and deliver mate in 1 with N×c2! Solved and done.

From rookie to king Hopefully this introduction has whetted your appetite for solving retrograde problems. A good place to start are R Smullyan’s books. Before you check out more problems, I suggest looking up the rather special move en passant and the rules of castling as they are frequently employed by composers and give rise to some fascinating puzzles. Perhaps you will be also interested in proof games, a special type of retrograde problem where a final position is given and the goal is to reconstruct the game from the starting position (in a specified number of moves). Before you go, I want to gift your two final problems to try your hand at. I promise that both are correct and solvable, but they are definitely on the challenging side! No solutions this time but I shall leave you with the words of Sherlock Holmes: “once you eliminate the impossible, whatever remains, no matter how improbable, must be the truth.” W Keym Die Schwalbe, 1979

R Smullyan Manchester Guardian, 1957

0Z0Z0Z0Z 7 Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 Z0Z0Z0Z0 4 0Z0O0Z0Z 3 OPj0Z0Z0 2 NM0Z0Z0Z 1 Z0J0Z0Z0

0Z0Z0Z0Z 7 Z0Z0Z0Z0 6 0Z0Z0Z0Z 5 ZrZbZ0Z0 4 BZ0Z0Z0Z 3 Z0Z0Z0Z0 2 0Z0Z0Z0Z 1 Z0ZkZ0Z0

s Problem 4: What was the last move?

s Problem 5: The white king is invisible. Where is he?

Dimitrios Roxanas Dimitrios is a mathematician at the University of Sheffield, where he teaches courses in analysis and financial mathematics. His mathematical interests are in analysis (pure and applied) and inverse problems. When he is not teaching or researching maths, he enjoys recreational maths and problem solving. Ah, and chess problems.

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Kimi Chen

n 1926, codebreakers in Room 40 of the Admiralty Old Building, Whitehall—the codebreaking centre of Britain at that time—were monitoring German communications after the first world war, when they started receiving baffling encrypted messages. This was the first time the British encountered messages enciphered by an electromechanical encryption machine first invented in 1918 by the German electrical engineer Arthur Scherbius. He originally marketed them for commercial purposes, such as encrypting communications between banks, but without much success. After some modifications and rebranding, in 1926 (shortly before his death in 1929) he finally found an interested client—the German navy. He called his machine Enigma. The Enigma cipher would of course become one of the most infamous ciphers in the history of cryptography. We all know stories about Enigma and the people who made a great combined effort to break it during the second world war. But this is an Enigma story you might not have heard. This is the story of how the original Enigma was actually deciphered in Poland in 1932— before the outbreak of the second world war, and long before the celebrated work of Alan Turing, his Bombe, and Bletchley Park. chalkdustmagazine.com

chalkdust

The Enigma machine An Enigma machine in the late 1920s consisted of a keyboard, a lightboard, and several scrambling elements: three rotors, a reflector, and a plugboard. A rotor looks like a cog with 26 electrical contacts on each side, so that each contact is labelled by a letter of the alphabet (or on some models a number 01–26). Each rotor contains 26 internal wires connecting electrical contacts on one side to those on the other side—connecting each letter to a different letter, essentially performing a permutation of the alphabet. On the plugboard, Enigma operators choose six pairs of letters to be connected by cables to switch around electric current, while the reflector can be thought of as a second plugboard with a fixed pairing-up of all 26 letters. Before an Enigma operator starts encrypting a message, they need to configure the machine by choosing some initial settings. First they choose the six pairs connected on the plugboard. Then they decide in which of the six possible orders they put the three rotors. And finally they rotate the rotors manually to some specific positions—there is a small window above the rotors so that exactly one letter label can be seen on each rotor. For example, the rotor position may be ABC, meaning the left rotor has A showing through its window, the middle rotor has B, and the right rotor has C (or equivalently on some models: 01-02-03).

Plug board: Bob Lord, CC BY-SA 3.0

s An early Enigma machine: three rotors

beneath a cover at the top, each with a window to see their position. Below is the lightboard then keyboard, and finally a top-down view of the plugboard. Below the main image is a face-on view of the plugboard. The reflector is internal, and a detailed view of a rotor can be seen on page 31.

A schematic of how an Enigma machine works is shown below. Once it is set up, every time the Enigma operator presses a key on the keyboard, an electric current is sent from that key to the corresponding socket on the plugboard. It may be switched to a different socket, or may be left unchanged, depending on the plugboard settings. Next, the current flows through each of the rotors from right to left, each one changing the letter as it goes. The current then enters the reflector and one more switch is made before being bounced back. It then travels again through the wirings of rotors and plugboard in reverse order by a different route, finally reaching the lightboard where it lights up a letter, indicating the letter which enciphers the original input. What about decrypting a received message? Decryption can be done easily since the Enigma cipher is symmetric, meaning that under the same machine settings, when you type in the ciphertext, the machine outputs the plaintext. 27

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chalkdust A

B keyboard

F lightboard

Reflector

Three rotors

Input from

Output to

Plugboard

s A simplified diagram of the wiring of an Enigma machine for an alphabet with just six letters, and

two plugboard cables: A–E and B–C. In this configuration a B is enciphered via B ↦ C ↦ A ↦ B ↦ A ↦ D ↦ C ↦ D ↦ F ↦ F to F. Symmetrically, an F is enciphered as a B.

Quintillions of combinations Scherbius’ design offered great security for two reasons. The three rotors, as you may suspect, are able to rotate. The right rotor rotates by one letter each time the machine performs an encryption. The middle rotor rotates by one letter each time the right rotor finishes a whole rotation (ie every 26 letters), and similarly the left rotor rotates by one letter each time the middle rotor finishes a whole rotation (every 𝟤𝟨𝟤 = 𝟨𝟩𝟨 letters). When the rotors rotate, so do their internal wirings, so the same letter will be encrypted to a different letter each time its key is pressed. This protects the encryption from methods like frequency analysis, which defeat substitution ciphers. Secondly, the number of possible initial settings was far too high for brute force attack. There were 𝟥! = 𝟨 different s Arthur Scherbius (1878–1929) rotor orders, and 𝟤𝟨𝟥 = 17,576 different initial rotor positions. Also, the number of plugboard connections was f ( 𝟤𝟨 𝟣𝟤 )×𝟣𝟣!! = 100,391,791,500. This gives a total of 10,586,916,760,000,000 different initial settings— a 17-digit number! Scherbius had hoped that his cipher machines would be lapped up by banks and other civilian organisations, but they did not appreciate the machines’ cryptographic strength and could not see the need for them. Finally, in 1925, the German navy saw Enigma’s value, and Scherbius’ company f Since first you must choose 12 of the 26 letters to be paired up; then the first letter has 11 choices for its pair, the next letter has 9 remaining choices for its pair, and so on, giving 𝟣𝟣!! = 𝟣𝟣 × 𝟫 × ⋯ × 𝟥 × 𝟣 ways to pair up the 12 letters.

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chalkdust started mass-producing them. They were then used in military and government organisations from 1926 onwards.

An encounter with Enigma Despite hearing about this mysterious new cipher, French and British leaders were unconcerned and did not attempt to break it, having just won the first world war. Through espionage, the French obtained two important documents about the Enigma: the German procedures for using the machines and the basic structure of the machines (ie the scrambling elements and how they worked together), but crucially the documents did not include the internal wirings. The French codebreakers found it difficult to deduce the internal wirings inside the three rotors, and did not attempt to build a replica Enigma machine. Instead, they handed the documents to the Polish government. Poland, however, feared German invasion at any time, which provided a powerful motivation to break the cipher. If they could read the German military’s communications, they would be in a much better position. The Polish cipher office invited twenty mathematicians from the University of Poznań in western Poland to a cryptography course. They selected the three most gifted— Marian Rejewski, Henryk Zygalski and Jerzy Różycki—and recruited them into the cipher office. Because the university was located in territory which had been ruled by Germany until 1918, all three codebreakers were fluent in German.

s From left to right: Henryk Zygalski, Jerzy Różycki, and Marian Rejewski (1932). To help break the cipher, they first built replicas of the Enigma machines. Using information from the documents recovered by French spies, along with many intercepted German messages, Rejewski was able to do what the French could not—calculate the internal wirings of the three rotors.

The Polish breakthrough The documents obtained by French spies detailed the procedure used by the German military for sending messages. The Enigma operators had a booklet that dictated the plugboard connections, rotor orders as well as the day rotor setting, each of which were different every day. They set their 29

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chflkdust machines up according to the booklet, chose a message rotor setting of their own (say CHA) and typed in the message rotor setting of their choice twice, CHACHA. This is called an indicator. They then rotate the rotors to their chosen message setting, typed in the actual message, and sent it—in particular the day rotor setting specified by the booklet was only used for the first six letters of each message. Messages were then received and read in a similar way. Knowing that repetition always leads to patterns, Rejewski exploited the fact that the first six letters of each message were two versions of the same triplet of letters. For example, if a message was intercepted beginning with the six letters AWESLD, then A and S were related because they were the encryptions of the same letter. Thousands of messages were intercepted every day which allowed Rejewski to construct a table like this to show how the letters were related: 1st letters: A B C D E F G H I J K L M N O P Q R S T U V W X Y Z 4th letters: S A B Q W E T Y U I O J P D F G R H K L Z X C V M N This can be thought of as a permutation of the 26 letters, and can be written using cycle notation: (A S K O F E W C B)(D Q R H Y M P G T L J I U Z N)(V X),

where the first cycle (for example) means that A is sent to S which is sent to K and so on until B is sent back to A. The three Polish codebreakers would then record (𝟥; 𝟤, 𝟫, 𝟣𝟧), meaning 3 cycles of lengths 2, 9 and 15. Different day rotor settings would lead to different numbers and lengths of cycles.

Understanding Enigma with permutation groups For expert readers who know a little of the theory of permutation groups, this viewpoint goes a long way to help understanding how Enigma machines work. Each scrambling element gives a permutation of the alphabet, so, writing 𝜇 for the permutation corresponding to the wiring in the reflector, 𝜌𝟣 for the combined effect of the rotors in their initial settings, and 𝜋 for the the plugboard, an Enigma machine applies the permutation 𝜋 −𝟣 𝜌𝟣−𝟣 𝜇𝜌𝟣 𝜋 = (𝜌𝟣 𝜋)−𝟣 𝜇(𝜌𝟣 𝜋).

Notice this is a conjugate of 𝜇 . Since 𝜇 is a product of 13 disjoint transpositions (see the wiring diagram above) it is (a) a derangement—no letter is unpermuted, and (b) of order 2. Any conjugate of 𝜇 also has these properties which is why an Enigma machine (a) never enciphers a letter to itself, and (b) is a symmetric cipher. If 𝜌𝟦 is the permutation for the rotors after they have clicked forward three positions, the permutation computed from the 1st and 4th letters of the the indicators is (𝜋 −𝟣 𝜌𝟦−𝟣 𝜇𝜌𝟦 𝜋)(𝜋 −𝟣 𝜌𝟣−𝟣 𝜇𝜌𝟣 𝜋)−𝟣 = 𝜋 −𝟣 (𝜌𝟦−𝟣 𝜇𝜌𝟦 𝜌𝟣−𝟣 𝜇 −𝟣 𝜌𝟣 )𝜋.

This is conjugate to a permutation which does not contain the plugboard permutation 𝜋 , and so its cycle decomposition does not depend on the plugboard settings. chflkdustmagazine.com

chalkdust One great advantage of the cycles was that if the plugboard cables were moved, but all the rotor settings left the same, the cycle decomposition data would be the same. This allowed Rejewski, Zygalski and Różycki to consider the effects of the rotors alone, without the plugboard. We have seen already that the total number of possible rotor settings, including both rotor order and rotor position, was 𝟥! × 𝟤𝟨𝟥 = 105,456—much less than the scary 10,586,916,760,000,000 that we had before! Next the Polish codebreakers catalogued all 105,456 settings with the patterns they produced (the number of cycles and their lengths). It took them about a year to complete the catalogue by hand, which included the cycles formed from the first and fourth letters, as well as those formed from the second and fifth, third and sixth letters. After that, they could find out the day rotor setting by simply checking the catalogues—there were about 2,400 different cycle patterns, so on average each pattern corresponded to 𝟦𝟦 different settings, a manageable number. Rejewski had already used a method similar to this when he deduced the internal wirings of the rotors. But what about the plugboard settings? This was easy to work out once they had the correct rotor settings: remove all cables from the plugboard first, then type in some intercepted messages. Something which almost makes sense would be produced, for example MUTEN GORMEN. It’s not hard to realise that if every G and M in that phrase is switched around, it would become GUTEN MORGEN, the German for ‘good morning’. So they would deduce that G and M were connected on the plugboard. In this way, the other connections could also be found. Now the Polish codebreakers could find out the rotor order, rotor positions, and plugboard connections on any given day, and so could read German messages easily.

The Bomba In 1938, the Enigma’s operating procedures were changed: all operators still used the same rotor order, and plugboard settings on a given day. However, instead of setting the machines to a day rotor setting specified in a cipher book, the Enigma operators choose their own day rotor setting first and sent it unencrypted at the beginning of a message. s Enigma rotor with an exploded view of its components, inThey then rotated the rotors to cluding the alphabet (or number) ring which could rotate reltheir chosen day setting, typed in ative to the wiring and then be locked in place. their chosen message rotor setting twice, and then rotated the rotors to the message setting, and typed in the message and sent it. The catalogues of cycle lengths relied on all operators using the same day rotor settings to encrypt the message rotor settings, so this change made all the cryptographer’s hard work redundant. 31

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chalkdust You may wonder what the big problem was—if the day rotor settings were sent unencrypted and the plugboard settings were easy to deduce once the rotor settings were known, hadn’t the German military shot themselves in the foot? Sadly not, because each rotor was actually made of two components. A core which contained the internal wirings; and an alphabet (or number) ring containing all the letters of the alphabet which labelled the electrical connections. This ring could rotate around the core essentially allowing each rotor to have 26 different wirings. All operators used the same so-called ring settings, that is the relative positions of rotor cores and alphabet rings were the same for all the Enigma machines on a given day. But this meant that knowing that the first rotor was in position F, say, didn’t tell Polish codebreakers anything about the position of the wiring in the core of that rotor.

s Part of a key sheet used to configure Enigma machines each day, used later than 1938 when the

German military made their procedures even more complicated. Each row from left to right gives the date, rotor order (three rotors chosen from five), ring settings, plugboard settings (ten pairs), and four day rotor settings.

Now the aim was to try to deduce the ring settings. To do this they exploited a rare phenomenon. Remember that the six-letter indicator, sent at the start of each message indicating the message rotor settings, was the same three letters encrypted twice using the chosen day rotor settings. Normally an Enigma machine will not encrypt a letter to the same letter twice in a row because the first rotor will have moved forward only a few times. This is clear from the description of how Enigma encrypts letters given above. In particular this applies to the first and fourth, second and fifth, or third and sixth letters of the indicator. But very occasionally this does happen, and an encrypted indicator like AERAPO will turn up. How is this possible? The only way is if at some point the first rotor caused the second rotor to tick over in between the first and fourth letters, and the wirings just happened to produce this pattern. Because this is so unlikely, when it does happen, it puts a lot of restrictions on the possible ring settings for the day. To be able to deduce the ring settings the codebreakers needed three of these rare indicators, for example AERAPO, WUFTUG, LICMBC. Now they just needed to find a rotor order and ring settings capable of producing this result. This was impractical to do by hand, so Rejewski invented an electromechanical machine called the Bomba. A Bomba consisted of six replica Enigma machines without their plugboards—two for each indicator—which could quickly cycle through all possible ring settings. Each Bomba machine checks one possible rotor order, so they used six Bomba machines running simultaneously to try all combinations as efficiently as possible. chalkdustmagazine.com

chalkdust Once they had found ring settings capable of producing each of the unusual indicators for that day, the codebreakers narrowed down the list by checking which settings were capable of producing all three indicators simultaneously. Then all they needed was to plug in ciphertext and use the same technique as before to deduce the plugboard settings.

Enigma strikes back Things changed at the beginning of 1939, however, when two more rotors were added. Instead of having three rotors, Enigma operators could choose any three out of the five rotors available, which gave them 60 possible rotor orders instead of six. In addition, ten cables were used to connect letters on the plugboard, rather than the six used previously, which made it a lot harder to use the old method to determine the connections even if the correct rotor settings were known.f f Clearly, more Bomba machines would be needed to test so many more rotor orders efficiently, however, the Polish did not have the resources to do this. As Germany withdrew from its non-aggression treaty with Poland in April 1939, it seemed inevitable that Poland would soon be invaded. Instead of destroying the Bomba machines and other secrets about Enigma decryption, the Polish cipher office arranged a meeting with French and British codebreakers in July, sending Enigma replicas and Bomba blueprints over to France and Britain, in hopes that the allies could build on their work.

Pnapora, CC BY-SA 3.0

s Statue in Poznań, Poland,

commemorating the work of Rejewski, Zygalski, and Różycki.

Famously, Alan Turing and many others at Bletchley Park would go on to use this information throughout the second world war to break each successively stronger version of Enigma employed by the German military. Turing’s design for a new electromechanical decryption machine was heavily influenced by the Polish Bomba and was called the Bombe in its honour. Although the contribution by Polish mathematicians remained unknown for decades after the war, their work was declassified towards the end of last century and people started to recognise their great achievements. In front of the Imperial Castle in Poznań, a bronze prism monument was dedicated to them, with the three faces bearing their names—Marian Rejewski, Henryk Zygalski, and Jerzy Różycki. Kimi Chen Kimi is a year 13 student who is interested in cryptology and abstract algebra, and would love to study maths at university. Apart from maths, she enjoys playing lacrosse and table tennis, but her favourite hobby is solving sudokus while holding a 20-minute plank. f f Later models even featured eight choices of rotor, up to three choices of reflector wiring, middle rotors which moved

forward twice during each full rotation of the right rotor, and extra scrambling elements. Moreover, each branch of the German government and other Axis powers used different models, setups, and procedures.

autumn 2021

HANDBOOK O

Tilly Pitt writes During my university interview I’m asked to do some graph drawing. Panic begins to set in and my mind goes blank as I can’t remember what an exponential graph looks like. But I have a sneaky trick up my sleeves—well, at the ends of my sleeves. 34

One of my maths teachers told us if you hold your right hand with the palm facing up you can see three lines quite clearly: 𝑦 = e𝑥

𝑦=𝑥

𝑦 = ln 𝑥

Admittedly, trying to subtly turn over your sweaty nervous right hand and examine your palm is harder than you may think. However, I think it’s a valuable gamble since I’ve found a plethora of other maths on hand. 35

chalkdust

chalkdustmagazine.com

chalkdust

◀ On the cover Madeleine Hall takes a brief dive into the world’s favourite set-relationship-representation diagram

ore often than I care to admit, I find myself sitting in the audience of a maths lecture or seminar completely and utterly lost as to what the speaker is going on about. What are they talking about? How does this relate to stuff I know about? Where does this fit within the sphere of mathematics as a whole? In fact, most of the time I am lost beyond the first slide of a presentation. In an endeavour to minimise the possibility that audience members would experience this feeling that I know all too well, I recently introduced myself at the start of a talk with this slide: my PhD

i d cs i am

fl dy n u

gen ic s

low Reynolds number fluid dynamics mathematics

stuff I learned in undergrad

stuff I learned in MSc

biology

behavioural genomics

You could be forgiven for remarking “My, what a beautiful Venn diagram you have there!” Indeed, I too was under the impression that what I had created was in fact a Venn diagram. You almost certainly will have come across Venn diagrams before. Though in case you have not, allow me to briefly introduce them. Venn diagrams are widely-used visual representation tools 37

autumn 2021

chalkdust that show logical relationships between sets. Popularised by John Venn in the 1880s, they illustrate simple relations between sets, and are nowadays used across a multitude of scenarios and contexts. And no one can be blamed for their wide usage—they truly are things of beauty. But who was John Venn? And how did he come to create such an iconic tool that’s used so broadly today? John Venn (1834–1923) was an English mathematician, logician and philosopher during the Victorian era. In 1866, he published The Logic of Chance, a groundbreaking book which advocated the frequency theory of probability—the theory that probability should be determined by how often something is forecast to occur, as opposed to ‘educated’ assumptions. The Victorian era in general saw significant shifts in the way that science and experimental measurements were thought about. It was at this point in history that science began to shift towards a new paradigm of statistical models, rather than exact descriptions of reality. Previously, scientists had believed that mathematical formulas could be used to describe reality exactly. But nowadays, we talk about probabilities and distributions of values, and not about certainties. We now interpret individual experimental measurements knowing that, no matter how precise or controlled an experiment is, some degree of randomness always exists. Using statistical models of distributions is what enables us to describe the mathematical nature of that randomness.

s Venn’s preface: “By the

general body... [probability is] regarded with indifference or suspicion.”

But I digress—back to Venn diagrams. When Venn actually first created his namesake diagrams, he referred to them as Eulerian circles, after everyone’s favourite Swiss mathematician Leonard Euler, who created similar diagrams in the 1700s. And this is where, I regret to inform you, the thing is—and I’m very sorry to be the one to tell you—that Venn diagram up there? Not actually a Venn diagram. By definition, in a Venn diagram, the curves of all the sets shown must overlap in every possible way, showing all possible relations between the sets, regardless of whether or not the intersection of the sets is empty: ∅. Venn diagrams are actually a special case of a larger group of visual representations of sets: Euler diagrams. Euler diagrams are like Venn diagrams, except they do not necessarily show all relations. When thinking about Venn diagrams, we normally picture something like my beautiful introductory slide above, right? Namely, there are circles, and they overlap. The interior of each of the circles represents all of the elements of that set, while the exterior represents things that are not in that set. For example, in a two-set Venn diagram, one circle may represent the group of all Chalkdust readers, while the other circle may represent the set of tea drinkers. The overlapping region, or intersection, would then represent the set of all Chalkdust readers that drink tea (a verifiably non-empty set). It is common for the size of the circles to represent relative size of the set that circle is representing (eg one’s undergraduate maths education being much (much) smaller compared with the sphere of mathematics as a whole). chalkdustmagazine.com

chalkdust We also commonly see nested Venn diagrams, where one set is completely situated within another set (again, one’s undergraduate maths education being entirely nested in the realm of mathematics as a whole [though this is debatable—I’ll save this for the next Chalkdust article]). But in a traditional, true-to-definition Venn diagram, every single possible combination of intersections of the sets must be visually displayed...

Chalkdust readers

Chalkdust readers cats

cats

tea drinkers

s Euler diagram

s Venn diagram

(While we cannot verify whether or not there exist cats that are also Chalkdust readers and/or tea drinkers, the Venn diagram insists we show all possible intersections.) It is actually mathematically impossible to draw a Venn diagram exclusively with circles for more than three sets. If we add a fourth set below, no matter how you move the four circles around, you can never find a region that isolates only the intersection of diagonally opposite sets—cats and tea drinkers (or Chalkdust readers and accordion players). Formal mathematical proof is LeF t aS An eXe Rc IsE fO r T hE rE aD eR .

Chalkdust readers

C h alk d u st r

ers

tea drinkers

accordion players

e ad

tea drinkers

accordion players

cats

s Actually a Venn diagram (Venn’s

s Not a Venn diagram

construction for four sets)

As you can see then on the right, for high numbers of sets (‘high numbers’ = greater than three), unfortunately some loss of symmetry in the diagrams is unavoidable. John Venn experimented 39

autumn 2021

chalkdust with ellipses for the four-set case in an attempt to cling onto some diagrammatic elegance and symmetry. He also devised Venn’s constructions which gave a construction for Venn diagrams for any number of sets. These constructions started with the three-set circular Venn diagram and added arc-like shapes which weaved between each of the previous sets to create every possible logical intersection. These constructions quickly become quite dizzying (see the loopiness of Venn’s construction for six sets):

s Four sets

s Five sets

s Six sets

So what can we do if we don’t want to weave back and forth so dizzyingly for higher numbers of sets? Enter: Edwards–Venn diagrams. Anthony William Fairbank Edwards (1935–) constructed a series of Venn diagrams for higher numbers of sets. He did this by segmenting the surface of a sphere. For example—as you can see on the right—three sets can be easily represented by taking three hemispheres of the sphere at right angles (𝑥 = 𝟢, 𝑦 = 𝟢 and 𝑧 = 𝟢). He then added a fourth set to the representation, by taking a curve similar to the seam on a tennis ball, which winds up and down around the equator, and so on. The resulting sets can then be projected back to a plane, to give cogwheel diagrams, with increasing numbers of teeth for more and more sets represented:

s Four sets chalkdustmagazine.com

chalkdust

s Five sets

s Six sets

So finally, to conclude this article, I present to you my new and improved introductory (actuallya-Venn) diagram:

low Reynolds number fluid dynamics

stuff I learned in undergrad

biology

stuff I learned in MSc genetics

behavioural genomics fluid dynamics

my PhD

mathematics

While I appreciate the poetic implication that the realms of my respective research groups (fluid dynamics and behavioural genomics) weave in and out of my PhD project, which is nested neatly in the centre, in reality the fact is that many many of these intersections are quite empty. Please consider this article in support of my upcoming petition to make it mandatory for academics to introduce themselves at the start of every talk with a Venn (or Euler) diagram. Madeleine Hall Madeleine is a pHd StUdEnT in mathematics and behavioural genomics at Imperial College London. She likes open water swimming, toast, the Oxford comma, and tHiS mEmE. She has still found none of her optimised strokes of any use in the Serpentine.

a @maddygracehall 41

autumn 2021

PUZZLES

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

1 2 4

3 5

6 7 8

Cylindrical crossnumber by Humbug This grid below can be cut out and wrapped into a cylinder so that the blue circles and blue lines match up. Although many of the clues have multiple answers, there is only one solution to the completed crossnumber. As usual, no numbers begin with 0.

Across

Down

4A − 6D − 𝟣.

(3) 1

2D × 6D × 7A.

(7)

(2D − 𝟣) × ((6D + 𝟫) × 6D − 𝟫).

(7) 2

𝟤 × 6D.

(3)

6D + 8D − 𝟤𝟢𝟢.

(3) 3

2A + 𝟥 × 6D × 6D.

(7)

1A + 6D + 𝟣.

(3) 5

1A − 𝟥.

(3)

6D × 6D × 1A + 6D.

(7) 6

7A ÷ 𝟦.

(3)

𝟤 × 2D.

(3) 8

𝟥 × 2D ÷ 𝟤.

(3)

Extra letters The words on the right are anagrams of words connected by a common theme with an extra letter added. When you write the themed words in the boxes to the left, and the extra letters in the extra letters column, two more words with the same theme will appear in the orange boxes. Extra letters

CAVES RAIN ANIME EDAM GIN RAVE GAME DAME O chalkdustmagazine.com

chalkdust 1

Spiral crossword The answers to the clues should be entered in the spiral in a clockwise direction. The last letter of each answer overlaps with the first letter of the following answer.

8 7

14 10 15

Once the grid is completed, the names of two shapes will be revealed along the diagonals in the directions marked by the arrows

11 4

12 5

1 100 and 100.

(8)

2 6.

(3)

3 12.

(3)

10 Hammer or Escher.

(3)

11 hypotenuse ÷ opposite.

(5)

5 We hope there are none of these (5) in Chalkdust.

12 A path in a graph whose first and last vertices are the same.

(5)

(6)

13 Counter

(4)

(3)

𝟨

eger or

egration.

crossword.

group or

algebra.

8 𝜂.

9 A simple, irrefutable truth.

(5) (1,1)

ple.

14 𝟣𝟢 . 𝟤

(3)

15 π𝑟 or

(4) 𝟣 (𝑎 𝟤

+ 𝑏)ℎ, for example.

Arrange the digits Put the numbers 1 to 9 (using each number exactly once) in the boxes so that the sums are correct.

= 𝟤𝟤

+ −

The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, 𝟦 + 𝟥 × 𝟤 is 14, not 10.

−

=𝟪

× ×

÷ = 16

(4)

= 8

=𝟪 = 48

Add operators Add an operator (+, −, ×, or ÷) into the gaps in each equation below to make them correct. The usual order of operations (× and ÷ before + and −) should be followed.

𝟥

𝟥=𝟥

𝟪 43

𝟣𝟢

𝟫

𝟨

𝟧=𝟨 autumn 2021

chalkdust

An odd card trick Michael Wendl dissects some variants of the magic separation, a self-working card trick

artin Gardner—one of history’s most prolific maths popularisers—frequently examined the connection between mathematics and magic, commonly looking at tricks using standard playing cards. He often discussed ‘self-working’ illusions that function in a strictly mechanical way, without any reliance on sleight of hand, card counting, pre-arrangement, marking, or key-carding of the deck. One of the more interesting specimens in this genre is a matching trick called the magic separation. This trick can be performed with 20 cards. Ten of the cards are turned face-up, with the deck then shuffled thoroughly by both the performer and, importantly, the spectator. The performer then deals 10 cards to the spectator and keeps the remainder for herself. This can be done blindfolded to preclude tracking or counting. Not knowing the distribution of cards, our performer announces she will rearrange her own cards ‘magically’ so that the number of face-ups she holds matches the number of face-ups the spectator has. When cards are displayed, the counts do indeed match. She easily repeats the feat for hecklers who claim luck. chalkdustmagazine.com

chalkdust

How the illusion works The magic separation was first devised by Bob Hummer in his pamphlet Half-a-Dozen Hummers back in 1940, and since has been the subject of discussion in books by senior conjurors such as Bill Simon in Mathematical Magic from 1964, Karl Fulves in Bob Hummer’s Collected Secrets in 1980, and by the master of mathematical puzzles Martin Gardner in his aptly named book The Scientific American Book of Mathematical Puzzles and Diversions from 1959. The unusual step of allowing the deck to be out of the performer’s control makes the trick particularly baffling, but this is simply a diversion to further obscure its deterministic workings. What might these workings be? Gardner explained the trick in terms of a conservation of mass property, borrowing from the classic riddle of equal containers of water and wine. Namely, if a litre of wine is transferred to the water container, which is then stirred, and a litre of the mixture is subsequently transferred back to the wine container, is there more wine in the water than water in the wine? Intuition often suggests there is, since the first transfer was pure wine, while the second, reverse transfer was a water and wine mixture. This intuitive, but incorrect supposition is what renders the magic separation trick so likewise remarkable. In fact, there is as much water in wine as wine in water: since the starting and ending volumes in the containers are equal, whatever wine is in the water container must be matched by whatever water is in the wine container. This conservation principle can be concretely demonstrated for the magic separation by starting with two separate piles, one of 10 face-up cards and the other of 10 face-down cards. Now, make a one-for-one exchange of cards between the piles any number of times you wish. You can make each selection randomly and you can even shuffle each pile after each transfer. The only condition is that transferred cards maintain their original orientations. At the conclusion, each pile still has 10 cards that will now be some mixture of face-ups and face-downs. What one necessarily finds is that the number of face-up cards in one pile (analogous to the wine) equals the number of face-down cards in the other pile (analogous to the water). The ‘magic’ part of the magic separation is that our performer simply—though with some theatrical flourish—turns her pile over to force matching numbers of cards. 45

s The water and wine riddle: a litre of wine is transferred to the water container, then a litre of mixture is transferred back. Is there more wine in the water than water in the wine?

autumn 2021

chalkdust In Hummer’s original 1940 pamphlet, he referred to this trick as “sure fire” and it is now easy to see how this is the case. Indeed, the deterministic outcome enabled by combining conservation with the ‘turn over’ manoeuvre and the elegant misdirection of spectator shuffling has made the magic separation a standard part of the close-up magic toolkit. The original trick has since expanded into a large family of variations, both stylistically and in terms of the numbers of cards used. The version above—10 cards up and 10 cards down—is popular and we shall call this a 10/10 configuration. Some other versions split a full deck evenly, while others use an uneven division of a deck, for example a 20/32 configuration. These variations go by many names within professional conjuring circles, including the match-up, the topsy turvy deck, and the gremlins.

Here come the mathematicians The fact that there are so many versions of the magic separation and that they all seem to work on the same combination of conservation and turn over manoeuvre beckons a deeper mathematical look: is there some sort of general law that governs the magic separation family? To start investigating this question, let 𝑇 be the total number of cards in the deck and 𝐹 be the number of face-up cards therein. Also, take 𝑆 as the total number of cards, face-up plus face-down, that the spectator is dealt. These are parameters that characterise a specific variation of the magic separation. However, separate from these are the variables, which, unlike parameters, are out of the performer’s control and which generally change each time the trick is performed because of shuffling. Following the Diaconis and Graham convention in their book Magical Mathematics, where an overbar indicates ‘face-up-ness’, let 𝜎 and 𝜎 represent the respective numbers of faceup and face-down cards the spectator is holding and 𝜇 and 𝜇 be the corresponding counts of the magician’s cards.

Magician’s cards

Spectator’s cards

3 2Q4♠ ♥♥♣

KA7♥ ♦♥ ♦♥ ♣♥♦ ♦ ♦♥ ♦ KING

♣ ♥ ♠♣ ♠ QUEEN

♥♠3♣♥♥ ♣♠ ♣

4 ♥♦ ♥♦♥♦♥♣

K 47A♣ ♥♦♥

4Q2

7AK

𝜎

3 2Q4♠ ♥♥♣

2 34Q♥ ♠♣♥

4 KA7♥ ♣♥♦

𝜎

𝜇

s Four classes of cards, tallied respectively by 𝜎 , 𝜎 , 𝜇 , and 𝜇 , result from two types of division: spectator versus magician and face-up versus face-down.

chalkdustmagazine.com

chalkdust The following tallies for cards held by each party are immediately implied: 𝜎 + 𝜎 = 𝑆 for the spectator and for the magician 𝜇 + 𝜇 = 𝑇 − 𝑆 . A third equation comes from noting that, although the cards are shuffled, the number of face-ups is fixed, no matter how the cards are distributed in any instance of the trick, meaning 𝜎 + 𝜇 = 𝐹 . The last, and most key ingredient is the observation that the number of face-up cards held by the spectator has to equal the number of initially facedown cards possessed by the magician because it is the conjuring ‘turn over’ manoeuvre that forces the matching effect into existence. In other words, the illusion only works if 𝜇 = 𝜎 . These ingredients are now baked into the following mathematical pie. Rearrange the face-up equation as 𝜇 = 𝐹 −𝜎 , substitute this into the equation tally for the magician’s cards, and then rearrange terms to find 𝜇 = 𝜎 + 𝑇 − 𝑆 − 𝐹 . The only way 𝜇 = 𝜎 can be satisfied is if the last three terms are self-cancelling, meaning that the mathematical condition enabling the magic separation is 𝐹 + 𝑆 = 𝑇.

For convenience, we will call this little result Hummer’s theorem.

Visualising the workings

Patient improved

Old drug

New drug

Hummer’s theorem is actually quite profound in the sense that it explains how all extant variations of the magic separation work and also shows how to immediately invoke numerous newer versions, as limited, it seems, only by the number of cards the magician can physically handle. For instance, following some quick calculations using the theorem, one could readily perform much grander versions of the trick that combine, say, several full 52-card decks.

40 21 61 6

Patient did not improve

31 37

46 52 98 s A Fisher table showing that a new drug is more effective than an old drug.

Magician

Spectator

Another interesting aspect of Hummer’s theorem t Visualising the magic separation. is that it immediately suggests how to visualise the workings of the magic separation at their most basic, mechanistic level by borrowing on 𝟤 × 𝟤 contingency tables. Such tables are ordinarily used in the context of assessing whether two categorical variFace-up 𝜎 𝜇 𝐹 ables, for instance medical treatment (old drug versus new drug) and clinical result (patient improves 𝜇 𝑇 −𝐹 Face-down 𝜎 versus does not improve), are statistically related. 𝑆 𝑇 −𝑆 𝑇 One of the most common statistical calculations executed on such tables is something called Fisher’s exact test, so it will be convenient to call these 𝟤 × 𝟤 structures Fisher tables. Now, in order to visualise the architecture of the magic separation, we use the columns for the people (spectator versus magician) and rows for the cards (face-up versus face-down). The diagram to the right shows the variables 𝜎 , 𝜎 , 𝜇 , and 𝜇 , and the parameters, 𝐹 , 𝑆 , and 𝑇 , in such a table. A 47

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chalkdust moment’s consideration should indicate that the only 𝟤 × 𝟤 tables consistent with Hummer’s magic separation are those in which the leading diagonal entries are equal. A bit more consideration suggests that all Hummercompliant tables are Fisher tables, but not all Fisher tables are Hummer tables. This is readily confirmed by example. The top two tables to the right satisfy diagonal equality, while the bottom table does not. Here, Hummer’s theorem is violated (𝟪+𝟣𝟤 ≠ 𝟤𝟦), so this instance lacks the required conservation property for the magic separation.

11 20

10 10 20

32 20 52

s A 10/10 Hummer configuration (left), a 20/32 Hummer configuration (right).

t A non-Hummer table that violates the These tables can be used to visualise the actual dyseparation condition. namics of the magic separation. Tables representing a trick possess a conservation property: the 3 5 8 margin totals are rigidly fixed. Because parame9 7 16 ters 𝐹 , 𝑆 , and 𝑇 are likewise fixed for a particular variation of the magic separation trick, we can ex12 12 24 ploit the conservation property here to visualise the 10/10 pile experiment introduced above as a series of 𝟤 × 𝟤 tables, one for each round of trades. The experiment starts with 10 face-ups for the spectator and 10 face-downs for the magician, resulting in the leftmost 𝟤 × 𝟤 below. round 0

round 1

round 2

last round

10 10

10 10 20

In the first exchange, the spectator cedes a face-up card to the magician, who in turn passes back a face-down card, resulting in the ‘round 1’ table. Importantly, while the boxed variable tallies show each person’s new holdings, the margin tallies have not changed. After shuffling both piles, the second iteration might see the same type of exchange, resulting in the table marked ‘round 2’. One could now step through many additional rounds of trades, sometimes exchanging cards of the same orientation, for which the table remains exactly the same, and sometimes exchanging cards of opposite orientation, whereby the boxed tallies are updated appropriately.

round 0

10 10 20 s A performance of magic separation, represented as a pile experiment with no card exchanges made.

Every table would display constancy of margin totals and strict satisfaction of Hummer’s theorem, no matter how the cards are distributed. One possible endpoint to this 10/10 experiment is familiar from above. chalkdustmagazine.com

chalkdust In a sense, the magic separation trick itself could actually be thought of as a special case of this process that consists of the magician’s deal, with zero rounds of card exchange.

Heightened mystery of odd-numbered decks In the original magic separation, Hummer emphasised that the deck size, 𝑇 , must be an even number. Fulves’ instruction manual repeats this condition, as do Simon’s and another of Gardner’s books on mathematical magic. But here is a question: can the magic separation work if 𝑇 is odd? The answer is not intuitively obvious. First, face-up cards must be matched one-for-one between spectator and performer, ie in pairs. Second, there are many tricks that do indeed require evennumbered decks (Diaconis and Graham discuss examples). However, since there were no parity conditions placed on 𝑇 in our derivation, we can immediately conclude that odd decks are indeed acceptable. A quick corollary to Hummer’s theorem evidently leads to a new and non-obvious extension of this 80-year-old trick! Because this aspect is somewhat more difficult to perceive, a truth table showing all possible parity combinations can be helpful: spectator

magician

case

𝐹

𝑆

𝜎

𝑇 −𝑆

𝜇

is 𝜎 + 𝜇 = 𝐹 satisfied?

even

odd

even

odd

even

odd

even

odd

even

odd

even

odd

even

odd

even

odd

even

odd

even

odd

even

odd

even

odd

even

odd

even

odd

even

odd

even

Half the cases—namely A, D, F, and G—are actually inaccessible because they violate 𝜎 + 𝜇 = 𝐹 . For instance, in case G, both people cannot hold an even number of face-ups if the total number of face-up cards in the deck, 𝐹 , is odd. For the admissible cases, B and C are the more pedestrian because the required parities are already established. Conversely, cases E and H are a little more interesting. Here, the magician’s own two subsets have different parities from one another and it is the conjuring ‘turn over’ that forces the parities of 𝜎 and 𝜇 to match because the latter now has become flipped. Hummer’s magic separation works even when there are an odd number of cards in a deck that itself starts with an odd number of face-up cards! This ‘double-odd’ configuration is surprising and does not seem to have been empirically discovered over the eight decades that the magic separation has been performed. 49

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chalkdust KC Cole remarked in her book The Universe and the Teacup that mathematics can produce a “literal expansion of consciousness” and it seems that Hummer’s theorem, namely that the spectator’s card count plus the face-up card count must equal the size of the deck, marks yet another example of this amazing and long-established phenomenon. Keep that in mind the next time you dazzle your friends and neighbours with the magic separation. Michael Wendl Michael is at Washington University in St Louis, Missouri, USA. Between his work in cancer bioinformatics and theoretical mechanics, he spends time playing the accordion.

c mwendl@wustl.edu

correction

Who is the best England manager? Following the Uefa European football championship this summer, I have an update to my issue 13 article including Gareth Southgate’s performance as England manager. In the England team’s entire history, it has only ever won 15 knockout games at major tournaments and Gareth Southgate was manager for 33.3% of those games. England have made it to the semi-final stage of a major tournament five times (excluding the 1968 Euros, which consisted of four teams, so getting to the semifinal wasn’t much of an achievement). Southgate has been in charge for 40% of those. England have reached the final of two major tournaments and Southgate is responsible for half of those. He is the best England manager ever (or at least in the top one, to paraphrase Brian Clough). In conclusion, I’d like to say... Looking back on when we first met; I cannot escape and I cannot forget, Southgate, you’re the one; you still turn me on, You can bring it home again.

(apologies to Atomic Kitten)

chalkdustmagazine.com

Paddy Moore

Кирилл Венедиктов, CC BY-SA 3.0

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

Talking Maths in Public 2021 A great opportunity to discuss public engagement and outreach, and meet up with other maths communicators to encourage them to write articles for issue 15. We’re already looking forward to TMiP 2023.

ggggg

= Ed Sheeran Sheeran’s first album received a very positive reception (the artwork was a big plus). His second was another popular product, but the third divided opinion. Now Ed is back again, with a new album which is equally as good as the last few.

Clopen Mic Night #1 A very entertaining live YouTube variety show.

ggggh

Maths Tricks to Blow Your Mind Kyle Evans Packed full of interesting and entertaining maths.

gghii

ggggg

No Time to Die 25 films in, and still the Fleming estate refuse to answer my emails about why Bond gives his code number to three significant figures! 004/5

Leopoldstadt Tom Stoppard (Wyndham Theatre) Trigger warning: contains references to death, tragedy, Riemann, Gauss.

ggggi

Taskmaster

Briefings: The Box Set

If you want to know what the relationship between a PhD student and their supervisor is like, look no further than Greg and Alex.

Jonathan Van-Tam Perfect binge watching.

Image contains public sector information licensed under the Open Government Licence v3.0.

ggggi

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Wikimedia commons user Yveltal, CC BY-SA 4.0.

A walk on the random side Tanmay Kulkarni

find subway maps and their connections fascinating. When I first saw the Tokyo subway map, I played a little mental game. I picked a random station, and followed a series of random routes, and each time I would end up at a different station. So, I thought to myself: if I was a passenger at a random station, and all I had was a list of stations and routes, could I reach any station I wanted without a map? How would I know if I could?

Subway maps and graph connectivity This problem turns out to be a basic question in graph theory: the problem of graph connectivity. A graph is collections of points called vertices connected by line segments joining them called edges. It asks, given some graph (I shall only look at finite graphs), is there an algorithm to determine whether or not it is connected (in other words, if every pair of vertices is joined by a sequence of edges). If we look back at the subway problem, the parallels between it and the graph connectivity chalkdustmagazine.com

chalkdust problem are immediately clear. At its essence, a subway map is a graph, made of stations (its vertices) and tracks between stations (its edges). There is a wide variety of algorithms to solve the graph connectivity problem: the most common being depth first search (DFS) and breadth first search (BFS). These algorithms both pick a starting vertex, and search all vertices which can be reached from that initial vertex; the difference between the two algorithms is the order in which the vertices are visited. DFS searches as far as possible along a path before backtracking and repeating along a new path, while BFS searches vertices in the order of how far they are from the starting vertex.

s On the left: depth first search; on the right: breadth first search. While doing these searches, we can simply check after each step whether we have reached every vertex, and if we have, then the graph must be connected. Of course, this seems trivial in the case of the graph above because it is so simple. But what about something slightly more complicated like the one on the right? It becomes harder to see at a glance, but it is still fairly simple. But what if we took a giant graph like a social network, with millions of people and billions of connections? At that point, only a computer could handle it! And remember, the computer does not see the graph like you or I do—all it has is a list of vertices and edges with no visual representation. That is why having algorithms like BFS and DFS is important. However, breadth first search and depth first search are both deterministic. What if we looked at something more probabilistic? Let us take a look at random walks on graphs, a theoretical model with a surprising variety of applications, and apply it to the problem of graph connectivity.

Random walks appear everywhere Random walks, as you might expect, involve ‘walking’ around a graph, starting at a vertex and randomly moving to one of its neighbours. These kinds of random walks appear all over the place. 53

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chalkdust Physics: When you take a random walk on an infinite grid-like graph, the motion you get appears jerky. If you contract the edges so the vertices get closer together, the limiting motion is similar to Brownian motion, so it can be used to model the motion of small particles suspended in water. Epidemiology: You can use random walks to model the spread of infectious diseases. If you think of each vertex as representing a community, and you have an traveller who visits the communities according to a random walk, you can model how the traveller infects others or gets infected as time goes on. Different arrangements of communities (ie different graphs) can create different infection rates and behaviours. Computer science: Consider a huge network like the world wide web. A graph of that size is unmanageable, but by taking a random walk, you can get a random set of connected vertices that you can use as a sample to estimate characteristics of the web. Economics: Burton G Malkiel is a proponent of the financial theory called the random walk hypothesis, which states that the stock market and its prices follow a random walk and are inherently unpredictable. However, many other economists have refuted this theory. Art: Quantum Cloud is a piece by British sculptor Antony Gormley, on display in London. It was designed using a random walk extending outward from points on a figure of Gormley’s body, and stands at 30 metres high. Chris McKenna, CC BY-SA 3.0

If random walks are so omnipresent in our world, then clearly they deserve a deeper look. Let us start by defining random walks on graphs more formally.

s Quantum Cloud by Anthony Gormley, outside the O𝟤 in London.

Random walks and connectivity A random walk on a graph follows four simple steps: 1) select a starting vertex; 2) choose one of this vertex’s neighbours at random; 3) step to that neighbouring vertex; 4) repeat steps (2) and (3) for as long as you want to. These random walks are actually a form of Markov chain (for more on these, check out Chalkdust issue 12), and the probability of moving from vertex 𝑣 to any particular neighbouring vertex, is one over the number of neighbours 𝑣 has.

chalkdustmagazine.com

chalkdust

s Converting a normal graph into the Markov chain which defines our random walk. All green arrows have transition probability 𝟣/𝟥, and all purple arrows have probability 𝟣/𝟤. There are a few properties of these Markov chains that we want to understand. First, given a pair of vertices, how many random steps do you need to take to get from one vertex to an other? Secondly, given a starting vertex, how many steps do you need to take to reach every other vertex? Because the walk is random, the exact answers to these questions change each time we repeat the random walk, so we can only look at the average number of steps (which is more accurately called the expected number of steps). We define the expected hitting time 𝐻 (𝑖 → 𝑗) to be the expected number of steps it takes to hit vertex 𝑗 for the first time starting from vertex 𝑖. We also define the expected cover time 𝐶(𝑖) to be the expected number of steps starting from vertex 𝑖 to reach every other vertex. But what do these definitions have to do with the connectivity of a graph? Let’s say we take a random walk across a graph. If the graph is disconnected, the walk will never reach some vertices and we can say that the expected cover time is infinite. Conversely, if the graph is connected, the expected cover time will be finite, and any random walk will eventually reach every vertex (with probability 1). Understanding expected hitting times will help us to bound the expected cover time for a graph.

Expected cover time and spanning trees The problem is, there are actually two reasons why our random walk might not have reached a vertex: either it could have, but didn’t; or it was not able to in the first place. The ‘not able’ case is the one we are looking for, because then the graph is disconnected. So, we need to remove the ‘could but didn’t’ case (or at least minimise the likelihood of it happening). This is where the expected cover time comes in. If we find an upper bound for expected cover time, and use that as the number of steps in our random walk, it decreases the likelihood of the ‘could but didn’t’ case happening. So how do we find an upper bound for the expected cover time? The full proof is complicated, so let us consider an example. When we work with cover time in this section, we will assume the graph is connected since the cover time is infinite when the graph is disconnected. Any connected graph has a minimal spanning tree, which is a smallest subgraph containing every vertex, and no cycles. It is called a tree because it can branch, but like botanical trees, it never contains a loop. An important fact about minimal spanning trees is that they always have 𝑉 − 𝟣 55

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chalkdust edges, where 𝑉 is the number of vertices in the original graph. We can see this for some spanning trees of our example graph: 𝟤

𝟣

𝟦

𝟣

𝟦

𝟥

𝟤

𝟥

s The subgraphs made up of the green edges are examples of spanning trees of this graph. We will just work with the first spanning tree. There is a (deterministic) walk restricted to this spanning tree which visits every vertex and travels across each of the green edges of this tree twice (once in each direction). We will call the vertices visited in this walk, in order, 𝑖𝟢 , 𝑖𝟣 , … , 𝑖𝟤𝑉 −𝟤 and the first and last vertices are the same, (ie 𝑖𝟢 = 𝑖𝟤𝑉 −𝟤 ). In the example on the left above the sequence could be 1, 2, 1, 3, 1, 4, 1. In a 1979 paper titled Random Walks, Universal Traversal Sequences, and the Complexity of Maze Problems, the authors point out that the expected time (number of steps) it takes for our random walk to cover the graph starting from 𝑖𝟢 cannot be greater than the expected time it takes for the walk to visit each 𝑖𝟢 , 𝑖𝟣 , … , 𝑖𝟤𝑉 −𝟤 in order (although it may visit other vertices in between as it is a random walk), since by definition this list already contains every vertex. We can express the time it takes for the graph to cover 𝑖𝟢 , 𝑖𝟣 , … , 𝑖𝟤𝑉 −𝟤 as the expected time for the walk to go from 𝑖𝟢 to 𝑖𝟣 , plus the time to go from 𝑖𝟣 to 𝑖𝟤 , and so on. So, we have 𝐶(𝑖𝟢 ) ⩽ 𝐻 (𝑖𝟢 → 𝑖𝟣 ) + 𝐻 (𝑖𝟣 → 𝑖𝟤 ) + ⋯ + 𝐻 (𝑖𝟤𝑉 −𝟥 → 𝑖𝟤𝑉 −𝟤 ).

The authors of that paper were further able to show that if 𝑖 and 𝑗 are adjacent vertices, 𝐻 (𝑖 → 𝑗) + 𝐻 (𝑗 → 𝑖) ⩽ 𝟤𝐸 , where 𝐸 is the number of edges in the whole graph. We will use this fact to bound expected cover time, but first let us see why it is true. Proving this in general requires some advanced probability theory, so we will try to understand it in a simple case. So, let us look at another example graph: 𝟣

𝟤

𝟥

If the walk reaches vertex 1 at some point, it must move to vertex 2 next, so 𝐻 (𝟣 → 𝟤) = 𝟣. If the walk reaches vertex 2, there are an infinite number of paths the walk can take, only to reach vertex chalkdustmagazine.com

chalkdust 𝟣 at the very end: 𝟤, 𝟣 (one step) 𝟤, 𝟥, 𝟤, 𝟣 (three steps) 𝟤, 𝟥, 𝟤, 𝟥, 𝟤, 𝟣 (five steps) ⋮

A step from vertex 𝟤 to any another vertex has a 𝟣/𝟤 chance of occurring, whereas a step from vertex 𝟣 or 𝟥 to vertex 𝟤 happens with probability 1. Thus, the path 𝟤, 𝟣 has a 𝟣/𝟤 chance of occurring based on the random walk. This also means the path 𝟤, 𝟥, 𝟤, 𝟣 has a 𝟣/𝟦 chance of occurring, and so on. Thus, the probability of a path occurring is 𝟤−(𝑁 +𝟣)/𝟤 where 𝑁 is the length of the path. I leave it to you to check that that the sum below converges to 3: 𝟣 𝟣 𝟣 𝐻 (𝟤 → 𝟣) = (𝟣 ⋅ ) + (𝟥 ⋅ ) + (𝟧 ⋅ ) + ⋯ = 𝟥. 𝟤 𝟦 𝟪

Using this fact, we see that in this case 𝐻 (𝟣 → 𝟤) + 𝐻 (𝟤 → 𝟣) = 𝟦 = 𝟤𝐸 . We can now return to the main goal (bounding expected cover time) and the inequality we concocted above, 𝐶(𝑖𝟢 ) ⩽ 𝐻 (𝑖𝟢 → 𝑖𝟣 ) + 𝐻 (𝑖𝟣 → 𝑖𝟤 ) + ⋯ + 𝐻 (𝑖𝟤𝑉 −𝟥 → 𝑖𝟤𝑉 −𝟤 ). There are 𝟤(𝑉 − 𝟣) terms on the right side, because a minimal spanning tree has 𝑉 − 𝟣 edges and we are traversing each edge twice. We can match up the terms on the right hand side into 𝑉 − 𝟣 pairs of the form 𝐻 (𝑖 → 𝑗) + 𝐻 (𝑗 → 𝑖) where 𝑖 and 𝑗 are adjacent. We can then substitute 𝟤𝐸 for those terms, giving us a bound on the expected cover time 𝐶(𝑖𝟢 ) ⩽ 𝟤𝐸(𝑉 − 𝟣).

A possible algorithm Now that we understand the expected cover time of connected graphs, we can apply this generally to all graphs (connected or disconnected). We begin by taking a random walk across the graph for 𝟦𝐸(𝑉 − 𝟣) steps and ticking off each vertex as we reach it. The number of steps may seem odd as we found the upper bound to be 𝟤𝐸(𝑉 − 𝟣), but by using 𝟦𝐸(𝑉 − 𝟣), we can ensure there is a less than 𝟣/𝟤 chance of failing to search the graph fully. But even then, less than 𝟣/𝟤 is not that great. What we can do, however, is run the random walk 𝟣𝟢 times. Each time, we tick off each vertex we reach. If every vertex is accounted for, we return true (the graph is connected). Otherwise, we return false (the graph is disconnected). If we generate a true in any of the 𝟣𝟢 iterations, we stop the loop immediately and return a final value of true. Otherwise, we let the loop keep running. If the loop finishes returning false each time, we return a final value of false. By doing this, we give the walk several chances to try and search the graph properly, which should increase the accuracy of the algorithm. However, it would be nice to know how accurate the algorithm actually is. It turns out that we actually need two values to understand the algorithm: the first tells us how accurate the algorithm is when it returns true (sometimes called the sensitivity), and the second 57

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chalkdust which tells us how accurate the algorithm is when it returns false (sometimes called the specificity). We can calculate these values using Bayes’ theorem (see Chalkdust issue 13), because the accuracies of our algorithm are 𝑃(graph is connected | true) and 𝑃(graph is disconnected | false). We define 𝑝 to be the proportion of input graphs which are actually connected. We also define 𝜆 = 𝑃(true | the graph is connected). Because there is a less than 𝟣/𝟤 chance the algorithm does not search the graph properly during one random walk, the probability of it searching it properly at least one of the 𝟣𝟢 times we run the random walk should be at least 𝟣 − (𝟣/𝟤)𝟣𝟢 . The first calculation is to figure out the accuracy of the ‘algorithm returns true’ case (sensitivity), but this is clearly just 𝟣 because of how the algorithm works. The ‘algorithm returns false’ case (specificity) requires Bayes’ theorem. However, the actual calculation is fairly simple: 𝑃(disconnected | false) =

𝟣−𝑝 𝑃(false | disconnected)𝑃(disconnected) . = 𝟣 − 𝑝𝜆 𝑃(false)

This is because 𝑃(false) = 𝑃(false | disconnected)𝑃(disconnected) + 𝑃(false | connected)𝑃(connected) = 𝟣(𝟣 − 𝑝) + (𝟣 − 𝜆)𝑝 = 𝟣 − 𝑝𝜆. 𝟣 𝟢.𝟪

Accuracy

When we graph the values we calculated in (3) and (4) against 𝑝 , we get the picture on the right. For the most part, it seems like our algorithm is very accurate. The sharp drop off in specificity as 𝑝 approaches 1 is to be expected, since if a graph is very likely to be connected, then we should be wary about trusting an output of false from our algorithm.

𝟢.𝟨

specificity sensitivity

𝟢.𝟦 𝟢.𝟤 𝟢

But how good is our algorithm? So far, we have taken a good look at the world of random walks and graph connectivity, and designed a connectivity algorithm using random walks which is actually quite accurate! It seems like we are done, right? Not quite—we still have one more question to answer. We already know there are deterministic algorithms like breadth first search and depth first search which can also be used to solve graph connectivity. So, how good is our algorithm relative to those?

𝟢 𝟢.𝟤 𝟢.𝟦 𝟢.𝟨 𝟢.𝟪 𝟣 Proportion of connected input graphs, 𝑝

s We can see how the sensitivity and

specificity change as 𝑝 increases. This graph uses 𝜆 = 𝟣 − (𝟣/𝟤)𝟣𝟢 , but the basic shape of the pink graph would be the same if 𝜆 was larger.

Our metric is computer time. It turns out that BFS and DFS are actually faster than our algorithm, as the time they take to execute are approximately proportional to 𝑉 + 𝐸 , whereas our algorithm’s time is approximately proportional to 𝑉 𝐸 . There is room to improve this, as a 2007 paper (Many Random Walks Are Faster Than One) showed that using multiple random walks in parallel (rather than in series as we did) could decrease the cover time (making our algorithm faster). chalkdustmagazine.com

chalkdust We have already seen that random walks arise everywhere from epidemiology to computer science to art. Because these objects are ubiquitous, studying them becomes very important to understanding other fields. The specific problem of graph connectivity is a core question, not just in graph theory, but also in complexity theory (the study of algorithms in general). In fact, one of the authors of the Random Walks paper (which studied cover time and hitting time), László Lovász, won the prestigious Abel prize this year for his work (including this paper) on graph theory and complexity theory! So although our algorithm may not be as fast as BFS or DFS, designing it provided us an insight into the surprisingly relevant world of graph connectivity, random walks, and complexity. I wish to thank Burton Newman, who is my teacher at Squares and Cubes (squaresandcubes.com). He introduced me to this weird and interesting problem, and guided me as I explored it. Tanmay Kulkarni Tanmay is a 13-year-old living in Seattle, USA. He loves exploring different types of maths (especially ones with a healthy dash of computer science in them). He spends his free time reading comics and playing the acoustic guitar.

a, ory of Enigm st y rl a e e th ciphers. e 26 telling ur favourite rticle on pag o a f o ’s n le e p h u C co i f Kim share a In honour o we wanted to

The solitaire cipher

Belgin Seymenoğlu

My favourite cipher needs only a deck of 54 cards (including two distinguishable jokers). Number the cards 1–54 by suit ♣♦♥♠ and value ace to king with jokers at the end. The sender and receiver first agree on an order for the deck. Every letter of the message is shifted forward in the alphabet by a different amount, determined by a complex sequence of shuffles and cuts of the deck of cards.

Vigenère cipher

Matthew Scroggs

Can you decrypt my message? Every letter of the plain text is shifted down the alphabet according to a repeating sequence of distances. There is a really neat technique for breaking this cipher: in this example, the letters WGCHF are repeated. Consequently, it would seem a reasonable guess that both of these strings encode alike sequences of letters. So the length of my encoding sequence is a factor of 30. The key to decrypting the message has been left in this paragraph... VHTPKLHOKDOMJEKKNZVOWGCHFEFAMXUSTIEBJOIGYHWEGLORGDWGCHFIGIIM 59

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#14 Set by Humbug

5 7

6 8

14 17

15 18

16 20

24 26

25 27

32 35

33 37

34 38 41

The down clues are provided as normal. The across clues are provided in alphabetical order without their clue numbers or lengths. Although many of the clues have multiple answers, there is only one solution to the completed crossnumber. Solvers may wish to use the OEIS, Wikipedia, Python, a book of log tables, etc to (for example) obtain a list of cube numbers, but no programming should be necessary to solve the puzzle. As usual, no numbers begin with 0. To enter, send us the sum of the across entries by 14 March 2022 via the form on our website ( d chalkdustmagazine.com). Only one entry per person will be accepted. Winners will be notified by email and announced on our blog by 1 May 2022. One randomly-selected correct answer will win a £100 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

chalkdust

Across

Down

(see instructions)

2 A square number.

(2)

3 An anagram of 16D.

(3)

4 Less than 38D.

(2)

? 217.

(?)

? 394.

(?)

5 An even number.

(3)

? 555.

(?)

6 Two times 12D.

(3)

? 777.

(?)

7 Equal to 5D.

(3)

? 790.

(?)

9 A square number.

(3)

11 A square number.

(3)

? 935.

(?)

12 Two times 18D.

(3)

? A factor of 2D.

(?)

13 Equal to 5D.

(3)

? A multiple of 51.

(?)

14 Less than 18D.

(3)

15 A multiple of 23 whose digits are all odd.

(3)

? A power of 2.

(?)

16 One less than 17A.

(3)

? A square number.

(?)

18 Two times 14D.

(3)

? A square number.

(?)

20 A palindromic prime number.

(3)

? An anagram of 25D.

(?)

22 A multiple of 23 whose digits are all odd.

(3)

23 A multiple of 23 whose digits are all odd.

(3)

? Equal to 18D.

(?)

24 Three times 30D.

(3)

? Equal to 34D.

(?)

25 A multiple of 23 whose digits are all odd.

(3)

? Five less than 1A.

(?)

27 21A squared.

(3)

? Four less than 1A.

(?)

29 A multiple of 25 whose digits are all odd.

(3)

30 Three times 32D.

(3)

? One less than 1A.

(?)

31 One more than 3A.

(3)

? One third of 6D.

(?)

32 Less than 30D.

(3)

? Three less than 1A.

(?)

33 A multiple of 23 whose digits are all odd.

(3)

34 A cube number.

(3)

? Two less than 1A.

(?)

36 10 less than 2D.

(2)

? Two times 4D.

(?)

38 A multiple of 5.

(2)

autumn 2021

chalkdust

Hollis Williams

n their most uncomplicated form mathematical models are essentially just mathematical descriptions of real-world systems. Stringing together variables and parameters into an equation we can attempt to describe complex behaviours that change with respect to time.

Today mathematical models are used for everything, from predicting exam grades, to the Earth’s climate a hundred years from now. But mathematical models don’t need to be complicated to be useful. Even simple models have the power to reveal insights into a problem, to guide us in the decisions we make and to reveal unexpected consequences of our actions. They even stand up surprisingly well to more sophisticated models, still managing to capture subtle dynamics, with far less computational expense. I think simple mathematical models are worth celebrating, so here I’m going to discuss three simple but very important models. You might be used to thinking of physics or engineering as the typical subjects in which mathematical models are employed, so let’s turn this on its head and describe some models from sociology and biology.

Survival of the fittest A key question in sociology and economics is how the size of human populations change over time, and when we talk about something changing with respect to time we use a differential equation to describe it. The simplest model of population growth comes from 1798: the Malthusian model, d𝑃 = 𝛼𝑃, d𝑡

where 𝑃(𝑡) is the size of the population at time 𝑡 , and 𝛼 is the constant growth rate of the population. This model argues that the rate at which the population changes over time depends linearly on 𝑃 , the number of people currently able to reproduce or die. If we use the initial condition, that at our commencement point in time the population size is 𝑃(𝟢) = 𝑃𝟢 , then this differential equation can be solved to give 𝑃(𝑡) = 𝑃𝟢 e𝛼𝑡 . chalkdustmagazine.com

chalkdust

global population (bn)

This is a very simple conclusion, but is the model any good? The graph on the right shows how global human population has changed over the last 220 years. The green circles represent the real numbers: data from the United Nations, and the blue line is the mathematical model’s prediction 𝑃(𝑡), where 𝛼 = 𝟢.𝟢𝟣𝟣 and 𝑃𝟢 is chosen so 𝑃(𝟣𝟪𝟧𝟢) is 1 billion people.

𝟨 𝟦 𝟤

𝟢 𝟣𝟪𝟢𝟢 𝟣𝟪𝟧𝟢 𝟣𝟫𝟢𝟢 𝟣𝟫𝟧𝟢 𝟤𝟢𝟢𝟢 This is a fairly good match; both prediction year and data show exponential growth. But what about the future? In Factfulness, the physician and academic Hans Rosling discusses how as countries get richer, citizens tend to have fewer children and growth rate slows, so 𝛼 , and the slope of our graph, should decrease over time.

So is the model perfect? No, but one simple equation, whose solution can be solved used a pen and a single side of A4 paper does a pretty good job of describing how human populations have changed for the last few centuries.

The lynx effect Now let’s look at a slightly more complicated mathematical model, how the populations of two different animal species interact. The Canada lynx is a wild cat that lives broadly in Canada and Alaska, with a diet consisting mostly of the snowshoe hare, which is native to the same geographical region. Let’s try to model how the numbers of each animal changes over time. We use variable 𝑃 to represent the hare population, and 𝑄 for the lynx. The Lotka–Volterra model, from the early 20th century, for how the sizes of these populations change is given below: d𝑃 = 𝛼𝑃 − 𝛽𝑃𝑄, d𝑡 hares reproduce proportional to current population size, assuming plenty of food

d𝑄 = 𝛿𝑃𝑄 − 𝛾 𝑄. d𝑡

hares die proportional to interactions with lynxes

lynxes reproduce proportional to amount of food, ie tasty interactions with hares

We can adjust the positive constants, 𝛼 , 𝛽 , 𝛾 and 𝛿 , to fit real-life measurements: in modelling we call these parameters. Unlike the Malthusian model, solving this system of equations has to be done numerically. The solution is periodic, both the predator 63

in absence of food, lynxes die proportional to population size

Why shouldn’t we necessarily want 𝛽 = 𝛿 ?

autumn 2021

chalkdust and prey populations oscillate, with the predators population trailing slightly behind that of its prey. Once more, we can compare our model to reality. There is plenty of data recording lynx and hare populations, or to be specific, there is plenty of real historic data about how many lynx and hare pelts were collected by fur traders in the area. The plot below shows the comparison between the solution of the system and pelt counts from between 1900 to 1920. I’ve found the best-fitting values for the parameters and the initial conditions by using a least squares method (if you want the details, Joseph Mahaffy’s lecture notes from San Diego State University take you through it and are easily found online).

pelts (thousands)

𝟪𝟢

s Hare today... t gone tomorrow

hares lynxes

𝟨𝟢 𝟦𝟢 𝟤𝟢 𝟢

𝟣𝟫𝟢𝟢

𝟣𝟫𝟢𝟧

𝟣𝟫𝟣𝟢 year

𝟣𝟫𝟣𝟧

𝟣𝟫𝟤𝟢 Eric Kilby, CC BY-SA 2.0

It might not seem intuitive why these populations should oscillate, but let’s have a think. When there is an abundance of tasty hares, there’s more than enough food to go around and the population of the lynxes swells. But a large and greedy population rapidly depletes the hare stash, and food shortages mean a large population can’t be supported for long, and the number of lynxes falls. Not bad for a couple of differential equations: the model isn’t too hard to understand, it describes this common predator–prey situation well, and it reveals this interesting regular periodicity to how their population sizes change.

Hot topic Moving on to our final model, we turn our attention to some of the hottest maths in the news. It would be difficult to have missed talk about the ‘R number’ in news reports on the ongoing Covid-19 pandemic. This number, called 𝑅𝟢 in the mathematical literature, is the basic reproduction number. Loosely speaking, 𝑅𝟢 tells us how many people we expect one person to infect, and it is typically used to refer to the spread of a disease prior to any government interventions to reduce transmission. chalkdustmagazine.com

chalkdust 𝑅𝟢 comes from the SIR model for the spread of infectious diseases. The key variables in this model come from how we split the initially susceptible population of the country into three groups:

• 𝑆(𝑡), the number of people still susceptible to the disease, • 𝐼 (𝑡), the number of infected people, and • 𝑅(𝑡), the number of people who have recovered from the disease and developed immunity. If we make the key assumptions that the total initially susceptible population size does not change over time, 𝑆 + 𝐼 + 𝑅 = 𝑁 , and that the population is completely homogeneous, this then leads directly to a system of nonlinear ordinary differential equations, d𝑆 = −𝛽𝑆𝐼 , d𝑡

d𝐼 = 𝛽𝑆𝐼 − 𝛾 𝐼 , d𝑡

d𝑅 = 𝛾𝐼. d𝑡 people moving to recovered proportional to number of infected

people getting infected because of interactions with infected people

We have parameters for the infection rate, 𝛽 , and the recovery rate, 𝛾 . The number 𝑅𝟢 is the rate at which secondary cases are produced, multiplied by the average infectious period, 𝑅𝟢 = 𝛽𝑁 ×

𝟣 𝛽𝑁 = . 𝛾 𝛾

What happens in virus outbreak when no measures are made to contain it? Let’s look at the solution curves for 𝑆 , 𝐼 and 𝑅. It’s possible to find these numerically if we choose some parameters. Fitting the model to data from the first wave of Covid-19 in Italy −𝟣 −𝟣 suggests a good match with 𝛽𝑁 = 𝟢.𝟣𝟪𝟢 day , 𝛾 = 𝟢.𝟢𝟥𝟩 day .

This is similar to Lotka– Volterra. There, 𝛿𝑥𝑦 represents the growth of the predator population since 𝑥𝑦 represents interactions. What do you think 𝛽𝑆𝐼 represents here?

individuals

If we start with one infected person, 𝐼 = 𝟣, and having everyone initially susceptible to start, 𝑆(𝟢) = 𝑁 , applying these parameters to the UK gives us plots for 𝑆 , 𝐼 and 𝑅 below on the left.

𝟤𝟢𝟢,𝟢𝟢𝟢 𝑆 𝐼 𝑅

𝟣𝟢𝟢,𝟢𝟢𝟢 𝟢 𝟢

𝟧𝟢 𝟣𝟢𝟢 𝟣𝟧𝟢 days since first infection

𝟤𝟢𝟢

That’s a peak of over 120,000 infections, 100 days after the first infection. A pretty scary picture, and clear warning that interventions need to be made! If you’re not convinced, at the start of 2020, the UK had just 5,900 critical care beds. Making an assumption that one in ten of those infected in the first wave would need a critical care bed in hospital, then the model predicts that without intervention the NHS would be overwhelmed less than three months after patient zero contracted the disease. autumn 2021

chalkdust Luckily, on a more positive note, this model also tells us what to do. We have a few different avenues to take action. We could try to change the number of people susceptible 𝑆(𝑡), by introducing a vaccine. Or we could try to decrease the average infectious period 𝟣/𝛾 , but this is tricky if we don’t know much about the virus. Possibly the cheapest and easiest solution we could try is to tackle our parameter 𝛽 . 𝛽 represents the infection rate; for 𝑅𝟢 to go down we need 𝛽 to be smaller. That means for every person who catches the disease we need them to interact with fewer others. The solution? Well perhaps we could consider quarantines, nationwide lockdowns, mask wearing or social distancing… do you see where we’re going here? For the parameters we’ve chosen, we have that 𝑅𝟢 ≈ 𝟦.𝟫. For England, with the initial strain in a fully susceptible population, the more sophisticated models from the team at Imperial College London (which were the models used to inform government policy), found 𝑅𝟢 at the time to be between 2.5 and 3.3. But if we have more sophisticated models, why bother with the simple ones at all? Even with modern computing power, simple models are less computationally expensive. This SIR model runs instantaneously on my desktop computer with a few lines of code; the Imperial models use a huge amount of code which needs considerable time to run on a large computer cluster. Even this simple SIR model correctly predicts that 𝑅𝟢 > 𝟣 and hence that the virus spreads, because each person infects more than one other person. Furthermore, a simple model like this helps the public to understand the epidemiological risks of a new virus. The solution curves are intuitive, and the figures arising from the model, like 𝑅𝟢 , are comprehensible enough to be discussed on primetime news channels. It’s rare for mathematical terminology to seep into public discourse: it takes a very powerful, but simple model to be able do that.

A happy conclusion Simple mathematical models are a great gateway into understanding both mathematical concepts and the workings of systems the mathematics seeks to describe. Even very simple models can provide powerful insights; insights which can be gained from more complicated models but only at the expense of elegant equations and quick computations. Yet there are plenty of real-world problems where a nice simple mathematical model is still lacking. If you’ve played with sand on a beach you’ll know that a collection of grains can flow like a liquid. In this area—known as granular flows—we are still lacking simple models for such a fluid which capture the observed behaviour. Why not consider your favourite physical system and see if you can come up with a simple mathematical model to describe it? Your model might just provide you with a new wealth of insight into what’s actually happening! Hollis Williams Hollis is a PhD student at the University of Warwick who occasionally dabbles in mathematics. He is a mediocre but enthusiastic squash player.

chalkdustmagazine.com

C R Y P T I C

10 11

12 13 14

15 16 17

C R O S S W O R D

21 22

Across 1 4 7 8 10 11 14

#2, Set by Humbug

15 17

20 22 24 25 26

∫ 𝟤𝑥 d𝑥 − 𝑥 𝟤 , 𝟨 − 𝟨, 𝟨 − 𝟣𝟢, 𝟣 − O, 𝑎 ÷ ℎ. Ape rambling by grand mountain. How much is contained inside Far East? Fifty-one first, million second, six third, until eventually tending to this. Limits of nonangle and torus can be folded up to make 3D shapes. 𝛁 ⋅ (kinetic energy − 𝑦 + 𝑡) is limitless. Knight on the telephone regularly called after band. Ball concealed by cusp, he realises. Romeo, for example, coming to grips with TEX archive, initially loads editing box. Boundary of zero-free area contains trivial origin‽ A principle of astronomy leads people home again. Contentless sudoku reworded to become irrational. Head left, alternatively split in two. Carla’s confused without direction.

Down (6) (4) (4) (5) (4) (9) (6) (6) (9)

(4) (5) (4) (4) (6)

1 Disoriented duck halts magazine. (9) 2 How quickly deep end fills up. (5) 3 50 and 5 dividing 𝟥 × 𝟤.𝟩𝟣𝟪, adding 𝑛 (6) makes a prime. 4 Endless DPhil leads to a solution of (3) 𝑥 𝟤 − 𝑥 − 𝟣 = 𝟢. 5 Quadrilateral built from potassium, (4) iodine and tellurium. 6 𝑥 , for example, starts after eleven (4) seconds. 8 Function that can be split into smaller (3) pieces using axes. 9 For example, [𝟢, 𝟧) ∈ 𝑇 precedes (8) perturbed real around five? 12 Calculates the area of misshapen (8) triangle. 13 Dreamer in disarray produces rest. (9) 16 Timeless tangles might be sharp? (6) 18 Third part of ratio mnemonic initially (3) takes odd approach. 19 Is a bit of curve quality? (5) 20 Heading for Zambia Egypt result: nil-nil. (4) 21 More complex than hyperbolic function. (4) 23 Shared divisor ends much algebraic faff. (1,1,1)

autumn 2021

TOP TEN

This issue features the top ten waves. To vote on the top ten matrices for issue 15, go to d chalkdustmagazine.com

At 10, it’s Walking on Punshine by Katrina and the Waves.

At 9, it’s the new single by Huey Lewis and the News: Hip to be a Square Wave.

At 8, and spending another week jumping up and down the charts: it’s a saw tooth wave.

At 7, Gravitational Pull vs. the Desire for an Aquatic Life by Stars of the Lid.

At 6, it’s Mex(ican wave) on the Beach by T-Spoon.

At 4, it’s Say Hello Wave (Function) Goodbye by Soft Cell (yes, they have songs other than Tainted Love).

At 5, it’s Electromagnetic Wave My Fire by The Doors.

At 3, it’s Bohemian Rhapsody by that wave the Queen does.

3 At 2 this issue, it’s Walking on Sunsine by Katrina and the Waves.

Topping the pops this week, it’s an electron behaving like a wave during the two slit experiment.

Pictures Katrina and the Waves: Jonn Leffmann, CC BY 3.0; One man Mexican wave: Pat Scullion, CC BY-NC-ND 2.0; Rydberg atom wave function: Hweimer, CC BY-SA 4.0; Queen: Raph_PH, CC BY 2.0; Double slit experiment: Jo Kalliauer, CC BY-SA 4.0.

chalkdustmagazine.com

Upcoming stocking ﬁllers

◼ WLOG: The Secret to Being a Happier Mathematician Dr N Joy Fuller GCSE PhD DBE, £12.99

◼ Counting to 100 Very Quickly by Not Saying the Odd Numbers Eve Ann O’Lee, £24.68

◼ Why You Can’t Solve Every Problem with Principal Component Analysis Michael Mosley, £19.01

◼ The Secret Life of Prime Numbers Marcus du Sautoy, £13.17

◼ The Mathematics of the Loch Ness Monster Smith & Nosebleed, £9.99

◼ Quantum Lamb Mick Herron, £18.99

◼ The Life-Changing Magic of Forgetting the +c Constance Lost, £10.00

◼ The Book of All Books That Don’t Contain Themselves Russell Square, £12.12

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