Chalkdust, Issue 15 by Chalkdust

IN THIS ISSUE 3

Page 3 model

Dear Dirichlet

Surviving the bridge in Squid Game E Adrian Henle, Nick Gantzler, François-Xavier Coudert and Cory Simon run the glass gauntlet

Puzzles

On the cover: Regular tilings Poppy Azmi needs help decorating her bathroom

Which Greek letter are you?

What’s hot and what’s not

The power of curly brackets Brace yourselves! Mats Vermeeren proves Noether’s first theorem

My favourite LATEX package

The big argument Whiteboards or blackboards?

My secret Julia Schanen hates being mistaken for a genius

The hidden harmonies of Hamming distance Chris Boucher is Hamming the time of his life

Significant figures: Gladys West Madeleine Hall navigates the wild West

Accidentally mathematical songs Goran Newsum always should be someone you really love

Cryptic crossword

The crossnumber Win a £100 goody bag

Opinions: An uncomfortable truth Lucy Rycroft-Smith and Darren Macey consider the legacy of statistics

4 In conversation with Sammie Buzzard Ellen Jolley talks to the extremely cool polar scientist

57 Slaying the dragon Callum Ilkuw delves into a Dungeons and Dragons dice dilemma

62 Covering infinity Sophie Bleau is a surjective map

Top ten: Matrices 1

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Welcome aboard issue 15. This issue will call at a range of excellent and entertaining mathematical content, before terminating at the top ten matrices.

The team Thùy Dương “TD” Ðặng Madeleine Hall Ellen Jolley King Ming Lam Vivienne Leech Sophie Maclean Jonathan Pain 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.

Along the journey, we’ll meet some really great mathematicians. Sammie Buzzard (pages 4–7) takes us to the ends of the Earth. We zoom out from the polar regions to meet Gladys West (pages 48–51) and look at the global navigation satellite systems in the atmosphere. Once we reach our destination, Emmy Noether (pages 29–35) will look back on the journey and relate the conserved quantities to the symmetries of our route. Sadly, not all mathematicians of the past are as great as those we’ll meet on this journey: there are difficult but important conversations to be had about the (at best) diminishment of contributions from, and (at worst) systemic oppression of minority groups (pages 70–75). As entertainment along the way, we take our chances diving (falling) into some probability underlying games of chance and survival. We try to survive the bridge crossing challenge from the Netflix series Squid Game (pages 10– 17), and attempt to defeat dragons, demogorgons, scorpions, and any other monster you can imagine (pages 57– 60). While onboard, you can use our free wifi to watch Have I Got News For You on BBC iPlayer, where you will spot Chalkdust making an appearance. Don’t worry, we have most certainly let the fame go to our heads. Now that we’ve hit the big time, we very much look forward to our invitation from the Ritz to host our launch party at the Rivoli Bar free of charge. Maybe it got lost in the post, or maybe we should check our spam folder again…

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; everyone at Maths Fest for inviting us along to give away magazines; Hat Trick Productions for choosing to feature Chalkdust as the guest publication on an episode of Have I Got News For You; Clare Wallace for her discussions with Professor Dirichlet; everyone at Achieve Fulfilment for their help with distribution. The emoji used throughout this issue are Twemoji, licensed under CC BY 4.0. 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 2022. 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

chalkdust

William Thomson—better known as Lord Kelvin—was a British mathematician who became interested in the temperature of the Earth around the late 19th century. Global temperatures have risen since his time, but our ability to predict the future of our climate has not, despite the enormous threat global warming poses to our planet. One of these is Earth…

Much like our current ignorance of the future, the 19th century was relatively ignorant of the past: Kelvin’s interest in global temperature was for a method to determine the age of the Earth.

His idea was roughly as follows: assume the Earth is like a baked potato left out on the kitchen table. While the skin has cooled completely, the hot interior mash is just waiting to scald the mouth of an unsuspecting potato thief. Over time, the parts of the potato near the surface will cool down more quickly, so the rate of cooling is proportional to the change in temperature gradient through the potato: 𝜕𝟤 𝑇 𝜕𝑇 , = 𝜅𝑝 𝜕𝑡 𝜕𝑥 𝟤

where temperature 𝑇(𝑥, 𝑡) depends on time 𝑡 and distance to the surface 𝑥 , and 𝜅𝑝 is a constant depending upon the potato. By assuming the skin is perfectly cool: 𝑇(𝟢, 𝑡) = 𝟢,

…the other is a baked potato.

and assuming a perfectly baked potato, ie the initial temperature is uniform: 𝜕𝑇 = 𝟢 everywhere at 𝑡 = 𝟢, 𝜕𝑥 Kelvin found an approximate solution to this equation, and was able to estimate the age of the Earth to around a million years. Of course, this is out by a factor of about a million, but was more accurate than many contemporary estimates.

It is a common misconception that this model gives inaccurate estimates because Kelvin was unaware of the decay of radioactive materials throughout the Earth’s crust, but these effects are relatively minor. A more sensible correction has to do with thermal convection currents in the Earth’s mantle, but anyone who thinks either of these happen in a potato is clearly baked. References FM Richter. Kelvin and the age of the Earth. Journal of Geology 94(3) (1986), 395–401.

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IN CONVERSATION WITH

Sammie Buzzard Ellen Jolley talks to the mathematician and scientist about Attenborough, arctan, and Antarctica

aths today is at the frontier of almost any scientific problem you can envision. Though still relatively unknown to the wider public, mathematicians are acutely aware of this and take great pride in it. Having said that, polar bears, rifles and immersive theatre shows are three things I would still have told you mathematicians have absolutely nothing to do with. Well, I would have been completely wrong, as I learned from my conversation with mathematician-turned-polar scientist Sammie Buzzard.

Countering collapse with calculus Sammie, as she tells me, is a glaciologist and climate scientist. “Glaciologists study all different kinds of ice. That can range from the frozen ocean in the Arctic, to my focus of Antarctica, which is mostly ice on top of land.” In particular, she is studying Antarctica’s ice shelves, which are floating bodies of ice that form at the edge of a land mass. They have found recent fame because of their unfortunate habit of collapsing, most recently in March of this year, with the loss of the Conger ice shelf making top headlines. This sudden disintegration is precisely what Sammie hopes to learn more about with her research. “They are areas that are particularly vulnerable to changes in climate because they’re affected by changes in the ocean below and the atmosphere above. I look at the above part, so I create numerical simulations or models of the melting of this ice on the surface of these ice shelves and then work out where that water is going and where that is moving around on the surface of the ice. We think water is key in the sudden collapse of ice shelves.” Losing our ice shelves could be disastrous for mankind because of the sea level rise it causes—however, not for the reasons one might naively expect. The ice is already in the sea so, according to Archimedes’ principle, it melting does not cause any rise. “But all the ice that’s behind them is held back by ice shelves. We call that a buttressing effect, where it’s holding back the ice on the land, so you take the ice shelf away, then all the ice on the land is free to accelerate and get into ocean, and that does contribute to sea level rise.” chalkdust

Perhaps the most famous example of ice shelf collapse was Larsen B, an ice shelf twice the size of Greater London which rapidly disintegrated over the first few months of 2002. The polar science community scrambled to understand why this had happened and how it could be prevented in future. A crucial clue lying on the surface of the ice would eventually become the focus of Sammie’s research. “What we’ve noticed about Larsen B is that it was covered in lakes before it collapsed, from melting on the surface of the ice. Essentially it’s all about water kind of getting into crevasses, and the pressure of that water freezing and unfreezing and adding load to the ice each year caused that collapse. So I am trying to work out if that could happen anywhere else.” Very interesting—but, of course, as Chalkdust readers, we need the gory mathematical details. “The model is coded in Python. It’s based on a one-dimensional column model, so there are lots of vertical columns looking through the ice and we calculate the surface energy balance at the top and then solve the heat equation moving down through the ice using a finite difference method.” The heat equation is a partial differential equation (a type of differential equation with more than one independent variable) which governs the flow of heat through a space as time passes. “Each column is solved independently, but then we can move water laterally between all these different columns both on the surface and within the snow at the top of the ice shelf. Thinking about water moving laterally is also really important, because you do get these lakes and rivers that we need to simulate.” Of particular concern is the George VI ice shelf on the Antarctic peninsula. “There are some satellite images from the start of 2020 that show the ice shelf is covered in bright blue water because it was melting so much. That’s a really interesting case study, even though the structure of the ice shelf maybe means it’s not the most concerning one in terms of preventing sea level rise. Just the sheer amount of processes that are going on there mean that’s one that I wanted to focus on, especially because there are people working on the field site, so you can get the validation and calibration from that.” Speaking of the field, don’t go thinking Sammie only ever gets to sit around on her computer all day—she has been to the Arctic herself. “I went on a field work training course while I was doing my PhD. We went up to Ny-Ålesund [the northernmost civilian settlement in the world], which is essentially just a big research town, so each country that works up there has different base that you can stay in while you do your field work. We went up to the glaciers and took some measurements and put all that training into practice, which was a great experience. You don’t expect a mathematician to end up in the Arctic.” ▲ An Arctic Buzzard Her trip was not without excitement; anyone doing fieldwork in the Arctic must undergo some pretty unique training: “I never thought as a mathematician I would be learning to shoot a rifle to potentially defend myself from polar bear attack.” She is keen to emphasise that she would never use the rifle unless absolutely forced—if even then. “I feel bad for the polar bears. I remember on the training course I said maybe if a polar bear came for me I would just let it eat me because I just really don’t want to shoot it, and they said ‘OK, tomorrow you’re not carrying the rifle!’” If what you’ve read so far wasn’t surprising enough, try this: it wasn’t cold. “It was really nice! We got given so much gear but I ended up just lugging around this massive coat. I got sunburn!” What is it called when you get a tan in the Arctic? An arctan! 5

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Access for all in the Arctic Luckily, I didn’t make the arctan joke in person so the interview continued. We discussed Sammie’s outreach work, which includes giving talks at events such as Maths Inspiration, and giving interviews to magazines such as Chalkdust. She says there are two reasons why she believes outreach is very important for her field. “First of all, because we’re studying areas that are so affected by climate change, it’s important to talk to people about that and let them know what’s going on, because it’s so disconnected from what people do every day and what they see. The majority of people are never going to go to the Arctic or the Antarctic but what’s happening there is going to affect all of us.” The second reason is to improve diversity in polar science. “The British Antarctic Survey had their Ben Sparks first women go to Antarctica in the 1983, which is ▲ Sammie talking at a Maths Inspiration event insane. When I was born, women weren’t allowed to go to Antarctica from the UK. I went to a fairly standard state school and the idea of being a researcher or scientist never crossed my mind, so I wanted to let people know those opportunities are out there.” She says this can be even worse for racial minorities or minority genders. But how can we improve the situation? “We are running a workshop on race and systemic bias within Arctic science, to essentially work out what we should be doing because it feels to me that there’s a lot more awareness now but we don’t really talk very much about solutions.” One of her outreach projects particularly piqued my interest: New Atlantis. “That was kind of crazy actually. It was an immersive theatre show. The organisation New Atlantis was like a United Nations for water so it was set in the future. We were in a water crisis and the audience had to decide how we would deal with it. So were we going to rely on science, would we share the water out nicely, or were we going to have military control? And as part of that there were lots of actual scientists playing the role of future scientists. We would talk a bit about our current work but put it in the context of how things might look in 100 years or so in the future. And the audience did generally vote to leave Antarctica alone.” I wonder why that was. “I think people generally recognise that Antarctica is one of our last pristine, untouched wildernesses. Everyone loves David Attenborough, watching Frozen Planet. You look at it and think, ‘Maybe we shouldn’t touch this, and we should just let the penguins carry on as they are.’”

From cats and dogs to stats and logs Many people assume that in order to have a career as a mathematician, you must have been incredibly gifted from a young age. Maths is something you either get or don’t, and that reflects something immutable about you that can’t be changed. Sammie is here to tell you that this is not true, and in fact hard work and determination can be far more important than ‘natural gift’ (whatever that is). “I think I was quite good at school, but I wasn’t the best. My best friend in secondary school would get maths much more quickly than me: she would just scribble it down and be like ‘OK, I’m done!’ Meanwhile, I would take five pages to make mistakes and work out where I was going, but… I persevered.” chalkdust

It was her maths teacher who told her that even though her friend seemed to have the more natural talent, her perseverance meant Sammie was more likely to get to the answer. “She would get it straight away or she would give up, whereas I would think, ‘No, I am going to get to these answers and I’ll take those five pages I need.’” But even with determination like Sammie’s, it was a while before she came to consider maths as a viable career option for her, and initially she went to university to study to become a vet. “I didn’t really know what the careers were in maths, which didn’t help. You could either become a maths teacher or an accountant and I didn’t really want to do either of those things. So I never really considered it until I actually went to vet school and realised it was not the right place for me. Then I looked more into applications like the maths of climate, cryptography and mathematical biology. I restarted my degree, and didn’t look back.” I think probably a lot of readers can relate to that— personally, I always found maths fun at school, but even so I struggled to envision what it would be used for in the real world. Because financial careers didn’t appeal to me, I was turned off from it until I was older and learned you could do all kinds of things with it. Applications of maths are generally not taught in school and for a lot of children, that obscures what makes it so cool. It was something Sammie had to go digging for to find out. “I did a lot of internet searches for maths careers and read a few popular maths books. I think generally universities have got a bit better at promoting where their graduates end up, so looking through prospectuses I saw people in cool jobs.” Perhaps including more about these applications that are largely unknown to the public in school curricula could go some way to encourage more children to take up maths— and maybe non-mathematicians wouldn’t keep asking us ‘What do you actually do?!’ (Just kidding. Obviously mathematicians never speak with non-mathematicians: it’s too scary!) ▲ Is going to Antarctica to avoid speaking After choosing to study maths at the University of Exeter, to non-mathematicians a step too far? Sammie then sought a PhD at the University of Reading. “I met with the person who turned out to be my supervisor, and I was actually quite concerned at the time because I didn’t know anything about ice, and he said it didn’t matter—as long as I had the maths, he would teach me the rest of it.” If any mathematicians are now feeling tempted by a career in climate science (I certainly am), that news should be reassuring. “Obviously it took a bit of work to fill in the gaps, but there was the support there because so many people come into it: it’s not like most people have a degree in glaciology.” If there is one thing she emphasises throughout this interview, it is that climate science is for everyone. “I feel like what I do is really cool, so I want everyone to be able to do it if they want to!” 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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Moonlighting agony uncle Professor Dirichlet answers your personal problems. Want the prof’s help? Contact c deardirichlet@chalkdustmagazine.com

Dear Dirichlet,

We’re getting married in Shetlan d next month and are having a local band come to play some folk tunes at the rec eption. Doubtless the weather will be good so we’ve hired a remote, unpowered, out door barn for the occasion. For ext rav agant reasons, some of our guests will only dan ce with certain people, and not with others. But I’ve only just got the breakfast seating sorted and now this…! Can you help with this dancing dilemma?

— Martha Tracy, on a boat

■

dirichlet says: From my considerable experience, the secret

to getting people to dance with those they wouldn’t usually is appropriate lighting. But your barn has no power? For peat’s sake! Thankfully I have a group of dieselpowered engines you can borrow. That should get you Unstuck. Before the party, get some Yellow and blue lightbulbs and hang them on the walls in the shape of leftwardpointing arrows. With my set of generators, the lighting should form a directed edgecolouring and once the band start playing you have… a Ceilidh graph!

Dear Dirichlet,

tiness of a brand of kons that he can predict the min My friend has this theory: he rec s done a load of ar relationship with the price. He’ chewing gum since it fits in a line ow am I supposed data for six different brands—h the me t sen and ting tes t duc pro esis? — Chop Choo, on a train to test this preposterous hypoth ■

dirichlet says: First, rank all of your brands according to

their mintiness and then according to their price. For each brand, find the difference between the mintiness rank and the price rank, and square it. Add them all up, divide by 35 and subtract the answer from l and there you have it the Spearmint rank correlation coefficient. chalkdust

Dear Dirichlet,

, highly fashionI’m currently planning a cheap holiday for me and an elderly g worn by budconscious companion who cares deeply about the designs of clothin and as I’m sure flight, cting conne a get airline cabin crew. It seems we’ll have to take are dressed ane aeropl you can understand, it’s important that the staff on the new offer this as a filter… exactly the same way as on the first one. Skyscanner doesn’t can you help? — Clare du Nord, on a plane ■

dirichlet says: It’s very simple: just park the planes next to

each other and roll out a carpet made from an atlas. As long as everyone walks on it, you’ll be fine: every continuous map be tween compact spaces preserves uniform continuity! (Related: why not protect your luggage in an exclusive Dear Dirichlet branded suitcase? Handbuilt by badgers; available until the end of the month, see left.)

Dear Dirichlet,

Recently while swiping \ot and \to on the mathematical dating app Remaindr, I met a man who seemed to have some really interesting though ts about connecting personality types to Greek letters . We had a few drinks together, but after I sent him £10,000 to help with accommo dation costs he stopped replyin g to my texts. What can I do??

— Rho-meo seeks Juli-eta, on foot

■ dirichlet says:

Bad luck, friend: it seems you’ve been scammed! There are more and more of these bad actors around today maybe you can recoup your losses by pitching your story to Netflix as a true crime/reality documentary. A few title suggestions: Inventing Alpha. The Beta Cheater. The Gamma Scammer…

Dear Dirichlet,

the sizes of molluscs I was at the beach and I noticed something strange about small snails than clinging to the rocks. Although each rock tended to have more of evened out. sort it rock each for large ones, when I looked at the average size Why? — Shellshocked , still on the beach

■

dirichlet says: Ah yes, a simple consequence of the Central Limpet Theorem.

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chalkdust

◀ Surviving the bridge in Squid Game E Adrian Henle, Nick Gantzler, François-Xavier Coudert & Cory Simon team up for a deadly challenge

n the popular Netflix drama series Squid Game, hundreds of adults in financial despair are recruited to play a sequence of games, in competition for a large cash prize. Though the games mimic those the characters played as children, ironically, these games are deadly. The players willingly participate in these games, risking their lives for the cash prize. Throughout, Squid Game makes powerful social comments on economic desperation, inequality and immobility. If that’s insufficient to convince you to watch Squid Game, episode seven (‘VIPs’) presents an interesting survival game amenable to probabilistic analysis. Sixteen remaining players, having already beaten four deadly playground games, are now told to number themselves 1–16. Having done so, the next challenge is presented to them: the bridge survival game. How will our players fare? Inspired by this episode, we can frame this game as a stochastic process and find the survival probabilities of the players.

The bridge survival game In the bridge survival game, 𝑁 players line up with the objective to cross an elevated bridge. The players are numbered 𝟣 to 𝑁 and have to attempt the crossing of the bridge in that order. The bridge is comprised of a 𝟤 × 𝐿 grid of glass panels. Half of the glass panels are strong (tempered) and can support the weight of a player; the other half are weak (untempered) and shatter under the weight of a player. The strong glass panels are randomly distributed on the bridge, under the constraint that each column contains exactly one strong glass panel. Owing to spacing between the columns, each player must traverse the bridge column-by-column, in 𝐿 hops.

→ → →

▲ An initial condition for a bridge survival game with 𝑁 = 𝟦 players attempting to traverse a bridge with 𝐿 = 𝟪 columns of glass panels. Strong glass panels are marked with

(unseen to the players).

To the players, the glass panels are visually indistinguishable. Therefore, for each advance to an unvisited (by any player) column, the player at the front is forced to make a random but existential decision 11

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of which panel to hop onto. If the player hops onto the strong glass panel, they proceed to hop onto the next column. On the other hand, if the player hops onto the weak glass panel, the glass shatters, they take a deadly fall, and the player behind then becomes the active player to attempt traversal of the bridge. The players behind observe the outcomes of the hops of players in front of them. Thanks to perfect memory and survival instincts, if a player in front has successfully traversed a column by hopping onto a strong glass panel, the remaining players will hop onto the same sturdy panel in their attempted traversal of the bridge.

→ → →

▲ The active player, player three, is traversing the bridge and, by observing the outcomes of the hops of the eliminated two players in front of them, has taken only two survival-risking hops to arrive at their current position.

The game proceeds until either all players are eliminated, or all columns of the bridge have been traversed and a subset of the players safely cross the bridge.

→ → →

▲ Continuing with the scenario above, in this outcome (end game), only player four successfully traversed the bridge, without taking any survival-risking hops.

In the Squid Game episode, there are 𝑁 = 𝟣𝟨 players and the bridge has length 𝐿 = 𝟣𝟪. Some dramatic elements are neglected in our version but as we’re generalising the number of players and the bridge length, we think this is a fair trade. Missing here is (i) the time limit to cross the bridge (which, with added pressure, affects the players’ capacity to perfectly remember which panels are tempered), (ii) social dynamics that cause players to act irrationally (eg two players breaking one glass panel in a single manoeuvre, hence not maximising the total survival rate), and (iii) the ability of a player later on in the line to visually discriminate between weak and strong glass panels.

Probabilistic questions We raise two questions about the stochastic process in the bridge survival game: How many of the players do we expect to make it safely across the bridge? What is the probability that any given player survives? We can see straight away that the survival of a player in the back of the line is dependent on the outcomes of the players in the front of the line. Players in the back of the line are more likely to survive because they know which panels are strong in the columns hopped on by the players in front of them. chalkdust

Let’s proceed with some mathematical analysis. Given the total number of players 𝑁 and bridge length 𝐿, we define: • 𝑆 𝑖 : the event that player 𝑖 survives

• 𝜃 : the number of players that successfully traverse the bridge (a random variable).

Our mathematically refined questions are:

What is the expected value of 𝜃 , 𝔼[𝜃]?

What is the probability of event 𝑆𝑖 , 𝑃(𝑆𝑖 ), for 𝑖 ∈ {𝟣, ..., 𝑁 }?

Overview of our approach A convenient approach to answer our two questions is to first find 𝑃(𝐹𝑛 ) for 𝑛 ∈ {𝟢, … , 𝑁 }, with • 𝐹 𝑛 : the event that exactly 𝑛 players fall to their death.

Note the events {𝐹𝟢 , … , 𝐹𝑁 } comprise the sample space of outcomes and are mutually exclusive. The expected number of survivors is

𝑁

𝔼[𝜃] = ∑(𝑁 − 𝑛)𝑃(𝐹𝑛 ),

probability of exactly 𝑛 players falling

𝑛=𝟢

( ) number of survivors in that case

since 𝑁 − 𝑛 players survive in the event 𝐹𝑛 . The probability that player 𝑖 survives is

𝑖−𝟣

𝑃(𝑆𝑖 ) = ∑ 𝑃(𝐹𝑛 ), 𝑛=𝟢

( ) 𝑖−𝟣

since event 𝑆𝑖 is the union of the mutually exclusive events {𝐹𝟢 , … , 𝐹𝑖−𝟣 }, ie 𝑆𝑖 = ⋃𝑛=𝟢 𝐹𝑛 . In other words, player 𝑖 survives if it is only players in front of them who fall to their death (𝑖 − 𝟣 or fewer players). For example, this is saying the probability of player four surviving is

What else can we say?

𝑃(𝑆𝟦 ) = 𝑃(𝐹𝟢 ) + 𝑃(𝐹𝟣 ) + 𝑃(𝐹𝟤 ) + 𝑃(𝐹𝟥 ).

Each hop by an active player onto a column not yet visited by any player constitutes a trial— specifically a Bernoulli trial—with an outcome of success (survival) with probability 𝟣/𝟤 and of failure (death) with probability 𝟣/𝟤. Regardless of the outcome, each trial reveals to the players behind which panel in that column is composed of strong glass. Only after 𝐿 trials is the entire state of the bridge known, which would reveal to the remaining players—if any—the path for safe bridge traversal. Since each Bernoulli trial is independent, the probability of any particular sequence of outcomes of 𝑐 trials—that is, any particular pattern of broken glass panels on the first 𝑐 columns of the bridge—is (𝟣/𝟤)𝑐 . 13

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We split our analysis of the bridge survival game into two cases: (i) at least as many players as columns, 𝑁 ⩾ 𝐿;

(ii) more columns than players, 𝑁 < 𝐿 (as is the case in the episode!).

Case 𝑁 ⩾ 𝐿 is easier to analyse because all columns of the bridge will certainly be visited, ie 𝐿 trials will take place. For case 𝑁 < 𝐿, in contrast, all players could be eliminated before the final column of the bridge is reached, so we could have 𝑁 , 𝑁 + 𝟣, … , 𝐿 trials.

More players than columns (or equal), 𝑁 ⩾ 𝐿

Suppose the number of players is greater than or equal to the bridge length, 𝑁 ⩾ 𝐿. Then, all 𝐿 columns of the bridge will be visited. Let’s find the probability of exactly 𝑛 players falling, 𝑃(𝐹𝑛 ). If we think about the maximum possible number of deaths, we can make an observation right away:

→ → →

Consider the player 𝑛 > 𝐿. At most, 𝐿 players fall to their death. This happens only if, as above, all of the first 𝐿 players choose to hop onto a weak glass panel in their first hop onto an unvisited column. This outcome would reveal the path for safe traversal over the bridge for the remaining 𝑁 − 𝐿 players behind them. Therefore, player 𝑛 is guaranteed to survive and 𝑃(𝐹𝑛 ) = 𝟢 for 𝑛 > 𝐿.

So let’s instead think about the player 𝑛 ⩽ 𝐿, and the event that 𝑛 players fall to their death. The key is to count the number of ways to distribute the 𝑛 broken glass panels among the 𝐿 columns of the bridge: 𝐿 𝐿! . ( )= 𝑛 (𝐿 − 𝑛)!𝑛!

𝐿

Since each of these distributions of broken glass panels occurs with probability (𝟣/𝟤) , 𝑃(𝐹𝑛 ) = ( 𝐿𝑛 ) ( 𝟤𝟣 ) for 𝑛 ⩽ 𝐿. Therefore: (𝐿) (𝟣) 𝑃(𝐹𝑛 ) = { 𝑛 𝟤 𝟢

𝐿

𝟢⩽𝑛⩽𝐿

𝐿 < 𝑛 ⩽ 𝑁.

( )

Another way to arrive here is to consider a binomial distribution: of 𝟤𝐿 equally likely outcomes, ( 𝐿𝑛 ) of them result in 𝑛 players falling to their death. Two sanity checks on this result:

(1) The probability that zero players fall to their death is 𝑃(𝐹𝟢 ) = (𝟣/𝟤)𝐿 , since then the first player must choose the strong glass panel in each column for each of their 𝐿 hops to cross the bridge; by similar reasoning, 𝑃(𝐹𝐿 ) = (𝟣/𝟤)𝐿 .

(2) Since the events {𝐹𝟢 , … , 𝐹𝑁 } are mutually exclusive and their union comprises the sample space, 𝐿 𝑁 we must have ∑𝑛=𝟢 𝑃(𝐹𝑛 ) = 𝟣, which holds since ∑𝑛=𝟢 ( 𝐿𝑛 ) = 𝟤𝐿 via the binomial theorem.

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More columns than players, 𝑁 < 𝐿

Suppose, as in the episode, that the number of players is less than the bridge length, 𝑁 < 𝐿. Then, not all 𝐿 columns of the bridge are necessarily visited, particularly in the event that all 𝑁 players are eliminated. Unlike the case above, no player is certain to survive, so let’s again find the probability that exactly 𝑛 players don’t make it, 𝑃(𝐹𝑛 ). A useful observation here is the following:

If exactly 𝑛 players fall, and 𝑛 < 𝑁 , then there must be players who made it! In which case, all 𝐿 columns must be visited and we’re in the same situation as our earlier case. Thus, equation ( ) for 𝑃(𝐹𝑛 ) holds for 𝑛 < 𝑁 .

So we’re just left with finding 𝑃(𝐹𝑛 ) for 𝑛 = 𝑁 . The event 𝐹𝑁 (all players eliminated) is the union of all the possible ways player 𝑁 could have died; ie the mutually exclusive events that player 𝑁 fell through at column 𝑐 = 𝑁 , … , 𝐿 of the bridge. For each column 𝑐 at which player 𝑁 could fall through: • there are ( 𝑁𝑐−𝟣 ) ways to distribute the broken glass panels of the preceding 𝑁 − 𝟣 players onto −𝟣 the previous 𝑐 − 𝟣 columns on the bridge;

• each of these distributions of broken glass panels (along with the broken panel at column 𝑐 ) occurs 𝑐−𝟣 𝟣 (𝟤)

with probability ( 𝟣𝟤 )

𝑐

= ( 𝟤𝟣 ) .

For example, if two players attempt to cross a bridge with four columns, the possible outcomes whereby both of them die (𝑁 = 𝟤, 𝐿 = 𝟦) are shown below, arranged vertically by 𝑐 , the column player 2 fell through. Each outcome occurs with probability (𝟣/𝟤)𝑐 . 𝑐=𝟤

𝑐=𝟥

𝑐=𝟦

𝑐

To obtain 𝑃(𝐹𝑁 ), we sum ( 𝑁𝑐−𝟣 ) ( 𝟤𝟣 ) over the possible columns 𝑐 ∈ {𝑁 , … , 𝐿} at which player 𝑁 could −𝟣 fall through. Therefore, 𝐿

(𝐿) (𝟣) 𝑃(𝐹𝑛 ) = { 𝑛𝐿 𝟤 𝑐−𝟣 𝟣 𝑐 ∑𝑐=𝑁 ( 𝑁 −𝟣 ) ( 𝟤 )

𝟢⩽𝑛<𝑁 𝑛 = 𝑁.

( ) spring 2022

𝑁

Two sanity checks on this result: (1) It reduces to equation ( ) when 𝑁 = 𝐿. (2) Again, ∑𝑛=𝟢 𝑃(𝐹𝑛 ) = 𝟣, a slightly more involved calculation included in the online version of this article.

Putting it all together Equations ( ) and ( ) give the probability that exactly 𝑛 ∈ {𝟢, ..., 𝑁 } players do not survive, 𝑃(𝐹𝑛 ), under cases 𝑁 ⩾ 𝐿 and 𝑁 < 𝐿. The expected number of survivors, 𝔼[𝜃], and probability that player 𝑖 ∈ {𝟣, ..., 𝑁 } survives, 𝑃(𝑆𝑖 ), then follow from equations ( ) and ( ). In the episode, then, how many of our 16 players do we expect to survive? On the right we plot 𝔼[𝜃] as a function of the bridge length, 𝐿, for 𝑁 = 𝟣𝟨 total players. As the bridge lengthens, fewer players are expected to survive. For a bridge of length 18, as in the episode, we’d expect seven survivors.

expected no of survivors, 𝔼(𝜃)

𝑁 = 𝟣𝟨

12 8

prob of survival, 𝑃(𝑆𝑖 )

What about the actual survival rate of each con4 testant? Below we show 𝑃(𝑆𝑖 ) under three different scenarios: (a) 𝑁 > 𝐿, (b) 𝑁 = 𝐿, (c) 𝑁 < 𝐿. 0 We also show 𝑁 = 𝟣𝟨 < 𝟣𝟪 = 𝐿 from the episode 𝟢 𝟤𝟢 𝟦𝟢 in (d). We see the players in the back of the line bridge length, 𝐿 have an advantage since they observe the outcomes of the players in front of them. For our episode scenario in (d), we observe our seven expected survivors are the ones with survival probability 𝑃(𝑆𝑖 ) > 𝟢.𝟧. 𝟣

(a) 𝑁 = 𝟣𝟨, 𝐿 = 𝟪

𝟣

𝟢.𝟩𝟧

𝟢.𝟤𝟧

prob of survival, 𝑃(𝑆𝑖 )

𝟢.𝟧

𝟢

𝟣

𝟢.𝟧

1 2 3 4 5 6 7 8 9 10 12 14 16 player, 𝑖 (c) 𝑁 = 𝟣𝟨, 𝐿 = 𝟥𝟤

𝟢

𝟣

𝟢.𝟩𝟧

𝟢.𝟤𝟧

𝟢.𝟧

chalkdust

𝟢

(b) 𝑁 = 𝟣𝟨, 𝐿 = 𝟣𝟨

1 2 3 4 5 6 7 8 9 10 12 14 16 player, 𝑖 (d) 𝑁 = 𝟣𝟨, 𝐿 = 𝟣𝟪

𝟢.𝟧

1 2 3 4 5 6 7 8 9 10 12 14 16 player, 𝑖 16

𝟢

1 2 3 4 5 6 7 8 9 10 12 14 16 player, 𝑖

But could it have all gone wrong for the organisers? Could all 𝑁 = 𝟣𝟨 players have been eliminated? For a bridge of length 𝐿 = 𝟣𝟪 we have that 𝟣𝟪

𝑐−𝟣 𝟣 𝑐 ) ( ) = 𝟢.𝟢𝟢𝟢𝟨𝟧𝟨, 𝟣𝟨 − 𝟣 𝟤 𝑐=𝟣𝟨

𝑃(𝐹𝟣𝟨 ) = ∑ (

so there was only a 0.07% chance that all players would have died in this round. Good news for the VIPs—99.93% chance of at least one potential Squid Game victor!

Extensions There are many possible extensions to the framework that we’ve presented here, which could make for interesting applied probability problems. For instance, inspired by one of the players in Squid Game who worked at a glass factory, incorporating an ability for some player(s) to discern between strong and weak glass panels would impact the probabilities in the Bernoulli trial hops for said players 𝑖 ∈ {𝟣, … , 𝑁 }. We could also account for imperfect memory of the players of which glass panels of a previously successfully traversed column is the strong one. Maybe this could be done by adding some stochasticity into the framework! Another option, to make the game even harder, could be to add more rows to the bridge (eg three instead of two) and/or vary the distributions of weak versus strong glass panels.

And finally, like in Squid Game, we could include a time limit for the players to cross the bridge and impose a distribution on the time per hop. We suspect this extension would show an optimal position in the line: the first player is unlikely to survive without the advantage of seeing the outcomes of other players in front of them; the last player is unlikely to survive owing to the time limit. The extra-rows extension may be an idea for an enhanced bridge survival game for season 2 of Squid Game! To help verify our formulas for 𝑃(𝑆𝑖 ) and 𝔼[𝜃], we wrote a Julia code that runs a Monte Carlo simulation of the bridge survival game. See d github.com/SimonEnsemble/SquidGame-BridgeGame E Adrian Henle Adrian is a doctoral student in chemical engineering at Oregon State University. He researches the application of novel machine learning techniques in predicting properties of nanoporous materials.

a @eahenle s adrian-henle Nick Gantzler Nick is a PhD student in physics at Oregon State University researching nanoporous materials and their applications. When off the clock, he likes to hike local trails and do hobby science projects. François-Xavier Coudert François-Xavier is a senior researcher at CNRS in France, and professor at PSL University in Paris. He teaches statistical physics, thermodynamics, and modelling of chemical systems.

d coudert.name a @fxcoudert Cory Simon Cory is an assistant professor of chemical engineering at Oregon State University. He enjoys applied mathematics and walks in the forest with his two Samoyeds.

d simonensemble.github.io a @CoryMSimon 17

spring 2022

PUZZLES

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

Mini crossnumber by Humbug As usual, no numbers begin with 0. Across

Down

1 2D ÷ 𝟤.

(3)

5 2D × 𝟤.

(3)

7 3A ÷ 𝟥.

(3)

3 1D × 𝟤.

(3)

6 3A × 𝟤.

(3)

1 3A ÷ 𝟤.

(3)

3 3D × 𝟣.

(3)

2 1A × 𝟤.

(3)

4 2D ÷ 𝟣.

(3)

Anagrams Each of the following sets of four phrases are all anagrams of connected words and phrases. Unscramble them all and work out what the theme of each group is. Group 1 INTEGRAL POLAR LEG ALARM GOOD DANCE GENTLE ARC

Group 2 I LAND LIQUOR TENT HIRE OWT HUNT SODA

Group 3 BEER SLUR MENU ALL FOUR RESUME DIET RYE UTENSIL LEISURE CAR ARCHITECT

Group 4 CARROT PORT POSSE SCAM ACRE PRINT GAP BENCH LIP

Group 5 I DON TUNIC ACCORDION TINT OXEN HIATUS FLAXSEEDS ARE THEREFORE CERTAIN chalkdust

Eight directions Each of the 36 words given below should be written in the grid in one of eight directions: left, right, up, down, or the four diagonals. The locations of the first letters of all 36 words are provided. In the completed grid, each square will contain a letter. Many squares form part of more than one word. R

A Q

MATHS MEAN MEDIAN MODE

ABELIAN ARCCOS ARCSIN ARCTAN AXES

BEARING

PARALLEL POLYGON

C F

M S

DICE EIGHT ELEVEN FIELD

M T

CHALKDUST CODE

GAME GRADIENT GROUP KITE

NINE

QED RADIAN RADIUS RATIO RHOMBUS RING SCALENE SERIES SET SQUARE SUM TERM TORUS

+ × = 98

÷ ÷ −

+ − = 0

− × ×

÷ + = 13

=𝟢 =𝟤 = 𝟣𝟪

Arrange the digits Put the numbers 1 to 9 (using each number exactly once) in the boxes so that the sums are correct. The sums should be read left to right and top to bottom ignoring the usual order of operations. For example, 𝟦 + 𝟥 × 𝟤 is 14, not 10.

spring 2022

chalkdust

◀ On the cover Poppy Azmi explores the patterns that are all around us.

ilings have been studied for thousands of years dating back to 3000 BC, and people are still finding out more and more about them now. They’ve been used in architecture for longer, and appear all over the natural world. I’m going to try and present a small subtopic within tilings that will hopefully be interesting and straightforward to understand.

Mathematica code by Sasho Kalajdzievski, CC BY-NC-SA 3.0.

▲ The rhombitrihexagonal tiling from around 200BC (left) and the Ammann chair tiling from AD1977 (right).

Second to the right and tile on till morning A tiling problem, in the context of this topic, is the following: given a set of tiles (closed simple curves and their interior) can we completely cover the plane without gaps or overlaps? The tilings which are easiest to visualise and draw are tilings by regular polygons, which I will call regular tilings here for simplicity. A lot of the results I’m going to mention can be extended to nonregular polygons, but for now let’s stick to maths which is a bit easier to deal with. Let’s even start with the simplest kind of regular tiling: tilings that are monohedral, which means only one type of tile is used to construct it. So how can we construct a monohedral regular tiling? Let’s start with the most common: a tiling of the plane by squares. It is easy to see how it is constructed and why it tiles the plane, simply because a square has four corners with four right angles. We know it is possible to place four squares together, edge to edge, so that all four of them have a common corner, simply because 𝟦 × π/𝟤 = 𝟤π, which is 21

spring 2022

the number of radians around a point. We can do this indefinitely so that each corner in the tiling is a corner common to four squares:

A slightly more complicated tile is the regular hexagon. See if you can work out how to prove that regular hexagons tile the plane. There is only one more monohedral regular tiling of the plane, and it can be constructed using the hexagon tiling. Take a regular hexagon and draw in its three axes of reflectional symmetry. This gives us a regular tiling with only equilateral triangles. These are actually the only three monohedral regular tilings of the plane! I will prove this later on. Now I want to discuss how to construct a vertex (ie a point where at least three tiles meet) in a regular tiling. We know there are 𝟤π radians around any vertex. Because we’re working with regular polygons, there’s a smaller number of angles that we have to choose from to make up this 𝟤π. Could we cut this down concretely?

It is a well known fact that the angle sum in a convex 𝑛-polygon, a polygon with 𝑛-sides, is (𝑛 − 𝟤)π (to prove this, consider splitting polygons up into triangles!). From now on, when I talk about polygons I mean regular polygons.

Counting corners If we consider a tiling where each vertex has 𝑟 polygons around it, polygons with 𝑛𝟣 , 𝑛𝟤 , … up to 𝑛𝑟 sides respectively, then these two facts combined give us the equation 𝟤π =

By cancelling π on both sides, we get

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(𝑛 − 𝟤)π (𝑛𝟣 − 𝟤)π +⋯+ 𝑟 . 𝑛𝟣 𝑛𝑟

𝑛 −𝟤 𝑛𝟣 − 𝟤 +⋯+ 𝑟 = 𝟤. 𝑛𝟣 𝑛𝑟

( )

So any polygons around a vertex in a regular tiling have to satisfy this equation. I think this is quite cool, because now we’ve got a concrete equation that we can work with algebraically! This equation can be used to prove there are only three monohedral regular tilings of the plane. Monohedral means that all 𝑛𝑖 = 𝑛, for some integer 𝑛 at least 3. Then ( ) simplifies to 𝑛−𝟤 𝑛−𝟤 +⋯+ =𝟤 𝑛 𝑛

Writing this in terms of 𝑛 gives

⇒

𝑟

𝑛−𝟤 = 𝟤. 𝑛

𝟦 . 𝑟 −𝟤 We know that 𝑛 is an integer, which means that 𝑟 − 𝟤 must divide 4. This implies 𝑟 − 𝟤 = 𝟣, 𝟤, or 𝟦, and so 𝑟 = 𝟥, 𝟦, or 𝟨. This is equivalent to saying, the only monohedral regular tilings of the plane are by triangles, squares, and hexagons! 𝑛 =𝟤+

A vertex by any other name would look the same One property about monohedral regular tilings is that all of the vertices ‘look’ the same. This leads onto the question: what about other regular tilings where all the vertices look the same? Equation ( ) seems like it could be quite helpful with figuring out how many different regular tilings we can form this way. I’m going to cheat a little with this part, and just tell you that there are exactly 21 ways to satisfy ( ). It is also noteworthy that the order of the 𝑛𝑖 ’s does matter in this case because reading off the polygons around a vertex in a different order makes a different tiling. I’ve not derived or included a proof of this result because it is quite long and tedious. There is actually a problem here. If for example we take the numbers 3, 4, 4, and 6 for the 𝑛𝑖 ’s. These fit the equation, but we can’t extend this arrangement of regular polygons around a vertex to a full tiling of the plane. This is because as shown below, at the black vertex we are forced to place the polygons in the order triangle-square-hexagon-square, but this is not in the order 3, 4, 4, 6. So not all vertices have the same arrangement of polygons, and this doesn’t extend to a tiling with the property we want.

▲ Trying to build the 3, 4, 4, 6 tiling by adding polygons around each vertex (indicated with a white dot) in the order triangle-square-square-hexagon.

Just because an arrangement of polygons fits around a vertex edge-to-edge, doesn’t mean this extends to a tiling of the plane where all vertices are the same type. Have a think about why this is. For this reason, it turns out there are only 11 ways to tile the plane in the way we want to. 23

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Showing this requires two steps. I’m not going to be super meticulous here and do this for all 21 different options, but I’ll do an example for each step. 1. We need to show that for 10 types of vertices, we can’t extend from the region around the starting vertex to a tiling of the plane, like we did for the numbers 3, 4, 4, 6 above. 2. We need to show that the remaining 11 types of vertices actually lead to tilings of the plane. We already showed this for the monohedral tilings, and in the picture below I show it for constructing a tiling with polygons 3, 6, 3, 6 from the monohedral triangle tiling:

▲ Constructing tiling with polygons 3, 6, 3, 6 starting with the monohedral tiling by equilateral triangles. Shaded regions are where the hexagons come from. These 11 tilings are known as the Archimedean tilings. So now we’ve shown that there are only 11 regular tilings where each vertex has the same arrangement of polygons around it! This, to me, is a good example of why maths is so interesting. We started with a problem that appeared to have an infinite number of solutions and managed to use only basic high school techniques and knowledge to deduce that there were only 11 different solutions instead!

Beyond Archimedean tilings The Archimedean tilings all have another property, and that is that they are periodic, invariant under translations in two non-collinear directions. I find periodic tilings really interesting as it’s fairly easy to see how they’re constructed. I also study aperiodic tilings, tilings that have no translational invariance. These can be studied using other branches of maths, such as topology. There is an object called the tiling space which is defined as all the translations of a tiling. For the grid unit square tiling, the tiling space is a torus. This is because the tiling is equivalent under translation by one tile horizontally or vertically, so opposite edges of a square are identified, which gives a flat torus. chalkdust

The tiling space of every periodic tiling is always the torus, so this topological approach isn’t so helpful for periodic tilings, but it is for aperiodic tilings. If you’re interested to learn more, a good article to introduce the subject is A primer of substitution tilings on the Euclidean plane by Natalie Frank. And, if you’re interested in geometry and tilings, then consider reading the book Tilings and patterns by Branko Grünbaum and Geoffrey Colin Shephard. You could also look into Platonic solids, which are an introduction to some of these ideas in three dimensions. Poppy Azmi Poppy is currently completing her master’s at Durham University. This article is based on material she learned during her final year project. In general she’s interested in geometry, algebra and topology.

The design featured on the cover of this issue is based on one of the tilings which can be found in the Alhambra, a fortress and palace near the southern coast of Spain. The building of the site was begun in the thirteenth century, and it is most famous, at least in mathematical circles, for its mosaic tilings. These are not only very beautiful, but also include examples of every, or nearly every—depending on who you ask—possible type of tessellation symmetry. The particular tiling on the cover exemplifies what is called the 333 symmetry, because the pattern is invariant under only translations and 3-fold rotational symmetry. In fact it has essentially three different types of 3-fold rotational symmetry (hence 333): one at the centre of each hexagon, one at the centre of each six-pointed star, and one at each place where three coloured wavy triangles meet. There is in fact an Archimedean tiling which has this same 333 symmetry, shown on the right (unlike in the main article, here the colouring is important). In addition to the translational symmetries, it has 3-fold rotation symmetry about the centre of each hexagon, the centre of each blue triangle which meets three different hexagons, and about the centre of each cream triangle which meets three different hexagons. 25

▲ This Archimedean tiling is called the snub hexagonal tiling, corresponding to 3, 3, 3, 3, 6. spring 2022

Which Greek letter are you?

START

Are you really a Greek letter?

Yes

Are you an angle?

Digamma Ϝϝ

No Epsilon Εε

A little Yes Do you care about any of this?

Maybe

Are you a Covid variant?

Not one

Upsilon Υυ

Are you a variant of concern?

Yes

No Yes

Tau Ττ

Yes

Are you the circle constant?

Can you be drawn without taking the pen off the paper?

Yes

Chi Χχ

What is the probability that I currently have you?

Should fluid dynamicists learn to stop using you in the same equation as p? Nu Νν

79%

10%

Alpha Αα

10%

Iota Ιι

Yes

Does your capital look like a Latin letter?

Yes

Rho Ρρ 26

Beta Ββ

Omicron Οο Covid free baby!

Theta Θθ PI Ππ

Are you the same letter as φ?

Yes

Phi Φϕ

Yes

Are you a human brain wave?

Yes

Are you the noise a cat makes?

Lambda Λλ

Omega Ωω

Are you in the Nato Phonetic alphabet?

Yes

Delta Δδ

Yes

Kappa Κκ

Does your mum think you’re a supplement for smartness?

Riemann’s

Dirichlet’s

Whose function are you?

Yes, but there’s something very wrong with my cat

Mu Μμ

Zeta Ζζ

Yes

Gamma Γγ Xi Ξξ 27

Yes

Are you an obscure programming language?

Eta Ηη

Yes

Psi Ψψ

Are you wavy?

Are you the circle constant?

Are you the president of China?

Sigma Σσ

Are you commonly confused with φ?

OT NFT hexagons N

WHAT’S

HOT NOT

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

& WHAT’S

Show your Twitter followers you’re an idiot by investing in a $3/month monkey

HOT NFT shapes of constant width Rolls down non-fungible coin slots, and comes free with every non-fungible copy of Chalkdust

TUE

OT Celebrating mathematical days Twosday Tuesday! Pi Day! Lots of numerical fun to be had this year.

Doing maths on these days

NOT

Who needs pi to more than 11 decimal places anyway?

Say goodbye and keep waving. Just… keep… waving… Behold! Your own private party!

HOT

T Selling your puzzle O to the NYT for $1m H Five pages of puzzles in this issue. One of them has to be worth something, right?

Chicken

NOT

Winning a Fields NOT medal for $15k

HOT

That’s like, a tank and a half of petrol?

Chalkdust launch parties Pizza! Quizzes! OK maybe we’ve had enough quizzes to last two lifetimes.

AGREE? DISAGREE?

Downing Street parties This issue currently racing Sue Gray for first to publish

Tell us @chalkdustmag Find more free fashion advice online d chalkdustmagazine.com

NOT

chalkdust

‘Whilst we thankfully live in a society where we can freely express our opinions’, these guys will spam your inbox then email your funders when you say on Twitter they have a silly name

Some pretty cool agentbased swarming modelling possible

Zoom chicken

Host has ended the meeting

NOT

Researching fish

HOT

Booooooooo

Researchfish

The power of curly brackets A simple {sketch of a} proof of Noether’s {first} theorem Mats Vermeeren relates symmetries to conserved quantities

Newton One of the most famous formulas in physics is Newton’s second law, 𝑭 = 𝑚𝒂.

It is named after Isaac Newton (1643–1727), who in all likelihood was never actually hit on the head by any apples, and states that ‘force equals mass times acceleration’. It shines in its simplicity, but contemporary mathematical physicists would much rather write it as d𝑧 = {𝐻 , 𝑧}. d𝑡 In this formula, 𝑧 can be any quantity related to the system and 𝐻 is the Hamilton function, which represents the total energy of the system. But what is this strange bracket? And why would any sane person write a simple idea like Newton’s second law in such an obscure way?

▲ Isaac kilogram metre per square second

There are a few possible answers to this last question. I could start singing praise for some abstract geometric beauty, but it also provides an easy explanation of Emmy Noether’s famous theorem on the relation between symmetries and conserved quantities. A conserved quantity, as the name suggests, is something that does not change. The most common example is probably the conservation of energy in physics, but there can be many other conserved quantities. Noether’s (first) theorem states that a system has a conserved quantity if and only if it possesses a related symmetry. Translation symmetry corresponds to conservation of (linear) momentum, for example, and rotational symmetry to angular momentum. 29

spring 2022

Hamilton Let’s consider a physical system consisting of one point particle of mass 𝑚 moving through space. The state of the system is given by its position 𝒙 = (𝑥𝟣 , 𝑥𝟤 , 𝑥𝟥 ) and momentum 𝒑 = (𝑝𝟣 , 𝑝𝟤 , 𝑝𝟥 ). In mechanics, momentum is simply the product of mass and velocity, 𝒑 = 𝑚𝒗 , so you could also say the state is given by position and velocity. But it turns out that using momentum leads to more elegant mathematics. If we know the state of the system at some point in time, then the states at all future (and past) times are determined by Newton’s second law. Suppose we can write the potential energy of the system as a function 𝑉 (𝒙) of the position. Then the force on the particle is given by minus the gradient of this function, 𝐹 (𝒙) = −𝛁𝑉 (𝒙), so Newton’s second law can be written as d𝟤 𝒙 d𝑡 𝟤

or, written in components,

𝟣 = − 𝛁𝑉 (𝒙), 𝑚

▲ William Rowan Hamilton (the musical)

d𝟤 𝑥𝑖 𝟣 𝜕𝑉 (𝒙) . =− 𝑚 𝜕𝑥𝑖 d𝑡 𝟤

Since the kinetic energy of the particle is |𝒑|𝟤 /(𝟤𝑚), the total energy will be given by 𝐻 (𝒙, 𝒑) =

𝟣 |𝒑|𝟤 + 𝑉 (𝒙). 𝟤𝑚

This is called the Hamilton function, after William Rowan Hamilton (1805–1865), who is also famous as the inventor/discoverer of the quaternions. One reason why this function deserves its special name, is that it gives the equations of motion in a very satisfying way. The time derivatives of position and momentum are, up to sign, equal to the partial derivatives of 𝐻 : d𝒙 𝜕𝐻 = d𝑡 𝜕𝒑

and

d𝒑 𝜕𝐻 =− . d𝑡 𝜕𝒙

We call these the ‘equations of motion’. By combining the two equations of motion, you can find an expression for the acceleration d𝟤 𝒙/d𝑡 𝟤 . You can check that this again gives Newton’s second law, so these equations really do describe the motion of the system. For our choice of Hamilton function, the first equation of motion will simply be d𝒙/d𝑡 = 𝒑/𝑚, but we prefer to write it in the form above. It is not just pleasing to have both equations of motion look so similar, but it also helps us find the time derivative of 𝐻 itself. By the chain rule we have d𝐻 𝜕𝐻 d𝒙 𝜕𝐻 d𝒑 = ⋅ + ⋅ , d𝑡 𝜕𝒙 d𝑡 𝜕𝒑 d𝑡

where the dot denotes the scalar product between vectors. If you prefer to write out the components, chalkdust

you can expand the first term as 𝜕𝐻 d𝒙 𝜕𝐻 d𝑥𝟣 𝜕𝐻 d𝑥𝟤 𝜕𝐻 d𝑥𝟥 ⋅ = + + 𝜕𝒙 d𝑡 𝜕𝑥𝟣 d𝑡 𝜕𝑥𝟤 d𝑡 𝜕𝑥𝟥 d𝑡

and the second term in a similar way. Using the equations of motion we can see the two terms in the expression for d𝐻 /d𝑡 cancel: d𝐻 𝜕𝐻 𝜕𝐻 𝜕𝐻 𝜕𝐻 = ⋅ − ⋅ = 𝟢, d𝑡 𝜕𝒙 𝜕𝒑 𝜕𝒑 𝜕𝒙

so we can call 𝐻 a conserved quantity, because it doesn’t change in time. Hence any system that fits into this Hamiltonian framework automatically satisfies conservation of energy.

Kepler When thinking about physics, it is always useful to have a particular system in mind. Consider for example a planet orbiting the sun (where we approximate both by point masses and assume that the sun is fixed at the origin). This is known as the Kepler system, after Johannes Kepler (1571–1630), who figured out the laws of planetary motion long before Newton came up with a theory of gravity.

The diagram below shows a planet (blue) orbiting the sun (yellow), tracing an ellipse. It is shown near its closest point to the sun, together with its velocity vector.

The Kepler system is governed by the Hamilton function, 𝟣 𝟣 𝐻 (𝒙, 𝒑) = |𝒑|𝟤 − . 𝟤𝑚 |𝒙| Its equations of motion are d𝒙 =𝒑 d𝑡

and

d𝒑 −̂ 𝒙 = 𝟤, d𝑡 |𝒙|

where 𝒙̂ denotes the unit vector in the direction of 𝒙 . This is Newton’s second law with the inverse square central force 𝒙̂ 𝐹 (𝒙) = − 𝟤 . |𝒙|

Rotational symmetry means that if we instantly rotate both the planet and its velocity vector through some angle around the sun, then the new orbit will be the same ellipse as before, rotated by the same angle.

If the planet is moving relatively slowly, its motion will consist of repeated orbits, tracing out an ellipse. If it is speeding, then its orbit will either be a parabola or a hyperbola, so it will approach the sun only once before rushing off into interstellar space. Let’s ignore the possibility of such a speedy galactic nomad and assume that we are dealing with an elliptic orbit. The Kepler system has some notable properties. One of these is that it is rotationally symmetric (see right). This can be seen from the formula for the Hamilton 31

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function: it only depends on the lengths of 𝒙 and 𝒑 , not their directions, so it will not change if we rotate these vectors. The conserved quantity corresponding to rotational symmetry is the angular momentum 𝑳 = 𝒙 × 𝒑 (see below). We can check this by verifying that its time derivative is zero. Using the product rule we find d𝒑 d𝑳 d𝒙 = ×𝒑+𝒙 × d𝑡 d𝑡 d𝑡 −̂ 𝒙 = 𝒑 × 𝒑 + 𝒙 × 𝟤, |𝒙|

which is zero because the cross product of parallel vectors is always zero. Hence 𝑳 is indeed a conserved quantity.

Our observations on the Kepler system illustrate Noether’s theorem, which states that symmetries and conserved quantities are in one-to-one correspondence. In particular, if a system is rota▲ Johannes Kepler tionally symmetric, then it conserves angular momentum. Other systems may have different symmetries. If a system has translation symmetry (eg billiards on an unbounded table) it conserves the total linear momentum. Even conservation of energy fits into this picture. It corresponds to the time translation symmetry: the laws of physics don’t change over time. To prove Noether’s theorem, we need one more item in our toolbox.

Angular momentum (𝑳 = 𝒙 × 𝒑 ) is a measure of how much an object is spinning (around a reference point). It increases both when the speed of rotation increases and when the distance to the reference point grows.

Conservation of angular momentum implies that the planet moves slower when it is further from the the sun. A terrestrial example of conservation of angular momentum can be observed when a figure skater performs a pirouette. They will spin more quickly when they move some of their mass closer to the rotation axis by bringing in their arms. chalkdust

{

, } Poisson

Earlier we checked that 𝐻 is always a conserved by calculating its time derivative. What if we want to know the time derivative of some other function 𝑧 of position and momentum? We can calculate d𝑧 𝜕𝑧 = d𝑡 𝜕𝒙 𝜕𝑧 = 𝜕𝒙

d𝒙 𝜕𝑧 d𝒑 + ⋅ d𝑡 𝜕𝒑 d𝑡 𝜕𝐻 𝜕𝑧 𝜕𝐻 ⋅ − ⋅ . 𝜕𝒑 𝜕𝒑 𝜕𝒙 ⋅

The expression in the last line is usually written as: {𝐻 , 𝑧} ∶=

𝜕𝐻 𝜕𝑧 𝜕𝐻 𝜕𝑧 ⋅ − ⋅ . 𝜕𝒑 𝜕𝒙 𝜕𝒙 𝜕𝒑

▲ Siméon Denis Poisson

This curly bracket is called the Poisson bracket, after Siméon Denis Poisson (1781–1840), whose name sounds a lot less posh once you remember that ‘poisson’ is French for ‘fish’. With this definition, we obtain our brackety formula, d𝑧 = {𝐻 , 𝑧}, d𝑡 as a reformulation of the Hamiltonian equations of motion. We can also go the other way and recover the Hamiltonian equations of motion from the brackety formula. Since 𝑧 can be any function of position and momentum, we can choose to set it equal to a component of either position or momentum and find d𝑥𝑖 𝜕𝐻 = {𝐻 , 𝑥𝑖 } = d𝑡 𝜕𝑝𝑖

and

d𝑝𝑖 𝜕𝐻 = {𝐻 , 𝑝𝑖 } = − . d𝑡 𝜕𝑥𝑖

We have now established that d𝑧/d𝑡 = {𝐻 , 𝑧} is equivalent to the Hamiltonian equations of motion, but why do we care about Poisson brackets? Well, they have some interesting properties: (A) A function 𝑧 of position and momentum is a conserved quantity of the system with Hamilton function 𝐻 if and only if {𝐻 , 𝑧} = 𝟢.

This follows immediately from the brackety formula: the vanishing of the Poisson bracket is equivalent to d𝑧/d𝑡 = 𝟢.

(B) The Poisson bracket is skew-symmetric, meaning that if we swap around its entries, we get the same result but with a minus sign: {𝑧, 𝑢} = −{𝑢, 𝑧} for any two functions 𝑢 and 𝑧 of position and momentum.

There are additional properties of the Poisson bracket which make sure that the identification of a time derivative with a bracket makes sense, but we won’t go into those technicalities here. Instead, let’s spin our attention towards Noether. 33

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⇔

Noether

Emmy Noether (1882–1935) was a mathematician of many talents. Much of her work was in abstract algebra, but she is most famous for her theorem stating that conserved quantities of a system are in one-to-one correspondence with its symmetries. She counts as one of the top mathematicians of the interwar period, a status she managed to achieve in the face of cruel discrimination because of her gender and descent. At the University of Göttingen, Germany, where she spent most of her career, she was refused a paid position, despite strong support from her colleagues David Hilbert and Felix Klein. In 1933, she emigrated to the US to escape the Nazi regime.

Daniel Schwen, CC BY-SA 2.5

▲ The University of Göttingen

Let’s put aside the grim history and step into Noether’s mathematical footsteps to find out what symmetries have to do with conserved quantities. Consider two Hamilton functions 𝐻 and 𝐼 and the corresponding dynamical systems, d𝑧 = {𝐻 , 𝑧}, d𝑡 d𝑧 = {𝐼 , 𝑧}. d𝑡

( ) ( )

In the Kepler problem, for example, the system labelled ( ) would describe the physical motion of a planet and the one labelled ( ) a rotation around the sun. The Hamilton function for a rotation is a component of the angular momentum vector, so in this example we would take 𝐼 equal to a component of 𝑳 = 𝒙 × 𝒑 .

Now what does it mean for ( ) to be a symmetry of the system ( )? It means that the motion of ( ) does not change the equation ( ). Since the dynamics of a system is fully encoded by its Hamilton function, this is equivalent to saying that the system ( ) does not change the Hamilton function 𝐻 . Hence

▲ Emmy Noether

( ) is a symmetry of ( )

⟺

𝐻 is a conserved quantity of ( ).

We can use property (A) to express this in terms of a Poisson bracket: ( ) is a symmetry of ( )

⟺

{𝐼 , 𝐻 } = 𝟢.

⟺

{𝐻 , 𝐼 } = 𝟢.

Next we use property (B): the Poisson bracket is skew-symmetric, hence ( ) is a symmetry of ( ) Or, using property (A) once again: ( ) is a symmetry of ( ) chalkdust

⟺

𝐼 is a conserved quantity of ( ).

This is essentially the statement of Noether’s first theorem: the symmetries of a system are related to its conserved quantities. There is an important thing that we have swept under the rug in this derivation. Not every possible symmetry is generated by a Hamilton function. Hence the correct formulation of Noether’s theorem is that conserved quantities are in one-to-one correspondence with Hamiltonian symmetries. This issue disappears when, instead of Hamiltonian mechanics, we consider Lagrangian mechanics, which is based on the calculus of variations. Within that framework a natural notion of symmetry leads to a one-to-one correspondence between symmetries and conserved quantities. In fact, Noether’s original paper dealt with the Lagrangian perspective. It included not just the case of mechanics, but also field theory, which deals with partial differential equations. Her main motivation was to understand conservation of energy in Einstein’s theory of gravity. This is a surprisingly subtle problem, because general relativity has an infinite number of symmetries.

▲ A page from Noether’s original paper

When a system has an infinite number of symmetries, the conserved quantities produced by Noether’s first theorem are trivial in some sense. For example, a function which maps position and momentum to a constant would be a trivial conserved quantity: it does not change in time, but that fact does not tell us anything about the dynamical system. Noether’s second theorem is relevant to these systems with infinitely many symmetries. Roughly speaking, it says that if a system has an infinite amount of symmetries, then the equations of motion must have a certain redundancy to it: some of the information contained in one of the equations of motion will also be contained in the others.

Noether’s legacy It was known before Noether’s time that conserved quantities are related to symmetries. And while her paper was the first one to make this connection precise, her main breakthrough was the lesser known second theorem. Noether’s insights were warmly welcomed by mathematical physicists of the time, including Albert Einstein himself, and are still a key part of modern physics. The power of Noether’s theorems lies in their generality: they apply to any system with the relevant kind of symmetries, and to prove them you don’t need to know the particulars of the system at hand. Similarly, Poisson brackets allow us to capture essential features of physics with an equation that takes the same form no matter what system it describes. Instead of having to work out all the forces between interacting objects, all you need to put into this framework is the total energy in the form of the Hamilton function. It’s no wonder that mathematical physicists often prefer d𝑧/d𝑡 = {𝐻 , 𝑧} over 𝑭 = 𝑚𝒂 . Mats Vermeeren Mats is a researcher at Loughborough University, working in the area of mathematical physics. He would not be able to illustrate angular momentum by performing pirouettes on ice. Instead he likes to get some linear momentum going as a long-distance runner.

spring 2022

My favourite ...LATEX package \usepackage{cleveref}

Adam Townsend

If you want cross-references in your LATEX file (spoiler: you do), then you need to learn about cleveref. Instead of filling your document with ‘theorem \ref’, ‘section \ref’, and ‘\eqref’, you can just write ‘\cref’ and cleveref will automatically write what kind of thing you’re referencing. It even does ranges! And if you decide later that you want to make ‘theorem’ start with a capital letter (spoiler: you don’t), simply add some cleveref settings to your preamble and they’ll all change.

ggggg \usepackage{booktabs}

Madeleine Hall

Second only to amsmath and graphicx, the package booktabs is my favourite due to its unbounded capacity to improve your tables in LATEX. No more \rule{0cm}{8pt} to get more space between your data and the horizontal line above. No more borders of uniform width. No more vertical lines! Throw out \hline and say hello to \toprule,\midrule and \bottomrule, and give your table data the space it needs to truly shine. Rating

Rating

Tables made without booktabs

ghiii

Tables made with booktabs

\usepackage{tikz}

ggggg David Sheard

Tikz is an amazingly versatile tool whose primary function is to draw 2D graphics in LATEX. While a little fiddly to get used to at first, it ends up being a really fun package to use (most of the time, except when it is infuriating—but that’s LATEX for you)! As an indication of its utility, it is used on about half of the pages of this magazine, sometimes in surprisingly hidden ways.

\usepackage{realhats}

Matthew Scroggs

Usually, the command \hat puts a circumflex above a letter: 𝑎̂ . When writing a paper, this can be confusing as hats can be used to denote many different things. Realhats can fix this problem by allowing you to put a huge range of hats over your letters: 𝑎 .

ggggg

\usepackage{graphicx}

Helena Millade

LATEX without pictures would just be rubbish. ***** chalkdust

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Mathematics is my secret— my secret weakness. I feel like a stubborn, helpless fool in the middle of a problem. Trapped and crazed. Also, thrilled. I’m gnawing on a bone and I fear for anyone who comes near and tries to make me stop. I growl. I refuse to drop that bone. I am dogged— did I mention that math is funny too? It’s hilarious how much time I secretly waste on it, when there are more productive things to do, like walk my dog, fold my laundry, or finish my actual homework. No one in my family thinks all this math is healthy. Get some sleep. Here, eat something. Go outside for some fresh air! And they’re probably right. But here’s another secret: I don’t care About rest or nutrition or any of it. blah blah blah I’m not even really listening. Too busy gnawing away. On a different day Back out in the blaring, glaring, glittering world I feel unfurled and chatty Eager for random conversations and observations Nosy for other people’s secrets, which surely must be as weird and wild as the ones that define the inside of me. And I notice something fascinating exasperating in all the talking talking talking. 38

Here’s what a lot of people say: I’m just not a math person. Insert shrug and sheepish smile. I can barely figure out the tip. Wait, you take extra math classes? What for? What do you mean, just for fun? Wow, you must be a genius or something. Here’s what no one says: I’m just not a words person. Insert shrug and sheepish smile. I can barely read a sentence. Wait, you read an entire book this weekend? What for? What do you mean, just for fun? Wow, you must be a genius or something. I feel silly being mistaken for a genius, when I’m really just a nerd. Struggling, stumbling, pushing, reaching for more of something that makes me feel mesmerised and uncomfortable, lost, hopeless, hopeful, real.

Julia Schanen from New Julia is a high school student ng linear lori exp s love who Jersey, USA, She also . ory the ber num algebra and hy, and enjoys bouldering, photograp while en listening to the Beatles (oft doing linear algebra).

The hidden harmonies of

Hamming distance Chris Boucher explores the secrets and symmetries behind a measure of the distance between binary strings

chalkdust

1010002

e think about the concept of distance all the time—how far is it to go visit a friend, is my phone close enough to the router to get wifi?—but in mathematics there are many ways of measuring the distance between things. Mathematics is sometimes thought to be a world of dense, dry formulae and esoteric logic, but its structures, especially when paired with modern computing, can generate images of arresting complexity like the one opposite. The goal of this article is to introduce and explore these images, which arise from attempts to visualise a notion of distance on the integers called Hamming distance. A distance function assigns a non-negative number to any pair of objects to serve as a measure of how far apart they are. In order to qualify as a distance, such an assignment needs three key features. First, the distance from an object to itself is zero. Second, the distance from object 𝑥 to object 𝑦 is always the same as the distance from object 𝑦 to object 𝑥 . Finally, a path from one object to another that goes through a third is never shorter than the direct path between the two objects. More formally, if we denote the distance from 𝑥 to 𝑦 by 𝐷(𝑥, 𝑦), then for any object 𝑧 , 𝐷(𝑥, 𝑦) ⩽ 𝐷(𝑥, 𝑧) + 𝐷(𝑧, 𝑦).

In what follows, we will consider two different ways of measuring the distance between positive integers.

Euclidean distance and Hamming distance The most common way to measure the distance between two numbers is using Euclidean distance, with which you may already be familiar. Given two numbers 𝑥 and 𝑦 , the Euclidean distance between them, expressed as 𝐸(𝑥, 𝑦), is simply |𝑥 − 𝑦|. In the mid-twentieth century, the American mathematician Richard Hamming developed the Hamming distance to measure the distance between sequences of zeros and ones by counting the number of positions in which the sequences differ, a measure that has the three features necessary for a distance. So, for example, the Hamming distance between 𝟣0𝟣𝟣𝟢 and 𝟣1𝟣𝟣𝟢 is 1 as they only differ in the second position from the left—these strings are close. Working in the early days of computer science, Hamming used this distance measure in his analysis of the effectiveness of error detection and correction in the transmission of messages encoded by strings of zeros and ones. The number of errors an encoding could correct was related to how ‘spread out’ the space of code words was as measured by Hamming distance. Hamming distance can also be thought of as a distance on the integers by using binary representations. For example, 𝟤𝟤 = 𝟣 × 𝟤𝟦 + 𝟢 × 𝟤𝟥 + 𝟣 × 𝟤𝟤 + 𝟣 × 𝟤𝟣 + 𝟢 × 𝟤𝟢 = 𝟣𝟢𝟣𝟣𝟢𝟤 ,

𝟥𝟢 = 𝟣 × 𝟤𝟦 + 𝟣 × 𝟤𝟥 + 𝟣 × 𝟤𝟤 + 𝟣 × 𝟤𝟣 + 𝟢 × 𝟤𝟢 = 𝟣𝟣𝟣𝟣𝟢𝟤 .

Here, the subscript 𝟤 indicates that the number is in base 2 rather than base 10. Writing 𝐻 (𝑥, 𝑦) for the Hamming distance between two positive integers 𝑥 and 𝑦 , 𝐻 (𝟤𝟤, 𝟥𝟢) = 𝟣 because the binary representations of 30 and 22 differ in one position. Of course, the Euclidean distance is 𝐸(𝟤𝟤, 𝟥𝟢) = |𝟤𝟤 − 𝟥𝟢| = 𝟪. The binary representation of the number six, 𝟣𝟣𝟢𝟤 , has only three digits, but we can measure the distance from 6 to a number like 22 whose binary representation has five digits by padding with extra zeros on the left, which has no effect on the underlying number. So 𝟨 = 𝟣𝟣𝟢𝟤 = 𝟢𝟢𝟣𝟣𝟢𝟤 , and 𝐻 (𝟨, 𝟤𝟤) = 𝐻 (𝟢𝟢𝟣𝟣𝟢𝟤 , 𝟣𝟢𝟣𝟣𝟢𝟤 ) = 𝟣.

1010012

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Hamming distance clearly measures distance between positive integers quite differently than the standard Euclidean distance does. For example, 𝐻 (𝟤𝟤, 𝟤𝟧) = 𝟦, so 22 is much further from 25 than it is from 6 as measured by Hamming distance, but 22 is much closer to 25 than to 6 on the number line. While we can ‘see’ the Euclidean distance between two numbers in their separation on the number line, it is not immediately clear how or whether we can visualise Hamming distance.

Euclidean distance in a square Distance is a function of two variables, so let us try to visualise 𝐸(𝑖, 𝑗) and 𝐻 (𝑖, 𝑗) as the pair (𝑖, 𝑗) ranges over a grid of integer pairs. We’ll define 𝐵𝑛 to be the set of positive integers whose binary representation contains 𝑛 or fewer digits, so 𝐵𝑛 = {𝟢, 𝟣, … , 𝟤𝑛 − 𝟣}. Our first strategy to visualise Euclidean and Hamming distance is to use the grid 𝐺𝑛 = {(𝑖, 𝑗) | 𝑖, 𝑗 ∈ 𝐵𝑛 } as our canvas and give each different value of 𝐸(𝑖, 𝑗) a different colour. For Euclidean distance, this gives a fairly simple picture. Points (𝑖, 𝑗) that satisfy 𝐸(𝑖, 𝑗) = 𝑘 lie either on the line 𝑦 = 𝑥 + 𝑘 or 𝑦 = 𝑥 − 𝑘 , and so the points whose coordinates are at Euclidean distance 𝑘 consist of two parallel lines. On the right, the squares are 𝟣𝟤𝟪 × 𝟣𝟤𝟪 cells, with the cell at position (𝑖, 𝑗) corresponding to that point in the grid. ▲ Euclidean distance visualised in a square. If While this picture is fairly simple, there are some 𝐸(𝑖, 𝑗) = 𝐸(𝑖′ , 𝑗 ′ ) then (𝑖, 𝑗) and (𝑖′ , 𝑗 ′ ) are the same features worth noting. First, the image is symmetcolour; cells with 𝐸(𝑖, 𝑗) = 𝟣𝟨 are outlined. ric about the line 𝑦 = 𝑥 because 𝐸(𝑥, 𝑦) = 𝐸(𝑦, 𝑥), a symmetry that would be present for any distance measure. Second, the image is unchanged if we translate along the line 𝑦 = 𝑥 because for any 𝑧 , 𝐸(𝑥 + 𝑧, 𝑦 + 𝑧) = 𝐸(𝑥, 𝑦).

( )

Hamming distance in a square Using the same strategy to visualise Hamming distance produces more interesting images. The picture below shows a few colourings of the 𝟣𝟤𝟪 × 𝟣𝟤𝟪 grid 𝐺 𝟩 using different values of the Hamming distance between cell coordinates. Colouring all the points in the grid 𝐺 𝟫 according to the Hamming distance between their coordinates gives the image at the start of the article. One thing to notice is that there are only 𝑛 possible values for the Hamming distance between two points in 𝐵 𝑛 , so there are only 𝑛 colours used in the grid, whereas there are 𝟤𝑛 possible values for the Euclidean distance between two such points. What is interesting about this picture is the nested structure of the grids 𝐺𝑛 —the image is a fractal in that the same patterns recur at progressively smaller scales.

The fractal looks like a quilt made of squares of similar structure. Moving from a point (𝑖, 𝑗) in some sub-square of the grid to a similarly positioned point in another sub-square involves translating one or both components by a power of two. To explore this, let’s use the notation 𝑥 + 𝐴 for the set of numbers chalkdust

1010102

𝐻 (𝑖, 𝑗) = 𝟣

𝐻 (𝑖, 𝑗) = 𝟤

𝐻 (𝑖, 𝑗) = 𝟥

𝐻 (𝑖, 𝑗) = 𝟦

𝐻 (𝑖, 𝑗) = 𝟧

𝐻 (𝑖, 𝑗) = 𝟨

▲ Colouring cells of 𝐺 𝟩 by the value of 𝐻 (𝑖, 𝑗). The cases 𝐻 (𝑖, 𝑗) = 𝟢 and 𝟩 are omitted as they consist of a single diagonal line of points corresponding to 𝑖 = 𝑗 and 𝑖 = 𝟤𝑛 − 𝟣 − 𝑗 respectively.

in the set 𝐴 shifted by 𝑥 . So formally, 𝑥 + 𝐴 = {𝑥 + 𝑎 | 𝑎 ∈ 𝐴}, and, for example, 𝟥 + {𝟪, 𝟫, 𝟣𝟢} = {𝟣𝟣, 𝟣𝟤, 𝟣𝟥}. This notation allows us to recognise 𝐵𝑛 as consisting of 𝐵𝑛−𝟣 together with 𝟤𝑛−𝟣 + 𝐵𝑛−𝟣 . For example: 𝐵𝟤 = {𝟢, 𝟣, 𝟤, 𝟥}

and

𝐵𝟥 = {𝟢, 𝟣, 𝟤, 𝟥, 𝟢 + 𝟤𝟤 , 𝟣 + 𝟤𝟤 , 𝟤 + 𝟤𝟤 , 𝟥 + 𝟤𝟤 } .

This means that the grid 𝐺𝑛 is built from four quadrants that are shifts of 𝐺𝑛−𝟣 (depicted below right).

If 𝑖 ∈ 𝐵𝑛−𝟣 , then the binary representation of 𝑖 + 𝟤𝑛−𝟣 is identical to that of 𝑖 except for having a 1 in the 𝟤𝑛−𝟣 th place instead of a 0. It follows that if 𝑗 is also in 𝐵𝑛−𝟣 , then 𝑖 + 𝟤𝑛−𝟣 and 𝑗 differ in all the same places 𝑖 and 𝑗 do, with an additional difference in the 𝟤𝑛−𝟣 th place. If, for example, 𝑖 = 𝟢𝟣𝟣𝟢𝟣𝟤 , 𝑗 = 𝟢𝟢𝟣𝟣𝟣𝟤 where 𝑖, 𝑗 ∈ 𝐵 𝟧 , then 𝑖 + 𝟤𝟧 = 𝟣𝟢𝟣𝟣𝟢𝟣𝟤 . Stated more precisely: 𝐻 (𝑖 + 𝟤𝑛−𝟣 , 𝑗) = 𝐻 (𝑖, 𝑗) + 𝟣.

𝐻 (𝑖, 𝑗

= 𝐻 (𝑖, 𝑗) + 𝟣.

(𝟤𝑛−𝟣 , 𝟤𝑛−𝟣 ) + 𝐺𝑛−𝟣

𝐺𝑛−𝟣

(𝟤𝑛−𝟣 , 𝟢) + 𝐺𝑛−𝟣

( )

This means that the cells in the lower right quadrant will have a colour value of exactly one more than the cell in the corresponding position in the lower left quadrant. A similar calculation shows that + 𝟤𝑛−𝟣 )

(𝟢, 𝟤𝑛−𝟣 ) + 𝐺𝑛−𝟣

( ) 1010112

▲ The four quadrants of 𝐺𝑛 . spring 2022

In a similar way, the upper left quadrant has the same relationship to the lower left quadrant as the lower right quadrant does. In particular, the upper left and lower right quadrants have identical colouring. If 𝑖, 𝑗 ∈ 𝐵𝑛−𝟣 , then 𝑖 + 𝟤𝑛−𝟣 and 𝑗 + 𝟤𝑛−𝟣 both have a 1 in the 𝟤𝑛−𝟣 th place and differ in precisely the same places as 𝑖 and 𝑗 . If we again take 𝑖 = 𝟢𝟣𝟣𝟢𝟣𝟤 , 𝑗 = 𝟢𝟢𝟣𝟣𝟣𝟤 ∈ 𝐵 𝟧 , then 𝑖 + 𝟤𝟧 = 𝟣𝟢𝟣𝟣𝟢𝟣𝟤 and 𝑗 + 𝟤𝟧 = 𝟣𝟢𝟢𝟣𝟣𝟣𝟤 . That is, 𝐻 (𝑖 + 𝟤𝑛−𝟣 , 𝑗 + 𝟤𝑛−𝟣 ) = 𝐻 (𝑖, 𝑗). ( ) This means the colouring of the upper right block will be identical to that of the lower left block. Since this block structure holds for 𝑛 = 𝟣, 𝟤, … , we can build the Hamming fractal starting with 𝐺 𝟣 . The first few steps are shown below:

𝐺𝟣 𝐺𝟤

𝐺𝟥

𝐺𝟦

𝐺𝟧

𝐺𝟨

▲ Building the fractal using the block structure, starting with a single cell.

The nested structure illustrated here causes the fractal nature of the grids 𝐺𝑛 to become increasingly apparent as 𝑛 → ∞. For larger values of 𝑛, the images contain finer detail and more complex selfsimilarity. In fact, equations ( ) to ( ), which cause the structure in these images, have generalisations that involve a type of addition called the XOR sum.

The block structure and XOR sums Hamming distance counts the number of positions in which the binary representations of 𝑖 and 𝑗 differ and is closely related to the binary operation of XOR addition. The XOR sum of two natural numbers 𝑖 and 𝑗 , which we will denote 𝑖 ⊕ 𝑗 , is the natural number whose binary representation has a 1 in any position where the binary representations of 𝑖 and 𝑗 differ and a 0 in any position where they are the same. For example, 𝟨𝟣 𝟣𝟣𝟣𝟣𝟢𝟣𝟤 ⊕ 𝟣𝟧 ⊕ 𝟢𝟢𝟣𝟣𝟣𝟣𝟤 𝟧𝟢 𝟣𝟣𝟢𝟢𝟣𝟢𝟤

If we use the notation ‖ 𝑖 ‖ for the number of ones in the binary representation of 𝑖, then ‖ 𝑖 ‖ = 𝐻 (𝟢, 𝑖). chalkdust

1011002

The binary representation of 𝑖 ⊕ 𝑗 has as many 1s as positions where the binary representations of 𝑖 and 𝑗 differ, so 𝐻 (𝑖, 𝑗) = ‖ 𝑖 ⊕ 𝑗 ‖. Because the binary representation of 𝟤𝑛−𝟣 contains all zeros except for a one in the 𝟤𝑛−𝟣 position, for any 𝑖, 𝑖 ⊕ 𝟤𝑛−𝟣 will have an identical binary representation to 𝑖 except in the 𝟤𝑛−𝟣 position, where it will differ.

Another way of thinking of the block structure in the images in the previous section is in terms of the ⊕ operation. If 𝑖, 𝑗 ∈ 𝐵𝑛−𝟣 , so both 𝑖 and 𝑗 have zeros to the left of the 𝟤𝑛−𝟤 th place in their binary representations, then adding 𝟤𝑛−𝟣 to either number affects their binary representations only by adding a one in the 𝟤𝑛−𝟣 th place. That is, equations ( ) and ( ) can be generalised if 𝑖 + 𝟤𝑛−𝟣 is replaced by 𝑖 ⊕ 𝟤𝑛−𝟣 , so that it becomes ‖ (𝑖 ⊕ 𝟤𝑛−𝟣 ) ⊕ 𝑗 ‖ = ‖ 𝑖 ⊕ 𝑗 ‖ + 𝟣

and

‖ 𝑖 ⊕ (𝑗 ⊕ 𝟤𝑛−𝟣 ) ‖ = ‖ 𝑖 ⊕ 𝑗 ‖ + 𝟣.

Moreover equation ( ) generalises to become ‖ (𝑖 ⊕ 𝟤𝑛−𝟣 ) ⊕ (𝑗 ⊕ 𝟤𝑛−𝟣 ) ‖ = ‖ 𝑖 + 𝑗 ‖.

Equations ( ) to ( ), as well as giving rise to this nested structure, can also be generalised using the XOR sum to explain more intricate symmetries of this image. Suppose the numbers 𝑖 and 𝑗 agree at some position of their binary expansions like 𝑖 = 𝟢𝟣1𝟢𝟣𝟤 and 𝑗 = 𝟣𝟢1𝟣𝟢𝟤 , which agree in the third position. Then ‘⊕-ing’ the same number to both doesn’t change this agreement. If, for example, 𝑎 = 𝟣𝟢𝟣𝟢𝟢𝟤 , then 𝑖 ⊕ 𝑎 = 𝟣𝟣0𝟢𝟣𝟤 and 𝑗 ⊕ 𝑎 = 𝟢𝟢0𝟣𝟣𝟤 both also agree in this position.

If, on the other hand, 𝑖 and 𝑗 disagree at the some position of their binary expansions like 𝑖 = 𝟢𝟣0𝟢𝟣𝟤 and 𝑗 = 𝟣𝟢1𝟣𝟢𝟤 , which disagree in the third position, then ‘⊕-ing’ the same number to both doesn’t change this disagreement either. Here, 𝑖 ⊕ 𝑎 = 𝟣𝟣1𝟢𝟣𝟤 and 𝑗 ⊕ 𝑎 = 𝟢𝟢0𝟣𝟣𝟤 both also disagree in this position. Thus, the binary expansions of 𝑖 ⊕ 𝑎 and 𝑗 ⊕ 𝑎 differ in precisely the same positions as the binary expansions of 𝑖 and 𝑗 , so 𝐻 (𝑖 ⊕ 𝑎, 𝑗 ⊕ 𝑎) = 𝐻 (𝑖, 𝑗). ( ) Comparing this to equation ( ), we see that the operation ⊕ is to Hamming distance much as the operation of ordinary addition is to Euclidean distance.

Visualisations on a circle We can get an alternative visualisation by identifying the points of 𝐵𝑛 with equally spaced points around a circle, and connecting points that are a given distance apart. We place the number 0 at the three o’clock position on the circle and spread the numbers 𝟢, 𝟣, … , 𝟤𝑛 −𝟣 evenly around the circle anticlockwise. If we connect points that correspond to numbers that are Hamming distance 𝑘 from one another, we get some interesting images. The image on the right corresponds to 𝑛 = 𝟦 and 𝑘 = 𝟤 and the image at the top, overleaf to 𝑛 = 𝟨 and 𝑘 = 𝟥.

The relationship in equation ( ) illuminates many symmetries of these images. If we ‘⊕-shift’ all the points on 1011012

1 0

8 9

15 10

14 11

▲ Connect 𝑖 and 𝑗 in 𝐵𝟦 if 𝐻 (𝑖, 𝑗) = 𝟤. spring 2022

▲ Connect 𝑖 and 𝑗 in 𝐵𝟨 if 𝐻 (𝑖, 𝑗) = 𝟥. the circle, that is, if we move every vertex 𝑖 to the vertex 𝑖 ⊕ 𝑎 for some 𝑎, then we don’t change any of the connections because such a shift doesn’t change Hamming distance. For example, if 𝑎 = 𝟤𝑛 −𝟣, the binary representation of 𝑎 consists of 𝑛 ones, and the binary representation 𝑥 ⊕ 𝑎 has ones precisely where that of 𝑥 has zeros, and has zeros precisely where that of 𝑥 has ones. That is, 𝑥 ⊕ 𝑎 = 𝑎 − 𝑥 , and the image is symmetric with respect to the interchange of 𝑥 and 𝑎 − 𝑥 , which amounts to a reflection over a line through the centre of the circle. In the image (below left) purple arrows connect each point 𝑥 with the point 𝑎 − 𝑥 . Choosing a different value of 𝑎, say 𝑎 = 𝟥, then interchanging 𝑥 and 𝑥 ⊕ 𝑎 corresponds to a more subtle symmetry of the graph as shown on the right. 5

15 10 12

15 10

14 11

▲ Symmetry with respect to swapping 𝑥 with 𝑥 ⊕ (𝟤𝑛 − 𝟣) = 𝟣𝟧 − 𝑥 (left), and 𝑥 with 𝑥 ⊕ 𝟥 (right). chalkdust

1011102

Where can we go from here? We have unlocked some, but by no means all of the mysteries presented by images like the Hamming fractal, and circle visualisations. There are many symmetries left to uncover, and fascinating questions to ask about these images. Why is there a white circle at the centre? How is the size of this circle related to 𝑛 and the chosen Hamming distance? Why does there seem to be more white space toward the centre of the circle pictures? What can be said about the structure of the white space around the centre circle, for example the eight ovals surrounding it? While it is not at all obvious from the circle picture for 𝑛 = 𝟦, 𝑘 = 𝟤, this graph actually has two connected components shown on the right. Is it always the case for these graphs? If not, under what conditions is it the case? Can the number of connected components be described in terms of the numerical properties of the vertex numbers they contain? When do these graphs have other interesting properties; for example, when are they bipartite? Just two examples to illustrate the variety of circle virtualisations are below.

▲ The circle visualisations for 𝐵𝟩 if 𝐻 (𝑖, 𝑗) = 𝟧 (left), and 𝐵𝟪 if 𝐻 (𝑖, 𝑗) = 𝟤 (right). The Wolfram demonstration projectsf referenced at the bottom of the page are freely available and offer a platform for quickly generating some of these images. I hope you share my enthusiasm for these pictures and will feel inspired to explore them yourself. Chris Boucher Chris teaches and learns at Salem State University in Salem, Massachusetts, USA. He has interest and some expertise in probability, mathematical computing, and data visualisation. He also has interest, but less expertise, in basketball and home improvement.

s chris-boucher-24aa6a f d demonstrations.wolfram.com/DistanceAndDissimilarityPatterns by Enrique Zeleny (2014).

d demonstrations.wolfram.com/PatternsProducedByHammingDistance by Chris Boucher (2020). 1011112

spring 2022

Significant figures

Gladys West Madeleine Hall looks at one of the minds behind GPS

hen was the last time you used a satnav in your car? Tagged your location on a social media post? Or used maps on your phone to figure out where to go? Allow me to introduce a significant figure to you, without whom none of those things would be possible.

Gladys West was a pioneer in geodesy—the branch of mathematics that deals with the precise shape and area of the Earth. Her work was critical in the development of global positioning systems (GPS). Specifically, her work was used to put together extremely precise models of the Earth’s shape, enabling GPS technology which impacts every single one of us on the planet.

Go West Gladys West, born 27 October 1930, grew up in rural Virginia on her family’s small farm. Her mother worked at a tobacco factory, and her father worked for the railroad while also being a farmer, growing mostly corn, cotton, and tobacco. The Great Depression in the late 1920s and early 1930s hit Virginia chalkdust

farmers especially hard, as smoking declined significantly during this time. When recalling her childhood, West describes waking early to perform work on the farm before going to school for the day, and upon returning from school, having to complete more farm work before doing homework and having dinner. Speaking about her childhood, she has remarked that her family were “rich when it came to having good food and good meals, but when it came to money, we were not so rich”. In her community the typical options for a young black girl’s future were continuing to farm or working at a tobacco processing plant, like her mother. West knew from an early age that these options were not for her, saying “I realised I had to get an education to get out”. She had an unwavering belief and determination that her life path was taking her far beyond her family’s farm. West graduated high school in 1948, first in her graduating class, which awarded her a full scholarship to Virginia State College (now University), of which there were only two available. Virginia State, founded in 1882, was a black college at the time. The institution did not become racially integrated until the American civil rights movement (1954–1968). At Virginia State College, Gladys West majored in mathematics, and was a member of the Alpha Kappa Alpha sorority, the first intercollegiate historically African American sorority.

Boston Public Library, CC BY 2.0

▲ Virginia State College in the 1930s

Gladys West graduated from Virginia State in 1952 with a bachelor’s in mathematics. She began working as a maths and science teacher immediately after graduation in the small town of Waverly, Virginia. She worked there for two years, before returning to Virginia State College to undertake a master’s in mathematics, from which she graduated in 1955.

In the navy A pivotal moment came in 1956 when she was hired by a weapons laboratory: the US naval proving ground in Dahlgren, Virginia. During the second world war the proving ground had received groundbreaking (pun intended) early computers to help with ballistic work—the field of mechanics concerned with the launching, flight behaviour and impact effects of projectiles and ranged weapons like bullets or bombs.

▲ A Fortran punch card

West was the second black woman ever hired there, and one of only four black employees. She was a human computer, tasked with solving complex mathematical equations by hand, before transitioning to programming the computers there to solve the mathematical equations for her. This programming involved punching holes in stiff pieces of paper termed ‘punch cards’ (typically only 80 characters or 29 centitweets long). These punch cards were then stacked into decks (this was the ‘computer program’), which were then fed into the machine to tell it what calculations to do. 49

spring 2022

One of West’s first major achievements was participating in the awardwinning astronomical study that proved the regularity of Pluto’s motion relative to Neptune in the early 1960s (which deserves its own whole Chalkdust article IMHO). Following this study, she became project manager for Seasat. Seasat was the first satellite designed to obtain data on a range of ocean features, including wave height, water temperature, currents, winds, icebergs, and coastal characteristics. ▲ Seasat

She went on to program an IBM 7030 Stretch computer, which was significantly faster than other machines at the time, to provide increasingly precise calculations to model the Earth’s shape. Many mathematical and physical factors needed to be taken into account for her models, including gravitational forces, tidal forces, as well as the Doppler effect. The Doppler effect needs to be taken into account because the relative movement between a GPS receiver and a GPS satellite causes the received signal to shift in frequency—if the satellite is travelling towards the receiver when sending the signal, the received signal is ‘squished’, whereas in contrast, if the satellite is moving away from the receiver, the received signal is ‘stretched’.

GPS is actually the name given to specifically the United States-owned global navigation satellite system (GNSS). As of September 2020, Russia’s GNSS (known as Glonass), China’s BeiDou NSS (known as BDS), and the EU’s Galileo are also fully operational GNSSs. That means there are four fully operational GNSSs in orbit, with the Indian Regional NSS (IRNSS) also planning to expand to a global version in the long term (too many satellites orbiting the Earth? technological overkill? politically necessary?)

Geoid heights (the shape of the ocean surface) derived from Seasat data were adjusted according to all of these factors, in order to determine a precise mean radius of the Earth. In her paper Mean Earth ellipsoid determined from Seasat 1 altimetric observations, published in 1982, she calculated the mean radius of the Earth to be precisely 6,378.1349 km. In a later paper, Determination of ocean geodetic data from Geosat (1987), satellite measurements corrected for orbital and environmental effects resulted in extremely precise geoid height contours and three-dimensional projections of the Earth’s surface. This detailed mathematical model of the shape of the Earth was a building block for what would become GPS technology. Of course, the satellites themselves fundamentally enable GPS technology to work. However, without the understanding of the Earth’s surface, it would be like building a record player without the vinyl being invented: the technology is useless without the subject of its analysis. During her time at the naval proving ground, Gladys West attended evening classes and gained a second master’s degree in public administration from the University of Oklahoma. West retired from the proving ground (at the time of her retirement named a naval surface warfare centre) in 1998, after 42 years of working there. chalkdust

ES Ince, F Barthelmes et al, CC BY 4.0

▲ Geoid undulation in false colour, to scale.

Though she was retired, she continued her education, and was awarded her PhD from Virginia Tech in public administration and policy affairs in 2000, aged 70. Her doctoral thesis, titled The effects of downsizing on survivors, analysed the effects and repercussions faced by those who remain employed, or ‘survive’ redundancy programmes in the workplace.

I wanna shake your hand West’s contributions to GPS technology went largely Don DeBold, CC BY 2.0 unremarked for many years. In the late 2010s, she sent ▲ Part of the console for the IBM 7030 a short autobiography to an Alpha Kappa Alpha sororStretch computer ity function, which was celebrating the achievements of members past and present. West’s sorority sisters were amazed by her story. “I just thought it was my work, and we’d never talk to our friends about work. I just never thought about it. I didn’t brag about what I was working on,” she has said. “But to see other people so excited about it, that was amazing.” One of her sorority sisters, determined to get West’s story told, shared the autobiography in early 2018 with the Associated Press, which kickstarted the domino effect of revealing West as among the so-called hidden figures that were computers for the US military in the 1960s.

▲ Gladys West being inducted into the the Space and Missile Pioneers Hall of Fame

In mid 2018, West was nominated and won the award for female alumna of the year at the annual Historically Black Colleges and Universities awards. In December 2018, she was inducted into the US air force’s hall of fame for space and missile pioneers in recognition of her work on the extremely accurate geodetic Earth model. This induction is one of the space command’s highest honours, paying tribute to the leaders of the early years of the programme, as well as the subsequent innovators. In 2020, Gladys West published her memoirs, It Began With a Dream, and in 2021, she was awarded the Prince Philip medal by the UK’s Royal Academy of Engineering, their highest individual honour.

In 2021, looking back on her life, she said “I keep changing my mind about the highlight of my career. I would think first, when I got that college degree and I knew that I was ready. Nowadays, I’m just thankful for being able to see the recognition that other people are giving me and how much I have affected other people. To me, that is a big thing.” Madeleine Hall Madeleine is a PhD student in mathematics and behavioural phenomics at Imperial College London. She is currently writing her thesis. It is going more slowly than she would like.

a @maddygracehall

spring 2022

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

Slow Horses (Apple TV+) An excellent series, based on the excellent first book in Mick Herron’s excellent series. Jackson Lamb is played by Gary Oldman, even though Bill Nighy (in a fat suit) or Robbie Coltrane (in a thin suit) were the more obvious choices.

ggggh Have I Got News For You

Baba is You

They pick some great magazines as guest publications in the missing words round.

Innovative puzzle game which has the player exploring a deep world of logical propositions and interactions in a fun and engaging way.

ggggh

ggggh British Applied Mathematics Colloquium

Clopen Mic Night #2: Clopen Winterval

Loving the Loughborough life and people’s attempts to pronounce Loughborough.

Another very entertaining live YouTube variety show.

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Matt Parker’s Humble Pi Cambridge Junction Very entertaining maths stand up show. We recommend getting the DVD once it’s out.

A Podcast of Unneccesary Detail Back for a second series. As good as the first series.

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Wordle

Bridgerton

ggggg ggggg ggggg ggggg

Back for a second series. As good as the first series.

iiiii

chalkdust

Chalkdust Book of the Year 2021 How to Think about Abstract Algebra by Lara Alcock How to Think About Abstract Algebra is a book aimed at a very specific target audience: people in the early days of an undergraduate maths degree. It isn’t an algebra textbook; instead it guides the reader through some of the things they should be doing to make more sense of the kind of pure maths they are about to meet. We wish we’d had this book when we were starting our undergraduate degrees. This book was selected by the editors of Chalkdust to be the Chalkdust Book of the Year 2021, based on our four judging criteria: style, control, damage and aggression.

Chalkdust Readers’ Choice 2021 Maths Tricks to Blow Your Mind by Kyle D Evans Maths Tricks to Blow Your Mind looks at a wide range of problems and puzzles that have gone viral on social media. As anyone who’s seen one of Kyle’s shows would expect, it’s both understandable and amusing. We very strongly recommend this book to mathematicians and nonmathematicians alike. This book was voted by our readers to be the Chalkdust Readers’ Choice 2021.

Shortlisted The winners were selected from our shortlist of seven books released in 2021. The five non-winning books are all also very good. They were:

Math Without Numbers by Milo Beckman; Danny Chung Does Not Do Maths by Maisie Chan; How to Read Numbers by Tom Chivers and David Chivers; Weirdest Maths by David Darling and Agnijo Banerjee; and At Sixes and Sevens by Rachel Riley. 53

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Accidentally mathematical songs ♭♯♮♭♯♮♭♯♮♭♯♮♭♯♮♭♯♮♭♯♮♭♯♮♭♯♮♭♯♮♭♯♮ + − ÷ × + − ÷ × + − ÷ × + − ÷ × + − ÷ × +𝟣 ♪♫♪♫♪♫♪♫♪♫♪

Girls & Boys by Blur

The chorus of Girls & Boys by Britpop band Blur goes: Girls who are boys Who like boys to be girls Who do boys like they’re girls Who do girls like they’re boys. Always should be someone you really love. If you take the sequence of ‘girls’ (G) and ‘boys’ (B), you get: GBBGBGGB. This follows the Thue–Morse sequence. The Thue–Morse sequence is represented as a string of binary digits. The sequence can be extended by taking the Boolean complement (take the opposite to each position) of the current string and adding this to the end of the string. We start with 0, and add the Boolean complement (1) to obtain the string 01. Now we have 01, and add the Boolean complement (10) to get 0110. Repeating this a few times, we get the sequence 0110100110010110… In fact, you could take any string of ones and zeros from the Thue-Morse sequence, and it won’t appear again until after the string finishes. For example, take the string 1101 which appears near the start. The next time this string appears is in position 14, without overlapping with the original string: 0110100110010110100101100… In fact, you could take any series of ones and zeros in the Thue–Morse sequence, and it too won’t overlap with the same series later in the sequence. To prove this in a hand-wavy wayf , suppose that the Thue–Morse sequence is overlap-free until a string 𝐴 that contains an overlap appears (for example, 𝐴 could be 10101 as it contains overlapping 101s). But by the nature of the sequence, the Boolean complement of 𝐴 (01010 in our example) must have already appeared in the sequence, which means there must have been an earlier overlap. This is a contradiction, so the Thue–Morse sequence must be overlap free. QED.

Replacing 0 and 1 with G and B, we see that the start of the Thue–Morse sequence is the same as the Girls & Boys sequence. As such, we can extend the chorus: Girls who are boys Who like boys to be girls Who do boys like they’re girls Who do girls like they’re boys Who need boys like they’re girls Who need girls like they’re boys Who have girls who are boys Who have boys who are girls Who choose boys who see girls Who choose girls who see boys Who meet girls who like boys Who meet boys who like girls

f Proving this fully is a little harder than this. Can you spot which bit of the proof needs a bit more work?

Who verb girls who verb boys Who verb boys who verb girls

Always should be someone you really love. chalkdust

Falling to Pieces by Faith No More In 1990, American rock band Faith No More, released the single Falling to Pieces, from their album The Real Thing. Within the song is the verse: From the bottom, it looks like a steep incline, From the top, another downhill slope of mine, But I know, the equilibrium’s there. Which makes me think of the intermediate value theorem. The intermediate value theorem states: ‘If 𝑓 is a continuous function whose domain contains the interval [𝑎, 𝑏] then it takes any given value between 𝑓 (𝑎) and 𝑓 (𝑏) at some point within the interval.’ A corollary of this theorem is that if you have the same conditions, then there must exist 𝑐 ∈ [𝑎, 𝑏] such that 𝑓 (𝑐) = (𝑓 (𝑎) + 𝑓 (𝑏))/𝟤. It looks like another downhill slope of mine.

𝑓 (𝑏) 𝑓 (𝑎)+𝑓 (𝑏) 𝟤

The equilibrium

It looks like a steep incline.

𝑓 (𝑎) 𝑎

𝑐

𝑏

But how does this relate to the song? Well, if we take ‘the bottom’ to be a point at 𝑥 = 𝑎 and ‘the top’ to be a point at 𝑥 = 𝑏 , then we can define a function between them. As stated, to use the intermediate value theorem we need the function to be continuous; since it is a slope, I think it is fair to say that the surface rarely breaks—you never have a hill where suddenly halfway up you find a huge gap in the floor! The corollary of the intermediate value theorem tells us that there must be a point exactly halfway up the vertical cross section. In other words, ‘the equilibrium’ is there; exactly as Faith No More’s conjecture stated. Goran Newsum Goran is a mathematician who likes 90s music almost as much as he likes maths. His other interests include railways, trains and gaming. Inspired by taking part in the online celebration for Chalkdust issue 12, Goran streams virtual Street Countdown on a semi-regular basis on YouTube.

a @GoranNewsum m youtube.com/channel/UCQ72lw4RlgmqreU93b_HSnw 55

spring 2022

Cryptic crossword

8 10

11 12

14 15 16

19 22

#3, set by Logisca

25 27

Across

Down

1 By morning, Somerville will be a fields medallist. (6) 4 It’s Mats Vermeeren’s favourite mathematician (7) that is debased. Not her! 8 In a direction somewhat antipodal to 17. (5) 9 Elects to decrease one degree by an order of (7) magnitude (ie a tenth). 10 Established time zone. (1,1,1) 12 A mark by a learner is situated on an axis. (4) 13 Magazine editors at heart regularly should (with (9) slack) organise. 15 Imaginary degree from learned society. (1,1,1) 16 The value approached is an extremity by the (3) sounds of it. 20 Two yields level best… (4,5) 22 …a single time produces no change nowadays. (4) 24 Starts of lemmas, examples, and theorems, and (3) the beginnings of some proofs. 26 European adopting one French mathematician. (7) 27 See 27. (5) 28 Bronze man is touching but not crossing. (7) 29 Easing treatment provided by the witch. (6)

1 Men lead current top three in science, causing results of tension. 2 Revolutionary hands around a mixed acid. 3 Race around a large area. 5 Campaign division, say. 6 Helium-59? According to Sophie Bleau, it’s a circle! 7 Outcome of Ulster riots. 11 This could be you? (leftmost of all of the answers ultimately). 12 A computing pioneer whichever way you look. 14 A measurement of American money. First nickels, then assorted lead-free coins 16 Algebra of untruth? 17 One initially Cartesian plane transforms to historic polar opposite of 8. 18 Othello’s ancestor is about the fifth king in Spain? Yes! 19 Nothing new, argued uninitiated. 21 About 45 inches?! Sounds like hell. 23 It’s about time to fix back gate by church. 25 Odd terming is used when discussing triangles?

(7) (7) (4) (9) (5) (6) (3) (3) (9) (3) (7) (7) (6) (3) (5) (4)

Slaying the dragon Callum Ilkiw takes us through a Dungeons and Dragons dice dilemma

gainst all the odds, your plan has worked. Your thief snuck in ahead of the team and got a surprise backstab, the paladin’s shield deflected the would-be deadly dragon’s fire breath and a well-aimed wizard’s frostbolt took out the columns of the building. You move your dwarven barbarian in for the killing blow. The trapped red dragon looks up with anger in its eyes. You know, all that stands between this creature and death is a single good hit of your axe. Even with all the work from the other players, the difficulty class of landing a hit on this creature is 16, meaning you will only hit it with a roll of 16 or higher on a d20 (a 20-sided die for those not in the know). That’s a 𝟣/𝟦 chance of hitting. Luckily, there is one last hope. The druid spent their turn giving you the help action. This means your next attack has advantage. Advantage allows you to now roll two d20s and choose the highest of those two values. This is obviously helpful, but what are the new odds that you will be able to give this dragon a one-way ticket to the afterlife? 57

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Sure, the odds that anyone reading this has even found themselves in that specific situation are pretty low. But over the course of lockdown, many new players found themselves taking to the famous tabletop role-playing game Dungeons and Dragons, known to fans as ‘DnD’. Whilst playing a typical weekly game of DnD, I had a similar situation to the one above. For anyone out there who has never played or watched a game of DnD, here is the main thing you need to know: almost every action in the game requires the roll of a 20-sided die to determine how well you did. This is called a skill check. Each of these skill checks has a difficulty class, which is the number you need to overcome to succeed in the roll. A higher difficulty class means the check is more difficult. For example, persuading a merchant to give you a discount might have a class of eight, but trying to stealthily steal his supplies would have a class of 20. A roll of 10 would pass the first case but fail the other. In some cases, the player might also have a situational bonus to reflect what is happening around them. For example, sneaking past from 1 to 20. In this picture, the merchant might be made considerably easier by having another a 2 has been rolled. player distract them. In the rules of DnD, this is called advantage. If, however, the situation is not in players’ favour, eg persuading a guard you insulted moments earlier, the player would have disadvantage. When applying advantage or disadvantage the player rolls two d20s instead of one. If the player has advantage, they take the highest of the two values shown. If the player has disadvantage, they take the lower of the two. The mathematically inclined may be wondering, how do your odds change when rolling with advantage or disadvantage? ▲ A d20 has faces numbered

Let’s start with a simpler version of this problem. The smallest die used in DnD is four-sided. When we roll two four-sided dice, there are 16 possible results: (𝟣, 𝟣)

(𝟣, 𝟤)

(𝟣, 𝟥)

(𝟣, 𝟦)

(𝟥, 𝟣)

(𝟥, 𝟤)

(𝟥, 𝟥)

(𝟥, 𝟦)

(𝟤, 𝟣)

(𝟤, 𝟤)

(𝟦, 𝟣)

(𝟦, 𝟤)

(𝟤, 𝟥)

(𝟤, 𝟦)

(𝟦, 𝟥)

(𝟦, 𝟦).

If you score the highest of the two rolled values, we can compute the following odds: 𝑛 1 2 3 4

𝑃(scoring 𝑛 with advantage) 𝟣/𝟣𝟨 𝟥/𝟣𝟨 𝟧/𝟣𝟨 𝟩/𝟣𝟨

Wintermute115, CC BY-SA 4.0

▲ A d4. As there is no face at the top, the number you have rolled is written around the top point.

Hazarding a guess, it looks as if each score 𝑛 has 𝟤𝑛 − 𝟣 ways of getting it. Let’s check this makes mathematical sense. Any pair (𝑎, 𝑏) will give a score of 𝑛 if at least one of 𝑎 or 𝑏 is equal to 𝑛 and the other is less than or equal chalkdust

to 𝑛. If 𝑎 = 𝑛, there are 𝑛 possible options for 𝑏 . If 𝑏 = 𝑛, there are 𝑛 options for 𝑎. That’s 𝟤𝑛 possible ways of getting 𝑛, but we counted the pair (𝑛, 𝑛) twice in those two lists, meaning the real number of ways to score 𝑛 is 𝟤𝑛 − 𝟣. So, when rolling advantage on a 𝑦 -sided dice, the probability is given by 𝑃(𝑁 = 𝑛) = (𝟤𝑛 − 𝟣)/𝑦 𝟤 . When rolling with disadvantage, this becomes 𝑃(𝑁 = 𝑛) = (𝟤𝑦 + 𝟣 − 𝟤𝑛)/𝑦 𝟤 . For a d20, in the case of advantage, this will give: 𝑛 1 2 3 4 5 6 7 8 9 10

𝑛 11 12 13 14 15 16 17 18 19 20

𝑃(scoring 𝑛 with advantage) 𝟣/𝟦𝟢𝟢 𝟥/𝟦𝟢𝟢 𝟧/𝟦𝟢𝟢 𝟩/𝟦𝟢𝟢 𝟫/𝟦𝟢𝟢 𝟣𝟣/𝟦𝟢𝟢 𝟣𝟥/𝟦𝟢𝟢 𝟣𝟧/𝟦𝟢𝟢 𝟣𝟩/𝟦𝟢𝟢 𝟣𝟫/𝟦𝟢𝟢

𝑃(scoring 𝑛 with advantage) 𝟤𝟣/𝟦𝟢𝟢 𝟤𝟥/𝟦𝟢𝟢 𝟤𝟧/𝟦𝟢𝟢 𝟤𝟩/𝟦𝟢𝟢 𝟤𝟫/𝟦𝟢𝟢 𝟥𝟣/𝟦𝟢𝟢 𝟥𝟥/𝟦𝟢𝟢 𝟥𝟧/𝟦𝟢𝟢 𝟥𝟩/𝟦𝟢𝟢 𝟥𝟫/𝟦𝟢𝟢

In the original scenario, our hero needed to roll a 16 or more, and only had a one in four chance of success. With advantage, the new odds of success become (𝟥𝟣 + 𝟥𝟥 + 𝟥𝟧 + 𝟥𝟩 + 𝟥𝟫)/𝟦𝟢𝟢 = 𝟣𝟩𝟧/𝟦𝟢𝟢 or about 44%. While still not a likely win, the chances of success have gone from one in four to almost one in two. With hope in their heart, the barbarian goes to swing down their axe, and you prepare to roll the two d20s in your hand. Wait!—your friend sitting next to you reminds you that your dwarven barbarian has the lucky feat. This feat allows you to roll three d20s and take the highest when rolling with advantage, instead of the pitiful two that most players would get. Let’s see if we can take our probability equation for advantage and generalise it for any number of dice.

Generalisation Generalisation is the inevitable step any maths problem reaches once the initial problem is solved. In this case, I want to extend our original problem to ask: If you rolled 𝑥 dice, each with 𝑦 sides, what is the probability of rolling any positive whole number 𝑛?

The problem now extends from one of probability to one of combinatorics. Combinatorics is best thought of as the study of how many ways you can combine a set of elements. In this problem, we can best express the result of the roll as a vector of scores (𝑛𝟣 , … , 𝑛𝑥 ), where 𝑓 (𝑛) = max(𝑛𝟣 , … , 𝑛𝑥 ) is the result of using advantage with 𝑥 dice. Each 𝑛𝑖 is drawn randomly, uniformly and independently from the set (𝟣, 𝟤, 𝟥, … , 𝑦 − 𝟣, 𝑦). There are many ways to begin a problem like this, but I like to start case-by-case. When 𝑛 = 𝟣, there’s only one way, all ones: (𝟣, 𝟣, 𝟣, … , 𝟣, 𝟣).

What about when 𝑛 = 𝟤? There are 𝟤𝑥 ways to get only ones and twos in our roll, due to there being only two options for each of the 𝑥 slots. This answer includes the option of all ones that we’ve already counted, hence there are 𝟤𝑥 − 𝟣 ways of scoring two. 59

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And when 𝑛 = 𝟥? This is achieved by any combination of all threes, twos and ones, excluding the combinations that don’t have any threes. This gives 𝟥𝑥 − (𝟤𝑥 − 𝟣) − 𝟣 = 𝟥𝑥 − 𝟤𝑥 ways of getting a three.

There seems to be a general rule emerging, so let’s hypothesise that the number of ways to score 𝑛 is 𝑓 (𝑛) ≔ 𝑛𝑥 − (𝑛 − 𝟣)𝑥 .

Proof by induction For the base case, we select 𝑛 = 𝟣: There is only one way of scoring one regardless of number of dice and 𝑓 (𝟣) = 𝟣.

Next, we assume that the number of way to score 𝑘 is given by 𝑓 (𝑘) and show that our formula is also true for 𝑛 = 𝑘 + 𝟣. If this proposal holds for all values 𝑘 and lower, there are (𝑘 + 𝟣)𝑥 rolls that involve 𝑘 only numbers 𝑘 + 𝟣 and lower. We also know that ∑𝑗=𝟣 (𝑗 𝑥 − (𝑗 − 𝟣)𝑥 ) = 𝑘 𝑥 of these contain no instance 𝑥 𝑥 of 𝑘 + 𝟣. Therefore, there must be (𝑘 + 𝟣) − (𝑘) ways of scoring 𝑘 , which matches our hypothesis.

We have proven that when you roll 𝑥 dice with 𝑦 sides each, and take only the highest result, your odds of getting any score 𝑛 (⩽ 𝑦) is given by 𝑃(𝑛) = (𝑛𝑥 − (𝑛 − 𝟣)𝑥 ) /𝑦 𝑥 .

Now, this is a good step, but in DnD we are rarely concerned with getting an exact score. So, we can use this equation to find the probability of getting at least 𝑛 when rolling 𝑥 𝑦 -sided dice with advantage: 𝑦

𝑥

𝑦 𝑥 − (𝑛 − 𝟣)𝑥 𝟣 𝑛−𝟣 𝑃 (at least 𝑛) = 𝑥 ∑ (𝑗 𝑥 − (𝑗 − 𝟣)𝑥 ) = =𝟣−( ) . 𝑥 𝑦 𝑗=𝑛 𝑦 𝑦

I know what you’re all wondering, what does this mean for our barbarian? Well, when rolling normal advantage in DnD, we always use a d20, so that means 𝑦 = 𝟤𝟢. This gives the equation of 𝑃(at least 𝑛) = 𝟣 − ((𝑛 − 𝟣)/𝟤𝟢)𝑥 . Therefore, the odds of rolling more than 16 when rolling three d20s with advantage are 𝑃(at least 16) = 𝟣 − (𝟣𝟧/𝟤𝟢)𝟥 = 𝟣 − 𝟤𝟩/𝟨𝟦 = 𝟥𝟩/𝟨𝟦 = 𝟧𝟩.𝟪%. This means the odds of success are suddenly more than 50%. You regather your composure and thank your friend for the advice. As you grip the three d20s in your hand, so too does your barbarian tighten his grip around the greataxe passed down through many generations of his bloodline. You roll a 6, an 11 and… a lucky 19 and slice the dragon in twain. This is a success. A success made easy with the help of combinatorial mathematics—the best kind.

How do you want to do this? With the conclusion of this story so too comes the conclusion of this article. For those of you who play DnD, I hope this advice comes in handy during your next game and spares your hero from an untimely meeting with death. For those of you that have never played, maybe seeing the maths working behind the scenes has persuaded you to pick up some dice and try the game for yourself. Either way, I wish you the best of luck in your next quest. But be wary adventurers, here be dragons. Callum Ilkiw Callum is a maths graduate from both Keele and Warwick, with an interest in all things nerdy. From DnD to drama games, he likes to find the maths hidden away in plain sight.

c clilkiw@gmail.com a @himecallum chalkdust

spring 2022

Sophie Bleau uncovers the secrets behind covering maps

’m sure some of you reading this may know about topology vaguely as “that one where the doughnut is the same as the mug”. And you would be right. In essence, topology is maths after you have forgotten what distance is, meaning that space is squishier than we’re used to. Then, the only defining feature of a mug is the hole between its body and its handle and the only defining feature of a doughnut is… well, its hole. A lot of topology is about saying whether one shape is the same as another (such as doughnuts and mugs). Today we’re going to be going a step further; to ask when two shapes look the same ‘up close’, despite not necessarily being the same overall. For example, if we zoom in closely on any part of a circle, it will look like a line. However these shapes are not the same, since a circle is not a line. We’ll discuss this in more detail soon. The question I want to answer is whether there is a way to intuitively create a shape that ‘covers’ another. I put it to you that there is. If shape 𝐴 covers a base shape 𝐵, there must be a way to map 𝐴 to 𝐵 such that • the base 𝐵 is entirely covered by 𝐴 (ie this is a surjective map); and • by zooomin chalkdust

g in on any small patch of 𝐵, the map looks like the identity map. 62

▲ The infinite plane covers the torus: the colouring has been added to illustrate how the covering map works. For instance, we can say that a plane covers a torus. To visualise this, imagine taking the plane, rolling it up into a cylinder, then rolling down the edges of the cylinder, sort of like rolling down a sock. As the edges loop round, the resulting surface is—lo and behold—a torus! This way, every part of the torus is covered (satisfying surjectivity), and every small part of the plane looks like the surface of a torus if you zoom in close enough. Another example of this is the circle. Any circle can cover another circle by winding round it some number of times. You can imagine that this is like wrapping an elastic band around your finger. On the right we have a circle mapping continuously onto a circle a third of its length. But we can think bigger; the real number line can cover a circle by twisting it round into a helix. This satisfies surjectivity since every part of the circle is covered, and, as mentioned above, every part of the helix looks like a circle if you zoom in close enough. However, a finite interval cannot cover a circle since if we zoom in on the interval’s endpoints, we can’t find any point on the circle that looks like this. Don’t worry though, our bounded interval can still play with other bounded intervals.

Infinity undercover So, we’ve seen how to cover lines and circles, but how does this covering business work when lines intersect? The infinity symbol, ∞—since I always like to be able to call a symbol something in my head, we’ll pronounce this ‘infty’—has two edges intersecting at one vertex. This is an example of a graph: a shape consisting of vertices and edges. Each vertex in a graph has a valency, ie the number of ‘arms’ it has. So if two edges meet at a vertex, they become four ‘arms’, and the intersection is a point with valency 4. Now we can reinterpret the way we think about covers; a circle cannot cover infty since zooming in on the infty at the intersection we have a point where four lines join together, whereas no point on the circle has this property. One idea is to cover this with another infty, using the identity map. But, to make other covers, let’s look at more graphs that are 4-valent at each vertex. To make sure that we are not mixing up the two arms of each edge, let’s give each edge a direction so 63

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we can easily differentiate the ‘source’ end of the edge and the ‘target’… end. Also colouring the edges is a simple way of making sure that we distinguish the two different edges.

A constraint for a graph to cover infty is that every vertex in the cover must have two red arrows (one source, one target) and two blue arrows (likewise). Let’s call the graph with arrows a directed graph, or a digraph for short. An example of such a cover is this:

From this we get infty by double-looping the blue-edged loop so that the two vertices end up on top of each other, with red edge on red edge and blue edge on blue. An example of a case where this does not hold is on the right. This does not work for a few reasons. You might have noticed that its vertex has too many arms, or that there is more than one blue arrow pointing towards/away from the vertex. Is there some other way to colour and direct the graph to make it a cover of infty? Well, whichever colouring we choose, we will always have more than two of one colour. So sadly, I topologise: this graph is not destined to cover infinity.

Inftys all the way down Another cover is the infinite grid (top of next page). A way to conceptualise this map is thinking similarly to the example of the plane rolling up to make the torus. First we roll the graph into a vertical cylinder, then we roll the edges of this cylinder down so that each rung is mapped onto the same one. chalkdust

Up until this point, taking a red path then a blue path in each cover will end up at the same point as taking a blue path and then red (try this out on the infinite grid). But can we break this pattern? Here we find a special case I call the fractal cross, where the cover is so intricate that the path between any two vertices is unique. Try starting at any vertex on the graph, and mark out the path travelling red then blue. Now go back and do the same with blue then red. Your destination depends on the order of your steps! I call this the fractal cross because it is infinitely detailed—like a fractal. The fractal cross covers infty in a similar way to the infinite grid. This, *drum roll please*, is the largest cover of infty. More precisely, it covers every other cover of infty and nothing else covers it.

From cross to cover All covers of infty can actually be derived from the fractal cross. As an example, if we take the central blue line of the graph, we can first wrap it around itself in the way that a line covers a circle; and then glue together all the branching red arms, and we will get this cover:

By repeatedly gluing and warping the fractal cross we can find more covers, and eventually get down to infty. For instance, let’s again take a blue-edged line on the fractal cross, but this time wind it around 65

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until every third blue edge in the line is glued together creating a loop. Now do the same with a rededged line intersecting this loop. Next glue the vertices of the loops together one by one. As you can see, this forms a triangle with vertices intersecting a circle. Finally this graph with six edges collapses down onto infty.

Illuminati confirmed? Every cover we’ve looked at so far has been 4-valent; in fact we have now recognised the 4-valency of every vertex in the cover as a necessary condition to be a cover. But is it a sufficient condition? If we have a 4-valent graph, will it necessarily cover infinity? We can generate complex, and often infinite covers of infty, all just by checking all our rules are satisfied. In this way, we have a surprising amount of flexibility to build up such a cover. You might enquire further; how much flexibility? Can we explore some artistic licence? For example, would a Celtic knot cover infty? A Celtic knot is a traditional weaving pattern, which is drawn like a rope tied in a knot, with the notable property that the two rope ends are stuck together. If we shine a torch on such a knot and take a picture of its shadow, we can imagine that the resulting silhouette is a graph. All vertices of this shadow graph will be 4-valent since it is the junction of two lines crossing. So… does this cover infty? A beautiful Celtic knot designed by Leonardo da Vinci is shown on the right. On the next page, I have drawn the digraph associated with the centre piece, verifying both that it indeed covers infty, and that I have far too much time on my hands. In fact, we find that a Celtic knot’s shadow will always qualify as a cover of infty. This is to say that there will always be a digraph of the knot satisfying the requirements—each vertex having a blue source, blue target, red source, and red target. chalkdust

For the interested reader, the fact that the shadow of any knot diagram can be made to cover infty follows from a few neat ideas in knot theory. First, the shadow of a knot diagram is, by definition a planar graph, and it is possible to ‘chessboard colour’ the complementary regions of this graph, ie colour those regions black or white so that no adjacent regions have the same colour. (There are several ways to prove this, one way being to prove that the graph ‘dual’ to the complementary regions contains no odd-length cycles by the definition of a knot.) Now choose an orientation of the knot, which gives an orientation of each edge in the shadow graph. Assign an edge the colour red if the region on its right is black, and blue otherwise. In this way the graph can be directed and coloured, and it is straightforward to check that every vertex has four arms, two red and two blue, and one incoming and one outgoing of each colour. This means the graph covers infty. In fact one can even prove that every regular 4-valent graph covers infty, but this is trickier, using the so-called 2-factor theorem.

It is fascinating that an artistic piece could relate to such a mathematical idea as a topological cover, and all of this stemming from the theory of a torus being considered topologically identical to a mug— who would have thought? The realisation that all knots (apart from the circle) cover infty is non-trivial and—in my opinion—really cool. However, as cool as topological space is, just make sure you don’t try pouring any tea in your doughnut! Sophie Bleau Sophie is a third year maths student at the University of Edinburgh. She spends her free time drawing Celtic knots and fangirling over Leonardo da Vinci. She has three cats (if Sophie has 𝑛 cats, the qualification for being a ‘crazy cat lady’ is always 𝑛 + 𝟣 cats) called Pushkin, Norma, and Idgie, who take great pleasure disrupting any mathematical progress—usually by sitting on it.

c sophie.bleau27@gmail.com 67

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The crossnumber sponsored by

#15, set by Humbug

0 0

0 0 0

Each of the letters 𝒓 , 𝒔, 𝒕 , 𝒖, 𝒗 , 𝒘 , 𝒙 , 𝒚 , and 𝒛 represents a different integer between 1 and 9 (inclusive). The clues are provided as normal, but the locations of black squares in the grid are not given. Instead, you are given the position of every zero in the completed crossnumber. The completed grid has order 4 rotational symmetry. The starting squares for the clues are numbered from 1 to 60 in the usual order: starting in the top left corner and proceeding left to right for each row. There is only one solution to the completed crossnumber. Solvers may wish to use the OEIS, Python, an abacus, 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 all the digits in the row marked by arrows by 30 September 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 November 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

Across 1 3

Down

𝒓𝟥

(3)

(𝟣𝟢 + 𝒓)𝟤

(3)

𝒓𝟤

(2)

𝟣𝟣𝒕

(3)

10 𝒛 × 𝒓 × 𝒔 × 𝒕

(3)

14 (𝟨𝟢 + 𝒖) × (𝟣𝟫𝟢 + 𝒕)

(5)

12 𝒛 𝟤

(2)

16 𝟤𝒛 × 𝒓 − 𝒔

(3)

20 𝟤𝟧𝟤𝟢 ÷ 𝒓

(3)

23 𝟥𝟥𝟩 × 𝒙

(4)

(3)

18 𝒕 × 𝒗 × 𝒘 × 𝒔 × 𝒛 21 𝟣𝟣𝟢 × 𝒗

(2)

3 𝒕 × (𝟣𝟢𝟢 + 𝒔)

(3)

5 𝟤𝟦𝟤𝟦 ÷ 𝒚

(3)

7 𝟤𝟫 × 𝒛

(3)

𝟤𝒔

(3)

4 𝒚×𝒛

(2)

6 𝒔×𝒙

(2)

9 𝟣𝟣𝟣𝟣 × 𝒕

(4)

(3)

5 (𝟪 + 𝒓) × (𝟤𝟢 + 𝒓)

1 𝒓 ×𝟥

11 (𝟣𝟢 + 𝒗) × (𝟥𝟥𝟢 + 𝒗)

(4)

14 𝒖 × 𝒔 × 𝒓

(3)

17 𝟣𝟣𝟣𝟣𝟣 × 𝒗

(5)

13 𝟣𝟢𝟢 × 𝒔 15

(3)

𝟣𝟥𝟥 × 𝒚 𝟤

(4)

𝟥𝒙

(3)

25 𝟥𝟥𝟩 × 𝒘

(4)

26 𝟩𝟫𝟤 ÷ 𝒚

(2) (3)

22 𝟣𝟣𝟢 + 𝒗

(3)

+ 𝒙)𝟤

29 𝟣𝟥 × 𝒔

(3)

26 𝟣𝟢𝟥 × 𝒔

(3)

+ 𝒔)𝟤

(3)

27 (𝒓

31 𝟤𝟢𝟤𝟤 × 𝒗

(4)

34 𝟣𝟣𝟣 × 𝒛

(3)

𝒘𝟥

36 (𝒖 38

+ 𝒛)𝟤

+ 𝒚)𝟤

(3)

28 (𝒕

(3)

30 𝟣𝟢𝟣 × 𝒛

(2)

33 𝟣𝟢𝟣 × 𝒘

(3)

(4)

39 𝟣𝟣𝟣𝟣 × 𝒕

(3)

27 (𝒘

(4)

32 𝟨𝟩𝟩 × 𝒘

24 𝟤𝟧𝟤𝟢 ÷ 𝒔

41 𝟨𝟩 × 𝒓 × 𝒛

(4)

𝒔𝒕

×𝒛×𝒙 ×𝒚

35 𝒙! 37

𝒛𝟦

𝒖𝒖

(3)

44 𝟤𝟧𝟤𝟢 ÷ 𝒘

(3)

45 𝟣𝟧𝟣 × 𝒘

(3)

40 (𝟧𝟢 + 𝒕) × (𝟧𝟢 + 𝒓)

47 (𝟤𝟧 − 𝒘) × (𝟤𝟧 + 𝒘)

(3)

49 (𝟣𝟣𝟢 + 𝒚) × (𝟣𝟣𝟢 + 𝒘)

(5)

54 56 57 58

𝟣𝟦𝒕

(3)

+𝒛

𝒓𝟥

×𝒕

𝒕𝟢

+ 𝒕𝟣

𝒖𝒕

(2)

59 𝟫𝟫𝟢 ÷ 𝒓

+ 𝒕𝟥

+ 𝒕𝟦

+ 𝒕𝟧

60 (𝟤𝟢 + 𝒚) × (𝟤𝟢 − 𝒚)

+ 𝒕𝟪

+ 𝒕𝟫

−𝟣

43 𝟫𝟫 + 𝒕

(3)

𝟤𝒛

(3)

42 𝟣𝟣𝟣 × 𝒖

(3)

46 𝟣𝟢𝟤 × 𝒘

(3)

𝟤𝒙

−𝟣

(2)

51 (𝟫𝟫 + 𝒕) × (𝒚 − 𝒓)

(3) (3) (2)

𝒕𝒖

(2)

57 69

×𝒓 +𝒚

53 𝒙 × 𝒚

(3)

(4) (4)

(3)

41 𝒙 × 𝒘 × 𝒔 × 𝒛 × 𝒓 ÷ 𝒕

(3)

(3) (4)

43 𝟨𝟢𝟨 ÷ 𝒙

52 𝒙 + 𝒚

(5)

𝟤𝒓

−𝟣

(2)

chalkdust

Opinions

◀ An uncomfortable truth Lucy Rycroft-Smith and Darren Macey unpick the legacy of some of the most ubiquitous names in statistics

tatisticians have the wholly unreasonable reputation of being the reprobates of the mathematical world: the drummers of the band, the viola players of the orchestra, the Chalkdusts of the scientific magazines. Much of this contempt seems to stem from statistics not being seen as a ‘real’ discipline within mathematics; statistics is so often the butt of mathematical jokes that it has spawned an entire joke genre of the ‘lies, damned lies and statistics’ type (for example, ‘A statistician is a person who draws a mathematically precise line from an unwarranted assumption to a foregone conclusion’, or ‘Did you know that 87.166253% of all statistics claim a precision of results that is not justified by the method employed?’). A power struggle as to whether statistics is a field of mathematics, an entirely separate field, or a science (or something else) has been ongoing for decades and shows no sign of abating. It may come as little surprise, dear readers, that many of of us with a love of mathematics seem to also have a love for forensic debate about boundaries, classification, and domains. But there is another, far more urgent power struggle that bridges the worlds of statistics and mathematics, too: the question of how we respond to the emergence of unpleasant historical details surrounding the lives of those we have previously considered heroes of our discipline. This is not a hypothetical question. Books and research papers from the last few decades have increasingly recognised that the founding fathers of statistics built the field on a dark and sinister foundation. The fact is: statistics as we know it is so intertwined with eugenics as to be inseparable. It could be said, in fact, that every time we speak the names of the men who gave their names to the statistical tests we might see as so innocuous, we are breathing life back into the memory of racists, obsessed with measuring ‘difference from the norm’, who built a ‘scientific’ reputation by metaphorically climbing on top of the broken bodies and reputations of entire peoples that they considered less than human. Don’t believe us? In Germany, an investigation into the workings of the organisation now called the Max Planck Society, before 1948 known as the Kaiser Wilhelm society, suggested scientists and academics there played a key role in legitimising Hitler’s ideology of racial purity. Not only did they contribute to the theoretical basis for Nazi politics, but in some cases they helped to set up concentration camps, gas chambers and conducted unimaginably cruel experiments on incarcerated humans. But this is Nazi Germany, you might say; nothing so evil could happen on British soil. (But you would be wrong.) Angela Saini writes in their bestseller Superior: “The truth: that it is perfectly possible for prominent scientists to be racists, to murder, to abuse both people and knowledge—doesn’t sit easily with the way we like to think about scientific research. We imagine that it’s above politics, that it is a noble, 71

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rational, and objective endeavour, untainted by feelings or prejudice…[but] science sits at the mercy of the personal political beliefs of those carrying it out.” Some of the scientists most influential to the thinking inherent in the Holocaust worked at UCL in Gower Street, London, and give their names to statistical measures we continue to teach in classrooms and lecture halls up and down the UK.

The genesis of eugenics Francis Galton was the cousin of Charles Darwin. He coined the term ‘eugenics’ (‘good breeding’) to describe the process of “improving stock, which is by no means confined to questions of judicious mating, but which, especially in the case of man, takes cognisance of all influences that tend in however remote a degree to give the more suitable races or strains of blood a better chance of prevailing speedily over the less suitable than they otherwise would have had.” As you can see, the word ‘improving’ is used here in a way that depends on a hierarchical framework of ‘suitability’ of ‘races’—in other words, white supremacy. The US National Human Genome Research Institute defines eugenics as “the scientifically inaccurate theory that humans can be improved through selective breeding of populations. Eugenicists believed in a prejudiced and incorrect understanding of Mendelian genetics that claimed abstract human qualities (eg intelligence and social behaviours) were inherited in a simple fashion.” While the scientific rebuttal alone should be sufficient to discredit eugenics as a discipline, there is a further ethical issue here: the underlying model of white supremacy. White supremacy is the idea that whiteness constitutes superiority and confers rightful power; but this concept of ‘whiteness’ is not just about skin colour. “Racism is based on the concept of whiteness—a powerful fiction enforced by power and violence. Whiteness is a constantly shifting boundary separating those who are entitled to have certain privileges from those whose exploitation and vulnerability to violence is justified by their not being white.” (Kivel, Uprooting Racism, 2011.) This means that various peoples throughout history have been marginalised and oppressed as part of white supremacy, even those who might in other contexts be considered ‘white’. This is because race is not a scientific concept—it is a social construct, and remarkably hard to pin down. It is, as Jenée Desmond-Harris says, neither “permanent, scientific, objective, logical, consistent, [nor] able to stand up to scrutiny.” So why does race loom so large in our society and our histories? Desmond-Harris also argues, along with many others, that the concept of race itself was invented by scientists to justify slavery.

Intersections with racism Galton’s father made money from the slave trade, which he inherited and which allowed him to work as an explorer and a scientist. Galton travelled to Africa, describing the people who lived there as ‘savage races’. In our office, we have a Galton board: a lovely object, a dynamic normal graph, showing as you rotate it hundreds of bouncing, jostling ball bearings which eventually settle into a beautiful bell curve (see opposite). But both the words ‘Galton’ and ‘bell curve’ (the name of a 1994 book by Herrnstein and Murray explicitly connecting race and intelligence) don’t feel right in our mouths. They taste horrible. Would we be better off ignorant, not worrying about these issues and just enjoying the mathematical ideas here? Perhaps, as individuals. But, as Aris Winger, assistant professor of mathematics at Georgia Gwinnett College, says, ‘How white people interrogate their own racial identity is a requirement for chalkdust

any meaningful racial, professional or personal progress in maths. The future of the discipline is tied to whiteness, how we understand whiteness, and white supremacy.’ Because we, as white people, have the luxury of not having to deal with racism in our professional lives—no one has ever assumed we are not mathematicians because of the colour of our skin. So we also have a painting in our office—the one you see displayed here— which helps remind us of some of these issues.

Colin Beveridge

▲ A Galton board in action: ball bearings are poured through the top and bounce through the pegs into a bell curve beneath.

Karl Pearson is often described as the father of statistics; working as recently as the 1900s, in many ways he founded the discipline. He was the first professor of national eugenics at UCL. He was racist and antisemitic, writing that Jewish immigrants were a ‘parasitic race’ and that those who were not white were ‘inferior’.

His namesake, the Pearson product-moment correlation coefficient (or PMCC) is something we have both taught, and used in our research—it is the first port of call for those investigating linear correlation, or asking ‘might these two numbery things have a relationship?’ This concept has many, many names— but the ‘Pearson’ part seems remarkably sticky. It’s almost as if someone is doing a PR job behind the scenes. It is remarkable how insistent people can be on proper credit and value for scientific work when said work involves white men, and how these principles seem to vanish without trace when minoritised people are involved. It’s almost as if we have a skewed (stats joke) idea of what constitutes authority and status based on kyriarchy (the social system that keeps all intersecting oppressions in place).

Implications and complexities In a world where toppling statues of slavers has become international news, it is worth considering what a scientist must do in order that we stop hallowing them in curricula, in assessment, in the simple act of speaking their name. Is a mathematical breakthrough ever neutral? Are there figures in our discipline for whom their ideologies are so distasteful that presenting their work to students, or in our research without proper contextualisation is in fact unethical, particularly in a subject like statistics where the two seem so very intertwined? It is impossible, of course, and indeed undesirable to remove mathematicians’ ideas from the wider picture; but at what point does it become reasonable to strike their names from their most famous methods? Remove their faces from our textbooks? Or at least include historical notes alongside their mathematical legacies telling the dark stories of their distasteful ideologies which, in some cases, allowed their work to continue? If Galton’s work was created on the backs of murdered and dehumanised slaves, should all those using it be aware of this fact? If Pearson’s legacy was built on the idea that humans can be used as data without regard for their humanity, should that history lesson not stand next to his statistical methods, as a warning for all future statistics students not to make the same mistakes? 73

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Indeed, these are not just hypothetical questions: in 2020, UCL announced that it was renaming buildings which had been named after both Pearson and Galton because of their connection with eugenics, and they have said this is one of many measures they plan to take in this vein. Of course the venerable institution with which we are associated, Cambridge University, does not escape critique; for example, there have been calls for the removal of the stained glass window in Gonville & Caius College which commemorates statistician Ronald Fisher, who was similarly enamoured of eugenics. In 2020, Daniela Witten, professor of statistics at the University of Washington, ran a successful campaign for the Committee of Presidents of Statistical Societies to remove Fisher’s name from the Fisher award and lectureship (now called the COPSS distinguished achievement award and lectureship), saying “If Quaker Oats can retire Aunt Jemima, then statistics can retire the Fisher lecture. I hope that this Caitlin Hobbs, CC BY 3.0 tiny little gesture of a name change can be ▲ The empty pedestal of the statue of Edward Colston just the beginning of a much larger converin Bristol, the day after its toppling in June 2020. sation in our field about how we can work to create a statistical community that is committed to our shared goals of diversity, equity, and inclusion.” The felling of one stone monument to the slaver Edward Colston signalled for many the beginning of the important work of questioning which other figures’ more metaphorical pedestalling might similarly be challenged. The Colston statue was toppled by protesters in June 2020 in the context of longstanding campaigns by Bristol residents to remove references to Colston throughout the city. They advocated to remove the statue, or even just to have a small plaque added, pointing out that the 84,000 African lives he helped sell into slavery were worth considering as a counterpoint to his much celebrated ‘philanthropy’. However, the veneration of history’s white men has a certain inertia, and multiple attempts to draft a plaque that satisfied everybody failed spectacularly, with one local councillor claiming that vandalising it would be an appropriate course of action should it ever appear. The plaque was never mounted.

Possible futures But what does this mean for mathematics, and in particular, statistics? Cases like that of Colston’s statue and others such as Cecil Rhodes’ statue outside Oriel College, Oxford, seem to demonstrate that long dead men can still count on the support of powerful institutions, individuals, and even government ministers whenever people have the temerity to threaten to peel a layer or two of varnish from their legacies and expose the rotten cores beneath. As mathematicians, maths communicators, and maths educators seeking to do the same within our discipline, perhaps we should expect the same forces chalkdust

to swing into action to defend the ‘good’ names of Galton, Pearson, Fisher and their ilk. How then do we develop the collective power of our community to call for change? What should our demands even be? And how do we counter the arguments that it was ever thus; that these people were born in different times; and that we should not seek to ‘vandalise’ our history by tit-for-tat retaliatory silences, overwrites, and heavy-handed edits? We do not have the solution. But we are mathematicians (and statisticians, and we do not differentiate between the two): to us this is familiar ground. We humbly suggest the same approach we take to our beloved discipline of mathematics when we are stuck: listen, learn, think, spend time with the problem; most of all, seek out greater expertise. One of the places we often turn to when navigating these issues is the excellent podcast Mathematically Uncensored. In the seemingly complicated and sometimes fraught discussion about inclusive practices in mathematics education, Aris Winger and Pamela E Harris, associate professor of mathematics at Williams College in Massachusetts and featured in Chalkdust issue 10, challenge us with a very simple question: what does it actually cost us to be inclusive? If we are only prepared to take action that is comfortable, that does not materially cost us in some way, then are we really making any meaningful challenge to the status quo? Ultimately, no matter how horrified we might be at reading some of these biographies, if we cannot even ‘give up’ somehow the names of some long dead men, or take action to prioritise the inclusion of those whose contributions have been historically marginalised or completely erased—do we really care at all?

The artwork at the start of this article is a painting by Lucy Rycroft-Smith. The name of the piece is Galton Under Fire (2021), so named in reference to the frustration the artist feels at the apparently fireproof reputation of some mathematicians, Francis Galton included. It poses the question: how unethical do someone’s actions have to be before we choose to take action against them as a mathematical community, and what should that action be?

Lucy Rycroft-Smith Lucy Rycroft-Smith is a writer, researcher, speaker and designer in mathematics education. She is studying for her PhD at the University of Cambridge, considering the idea of knowledge brokering—how, what, when, where and why we communicate research to mathematics teachers. Her other research interests include gender, sexualities and identity, feminism, decolonisation, curriculum design, board gaming, teaching discrete mathematics, representations of mathematics, maths and the arts, and professional development. She is the author of The Equal Classroom: LifeChanging Thinking About Gender (2019) and co-editor of Flip the System UK: A Teachers’ Manifesto (2017).

a @honeypisquared Darren Macey Darren Macey is a writer, researcher, speaker and designer in mathematics education. He is studying for his PhD at the University of Cambridge, researching how neoliberal drivers in education can be subverted to enhance teaching and learning. His other research interests include teacher agency, professional development, curriculum design, and decolonisation. He is the co-author of Teaching Statistics (2018).

a @darrenmacey 75

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