Chalkdust, Issue 09

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In this issue... Features 4 Nerding out with Matt Parker Interviewing Ma was a mistake

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Playing billiards with cue-bics Yuliya Nesterova misses all the pockets

16

When Truchet met Chladni Stephen Muirhead meets neither

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The phantom parabola Zoe Griffiths talks to ghosts

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Counting caterpillars Peter Rowle plays with a toy

38

Hiding in plain sight Axel Kerbec gets locked out

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Striking the right chord Paula Rowińska goes round in circles

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Bells, braids, and taxicabs Andrew Stacey links three strands

38 Regulars Page 3 model 3 8 What’s hot and what’s not 14 Dear Dirichlet 24 Which symbol are you? 36 Horoscope 46 Puzzles 53 Reviews 54 Oπnions: Erdős numbers

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by Lucy Rycro-Smith

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54 1

59 60

Academic webpage checklist

62 64 72

Prize crossnumber

On the cover by Mahew Scroggs and Edmund Harriss Letters Top ten: Chalkdust regulars spring 2019


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chalkdust The team Lara Anisman Carmen Cabrera Arnau James Cann Hugo Castillo Sánchez Atheeta Ching Thùy Dương “TD” Ðặng Eleanor Doman Evita Gamber Sean Jamshidi Nikoleta Kalaydzhieva Jean Lagacé Emily Maw Sam Porri Tom Rivlin Mahew Scroggs David Sheard Belgin Seymenoğlu Adam Townsend Cover adapted from Harriss spiral by Edmund Harriss

d chalkdustmagazine.com c contact@chalkdustmagazine.com a @chalkdustmag b chalkdustmag l chalkdustmag n @chalkdustmag@mathstodon.xyz e Chalkdust Magazine, Department of Mathematics, UCL, Gower Street, London WC1E 6BT, UK.

Welcome to Chalkdust issue 09! Did you miss us? Once upon a time, in central London, a team of friends decided to write another magazine. They put out the call, and in came articles from across the land—a mix of new ideas and old favourites. For the first time they published an Oπnions piece (pp 54–58), to beer represent the community that they serve. Other articles came from old companions whom they had worked with before, and one (When Truchet met Chladni, pp 16–23) united two heroes from adventures past. In this issue, they extended an invitation to their readers to take part in competitions, puzzles and challenges. To win a book (In conversation with Ma Parker, pp 4–7), a bag of loot (the crossnumber, pp 62–63) or a T-shirt (Striking the right chord, pp 48–52). They also dabbled in occult maers, to revive a mysterious tradition from days gone by (horoscope, pp 36–37). But! The friends did not just produce a magazine. They also travelled around the city, sharing their love of numbers. Three of the team ventured into the land beyond the river, to an event promoting mathematics for girls. They baled dragons, derivatives and deadlines to spread the word, and returned home triumphant. On their return they were granted three wishes: that their proofs come quickly, there typos are few, and that the mailbag continues to be filled with articles from you, the fairy godparents who make sure the friends always have more wonderful content to share. And they all lived mathsily ever aer. The Chalkdust team

Acknowledgements Massive thanks to: all our authors for continuing to send us the highest quality content; our sponsors for allowing us to continue making this magazine; Albert Wood for his help with graphics; Helen Wilson, Helen Higgins, Luciano Rila, and everyone else at UCL’s Department of Mathematics; Ma Parker and Sarah Cooper for their help with distribution; Gulnaz Abdullah for organising the girls in maths event; Nazar Miheisi, David Lammy, Nira Chamberlain, and Angus Grogono for their contributions to the Black Mathematician Month event; Jo Morgan for inviting us to #latemaths and the Humble Pi book launch. To calculate how weighty our massive thanks are, multiply them by 9.8 ms−2 . 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 2019. All rights reserved. If you wish to reproduce any content, please contact us at Chalkdust Magazine, Dept of Mathematics, UCL, Gower Street, London WC1E 6BT, UK or email contact@chalkdustmagazine.com

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Be warned: this article is dark and full of spoilers.

Eleanor Doman

O

NE of the best parts of getting into a series is getting to know and love the main characters. However, in Game of Thrones (or A Song of Ice and Fire, for you purists), this can be a heart-breaking activity. Who will survive to the end and who will bite the dust? No one knows, but perhaps maths can lend a hand.

Andrew Beveridge and Jie Shan used network theory to investigate who the main characters of Game of Thrones are. The diagram above shows all the interactions between characters during the seventh series: the larger characters are more central, as determined by the PageRank algorithm. However, it only takes one swing of an axe to drastically change the network... Pictures Welsh dragon: Wikimedia commons user Sodacan, CC BY-SA 3.0; Ice road: Ian Mackenzie, CC BY 2.0; GoT S7 network: Andrew Beveridge.

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In conversation with. . .

Matt Parker

TD Ðă.ng and Sean Jamshidi

M

 Parker is a nerd, and proud of it. So nerdy that, in a list of the nerdiest things he’s ever done, hiking for three days through the Australian outback to document a ‘confluence point’—an integer intersection of a line of latitude and a line of longitude—“might not even make the top ten”. A quick scroll through his YouTube channel (m standupmaths) shows videos with titles like Back to the fax machine, Speed Rubik’s cubing for drunk people and Standup comedy about equations that correspond to vortex motion. In the laer, he makes jokes about integration and pokes fun at physicists for calling a torus a doughnut. The video also has over 300,000 views, so clearly there is a huge audience for his brand of nerdiness. When we met Ma, on a grey evening in late January, he was about to embark on a week of talks around the country where he would share his enthusiasm for maths at festivals and in schools. “I want people to learn stuff, sure, but I’m also trying to do a bit of maths PR. I want to make the nerdy kids cool for the day.”

An unexpected journey Although Ma has a clear mission now, he admits this wasn’t always the case. “Basically, I didn’t have a career plan. I’d done a bit of tutoring at university and loved that, so I got my teaching qualification and moved to the UK.” He had always been interested in making videos and writing comedy, and while working as a teacher he realised he was missing that creative outlet. So, the stand-up mathematician was born. At first, his comedy routine played off the fact that he was a chalkdustmagazine.com

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chalkdust mathematician “who expects life to be logical.” Being a teacher also helped: “When you say you’re a maths teacher then, for beer or for worse, people have an image of that. And you can play with it.” Gradually the comedy became nerdier, he started being more involved with university engagement programmes, and eventually in 2012 he le teaching and made the transition to being a full-time outreach mathematician. These days, Ma gives talks to nerds and non-nerds alike. “If I can make a room full of nerds laugh at a joke about spreadsheets, that’s great. But in some sense, that’s the audience.” Working with kids, on the other hand, is much more challenging and requires careful judgment. “The moment they think you’re trying to impress them, it’s all over.” You have to pick your jokes wisely, and make sure that the educational content still comes through. If it works though, and you can get teenagers who don’t want to be there accidentally enjoying themselves while they learn about mathematics, then it’s a “much bigger achievement.”

Ma grew up in Australia, which famously celebrates Christmas at the wrong time of year.

It’s clear that Ma’s time in teaching has had an effect on the work he does today. His goal is to keep his sessions relevant to what teachers are doing at school, and make them a part of the experience, which “keeps the excitement going” once the day is over. “I don’t want to do shock and awe mathematics, where they go back to the classroom and say, why aren’t our lessons like that guy’s?” This is also the reason that he started MathsJam, the monthly meet-up series that brings together mathematicians of all stripes in pubs around the world. “It is really important for maths outreach people to keep in touch with teachers, to find out what they need and whether what we’re doing is useful.” It’s a great approach to outreach: something that fits within the system and complements the everyday job of educators, rather than creating a division between ‘fun maths’ and ‘school maths’.

It is really important for maths outreach people to keep in touch with teachers…to check what we’re doing is useful.

Video made the mathematician In recent years, a lot of Ma’s most well-known output has been through YouTube videos, either from his own account or as a guest on the mathematics channel Numberphile. Here again we find him carefully balancing his different audiences. “I would be disappointed if my videos were only watched by maths people, but likewise I would be disappointed if I don’t occasionally put out videos for that demographic.” Sometimes Ma can tell that a topic is going to provoke interest among non-mathematicians. One example of this is a video with Hannah Fry titled The mathematics of winning Monopoly, where they crunch the numbers to work out which squares are worth buying. At other times, the internet doesn’t react the way you would expect and a video that took a lot of effort to produce barely makes a splash. 5

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chalkdust So some videos land well and others don’t, but being mathematicians we felt the best way to understand Ma’s career as a YouTuber was to take an average. Specifically, to watch his median video, according to quality. “I made a video once about ordinal and cardinal numbers. And it was fine. Just, fine.” Cardinal numbers are used to count how many of something there are (you can have zero apples, one orange, three pineapples, etc) and ordinal numbers order items into first, second, third. Ma wanted the video finished in time so that he could show it at a talk, where he would test his audience’s powers of self-organisation. Everybody who came to the talk was given a slip of paper with a number wrien on, and the challenge was for them to comment on the video online in the correct order. “It got to over a hundred, before I released the video to the public and they broke it.” Over the course of more than one hundred videos, Ma’s YouTube channel has covered all sorts of mathematical and mathematically-related topics in an entertaining, accessible way. But there are some topics that he won’t ever make a video on. The Riemann hypothesis is a classic example of a popular, famous topic that he feels just won’t work on YouTube. “First of all, you’re going to have to do Matt Parker ploing a complex function… and the Basel problem, and “Cardinal numbers are so-called bewhat the zeta function is generalising. Suddenly you recause they are holy. As in, they are alise there’s so much background stuff, that a video is just whole numbers.” not the right format.” This desire to tell a story all the way through is what led to Ma’s first book: Things to Make and Do in the Fourth Dimension, which was published in 2014. Although the book deals with lots of different mathematical topics, ranging from knots to different size infinities, Ma was “seing up everything he needed to do the Riemann hypothesis properly.”

Making mistakes Ma’s new book, released in March 2019, is called Humble Pi. The book is framed as a collection of errors in mathematics: times when people made a mistake and faced real-world consequences. But actually, Ma says, the real aim was to show people how maths is “incredibly useful and underpins society.” By telling stories and anecdotes about these mistakes, he hopes to reach people who “wouldn’t normally pick up a maths book” and show them that maths is everywhere… just you might only notice it when it goes wrong. The stories contained in the book range from the serious to the sublime. For example, in 1997 the cruiser USS Yorktown was le powerless during training Penguin books manoeuvres aer a crew member tried to divide by zero and crashed all the computer systems that controlled the engines. Another chapter tells of a flight in Canada that had to make an emergency landing aer ground crew twice used the wrong amount of fuel. Their mistake? To calculate the amount needed in kilograms, and then load the tank with that many pounds. But Ma was cautious not to fill the book with disaster tales. “Nobody dies in any of chalkdustmagazine.com

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chalkdust the [aviation] stories. It’s safe to read if you’re scared of flying.” Some of them—like the tale of Steve Null, whose unfortunate name was incompatible with the company database—are just there to tickle you. And yes, the book does include a mistake that Ma himself has made. The ‘Parker square’ is an aempt at making a 3-by-3 magic square using only square numbers. He suspected that his methods weren’t perfect, but he thought it would be fun so he gave it a go. And, lo and behold, his answer included a mistake (check the diagonal sums in the image on the right). Ma uses the Parker square to teach us another important lesson: mathematics is oen about making mistakes. “Mathematicians are not people who find maths easy, they’re people who enjoy that it’s difficult… we make mistakes all the time. People wear T-shirts that confirm this.”

292 12 472 412 372 12 232 412 292 “A metaphor for something that is almost right, but a lile off.”

Another part of Ma’s life these days is that he regularly signs calculators. What began as a lighthearted joke now sees hundreds of school children around the country with their names Sharpie-d in the encoding standard Ascii on their devices. Ma takes great pleasure in doing this; not only is it an ironic celebration of nerdiness, and gives the students some maths to do if they want to read his message, it serves as a reminder of their experience that will last for months or years aerwards. “If they get so involved that they want to get their calculators signed, then I think that’s hilarious.” As a mathematician and comedian, Ma seems to be in the middle of two very different worlds. But he’s happy there, describing it as a “stable equilibrium”. He has many outlets for his creative side, and says that he could “probably find something else to scratch that itch if he wasn’t doing comedy.” But as for maths, there’s nothing that can replace it. “It’s much more pervasive in your life. It’s something that you’re always thinking about.” Even when you make mistakes, “it’s worth puing the effort in, because when you get it right it’s just so useful. And so fun.” Would you like to win a signed copy of Humble Pi? Tell us about a time when you’ve made a mistake in mathematics by sending an email to c contact@chalkdustmagazine.com before 9 September 2019, and Ma will pick a winner! TD Ðă.ng and Sean Jamshidi TD and Sean are students at UCL. They have both made many, many mistakes in mathematics.

a @televisionduck a @sean_jamshidi My favourite symbol Maths wouldn’t be maths without algebra. And algebra wouldn’t be algebra without symbols. We’ve spread some of our favourite and not-so-favourite symbols throughout this issue.

ε

Sam Porritt

It’s always positive but not in an over-the-top, in-your-face kind of way. Rating: (10 − ε)/10 7

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WHAT’S

& WHAT’S

HOT NOT The 10-issue challenge

We challenge you to write an article for issue 10 of Chalkdust.

HOT

The 10-year challenge

NOT

What’s so special about base 10 anyway?

Maths is a fickle world. Stay à la mode with our guide to the latest trends. Agree? Disagree? a @chalkdustmag b chalkdustmag l chalkdustmag f chalkdustmag

HOT Conferences in exotic places

Conferences in Milton Keynes

A two-week conference on the isle of Saint Marie? Yes please!

NOT

How about a week in MK? Not in the slightest, Florence.

NOT

A new largest prime number

As Brenda from Bristol brilliantly put it: “You’re joking—not another one!”

A new largest compound number

You could fill a warehouse with people looking for a new largest composite number and call it a factory.

HOT

Puzzles on Twitter

HOT

Twitter is full of excellent puzzles thanks to a @CShearer41, a @solvemymaths, a @panlepan, a @puzzlecritic and others.

Top ten

Puzzles on Facebook

No one cares that only a genius can do BIDMAS.

NOT

HOT

See page 72.

What’s hot and what’s not

See page 8.

NOT

More free fashion advice online at d chalkdustmagazine.com

Pictures MK roundabout: Chris Nyborg, CC BY-SA 3.0; Puzzle: a @CShearer41

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Playing billiards with cue-bics Yuliya Nesterova

I

 patches sit in wait and bare trees shiver in anticipation but spring is still dawdling, likely enjoying a final cuppa before going back to work. What are we to do while days are dreary and the wind rummages through empty branches? Why, stay inside and play with geometry soware, of course! Yet winter was long and snowy, and aer one too many hours indoors the treasure trove of maths is perhaps running low. What to do once ancient theorems are re-proved and the excitement of a parametric plot has wilted? Well, there is still a nugget of algebra that could flourish into hours of exploratory fun. So grab your Geometer’s Sketchpads, your d geogebra.com, or simply some transparencies and coloured markers, and let’s play billiards with polynomials. In the vast sea of beautiful mathematics it is quite easy to drown, finding more and more interesting papers on a subject, eventually geing lost in the plethora of niy theorems and charming proofs. It’s best to restrict ourselves early on in the game if we are to make an aernoon of it. Let’s say for today that we consider only the polynomials of degree 3. Armed with our restriction—cubics only!—let us dive into the deep end of adventure. 9

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Build a billiards table To play a good long game of billiards, you must first build a pool table (as Carl Sagan might have it). Look at your cubic polynomial, for now remaining mysteriously general: a3 x 3 + a2 x 2 + a1 x + a0 . First, from the origin in the positive x-direction, draw a line segment a3 units long. At the end of it, turn 90° anticlockwise and draw a line segment of length a2 . Turn 90° anticlockwise again to plot a1 , and do the same for a0 . Call the final point you arrive at Ω, where you build a corner pocket. If ever a coefficient is negative, you ‘backtrack’, ploing out a segment backwards but still remembering the direction in which you face for the next turn. Now for the playing part: the segments constructed, and the lines extending them, have become walls. Our objective is to launch a ball from the origin and hit three walls, sinking it into Ω, much like in three-cushion billiards. In the words of our indices, “3, 2, 1, sc0re”! But on this special table, the physics is peculiar: our ball makes a 90°-turn each time it bounces off a wall. Ω

+

a0 = −13

+

a3 = 1

a2 = −5 a1 = 17 Here is a path on x 3 − 5x 2 + 17x − 13. We try different launch angles α3 from the origin (3 is for cubics), bouncing against walls (hiing the corners if necessary), until we find the winning path to Ω. Once a path is found, − tan α3 is the real root of the polynomial! The path in the picture corresponds to the root − tan α3 = − tan(− π2 ) = 1. A visual way to factor polynomials through a geometric game.

You can handle the truth Perhaps it is far more rewarding to know why this method works than it is to practise it. If you want the magic and mystery to remain, pause now until you’re ready. Otherwise, let’s dive deep into geometry and fish out some nuggets of truth. chalkdustmagazine.com

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chalkdust With the choice for each coefficient of whether to go forwards (+), stay and turn (coefficient of 0) or to backtrack (−) along each segment, we need casework to do the proof properly. But with long cold nights in our near future, we will talk business here and leave the details to sele into place in due time with some well-chosen parameters. Let’s consider first the ideal case: all coefficients positive, a3 x 3 + a2 x 2 + a1 x + a0 , our billiard ball makes two right angles to land perfectly at Ω, the endpoint of a0 . In our 90°-bounce universe, the billiard ball path forms a series of 3 similar right-angled triangles. In each triangle, call sk the side opposite angle α3 , starting with s3 for the triangle at the origin. Let’s trace out the trigonometry.

a1

s2

α3

α3

a0

a2 s3

α3

Ω We start with s3 = a3 tan α3 . In the next triangle: s2 = (tan α3 )(a2 − a3 tan α3 ) = a2 tan α3 − a3 tan2 α3 , and s1 = a0 = (tan α3 )(a1 − s2 ) = a1 tan α3 − a2 tan2 α3 + a3 tan3 α3 .

a3

Replacing x = − tan α3 , we get exactly our polynomial, a0 + a1 x + a2 x 2 + a3 x 3 = 0, so x = − tan α3 is indeed a root of the original polynomial. Simple enough. Ever catch a side length of zero? A length, you will agree, unacceptable for a stable pool table (or a pool). And yet the maths works out! If an ak = 0, we generalise: draw a line where segment ak would have been: a ball hiing a line-wall outside the segment ak will go right through the wall, still making a 90° angle with the path coming in. Not much of a table in this case and not much is le of billiards but the procedure still delivers a visual way to find real roots. Three examples with zero coefficients are showcased below. s3 = a3 tan α

s3 = a3 tan α

s2 = s3 tan α

s2 = (s3 − a2 ) tan α

= a3 tan2 α a0 = (a3 tan2 α + a1 ) tan α

= a3 tan3 α + a1 tan α

a0 = s2 tan α

= a3 tan3 α − a2 tan2 α

The proof for a triple-root cubic is too short to mention, though a beautiful picture would do it credit: a0 = a3 tan3 (α)

Ω 1 1

−3

9

Ω x 3 − 3x + 2

2

1

2

−2

Ω

x 3 − 2x 2 − 9 11

x3 + 2

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How the tables have turned Good news for the completionists in the audience: should you chance upon a cubic with three real roots, you can extend the process, building the walls of a new billiard table on the path of the ball’s old trajectory, continuing onwards: the ball’s path you have drawn becomes the walls for a new polynomial. D

C

G α3 α2

α3

H

F α1

α2

α3

A B

Ω

In the diagram above, I have done just that with (x+3)(x+1)(x+2), beer known as x 3 +6x 2 +11x+ 6. The original rectangle of a pool table can be played on via the red path. Its angle α3 (measured in the usual manner: anticlockwise about the origin from a3 ) respects − tan α3 = −3, one of our roots. Build a second billiards table! For the three similar right-angled triangles with hypotenuses on the red path, we find the relationship of hypotenuses to be AF : FG : GΩ = 1 : 3 : 2. The path AF, FG, ΩG itself represents a polynomial (x 2 + 3x + 2)! Well, we couldn’t quite tell that it 299 + 2x298 , or, for that maer, 9x 2 + 27x + 18. In fact, wouldn’t been x 3 +√3x 2 + 2x, or√ x300 + 3x √ have √ √ it is 10x 2 + 3 10x + 2 10, where 10 is 1 + tan2 α3 , or a3 / cos α3 . You can find this from the diagram, or directly from the polynomial with a bit of work. Now, we jolly good shots have found an orange path to sink the ball: notice that − tan α2 = −FH/AF = −1: another one of our roots! Triangles AFG and FGH are once again similar, with the laer twice the size of the former, so AH/HΩ = 12 : the yellow path describes the path of x + 2 (up to multiplication by cx n , c ∈ R, n ∈ Z). We end by firing off a shot down the x-axis from A, to E, to victory, with − tan α1 = −2. Our roots are indeed, −3, −1, −2. For our friend from earlier, x 3 − 5x 2 + 17x − 13, there are no other paths to victory. This is a method for real roots, you see: √ there just aren’t triangles around whose opp/adj ratio is 2 ± 3i, for i = −1. But there are plenty of polynomials in the maths sea! The method works for polynomials of any degree, you will agree: nothing in our proofs was specific to cubics. One can factor continuously, perching a precarious sequence of billiards games one on top of the other in a whirlwind of a billiard tournament, until the real roots are all encoded by triangles for a straight and easy parting shot. The billiard game is an inefficient method, albeit a fun one, but with practice one gains enough intuition to seriously improve one’s game, and this is a valuable skill of visual factorisation: give it a try! chalkdustmagazine.com

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fo ch lo Be

This is the pastime which Austrian engineer Eduard Lill gied the maths world in 1867. Felix Klein publicised the cubic version in his 1926 book, Elementarmathematik vom höheren Standpunkte aus, II: Geometrie and for a while it seemed to feed its own craze, but eventually became forgoen. Interest was revived again in 2011 by Thomas Hull. His article tells also of Margherita Beloch, the Italian mathematician with a fascinating pastime, and of origami, and forgoen manuscripts, and much intricate maths besides. Within that paper, you’ll read that x 3 + 2 (pictured right) produced a Beloch square, which is itself constructible from a Beloch fold, itself one of three mutual tangents of two parabolas… But hark! Here there be quadratics.

ld

More to explore

x3 + 2

Ω

We have commied to a cubics-only rule in our dive and so we must emerge, niy pastime in hand. We have yet to exhaust the topic and if you wish to dive deeper, there are beautiful and obscure results stretching far and wide. A topic for another deep dive, I’m sure. Yuliya Nesterova Yuliya is a master’s student of algebraic geometry at een’s University, Canada. She likes to celebrate the informal mathematics that happens in the cracks of academia. She volunteers at Mathest camp, a mathematics camp for girls, creates Pi Day celebrations, and hosts Oawa’s MathsJams.

a @oawamathsjam d mathquest.ca My favourite symbol

φ

Hugo Castillo Sánchez and Atheeta Ching

We like this symbol because it looks fancier than the average leer and it’s fun to write! In LATEX, this is known as varphi, φ (as opposed to phi, ϕ) and we always go out of our way to make sure we use this one! Atheeta uses it for a surface which comes from the solution of a PDE and Hugo tends to use it in non-Newtonian fluid mechanics to represent fluidity. Rating: 10φ/10ϕ

We have a podcast Are you stuck in traffic? Are you about to start a long train journey? Are you looking for some maths to play loudly at 2am to annoy the neighbours? We have a solution for you: Talkdust, a podcast for the mathematically curious. You can find all the episodes so far at d chalkdustmagazine.com

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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, My husband and I have found our selves in a long-distance relation ship. His company offered him a large promo tion if he moved to Canada for six months, but it seems now that the position wil l require more time. I’m not sur e that I want to move out there with him, or tha t I could be happy knowing he had to move back here. Conversation with the tim e difference is hard as well. Any tips?

— Getting tensor, Four Oaks

DIRICHLET SAYS: This has been a common theme among correspondents

to my column, but thankfully technology offers some options. Part of your problem is that you are measuring time and distance using the flat spacetime metric, ds2 = -c2 dt2 + dx2 + dy2 + dz2 . Instead, get out your iPad and try the flat FaceTime metric. In my experience, this works better for special relatives.

Dear Dirichlet,

It has the department common room. in rk wo n ofte and e uat rad I’m an underg the smell of which is great – except I hate ne, chi ma ee coff a and ks boo desks, others. Reckon I s but get frosty looks from the coffee! I try to open the window r drink? can get them to switch to anothe e Ways

— Sick Paul, Fiv

■ DIRICHLET SAYS: Clearly the students have failed the tea-test. You could try getting in early and setting up a tea distribution. No promises though, it might just be that you and the others are too significantly different from each other. In which case, consider testing another location.

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Dear Dirichlet,

Did you know that it is ILLEGA L to sell triangles whose angles don’t add up to 180o ???? When I was growing up, our triangles had all sorts of inte rior angle sums and no-one got hurt. Now we’re being told that HEALTH & SAFETY laws mean that parallel lines can NEVER meet! Time for the fractal generation to grow up!

— Relative on Facebook, Essex

■ DIRICHLET SAYS: I see you look forward to leaving EUclidean geometry. But if you enjoy being obtuse, you could have just summed the Brexterior angles, for Brexample, instead of resulting to Brextrema. If you want me to Brexplicitly Brexpress what I Brexpect to come from this Brexercise, I fear that by Brextemporising, we are Brexposing ourselves to Brextinction. To Brexplain using predicate logic, Brexponential decay.

BRE

Dear Dirichlet,

fun and ld son. As a boy, he was always r-o yea 13a of ent par ned cer I am the con going through l in the box. But he seems to be too st rpe sha the not if us, uro advent always returning se for large periods of time, not a phase where he leaves the hou think he will be e strain on the household: do you until quite late. It’s putting a larg ters — Thirteen, THIRTEEN?!, Seven Sis done with this soon? ■ DIRICHLET SAYS: I am pleased to hear that your son is a bit thick and that he often doesn’t come back. As any rheologist will tell you, if he’s elastic, the stress you experience will be proportional to the strain; but if he’s viscous, the stress will be proportional to the strain rate. As long as you keep the strain at a constant level, therefore, you will feel considerably more relaxed.

Dear Dirichlet,

es? Any ideas for Halloween costum

I’m getting it sorted in advance

this year.

d

— David S Pumpkins, Leighton Buzzar

DIRICHLET SAYS: Captain Hooke’s law. Franken-spline’s monster. A scary badger. More Dear Dirichlet, including two seasonal specials, online at d chalkdustmagazine.com 15

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When Truchet met Chladni Waves, tiles and percolation theory Stephen Muirhead

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 catching up on past issues of Chalkdust over the Christmas break, I was intrigued to see that issue 07 featured an article on Chladni figures (On the cover: Chladni figures of a square drum by James Cann), followed in issue 08 by an article on Truchet tiles (Too good to be Truchet by Colin Beveridge). This coincidence fascinated me because, although not the focus of the articles, there is a remarkable conjecture that links these two apparently unrelated objects, via the mathematics of percolation theory.

Chladni figures and random waves As explained in issue 07, Chladni figures are the complex nodal paerns created by the zero-set of standing waves, illustrated most famously by Ernst Chladni in his 18th-century experiments with sand sprinkled on a vibrating plate. Chladni figures are of interest in many different domains of mathematics and science (for example, in quantum mechanics, oceanography, astrophysics etc). One problem faced in applications is that, as the frequency of the standing waves gets larger, the Chladni figures get more and more intricate and chaotic. A useful approximation is to consider ranchalkdustmagazine.com

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Chladni figures created by standing waves on a square plate.


chalkdust domised versions of standard waves, but this still raises the question of how to understand and describe the zero-set of these random waves (which we can think of as random Chladni figures).

A high frequency Chladni paern and a contour plot of the associated standing wave.

Random Truchet tilings As introduced in issue 08, Truchet tiles are a set of polygonal tiles, which can be square, hexagonal, or some other shape, coloured black and white such that the corners are alternating in colour. These tiles can be arranged in various ways to create complex and beautiful tilings, and were first described in the 18th-century by French polymath and priest Father Sébastian Truchet. In issue 08, the combinatorics of these tiles were considered, but it turns out that they are a useful tool with which to understand Chladni figures. In fact, there is an amazing conjecture, due to the physicists Bogomolny and Schmit in 2002, that says that random Chladni figures look identical, on large scales, to the paerns created by a random tiling of square Truchet tiles. To form this tiling, we start with a square laice (or grid) and assign to each vertex a small disk coloured black or white in an alternating paern.

The two kinds (up to rotation) of square Truchet tiles and the three kinds (up to rotation) of hexagonal Truchet tiles.

Notice that, for each face of the square laice, there are two possible square Truchet tiles that can be placed whose colours are compatible with the disks. Choosing these two possibilities independently at random with equal probability (which has the appealing property of retaining black/white symmetry), we have created a random Truchet tiling of the entire plane. The Bogomolny–Schmit conjecture says that the statistical properties of the interfaces of the tiling (ie the curves that separate the black and white regions) match, on large scales, the statistical properties of the curves 17

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The square laice with alternating coloured disks and the two kinds of square Truchet tiles that are compatible with each face of the laice.

created by random Chladni figures. To be concrete, examples of such large-scale statistical properties are the crossing probabilities, that is, the probability that a curve crosses a large rectangle from le-to-right.

Comparison of large random Truchet tiling and large Chladni paern.

Where does this conjecture come from? The reasoning is actually quite simple. We first observe that the local maxima of standing waves are all positive, and the local minima are all negative (mathematically, this is a consequence of the standing waves being solutions to the Helmholtz equation). Then we suppose that a random standing wave has roughly as many local maxima as local minima and that these are arranged in a square laice, with saddle points between every pair of maxima/minima. Finally, we assume that the saddle points take positive or negative values independently with equal probability; using the sign of the saddle points to connect up neighbouring maxima/minima, and identifying ‘+’ with black and ‘−’ with white, this yields a model for random Chladni figures that is equivalent to the random Truchet tiling. So does this conjecture help us understand Chladni figures? Doesn’t it just tell us that one complicated object—a typical high-frequency Chladni figure—is well-approximated by another comchalkdustmagazine.com

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If the saddle is negative

If the saddle is positive

A grid of maxima/minima/saddles, and replacement rules to form the random Truchet tiling.

plicated object—a random Truchet tiling? Well yes, but this is actually an immense improvement, because in fact random Truchet tilings are just a thinly-disguised version of a well-understood object in probability theory—critical percolation.

Truchet tilings as a model for percolation Percolation is the mathematical study of porous media, like sandstone or Swiss cheese. There are many models of percolation, but perhaps the simplest is to start with an infinite periodic laice, such as the square or hexagonal laice, and colour every edge of the laice black or white independently with respective probabilities p and 1 − p for some fixed p ∈ [0, 1] (this model is called bond or edge percolation; one could alternatively colour every face black or white, which is called face percolation, or every vertex black or white, which is called site percolation). One then considers the subgraph formed by the black edges, which is a simple model for an inhomogeneous porous medium. There are two crucial facts about percolation. The first is that, for essentially any model of percolation, there is a critical probability pc such that if p < pc then the black edges form only finite clusters (ie finite connected subgraphs), whereas if p > pc then there is a black cluster that is infinite. In other words, there is a phase transition in the global connectivity of the model, from a graph that is composed of finite clusters, to a graph that is globally connected (although it will also contain some small disjoint clusters). The second fact is that, whereas certain local features Bond percolation on the square laice of percolation depend heavily on the particular percolation model being considered, many of the large-scale properties of the model are universal, that is, they behave in essentially the same way in the three regimes p < pc , p = pc and p > pc for every percolation model. Examples of local or non-universal properties include the value of pc itself, and also the number of black clusters contained in a given 19

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chalkdust large region, whereas examples of universal properties include the crossing probabilities mentioned earlier. Understanding the universal properties of percolation is an active area of research, especially in the critical regime p = pc , and two Fields medals have been recently awarded for progress in this area (Wendelin Werner in 2006 and Stanislav Smirnov in 2010).

Bond percolation on the square laice in the sub-critical (p < pc ), critical (p = pc ) and supercritical (p > pc ) regimes. For this model pc = 1/2.

Let’s make the link to random Truchet tilings. As we indicated, the random Truchet tiling is just a disguised version of bond percolation on the square laice with parameter p = 1/2. To see why, start with the percolation model, rotate the laice by 45°, draw a red vertex in the centre of every face, and then replace the black and white edges with the corresponding Truchet tile so that the models coincide.

Bond percolation on the square laice, a rotated version, and the replacement rules that create the corresponding Truchet tiling.

So which percolation regime does the random Truchet tiling correspond to? It turns out that the critical probability for bond percolation on the square laice is precisely pc = 1/2 (this can be understood, in terms of the Truchet tiling, as a consequence of the black/white symmetry), so the random Truchet tiling should behave on large scales like critical percolation. Hence if the Bogomolny–Schmit conjecture is true, random Chladni figures also behave, on large scales, exactly like critical percolation. Since the properties of critical percolation are well-understood, this gives us a way to analyse, and make predictions about, the more complicated random Chladni figures.

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So is the Bogomolny–Schmit conjecture true? There are a few aspects of the reasoning of Bogomolny–Schmit that are hard to swallow. In particular, the argument relied on the assumption that the local maxima/minima of a random standing wave are arranged in a square grid, which is simply not true. So it is natural to be sceptical. While there hasn’t been much (to be fair, barely any!) mathematical progress on proving or disproving the conjecture, numerical studies (by Belyaev, Kerata, Nastasescu and others) have shed some light on the situation. What they show is that, for the non-universal features of critical percolation, the conjecture is wrong. This is unsurprising, since non-universal features should, by definition, depend on the specifics of the model (such as the choice of the square laice over the hexagonal laice), and as we just said, there are reasons to believe these specifics are not wholly accurate. On the other hand, numerical studies have confirmed that the universal properties of critical percolation do match exactly the equivalent properties for random Chladni figures. This is remarkable, and it remains a genuine mystery why it is true.

A percolation model with hexagonal Truchet tiles In issue 08, different shapes of Truchet tiles were discussed, and we can easily generalise our random Truchet tiling to other shapes. Let’s consider the case of hexagonal tiles. Starting with a hexagonal laice, we again assign to each vertex a small disk alternating black and white. There are now five choices of Truchet tiles that we can place in each hexagonal face that are compatible with the disks.

A hexagonal laice, and the five tiles that are compatible with each face.

What probability should we assign to each choice of tile to get a sensible model for a random tiling? Well, since three of the five tiles are rotationally equivalent, it would be natural to assign them an equal probability p. To get a critical percolation model we also want to impose black/white symmetry, so the remaining two tiles should have equal probability q = (1 − 3p)/2. Since each of these probabilities depends on a single parameter p ∈ [0, 1], this defines a one-parameter family of percolation models, all of which will be critical. In his seminal work in 2006, Smirnov considered the critical percolation model in which the faces 21

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Appear with probability p = 0.1

Appear with probability q = 0.35

Depiction of a large-scale random hexagonal tiling and a key to indicate which value of p in the one-parameter family of models is being depicted.

of a hexagonal laice are coloured black and white with equal probability; by exploiting the symmetries of the hexagonal laice, he was able to establish rigorously, for the first time, many of the conjectured properties of critical percolation. Notice that this model is really just a degenerate case of the hexagonal Truchet tilings we constructed above, corresponding to the case p = 0, and hence q = 1/2 (to see this, simply remove the black/white circle segments from the Truchet tiles, which doesn’t affect the connectivity of the model). A few years later Chayes and Lei studied exactly the one-parameter family of critical hexagonal Truchet tilings defined above, and showed that they could also be studied with Smirnov’s techniques. It is a surprising fact that, 13 years aer Smirnov’s work, these are (as far as I know) still the only critical percolation models which have been rigorously studied in detail. In other words, hexagonal Truchet tilings seem to possess a special structure that is unique among the (enormous!) class of critical percolation models.

Smirnov considered the percolation model using the faces of the hexagonal laice rather than hexagonal Truchet tiles. In this model, the conjectured properties of critical percolation have been established. chalkdustmagazine.com

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So did Truchet ever meet Chladni? Let’s conclude by indulging in some idle speculation. Sébastian Truchet died in 1729, 27 years before the birth of Ernst Chladni in 1756, so they never had the opportunity to meet. However, as it turns out, Chladni was not the first scientist to investigate nodal paerns of standing waves—this was done half a century earlier by English scientist Robert Hooke. So did Truchet ever meet Hooke? Given their standing in the scientific communities of Paris and London (in 1700, Truchet and Hooke were simultaneously members of the Académie des sciences and the Royal Society respectively), it is quite possible that they did. Stephen Muirhead Stephen is a lecturer in mathematics at een Mary University of London. Like any mathematician, he is always delighted to find unexpected connections between mathematical objects.

d stephenmuirhead.wordpress.com My favourite symbol

ℵ0

Tom Rivlin

My favourite symbol is ℵ0 , or “Aleph-null”, which represents a type of infinity. Georg Cantor probably decided to use this symbol, the first leer of the Hebrew alphabet, because he was tired of overlapping meanings for every lower and upper case Greek and Latin leer. I think we should use more characters from other alphabets in maths, so in the spirit of Cantor, here’s an equation for you to solve: и+

×♂=

Rating:

/10

Did you know... …that the Fahrenheit and Celsius scales have one point at which they intersect? −40℃ is equal to −40℉.

My least favourite symbol

εijk

Eleanor Doman

My least favourite symbol is the Levi–Civita symbol, also known as the epsilon-i-j-k-thing. I first encountered this symbol seven years ago and I’m still not sure why you would ever willingly want to use it. We’re of course told it’s useful when writing the cross-product with using Einstein summation notation and simplifying δij blah, but I’m yet to meet anyone who would rather use this than say vectors, or anything else for that maer. If you truly find yourself in a situation where using this symbol is preferable to any other method… Rating: εijk δij /10 then you’re a stronger soul than me. 23

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Which symbol are you? START VANILLA

How conventional are you feeling?

Are you feeling insignificant?

I PUT MILK BEFORE WATER IN MY TEA

YES NO YOU ARE

OF MUSIC

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Are you a doctor ?

Are you in a library?

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Do you like primes?

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Are you on good terms with your !?

Do you watch Only Connect?

Do you work in fluids?

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Pictures: Cartesian coordinate system by Crystoph., CC BY SA 2.0

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YOU ARE


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The phantom parabola Image of ghost produced by double exposure, The National Archives UK

Zoe Griffiths

I

originally wrote this piece as my semi-final entry for the Big Internet Math-Off, an online competition run by the Aperiodical. The knock-out tournament saw 2 4 mathematicians go headto-head in presenting ideas they found particularly interesting. Each round readers voted for their favourite of 21 entries, with the winner advancing to the next round, until only 20 participant emerged victorious. The audience was given the brief to vote for the entry that made them ‘do the loudest “aha!”’ This ‘aha!’ success criterion seemed entirely fiing since our response to mathematics is oen exactly this; involuntary, emotional and beyond the reach of words. As such, when asked ‘why do you like mathematics?’, I always feel words cannot fully capture quite what it is to find something mathematically satisfying, or mathematically beautiful. But mathematics itself is the opposite; using maths we can convey huge ideas, completely, and efficiently. When we look at things mathematically it is like we are casting a perfect light on to the world that allows us to see the very structure and essence of things. I chose to write about the phantom parabola because it demonstrates this power; by using mathematics (in a simple, ‘aha!’-inducing way) we reveal something that was otherwise hidden. I introduce the phantom parabola as a ghost story, in a blatant clickbait-esque aempt to make people read on and vote. But as this is a story about invisible mathematics that genuinely gives me the goose bumps, as well as an ‘aha!’, I think my claim is justified. 25

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The phantom parabola This is the closest you’ll ever get to a maths ghost story—meet the phantom parabola that hangs, unnoticed, perpendicular to every quadratic and allows us to visualise complex solutions. Roots of quadratics with real coefficients come in one of three situations; either the parabola intersects the x-axis twice giving us two real solutions, or it turns on the x-axis making one repeated real solution, or it never reaches the x-axis. In this final situation, we know that instead of real solutions we have a (conjugate) pair of complex solutions that cannot be represented on our real x-axis.

y

x The parabola y = x 2 + 2 has no real roots.

It seems incomplete to not have the corresponding visualisation of complex solutions. This, aer all, is an area of mathematics that showcases the powerful and elegant unity that exists between the algebraic and graphical representations, as dreamt up by Descartes all those years ago. But then it breaks down when those solutions are complex. How dissatisfying! Enter the phantom parabola! The complex solutions of the quadratic sit on this phantom which hangs ‘upside down’ and at right angles to the original, with the two parabolas meeting at their extremes. And we can plot the phantom parabola if we represent the values not on a one-dimensional x-axis, but on a twodimensional x-plane, that has a real axis and an imaginary axis. The y-axis is then perpendicular to this plane. By following the steps below, we can work out what the phantom looks like: We are finding the roots of the quadratic y = ax 2 + c where the constants are real. To find the roots of a general quadratic, write it in the form y = a(x − b)2 + c and set v = x − b. You can then follow the rest of these steps with v in the place of x. Since x is complex, let x = m + ni, where m and n are the axes of the complex x-plane. We will now look for the values of x when y = 0.

n

y is real and therefore x must either have only a real component or only an imaginary component (which when squared would become real).

x = m + ni

m

If x is real, then n = 0 and y = am 2 + c. (This quadratic is the original one.) If x has only an imaginary component, then m = 0 and y = a(ni)2 + c. This simplifies to y = −an 2 + c. This is the phantom parabola! chalkdustmagazine.com

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chalkdust We can understand that the parabola and its phantom are perpendicular to each other because the original only varies over the real m dimension, whilst the phantom is fixed to take only one value on the real m-axis but can take any value on the imaginary n-axis. Equally, the phantom is an inversion of the original because its coefficient a has been multiplied by i2 . The plot below shows a specific parabola and its phantom. y

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The parabola y = m 2 + 2 (red) y = −n 2 + 2 (blue) viewed from three different √ and its phantom √ angles. The roots at n = ± 2 (or x = ± 2i) are shown.

Finally we glimpse the complex solutions! The quadratic in this case √ (y = x 2 + 2) has roots at x = ± 2i which can be found at the points marked above where the phantom curve crosses the plane with equation y = 0 (the complex x-plane). This is of course analogous to where real solutions of a quadratic appear on a 2D plot.

y

Every parabola has a phantom and if we plot both it and the original parabola we will always be able to visualise the two roots of the quadratic. In the case when the two roots are real and distinct the phantom turns before the complex x-plane, with the original curve crossing the plane instead.

y

n

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Le: The parabola y = x 2 − 0.4 has two real roots; its phantom is below the x-plane. Right: The parabola y = x 2 has a repeated root; it meets its phantom at this root.

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chalkdust And in the case when we have what we call a repeated real root? Well, how befiing is it that in this situation we have two curves, the original and the phantom, both touching the x-plane in the same place but from opposite directions? At d geogebra.org/m/wuwekaxy, you can find an interactive GeoGebra plot of the general quadratic and its phantom. Vary the constants a and c and see for yourself the two solutions of the quadratic in each different situation. Never again must we take it blindly on trust that every quadratic has two solutions! I was first introduced to the phantom parabola whilst on a course run by MEI (Mathematics in Education and Industry). They in turn referenced the work of maths teacher Philip Lloyd. Philip developed this idea to help his students understand complex roots. He was the first to describe these curves as phantom. But it is not only the quadratic that has a phantom! For even more excitement, check out Philip’s blog d phantomgraphs.weebly.com, where he leads us through the derivations and properties of the phantoms of other functions—from phantom cubics to phantom hyperbolas. It is beyond pleasing.

y

y

n

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m

Le: The cubic y = x 3 + x 2 − x + 2 and its phantom. Right: The hyperbola y 2 = x 2 + 1 and its phantom.

Mathematics has once again made the invisible, visible. With many thanks to Ben Sparks for making my ambitions of an interactive Geogebra plot a reality. You can read more about the Big Internet Math-Off at d aperiodical.com Zoe Griffiths Zoe is a maths and philosophy graduate and ex-maths teacher, who now works as a maths communicator with Think Maths. Her work has seen her give talks and deliver workshops for students, families and teachers in schools both in the UK and internationally, in the Royal Institution’s world-famous lecture theatre, at science festivals, teacher conferences and at comedy nights.

a @ZoeLGriffiths d think-maths.co.uk/presenters Did you know... …that 206156734 = 26824404 + 153656394 + 187967604 ? This is an integer solution for the equation w 4 = x 4 + y 4 + z 4 found by Noam Elkies. chalkdustmagazine.com

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chalkdust

Have you been taking your weekly dose of chalkdust?

Unearthing roots

Talkdust, our new podcast

Black Panther’s indestructible suit

Let’s think rationally

Partitions of a tree

Unlocking diversity

Between nothing and something

Deal or no deal

All these topics, and more, have been covered in our weekly online articles, alongside more including cosmology, sport, fluid dynamics, politics, great puzzles, and mediocre jokes. Read it every week and sign up to our monthly newsletter at

chalkdustmagazine.com Roots: Nick Bonzey, CC BY-SA 2.0. Safe: Brook Ward, CC BY-NC 2.0.

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Counting caterpillars Didier Descouens, CC BY-SA 4.0

Peter Rowlett

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 bought a new toy for our three-year-old, a robot caterpillar. This has a head and eight segments which tell the caterpillar to perform actions like moving and playing music. A caterpillar is formed by the head, always at the front, followed by at least one segment. These form an ordered set of commands which are executed in sequence. This is a fun toy which principally involves colourful lights and music, while also developing some logical thinking and proto-programming skills in 3–5 year olds. But that’s not why I’m writing about it. On the packaging is made the following claim, acting for me like a red rag to a bull:

8 pieces. Endless combinations! This cannot possibly be true. You cannot arrange eight pieces endlessly and never repeat the same combination. This isn’t how infinity works. My son is quite into counting, though he becomes a lile inconsistent around “five-teen” so perhaps swapping segments into place and counting them isn’t the way to go about this. Thankfully, mathematics has an answer: combinatorics. chalkdustmagazine.com

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A robot caterpillar


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Some combinatorics fundamentals Imagine you have three tiles containing leers A, B and C, and you want to know how many different ways you can arrange these. A

A

B

B

C

C

B

C

A

C

A

B

C

B

C

A

B

A

Six rearrangements of three different tiles labelled A, B and C.

Notice that there are three options for which tile to place first: A, B or C. Once one of these is chosen, there are two options for which tile to place second. In the first two rearrangements A is placed first leaving two options for which to place second: B or C. Once either is chosen, there is one option remaining for the third and final tile. This means that there are 3! = 3 × 2 × 1 options for arrangements of three different tiles. This logic continues for any number of tiles, indeed any number of objects. The number of ways of arranging n objects is n ! = n × (n − 1) × (n − 2) × . . . × 3 × 2 × 1. Let’s consider a different problem: say you have a bag of three balls labelled with leers A, B and C, and you want to know how many different ways there are of pulling two balls out of the bag. In particular, imagine you aren’t interested in which order the balls came out of the bag (as is the case in, say, a loery). At first, this may appear very similar to the tiles. There are three choices for the first ball, then two choices remaining for the second ball, then one choice for the ball le in the bag. That’s n !, and for three balls the 3! = 6 options are listed here: A

A

B

B

C

C

B

C

A

C

A

B

Six ways of choosing two balls from three labelled A, B and C.

However, if you look at the pairs of balls just above, you will notice that we are counting some twice. Whether you draw A and then B or B and then A, you are still holding A and B at the end of the draw. Actually, there are only three different ways to draw two balls from three. 31

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chalkdust To see how this works for a larger case, consider a bag with ten unique balls from which you desire to draw three, and you aren’t interested in the order of the three. You could consider drawing balls to be an act of arranging objects: arrange ten balls into a line and then ‘draw’ the three lemost balls. Doing so, you arrange the balls into two groups: the three you have drawn and the seven you have not. 7 balls le in bag

3 balls drawn

Arranging ten balls and drawing three of them.

The birthday problem The birthday problem says that you need only 23 people in a room for a beer than 50% chance that at least two of them will share a birthday. One way to think about this is to think about pairs of people. Given the birthday of one member of a pair, if the other member of the pair is going to have a different birthday then (ignoring leap years) there are 364 days le for their birthday. This yields a probability (assuming all birth days 364 . of a 365-day year are equally likely) of 365 (23) Now we notice that there are 2 = 253 ways to choose a pair from the group of 23 people. If we assume that each pair of birthdays is an independent event, then we have an approximation for the probability of each and every pair not sharing ( 364 )253 ≈ 0.4995. The probability, a birthday as 365 therefore, of at least two of them sharing a birthday is approximately 1 − 0.4995 = 0.5005, slightly greater than 50%.

There are 10! ways to arrange these ten balls. But within these 10! arrangements, there are 7! that simply shuffle the balls le in the bag, which doesn’t affect which ! three are drawn. So we are down to 10 7! arrangements. This represents all possibilities for drawing three balls from the bag. If we aren’t interested in the order that these three are drawn, then we can further reduce this number by 3!, the number of ways of arranging them. This gives a formula for the number of ways of drawing three balls from ten when the order of the three drawn isn’t important: 10! = 120. 3! × 7! In general, the number of ways of choosing r objects from n without repetition when we don’t care about the order of the objects chosen is

n! . r !(n − r)! ( ) n . This is the binomial coefficient and can be wrien using the notation r The binomial coefficient, and the fact that there are n ! ways of arranging n objects, are key tools at the heart of combinatorics. chalkdustmagazine.com

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chalkdust We can get very surprising and unintuitive results through combinatorics. I used to give a talk about coincidences which started with two counting problems—the birthday problem and shuffling cards (see the sidebars on this page and the previous page for descriptions of these problems). Though I know the methods involved and intellectually it makes sense, I still find it hard to genuinely, intuitively appreciate that if I collect 23 unconnected people, there is a 50% chance that two of them will share a birthday. Meanwhile, I find it really hard to (genuinely, intuitively) believe that a bunch of cards I’ve just thrown on a table are arranged in a way such that it is extremely unlikely any pack of cards has been arranged in the history of the universe. Such is the nature of combinatorics, which offers techniques to address counterintuitive counting problems.

Back to the caterpillar With eight segments to arrange, it feels to me like there can’t be very many different ways to arrange these. But, as I know, numbers in these sorts of problems are counterintuitively large. Let’s put an upper bound on the number, anyway, to dispute the ‘endless’ claim. If all eight segments were different, for sequences using k ⩽ 8 positions, we choose k segments from the 8 possible segments and there are k ! ways to arrange these, so the number of possible sequences would be ( ) 8 8! k! = . k (8 − k)!

Shuffling cards While I was explaining the birthday problem, I would slowly shuffle and deal all 52 cards from a standard pack of playing cards onto the table in front of me. I’d observe that 52 different playing cards can be arranged in 52! = 8 × 1067 ways. So the exact set of cards arranged on the table in front of me (assuming a well-shuffled deck) represented 1 case in 52!, with probability approximately 1.24 × 10−68 . The fact we got to witness this hand of 52 cards is a truly astonishing coincidence. I’d invite members of the audience down to the front to view the astonishing layout, but for some reason I’d never get any takers.

Notice that for k = 8, this reduces to 8!, as you might expect. The number of possible sequences using any number from 1 to 8 segments would be the sum of this arrangement for all possible lengths, ie 8 ∑ k=1

8! = 109600. (8 − k)!

I think you’ll agree with me that 109,600 is significantly smaller than infinity. For this problem, though, it’s definitely too large.

But it isn’t that simple The caterpillar has three identical segments which instruct it to move forwards, two to turn le, two to turn right and one to stop and play music. 33

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chalkdust For the avoidance of doubt, some other features and restrictions: • there are eight positions that can hold segments, the head is always present; • any number of segments from 1–8 may be used, and not 0; • the order of segments maers, because it is a sequence of commands; • segments are connected (by USB), so there can be no gaps in a sequence—if a position is unfilled, the caterpillar ends at the preceding segment. Let the forwards segments be labelled F, the le L, the right R and the music M, and let a sequence be denoted by a string of leers from le to right (eg the head, not denoted, is on the le of the string). Then some examples of valid caterpillar sequences would be FFFM, FFMF, FLRFLRFM, LRL, F, and so on. In the simple analysis that led to the upper bound 109,600, we were over-counting the cases where multiple F, L and R segments are used. For example, the sequence FF would be counted twice, even though they produce functionally identical caterpillars. There is a neat trick in combinatorics for counting in this kind of circumstance—generating functions.

Generating functions—an example Say I have three cards, two blue and one green, and I want to know how many ways I can arrange them into sequences of length 1–3. This is fairly straightforward to enumerate by hand, as you can see here. For a generating function approach, I assign two functions. For green, I use 1 + g. Think of this as the 1 representing no green card being selected and g representing one green card being selected. For blue, I use 1 + b + b 2 . Similarly to green, 1 and b represent no blue cards and one blue card, with b 2 representing two blue cards being selected.

B

B B

G

B G

G B

Now we simply multiply the two functions together.

(1 + g)(1 + b + b 2 ) = 1 + b + b 2 + g + gb + gb 2

B B G

B G B

G B B

Arrangements of two blue cards and one Each term of the expansion represents a selection green card of length 1, 2 and 3. from the set of three cards. For example, g represents choosing one green card only for the sequence ‘green’, and b 2 represents choosing just the two blue cards for the sequence ‘blue, blue’. The term gb represents choosing one green and one blue card, and there are 2! = 2 ways to arrange these two cards: ‘green, blue’ and ‘blue, green’. We didn’t have this rearrangement problem with b 2 because the two blue cards are identical. The term gb 2 represents choosing all three cards. There are 3! ways to arrange three cards, but 2! of chalkdustmagazine.com

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chalkdust them are the same, because the two blue cards are identical. So we can arrange gb 2 in 32!! = 3 ways. The final term in our expansion is 1, which represents choosing none of the cards. Whether this is a valid sequence depends on the context and the rules you are following.

Generating caterpillars Since there are three F segments, we can use the function 1 + f + f 2 + f 3 to represent these. For the two R segments, use 1 + r + r 2 , and for the two L segments use 1 + ℓ + ℓ 2 . Finally, for the one M use 1 + m. To find out how many different ways there are of selecting from these segments, expand (1 + f + f 2 + f 3 )(1 + r + r 2 )(1 + ℓ + ℓ 2 )(1 + m). The expansion has 72 terms, each representing a different caterpillar. One of these, however, is formed by multiplying the four 1s together to get 1, which represents selecting no segments. Since this isn’t a valid caterpillar (the toy doesn’t work using only the head), there are 71 different ways of forming caterpillars up to length 8. For example, one of the terms of the expansion is f 3 rℓ 2 m. This represents choosing F, F, F, R, L, L and M for a caterpillar of length 7. A caterpillar of length 7 can be arranged 7! ways, but some of these are duplicates; how many depends on which segments are involved. In this case, a caterpillar involving three Fs, an R, two Ls and an M can be rearranged 37!2! ! = 420 different ways, with 7! divided by 3! for the three identical Fs and 2! for the two identical Ls. We complete this calculation for all 71 valid selections of segment arrangements. The total number of possible caterpillars is 5023. So the “endless combinations” claimed on the packaging is, in fact, a lile over five thousand possible caterpillars. Not endless, but certainly enough to keep us busy for a while! Oh, and the manufacturer sells add-on kits containing extra and different segments, which opens up whole new worlds of combinatorics possibilities… If you are interested in a gentle-but-thorough introduction to combinatorics, I heartily recommend Robin Wilson’s Combinatorics: A Very Short Introduction. Peter Rowlett Peter is a reader in the mathematics group at Sheffield Hallam University, where among other things he teaches a module called Game theory and recreational mathematics. His research (d shu.ac.uk/about-us/our-people/staff-profiles/peter-rowle) investigates teaching and learning of mathematics at university level. He blogs at The Aperiodical (d aperiodical.com).

a @peterrowle d peterrowle.net Did you know... …that the number 365 is equal to the sum of three consecutive squares and two consecutive squares in which the five squares are also consecutive? 365 = 102 + 112 + 122 = 132 + 142 35

spring 2019


Mystic Mug’s famous mathematical

Horoscope Aries

21 Mar–19 Apr

Your research output doesn’t look so bright until a tall, dark, handsome stranger presents you with a proof of the Riemann hypothesis.

Torus

20 Apr–20 May

The heavens are not in your favour, and people may try to take advantage of you. Don’t let them take you for a mug.

Gemini

21 May–21 June

Tomorrow you will wake up in a parallel universe which is identical to this one, except for the fact that no one ever invented the Banach–Tarski paradox. So you get to invent it! Good for you!

Cancer

22 June–22 July

Jupiter is in retrograde today, so you should walk backwards. Avoid proof by induction until the end of the month.

Leo

23 July–22 Aug

Writing a horoscope for Leo is left as an exercise for the reader.

Virgo

23 Aug–22 Sep

Today you will narrowly miss being nominated for a Fields medal, but you’ll still be lauded for your contributions to astrology. I guess that’s a pretty good constellation prize! Eh? Eh? 36


Libra

23 Sep–23 Oct

An alignment of heavenly bodies can bring you great fortune... if you manage to get a Nature paper out of the exoplanet transit you just observed.

Scorpions

24 Oct–21 Nov

A small, silly mistake in your past can come back to haunt you, but only if you allow it to by making it a running joke in your maths magazine for four years. f

f f

fff

fff

Sagittarius

f

22 Nov–21 Dec

Today you will feel a wave of positive energy. Why? You’re a Sagittarius! Your sign is a rad-as-heck centaur archer! You should feel good about yourself. What do the other signs have? A fish? Come on... 62 u Your wave of positive energy is a solution of 2 = c2 Δu. 6t

Capricorn

22 Dec–19 Jan

You have been feeling a sense of unease building up for a while now, but you’ve been unsure why. It may be that a theorem you’ve relied on is based on a shaky proof. Take the time to re-evalute your hypotheses.

Aquarius

20 Jan–18 Feb

An adverse pressure gradient will cause an obstruction to your flow. This month, green is your lucky colour.

Pisces

19 Feb–20 Mar

See Sagittarius, but in reverse. Seriously, a fish?!

37


chalkdust

Hiding in plain sight:

The Diffie–Hellman key exchange Thomas Hawk, CC BY-NC 2.0

Axel Kerbec

F

 a very long time now, people have been coming up with ingenious ways of exchanging messages securely. From the basic Caesar cipher to the Enigma, however, there has been a universal caveat: how to agree on the encryption/decryption key without meeting face to

face?

To put it simply, consider the following situation: Alice and Bob want to communicate with one another using an Enigma machine. Alice adjusts the seings of her machine, tells Bob what those seings are, types a secret message, and sends the encrypted text to Bob, who then adjusts his own machine’s seings and proceeds to decrypt the message. But hang on a minute. How exactly did Alice give her seings to Bob? Alice can’t send encrypted messages unless Bob has those seings, but at the same time Alice has no way of secretly passing those seings on to Bob. Is there a way for them to somehow come up with a shared secret key (ie the seings for the machines) by exchanging messages visible to anyone listening to their conversation? The Diffie–Hellman protocol allows us to do exactly that. Let’s dig in and see how it works!

Some modular arithmetic We are all used to the regular addition and multiplication of numbers. Indeed no one would ever question that 4 + 11 = 15. In modular arithmetic, we change the rules slightly, by saying two chalkdustmagazine.com

38


chalkdust numbers a and b are the same mod p, or congruent mod p, if their difference is an integer multiple of p, and we write a ≡ b mod p. For example, since 15 − 1 = 2 × 7, we would write 15 ≡ 1 mod 7. This is sometimes know as clock arithmetic because we kind of wrap around every time we reach p. Let’s think of what happens when we take the integers mod 5. Observe that any integer will be congruent to one of 0, 1, 2, 3, 4, and that none of 0, 1, 2, 3, 4 are congruent to each other. The set {0, 1, 2, 3, 4} is therefore enough to “represent” all the integers mod 5, and we call this set Z5 . Just as with the regular integers, we can add and multiply integers mod p together. Of particular interest to us is the multiplication, because it gives an interesting structure to the set. Notice that almost every element in the set Zp has what we call a multiplicative inverse. For example in Z5 , 2 × 3 ≡ 1 mod 5, so we say that 3 is the inverse of 2 mod 5 (and vice-versa that 2 is the inverse of 3 mod 5). Similarly, 4 × 4 ≡ 1 mod 5, so we say that 4 is its own inverse mod 5. I did say almost every element, because 0 never has a multiplicative inverse. This could cause some problems down the road, so we decide to do away with zero entirely, and call the resulting set Z∗p . Now, Z∗p together with multiplication has a lot of structure, and it is in fact an example of a group. We will see another example of a group later on in the article. 1 6

2

There is something special about Z∗p that makes it very useful in cryptography. Let’s look at Z∗7 and enumerate the powers of 3 mod 7. We have: 31 ≡ 3 mod 7 32 ≡ 2 mod 7

5

33 ≡ 6 mod 7

3

34 ≡ 4 mod 7

35 ≡ 5 mod 7

4

36 ≡ 1 mod 7

The elements of Z∗7 wrien in a circle, with arrows showing multiplication by 3

What do we notice? Every one of 1, 2, 3, . . . , 6 mod 7 can be expressed as a power of 3 mod 7! Because of this, we say that 3 is a primitive root mod 7 and that Z∗7 is a cyclic group. Note that not every element of Z∗7 is a primitive root: if we instead looked at powers of 2 mod 7 for example, we would only get half of all possible values. Because 7 is prime, however, we are guaranteed the existence of primitive roots in Z∗7 . This simple fact will be paramount in making sure our Diffie–Hellman protocol is secure, but before we get there, let me give you a small problem: can you list all the primitive roots mod 7? It shouldn’t take very long to determine that these are 3 and 5. But 7 is a small prime; what about finding all the primitive roots mod 23? This is a bit tedious if you try every number up to 23, but you should find exactly 10 distinct primitive roots. Notice that 10 is exactly the number of integers smaller than 22 which are coprime to 22. This is not a coincidence! In general, there are exactly φ(p − 1) distinct primitive roots in Z∗p , 39

spring 2019


chalkdust where φ(n) is the number of integers between 1 and n that are coprime to n. Although this doesn’t tell us how to find primitive roots, it does give an upper bound on how many numbers we have to try before finding a primitive root. Computers are reasonably good at finding primitive roots, but the larger the prime the longer it takes. Because of this, we tend to reuse primes whose primitive roots we’ve already computed when doing cryptography.

Discrete logarithms and the Diffie–Hellman key exchange Now let’s think about another problem: if I give you some k, say k = 3, can you compute 5k mod 7? This is not too difficult, and computers are quite fast at computing powers of elements mod p. The inverse problem, though, is not so straightforward: given that 5k ≡ 3 mod 7, can you find k? Yet again, because 7 is a small prime, you shouldn’t have a hard time figuring out the answer. I invite you to think about your process in solving the problem. My guess is that you raised 5 to different powers until you found that 55 ≡ 3 mod 7. What if I told you that 5k ≡ 48 mod 53? Can you find k then? Going about it the same way as before, it might take you a while. The general problem of finding k such that g k ≡ m mod p, where g is a primitive root, is known as the discrete logarithm problem, and it is a very difficult problem indeed when p is very large. Hence the function k 7→ g k mod p is an example of a trapdoor function: it’s easy to compute g k mod p, but hard to find k if you only know g and g k mod p. We can use this trapdoor function to publicly exchange keys. Let’s see how:

chalkdustmagazine.com

40


chalkdust That’s it! I personally find it hard not to be taken aback by the simplicity of the procedure: because Alice knows n, she can raise g m , which she received from Bob, to the nth power and get g nm , and similarly for Bob. Aer an exchange of information taking place entirely in plain sight, Alice and Bob now have a secret key that they can use to encrypt all further communication! This is what made the Diffie–Hellman protocol revolutionary. To understand why this is secure, let us put ourselves in the shoes of an aacker. In the course of the whole exchange, what do we see? Because the communications are not encrypted, we will know the values p, g, g n mod p, and g m mod p. To obtain g nm mod p, one essentially needs to solve the discrete logarithm problem to find n or m. Ironically enough, there is no formal proof that the discrete log problem is hard to solve. We do have quite a few algorithms that compute discrete logarithms, but none of them performs particularly well. There is an efficient quantum algorithm due to Peter Shor: it is an algorithm that would run on a quantum computer, which could pose a threat to current systems should practical quantum computers ever hit the market. But quantum computing aside, Diffie–Hellman is not perfect. As mentioned before, choosing such large primes and finding their primitive roots is not something you want to do every single time you initiate a link, and, in practice, the same primes are reused. This is fine in general, as the discrete logarithm problem is considered hard enough for very large groups, even if those groups are known in advance. In 2015, however, a group of computer scientists reported that pre-computations were feasible for those primes which we know are used—essentially building a lookup table of the Another type of logjam powers of primitive roots for those specific primes. These precomputations would take a lot of resources, and the estimated cost of such an operation is in the hundreds of millions of dollars, which has been pointed out to be well within the budget of some national agencies. This vulnerability is known as logjam. The ways to circumvent it are to either use larger primes, or to use a variant of the Diffie–Hellman protocol known as elliptic-curve Diffie–Hellman.

What is an elliptic curve? An elliptic curve is the solution set of an equation of the form: y 2 = x 3 + ax 2 + bx + c. For us, the numbers a, b, and c will be integers. We also require, as a technicality, that the curve has no cusps and does not cross itself. We have to specify which values of x and y we allow: rational numbers? real numbers? complex numbers? We will denote the set (field) we draw our values from by k and the elliptic curve by C(k). On the next page, there are some examples of elliptic curves over the real numbers. 41

spring 2019


chalkdust y

y

3

3

2

2

1

1

−3 −2 −1 −1

1

2

3

x

−3 −2 −1 −1

−2

−2

−3

−3

y2 = x3 − x + 4

1

2

3

x

y2 = x3 − x

Later on, we will be considering curves over finite fields, in particular the Zp we talked about earlier. Here’s what our curves look like over those fields: y

y 90 80 70 60 50 40 30 20 10 0

6 5 4 3 2 1 0

0

1

2

3

4

5

6

x

Points on y 2 = x 3 − x + 4 over Z7

0 10 20 30 40 50 60 70 80 90

x

Points on y 2 = x 3 − x + 4 over Z97

You might be thinking that these don’t exactly look like curves, but they are: they’re the set of points that satisfy the equation over Zp . What we’re interested in is the structure of the points on those curves, and in practice we may have lots of questions about those points: how many are there? Is there an effective way of finding all of them? A theorem of Hasse tells us that in general, there should be about p + 1 points on an elliptic curve over Zp , and another result by Mordell and Weil tells us that starting with finitely many points on an elliptic curve, we can generate all the points on the curve. We won’t understand the proof of this theorem by the end of this article, however it would be nice to understand what it is saying. How can we generate new points from old points? It turns out that’s not particularly complicated. Take two points P and Q on the elliptic curve. Draw a line through P and Q. This line will intersect the curve at some third point, which we name P ∗ Q. chalkdustmagazine.com

42


chalkdust y

This point is on the elliptic curve, so we could stop here, but we go one step further, and for good reasons. Draw a vertical line through P ∗ Q. It will intersect the curve at one other point, and this point we name P + Q.

P∗P P

Q P∗Q

If P and Q are the same point, then we do the same thing except that we draw a line tangent to P instead of through P and Q at the beginning.

x

This new “addition” of points is the first step to giving some structure to the set of points on an elliptic curve.

P+Q

Let’s make another definition: starting from a point P , draw a vertical line 2P through it. Just as before this vertical line intersects the curve at one other point, y 2 = x 3 − x + 4 with point addition constructions and we call this point −P . Now I want you to add P and −P together. What happens? It seems like the whole thing breaks down: the line through P and −P doesn’t intersect the curve anywhere else, so we don’t get a third point at all. We can circumvent this by thinking of there being an extra point on the elliptic curve, a point which exists at infinity and where all vertical lines meet. To make this notion clear requires some projective geometry, and this might be a good moment to ask for a leap of faith—it works. To give you some intuition, you could think of our flat plane as being embedded on some very large sphere. Then any two vertical lines we draw will look parallel as seen from within the plane, but if we pull back and see them as being on the sphere, we will see that they all intersect at the north pole of the sphere. Therefore we can think of the north pole as our extra point at infinity, and we call it O. This analogy is very limiting and should really be taken with a grain of salt, but I hope it gives some intuition for what O is. Now it becomes clear that P + (−P ) = O. Furthermore P + O = P , so O serves as an additive identity. Lastly, addition of points on elliptic curves defined like this is associative. The boom line is that the set of points on an elliptic curve with addition is quite similar to the set Z∗p with multiplication: it’s a group! Though the geometric explanation for the group law makes sense, it has its shortcomings: it doesn’t tell us how to compute the coordinates of P + Q in a way a computer could follow, and it doesn’t work when we consider our curve over Zp . These issues are easily taken care of: since all we’re doing is playing with equations of lines and cubics, we can derive formulae for the coordinates of P + Q in terms of the coordinates of P and Q (I invite the motivated reader to try and find these formulae). Once we have the formulae, we can use them to define addition of points and forget altogether about the previous geometric description. 43

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chalkdust

Elliptic curves over finite fields and our trapdoor function Let’s actually go through an example of an elliptic curve over a finite field to get a sense of what’s going on. Consider the curve y 2 = x 3 + x over Z5 , the integers modulo 5. We are going to compute all the points on this curve. The way to do this is to plug values of x into the right-hand side and to see if we get a square modulo 5, in which case we get a solution. For example plugging x = 0, we get y 2 ≡ 0 mod 5, which has solution y = 0, so (0, 0) is a point on our curve. Plugging x = 1, we get y 2 ≡ 2 mod 5, which doesn’t have any solutions. Plugging x = 2, we get y 2 ≡ 0 mod 5, so (2, 0) is on our curve. Continuing like this, we find (3, 0) and no other points. We also shouldn’t forget O (it’s there even over finite fields!). Hence in this case C(Z5 ) = {O, (0, 0), (2, 0), (3, 0)}.

A slightly more interesting example is the curve y 2 = x 3 + 2x + 2 over Z17 . I invite you to check that P = (3, 1) is a point on this elliptic curve. If we do some computations (most likely with a computer), we find that 19P = O and that P, 2P, . . . , 18P ̸= O. We say that P has order 19. We also find that P, 2P, . . . , 18P are all distinct. An interesting question is now: given the point P and a point kP on our elliptic curve, can we find k? The problem is analogous to the discrete logarithm problem, and is called the elliptic curve discrete logarithm problem. Even though it is relatively easy to state, it is very hard to solve, and our methods for tackling it are only marginally beer than brute-forcing. This is why we can use smaller prime numbers in the protocol without fear of geing hacked. Choosing a point G on our elliptic curve, our trapdoor function is then k 7→ kG, and we can implement Diffie–Hellman as before:

chalkdustmagazine.com

44


chalkdust

How much better are elliptic curves? The whole point of using elliptic curve cryptography instead of ‘classical’ algorithms is that we are able to achieve comparable cryptographic strength with much smaller key sizes. For example, a 228-bit elliptic curve key is equivalent to a 3072-bit Diffie–Hellman key. This means that choosing a 69-digit prime p for our Zp in elliptic curve Diffie–Hellman gives us roughly the same security as using a 925-digit prime for regular Diffie–Hellman. As you can imagine, computations are much faster with smaller numbers. Personally, I think the sheer fact that elliptic curves, which are objects used extensively in pure mathematics, can be impactful in everyday applications is amazing in and of itself! Axel Kerbec Axel is an undergraduate maths student at UCL. He is quite fond of number theory, and can oen be found rambling on about elliptic curves to his friends.

Extremely fashionable T-shirts

Visit d chalkdustmagazine.com to order one. My least favourite symbol

ϖ

David Sheard

A varpi is just an ω wearing a silly hat, invented purely to confuse the audience in a seminar or lecture. Rating: 0/10 [Ed: No David, this is an ω wearing a silly hat: ω] 45

spring 2019


Puzzles Word wheel 1 9

2

8

3

E 7

4 6

5

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

Each of the clues gives a six-leer word ending in E. Write these words in the circle, starting on the number and ending with the E at the centre. Once completed, the leers in the yellow squares will spell one of your favourite things. 1 Adjacent ÷ hypotenuse. 2 Due to this, a mug has genus 1. 3 SI unit of current. 4 Small. 5 Turn this on before making tea. 6 Multiply by 2. 7 Meaning of ! in ∃!x ∈ R. 8 {x ∈ R3 : |x| = 1}. 9 Multiply by 3.

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 le to right and top to boom ignoring the usual order of operations. For example, 4 + 3 × 2 is 14, not 10.

− × × −

Mini crossnumber by Humbug Across 1 9 times 2D. 3 A multiple of 6A. 1 2 5 6A less than a square number. 6 6D more than a prime number. 7 Equal to 1D. 3 5

4

6

7

= 40

× ×

= 10

+ = 13

×

+

+

=5

+

= 8

= 112

Down 1 A palindrome. 2 1A wrien backwards. 4 A square number that is equal to the sum of the factors* of 6D. 6 A square number.

(5) (3) (2) (2) (5) (5) (5) (3) (2)

* Including 1 and the number itself as factors.

chalkdustmagazine.com

46


chalkdust

Cryptic crossnumber by Humbug Across 1 Triangular one then square. (3) 3 Audible German no between tutus, for one square. (5) 5 Irreducible ending Morpheus halloumi fix, then (3) Trinity, then mixed up Neo. Down 1 Inside Fort Worth following unlucky multiple of (3) eleven. 2 Palindrome two between two clickety-clicks. (5) 4 Confused Etna honored thundery din became (3) prime. Wordsearch L E V Y N B L S A G S I D E F

T A A A R O T L O G E N A C A

T R D O S E S L E L S A U A U

G H W Y C I D N A B V H B L L

G N G K Z W L G I O C K E E K

E E L I A H N K R B O A C V N

N E R S R I E E O E O Z H O E

S E S M T W T N N V N R I L R

F E A H A S T O S A A I E C E

R M G L E I S R M K R M S H P

Y I S N E K N L A I A I O E P

N L E H C S R E H C P Y N N O

K O V A L E V S K A Y A A G H

H E J S P E L L I V R E M O S

G R A S E G R I F F I T H S S

2

4

3

5

Find the names of the mathematicians below in the grid to the le. The unused leers will spell a hidden message. Jackson Kovalevskaya Ladyzhenskaya Lovelace Mirzakhani Neale Nesterova Nightingale Perlman Robinson Somerville Steckles Vasilkova

Bell Browne Cartwright Cheng Daubechies Faulkner Fry Germain Goldwasser Grase Gray Griffiths Herschel Hopper

Arrange the digits II 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 le to right and top to boom ignoring the usual order of operations. For example, 4 + 3 × 2 is 14, not 10.

+

− =7

÷

+ ×

= 18

+ −

+

× ×

= 60 47

1

− = 53

+ = −16

= 3 spring 2019


chalkdust

Striking the right chord Phillip Maiwald, CC BY-SA 3.0.

Paula Rowińska

G

 ahead and try to solve this lile problem! Or, even beer, grab a friend (or enemy) and think about the problem together.

A curious problem

Suppose we have an equilateral triangle inscribed in a circle with radius 1. What is the probability that if we choose a chord of the circle at random, it will be longer than each of the sides of the triangle? The first thing you might try to √ do is compute the length of the sides of the inscribed equilateral triangle: you should find it is 3.√Let’s denote the length of the random √ chord by L. To find the probability that L is bigger than 3, which we shall√ denote P(L > 3), we need to divide the number of ‘desired’ outcomes (chords such that L > 3) by the number of all possible outcomes (chords of any length). I would not recommend drawing all possible chords of the circle; for problems with infinitely many possible outcomes we need a slightly different approach than simply counting infinitely many objects.

Geometric probability Luckily, we can use a very handy tool called geometric probability. It allows us to calculate probabilities in terms of size: length, area, volume etc. chalkdustmagazine.com

48


chalkdust For example, imagine that we want to calculate the probability that a real number M chosen at random between 0 and 1 is bigger than 0.6. Remember that we can represent real numbers on a line, so we identify picking the number M with picking a point on a line segment of length 1. In this case we need to divide the length of ‘desired’ outcomes by the length of all outcomes: P(M > 0.6) =

0.4

0.4 = 0.4 1.0

0

1

0.6

Too easy?

A harder problem Now imagine that you join forces with two of your friends and you each pick a random number between 0 and 1. What is the probability that the sum of the squares of these three numbers is smaller than 1? Give it a try before you continue to read! z

Your chosen numbers A, B and C represent the coordinates of the point (A, B, C) in the region [0, 1]3 , ie in the solid unit cube with one vertex at (0, 0, 0). Hence the volume of all possible choices of three numbers equals the volume of the cube: exactly 1. What is the volume of desired outcomes? We want A2 + B 2 + C 2 ⩽ 1, which describes the solid unit sphere centred at (0, 0, 0) with volume 4π/3. Since A, B and C are bigger than 0, we actually take only (1/2)3 = 1/8 of this sphere, the part which intersects the cube, so the ‘desired’ volume equals 1/8 · 4π/3 = π/6. Now we can compute our probability: P(A2 + B 2 + C 2 ⩽ 1) =

π/6 π = ≈ 0.524 1 6

y x

Unit cube intersecting the unit sphere

Striking a chord Now we are ready to tackle the original problem. Because we only care about the length of the chord, the problem is rotationally symmetric, which means that we can rotate the circle and the triangle however we fancy without changing the answer. To make things easy, let us assume that the chord starts at one of the triangle’s vertices. We pick randomly its other end X somewhere on the circumference of the circle. The vertices of our triangle divide the circle into three arches of equal length. The chord is longer than the side of the triangle if and 49

3 L

X We can pick endpoints of the chord at random spring 2019


chalkdust only if X lies somewhere in the blue arc, so P( L >

1 3) = . 3

Maybe you chose your random chord by a different method. To be safe we should check other methods to ensure we get the same answer.

Discord to our harmony Every chord is perpendicular to a unique radius (except a diameter which is perpendicular to two radii), so using the symmetry of the problem let us draw the top vertical radius and inscribe an equilateral triangle with one side perpendicular to this radius. We pick a random point X on the radius and draw the chord parallel to this triangle’s side, ie perpendicular to the chord. The chord is longer than the side of the inscribed triangle if and only if it intersects the radius somewhere in the blue segment between the triangle’s side and the centre of the circle. This segment is exactly half of the radius (check!), so P( L >

3 X L

We can pick a chord at random perpendicular to a radius

1 3) = . 2

Wait… it seems that we made a mistake, because we got a different answer! Well, luckily we can think of more methods of choosing a random chord, surely if we try another one we will find out which answer is correct.

3

L X 1 2

We can pick the midpoint of the chord at random

What if we pick at random a point X inside the circle, which will become the midpoint of the chord? We can draw a radius going through this point and construct the unique chord perpendicular to this radius. OK, you got me! Uniqueness has one exception: if we happen to choose the centre of the circle, this is the midpoint of infinitely many chords (diameters, in fact), but they all have the same length, so this does not cause a problem. Our chord is longer than the side of the triangle if and only if its midpoint is closer to the centre of the circle than to its circumference, ie it lies in the blue disk. Look again at the previous method to understand why this is the case. Now we just need to divide the area of the blue disk by the total area of the circle, which gives us P( L >

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3) =

π · (1/2)2 1 = 2 π·1 4


chalkdust This has not fixed our problem at all; instead of confirming whether 1/3 or 1/2 is the right answer, we have got yet another different solution.

What is going on? Do not waste your time trying to find mistakes in these three solutions. Since 1889, when the French mathematician Joseph Bertrand described this paradox in Calcul des probabilités, generations of mathematicians have come to the same conclusion: the probability that a random chord is longer than a side of an inscribed equilateral triangle equals 1/2, 1/3, 1/4, or maybe something else, depending on the method of generating the random chord. This example warns us that analysing probabilities without a clear statement of the problem can lead to disaster. To work with random events, we need to build models, which can be simple or very complex. Different models can give us different answers, as we just observed. ‘Choose a chord of the circle at random’ is not a strict, mathematical formulation. For mathematiJoseph Bertrand cians the phrase ‘at random’ must be followed by specification of some probability distribution, a function assigning a number between 0 and 1 to all possible outcomes. The formulation of our question allows for many possible probability distributions, we have just mentioned three of them. However, we can state the problem differently: ‘Pick independently at random two points on the unit circle, uniformly distributed on the circumference, √ and draw a chord between them. What is the probability that the chord is longer than 3?’ Now we can obtain the unique answer 1/3. Different formulations would give us different answers, as we saw before. Which solution is correct then? All of them. This is the beauty of mathematics: we can model the same phenomenon in many different ways, depending on our needs, as long as we clearly state our assumptions. We get different answers to different questions, so we do not have any contradictions here. If 1/2 were equal to 1/3 and 1/4, we would face the beginning of the end of mathematics. Thankfully, long ago, mathematicians realised the importance of making sure their problems were strictly defined and well posed. I believe that everyone should see this example at the beginning of their mathematical journey. I used to think that maths teachers developed a weird obsession with unnecessary formalities and I could not understand why I had to waste so much time writing down all the ‘obvious’ facts. Aer geing a few different answers to Bertrand’s problem I never complained about mathematical rigour again. 51

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Competition It is great fun trying to come up with other methods of choosing a random chord. Here are a couple of problems for you to think about. If you submit your answer to either of them on our website by 9 September 2019 we’ll choose at least two of the best to win prizes! √ 1. Come up with your own way to choose a random chord and calculate P(L > 3). 2. A random chord of the unit circle forms a side of a unique inscribed rectangle R. Choose a method (from the article, or your own) of selecting a random chord and calculate the probability that the area of the inscribed rectangle is greater than the area of an inscribed equilateral triangle T: P(area(R ) > area(T )). The best solution is not necessarily the most difficult or complicated. Depending on the method you use it can be easy or hard to calculate the probability, so anyone can have a go.

Paula Rowińska As a PhD student at Imperial College London, Paula uses maths and stats to study the impact of wind energy on electricity prices. The title of her TEDx talk ‘Let’s have a maths party!’ seems to summarise her two favourite activities.

a @PaulaRowinska d paularowinska.wordpress.com My favourite symbol

ζ

David Sheard

It takes practice to master writing ζ by hand, but the satisfaction from executing the perfect twirl-swish-flick is sublime! Also it is reserved for some of the coolest functions in maths. Rating: 10/10

My least favourite symbol

ζ

Sean Jamshidi

My least favourite symbol is ζ, also known as squiggle. I am still not sure what it’s supposed to look like. Rating: 0/10

Bases by Matthew Scroggs In base 2, 1/24 = 0.000010101010... = 0.0000̇1̇. In base 3, 1/24 = 0.00̇1̇. In base 4, 1/24 = 0.002̇. In base 5, 1/24 = 0.01̇0̇. In base 6, 1/24 = 0.013. Therefore base 6 is the lowest base in which 1/24 has a finite number of digits. Let n be an integer. What is the lowest base in which 1/n has a finite number of digits? chalkdustmagazine.com

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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

Hello World Hannah Fry A really enjoyable non-technical book, narrated in an anecdotal manner. Fry provides her view of why algorithms are so crucial in some disciplines, whilst being very critical about them. Algorithms have their biases, errors and misleading results, and it is up to us to use them responsibly. Reading Hello World has definitely made me more careful about accepting cookies or allowing mobile apps to access my device.

ggggg Humble Pi

Laws of Thermodynamics

Matt Parker Reading this book is definitely not a mistake.

Entertaining Spanish nerdy romantic comedy. Recommend watching it to understand the importance of these laws in our world.

ggggg

Wrong, But Useful

ggghi

The second best maths podcast.

ggggh

Odds and Evenings The second best maths podcast.

The Numberphile Podcast

ggggh

The second best maths podcast.

ggggh

Footnotes socks

Star Trek: Voyager

Avoid awkward first date moments by whipping o your shoes and debating which of the sixteen equations is your favourite.

The two-parter that ends the 3rd & starts the 4th series is the best.

ggggi

ggghi

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Oπnions Can a horse have an Erdős number? Lucy Rycroft-Smith

A

 you a mathematician? Care to quantify that? For decades, mathematicians have been having fun by measuring the collaborative distance between themselves and the most prolific mathematician (in terms of number of individual papers published) to date, Paul Erdős (although he may only be number two; this is disputed). Your Erdős number is calculated by the shortest collaborative distance between you and Erdős, where 0 is ‘you are Erdős’ and 4 is ‘you published a paper with someone who published a paper with someone who published a paper with someone who published a paper with Erdős.’ He was a favourite amongst people who tell colourful stories about mathematicians and people who write mathematical quotes on coffee mugs alike, widely quoted as saying: “A mathematician is a device for turning coffee into theorems.” Erdős died in 1996, meaning the number of people with an Erdős number of 0 or 1 is now finitely bounded. (Unless you believe in ghost writing.) The criteria used to define ‘paper’ and ‘published with’ vary, as you can imagine, but the Erdős Number Project at Oakland University in Michigan, USA, suggests:

Our criterion for inclusion […] is some research collaboration between them resulting in a published work. Any number of additional coauthors is permied. Not normally included are joint editorships, introductions to books wrien by others, technical reports, problem sessions, problems posed or solved in problem sections of journals, seminars, very elementary textbooks, books on history, memorial or other tributes, biography, translations, bibliographies, or popular works. Erdős numbers Erdős wrote or co-wrote 1475 papers by this definition. How many people do you think there are with Erdős number 1? What about 2? See if you can make a reasonable conjecture before I tell you.

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chalkdust There are approximately 511 people with Erdős number 1, and around 11,002 people with exactly Erdős number 2. Were you close? Now, what do you think happens to the distribution of these values as the Erdős numbers increase? See if you can predict the results for Erdős numbers 3 and 4. What do you predict to be the mean and median Erdős number for published mathematicians? What about non-mathematicians (however you define them)—what sort of numbers would you expect from them?

Number of mathematicians

88,000

66,000

44,000

22,000

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9 7 8 6 Erdős number

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A plot of the ‘Erdős function’ using approximate (and restricted) data from the Mathematical Review database in the US. Was it the sort of distribution you anticipated? The mean Erdős number from this data comes in at 4.65 and the median Erdős number is 5.

The Erdős Number Project at Oakland University has some intriguing things to say, including the revelation that at least one non-mathematician (a medical doctor) has an Erdős number of at most 9—and that someone once auctioned off an Erdős number of 5 on eBay. There is even a suggestion that there may be a horse with an Erdős number of 3… I went in search of that horse, of course.

Recent work by animal psychologists showing that not only primates but even birds have surprising intellectual capacities, opens the question as to whether non-humans can boast an Erdős number. The answer is indeed yes; the story is however curious. Jerry Grossman at Oakland University contributed an article to a Bridge magazine, jointly with Smarty, his wife’s horse. As Grossman (Grossman, Jerrold Wayne) has an Erdős number of 2, the horse achieved an Erdős number of 3.’ — The Extended Erdős Number Project

So: if we relax the criteria enough to include almost any publication, then yes, at least one and probably more horses have an Erdős number (please write in if you know of more—the Extended 55

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chalkdust Equine Erdős Number Project sounds like a brilliant plan). This would also mean that the numbers quoted previously would be significantly higher—but only, presumably, above the Erdős number of 1, because Erdős himself didn’t publish much in ‘lowbrow’ places. Does the measure become meaningless if we relax the constraints to this degree? We can turn to several parallels for ideas. The Bacon project computes an actor’s ‘distance’ from Kevin Bacon in the much the same way, using IMDB. This means that it uses data from a single database of credits on films and TV shows— an imperfect but fairly defined line (does it include cameos? singers? writers and directors?). Looking further afield, we come to the general principle of ‘six degrees of separation’, originally coined by Frigyes Karinthy in 1929—the idea that in any kind of highly connected network of human relationships or endeavours, the ‘average’ shortest path is six. Was and Strogatz showed that in ‘small-world networks’, the average path length between two nodes is equal to ln N/ln K where N is the total number of nodes, and K is the number of acquaintances per node. For example, in 2011 the average distance between people (by number of connections) was 4.74 on Facebook, and 4.67 on Twier. Of course, here we have prey discrete rules about connection: the idea of ‘friending’ or ‘following’ on social media is slightly less nebulous than ‘knowing’ in real life. Should we use some kind of index to tell people how connected they are to an arbitrary person of value on social media? It’s likely this would have very lile meaning; those sorts of connections, unlike publication credits, are easily obtained and cheaply won. So it seems that by most reasonably strict definitions, any mathematician worth their salt would expect an Erdős number of around 4 or 5, right? (I’ve been working up to something. Did you notice?)

I can remember exactly when I first heard of Erdős. It’s February 2000. I’m 16 years old and for my birthday, someone gets me a book with a weird title: The Man Who Loved Only Numbers. I’m not particularly into men and feel prey much the same way about numbers (perhaps I’m pi-curious), so I put it into a pile and forget about it for a while. Then, one evening, casting around for new reading material, I pick up the purple and turquoise paperback out of idle curiosity. I don’t sleep that night. I devour the book like it’s chocolate. I’m thrilled by the portrait of the highly eccentric, highly prolific mathematician who just arrives on each international doorstep with a suitcase and a catalogue of problems to solve, sure of a welcome and a cabal of willing minds. I’m uerly intrigued by the religious overtones of the book; to Erdős, mathematics is inseparable from a kind of fanaticism that inextricably connects him to a higher being who he suggests has wrien a book of all the perfect proofs in the world (called, somewhat underwhelmingly, The Book). His life’s mission: to discover as many of them as possible. His moo: chalkdustmagazine.com

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chalkdust “my mind is open”.

When he said someone had ‘died’, Erdős meant that that the person had stopped doing mathematics. When he said someone had ‘le’, the person had died. A teenage me, proficient but not amazing at mathematics, dreams of an Erdős number to call my own and all that it might bring: validation, acceptance into a community, that magic label “mathematician”. I’m tired of people not taking me seriously, of calling me a “silly girl”. I have no authority in my lile world, and the tales of Erdős and the reverential welcome he gets wherever he goes seem far, far out of reach.

It’s October 2017. Winter is seing in. I’m cold and grumpy. I stumble into work and fire up my computer, opening my email inbox. Then I sit up straight in my chair and punch the air, thinking ‘what a time to be alive’. This is the email I received on that fateful day:

I have just received my copy of Mathematics Today and see that you have very kindly added my name to your brilliant article. You now have the privilege of having an Erdős number 3, so shout about that! My PhD supervisor published with Erdős when the laer visited Reading many years ago. I have a couple of publications with my supervisor, so I’m a 2! Let me be clear—this was not a ‘brilliant’ mathematics research paper—it was a very ordinary review article from a conference, where I happened to have credited a kind colleague who read it through and suggested amendments with an entirely earned co-authorship. Nowhere in my wildest dreams would I have imagined that this would be the result. Thrilled is something of an understatement. Somehow, the time and space between me and this ‘real’ mathematician, respected and revered, had been shrunk. I was breathing the same air as Erdős. (Please don’t write in about that, either.) But of course, aer the euphoria had died down, I saw the real truth: while I might feel happy about this numerical value placed on my metaphorical value as a mathematician, it’s in many ways meaningless. The stuff that suggests a person has contributed to the field can be as intangible as a patient explanation to a child or a quick sketch of a unifying structure in the air with a finger. While formal publication is important, it’s not all or even most of what mathematicians do. The focus on geing this type of validation is prey telling of a group of people who feel insecure, and I’d put actors and mathematicians right at the top of that list, in my humble experience. What about the veritable army of maths teachers out there painstakingly explaining quadratic relationships for the fiieth time to the same class? What about the pubfuls of people playing with puzzles and games and origami every month? What about the women and LGBT people and BAME people and disabled people and countless others who don’t get to study to the highest level they like, don’t get picked for the jobs they’re qualified for, don’t get to publish their ideas where they deserve? 57

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chalkdust This isn’t just an ‘only measuring in one dimension’ problem—it’s a problem of thinking that the product, not the process, is the thing. Translated into education contexts, it’s easy to see why we have generations of children thinking not geing a perfect score on a mathematics exam means they are not a (‘natural’) mathematician. Put simply: you are not a mathematician because of an Erdős number. Sometimes, you are a mathematician in spite of it. Do we need a beer measure? Nope. We need to stop measuring. Erdős is widely quoted, and oen misaributed—see, for example, the first page of this article, where I suggested he said “A mathematician is a device for turning coffee into theorems”. Another famously misaributed quote (this time to Einstein) would seem to apply here:

Not everything that can be counted counts, and not everything that counts can be counted. — William Bruce Cameron Lucy Rycroft-Smith Lucy Rycro-Smith is a former maths teacher and now writes and researches for Cambridge Mathematics. She is the editor of Mathematical Salad, the weekly Cambridge Maths blog, and writer of the monthly Espressos: filtered maths education research for teachers.

a @CambridgeMaths d cambridgemaths.org Hungry spiders by Alaric Stephen and Alex Mayall An ant and three spiders are walking around a wire frame in the shape of a tetrahedron. The spiders are trying to catch the ant. The ant is very small, so the three spiders can’t see it, although the ant can see its aackers and plan accordingly. The ant is slightly slower than the spiders. Can the spiders find a strategy to be certain of catching the ant? Or can the ant always escape? Pictures Ant: Jacob Eckert from The Noun Project, CC BY 3.0.

You can hear Alaric and Alex discuss puzzles like this on their superb podcast Odds and Evenings. Find out more at d oddsandevenings.com.

Did you know... …that Ancient Babylonians did maths in base 60 instead of base 10? That’s why we have 60 seconds in a minute and 360 degrees in a circle. chalkdustmagazine.com

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Academic webpage checklist Are you an early-career researcher? Do you want to improve your online presence and boost your career? Just follow this simple checklist to make a webpage that looks like everyone else's!

Things to do today Picture of you on a hike, the first and last time you went outside Invalid email address in the form [my name]AT[my uni]DOT AC DOT UK 'Contact me' link to a different, but also invalid, email address Slides from the blackboard talk you gave last term A fax number A design scheme that only uses block primary colours A list of papers 'awaiting publication' Mismatched fonts

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chalkdust

On the cover

Harriss spiral

Matthew Scroggs and Edmund Harriss

T

 golden ratio (1.6180339…) has a rather overblown reputation as a mathematical path to aesthetic beauty. It is oen claimed that this number is a magic constant hidden in everything from flowers to human faces. In truth, this is an exaggeration, but the number does however have some beautiful properties. √ ), and is one of the solutions of the equation The golden ratio, oen wrien φ, is equal to 12 (1 + 5√ x 2 = x + 1. The other solution of the equation is 12 (1 − 5), or −1/φ. One of the nicest properties of the golden ratio is self-similarity: if a square is removed from a golden rectangle (a rectangle with side lengths in the golden ratio), then the remaining rectangle will also be golden. By repeatedly drawing these squares on the remaining rectangle, we can draw a golden spiral.

Le: The large rectangle is golden. If a square (blue) is removed, then the remaining rectangle (green) is also golden. Right: A golden spiral. chalkdustmagazine.com

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chalkdust Numbers that are a solution of a polynomial equation with integer coefficients are called algebraic numbers: the golden ratio is algebraic as it is a solution of x2 = x + 1. At this point, it’s natural to wonder whether you can create interesting spirals like this with other algebraic numbers. Unsurprisingly (as otherwise, we wouldn’t be writing this article), there are other numbers that lead to prey pictures. The plastic ratio, ρ = 1.3247179..., is the real solution of the equation x 3 = x + 1. Its exact value is √ √ √ √ 69 69 3 9 + 3 9 − ρ= + . 18 18 A plastic rectangle—a rectangle with side lengths in the plastic ratio—can be split into a square and two plastic rectangles. If this spliing is repeated on the smaller plastic rectangles and two arcs are drawn in each square, a spiral is formed. These particular arcs are chosen so that they line up with the corresponding arcs drawn in the smaller rectangles.

Le: The large rectangle is plastic and can be split into a square (blue) and two plastic rectangles (red and green). Centre: The two arcs drawn in each square. Right: A Harriss spiral.

This spiral is called the Harriss spiral, and is named aer its creator Edmund Harriss. It is the shape that appears on the cover of this issue of Chalkdust, and we think its resemblance to a tree in bloom makes it perfect for spring-time. We also believe that its beauty shows that the golden ratio is a gateway into a world of mathematical creativity, not an end point. There must be other nice algebraic spirals out there, buried in the roots of polynomials. If you unearth a prize-winning specimen, let us know. You may even see it on the cover of a future issue! Matthew Scroggs Mahew is a PhD student in the department of mathematics at UCL, who works on boundary element methods. He is one of the co-authors of the realhats LATEX package.

a @mscroggs d mscroggs.co.uk n @mscroggs@mathstodon.xyz f mscorps Edmund Harriss Edmund is a Clinical Assistant Professor in the Mathematics Department of the University of Arkansas, and author (with Alex Bellos) of the mathematical colouring books Snowflake Seashell Star and Visions of Numberland (called Paerns of the Universe and Visions of the Universe in the US).

a @Gelada d maxwelldemon.com 61

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#9 Set by Humbug 1

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Rules Although many of the clues have multiple answers, there is only one solution to the completed crossnumber. As usual, no numbers begin with 0. Use of Python, OEIS, Wikipedia, etc is advised for some of the clues. To enter, send us the sum of the across clues via the form on our website (d chalkdustmagazine.com) by 9 September 2019. Only one entry per person will be accepted. Winners will be notified by email and announced on our blog by 28 September 2019. One randomly-selected correct answer will win a ÂŁ100 Maths Gear goody bag, including non-transitive dice, a Festival of the Spoken Nerd DVD, a dodecaplex puzzle and much, much more. Three randomly-selected runners up will win a Chalkdust T-shirt. Maths Gear is a website that sells nerdy things worldwide. Find out more at d mathsgear.co.uk chalkdustmagazine.com

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chalkdust

Across 1 All the digits of this number are (15) equal. 8 A factor of 9D.

(2)

10 A palindrome.

(3)

12 A prime number.

(2)

14 The product of the digits of this (4) number is 70. 16 All the digits of this number are (15) equal. 18 This number is equal to the sum (4) of the fourth powers of its digits. 19 Equal to another eight-digit num- (8) ber that appears in this crossnumber. 25 An anagram of a palindrome.

(5)

27 This number is equal to the sum (3) of the cubes of its digits. 28 This number is equal to the sum (3) of the cubes of its digits. 30 Twice the fourth power of 12A.

(5)

32 A multiple of 3.

(8)

1 2 3 4 5 6 7 9 11 13 15

17 20 21 22 23 24 26

29 31

35 This number is equal to the sum (4) of the fourth powers of its digits.

33

38 All the digits of this number are (15) equal.

34

42 This number is equal to the sum (4) of the fourth powers of its digits. 43 Not equal to 41D.

(2)

44 This number is equal to the sum (3) of the cubes of its digits. 45 A power of two that is one more (2) than a multiple of 9. 46 All the digits of this number are (15) equal.

36 37

39 40 41 45 63

Down A multiple of 12A. An anagram of 3D. A multiple of 3 whose second digit is 4. An anagram of a palindrome. A factor of 10A. A multiple of 3. Each digit of this number is one more than the previous digit. Equal to 10A. A multiple of 8A. The sum of the digits of 1D. Each digit of this number is either one more than or one third of the previous digit. A multiple of 5. A multiple of 2D. Not a multiple of 3. A square number. The square root of 1 less than 30A. Equal to 23D. A five-digit number that appears in this crossnumber followed by a three-digit number that appears in this crossnumber. Less than 43A. A prime number that is equal to the average of the previous and next prime. Each digit of this number is either one more or three more than the previous digit. Each digit of this number is either one less than or three times the previous digit. Not an anagram of a palindrome. A three-digit number that appears in this crossnumber followed by a dierent three-digit number that appears in this crossnumber. This number is equal to the sum of the cubes of its digits. An anagram of a palindrome. A factor of 43A. The sum of the digits of 6D.

(6) (7) (7) (6) (2) (9) (8) (3) (3) (6)

(3) (9) (6) (2) (3) (3) (8)

(2) (3) (7)

(7) (6) (6)

(3) (3) (2) (2)

spring 2019


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

Dear Chalkdust, It’s Stephen here (I wrote an article for issue 04). I was catching up on some past issues over Christmas and I noticed that you featured an article in issue 07 on random waves (the On the cover article by James Cann) and an article in issue 08 on Truchet tiles. Now, there is an important conjecture in physics that says that these two things are connected… so I thought it might be interesting to write an article explaining this conjecture. Stephen Muirhead, London

Dear Chalkdust, TL’s Super Secret Math [sic] Club was delightfully stumped with Chalkdust’s crossnumber over lunch. They were happy to do something really hard! Halcyon Foster, USA a @halcyonfoster

Second Chalkdust podcast just popped up on my phone! That’s tonight’s drive home sorted!

Yay! Nothing like a good puzzle to keep the brain ticking over during vacations! Rachel Long

James Allen, London a @MeetJamesAllen

This is just a quick note to say thank you for all your work in producing Chalkdust. I have only recently come across it, and am really enjoying reading them Russell Brooks

Chalkwho? Sam Holloway, Cambridge a @samholloway

chalkdustmagazine.com

Excellent event this morning with a @DavidLammy and a @ch_nira organised by Chalkdust and a @DrTrapezio. Inspiring stuff! #BlackMathematician Helen Wilson, London a @profhelenwilson

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chalkdust

Bells, braids, and taxicabs Braids: Wikimedia commons user Stilfehler, CC BY-SA 3.0.

Andrew Stacey

A good puzzle leads you to explore unexpected places and discover hidden gems of mathematics. — Me, just now

B

 that reasoning, Catriona Shearer’s (a @CShearer41) puzzles rate extremely highly. Many of them have sent me on an intriguing journey through mathematics, right from the very first that I encountered. The resulting magical mathematical expedition is what I hope to show you in this article. The puzzle itself is on the next page. For such a lovely puzzle (as we’ll see), it didn’t get much traction. Indeed, the next day Catriona posted an update with some of her ideas about solving the problem. That second tweet was sufficient to pique my curiosity. Aer a couple of back-and-forth tweets, I learnt that the problem had been inspired by the bell ringing paern ringing the Wikimedia commons user Dougsim, CC BY-SA 4.0. Bell ringers changes. With a lile time to spare (these were posted in the school holidays!), I set to and investigated. I was pleasantly surprised to find myself embarking on an adventure that started with a morning of bell ringing, continued with a merry aernoon of braiding, and ended the day with a mad dash in a taxi. 65

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Dotty puzzle by Catriona Shearer Change 1 Change 2 Change 3 From one row to the next, each point swaps with the neighbour it didn’t swap with last time.

Change 4 Change 5

What percentage of the grid is enclosed by the red and orange lines? By red and yellow? By other pairs of colours?

Change 6 Change 7 Change 8 Change 9 Change 10

Let’s ring in the changes Those around me have goen used to me declaring “but that’s mathematics” to the point that they are quick to reply “yes, but you say that everything is mathematics”. Which is fair enough: I do have a tendency to think that. But with bell ringing, I’m not on my own in this claim. In bell ringing, the bells are rung in changes or rows. Each change consists of every bell being rung once in some order. From one change to the next, bells can swap position with a neighbouring bell. We can draw this in the fashion of Catriona’s pictures wherein each row is a change and each bell is a dot in that row. The first diagram below shows three such changes. A picture that doesn’t paint a thousand words is missing something. In this case, it’s very tempting to ‘join the dots’ and that’s a helpful thing to do as it shows the progression of each bell. Of course, colour is always a welcome addition as in Catriona’s original picture. Indeed, with the colours we don’t need the labels. The second diagram shows the end result. Change 1 Change 2 Change 3

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3 Joining the dots

Ringing the changes

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All tangled up Images such as the above are reminiscent of braid diagrams. A braid in mathematics is very like a braid in ‘real life’: it consists of strands travelling in a given direction which can cross over or under each other. The picture below gives a simple example. If you haven’t encountered braids before, here is a quick precis. They form a part of both topology and algebra. In topology, they are closely related to knots and links. Knots in mathematics model knots in the ‘real world’, the main difference being that we join the ends to stop them undoing themselves. Links generalise knots in that they can have multiple strands. A braid can be made into a knot or a link by joining the strands from the boom back up to the top and the crossings of the braid then create the knoedness of the resulting knot. Braids also find a place in algebra. The set of braids on a fixed number of strands forms a group, with the group operation being concatenation: placing one braid aer another. These groups have a lot of structure which provides a rich source of study. Knots, links, and braids have many applications both in wider mathematics and further. One such application is in the study of DNA: the helix structure of DNA makes it naturally braided. When DNA is replicated as part of cell division, it remains tangled so has to be untangled for it to properly separate A simple braid and be able to pass into the two new cells. There is a particular enzyme that does this, so a mathematical concept—unlinking—is realised by a particular biological process. The bell ringing diagrams are not quite braid diagrams, however. The difference is that in bell ringing there is no information as to which strand crosses over which. Despite the temptation of modelling the swaps by actually intertwining the bell ropes, this would quickly lead to them being unringable. There is no movement in the swaps in bell ringing and so there is no obvious choice of which strand should go over which. There are a variety of ways that the choices could be made, and readers might find it an interesting diversion to come up with rules to follow and to see what different braids result from the same bell ringing diagram. In what follows we will be most interested in something called the linking number of two −1−1 −1+1 1−1 1+1 = − 1 = 0 = 0 = 1 strands. This is a concept on loan from the 2 2 2 2 theory of links which measures how interCalculating the linking number twined two strands are. It is straightforward to calculate with braids: pick two strands and follow them down the braid diagram. Each time they cross count 1 if the right crosses over the le and −1 if the le crosses over the right. At the end, halve the number. This is the linking number of the two strands. Some examples are given above, showing why it measures their entanglement. 67

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chalkdust A very simple rule to use when converting a bell-ringing diagram to a braid diagram is to make every crossing ‘right over le’. This would ensure that the linking number between any two strands is as large as it can be. A possible line of enquiry would be to see how the linking number is affected by other choices of rule. With this rule, Catriona’s original puzzle translates to the braid shown to the right.

Enter the taxi The original problem asked about the area swept out between two strands as the braid progresses. (It is worth pointing out that this doesn’t depend on the choices of crossings in the braid.) There was also the ‘must swap if possible’ rule but we can equally well work with patterns that don’t have that property. We shall, however, assume that our braids are drawn in the more angular fashion of the original braid, which matches the style of the original problem. To talk about area, we need to decide on some units. We’ll assign length 1 to everything atomic: the horizontal distance between adjacent strands and the vertical distance between the lines corresponding to the bells being played. Now let’s look at two strands, say the red and orange strands. The first part of their journey looks like a trapezium, as in the following picture.

A braided bell ringing

Our choice of units make the area easy to calculate: it is simply the average of the two horizontal sides. More generally, for any two strands then the area of one segment where they don’t cross is the average 1 of their horizontal distances at the start and end of that segment. The crossings behave differently, see the picture below. In this case, 3 the area given by the above rule is 1 but the actual area is 12 . As is common in mathematics, we’ll make that 12 look more complicated to A single piece make life simpler later on. We’ll write it as 1 − 12 where we think of the 1 as being the area given by the trapezium rule and the − 12 a correction for the crossing. 1

To see what this looks like, let’s establish some convenient notation. We’ll convert each thread in the braid into a list of numbers. These will record the position of each strand as it passes down the braid. Equivalently, the list is of the position of the bell in each change. The lists for the braid above are (from le to right):

1 1

1

2

3

4

5

5

4

3

2

1

1

1

2

1

1

2

3

4

5

5

4

3

2

3

4

5

5

4

3

2

1

1

2

3

A single crossing

4

3

2

1

1

2

3

4

5

5

4

5

5

4

3

2

1

1

2

3

4

5

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chalkdust Notice that the last change is the same as the first. Let’s look at the calculation of the area between the first two strands. Separating it by change, it is:

|2 − 1| + |1 − 2| 1 |1 − 2| + |1 − 3| |1 − 3| + |2 − 4| |2 − 4| + |3 − 5| − + + + 2 {z 2} | 2 2 2 | {z } | {z } | {z } Change 1 (crossing)

+

Change 3

Change 4

|3 − 5| + |4 − 5| |4 − 5| + |5 − 4| 1 |5 − 4| + |5 − 3| |5 − 3| + |4 − 2| − + + + 2 2 {z 2} | 2 2 | {z } | {z } | {z } Change 5

+

Change 2

Change 6 (crossing)

Change 7

Change 8

|4 − 2| + |3 − 1| |3 − 1| + |2 − 1| + 2 2 {z } | {z } | Change 9

Change 10

What is important to note in this is that the second term in each fraction is the first term of the next. This even works with the last and first terms. Writing ua and v a for the terms in each list, we therefore have that this part simplifies to 10 ∑ a=1

|ua − v a |

(∗)

Note that we’ve used the fact that the first and last terms in our lists are the same here. The other part is the linking number of these two strands, at least the way that we’ve drawn the crossings. As the area doesn’t depend on the choices of crossings then this might not always be the linking number of the given choices. What we can say is that it is the maximum linking number. Let’s write this maximum linking number of the strands which start at positions i and j as χi,j . Now let us return to the first part of the expression for the area, given in equation (∗). This is where the taxis arrive.

The taxicab metric—more formally known as the ℓ 1 metric—is a way of measuring distance between points. In 2D, the image is of points laid out on a grid, much like a city such as New York, as in the next diagram. You can move along the lines on the grid but you can’t move across them. The distance between two points is therefore not the standard Euclidean distance but a different formula. More prosaically, it is not ‘as the crow flies’ but rather ‘as the taxi drives’. The distance between the two points in the diagram to the right is 2 + 1 + 2 + 2 = 7. No maer what choice of route, no shorter distance can be found. This gives rise to the formula 69

The distance between these two points in the taxicab metric is 7. spring 2019


chalkdust for the taxicab distance between two points a = (xa , ya ) and b = (xb , yb ) as:

∥a − b ∥1 = |xa − xb | + |ya = yb | There’s nothing special about 2D here. The analogous formula holds with arbitrary length vectors. Thinking of a vector as just a list of numbers (ie without trying to assign any physical meaning to the directions), our listing for each strand of our braid gives a vector. The expression in equation (∗) shows that we should drop the repeated ending position and so the vectors from the braid in Catriona’s original puzzle are of length 10. Our vectors from that braid are therefore: v1 = (1, 2, 3, 4, 5, 5, 4, 3, 2, 1) v2 = (2, 1, 1, 2, 3, 4, 5, 5, 4, 3) v3 = (3, 4, 5, 5, 4, 3, 2, 1, 1, 2) v4 = (4, 3, 2, 1, 1, 2, 3, 4, 5, 5) v5 = (5, 5, 4, 3, 2, 1, 1, 2, 3, 4) The area formula then simplifies to A(vi , vj ) = ∥vi − vj ∥1 − χi,j Calculating this for the above vectors (exploiting a bit of symmetry to avoid too many calculations), we get: A(v1 , v2 ) = A(v2 , v4 ) = A(v4 , v5 ) = A(v5 , v3 ) = A(v3 , v1 ) = 15 A(v1 , v4 ) = A(v4 , v3 ) = A(v3 , v2 ) = A(v2 , v5 ) = A(v5 , v1 ) = 23 In conclusion, the area between two strands is given by taking the taxicab metric applied to the corresponding vectors and then subtracting their maximum linking number.

And finally As I said at the outset, the mark of a good problem is that it gives you the opportunity to travel in mathematics. This particular problem took me on a meandering wander through some parts of mathematics that I know well, but that I hadn’t realised were so closely connected. It is easy to forget how intricately connected mathematics is, and I find expeditions such as this one help to remind me of this. While I knew of the connection between bell ringing and braids, I certainly wasn’t expecting to see the taxicab metric making an appearance. Moreover, its role here is to calculate an area whereas its natural home is as a way to measure length, so it also served to remind me that area and length are more closely tied together than I oen allow. This journey has also led me to places that I’d like to revisit when I have a bit more time. For example, the relationship between bell ringing and braids is not quite as I’d thought: I hadn’t realised chalkdustmagazine.com

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chalkdust the significance of the crossings. So there’s plenty to explore further with trying out different crossing rules to see what braids appear. Plus it gives me an excuse to draw some more braid diagrams, which is always a bonus. This article is inspired by a puzzle from Catriona Shearer ( a@CShearer41). The central council of church bell ringers is always looking for mathematically-minded recruits! Find them at dcccbr.org.uk. Andrew Stacey Andrew is a Maths and Computer Science teacher at Oxford High School, part of the Girls’ Day School Trust. Prior to that, he was an academic mathematician specialising in differential topology. In his copious spare time he writes packages for LATEX for drawing various mathematical objects including knots, links, and braids.

a @loopspace d loopspace.mathforge.org c loopspace@mathforge.org o loopspace My favourite symbol

Ш

Nikoleta Kalaydzhieva

My favourite symbol is Ш. Now, you may notice it does not look like your usual Greek leer, well that’s because it is not! It is one of the few Cyrillic leers used in mathematics. In analysis it is used to represent the Dirac comb, get it, because it looks like a comb, [insert head exploding emoji here]. In arithmetic geometry/number theory it is a very important leer used to denote the Tate-Shafarevich group which is one of the ingredients necessary to proving the BSD conjecture and consequently earning yourself $1,000,000. Rating: 1,000,000/10

The blue square by Catriona Shearer

?

16

The three red squares have areas of 16. What is the area of the blue square?

16 16

You can find Catriona and loads more of her puzzles on Twier a @Cshearer41.

My least favourite symbol

∂ Belgin Seymenoğlu

My least favourite symbol is the one you see in partial derivatives, because no-one actually knows how to pronounce it! Rating: ∂z∂ (x 2 + y + 3)/10 71

spring 2019


This issue features the top ten Chalkdust regulars. To vote on the top ten issues of Chalkdust, go to d chalkdustmagazine.com

At 10, it’s the page you probably didn’t use to find this page: the contents page.

Did you know that the online vote is the reason that did you know made it to number 8?

At 7, but about to move to 14 as an infinite number of people have arrived, it’s Hilbert’s hotel: the game.

At 6, it’s the puzzles page. Can you work out why it’s so popular?

At 5, and causing a moderate amount of recursion, it’s top ten.

Storming back into the top ten after not being seen since issue 03, it’s the horoscope.

At 4, despite being deemed not hot, it’s what’s hot and what’s not.

Dear Dirichlet,

what is the second most popular Chalkdust regular? ■

DIRICHLET SAYS: No idea.

chalkdustmagazine.com

At 9, it’s the big block of text on the page that no-one looks at: the editorial.

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Topping the pops this issue, it’s the crossnumber.




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