In this issue... Features 4 A nice warm Bath... …with Chris Budd
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Game, set, maths No more Katie Steckles
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e to thee x Zoe Griffiths on the life of e
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Counting palindromes Alex Burlton not l rub xela
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Adventures in Chance-ylvania High stakes gambling with Paula Rówinska
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Thinking outside2 the box Rob Eastaway joins the dots
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Primes à la Euler Sam Porri ends in a 3
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Domino tiling & domineering John Dore and Chris Woodcock also join the dots
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Machine learning We enslave/destroy humanity
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Baking a Menger sponge sponge Sam Hartburn bakes your favourite fractal
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Regulars Page 3 model 3 8 What’s hot and what’s not 17 Conference bingo 28 Dear Dirichlet 30 Signi cant gures
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Belgin Seymenoğlu tells us about Sir Christopher Zeeman
18 1
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On the cover by James Cann
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The comic by Tom Hockenhull
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How to make... a catastrophe machine
52 60 61
Prize crossnumber
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Top ten: mathematical days
Reviews Routes: Edgser Dijkstra by Emma Bell
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chalkdust The team Rob Becke Chris Bishop Hugo Castillo Sánchez Atheeta Ching Thuy Duong “TD” Dang Eleanor Doman Ed Goldsmith Sean Jamshidi Nikoleta Kalaydzhieva Antigoni Kleanthous Emily Maw Sam Porri Tom Rivlin Sally Said Mahew Scroggs Belgin Seymenoğlu Yiannis Simillides Alex Stokes Adam Townsend Cartoonist Tom Hockenhull
d chalkdustmagazine.com c contact@chalkdustmagazine.com a @chalkdustmag b chalkdustmag l chalkdustmag f chalkdustmag e Chalkdust Magazine,
Hello there! Welcome to issue 07 of Chalkdust! Once again, we have a ton of great quality content. We are probably most excited about you reading Zoe Griffiths’s poem, e to thee x (pages 18–19). Although we’ve featured poems on our leers page before, this is the first time we’ve printed one proper, and we had great fun working out how to design the pages to bring the most out of it. We hope that you decide to write something for our next issue. Even if you’re thinking of something that we’ve never published before—like the poem—you will almost certainly find us very happy to publish it… like the poem. As well as finding lots to read inside, we hope you’ll find lots to do. As always, you can find our notorious prize crossnumber, as well as other puzzles, things to make, and a couple of games to play. Next time you’re at a conference, make sure you take along a copy of conference bingo (page 17) to keep you entertained. In Katie Steckles’s article (pages 9–16), she tells us about some maths related to another game, No More Women. In this game, you name a thing and a category it belongs to, which is then disallowed. We decided to use the rest of this editorial to demonstrate… I (no more vowels) xpct ths txt s vry hrd t rd, bt myb sm f y wll njy dcphrng t. s Kt sys n hr rtcl, w wld clssify bnnng vwls s prtclrly mn mv. Bt r nxt n s prbbly mnr… W’r (n mr cnsnnts)
Department of Mathematics, UCL, Gower Street, London WC1E 6BT, UK.
,
’! ! The Chalkdust team
Acknowledgements We would like to thank: our sponsors for allowing us to continue making Chalkdust; Robb McDonald, Luciano Rila, Helen Wilson and the rest of the staff at the UCL Department of Mathematics for their continued support; Colin Wright, Katie Steckles, and the other MathsJam organisers for yet another excellent MathsJam Gathering in November; Hannah Fry, Thomas Oléron Evans, Helen Arney, Steve Mould and Bletchley Park for providing prizes for our advent competitions; Márton Mester for solving the triangle puzzle for Sean; Nira Chamberlain for his support during Black Mathematician Month; all this issue’s authors for sending us yet more excellent articles; and you for bothering to read the acknowledgements. 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 2018. All rights reserved. If you wish to reproduce any content, please contact us at Chalkdust Magazine, Dept of Mathematics, UCL, Gower Street, London WC1E 6BT, UK or email contact@chalkdustmagazine.com
chalkdustmagazine.com
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Eleanor Doman
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F I had a pound for every time someone assumed I studied maths because I wanted to be an economist without writing essays, I’d have enough to make it worth following the stock market. However, once the indignation fades, I can see the attraction— there are a lot of interesting uses of mathematics in economics. One of the most basic, yet most important, is modelling unemployment. Unemployment might be caused by too few jobs in an area. Or, it may also be due to a lack of information being provided to employers or potential workers: there may be perfectly good jobs available that quali ed workers simply don’t know about. This sort of unemployment is called frictional unemployment. We split the labour force L into two separate populations: employed (N) and unemployed (U). We then de ne s and f to be the rates at which people gain and lose employment: Rate of nding employment f Unemployed U(t)
Employed N(t)
Rate of job loss s The rate of change in unemployment is: dU = number becoming unemployed − number entering work dt = sN(t) − fU(t) If we assume that the total size of the labour force is constant, then this leads us to: du + (s + f)u = s, dt where u is the proportion of the labour force that is unemployed. A lovely rst order ODE, which can be solved using the integrating factor method (an exercise left for the reader). Simple enough that even an economist would understand! Pictures Desk work: Christian Heilmann, CC BY 2.0; Jobcentre Plus, Wikimedia Commons user Mtaylor848, CC BY-SA 3.0.
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In conversation with. . .
Chris Budd Ed Goldsmith, Emily Maw, Sam Porritt and Sean Jamshidi
“L
’ chat any time, I’m fairly free.” Coming from Chris Budd, a professor of mathematics at both the University of Bath and the Royal Institution, as well as a board member of several of the most influential mathematics organisations in the UK, this is somewhat of a surprise. But he is true to his word, and one Friday aernoon we sat down for a conversation with one of the UK’s most experienced voices in mathematics communication. A quick glance at Budd’s website reveals, through a CV that runs to 25 pages, the diversity of his interests and professional experiences. At various times, he has advised on seing A-level examinations, held highranking positions in professional societies like the In- Wikimedia commons user Mintchocicecream, CC BY-SA 3.0 The University of Bath stitute of Mathematics and its Applications, directed the Bath Taps into Science festival and been part of the Vorderman Commiee, which produced a report in 2011 about recommendations for mathematics education. All of this is in addition to a research career that has seen him tackle problems in nonlinear partial differential equations, high performance computing, weather forecasting and cancer treatment. Speaking to Budd, however, it is clear that all of these are underpinned by one thing: a love of mathematics, yes, but more importantly a passion for communicating it, and a willingness to engage with those in industry, politics and education who might not otherwise support mathematical research. chalkdustmagazine.com
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Maths in the real world Such a noble undertaking cannot, however, be faced alone. Since 2001, Budd has run an accredited course at Bath about maths communication, which trains up mathematics students and has them going into schools, running workshops and working at science festivals. Although the programme was initially designed to provide a formal structure and training for those who would be doing these things anyway, Budd quickly saw another advantage. “It turns out that some of the best ambassadors for maths are students. They’re closer in age to the audience, and they’re oen beer at communication than older generations.” The course now aracts around 25 students per year and he has just wrien a comprehensive guide about it. King’s College London and the University of Bristol, among others, have independently begun running similar courses. “I think all universities should run this course,” Budd says. “It leaves a lasting impact on the students, teaches them transferable skills and helps the wider community as well. I can see no downside to it really.” The idea of talking about mathematics outside of schools, allowing people to have fun with it and see it as the ‘hands-on’ science that it is rather than something bound to books, is surely an invaluable way to boost the profile of the subject. The fact that undergraduates are taught to do this—that it is seen as a valuable asset for a mathematician—is similarly revolutionary, and with Budd at the helm we have no doubt that the course produces talented communicators.
It turns out that some of the best ambassadors for maths are students… they’re beer at communication than older generations.
Building transferable skills, doing things during a maths degree that are not just mathematics, crops up again when we speak about Budd’s own research, which he describes as “industrial applied mathematics”. Again, even in his academic work it is clear that Budd enjoys breaking out of the university bubble and geing his hands dirty in the real world, working on problems that he hopes can “benefit society directly”. He has had several students funded by CASE awards— industry-supported PhD positions that require the student to spend a portion of their time with the organisation that sponsors them. “It broadens your overall experience, and it’s a lot of fun. I had a student working with the food industry, looking at the design of microwave cookers. It’s just great!” Other students on similar awards have worked with the electricity board, and have worked on designing sensors for whale conservation. Today, however, the meaning of industrial matheChris has previously worked on matics is changing somewhat. The importance of cybersecumodelling microwave cooking rity in the modern economy has driven a boom in industrial number theory, and pure mathematics is also playing a big part in hot topics such as big data and machine learning (see pages 69–71). Hearing this is somewhat surprising, and perhaps comes from a slightly snobbish aitude towards industrial mathematics that Budd believes to be completely unjustified. “Industrial mathematics is not about dumbing things down. When you start working on a real problem, you very quickly find that you run out of mathematics that you know. Industrial 5
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chalkdust problems are, and always have been, a marvellous way of generating new mathematics.” But surely, even with this broad definition of what is industrially useful, there will be areas that are le out? Surely research driven by purely mathematical interests rather than commercial ones is still valuable? Budd agrees: “I would always support abstract mathematics”. He also points out that, for every graduate student who has their research funded by, say, the aviation or toothpaste industries, universities can spend more of their own money on funding people to research areas of mathematics that are not (currently) applied in industry.
Industrial problems are, and always have been, a marvellous way of generating new mathematics.
Maths in the future As well as training the next generation of research mathematicians, Budd has invested a lot of time in trying to broaden the appeal of elite mathematical events to new audiences, particularly those from state schools. He is chair of the UK Mathematics Trust, an educational organisation that reaches 600,000 pupils a year—providing many of them with the joy of problem solving that is, “on some level, why we all do mathematics”. In particular, Budd is keen that competitions like the International Mathematical Olympiad aract a more diverse team. “If you look at the make-up of the team, it’s been mostly male and mostly they come from independent schools… if I can achieve more diversity in my time as chair I’ll be rather pleased.” He points to the recent launch of the European Girls’ Mathematical Olympiad, as well as team challenges that include a poster round, as ways of making the competitions more approachable. Charitable organisations like the UKMT could be even more vital in the near future, as the full effects of educational reform become apparent. Mathematics Alevels were changed from modular to linear in 2017, meaning that students have to sit exams on subjects ranging from calculus to statistics all in the same season at the end of a two year period. The fear is that this will scare people away from taking additional (further) mathematics courses: with lile opportunity to test the water via modular exams, they could feel like UK parliament, CC BY-NC 2.0 Few members of parliament have mathit is too much of a gamble to take two A-levels in the ematics degrees same subject. “I fear that this will affect the state sector disproportionately… teachers at state schools are just as dedicated, but they don’t have the time or the resources.” But why is mathematics geing a short shri in schools? Budd has previously spent time in Singapore, a country which tends to do very well in mathematical league tables. One reason for this could be that Lee Hsien Loong, the prime minister, holds a mathematics degree from Cambridge. “Mathematics has support from the top down, and that goes right the way through the system. In the UK, I don’t know of a single mathematician in the House of Commons.” (In fact, since the 2017 general election, there are three members of parliament with mathematics degrees—Stephen chalkdustmagazine.com
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chalkdust Timms, Karen Bradley and Bill Esterson.) The point, however, is still that mathematics does not enjoy the same political profile as it might in other countries. As an anecdotal example, Budd recalls applying to a scheme organised by the Royal Society, where 30 research scientists each year are partnered with a politician in order to learn about how research findings can have an impact on policy. Budd’s application was successful on the Royal Society’s side, but was not chosen by any of the participating MPs. “That was rather disappointing, actually.”
Maths on the television? If politicians are not reliable conduits for mathematical knowledge, where else can we look? “My advice is to work with the media. TV is a hard nut to crack, but people have done it.” Throughout our conversation, Budd namedrops previous Chalkdust interviewees Hannah Fry, Marcus du Sautoy and Ian Stewart, as well as Eugenia Cheng, as exemplars of mathematics communication. “Hannah is excellent. She’s out there, prepared to take on the challenge of going public. It’s not easy being in that position.”
TEDxGeorgeMasonU, CC BY 2.0
Chris encourages mathematicians to work with the media
Although this route may not be for everyone, Budd is insistent that mathematicians should support their colleagues who spread mathematics outside of the classroom, whether that be through the media or industry. “Don’t give up. Remember that most journalists think of themselves as ‘English people’ rather than ‘maths people’ and be sensitive to that.” From someone who has built a reputation through speaking about mathematics to those who aren’t necessarily ‘maths people’, we think this is advice worth listening to. “People really are fascinated by mathematics. They love the idea that there are problems that can take years to solve. Get out there. Be bold!” Ed Goldsmith, Emily Maw, Sam Porritt and Sean Jamshidi Ed, Emily, Sam and Sean are PhD students at UCL and members of the Chalkdust team. Ed and Sean work in geophysical fluid dynamics, Sam works in number theory, and Emily in geometry.
Inscribing a triangle 75◦
75◦
The diagram on the le shows a triangle inscribed inside a square. Can you prove that the shaded triangle is equilateral?
(There is at least one proof that doesn’t require the use of any trigonometric functions.) 7
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WHAT’S
& WHAT’S
HOT NOT HOT
The new Chalkdust T-shirt
Available to buy at d chalkdustmagazine.com
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Agree? Disagree? a @chalkdustmag b chalkdustmag
Available in charity shops throughout the midlands.
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Maths is a ckle world. Stay à la mode with our guide to the latest trends.
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NOT 274 207 281 − 1 Past its prime.
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Mathematical poems To see why these are hot, see Chalkdust pages eighteen to nineteen.
HOT 13 March
An excellent maths-themed occasion, and a great day to launch a maths magazine.
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Papers with multiple pictures
You might actually have a chance to understand one of these.
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Game, set, maths (no more tennis puns)
Flickr user Spiralz, CC BY 2.0
Katie Steckles
A
while ago, my friends and I learned of a brilliant and simple game, played by comedians Alex Horne and Tim Key. We discovered a clip of them playing the game backstage at a show, and were immediately hooked. It’s a naming game, requiring a vague knowledge of a bunch of famous people, and anyone can play. The game is called No More Women (reasons for which will become clear later) and I’m taking the opportunity both to share the rules of the game with you, and explore a lile of the maths behind it.
No More Women Each move in the game consists of two phases. First, you name a celebrity (for example, “Simon Singh”). Then, you exclude a category of people—and primarily, the category you give must include the person you’ve just named (for example, “no more authors”). Play then passes to the next person, who must name another celebrity—but crucially, they can’t name anyone who falls in a category that’s already been excluded. Play continues until someone’s stated celebrity is proven disallowed, or they can’t think of someone. There are a few other rules—for example, you’re not allowed to disallow everyone. There must always be at least one person le in the remaining chunk of people that are allowed—for example, having heard the move described previously, you wouldn’t be allowed to say “Courteney Cox; no more people who aren’t authors”—as the two categories, ‘authors’ and ‘non-authors’, together make up the whole set of people. 9
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chalkdust If the people you’re playing with suspect that your most recent exclusion has made the next move impossible, they may challenge—and you’d have to name someone you can think of who’s still allowed. Challenging like this is a risky move, as you’re then not allowed to say the example just given, and you have to think of yet another one.
Richard Cooper, CC BY-SA 3.0
“Simon Singh, no more authors.”
Alan Light, CC BY 2.0
“Courteney Cox, no more actors.”
It’s also generally agreed that the people you’re naming should be famous people—as in, people everyone might reasonably be expected to have heard of. If you name someone that nobody has ever heard of, your fellow players can insist you think of someone else, or decide whether a quick cheeky online search is allowed, to verify that they are indeed a tennis player (or whatever you’ve claimed them to be). It’s a maer of opinion whether someone who’s not a real celebrity but is known to everyone in the group playing should be allowed, although this does make it a bit more fun if you allow it. Also, the properties you give should be ones which people might reasonably be expected to know the truth value of for the majority of people you might name: while it does exclude a sensible proportion of people, “no more June birthdays” is harsh unless you are prepared to let people look up every single name then given to find when their birthday is—and that kind of thing leads to cheating. Impromptu discussion can also break out, for example around what exactly constitutes an author—does it have to be someone who makes a living writing books, or does a presenter who’s wrien an autobiography count as an author, since they have a published work? The title of the game is obviously a nod to one massive move you could make in the game, excluding basically half of all people (although is it half of all famous people? Smash the patriarchy etc). As the game continues, the set of people you’re allowed to name gets increasingly small, and the game becomes more difficult. The rate of increase in difficulty depends on how cruel your fellow players are with category choices—“no more brunees” is much harsher than “no more people from Bristol”, but if that’s the only fact you know for sure about your given celeb, you’ll have to go with that.
Particularly mean moves include “no more currently living people” and “no more people whose first and second name begin with different leers”. The trivial move is to say, for example, “Pat Sharp; no more people who used to present the kids’ TV show Fun House”, excluding precisely one celebrity (this is valid, but boring). “John Venn, no more logicians.”
This is an excellent game to play when you’re all, for example, siing around on public transport, or somewhere you don’t have access to conventional game-playing equipment and/or tables. Part of the challenge is remembering all the categories that have been counted out already, although if you prefer you could write down the categories on a board or piece of paper everyone can see as they’re named, and be real sticklers. But is writing down the names in a list the best way? chalkdustmagazine.com
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No more lists A more interesting way of writing down the moves could make the game more concrete and help us understand it mathematically. Since each move divides the space cleanly (assuming the definition of your category is precise enough), you could imagine the people all on a page, then draw a circle around a set of people and everyone inside the circle could be in the category, and vice versa. This leads us to a natural way to describe moves in the game—using Venn diagrams. Everybody knows Venn diagrams—the overlapping circles of catStephen egories described by John Venn in 1880, now a staple of set theno more aut King ho rs ory and a lovely way to visualise sets of things with certain properties. In this game, by definition, every person falls neatly in or Antonio out of a set. For example, you could have “no more authors” Banderas which would exclude everyone inside the set ‘authors’. In terms Lupita of Venn diagrams, the nicest way to display this would be to call Nyong'o the outside of the circle ‘authors’ and the inside ‘non-authors’, so the circle contains all the people still allowed (see diagram). By our rules, there must be at least one person outside the circle JK Rowling (in particular, the one you just named on your turn), and one person inside the circle (which you’d need to be able to name if challenged). uth
or e
a A second move (eg “no more brunees”—agreement will have to be reached whether you count natural no more bru rs ne tho au brunees only) could then overlap with this, restricting the next move to those in the intersection of those Antonio Banderas two circles. You’d also need at least one person to Lupita Nyong'o be in the region you’d just excluded (the non-author brunee you just named) and at least one still alJK Rowling lowed (one non-author non-brunee you can name Stephen King if challenged). However, it’s not necessary for there to be a non-brunee author (no offence, JK Rowling), as that category has already been excluded, so we don’t care what happens there.
no more bru ne
ors
es
Ed Miliband
Barack Obama
Charles Kennedy Lupita Nyong'o
JK Rowling
Antonio Banderas
Jeffrey Archer Stephen ff King
po
cians liti
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no m or e
no m or e
es
ff
th au
no m or e
Each move of the game adds another circle to this diagram—it’s easy enough to draw a third circle to create a three-way Venn diagram. Let’s say, for example, someone says “no more politicians”; this would overlap with all four existing categories (brunee authors, in the outer region; brunee non-authors; non-brunee authors; and non-brunee non-authors, which is where allowable moves still lie). You’d need to have named a non-brunee non-author politician as your turn (we went with Charles Kennedy, confirmed redhead) and you should also be able to name a non-brunee nonauthor non-politician (such as Lupita Nyong’o, who
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chalkdust was in The Jungle Book, but hasn’t published one yet).
no more bru ne
ors
p eo all
no m or e
ple
th au
es
no m or e
A four-way Venn diagram using only circles is not possible—in the example shown on the le below, not all sets of intersections are present. For example, there would be nowhere to place Steven Tyler (a brunee non-author who’s tall and not a politician). However, a four-way diagram can be constructed using ellipses (shown on the right), and was designed by John Venn himself.
po
no m ore t
cians liti
Two aempts to draw a four-way Venn diagram. In the aempt on the le, you can dream on if you think there’s a space for Steven Tyler. le, devised It is also possible to construct Venn diagrams with five identical pieces—the one on the by Branko Grünbaum has rotational symmetry. The five-way Venn diagram shown next to it is called an Edwards–Venn diagram, and ffwas devised by Anthony Edwards. They represent a way to create Venn diagrams with arbitrarily many regions, using sections of a sphere projected back down onto a plane, and were devised while designing a stained-glass window in memory of Venn. Starting with the le and top hemispheres, given by the two rectangles, and the front hemisphere represented by the circle, the next set is the shape made by the seam on a tennis ball (winding up ff and down around the equator) and further sets are made by doubling the number of oscillations on subsequent winding lines. They’re sometimes called cogwheel diagrams, due to the shape.
A ABD AB DE
BE
ABC E BCE
DE
BCDE BDE
ABCDE
BD
D BC
D
B
ABE
CD E
E
AD
ADE
AB
AE E AC DE AC CE
BC ABCD CD ACD
ABC
AC
C
Two five-way diagrams: Grünbaum’s diagram (le) and an Edwards–Venn diagram (right) ff
Technically any number of segments is possible, although the diagrams get much more complex as chalkdustmagazine.com ff
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chalkdust you go on. There’s a lovely interactive seven-way Venn diagram by Santiago Ortiz (on his website: d moebio.com/research/sevensets—neither he nor I am able to determine the original author of this shape). Of course, this would still only get you seven moves into a game of No More Women, and representing a full game in diagram form would be challenging and likely uninformative.
No more Venn diagrams Another way to interpret the game’s structure would be to use half-planes. If you could theoretically arrange everyone in the (so far useful) 2D plane of all celebrities—presumably a private jet—then a category excluded could be represented by a line cuing the plane in half, with all the allowed persons on one side and the disallowed persons on the other side. For example, the line x = 2 describes a vertical line on an xy-plane running through the point 2 on the x-axis, and everyone to the le of the line might be an author and everyone to the right not an author. Then, a second line at x + y = 1 might cut diagonally across, with brunees above and non-brunees below. It reminds me slightly of the loci problems we used to play with at school—a way to visualise solutions to sets of linear equations, or to determine which bits of a field a tethered goat can reach (for some reason, it’s always a goat).
A goat acting out a maths puzzle
We can intersect arbitrarily many half-planes to define the space of allowed people. In fact, any convex set can be described as an intersection of half-planes. But this is probably not useful either— firstly, the sections will become arbitrarily tiny and hard to see, much as in the Venn diagram case; secondly, you’re kind of imagining the people all standing in the room (or on a very, very big plane) and each time you define a new line you’d need everyone to be miraculously standing on the correct side of it. This is leading me to imagine celebrities looking upwards, then sprinting across so they’re on the correct side before a giant imaginary looming line comes crashing down, which is fun to picture, but not hugely helpful. And finally, the lines are defined in a prey arbitrary way—the categories we’re using are not quantitative (unless you’re using a category like “no more under-25s”, in which case you can plot that on an axis), and otherwise there’s no natural way to assign an equation to a category, so it’s a bit unsatisfying.
No more diagrams So I guess we’ll have to fall back on classic set theory. Developed by Cantor, while he was attempting to work out a way to compare the magnitude of different infinite sets, the theory of sets underpins a huge amount of mathematical rigour and thinking, and contains parallels with algebra and logic in ways that illustrate the beauty of mathematics in its purest form. But we just want to play a stupid game about famous people, so here goes. In set notation, we might define: Authors = {Simon Singh, Stephen King, JK Rowling, ...} The set is specified by the list of things in brackets, so this set equals this collection of people. Then our game would consist of moves as follows: 13
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Simon Singh ∈ Authors Authors ̸= ∅ Here the ∈ symbol means ‘is an element of the set’. We use a line above the name of the set to mean ‘the complement of this set’, or the set of all things in the universe that aren’t in this set. In the Venn diagrams we defined earlier, Authors would be the inside of the ‘no more authors’ circle, and the set Authors would be everything outside this circle. We’ve also specified, in the second line, that the set Authors—the set of all non-authors—is not equal to ∅, where this symbol denotes the empty set. This means that set is not empty, because something exists in it. Play continues: Courteney Cox ∈ Brunees ∩ Authors Brunees ∩ Authors ̸= ∅ Here we’ve used the ∩ symbol to define an intersection—this is the set of all things that occur in both the given sets—here, brunees who are also not authors. Along with the ∪ symbol for a union (the set of all things in either or both sets), this is one of the two main operations you can do with sets, and they correspond respectively to the and operators in Boolean logic, in a rough sense. On the Venn diagram, the intersection is the part of those two sets which overlaps—here, Brunees is the outside of one circle, and Authors is the inside of another circle, so this is the portion of the non-Authors circle that doesn’t overlap with the non-Brunees circle. Again, the requirement of being able to name an example means that the set of non-brunee nonauthors now has to definitely not be empty, and must contain something. Ed Miliband ∈ Politicians ∩ Brunees ∩ Authors Politicians ∩ Brunees ∩ Authors ̸= ∅ And so on. Sets can intersect arbitrarily, and while this doesn’t necessarily give us a nice visual way to imagine the celebrities, it does give us a formal structure. Properties of set intersections can be considered as they apply to the game—for example, set intersection is commutative: A ∩ B = B ∩ A. This means it’s independent of ordering, which feels obvious—brown-haired women are women with brown hair, duh—but in mathematics you have to be careful whether the order in which you do things maers (multiplying by three then adding four is different to adding four then multiplying by three, and oen on Facebook only a real genius can work out the correct answer to this math problem, so keep your wits about you). Intersection of sets is also associative: A ∩ B ∩ C = (A ∩ B) ∩ C = A ∩ (B ∩ C). chalkdustmagazine.com
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chalkdust Intersecting three sets is the same as first intersecting two of them, then intersecting the result with the third one. Because the action of intersecting sets is associative, it doesn’t maer which order you do this in, you’ll get the same result. Someone who’s a brunee author and also a philanthropist could also be considered to be an author/philanthropist with brown hair.
These two properties together mean that if you’re playing the game, and the categories defined are given in a different order—say, you’re playing against two others and they each give a specific category—they could occur in either order and still leave you with the same challenge aerwards. Non-brunee non-authors are just as hard to think of, and just as numerous, as non-author nonbrunees. It will, however, affect the choice of named celebrity each of the two other players is allowed to give. There are other properties of sets you could think about in terms of gameplay—for example: A ∪ B = A ∩ B. This says, the complement of the union of twoff sets is the same as the intersection of the complements. This is shown in Venn form to the right, and it essentially means that if you consider the complement of each set (the things outside it) ff individually, the things they have in common will be the same as just the things that are in the complement of the union of these two sets—consider them together as an overlapping shape, and look outside of that.
A
B
In the context of the game, you can consider this to mean that if you’re looking for someone who’s not a brunee not an author (because of two successive turns that have occurred in the game), you need to think about the set of people who are either a brunee an author, and look for someone not in that set. Again, this feels obvious when you think about it in the context of naming celebrities, but the set theory confirms it. Maybe formalising the game in this way will help you to get your head around the ideas of sets and set theory, if you’ve only recently encountered it. As someone who studied it—cough—years ago, it’s become part of my way of picturing any kind of problem like this, and a natural language with which to describe intersecting sets. I’ve been trained through years of maths to take any fun thing and abstract it into some symbols on a page. Yay!
No more No More Women Of course, the mathematician’s real job is abstraction then generalisation—so this game could be played using any well-known set of things that can be categorised according to properties, and we have piloted among ourselves a much more niche version of the game, provisionally titled No More Integers. 15
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It’s also totally possible for someone to say something that’s wrong/has already been excluded but nobody notices because their brains are all fried from factorising numbers in their head. It might take a few plays through before you all learn the best way not to end up in a mass brawl with whomever excluded the primes, but hopefully you can find some enjoyment in it. Or just play the version with celebrities—it’s a lot of fun. Katie Steckles Katie Steckles is a mathematician based in Manchester, who gives talks and workshops on different areas of maths. She finished her PhD in 2011, and since then has talked about maths in schools, at science festivals, on BBC radio and TV, at music festivals, at comedy shows and on the internet. She blogs at d aperiodical.com and has a YouTube channel at m youtube.com/katiesteckles.
a @stecks d katiesteckles.co.uk m youtube.com/katiesteckles My favourite equation Maths is full of equations. We’ve spread some of our favourite, and not-so-favourite, throughout this issue. We’d really love to hear yours! Send them to us at c contact@chalkdustmagazine.com, a @chalkdustmag or b chalkdustmag and you might just see them on our blog!
Euler’s polyhedral formula Emily Maw
Euler’s polyhedral formula says that the number of vertices, V, edges, E, and faces, F, of a convex 3D polyhedron satisfy V − E + F = 2. This is one of my favourite equations, because it’s arguably the first theorem in topology! Despite the name, Euler couldn’t actually prove it; the first of many proofs were found by Leibniz and Cauchy. It can be used to obtain strong constraints on the existence of shapes, for example in the proof that there are only five Platonic solids, or that all polyhedra built from pentagons and hexagons (such as footballs) must contain exactly 12 pentagons! The reason it holds for convex polyhedra is that they are all homeomorphic (topologically equivalent, by smoothing off the corners) to a sphere. In fact the formula can be generalised: for nonconvex polyhedra, which can be homeomorphic to surfaces with holes, the formula becomes V − E + F = 2 − 2g, where g is the genus (number of holes). This is because the le hand side is secretly calculating a topological invariant of the space, called the Euler characteristic, which is equal to 2 − 2g for surfaces.
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I have a truly marvellous picture of a scorpion which this margin is too narrow to contain.
Played on the set of all numbers, starting with the complex plane, it’s basically a free-for-all and in general we’ve found it becomes very difficult very quickly, not just to name a number that hasn’t been excluded, but to think of a category that works and doesn’t just make the game totally boring and impossible. If non-integers are excluded early on, and then numbers with given factors start geing thrown out, it can get a bit like the coding/drinking/coding-while-drinking game Fizz Buzz.
How to play
Next time you’re at a conference, you and a friend/coworker each pick one of the grids below. When you see something happen cross it o . The rst player to cross o ve boxes in a row wins.
Grid A A proof is left as an exercise
Someone reading Chalkdust
You fall asleep in a talk
Drunk Fields medallist
One of everything for breakfast
Slides projected in the wrong aspect ratio
Laser pointer gets shined in your eye
Fire drill
You run out of tea bags in your room
Someone else playing conference bingo
Run out of co ee
Your free pen runs out
Ugly PowerPoint slides
A video doesn’t work
Board pen runs out
Someone cites your paper
A talk overruns into the break
Ugly Beamer slides
You understand a talk
The roof is leaking
A question that’s actually a comment
Someone forgot their Mac adaptor
Spill ketchup on your last clean shirt
Someone uses confusing notation
Someone uses chalk
Grid 1 No clock in the room
You smash your conference mug
Awkward forced socialising
You forgot toothpaste
Too cold in room
Permanent marker on a whiteboard
Skip talks to go sightseeing
Person in front of you playing Minesweeper
You get someone’s name wrong
Too hot in room
You sleep through rst talk
Only bad biscuits left
You can’t connect to eduroam
Someone gets your name wrong
Projector overheats
Slide clicker batteries run out
You lose your name tag
Your supervisor is drunk
You get on the wrong German train
Windows needs to install updates
PDF slides not full-screened
No milk for tea
You lose your programme
You’re the only one using the hashtag
Someone’s phone goes o in a talk
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I thought the three of us Were meant to be, We were all part of the same identity.
I’m e, If you don’t know me I live Between 2 and 3. This is my story.
But then we grew up.
I’m misunderstood. Everyone thinks I’m just 2.71 But there is more to me That they don’t see I wish they would.
I started to see Pi from a new angle, She had nice legs, I loved that rest of her body was basically a rectangle. We went on a couple of dates.
And I’m named aer a leer Which makes things worse! The other numbers laugh at me for that too.
But then came Pi Day: Three, Fourteen. She let them approximate her!
So I say to them—look at the things I can do, I mean, surely I’m the only number To have walked this earth And expressed themselves in verse?
And she became An overnight sensation, A household name, One of those faces Everyone knows.
But that’s not cool, that’s not fashionable, It’s all about being able to Express yourself as something fractional, instead.
And lost in that world of meaningless approximation She wouldn’t let me take her to more than two places.
Otherwise they say there is something wrong with you You’re crazy, Irrational.
Time passed, And even my old friend i and I grew apart, The headiness of youth Replaced by the steadiness Of age, I began to see what others had told me And I’d refused to believe: i was imaginary!
Natural numbers can count themselves Lucky. They can easily find their space Because our place On that line Is defined By our digits And I don’t know all of mine.
You might be wondering what about Tau? She is twice the number Pi will ever be, But Pi and Tau they are similar And for me it was all a bit familiar, Tau, she’s just too Pi.
But there was a time When there were three of us who didn’t fit. Pi also didn’t know, Exactly where to stand, or sit, And next was i, She was in a dimension of her own! chalkdustmagazine.com
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e to thee x, this poem I’d wrien, This part of myself I’d given Was supposed to feel just right, But it was the opposite It was the inverse of natural (Logarithm?)
So that’s it, back to lonely me. It was hard And I’m not a negative number. But then I met her, x.
Actually, that poem, didn’t happen. I just dreamt that. It’s the 21st Century I sent her a Snapchat.
She said ‘I’m x’, I asked, ‘are you a multiplication sign?’ She laughed, ‘I get that all the time’, ‘No, I’m curly x,’ She said.
I sent it and then screamed inside Until she read it, and she replied.
She’s curly x, she’s sort of curvy x, But that’s not it, I’m not one to make judgements Based on digits or figures, She’s different, She’s fun.
I won’t tell you what she said But the bit that is etched in my head Is the final line, a string of Xs. The first was curly x, That’s her name, The rest were to say She’d like to see me again.
And it’s true I can’t always work her out But I like that, too. With her is where I always want to be, I want her to be my unknown quantity!
And if you’re worried that what is essentially A joke about numbers Just did something unexpected to your heart, Then shame on you! Numbers can have feelings too.
So I wrote her a poem ‘e to thee x’, (It was beer than this).
So this story is about me and x And the number she’s shown me I can be, I’m e, I live between 2 and 3, I don’t know where exactly I don’t care!
She opened it, tentatively, She read it, awkwardly, Aloud, Other numbers could hear! Never have I wanted so much To just Disappear.
With x, I see things from a different view, I laugh in the face Of the quite frankly ridiculous number queue.
I’m me, I’m e! Zoe Gri ths Zoe is a maths communicator who visits schools with Think Maths to give engaging talks and workshops. She likes poetry!
a @ZoeLGriffiths 19
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Counting palindromes Flickr user Chris, CC BY 2.0
Alex Burlton
A
a year ago, I was walking to work when I started to think about palindromes. This was partly because I’d seen one at work the day before, but mostly because my iPod had packed up on me, leaving me with a good half an hour with nothing beer to do.
A palindrome is a word like LEVEL, which reads the same forwards and backwards. Specifically, I was thinking about 5-digit numbers, so my palindromes were things like 01410. I’d seen one like this at work the previous day, which got me thinking—what were the chances of that happening? This turns out not to be a very interesting question to ask—for a random 5-digit number, the answer is simply 1 in 100. The first 3 digits can be whatever they like, and then the remaining two must match up with a 1-in-10 chance for each. M Disdero, CC BY-SA 3.0
I was still 25 minutes from work at this point, so I needed to make The Sator Square (AD79), a word square containing a fivethings harder. Next, I thought about what would happen if you word Latin palindrome threw re-arrangement into the mix. If a well-meaning person presented you with a bag of five numbers and wanted you to make a palindrome, what are the chances that you could do it? To answer this, we need to first define exactly what is meant by one of these ‘bags’. In mathechalkdustmagazine.com
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chalkdust matical terms, I’m talking about unordered multisets. Unordered simply means that, for example, {1, 2, 3, 4, 5} and {5, 4, 3, 2, 1} are equal—they are considered to be two representations of the same set. And multiset means that the sets may contain repeated elements, so {1, 1, 2} is a different set to {1, 2}. If it’s possible to make a palindrome from the individual elements, then I’ll call such a set palindromic. To see how complicated this makes the problem, let’s think about something simpler for a moment. How many of these sets are there? When we were talking about 5-digit numbers, this was easy— there are 100,000 (it’s simply all the numbers from 00000 to 99999). However, there are far fewer unordered sets of 5 digits. To see why this is true, consider the set {0, 3, 3, 8, 8}. This is a single set, but can produce 30 different 5-digit numbers (03388, 33088, 83830 to name a few). So, do we just divide by 30? Nope—the relationship isn’t even linear. For {1, 2, 3, 4, 5} there are 120 rearrangements, and for {0, 0, 0, 0, 0} there is only 1. Clearly this isn’t simple, although there is a known solution to this question which we’ll come to a bit later.
Breaking the problem down For the remainder of my commute I didn’t calculate the answer, but I did sele on an approach to complete when I returned home that evening. To solve the problem, I broke it down into cases. In fact, as I was thinking about 5 ‘things’ and how oen they were repeated, I thought about it like a game of poker. As an example, let’s think about the bags where all 5 digits are the same (“five-of-a-kind”). There are clearly 10 of these, and for any one of them you can make a palindrome. Similarly, there are 90 “four-of-a-kind” bags (of the form AAAAB: 10 choices for A, then 9 for B: 10 × 9 = 90) and 90 “full house” bags (of the form AAABB), all of which also produce palindromes. Continuing in this way, the remaining cases are AAABC (“three of a kind”), AABBC (“two pair”), AABCD (“one pair”) and ABCDE (“high card”). Crunching the numbers reveals that 550 out of a total of 2002 bags are palindromic—the chance of finding a palindrome in the rephrased problem is now a lile over 25%. Coincidentally, the total number of bags is itself a palindrome in this case! Well, that was a fun diversion—nothing more to see here, right? Wrong. Now I wanted to solve the general problem, by which I mean two things. Firstly, let’s throw away the digits 0–9 and have an arbitrary alphabet of b characters. That could be A–Z, it could be base 64—whatever we feel like. And instead of bags of size 5, let’s make that arbitrary as well and call it n. Is there a general solution for this probability in terms of b and n? It turns out that yes, there is—but it would be a couple of weeks before I found it.
The general problem The first step was to realise that counting unordered sets was something I’d encountered before, while studying combinatorics at university. I dug out my lecture notes and found the multinomial 21
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chalkdust coefficient, which gives an expression in terms of n and b for how many unordered multisets there ( ) are in total: n+b−1 (n + b − 1)! = . b−1 (n )! (b − 1)! Plugging in n = 5 and b = 10 gives us 2002, which matches the number we manually worked out earlier on. To see why it’s true in general, we must first rephrase the problem—something which happens a lot in combinatorics proofs. Suppose that we choose to represent our sets in a different way. We’ll agree an order for our alphabet (eg A–Z, or 0–9), and use it to write out a particular set as follows. Starting with the first character in the alphabet, we’ll draw a star for any that are present in our set. Then, we’ll draw a single line to show that we’re onto the next character, and repeat the process. This is much easier to demonstrate with an example:
With a lile mathematical rigour, you can show that these two representations are equivalent, meaning we can instead choose to count how many ways we can draw out n stars and b − 1 lines. The total number of characters is n + b − 1, and we must choose b − 1 of them to be lines. This is our answer! This was brilliant—most of my work was done for me. Now I just needed to work out how to count the palindromic ones for the general case.
Two tricks There were two ‘tricks’ that got me to the solution. The first was something I spoed almost straight away, but its significance didn’t become clear to me until later. There is a noticeable paern in the number of palindromic sets as n oscillates between odd and even. To see what I mean, here are the numbers where b = 10 (the digits 0–9, say) and we let n increase: Bag size, n
2
3
4
5
6
7
8
9
Palindromic sets
10
100
55
550
220
2200
715
7150
The number of palindromic sets multiplies by 10 every time we increase from an even number to an odd number. More generally, it multiplies by b, and when you think about it the reason becomes clear. In an even-sized set, every element must be paired for a palindrome to be possible. When we increase the size by one, we can put any of our b elements in there and it’ll still be palindromic—the new element can just go in the middle! Mathematically, if we label the set of palindromic sets over n and b as Pbn then the result is: chalkdustmagazine.com
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|Pb2n + 1 | = b |Pb2n |
∀n, b.
This fact is interesting, but it feels like we still have the bulk of the work to do. All we’ve shown is that if we can solve one of the even or odd cases, then we’ve finished the problem. And that’s where the next trick came in, focusing on the even case. Again, I’ll illustrate this with a simple example. This time I’m going to restrict us to an alphabet of {0, 1} to make things more manageable:
(a) All bags for n = 3
(b) All palindromic bags for n = 6
What’s illustrated here is that the number of palindromic bags of even size is the same as the total number of bags that are half the size. Just take every bag from the space half the size and double up all the elements—everything you’ve just created is unique and palindromic, and you can show that every palindromic set can be created this way. And, crucially, we know the total number of bags in the smaller space because that’s the multinomial coefficient. We’ve done it! The final form is as follows, where M(n, b) is the multinomial coefficient for n and b: n M( 2 , b) , if n is even, M(n, b) M(n, b) = M( n−2 1 , b) , if n is odd. b M(n, b)
Convergence
My final area of interest before puing this whole thing to bed was to explore the limits of my closed form in n and b. I was able to graph the function (pictured on the next page), and then thought I might as well do some analysis too. As it turned out, I was in for one last surprise. The limit as b increases isn’t very interesting—if we think about it logically for a second it’s obvious what should happen. For a fixed size of bag, if we just keep increasing the number of characters in our alphabet then obviously the probability of being able to make a palindrome converges to 0 (and prey quickly, I might add). However, things weren’t so simple when looking at the n case. I expected this to go to 0 as well— it feels like by making the bag larger you’re making it increasingly difficult to find palindromes. This intuition is correct, but only partially. The probability does decrease as the bag size increases (ignoring the fluctuation between odd/even, of course). However, it doesn’t decrease to 0—there is 23
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Plots of the palindromic density function
a positive limit! This was exciting for me as it was completely unexpected. It also explained why my aempted proofs that it converged to 0 hadn’t been going very well… The limit in n comes out as follows—something that can be shown using L’Hôpital’s rule: 1 2b−1
b
, if n is even,
2b−1
, if n is odd.
But why is this? Can you come up with a logical reason for why this should be the case?
An alternate problem To conclude, I’ll leave you with an alternative version of the same problem. At the time of writing, I haven’t delved into it myself yet, so there may be other interesting results to be found. The re-phrased problem is this:
Challenge Given an alphabet of b characters, pick n elements at random (allowing duplicates). What is the probability that you can make a palindrome with the resulting set of elements? There is a subtle difference here, which I’ll leave for you to figure out. Is there a general solution for this version? Does it have interesting limits? Enjoy!
Alex Burlton Alex studied maths at the University of Warwick, graduating in 2013. He is a soware developer for a clinical soware company based in Leeds. He is a qualified day skipper—when he’s not working or solving puzzles, you can find him sailing the Caribbean!
c alexburlton@hotmail.com My least favourite equation
Euler’s identity Matthew Scroggs
My least favourite equation is eiπ + 1 = 0. People rant about how it’s the most beautiful equation in maths, but it’s not. Overrated. chalkdustmagazine.com
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Adventures in Chance-ylvania Paula Rowińska
“I
’
really sorry, the vampirograph indicated that you are a vampire.” Imagine that you (or your mother/brother/girlfriend/pet scorpion) received such a message. You’re probably terrified, worrying about the future, thinking about the upcoming treatment. Wait a moment! Before you start panicking, consult… a mathematician. Medical tests aren’t perfect. Testing positive for an illness doesn’t necessarily mean that you’re sick; for many reasons, tests can detect things that aren’t really there. On the other hand, a negative test result doesn’t exclude the disease without any doubt. The question is: can we quantify the level of uncertainty linked to a particular test? Or in this case: if you test positive on a vampirograph, what’s the probability that you’re really a vampire? A similar question was tackled in the 18th century by a Presbyterian minister, Reverend Thomas Bayes. Bayes’ theorem became a basis for statistical inference, even though conclusions drawn from it are sometimes counter intuitive. His result gives us an explicit formula to update our prior belief about the probability of some event of interest based on additional evidence: P(event|evidence) = P(event)
P(evidence|event) . P(evidence)
Let’s get some intuition about this equation. I assume you’re familiar with the notion of the probability measure P; don’t worry about a rigorous definition, a common interpretation—ie how likely 25
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chalkdust the event is—will suffice. The mysterious symbols P (something|something else) denote a conditional probability—how likely something is given that something else happened. Now we’re ready to take a look at Bayes’ theorem again. In the beginning we have some vague idea about the probability of the event, a so-called prior probability P(event). For example, let’s say we’re interested in the probability that this cute guy we just met is a nice person. We might base our estimation on our previous experience with similar people or the fact that he’s a mathematician (mathematicians are usually nice people, aren’t we?), but our knowledge is prey limited. However, later we gather some additional observations (he smiles a lot, he helps us solve a ridiculously difficult equation etc) and we keep updating our prior probability. In the end we’re le with a posterior probability P(event|evidence) of him being a nice person given the evidence we have.
P(this picture contains a vampire) = 1
Hold on, how did we get from vampires to estimating if a cute guy is also nice? (No, I’m not a Twilight fan.) Bayes’ theorem has many applications! Before we approach our vampire problem, we need to make a few assumptions—all numbers come from my imagination [citation needed]. The scenario is as follows:
k k
Of the vampirography participants, approximately 2% are in fact vampires.
k
When someone is an actual human being, they have 0.1 chance of being falsely “detected” (so the remaining 90% of vampirographs give legitimate negative results).
When someone is a vampire, they have 0.85 chance of being detected, ie geing a positive result from a vampirograph (so there is a 0.15 chance they remain undetected).
In other words numbers: positive result negative result
vampires (0.02) 0.85 0.15
humans (0.98) 0.1 0.9
Now assume that you tested positive on a vampirograph. What are the chances that you’re a genuine vampire? Time to ask Bayes for help. We’re interested in P(vampire|positive result)—the probability that you’re a vampire if you tested positive. So what do the numbers tell us?
j
The probability that you’re a vampire based only on the fact that you’re geing a vampirography: P(vampire) = 0.02.
j
The probability that you’re a human based only on the fact that you’re geing a vampirography: P(human) = 0.98.
j j
The probability that you test positive if you’re a vampire: P(positive result|vampire) = 0.85. The probability that you test positive if you’re a human: P(positive result|human) = 0.1.
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chalkdust We also need the probability that you test positive regardless of what you are (you’re either a vampire or a human, we assume no other possibilities). This is a bit more tricky, but let’s see what we can squeeze out of our data. We’ll need the law of total probability, which might be interpreted as a weighted average of probabilities, where we average over all possible cases. In our example we have only two possibilities—someone is either a vampire or a human. A vampirograph gives a positive result, in each of these cases: rightly when we deal with an actual vampire and falsely when the participant is human. Therefore we can split our calculation of the probability of the positive result into these two separate cases. P(positive result) = P(positive result|vampire)P(vampire) + P(positive result|human)P(human)
= 0.85 × 0.02 + 0.1 × 0.98 = 0.115, where we have used the law of total probability. Now we’re ready to plug everything into Bayes’ formula: P(vampire|positive result) = P(vampire)
P(positive result|vampire) 0.85 = 0.02 · = 0.148. P(positive result) 0.115
Yes, even though vampirography seems to be prey good at detecting vampires, if you test positive the chance that you’re actually a vampire is only 14.8%! No need to panic yet, I guess. Why is this number so small though? This is always the case with very rare conditions, when the prior probability has a big influence on the posterior. Before the test, the chance that a randomly chosen participant prefers human blood to ketchup is very small, only 2%, because this is the proportion of vampires in the population. Geing a positive result significantly increases this value, but we started from a low level, so the final probability remains quite low. Luckily most dangerous diseases, such as different types of cancer, tuberculosis or AIDS, are relatively rare, which means that conclusions of our study would be similar if you replaced being a vampire with a real illness. This means that a worrying test result was most likely a mistake, not a real problem, and that you should follow up with a doctor and possibly repeat the test. Conclusion? Take care of yourself and get tested regularly (this article isn’t sponsored by the NHS in case you’re wondering). However, if you test positive, don’t panic and consult with a doctor… or a clergyman. Preferably Thomas Bayes. Paula Rowińska Paula is doing her PhD at Imperial College London, where she’s using maths and stats to predict electricity prices. She suffers from a mild theatre addiction.
a @PaulaRowinska d paularowinska.wordpress.com Did you know... …that 40 is the only number in English that is spelled with its leers in alphabetical order? 27
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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, I’ve recently had the good fort une of winning three pigs at the village fete. However, I’m not sure whether my tria ngular garden is big enough for them as well as my collection of metal, woode n and other deckchairs. The pig s are of substantial size and my tape measure is not long enough to measure the lon gest side of the garden. I’ve also heard that pig s are very intelligent and would like to hear suggestions for entertaining them.
— Pearl among swine, Lower Brailes DIRICHLET SAYS: It seems as though you have an issue with these pigs hogging your space. If your garden is right-angled, you can use Stythagoras’ theorem. Otherwise, I recommend pigonometric functions: the swine and coswine rules will be helpful. I shan’t boar you with the details. If they are math-ham-atically inclined, perhaps you could introduce them to Porkdust. Maybe skip the article about the ham sandwich theorem. My Bacon number is 2. ■
Dear Dirichlet,
competitive sport. I have been inspired to take up After a thrilling winter Olympics, us and I have no lie mostly in multivariable calcul However, my previous interests anything I’ve le. It’s completely di erent from sty life g rtin spo a ow foll to clue how erience in this area? done before. Do you have any exp — Mr Kim, Pyongyang
■ DIRICHLET SAYS:
Congratulations on your change of variables. On the surface, it might just seem a bit of fun and games, but exercise is integral to a healthy life. I recommend heading down to the gym to see if you can join a combined aquatic and winter sports team. Once a member, you can expected to be ∇ed on your ∇×ing and ∇·ing.
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Dear Dirichlet,
Thanks to your helpful advice in Chalkdust issue 06, I am now the pope! The rst ever pope, in fact, to also unders tand nite element methods. Un fortunately, I went for a stroll the other day to purcha se some badger feed and, being new to the area, I got completely lost. How can I get home?
— Benedict Cumberpope, Location una
vailable
■
DIRICHLET SAYS: Never fear! If you’re lost in Italy, just speak to
Anna (my pal-in-Rome). She cannoli point you in the right direction. For future sojourns, however, I have one pizza advice. The Rome-bus will take you directly to St Peter’s Square. From there, it’s a short hop to the numerical-analysistine chapel. Make sure you get off at the right stop though — otherwise you’ll be pasta point of no return.
Dear Dirichlet,
d known , I decided to participate in the tren son sea tive fes the of es ess exc After the ts, in search of acwent all animal-based produc as Veganuary. For 31 days I fore have decided to pere. Now that the month is over, I ceptance on my Instagram pag king. Would am looking to diversify my coo and le, sty life an veg a pt ado manently d recipes? you happen to know of any goo l Metcalfe, Winchester
— Pau
■
DIRICHLET SAYS:
My dear child, it seems you are limiting yourself to s-kale-r products as you are cross with yourself. I consulted on this matter with my friend William Hamiltomatoes and my work colleague Henri Poincarrots, with whom I commute. I am afraid to report that your choice of ingredients will be limited to vegetabelian groups. Furthermore, you will no longer be able to eat duck a Lagrange (as we have realised that Lagrange is an animal). If you decide to weaken your constraints, there are stiltons of vegetarian options. I myself enjoy macaroni cheese, or for something actually Italian, ris-8. If you can’t find rennet-free parmigiano-reggiano, my briemann hypothesis is that any other hard cheese is a goudapproximation.
More Dear Dirichlet, including two seasonal specials, online at d chalkdustmagazine.com 29
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Signi cant gures
Sir Christopher Zeeman Nicholas Jackson, CC BY-SA 3.0
Belgin Seymenoğlu
A
the Chalkdust issue 06 launch party, we brought along a challenge for our guests: we connected two people up with ropes and challenged them to separate themselves. To make things interesting, they weren’t allowed to remove the ropes from their hands, cut the ropes or untie the knots. Although the trick is 250 years old, it was made popular by Sir Erik Christopher Zeeman who used it as an interactive way to demonstrate topology, hence the challenge became known as Zeeman’s ropes. Erik Christopher Zeeman was born in 1925 in Japan to a Danish father, Christian Zeeman, and a British mother, Christine Bushell. A year aer his birth they moved to England. Zeeman was educated at Christ’s Hospital, an independent boarding school in Horsham, West Sussex. He did not enjoy the experience, feeling it was a prison in which he lost his self-esteem. In 1943–1947, Zeeman served in the RAF as a flying officer. In his own words:
I was a navigator on bombers, trained for the Japanese theatre, but that was cancelled because they dropped the atomic bomb a week before we were due to fly out. Since the death rate was 60% in that theatre it probably saved my life, but at the time I was disappointed not to see action, although relieved not to have to bomb Japan, the land of my birth. During his service, Zeeman forgot much of his school maths. But this didn’t stop him from going chalkdustmagazine.com
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chalkdust on to study maths at Christ’s College, Cambridge, where he earned his MA. Zeeman stayed on in Cambridge for his PhD, in which he wrestled with unknoing spheres in 5 dimensions, spinning knots in 4 dimensions, as well as trying (and failing) to solve the Poincaré conjecture (which would only be resolved in 2005 by Gregori Perelman). He was supervised by Shaun Wylie, who had worked with Alan Turing at Bletchley Park during the war on projects including deciphering a German teleprint cipher called Tunny.
Founding Warwick In 1963, Zeeman was invited to join the newly established University of Warwick as the foundation professor of mathematics. He initially declined, since he believed Cambridge to be “the centre of the mathematical world”. However, aer “a sleepless night”, Zeeman changed his mind and made the biggest move of his life in 1964. At Warwick, Zeeman was determined to “combine the flexibility of options that are common in most American universities, with the kind of tutorial care to be found in Oxford and Cambridge”. Initially, he recruited lecturers in three main branches of mathematics: analysis, algebra and topology. Legend has it that those he invited to Warwick all declined their offer; his response was to encourage them by telling them that all the others had accepted his invitation. Later on, Zeeman also appointed six lecturers in applied mathematics. His leadership style was informal, which helped produce an atmosphere in which mathematical research flourished. By the time Warwick accepted its first students in October 1965, the department was already competing with other universities at an international level. The glass building it is now housed in is named aer Zeeman in honour of his tremendous effort in founding the department. Zeeman le Warwick in 1988, and was made an honorary professor there upon his departure. He moved on to become the principal of Hertford College, Oxford and Gresham professor of geometry at Gresham College, London. He retired from these two positions in 1995 and 1994 respectively. The Zeeman building, home to the University of Warwick’s maths department
Catastrophe From 1966 to 1967, Zeeman was a visiting professor at the University of California, Berkeley. Shortly aer his return to Warwick, a dynamical systems symposium was held, aended by many of the world leaders in dynamical systems, including Stephen Smale and René Thom. They inspired his change of discipline from topology to dynamical systems, and prompted Zeeman to spend a sabbatical with Thom in Paris, where he grew fascinated with what came to be known as catastrophe theory. 31
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chalkdust He is famous for inventing a catastrophe machine, consisting of a circular disc that can rotate freely about its centre, and two elastic bands of identical length aached on the edge of the disc (see page 43). The other end of one piece of elastic is fixed, while that of the second elastic is free to move on the plane. Zeeman’s machine has some surprising behaviour: as the free end moves around, the disc would do something unexpected: it flips to a drastically different position. The flipping action is a vivid example of a catastrophe: a discontinuous effect resulting from a continuous change of forces. You can plot the set of points at which the disc flips, called the bifurcation set, which takes on a diamond-like shape consisting of four concave edges and four cusps. According to Hirsch, Zeeman once tried to take his machine with him to the USA. As soon as he mentioned the name of his machine, US customs cleared the room and had Zeeman arrested!
The bifurcation set on my cardboard catastrophe machine
Zeeman played a huge role in making catastrophe theory a hot topic in the 70s. He was keen to apply it in numerous contexts such as nerve impulses, the collapse of bridges, stock markets and even prison riots. On returning to Warwick, he taught a course in catastrophe theory for undergraduates, which soon became extremely popular.
Outreach and the Royal Institution Zeeman was not only passionate about his research, he was also heavily engaged in promoting mathematics to the general public. He was the first mathematician to present the Royal Institution Christmas lectures, in 1978. Going by the title of Mathematics into pictures and including a mix of pure and applied mathematics, Zeeman inspired his live audience with the aid of various demonstrations, including his own catastrophe machine. His lectures sparked plenty of enthusiasm; among the live audience was Marcus du Sautoy, a budding young mathematician who would go on to deliver his own Christmas lectures in 2006. But that was not the only result from Zeeman’s Christmas lectures—they also served as the inspiration behind the Royal Institution masterclasses for both mathematics and engineering. Starting from 1981, the masterclasses were designed to inspire keen schoolchildren across the UK. When I aended them as a schoolgirl in 2005, I had no idea about the history behind the masterclasses at the time!
Awards and positions In 1975, Zeeman was elected a fellow of the Royal Society and was president of the London Mathematical Society (LMS) from 1986 to 1988. He also took up many other positions and received various awards—too many to list in one article. Zeeman was knighted in 1991 for his “mathematical excellence and service to British mathematics chalkdustmagazine.com
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chalkdust and mathematics education”. More recently, the Institute of Mathematics and its Applications (IMA) and the LMS jointly set up the Zeeman medal in his honour to recognise those who “have excelled in promoting mathematics and engaging with the general public”. And you don’t have to be a seasoned professor to earn the medal—the 2016 Zeeman medal was won by author Rob Eastaway (see pages 38–42). Solomon Lefschetz
Epilogue Shaun Wylie
Zeeman had three daughters and three sons from two marriages. One of his daughters, Mary Lou Zeeman, became a mathematician herself, eventually collaborating with her father in mathematical ecology. In total, Zeeman had 29 PhD students, including David Epstein, Terry Wall and Jaroslav Stark, and over 700 other descendants, including me! It is now about two years since Zeeman passed away aged 91, on 23 February 2016. I never met him in person, but I have seen and felt the effects of his legacy, and am proud to be (academically!) descended from him. Some of his methods in dynamical systems which he applied in mathematical ecology came in handy for my research in population genetics.
Christopher Zeeman Jaroslav Stark
Terence Wall
David Epstein
Steve Baigent Belgin Seymenoğlu
Atheeta Ching
Part of Zeeman’s (academic) family tree
Now I want to share my love of mathematics with the rest of the world, like Zeeman did, and there is no beer place to start than Chalkdust. If you’re equally inspired, you should write for Chalkdust too! I look forward to reading it. Belgin Seymenoğlu Belgin is a PhD student at UCL, working on population genetics. When not working, you can usually find Belgin either playing the piano or playing Math Blaster. She is pictured here standing next to her copy of Zeeman’s catastrophe machine.
a @SmokyFurby
Sharing a cake by Daniel Griller
Isaac and Gofried are spliing a cake using a cut-cut-choose-choose method: Isaac cuts the cake into 2 pieces, then Gofried cuts one of these pieces, so that there are 3 pieces in total. What fraction of the cake can Isaac guarantee for himself if: a) Isaac chooses the first piece, Gofried picks the next piece, then Isaac takes the last one? b) Gofried chooses the first piece, Isaac picks the next piece, then Gofried takes the last one? You can find more puzzles by Daniel in his (highly recommended) book Elastic Numbers, on his blog d puzzlecritic.wordpress.com, and on Twier a @puzzlecritic. 33
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On the cover
Chladni figures of a square drum
T
James Cann
18th century—the age of enlightenment. Ernst Chladni travelled around Europe demonstrating his ‘musical curiosities’. The star araction was a novel technique to expose the various modes of vibration of a rigid surface.
Chladni’s original method to expose vibrational modes of a rigid plate.
A plate or membrane, when resonating, divides into regions that vibrate in opposite directions. These are bounded by fixed lines where no vibration occurs (nodal lines). Chladni would lightly scatter sand on a metal plate, then violin-bow the plate to the point of resonance. The vibration encouraged the sand to accumulate upon the nodal lines, and paerns—now called Chladni figures—emerged. Forget rigid plates (they’re so 18th century). Let’s clamp the edges of a flexible square membrane to make a square drum. Suppose you scaer fine sand (or…chalkdust) on it and then vibrate at prescribed frequencies. Equally fascinating nodal paerns emerge— the Chladni figures of a square drum.
So far this all seems dangerously close to real life. This is a maths magazine. Where’s the maths? The modes of vibration for the square can be uniquely described by the surfaces that they draw out at their maximal amplitude. By analogy think of a vibrating (one dimensional) string. When vibrating, the whole string oscillates from a sinusoidal curve, to the negative of the curve, and back again. The situation is very similar in two dimensions, except we have a surface on a square domain, instead of a curve on an interval. The modes of vibration are stationary solutions of a differential equation called the wave equation. When you solve this equation, it turns out that there are only special, isolated, frequencies at which standing waves can form on the drum. We call the set of these special values the frequency spectrum of the square drum.
A selection of Chladni figures.
Remarkably, for each fixed frequency in this spectrum there are multiple basic vibrational modes. By superimposing amplified versions of these basic modes, we can form any and every vibrational mode of that frequency. Even more remarkably, the multiplicity is intimately related to a tricky arithmetic function which is notorious in number theory. Now, back to nodal lines. The Chladni figure of each basic vibrational mode for the square is, by itself, easy to understand. It looks similar to the simplest of the experimental paerns we chalkdustmagazine.com
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chalkdust saw above—a grid of lines. However, a superposition of equal-frequency vibrational modes can produce strikingly intricate and complex nodal paerns. Even the most basic geometric question— How many distinct nodal lines are there in a ‘typical’ paern?—is considerably difficult to answer with mathematical rigour. The first step towards answering this question is to put ‘typical’ on firm mathematical footing. And that brings us to (possibly) the only question you really care about right now: So what exactly are those paerns, tiling the cover of issue 07? Pick a frequency λ from the spectrum of the square drum, list all of the basic vibrational modes of frequency λ, amplify each of them by a random (independent and standard normallydistributed) coefficient, sum them together, and what comes out looks something like the contour plot below right. The superposition of randomly-amplified basic vibrational modes is usually called a (boundary-adapted) arithmetic random wave, and the plot is a sample from this random wave. My research focuses on studying the statistical properties of the number of distinct nodal lines—a random variable which is defined by the random wave.
Chladni figure with nodal lines traced out in white.
We could think of this coloured square as a (randomly generated) sinusoidal landscape with a preference for oscillating around sea-level. From this topographical perspective, the nodal lines are the contour lines at sea-level of the landscape. The image to the right is a sample from the boundaryadapted arithmetic random wave with λ = 1105π2 . Why study such an obscure mathematical object? Well: they have interesting and subtle statistical properties, not to menContour plot coloured red > 0, tion that the sample waves look prey damn fine. But if that white = 0, blue < 0. isn’t enough for you, the nodal paerns of boundary-adapted arithmetic random waves are of interest to a much wider group than mathematicians. antum physicists like them too: they capture properties of the wavefunctions of a ‘quantum billiard ball trapped in a box’ with a fixed energy corresponding to the frequency of the wave. And the community of interest is even wider than the combined esoteric worlds of quantum and mathematics. Earthquake scientists; makers of fine musical instruments; all are interested in the properties of nodal lines—the sets which remain stationary when a domain is vibrating at resonant frequencies. More than just prey paerns, I hope you agree. James Cann James Cann is a PhD student at UCL and is aached to the London School of Geometry & Number Theory. He likes maths best when explained with pictures.
c james.cann@ucl.ac.uk 35
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Thinking outside outside the box Todd Lappin, CC BY-NC 2.0
Rob Eastaway
T
aren’t many puzzles that became famous as a corporate cliche, but the nine dots problem is certainly one of them.
There are nine dots arranged in a square, and your challenge is to join all the dots using only four straight line strokes of a pen, and with your pen never leaving the paper. No tricks are needed, no folding the paper and no using an ultra-fat pen. You’ll have no problem finding a solution with five lines, but in the unlikely event that you’ve never seen this puzzle before, four lines might prove to be a struggle.
The nine dots problem
The puzzle first rose to notoriety when Sam Loyd included it in his 1914 cyclopedia of puzzles. He called it the Columbus egg puzzle, aer an apocryphal story about Christopher Columbus challenging his colleagues to get a boiled egg to stand on its end—a challenge that seemed impossible until Columbus crushed the base of the egg and rested it on its now flat boom. Easy when you know how. The solution to the nine dot puzzle led to the cliche of ‘thinking outside the box’ (that’s a massive clue to the solution, but if you still can’t find the answer, you’ll find it the boom of page 42). The solution is a useful metaphor for the challenge of problem solving, when we oen become chalkdustmagazine.com
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chalkdust restricted by constraints of our own making. Deliberately challenging your assumptions and allowing yourself to be ‘silly’, at least for a short period, can be a handy strategy for creative problem solving in all walks of life. What I find interesting, however, is that the lessons of the nine dots puzzle are oen quickly forgoen when the puzzle is modified.
The 16 dots problem Take the 16 dots problem. It’s exactly the same idea, but this time instead of using four straight lines, you are allowed to use six straight lines. Can you find a solution? And if you want a proper challenge, can you find a solution that nobody else is likely to find? I strongly encourage you at this point to have a go at the 16 dots problem, and allow yourself at least ten minutes on it before returning to this article, because there are spoilers aplenty coming up.
The sixteen dots problem
This article continues on the next page, where the answer to this puzzle is revealed. Only turn the page once you’re ready for spoilers…
My least favourite equation
Incompressible Euler equations for inviscid uids Hugo Castillo Sánchez
) ∂u + u · ∇u = −∇p ∂t These equations, first published by Euler, are a set of coupled quasilinear partial differential equations widely used in fluid mechanics. They allow you to calculate the velocity and pressure fields in a moving, inviscid fluid. I dislike them for the simple fact that they are only valid for water and are completely useless for cases when the fluid viscosity is highly important (eg my PhD research and 95% of the fluids present in this world). ∇·u=0
ρ
(
Odd arrangement 1 7
33
11 17
35
What is the sum of the numbers in the 23rd row? 19
27
25
23
21 31
9 15
13
Write the odd numbers in a triangle, as shown to the le.
5
3
37
What is the mean of the numbers in the 23rd row?
29 39
41 39
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chalkdust How did you get on? If you gave the puzzle a serious go, then you probably did find a solution. Take a look at your aempts. Of course, you allowed yourself to go outside the box this time, but what angles did your lines go at? The majority of solvers try lines that go horizontally, vertically, or at a 45° angle. They do this because (a) that’s the paern they’ve seen already, and (b) those are the most ‘efficient’ lines to take out dots. A diagonal line can take out up to four dots, whereas a line at a different angle will take out two dots at most. If you took the 45° approach, then most likely your solution looked like one of these two:
Liverpool cathedral
The Archbishop’s hat
Both solutions are just extensions of the 9 dot solution. When it comes to categorising solutions, however, these two are in the same family. If you extend all of the lines off the page, you can see the two solutions use the same set of lines. Are there other solutions, that don’t belong to this family? Yes there are. But to find them, you have to overcome the selfimposed constraint of 90 and 45 degree angles. And to stand out from the crowd, you also have to allow yourself to think further outside the box. Literally.
Turning a hat into a cathedral
If you allow one line to go at a more jaunty angle, you’ll get some interesting solutions (below, le). But if you really begin to let your hair down, you can find even more zany solutions (below, right):
3 units
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chalkdust If we call the distance between two horizontal dots one unit, notice how the right-hand solution has at one point travelled three units outside the box before turning back—that’s one whole box width beyond the boundary of the dots. Which might make you wonder, are there any solutions that push even further outside the box? And, happy days, yes there are! For example, this one to the right has gone four units, or 133%, outside the box. I believe that’s the furthest you can get out 4 units of the box when solving the 4 × 4 puzzle, but I know that for the 7 × 7 square there is at least one solution that is 183% out of the box. For larger N × N squares there are surely solutions that go more than 200% out of the box—but nobody (yet) knows how big N has to be for this to be possible.
Symmetrical solutions Some of the solutions to the 16 dot puzzle look a bit messy, but out of mess sometimes comes beauty. It was only by exploring messy solutions that I found these four symmetrical beauties: Frog
Anvil
Star
Chubby pentagram
To my knowledge, this is the first time the ‘Chubby pentagram’ solution has ever been published, and possibly the first time that the words chubby and pentagram have been used in the same sentence.
Inside the box I sometimes wonder if we give too much credit to thinking outside the box, while down-playing the importance of less glamorous inside-the-box thinking. Sometimes in life we have no choice but to think inside the box we’ve been given. So let’s apply this thinking to dot puzzles. We’ve seen that for the nine dot puzzle there is no solution with four lines that are all inside the box, though a solution with five lines is possible, for example on the le here. In fact by continuing this spiral outwards, we can see that for the N × N square there will always be a solution inside the box, as long as you are allowed 2N − 1 lines. 41
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chalkdust But what if you can only use 2N − 2 lines (so, six lines for the 4 × 4, eight lines for the 5 × 5 etc). No such inside-the-box solution is possible for the 4 × 4 dot puzzle, but if we take our old friend The Archbishop’s hat and extend it, it’s easy to find an eight line inside-box solution to the 5 × 5 puzzle (see right), which we can then spiral outwards for larger N × N. So we’ve shown that for the N × N puzzle (where N > 2), not only is there always a solution that uses 2N − 2 lines, but for N > 4 there is also always at least one solution that is inside the box.
Dot puzzle theorems For an N × N square of dots: • There exists at least one solution inside the box using 2N − 1 lines, for all N. • There exists at least one solution using 2N − 2 lines for all N > 2. • There exists at least one 2N − 2 solution that is inside the box for all N > 4.
The dot puzzle conjecture Without bending the rules, or bending the paper, there’s no solution to the 9 dot puzzle using three lines. Nor is there any known solution to the 16 dot puzzle using five lines. In fact there is a conjecture, not yet proved as far as I know, that there does not exist any solution to the N × N puzzle that uses 2N − 3 lines. There may not be a million-dollar prize on offer, but there’s still some kudos to whoever comes up with the proof.
And nally... There is one solution to the 16 dot puzzle that stands above all others. It’s the only solution that has two lines of symmetry and the only one with rotational symmetry. It also starts and finishes at the same point, with no dot having more than one line passing through it. Here it is. For its elegance and the fact that it’s two intertwined zigzags, I call this solution DNA:
Rob Eastaway Rob Eastaway is an author and speaker best known for his popular maths books, including the bestselling Why Do Buses Come In Threes? He is the director of Maths Inspiration, the national programme of theatre-based maths lecture shows for teenagers.
a @RobEastaway d robeastaway.com The solution to the 3 × 3 puzzle on page 38 →
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chalkdust
a catastrophe machine You will need cardboard, scissors, pins, rubber bands
Instructions pins
large cardboard rectangle
rubber bands
cardboard circle
put your nger in here and move it around
1
Cut out a card circle whose diameter is the length of an unstretched rubber band. Use pins to attach everything together, as shown above.
Drag the lower rubber band around to investigate catastrophe theory (see pages 30â&#x20AC;&#x201C;33).
2
Tube map platonic solids, FrĂśbel stars and slide rules: more How to make at d chalkdustmagazine.com 43
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Primes Ă la Euler Sam Porritt
A
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If we highlight them in a grid like this, it is clear that the last digit of any prime must be either 1, 3, 7 or 9 (except for 2 and 5 of course). But there seems to be no good reason for primes to prefer any one of these four possibilities over any other. chalkdustmagazine.com
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chalkdust It is natural therefore to conjecture that There are infinitely many primes whose last digit is b for b equal to any of 1, 3, 7, or 9. Unless you have seen this before you probably don’t know how to prove it. When stuck trying to prove something and you don’t know what to do, it is oen a good tactic to first try to prove something simpler than, but related to, what you really want to prove. 1 2 3 4 One way to simplify this particular problem is to notice the following: 5 6 7 8 besides the exceptions 2 and 5, when we divide a prime p by 10 there are four possibilities for the remainder, 1, 3, 7 or 9, which we write as 9 10 11 12 13 14 15 16 p ≡ 1, 3, 7 or 9 (mod 10). When we divide an odd prime by 4 though, there are only two possibilities, 1 or 3. So we might start by trying to prove that there are infinitely many primes p such that p ≡ b (mod 4) for any b equal to 1 or 3. This just means that the remainder when we divide p by 4 is b. Let’s now see how Euler might have thought about this problem by looking at how he proved that there are infinitely many primes.
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The odd primes are all 1 or 3, mod 4
Primes, in nite series and π We start with the following remarkable identity ∞ ∑ 1
n=1
n
=1+
1 1 1 1 + + + + ··· = 2 3 4 5
1 . 1 − p1 primes p ∏
Knowing that the le hand side diverges to ∞, we see immediately that the right hand side cannot be a finite product. And that’s it! The right side cannot be a finite product–so there are infinitely many primes! That’s Euler’s proof. But how do you prove the identity? Simple. Expand each factor of the right hand side into a geometric sum to get ) ∏( 1 1 1 1 + + 2 + 3 ··· . p p p p Multiplying out the product involves choosing a 1/p k from each factor, multiplying these 1/p k together and summing over all the ways of choosing the value of k. The only resulting terms which are non-zero are those in which we pick k to be non-zero for only finitely many primes and k = 0 for all other primes. Doing so, we get 1/n for each positive integer n precisely once, just as we wanted. This is a consequence of the fundamental theorem of arithmetic which says that each integer can be wrien as a product of primes in a unique way (up to the order of the primes). It is well worth thinking this through so that it makes sense because it is the most crucial part of the argument. Euler used his product formula to prove even more. By taking the logarithm of ∞ to get ∞ and converting the infinite product into an infinite sum (this is not rigorous mathematics by today’s 45
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chalkdust standards but Euler wasn’t one to let pedantic details spoil a beautiful argument!) ) ( ∏ ∑ 1 1 ∞ = log . =− log 1 − p 1 − p1 p primes p Aer expanding log using the power series
− log(1 − x) = x +
x2 x3 x4 + + + ··· , 2 3 4
valid for all −1 < x < 1, Euler gets
∞=
∑1 p
p
+
1∑ 1 1∑ 1 + + ··· . 2 p p2 3 p p3
Most of these sums are bounded. In fact, ∞
∞
∑∑ 1 ∑ ∑ 1 1 1∑ 1 1∑ 1 + + ··· < = < < ∞. 2 3 n 2 p p 3 p p p p (p − 1) n2 p n=2 p n=1 So we have the astonishing conclusion that ∑1 p
p
= ∞.
Remember that we would like to show that the primes are in some sense split evenly between the two classes 1 (mod 4) and 3 (mod 4) - and in particular that there are infinitely many of each type. Inspired by Euler’s argument involving infinite series, we next introduce the following alternating series ∑ (−1)(p−1)/2 1 1 1 1 1 1 A=− + − − + − + ... = 3 5 7 11 13 17 p p̸=2
which has a plus sign in front of those primes which are 1 (mod 4) and a minus sign in front of those primes which are 3 (mod 4). The prime 2 doesn’t really feel like it belongs in this series and in particular it is not clear what the sign for 1/2 ought to be. To avoid having to make an arbitrary choice we will just not include the prime 2.
Luciano Rila
A pencil showing the largest ‘le-truncatable’ prime, created by Maths Inspiration chalkdustmagazine.com
Our strategy then is to prove that the series A∑ converges. If we could, it would then follow immediately from 1/p = ∞ that there are infinitely many primes of each type, 1 and 3 mod 4. Because if there were only finitely many of one of these types, the series A would diverge. Taking a cue from Euler’s original argument, we can add in the prime powers without changing whether or not the series converges (because, as we have seen, the sum of the reciprocals of 46
chalkdust the prime powers converges). So expanding the logarithm into an infinite series, it follows just as before that A converges if and only if
−
∑ p̸=2
( ) (−1)(p−1)/2 log 1 − p
converges. Exponentiating, this converges if and only if ∏ p̸=2
1 1−
(−1)(p−1)/2 p
converges to something non-zero. We can expand this infinite product much like before by first writing each factor as a geometric sum ∏( p̸=2
(−1)(p−1)/2 1 (−1)(p−1)/2 1+ + 2+ + ··· p p p3
)
.
Expanding the product is a bit harder this time but not too bad. It is ∏( p̸=2
(−1)(p−1)/2 1 (−1)(p−1)/2 1+ + 2+ + ··· p p p3
)
=1−
1 1 1 1 1 + − + − + ··· 3 5 7 9 11
now with only odd terms and every other term having a minus sign. This one is more difficult because we need to keep track of all the minus signs. Multiplying out the product, we get ±1/n for each odd integer n exactly once for some choice of sign. There are no even integers as we have excluded the prime 2. An integer appears with a minus sign if and only if it has an odd number of prime factors which are 3 (mod 4). This is exactly the right hand side though since such integers are exactly those which are themselves 3 (mod 4). You might want to stop to check this last point makes sense. But wait. We recognise this (don’t we) as Leibniz’s famous series 1 1 1 1 1 π =1− + − + − + ··· 4 3 5 7 9 11 which is certainly finite. So, there we have it ∑ (−1)(p−1)/2 p̸=2
p
and we proved above that ∑
p≡1 (mod 4)
= “something finite”
1 + p
∑
p≡3 (mod 4)
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1 =∞ p spring 2018
chalkdust because leaving out the prime 2 doesn’t change whether or not the series converges. By adding and subtracting these two equations we conclude that. ∑
p≡1 (mod 4)
1 =∞ p
and
∑
p≡3 (mod 4)
1 = ∞. p
What an astounding thing to follow from an infinite series for π!
The proof As pleased as we are with this dramatic conclusion, it doesn’t solve our original problem. We wanted to show that for each b = 1, 3, 7, 9, there are infinitely many primes whose last digit is b. We can rewrite this condition by noticing the following equivalences which hold for all odd numbers p, p ≡ 1 (mod 10) ⇔ p ≡ 1 (mod 5) Leo Reynolds, CC BY-NC-SA 2.0
Analogue clocks tell the time modulo 12
p ≡ 3 (mod 10) ⇔ p ≡ 3 (mod 5) p ≡ 7 (mod 10) ⇔ p ≡ 2 (mod 5) p ≡ 9 (mod 10) ⇔ p ≡ 4 (mod 5).
Convince yourself this is true by trying a few examples. So can we do for 5 what we just did for 4? It will be a bit harder this time because there are 4 remainder options for our primes rather than just two. Let’s first take some time to reflect on what made the argument work before we try to generalise it. 1. We had an infinite convergent series that factored into primes and could somehow distinguish between primes in different remainder classes modulo 4. ∑ ∑ 2. We took logarithms and “solved for” the two sums p≡1 (mod 4) 1/p and p≡3 (mod 4) 1/p by taking a linear combination with the logarithm of Euler’s product formula. 3. To deduce that these sums over primes were infinite, it was important that the original series not only converged, but that it converged to something non-zero. We needed it to be nonzero so we could take the logarithm and get a finite answer. It wasn’t important what the answer was, just that it was finite. Point 1 suggests (or rather, it suggested to Dirichlet, see pages 28–29) that we assign coefficients χ(n) to each integer n which only depend on the remainder of n when divided by 5. These coefficients are to be used to detect which class the primes belongs to. They should also give rise to a nice convergent series which can be factored into a product over all primes ∞ ∑ χ(n) ∏ = factor(p). n p n=1
chalkdustmagazine.com
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chalkdust We will be able to factor the series just as before if the coefficients have the special property χ(mn) = χ(m)χ(n) for all positive integers m and n. If χ has this special property then we can write ∞ ∑ 1 χ(n) ∏ . = χ(p) n p 1− n=1
p
The proof is almost exactly the same as before. This multiplicativity property is quite restrictive, especially if we also require χ(n) to only depend on the remainder of n by 5. First of all, if we don’t want our χ(n) to equal zero for all n then we must have χ(1) = 1 since χ(n) = χ(n · 1) = χ(n) · χ(1) so χ(1) ̸= 1 implies χ(n) = 0 for all n. If we set χ(2) = α, say, χ(4) is determined to be χ(4) = χ(2 · 2) = χ(2) · χ(2) = α2 . Similarly, the value of χ(3) is determined by χ(3) = χ(4 · 2) = χ(4) · χ(2) = α3 . Here we used that 4 · 2 = 8 ≡ 3 (mod 5). So we just need to decide on a value for α and χ(0). Recalling what happened before, we will set χ(0) to be zero. 5 is the only prime in this remainder class anyway so if the point of these coefficients is to detect primes, they ought to be zero on this class. We don’t actually have much choice for α since 1 = χ(1) = χ(3 · 2) = χ(3)· χ(2) = α4 . Therefore, α must equal 1, i, −1 or −i. Each of these choices of α gives a different χ which we will denote χ α . In table format: n (mod 5) 0 1 2 3 4
α=1 χ1 (n) 0 1 1 1 1
α=i χi (n) 0 1 i −i −1
α = −1 χ−1 (n) 0 1 −1 −1 1
α = −i χ−i (n) 0 1 −i i −1
Each χ gives us a different series ∞ ∑ χ1 (n) 1 1 1 1 1 1 1 1 = + + + + + + + + ··· n 1 2 3 4 6 7 8 9 n=1 ∞ ∑ 1 i 1 i 1 i 1 i χi (n) = + − − + + − − + ··· n 1 2 3 4 6 7 8 9 n=1
∞ ∑ χ−1 (n) 1 1 1 1 1 1 1 1 = − − + + − − + + ··· n 1 2 3 4 6 7 8 9 n=1 ∞ ∑ χ−i (n) 1 i i 1 1 i i 1 = − + − + − + − + ··· n 1 2 3 4 6 7 8 9 n=1
This ∑ is nice because point 2 above suggests that we need four series to solve for our four unknowns p≡b (mod 5) 1/p with b = 1, 2, 3 or 4. Let us look at each of these four series in turn and see if they 49
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chalkdust have the properties we want. ∑ Case 1: α = 1. This one diverges and is the one that tells us that p 1/p = ∞. In some sense this is the most important one because it is the one that tells us we have infinitely many primes when we include all remainder classes. Case 2: α = i. This one involves the complex number i and it is natural to look at its real and imaginary parts separately. Remember that we want to show that it converges to a finite non-zero number but we are not overly interested in what that number is, so long as it isn’t 0 or ∞. This will happen provided the real and imaginary parts both converge to some finite limits that are not both zero. This in turn is fairly easy to prove by grouping the positive and negative terms together. We will do the real part and invite the reader to fill in the details for the imaginary part. ) ∑ ∞ ( ∞ ∞ ∑ ∑ 1 1 1 1 1 1 1 2 1 2 − + − + − +· · · = − < < ∞. = 2 1 3 6 8 11 13 5k − 4 5k − 2 (5k − 4)(5k − 2) n n=1 k=1
Notice that each term
2 (5k−4)(5k−2)
k=1
is positive so the sum is certainly not zero.
Case 3: α = −1. We can do this by collecting the terms into groups of 4 and using a lile algebra. ) ∞ ( ∑ 1 1 1 1 1 1 1 1 1 1 1 1 − − + + − − + + ··· = − − + 1 2 3 4 6 7 8 9 5k − 4 5k − 3 5k − 2 5k − 1 k=1
= =
∞ ∑ (5k−3)(5k−2)(5k−1)−(5k−4)(5k−2)(5k−1)−(5k−4)(5k−3)(5k−1)+(5k−3)(5k−2)(5k−1) (5k−4)(5k−3)(5k−2)(5k−1)
k=1 ∞ ∑ k=1
20k − 2 < ∞. (5k − 4)(5k − 3)(5k − 2)(5k − 1)
Again, each term
20k−2 (5k−4)(5k−3)(5k−2)(5k−1)
is positive so whatever this sum is, it is positive.
Case 4: α = −i. This one is very similar to case 2 so we will leave the details to the interested reader to fill in for herself. Now that we know our series converge to non-zero limits we can take logarithms to convert our products over primes into sums over primes. Expanding log into an infinite series and bounding the terms involving prime powers exactly as before we get the following four equations:
Case 5
∑ χ (p) ∑ χ−1 (p) ∑ χ−i (p) ∑ χ (p) i 1 = ∞, = “finite”, = “finite”, = “finite”, p p p p p p p p chalkdustmagazine.com
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chalkdust which we can write more explicitly as ∑
lim
n→∞
p⩽n p≡1 (mod 5)
∑
lim
n→∞
p⩽n p≡1 (mod 5)
∑
lim
n→∞
p⩽n p≡1 (mod 5)
lim
n→∞
∑
p⩽n p≡1 (mod 5)
1 + p 1 + p 1 − p 1 − p
∑
p⩽n p≡2 (mod 5)
∑
p⩽n p≡2 (mod 5)
∑
p⩽n p≡2 (mod 5)
∑
p⩽n p≡2 (mod 5)
1 + p i − p 1 − p i + p
∑
p⩽n p≡3 (mod 5)
∑
p⩽n p≡3 (mod 5)
∑
p⩽n p≡3 (mod 5)
∑
p⩽n p≡3 (mod 5)
1 + p i − p 1 + p i − p
That’s a bit messy so let’s introduce the notation Sb (n) = rewrite it as
∑
p⩽n p≡4 (mod 5)
∑
p⩽n p≡4 (mod 5)
∑
p⩽n p≡4 (mod 5)
∑
p⩽n p≡4 (mod 5)
∑
1 =∞ p 1 = “finite” p 1 = “finite” p 1 = “finite” p
1/p p⩽n p≡b (mod 5)
for b = 1, 2, 3, 4 and
lim S1 (n) + S2 (n) + S3 (n) + S4 (n) = ∞
n→∞
lim S1 (n) + iS2 (n) − iS3 (n) − S4 (n) = “finite”
n→∞
lim S1 (n) − S2 (n) − S3 (n) + S4 (n) = “finite”
n→∞
lim S1 (n) − iS2 (n) + iS3 (n) − S4 (n) = “finite”
n→∞
This looks like a simple system of four linear equations in four unknowns, but it involves limits and infinity and one must not be tempted into taking the limits too soon and writing ∑ nonsense like ∞ + i∞ − i∞ − ∞ = “finite”. We leave the task of carefully deducing that p≡b (mod 5) 1/p = limn→∞ Sb (n) = ∞ for each b = 1, 2, 3, 4 as an exercise. If successful, the reader will have finished the proof of our original conjecture, and proved that there are infinitely many primes ending in each of 1, 3, 7, and 9! Sam Porritt Sam is a PhD student at University College London, who works on number theory.
Did you know... …that you can buy Chalkdust T-shirts, read our weekly blog posts, and read all of our back issues at d chalkdustmagazine.com?
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#7 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 1 August 2018. Only one entry per person will be accepted. Winners will be notified by email and announced on our blog by 19 August 2018. 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 utilities puzzle mug 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, with free UK shipping. Find out more at d mathsgear.co.uk chalkdustmagazine.com
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chalkdust
Across
1 The first digit of this number is equal to the last digit of 34A. 9 The product of the digits of 30A. Also the sum of the digits of 1A. 10 The product of this number’s digits is a multiple of 1,000,000. 13 None of the digits of 34A is equal to the middle digit of this number. 14 All the digits of this number except one are the same. 17 The sum of two positive cube numbers. 18 The product of the digits of 16D. 21 All the digits of this number except one are the same. 22 Greater than 13A. 23 The product of the digits of 6D. 25 None of this number’s digits are 0, and the sum of this number’s digits is greater than the product of its digits. 26 A number n such that the last three digits of 2n are equal to n. 27 One more than the product of the digits of 23A. 28 A multiple of 9. 30 The product of the digits of one less than 27A. 31 The product of this number’s digits is 7D. 32 The sum of the digits of 34A. 34 The product of this number and 1A is a multiple of 1,000,000,000,000.
(16) (2) (14) (3) (10) (3) (6) (5) (3) (4) (4) (3) (3) (10) (3) (12) (2) (16)
Down 2 This number is equal to the sum of the cubes of its digits. 3 The first three digits of this number are all even. 4 The sum of the cubes of the digits of 5D. 5 This number is not equal to 4D, but it is equal to the sum of the cubes of the digits of 4D. 6 The product of the digits of 18A. 7 A triangle number whose digits add to 18. 8 A power of 13A. 10 A power of 22A. 11 This number is equal to the sum of the cubes of its digits. 12 A power of 13A. 15 Equal to the last 8 digits of 28A. 16 The product of the digits of 3D. 19 The product of this number’s digits is a multiple of 9. 20 Each digit of this number is a factor of the previous digit. 24 This number is equal to the sum of the fourth powers of its digits. 29 This number is equal to the sum of the fih powers of its digits. 33 The product of the digits of 9A.
(3) (14) (3) (3)
(5) (3) (16) (14) (3) (14) (8) (12) (4) (6) (4) (5) (2)
My favourite equation
Euler’s continued fraction formula Nikoleta Kalaydzhieva
Euler’s continued fraction formula connects some infinite series with an infinite continued fraction. For example, it gives the following result for ez : 1 ez = z 1− z 1+z− 2z 2+z− 3z 3 + z − Isn’t it prey⁉ 4 + z − ··· 53
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We love it when our readers write to us. Here are some of the best emails, tweets and letters weâ&#x20AC;&#x2122;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, My students took on this problem in our math [sic] club. Two of them built the well in Minecraft and are using that to reason. I thought you might enjoy knowing that.
Halcyon Foster, USA a @halcyonfoster
Amidst the excitement of nishing the Chalkdust crossnumber is a tinge of disappointment that I have some time to wait for the next one.
I was recently handed a copy of Chalkdust #5 (at the LMS Popular lectures in Bâ&#x20AC;&#x2122;ham) and have gotten around to reading it.
TAS, Nottingham a @NHSMaths_TAS
Loved it.
What is Chalkdust?
Tim Cross, Birmingham
Alison Kiddle, Cambridge a @ajk_44
Dear Chalkdust, I want to say to the young writing team at Chalkdust that your series on black mathematicians was amazing and thought-provoking. Awesome reporting! Habib Hourani,
chalkdustmagazine.com
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a @habib_hourani
chalkdust
Baking a Menger sponge sponge Flickr user fdecomite, CC BY 2.0
Sam Hartburn
I
you’ve been following Chalkdust for a while, you’re probably already aware of the MathsJam gathering. You might even have been yourself. For those who aren’t aware, it is an annual event where maths lovers get together to talk, play, sing and generally have fun with maths. One feature is a baking competition: aendees bring a maths-related cake and there are prizes for the preiest cake, the tastiest cake and the best maths in a cake. This is the story of one of those cakes: the diagonal cross-sections of a Menger sponge.
Why a Menger sponge? I was looking around for a cake subject, and had considered a famous fractal known as the Menger sponge. Then I came across an article that showed what it looks like when you make a diagonal cut through the middle. It was totally unexpected! I decided then and there that I had to make this cake. A Menger sponge is a self-similar fractal made from a cube. To make a level 1 Menger sponge, divide a cube into 27 smaller cubes and remove the centre cube from each face and the cube in the very centre. To make a level 2 sponge, carry out the same process on each of the 20 remaining small cubes. And so on. 55
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Level 0, level 1 and level 2 Menger sponges
As this process is repeated, the volume of the sponge decreases, because part of it is removed at each stage. Conversely, the surface area increases, going from 6 square units in the original cube, to 8 square units in a level 1 sponge to 13 13 in a level 2 sponge. If the process is repeated indefinitely, the volume approaches zero and the surface area approaches infinity. This makes an interesting prospect for a cake: the more levels you do, the less cake there will be, but the more icing you will need to cover it. This odd behaviour comes about because the Menger sponge cannot comfortably fit into two dimensions, but neither can it fully occupy three; fractals such as this are somewhere in between, which can’t be measured by the usual concept of dimension. One way to measure the dimension of such a space is to calculate its Hausdorff dimension, which should be somewhere between 2 and 3 for a Menger sponge. If c is the number of copies of itself that a fractal contains, and 1/s is the scale of those copies compared to the original, then the Hausdorff dimension of a self-similar fractal is given by log c . log s A Menger sponge contains 20 copies of itself (we divided the original cube into 27 sub-cubes, then removed 7 of them) and each copy is scaled by 1/3, so the dimension of a Menger sponge is log 20 = 2.727, log 3 (to 3 decimal places). However, a cake sits firmly in three dimensions, so my Menger sponge cake required some compromises.
Compromise 1: In nity It was clear from the outset that I wasn’t going to produce an infinite Menger sponge cake. It would have been stale long before I got anywhere close. So I had to make a decision about what level I could go to. One restriction was the size of the smallest cube of cake that can reasonably be cut and handled without crumbling. In trials, I found that cubes with 1 cm sides were too small to deal with, but chalkdustmagazine.com
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chalkdust 1.5 cm sides were just about comfortable. For a level n cake, this would give an edge length of 1.5 × 3n cm, where n is the number of times the fractal process is carried out. The following analysis made the decision easy. Level 5
Edge Length (cm) 364.5
4 3
121.5 40.5
2 1 0
13.5 4.5 1.5
Comments Higher than the ceiling at Yarnfield Park, where the MathsJam gathering takes place. Wouldn’t fit through my front door. Would require baking at least 32 cakes in my standard 20 cm square tin, not allowing for waste. Not too big, not too small. Not very impressive. That’s just a cube.
Verdict No No No Yes No No
Compromise 2: Holes The problem with holes in a cake is that, with nothing to support the parts above the holes, the cake is likely to collapse. I made my holes with chocolate cake, so even if I had been able to produce an infinite sponge it would not have had volume approaching zero (sorry, calorie counters). Note, for the seven level 1 holes I used bigger 4.5 cm chocolate cubes instead of building them from the smaller 1.5 cm cubes
Compromise 3: Construction Officially, the construction method for a Menger sponge starts with a large cube and removes parts of it, as described above. I chose instead to build the cake up layer by layer using cubes. An unfortunate side effect of making the cake from smaller cubes is that there are visible lines where the cubes are joined together, where there would be no lines in a true Menger sponge. However, this also meant that there was plenty of icing throughout the cake to stop it being too dry, so I decided to live with it.
The cake was built up layer by layer
The cut All the artistry in this cake came directly from the maths. All I did was cut cubes and stick them together. 57
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chalkdust The main cut went through the midpoints of six edges, at an angle of 45° to the top face. I have to confess that I was shaking when I made this cut. I’d spent more hours than I care to admit constructing the cake, and there was every possibility that it would fall apart as I cut through it. Either that, or I would get the cut in the wrong place. The relief I felt when the star appeared in the centre was tremendous! Nobody has been mean enough to point out that the cut was slightly off— the small chocolate triangles around the edge should have been small stars, but only the boom right one made it. Regardless of that, I was pleased with the result.
A slice throught the finished cake, compared to a slice through a computer generated model
It’s fun to work out where the stars come from, so I won’t say too much about it. Think about the shape of the hole inside a Menger sponge, and about the shapes you can get by cuing a cube or cuboid at different angles.
Some slices through a cube. What other shapes can be produced?
I made two more cuts, both parallel to the first. The exact placement of these cuts wasn’t important— the diagonal cross-sections make interesting shapes wherever they are. There is a good animation to show this at Zachary’s blog (d blog.zacharyabel.com/2012/02/seeing-stars; spoiler alert: the site also shows where the stars come from).
Now it’s your turn! If you would like to make your own Menger sponge sponge, here are a few tips to help you on the way. I found the best type of cake for cuing to be a Madeira cake (which is a sponge cake, so we can officially call it a Menger sponge sponge). There are lots of recipes online—find one that suits you. chalkdustmagazine.com
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chalkdust Madeira cake is stronger and less crumbly than a Victoria sponge; chilling the cake first makes it even easier to cut it neatly, without making crumbs. The level 1 holes require cubes with an edge length of 4.5 cm. This meant baking a deep cake—at least 5 cm, to give some allowance for cuing off the crust. Deep cakes can be difficult, as the outside cooks a lot quicker than the inside, so you can end up with burnt edges and a doughy centre. I recommend using a flower nail, which is a nail with a big flat head, normally used for mounting icing flowers; they’re available cheaply from eBay. Stand it upside down in the middle of the cake tin before pouring the dough in and it will conduct heat directly into the centre, helping it to cook at a similar rate to the sides. As a bonus, this also makes the cake rise more evenly and gives a flaer top.
The finished cake
This cake requires 7 large cubes (chocolate) and 540 small cubes (400 vanilla and 140 chocolate). It would therefore be helpful to have a quick and reliable method of cuing accurate cubes. Unfortunately, I’ve yet to find one! You need something strong and firm to stick the cubes together. I used buercream, and chilled the cake overnight before cuing it, to make the buercream as firm as possible. This meant that when I made my first diagonal cut, the cake stayed in one piece. In an early trial I used jam, and that definitely didn’t work! Finally, make sure that you have plenty of time set aside. The construction took nearly a whole day. When I finished making the cake, my joints were stiff from standing in the same position for so long, and I had ten minutes to gobble down a very late lunch before running to get the kids from school. Was it worth it? Well, looking back at the final result, I would have to say yes. Sam Hartburn Sam spends a lot of time trying to resist eating the cakes and biscuits she bakes. To take her mind off food, she proofreads and copy-edits maths books.
c sam.hartburn@qbfproofreading.co.uk a @SamHartburn d qbfproofreading.co.uk My least favourite equation
Euler’s method Sean Jamshidi
f (x + h) − f (x) h As accurate as a sledgehammer, and about as good at maths. What am I? An engineer? f ′(x) =
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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
Closing the Gap Vicky Neale “A beautiful tale about an age old conjecture about the gaps between primes.”
MathsJam Gathering “The best weekend of the year. Better than Weetabix.”
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Galois Knot Theory
The Element in the Room
J Maruyama “Definitely going to Fourier series to Fourier series to a nonspecialist.”
Helen Arney & Steve Mould “Fun and easily accessible to a nonspecialist.”
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Math Blaster: Episode 1
Dr Euler’s Fabulous Formula
“Definitely going to play it again.”
Paul J Nhin “A whole book about applications of eiθ = cos θ + i sin θ, from pure maths to Fourier series to electronics and more.”
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The Maths Behind
Colin Beveridge “Colin has covered every possible topic we can ever blog about.”
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Chalkdust issue 07 “The best magazine ever.”
∼ ∼ ∼ ∼ ∼ ∼ ∼ ∼ ∼ ∼ ∼
New Chalkdust T-shirt
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“You need this in your Life.”
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PBS In nite Series (YouTube)
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“Very in-depth maths, with excellent presenting and animations.”
Star Wars: The Last Jedi “No numbers in title =⇒ no stars.”
ggghi chalkdustmagazine.com
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Edsger Dijkstra Emma Bell
I
a 2001 interview with Philip L Frana (a @ArtificialOther) for the Charles Babbage Institute, Edsger Dijkstra succinctly demonstrated the importance of his shortest path algorithm: Dijkstra: If, these days, you want to go from here to there and you have a car with a GPS and a screen, it can give you the shortest way from your hotel to Robbie Creek Cove. Yes? Frana: I did generate a map, yes, on line. Dijkstra: Did you use it? Frana: Yes, I did use it. Dijkstra: Then you used my algorithm this morning.
The roots of mathematics reach back through time for centuries, providing a solid foundation by anchoring our knowledge in the ages of Pythagoras, Euclid, Fibonacci, Galileo, and Newton. However, some areas of mathematics are new. Their roots are not deep, yet they still carry the huge weight of technological advances that we use, and take for granted, every day. Edsger Wybe Dijkstra was born in Roî&#x20AC;źerdam, the Netherlands, in 1930. Dijkstraâ&#x20AC;&#x2122;s father, Douwe, taught chemistry at a local secondary school, whilst his mother, Brechtje, was a trained mathematician. 61
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chalkdust When describing his mother, Dijkstra states: “she had a great agility in manipulating formulae and a wonderful gi for finding very elegant solutions”. This elegance permeates his own work and is evident throughout his writings. Dijkstra’s catalogue of carefully numbered manuscripts, known as EWDs, which document his ideas, reports, lectures, proofs, and leers, are fascinating to read. Dijkstra wrote a great deal of them by hand, in fountain pen. Most of them can be found at the archive hosted by the University of Texas where he worked from 1984 until his retirement. As a teenager, Dijkstra had career aspirations to represent his country— he believed that he could study for a degree in law and follow this path Hamilton Richards, CC BY-SA 3.0 Edsger W Dijkstra in order to serve on the United Nations council—however, following outstanding results in maths and the sciences, Dijkstra was persuaded to enrol at the University of Leiden in 1948 to study maths and physics. It was the result of what Dijkstra called “a long series of coincidences” that led to him becoming the Netherlands’ first programmer in 1952. In The Humble Programmer, his acceptance speech when receiving the Turing Award in 1972, Dijkstra tells how his father enrolled him on a short programming course at the University of Cambridge in 1951. It was felt that this course would help Dijkstra in the long term as a theoretical physicist. The head of the Amsterdam Mathematical Centre heard that a fellow Dutch citizen had taken the programming course, and in light of the fact that the centre was building computers, offered Dijkstra a job. Dijkstra was convinced that programming could become a respectable discipline, and so finished his physics studies in Leiden as quickly as possible while working part time in Amsterdam. Dijkstra became the first programmer in the Netherlands, representing his nation. The role was so new that he was not permied to use it as his occupation on his marriage certificate, being told to write “theoretical physicist” instead. The new Armac computer was completed in 1956, based on Dijkstra’s programming. For its “festive opening”, as Dijkstra described it, the machine would need to run a programme which would show off its capabilities to an audience of non-specialists—Dijkstra told Frana in the 2001 interview: “you have to have a problem statement that non-mathematicians can understand; they even have to understand the answer”. Whilst taking a break from a shopping trip in Amsterdam, Dijkstra came up with the problem, and the answer. He sat, drinking coffee with his fiancée, later wife, Ria (herself a mathematician and programmer). Over the course of twenty minutes, without pencil and paper, Dijkstra invented the shortest path algorithm to find the shortest way to travel between two places. He used cities in the Netherlands as a guide, reducing the number of cities to 64 to make the coding more efficient. 1. Mark starting vertex as 0. 2. Consider all neighbours of this vertex, calculate their distance away from the start. 3. Now visit every one of these neighbours and repeat the process, updating distances if they become shorter, filtering your way across the network. chalkdustmagazine.com
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chalkdust 4. Once your destination vertex is visited in this way, you can stop. You’ll have the shortest distance from the start.
A fortuitous by-product of this algorithm, which was eventually published in 1959, is that the shortest distance to every vertex from the start is calculated along the way. This algorithm is used today in such technologies as GPS, route planning and phone networks. He was ahead of his time. Dijkstra appreciated how his work, and the work of those that followed him, would impact on society. He wrote programs for computers that were still being built, and encouraged programmers to write code without bugs so that debugging would not be needed! In his acceptance speech for the Turing Award, Dijkstra prophesised about the future of computing:
In their capacity as a tool, computers will be but a ripple on the surface of our culture. In their capacity as intellectual challenge, they are without precedent in the cultural history of mankind. — Dijkstra, The Humble Programmer (1972) Dijkstra’s work spanned over 40 years. He gained a PhD from the University of Amsterdam in computer science in 1959 with his thesis Communication with an automatic computer. From 1962 to 1984, Dijkstra was a professor of mathematics at the University of Eindhoven. Here he further developed what came to be known as structural programming—a method for writing code which was clear, logical and elegant—a personification of Dijkstra himself. He also developed THE operating system (named aer Technische Hogeschool Eindhoven, the Dutch name for the University of Eindhoven) which was the precursor to many operating systems in use today. Dijkstra moved to the University of Texas in 1984, and remained there until his retirement in 1999. He died in the Netherlands in 2002. Dijkstra’s twenty-minute coffee break, another in his long series of coincidences, changed the world: his algorithm providing roots for new branches of mathematics and computer science. Emma Bell Emma is the maths enhancement manager at Grimsby Institute of Further and Higher Education.
c belle@grimsby.ac.uk a @El_Timbre 63
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Domino tiling & domineering Chris Woodcock and John Dore
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everyone reading this article has, perhaps some time ago, played dominoes with the standard set of 28 pieces or even with the larger set of 55 pieces. But do not fret if you haven’t: we will not be examining these sets, but merely taking the concept of a domino as a couple of squares joined along an edge. Asking how we may arrange a set of pieces to cover a plane rectangular board leads to an interesting mathematical problem. If we add some orientation restrictions, we can make an interesting twoplayer game.
Domino tiling There is a standard problem in many recreational maths books in which two opposite corners of a chessboard are removed and the reader is asked to place dominoes to cover all the remaining squares of the board. It is, of course, impossible because the two squares removed have the same colour and each of the dominoes must cover one black square and one white square. However, there is no need to introduce this somewhat contrived condition in order to obtain some relatively simple, but interesting mathematical problems. chalkdustmagazine.com
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A chessboard with two opposite corners removed
chalkdust First, we ask the following:
Problem 1
How many different ways can eight dominoes be used to cover a 4 × 4 square? Alternatively, tilings may be considered the same if one is a rotation of the other, but still regarded as different if they are mirror images. This change in counting the distinct tilings obviously leads to a smaller number due to symmetry, but there is no simple means of calculating the different possibilities as some paerns have two-fold rotational symmetry and others involving squares may have four-fold rotational symmetry. However, it does provide a nice problem to be examined through a systematic ‘trialand-error’ approach:
For problem 2, these two tilings are considered the same, as one is a rotation of the other
Problem 2 How many different ways can eight dominoes be used to cover a 4 × 4 square, if rotations of the same paern are excluded? We now return to including all the rotations, as in problem 1, take a more general m × n rectangle and ask a similar question. This problem has troubled mathematicians and the full answer is quite complicated. It was solved in 1961 by Kasteleyn in the context of covering a surface with molecular dimers, and independently solved by Temperley and Fisher in the same year. The resulting equation for the total number of domino tilings of an m × n rectangular board can be shown to be ( 1/4 ( ( )2 )2 ) n m ∏ ∏ πj πk . Am,n := 4 cos + 4 cos m+1 n+1 k=1
j=1
Perhaps surprisingly, this product must therefore always be a non-negative integer! Although we may use this expression to calculate the numbers for various rectangles, it is perhaps more interesting to examine a few specific cases. We begin by looking at the problem for a 2 × n rectangle. As you will see, this has a surprisingly familiar answer. The first few cases with small n can be easily worked out and the answers are: n A(n)
1 1
2 2
3 3
4 5
5 8
6 13
Let us now assume that we know the answer, A(n), for some small values of n. We see that we can gen65
The 5 ways of tiling a 2 × 4 rectangle spring 2018
chalkdust erate the ways of tiling a 2 × (n + 1) rectangle in two different ways: either by adding a vertical domino to one of our ways of tiling a 2 × n rectangle; or by adding two horizontal dominoes to one of our ways of tiling a 2 × (n − 1) rectangle. A(n + 1)
=
A(n)
+
A(n − 1)
Consequently, we may write A(n + 1) = A(n) + A(n − 1) and, since we already know A(1) = 1 and A(2) = 2, we have a recurrence relation that generates the sequence that we saw above, and that you will immediately recognise as the Fibonacci sequence. We suspect that you didn’t expect it to make an appearance in this context. Of course, it’s not necessary to restrict ourselves to rectangles, so here are a few more shapes to consider:
Problem 3 Which of the following figures are capable of being completely tiled by dominoes?
Returning swily to rectangles, we note that the 2 × n rectangle has an interesting solution. This immediately raises a further question: is it possible to similarly analyse the 3 × n rectangle? This question is more difficult to address as there are more configurations for the dominoes in the first few columns. One approach has been made by Brundan, but to our knowledge no detailed analysis of the complete solution has been previously published. The way of approaching the problem, along similar lines, is to think in terms of the arrangement of dominoes on the le-hand side of the block and to investigate how these define the number of configurations as the further dominoes are placed on the right-hand side. If m = 3, then clearly there are no tilings unless n = 2k is even. Let S(k ) denote the number of chalkdustmagazine.com
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chalkdust tilings of a 3 × 2k rectangular board. Then S(0) = 1 (as there’s only one way to tile a 0 × 0 board: do nothing!) and it is also easily seen that S(1) = 3. In fact, by considering the possible configurations of the dominoes in the lemost 3 × 2 and 3 × 4 rectangles, it can be shown that the recurrence relation S(k) = 4S(k − 1) − S(k − 2) holds for all k > 1. Thus S(2) = 11, S(3) = 41, S(4) = 153, and so on. This recurrence relation allows one to determine the number of tilings for any value of k. It follows that there are 571 ways to tile a 3 × 10 rectangle. Perhaps you’d like to try drawing all of these to check? If we only count up inequivalent tilings of an m × n rectangular board (as we did in problem 2, where we regarded two such tilings as being equivalent if one is a rotation of the other), then the calculation becomes even more intricate! For example, if m = 2, then the number of inequivalent tilings is 1 if n = 2, F (n + 1) + F ( n+2 4 ) 2 F (n + 1) + F ( n+2 1 ) 2
if n > 2 and n is even, and if n is odd,
where F(n) is the nth Fibonacci number. This result, involving an averaging process, is a special case of Burnside’s lemma from group theory (in this case the group involved is the rotational symmetry group of a rectangle). Thus for a 2 × 6 rectangle, there are F(7) = 13 tilings but only 1 2 (F(7) + F(5)) = 9 inequivalent tilings. It is, of course, possible to extend this method to the general case with m = 4. It is obviously more difficult to derive the recurrence relation for this condition but it can be done. Various surprising factors occur and the procedure, which is a lile more technically demanding. For the keen reader, an outline of this method can be found on the Chalkdust website. Our method gives the answer for a 4 × 10 rectangle as 18,061 and for a 4 × 15 Two of the 3,335,651 ways of tiling a 4 × 15 rectangle rectangle as 3,335,651. If you disagree with this result please let us know, but please don’t send all the possible tiling paerns to Chalkdust for checking!
The game of domineering We now move to a two-player game that uses dominoes. It was invented in 1974 by Goran Andersson, who originally called it crosscram, although it is now more widely known as the game of domineering. 67
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chalkdust In this game, players take turns to place a domino on a rectangular board but one is restricted to placing them in a vertical orientation (player V) and the other (player H) is restricted to placing them in a horizontal position; it is conventional for player V to start with square boards and this causes no restriction as the board can obviously be rotated by 90°. The first person who is unable to play has lost the game.
An example game of domineering on a 5 × 5 board. Player H cannot make another move so has lost this game.
The rules are therefore very simple but the game has a number of interesting aspects that relate to the tiling problem. A quick (around 5 minute) game can be played on a 5 × 5 board in order to learn the tactics. Play on a 6 × 6 board provides some interesting possibilities that can be easily analysed. So, we now present you with problem 4.
Problem 4 With best play, who wins (a) on the 5 × 5 board and (b) on the 6 × 6 board? The answers to the puzzles will be given in a few weeks’ time on the Chalkdust blog, when we will reveal a lile more about this game. If you require a more challenging game to play and analyse, try domineering on an 8 × 8 board… Chris Woodcock and John Dore Chris and John are both at the University of Kent in Canterbury: Chris is a senior lecturer in the School of Mathematical Sciences, and John is a retired physicist, now an emeritus professor in the School of Physical Sciences. They both have an interest in various aspects of recreational mathematics and are keen to introduce others to the fun aspects of mathematics through games and puzzles. They particularly enjoy encouraging people to reach that ‘Aha!’ moment, when the correct approach suddenly becomes apparent. Both were keen chess players in their youth but don’t have time to play competitively, these days.
Did you know... …that in Thai, the number 5 is pronounced as ‘ha’, and 555 represents laughter? chalkdustmagazine.com
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Machine learning 101
Tom Rivlin
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a @TomRivlin you want to build an artificial intelligence. Maybe you want to automate a repetitive task. Maybe you want to enslave/destroy humanity. Either way, here’s a quick guide to offloading all your thinking from your fallible, squishy brain to your unstoppable silicon automaton.
Step 1: Give it a goal First, you need to define a phase space for the machine and its task. This is something like ‘all possible images’ or ‘all chess moves’. Within that phase space, you then need to give it an objective. This could be something like ‘identify cats in this image’, ‘win at chess’, or ‘kill all humans’. The more clearly you define the end state, the easier it will be for the AI to find it.
Exercise for the reader: train your AI to identify this cat
Step 2: Define a score function
A knight to remember
Your AI needs to be able to distinguish good answers from bad ones: to do this it will need a score function, which tells it how good a position is. Modern chessplaying computers are much beer than ones from 20 years ago not just due to increased processing power: they’re also beer because of more sophisticated score function algorithms. It’s not enough to tell your AI to ‘checkmate the opponent’ because for most of a chess game that’s not your goal—your AI will also need learn to secure strong positions on the board and win pieces.
Step 3: Give it a way to evolve Machine learning works by assigning scores to various states the bot can be in, and then updating its score to inch closer to its goal. You could use a genetic algorithm, inspired by biological evolution: randomly mutate your bot, and keep the mutants that perform beer; or, you could use a neural network, like the new champion-crushing chess AI AlphaGo Zero: create a network of neurons, and make your bot change the connections between them to improve its score function.
Step 4: Throw processing power at it Even if you optimise your algorithm to need as few iterations as possible to achieve its goal, it never hurts to get some time on a supercomputer. AlphaGo wouldn’t have been able to master chess in a maer of hours without Google’s fancy new hardware with unparallelled speed.
Step 5: Have fun!
Buy lots of these
As this is just the bare-bones introduction, I hope I’ve piqued your interest enough to learn this all in more detail. It’s a hot topic right now, so there’s a plethora of options for self-teaching the niy-griy of it all. Just remember to show me mercy when your unstoppable swarm intelligence AI ends our feeble human civilisation. 69
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Mike Jordan on machine learning Sean Jamshidi, Nikoleta Kalaydzhieva, Sam Porritt
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you’ve had a cursory eye on the news over the past few years, chances are you’ll have noticed that machine learning and artificial intelligence are geing lots of air time. Whether it’s spooky algorithms employed by your favourite social networking site or fears of a robot rebellion, machine learning has never been more present in the public consciousness. To find out more, Chalkdust aended a lecture by Mike Jordan—undoubtedly the most influential voice on this hoest of hot topics. Jordan, an award-winning academic who has also advised several machine-learning related companies and developed toolkits that are now the industry standard, is the perfect person to provide a level-headed response to the hype. He has been involved in machine learning for over 30 years and his broad experience gives him authority in both the academic and commercial spheres. Given the recent explosion in news coverage, you might be forgiven for thinking that machine learning is undergoing something of a revolution. From an academic perspective, this is not quite correct. “Really, the techniques have not dramatically evolved over the last twenty years. The introduction to machine learning class I taught in 1999 is still prey much the same today.” Instead, Jordan suggests that we are being more creative in our applications, with “every big business in the world employing Machine learning is changing how we machine learning on some level”. Previously, these methods had been shop online used in the back-end of businesses, for example by Amazon to reduce their levels of credit card fraud and by shipping companies to predict how many of a product would be needed in a certain part of the world. Now, however, machine learning algorithms are in our homes and our pockets in the form of targeted advertising, personalised recommendations and (soon, according to Jordan) domestic robots. Although this is an exciting development, as your fridge, television and car all become smarter and more personalised, it isn’t without risks. “If Google messes up and shows you a rubbish search result, it’s frustrating. If machine learning is applied to healthcare and it goes wrong, then lots of people could die.” Like all of mathematics, development of the theory is slow. “People are using these models, and we haven’t got good ways of quantifying the error. We don’t know how well these things scale, and algorithms are being applied in areas that they were never intended for.” Jordan also suggests that journalists, hungry for clickbait, are partly at fault. “I’d love to see people write articles who know about the stuff, and have spent time geing involved in it. But at the same time, I realise people want views and clicks.” (Rising to the challenge, we have given a quick primer on machine learning on page 69.) chalkdustmagazine.com
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chalkdust So if the pace of change is slow, does that mean we needn’t fear a robot revolution? “Anybody who tells you that we’ll be making machines smarter than we are, doesn’t know what they’re talking about. Not in our lifetime, not for a long time.” Several times during his lecture, Jordan makes the distinction between learning and intelligence. Machinelearning algorithms are simple and dumb, and we have no idea how to achieve the genuine understanding that constitutes intelligence in a human being. As a concrete example, Jordan talks about natural language processing (“the most challenging problem in the field”). “If I make up a word, and say ‘the gretch walked from London to Campopculturegeek.com, CC BY 2.0 Don’t expect a robot bridge in an hour’, you understand lots about this gretch thing. If I add rebellion soon (or ever) the word ‘gretch’ to a machine learning algorithm, it shis its knowledge by a miniscule amount.” Thus, flexibility would seem to be the thing missing from machine learning. The ability to improvise, hypothesise and to infer genuine meaning (rather than statistical likelihood) from a sentence are all natural features of human communication that seem impossible from the current perspective of large data sets and finely tuned parameters. Although progress to true intelligence seems like a slow road, Jordan is adamant that machine-learning enabled devices will fill our homes in the near future. “In ten years, I expect most people to own a household robot. It’ll be like having a cellphone, but it will move around.” It’s important to be prepared for these changes, to teach statistics and computer science together and to make sure that people can retrain as the workplace becomes more automated. “In previous tech revolutions, you had thirty years to adapt. This time it’ll be faster.” Jordan is insistent that these advancements are distributed equitably, regularly US Army photo by Maria Pinel Applying machine travelling abroad to spread the word about machine learning, and delearning incorrectly to veloping his soware as open source, so that anybody can make use of healthcare could affect its incredible power. The transformative effect of machine learning is thousands of lives already starting to be felt in China, a country that Jordan has visited several times. Here, banks have traditionally been hesitant about lending money to individual citizens, and access to cash can oen be difficult for those living in rural communities. Now, machine learning methods are being applied by Ant Financial to lend money based on people’s online shopping history. Loans are thus easier to come by, and less risky for the lenders, than ever before. In this case it seems like everybody wins, and that the hype around the impact of ‘dumb algorithms’ is justified aer all. Just don’t expect to hold a conversation with your toaster any time soon. Mike Jordan was speaking as part of the G-Research lecture series, in which world experts in science and technology give accessible talks about cuing-edge research. Sean Jamshidi, Nikoleta Kalaydzhieva, Sam Porritt Sean, Niki and Sam are PhD students at University College London and part of the Chalkdust team.
Did you know... …that 111,111,111 × 111,111,111 = 12,345,678,987,654,321? 71
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This issue features the top ten mathematical celebration days. To vote on the top ten units go to d chalkdustmagazine.com At 9 this issue, the Reentering the top ten, it’s favorite day of fans of the day of monthly the American date MathsJam: the second last format and decimals: Pi Tuesday of the month. day (14 March). At 7, it’s the day commemorating physicists trying to do in nite series: sum of all the negative integers day (1 December).
A new entry at 8: Hannah Fry’s birthday, or Fryday (21 February).
At 5, it’s the favourite day of people who like good date formats and fractions: Pi approximation day (22 July).
At 6, it’s a celebration for people who’d rather have two pies: Tau day (28 June).
At 3, it’s days like 15 August 2017 (152 + 82 = 172 ): Pythagorean triple days.
At four, it’s 25 December, the day we celebrate the birth of our lord and saviour: Sir Isaac Newton’s birthday.
There’s an approximately 1 in 365.25 chance that the picture for this issue’s number one is relevant: it’s today, because every day is a great day for maths.
At two, it’s the Chalkiest day in March: 13 March.
Pictures Knight’s Templar: Ewan Munro, CC BY-SA 2.0; Pi pie: Paul Smith, CC BY 2.0; Two pi pies: Amit Patel, CC BY 2.0; Hannah Fry: Sebastiaan ter Burg, CC BY 2.0.
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