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Too good to be Truchet

It is altogether too hot, it is altogether too full of people, and it is altogether too lunchtime for what feels like 8pm. Fortunately, the organisers have fed mathematicians before, and have thoughtfully provided plain paper tablecloths and pens with which to postulate, puzzle and prove while we eat. We’re in Atlanta, Georgia for the 13th Gathering 4 Gardner, or G4G: every couple of years, mathematicians, magicians, sceptics, jugglers and assorted others gather to honour the work and memory of popular science writer Martin Gardner with a week-long conference.

An idea from an earlier talk had lodged in my head. Cindy Lawrence of MoMath—New York’s Museum of Mathematics, where one can ride a square-wheeled tricycle or explore the inside of a Möbius strip, the kind of thing that Gardner would certainly have written about—had raved about Truchet tiles. What are they? Well, start with a square, coloured either black or white. Pick two diagonally-opposite corners and shade them with a quarter-circle of the other colour, as shown. Make several. Then place them however appeals to you! The way they’re set up, it’s practically impossible not to start making patterns: blobs and whorls that seem almost alive.

Of course, you don’t need the tiles themselves. You can just as easily doodle them on a convenient sheet of paper, such as a tablecloth. And you can just as easily start asking yourself questions like, what kinds of tiles can you get if you remove the restriction of squareness? What happens if you move into three dimensions? And is there any pretty maths underlying the pretty patterns?

Sébastian Truchet and his tiles

The tiles Lawrence talked about are not, strictly speaking, due to Truchet. In 1704, Sébastien Truchet published A Memoir On Combinations, in which he discusses squares split diagonally into triangles (pictured right), giving four possible orientations for each tile.

It’s an interesting read. Truchet methodically looks at the number of ways you can place two such tiles next to each other, edge-to-edge. (He says that he’s started work on three tiles, but isn’t happy with it yet. I’ve seen nothing to make me believe he ever was happy with it.)

In the paper, Truchet carefully reasons that there are:

  • four possible orientations for a single tile;
  • four positions to place a second tile next to a first (north, south,
    east, or west); and
  • four orientations for each of the second tiles

… making a total of 64 possible arrangements.

He then notes that some of the arrangements are indistinguishable from others: placing a tile in orientation A to the left of a tile in orientation B is the same as placing B to the right of A, reducing the number to 32. Furthermore, some arrangements are rotations of others—for example, arrangement AA (pictured below) is a rotation of arrangement CC (not pictured below unless you’re reading Chalkdust upside down).

In all, he reduces the 64 original possibilities to 10 (six appear in eight configurations each, and four—those with `stripes’ across the middle—appear four times apiece).

It turns out to be worthwhile to consider the overlying structure of these arrangements. Of the ten possibilities, eight appear in pairs: swapping the colours of one gives the other. The other two are self-inverse: swapping the colours gives a rotation of the original arrangement.

And that’s where Truchet tiles remained until the 1980s, when CS Smith, writing in Leonardo, took a deeper dive into the topology of Truchet tiles.Among other things, he suggests several possible changes to the design—for example, removing colour from the equation altogether and simply tiling with diagonal lines, or—rather nicely—by a pair of arcs in opposite corners. This is, in terms of symmetries, just the same as a diagonal, but placing large numbers of them together makes for much more appealing patterns—which you should totally play around with, but after you’ve finished reading this article.

Smith also suggests extending these tiles to include colours: if you make the cut-off corners one colour, and the remaining strip of the tile the other, you have exactly the blobby-cornered tiles MoMath brought to G4G.

But why limit yourself to squares?

Extending tiles to 2n-gons

Don’t get me wrong, I have nothing against squares. They’re certainly in my top ten favourite shapes. Just… even if the tessellation patterns are neat, the configurations of the tiles themselves are not all that interesting. When you go beyond four sides using blobby corners and two colours, though, fascinating things begin to emerge, without even having to put the tiles next to each other. (From here on, you can assume that all tiles are blobby-cornered and two-coloured.)

Indeed, any regular polygon with an even number of sides can be turned into blobby Truchet tiles (although hexagons are the only ones that would tile the plane alone). Odd numbers of sides don’t work with this kind of colouring, because the vertices need to alternate between two colours. How many hexagonal tiles are there? Well, it depends how you count.

I choose to count rotations of the tile as the same tile—so there exactly two square tiles, one with two isolated white corners and one with two isolated black corners.

With hexagons, it’s simple enough to do the counting: let’s start by considering the three white corners and how they connect (or don’t connect) to each other. Either each corner is on its own, all three are connected, or a pair of corners is connected and the other isolated. (Note that if any member of a group of corners connects to another corner, all members of that group must connect—we can’t have a case where the first corner connects to the second and the second to the third unless the first and third are also connected.)

These correspond to the figures to the right—three isolated white corners and three interconnected white corners, followed by—more interestingly – an isolated white corner, followed by a black band that ends halfway across the hexagon, followed by a white band and an isolated black corner.

I’ve paired them like this for a very good reason: inverting the colours on the black dots tile gives the white dots, and vice versa; inverting the colours on the split hexagon gives… a rotation of itself! I count the hexagons as having three possible tiles, one of which is self-inverse.

What possible patterns are there for an octagon?

Obviously, octagons don’t tile the plane on their own (although there’s nothing to stop you filling in the gaps with square tiles!) Independent of how we’re going to arrange them, we can still consider the viable patterns.

Again, it’s good to start by considering the four white corners and how they connect. When none of them connect, we have another white dots pattern; when all four connect, we have a corresponding black dots pattern. These two are inverses.

If two neighbouring corners are joined, and the other two left unattached, the resulting shape looks like a pair of black pants. Its inverse pattern, the pair of white pants, arises from any three white vertices being connected.

We could also connect a pair of diagonally-opposite corners to give a white stripe; the black stripe comes from connecting one pair of adjacent vertices, then connecting the remaining pair.

The octagons therefore have six possible tiles, none of which is self-inverse. (Or do they? Have I counted correctly? How do you know?)

Decagons are where (for me) it gets interesting: but perhaps you want to try working out how many decagon tiles there are yourself first. Scroll down once you’ve had a go…



There are ten decagon tilings, up to rotation—of which two are self-inverse. That’s a structure we’ve seen before: it’s the same as the structure behind Truchet’s pairs of adjacent tiles. Don’t you think that’s neat? Well, hold my coffee.

Truchet tiles in three dimensions

The following day, I pick a table with the MoMath people, who have a set of the square blobby tiles out to play with, black with white corners one side, white with black corners on the other. “How many ways can you make a cube?” asks Tom. We pick up the tiles and try to arrange them, one per face.

Let’s adopt the reasonable rule that the corners of each tile that meet a vertex must be the same colour. Each face then has two diagonally-opposite black vertices, and two diagonally-opposite white vertices, and there are only two ways to place a blobby tile to satisfy that: either the black vertices are joined by the stripe, or the white ones are. We can therefore think of each face as either black (if its black vertices are connected), or white (if the white pair is joined).

The puzzle reduces to finding how many ways there are to colour the faces of a cube using only two colours.

If all six faces are white, there’s only one possibility. The same goes for no white faces.

If five are white, there is again only one possibility (up to rotational symmetry); this is also true for one white face.

With four white faces, there are two possibilities: either the black squares are on opposite sides, or they are on adjacent sides. As you might expect, the same goes for two white faces.

With three white faces, there are again two possibilities: either the three faces meet at a vertex, or they wrap around like a tennis ball.

Now, simply listing the possibilities isn’t all that interesting. However, the structure behind the list is: of the ten arrangements, eight have a natural colour inverse, and the other two are self-inverse – switching the colours gives you a rotation of the same cube.

That’s precisely the same as the structure of the decagonal tile arrangements—and Truchet’s original pairs.

Coincidence?

It’s not clear to me whether this repeated structure is a coincidence, or some deep property of Truchet tiles. It’s not unnatural for a ten-element set to have that same structure—two elements that are their own inverses, and eight that form inverse-pairs.

Indeed, if you consider the group of integers from 0 to 9 under addition modulo 10, you also get that structure (0 and 5 are self-inverse). However, to consider the tiles or cubes as a group, we’d need a way to combine them in pairs, and if there’s a simple operator for any of the Truchet sets, it’s not obvious to me.

So, I’m opening it up to you, knowing that the readership of Chalkdust has, collectively, far more insight than I do: is this common structure a coincidence, or something deeper?

While you’re working it out, draw some Truchet tiles of your own. You’ll be glad you did. See you later, tessellator!

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Topological tic-tac-toe

Tic-tac-toe also known as noughts and crosses) is a classic game known for its simplicity, and has been popular since ancient times. You and a friend (or enemy!) take turns to mark the squares of a $3 \times 3$ grid. The winner is the first to get three of their symbols in a row (horizontal, vertical or diagonal). A winning game for X is shown below right.

A winning game for X

It’s not difficult to work out how to play optimally on a standard board, where you’re guaranteed to at least draw with your opponent. However, what if you’re not restricted to the standard two-dimensional square grid? How would you play then?

To present a fresh challenge and to make tic-tac-toe exciting again, here is a collection of puzzles where you will be swapping your standard square board with one on various topological surfaces. Have fun!

The cylinder

The first new board to consider is the cylinder. To form a cylinder as in the diagram below, imagine that the board is wrapped around so that the left and right edges of the board are connected to each other, like a piece of paper that has been rolled into a tube. Matching edges are denoted with a $\uparrow$.
In all the subsequent puzzles, it is X’s turn to play, and it is possible for X to win in some number of moves, even if O plays optimally.

Puzzle 1: cylinders

How does X win both of these games on cylindrical boards? It is possible to win the first game in one move.

The first game can be won in one move. A demonstration of this, along with the folded cylindrical board, is shown below.

Bending the board into a cylinder. The winning move for the first puzzle is shown in red and the winning line is marked in blue.

 

The Möbius strip

A Möbius strip. Image: David Benbennick, CC BY-SA 3.0

How about a Möbius strip? Imagine that the right edge of the board is wrapped around the back of the board and given a half turn, so that it connects to the left edge of the board. The half turn means that top and bottom become flipped as we move off the left or right edge of the board. The edges with a half turn are denoted by a $\uparrow$ and a $\downarrow$. (If you want to actually construct this board, draw each grid square as a wide rectangle so that the grid is wide enough to wrap around with a half twist.)

The figure on below shows adjacent squares in the new board by making copies of the Möbius strip board, and puzzle 2 gives two puzzles on the Möbius strip board.

A figure showing which grid squares are adjacent on a Möbius strip board. For example, if you go one place right from the top-right square 3, you will go to the bottom-left square 7.

Puzzle 2: Möbius strips

How does X win both of these games on Möbius strip boards? It is possible to win the first game in one move.

The torus

There’s no need to limit ourselves to only connecting the left and right edges—we can also connect the top and bottom edges. If we return to the cylinder formed by wrapping the right edge of the board round to meet the left edge, we can now connect the top and bottom edges together to form a torus. Now we have two pairs of connected edges, denoted by $\rightarrow$ and $\uparrow$.

For tic-tac-toe purposes, the cylindrical and toroidal boards are identical, and this holds even if we change the game to be `make a line of length $n$ on an $n \times n$ board’ for any $n$. However, for puzzle 3, you need to find a line of length 3 on a $4 \times 4$ board.

Puzzle 3: torus puzzle

What should X do to make three in a row on a torus?

On the torus, we can think of any row (or column) as being the central one, so it’s easier to prove facts about the torus than for other boards. Have a go at the following challenges.

Puzzle 4: more torus puzzles

Can you show that making any line of length $n$ on an $n \times n$ cylindrical board is also a line of length $n$ on an $n \times n$ toroidal board and vice versa (so a game on a torus is equivalent to a game on a cylinder)?

Are any starting positions better than others on a $3 \times 3$ torus?

Is it possible for the game on a $3 \times 3$ toroidal board to end in a draw, with neither player getting 3 in a row (the previous question gives you a shortcut to solving this one)?

The Klein bottle

A figure showing which grid squares are adjacent on Klein bottle board.

Now we will think about playing on a Klein bottle, a shape that cannot be constructed in three dimensions without it intersecting with itself. Fold the top edge of the board over to touch the bottom edge, and connect the left and right edges with a half twist like for the Möbius strip.

The figure to the right shows which squares are adjacent on the Klein bottle board, and puzzle 5 gives two puzzle.

 

Puzzle 5: Klein bottle puzzles

How does X win both of these games on Klein bottle boards? It is possible to win the first game in one move.

The projective plane

Another shape that it is not possible to construct in three dimensions without the shape intersecting itself is the projective plane. This is a shape where the top and bottom edges are connected by a half twist, as are the left and right edges. To imagine how it would be made, think about connecting the single edge of the Möbius strip to itself. (You’ll have to be very dextrous to actually create this from paper!) Puzzle 6 gives two puzzles using the projective plane. Top tip: construct an adjacency map like the one for the Klein bottle.

Puzzle 6: projective plane puzzles

How does X win both of these games on projective plane boards?

(Note: I’m assuming that a valid line comprises three distinct squares, so a single square does not appear more than once in a valid line.)

 

That brings us to the end of our foray into topological tic-tac-toe. I hope you enjoyed these mind-bending puzzles! The inspiration for this article came from Across The Board by John Watkins, the most complete book on chessboard and other grid puzzles. I would recommend this book for some further interesting puzzles, eg how many queens are needed so that every square on a chessboard is targeted or occupied by one of the queens? The author gives an interesting history of chess problems and shows that they have inspired important advances in maths.

For solutions to the puzzles above, see this article.

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Somewhere over the critical line

Maxamillion Polignac was a number. He was prime, and proud of it. One sleepy Sunday morning, with a cup of tea in hand, he opened the newspaper. The major headline shouted, prime club vandalised: composites blamed! Primes had been subject to prejudice for so long even though primes had founded the society. The strong dislike stemmed from primes being factors of composites. Maxamillion sighed with despair, but resigned, he continued reading.

He aimlessly scanned the newspaper until something caught his eye: the critical line: fact or fiction? The article speculated that the Critical Line was a fabled golden brick road that could lead to the Formula connecting all primes. It was supposedly hidden in the zeta landscape: an untouched land that defined the real and imaginary axes. The zeta landscape was complex and hard to navigate, and hence, the perfect place to hide the Formula. With the Formula, all the secrets underlying primes and how to find them would be revealed. Being a prime, Maxamillion did not have siblings, and living in a conflicted society, he felt alone, despite his many friends. But the Formula would give him a chance to find his twin prime who would make him feel complete.

With this idea on his mind, he went to his friend, another prime named Bernhard Oblong. Maxamillion said “I think we should try and find the Critical Line—we both get something: you’re a factorial prime, hence with the Formula, you could find out what your $n$ is; and I could find my twin prime!”

“Are you sure the Formula exists?” Bernhard asked.

“We have nothing to lose and everything to gain! Just like Pascal’s Wager.” Maxamillion reasoned.

“I hope you are right because I really want to know what $n$ is!” Bernhard agreed enthusiastically.

They embarked on their journey, oblivious to the dangers ahead. Starting in their city, the Number Line, they eventually reached a large building, the RSA bank in the outskirts of town. They noticed a sign with big black letters: no primes allowed.

“Let’s withdraw some money for the journey to the zeta landscape,” suggested Bernhard.

“Sure. But we better be careful,” replied Maxamillion.

A man with a baton stopped them. “Halt!” he said. “Can’t you read? no primes allowed! We don’t have primes coming this side of town.”

Maxamillion and Bernhard had no choice but to leave and try the next village. Another man, wearing a smug smile, saw the commotion through the glass doors of the bank.

“Hmm… primes. Interesting! Primes usually avoid banking with us because we use them to encrypt our credit cards. I know an opportunity when I see one. Let me see if I can capture them,” he thought to himself.

The slippery man phoned the most notorious prime-hunter in the zeta landscape—the Mersennary.

“Hello?” Mersennary crackled.

“I want you to capture two primes headed your way. Bring them back dead or alive. \$1,000,000 in credit cards,” the man ordered.

“It’s a deal!” Mersennary replied.

Their pockets empty, Maxamillion and Bernhard soon reached a small village. A sign hung over the entrance: \sign{this town is composite-free}.

“Wow. Those are some extreme views!” exclaimed Bernhard.

Not soon after, they found themselves surrounded by hundreds of primes who Maxamillion recognised immediately: Sophie Germain primes. They resisted any composite prejudice. They were well-built primes $(2p + 1)$ rebelling against the system and hoping to teach composites that primes are the building blocks of society, deserving equality.

“Who are you? Why are you here? Are you really primes or are you composite sympathisers?” The leader pelted questions faster than the duo could process.

“We are Maxamillion and Bernhard. We are trying to find the Critical Line. We are primes and certainly not composite sympathisers.” Maxamillion responded swiftly.

“Then you must be admitted into the UPS at once,” the leader proclaimed. “Follow Friedrich into the tent.” The two friends did as they were told.

As they were walking, Friedrich explained to them: “The UPS stands for United Prime Service. We are dedicated to protecting the rights of primes against the relentless prejudice of the composites. Join us. We are with the primes, we will continue to be so until the end.”

“Consider us members,” Maxamillion said. “Even though you are not as sturdy as us, we could use your help. Find the Formula and put an end to this!”

Before she sent them on their quest, the leader gave them a fascinating relic: $\zeta(s)$. “This is the zeta function,” the leader explained. “Use this once you reach the zeta landscape: it will help you navigate your way along the Critical Line.”

Progressively, the scenery morphed into a barren land: the zeta landscape. The two dimensions defined the real and imaginary axes. Using the relic, they navigated across the imaginary hills and the complex terrain. Even with hypothetical fog layering the land, they could clearly make out the glowing pathway. There it was—the Critical Line! “Whoa. It really is real. Really real.” Bernhard gasped. However, there was a dilemma ahead, for the Critical Line split into three paths.

Suddenly, a prime emerged from the fog. His sunken eyes added to the barren landscape. He whipped out his weapon, the square function. With it, the prime could square Maxamillion and Bernhard and turn them into composites. He advanced towards them, armed and dangerous. The duo trembled as beads of icy sweat trickled down their backs.

“I am the Mersennary. I am paid to hunt down primes like you,” he rasped.

Maxamillion tried to plead with him: “Why are you trying to break something that can’t be broken? We are all primes here. We have a rich history. Primes have been the dominant species in the whole of maths for hundreds of years. We ruled because we could not be broken down into other numbers. When we multiplied ourselves together, we created composites. Even though the composites have oppressed us, we remain strong and resistant. Primes will never be split. You are one of us, so are you a traitor?”

“Sorry. It is nothing personal, just business.” Mersennary responded coldly and inched closer.

In a desperate attempt, Maxamillion tried again: “Wait! You are in it for the money, right? War is not a steady business, and I am sure you would earn more at a new job. We want to get the Formula, which could help you learn more about yourself and other primes! You could use that to your financial advantage, eh?” Mersennary pondered and realised he had the wrong end of the number line. He decided to join them in the search.

Together they looked at the three new paths: $\operatorname{Li}(x)$, $\pi(x)$, and $x/\!\ln(x)$. $\operatorname{Li}(x)$ looked the most promising, because it went the highest, and it looked daunting enough to hide the Formula. $x/\!\ln(x)$ was very low, and it seemed like a place to start.

“Let’s go $x/\!\ln(x)$!” Mersennary said.

“No! Let’s go $\operatorname{Li}(x)$!” Bernhard replied.

Maxamillion urged: “Stop arguing! How about we compromise? Let us go explore the stairway $\pi(x)$. Maybe the Formula is hidden in the middle to stop people who aim too high or too low!”

They began climbing the never-ending stairs.

They were about to give up hope when they saw the Formula. It was $\pi(x)$. When Maxamillion touched it, it surrounded him with a blue light. Full of excitement, he asked the Formula to find his twin prime. But his enthusiasm didn’t last long as the formula would not give an answer. He sighed in desperation, but then he had an idea. He asked it the value of Bernhard. It answered 26951. Then he asked it the value of himself. It answered 26953.

“What?! We were twin primes all along?!” Maxamillion shouted.

“Wow!” Bernhard exclaimed. He then proclaimed: “With this, we can end prejudice! We could change the composites’ opinion of us by explaining all the secrets behind the primes and how intricate and beautiful we are!”

“We could also start a bank that serves all number-kind! Then I would have a steady source of income!” declared Mersennary excitedly.

As soon as they got back to Number Line, the trio started a bank: the Riemann bank. Soon it was booming and bought over the RSA bank. The first thing Maxamillion did as CEO was to demolish the sign saying, ‘no primes allowed‘. The law that barred the primes went down with the sign and they both crashed to the ground with a satisfying BANG!

The Formula and the relic were placed in Museum Polytechnique. The conflict between the two sets was finally resolved as the composites realised that the Formula revealed the complexity behind the primes. They realised that primes are so complex that they deserve to be treated better. Thus, the numerical landscape was changed forever!

Glossary

Composites Numbers that can be written as the product of 2 or more primes.
Logarithmic integral, $\operatorname{Li}(x)$ An approximation of the number of primes until a certain given number, formulated by Gauss.
Mersenne primes Primes of the form $2^n – 1$.
Factorial prime Primes of the form $n! – 1$.
Natural logarithm A logarithm with base $\mathrm{e}$, not base 10.
Pascal’s wager A wager that states that if you believe in God and God does not exist, you have nothing to lose. If God does exist, you have everything to gain.
Primes Numbers that do not have any factors beside 1 and themselves.
Riemann hypothesis Riemann’s conjecture deals with the locations of the solution to Riemann zeta function. It is the holy grail of mathematics.
Riemann zeta function An infinite series used to investigate properties of prime numbers.
Sophie Germain primes Primes in the form $2p + 1$ where $p$ is a prime.
Twin primes $n$ and $(n + 2)$ are primes.
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Mathematics and art: the ELHP

Some people like to hear about mathematics being used to address real-life problems. I am going to claim that the problem I describe in this article is real-life because it arises from a conversation between two non-mathematicians.

The logo of the ELHP.

Specifically, one of my sisters-in-law did an art degree, and as part of a project she did for this she visited Sardinia to interview the sculptor Pinuccio Sciola. At least some of his works are quite large, by which I mean maybe 3m high or more, based on things I see on the web. During their conversation, he said something about wanting to install one of his sculptures on a named mountain somewhere in the Catania area, for the benefit of the residents. I don’t know his exact words, but my sister-in-law found this remarkable enough that she reported it to me and other members of the family. I’m not naming my sister-in-law here because she is not a public figure, and feels that this article is not the way she would choose to become one.

It may be important to note that my sister-in-law lives in a small town near Catania, and this may be what prompted him to say this. It seems entirely likely that he had not spent any real time looking into this idea. Indeed, I am told that when he later visited Catania he immediately realised that his idea was probably unrealistic.

Let’s use maths, and some other disciplines, to consider his idea, pretending for the sake of discussion that he or someone else really does want to go ahead with it. You may notice that even though Sciola named the mountain, I have chosen not to do so. I shall do this later. For starters, let’s have a look at the mountain. I took this photograph from just outside Catania airport.

The view from Catania airport.

Continue reading

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What is the point of intersection?

Several months ago, one of my A-level students shared a puzzle with me. It’s a neat puzzle in its own right, but what came out of our conversation is even neater. My student made a fantastic insight into understanding what a logarithm was trying to say.

Let $y=b^x$ where $b > 0$ and $b\neq 1$. Consider the line which passes through the origin and is tangent to this graph at point $T$. What are the coordinates of $T$?

This is great fun to slap into your graphing program of choice (all hail Desmos!). Look at $y = 2^x$, $y = 3^x$, maybe $y =(1/2)^x$. Draw the line $y = mx$ and twiddle its gradient until the line just meets the exponential. The program will give a good approximation to the coordinates of $T$, and voilà, you’re off and away to happy sandboxing and conjecturing.

Pause the article here to have a play for yourself.



It’s a pretty neat result. But—cracks knuckles—there is much satisfaction to be had in proving a conjecture. My student and I brandished our pencils, and this is what we wrote.

Take $y = 3^x$. The tangent line through the origin intersects the graph at point $T$ with coordinates $(t,\ 3^t)$. There are two ways to calculate the gradient of this tangent line.

Because the line goes through $(0,\ 0)$ and $(t,\ 3^t)$, $\Delta y/\Delta x = (3^t – 0)/(t – 0) = 3^t/t$.

Also, as $y = 3^x$, $\mathrm{d}y/\mathrm{d}x = \ln 3 \times 3^x$. At point $T$, the gradient of the tangent is $\ln 3 \times 3^t$.

Equating these two expressions and solving for $t$ gives $t = 1/\ln 3$. Thus the coordinates of $T$ are $\left(1/\ln 3 , 3^{1/\ln 3}\right)$.

In general, for $y=b^x$, the coordinates of $T$ are $\left(1/\ln b , b^{1/\ln b}\right)$.

Ta-dah! We sat back in our chairs.

Except… what kind of number is $3^{1/\ln 3}$?! Our conjecture had strongly hinted at what to expect, and this wasn’t quite it.

I leaned forward again, scratching my pencil across the paper to find a simplification. My student remained still, regarding the expression $3^{1/\ln 3}$ thoughtfully. Then he observed, “That’s $\mathrm{e}$. It’s just $\mathrm{e}$.”

! How did he recognise it on sight?

One of my mathematical mantras is logs are powers (ommm). That is, $\log_bc$ is the power to which you raise $b$ in order to get $c$. The number $\log_bc$ is named by the description of what it does, much like how writing $\sqrt{5}$ means the number such that when you square it, you get 5. We’re identifying a specific number by its property rather than by its explicit value. This mantra directly generates the delightful construction
$$b^{\log_bc}$$
which now clearly shakes out as $c$. See? $\log_bc$ is the power you raise $b$ to, in order to get $c$. We’ve raised $b$ to that very power, so what do we get? $c$. Hurrah!

My student had taken this mantra to heart. And then he took it one step further. He reasoned as follows.

We know, by the mantra, that $\mathrm{e}^{\ln3} = 3$.

Also, the expression $3^{1/\ln 3}$ means take the $(\ln 3)^{\text{th}}$ root of $3$.

If we take the $(\ln 3)^{\text{th}}$ root of both sides of $\mathrm{e}^{\ln3} = 3$, then on the LHS, we are left with just $\mathrm{e}$.

Hence, $3^{1/\ln 3}$ equals $\mathrm{e}$.

I think this is a beautiful insight. It comes from viewing a power as a root and then seeing the power of that root, as it were. Until my student pointed it out, I hadn’t realised this elegance.

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An invitation to category theory

Early in our mathematical education, we learn about a strong interplay between algebra and geometry—algebraic equations give rise to graphs and geometric figures, and geometric features can be encoded in algebraic expressions. It’s almost as if there’s a portal or bridge connecting these two realms in the grand landscape of mathematics: whatever occurs on one side of the bridge is mirrored on the other. Continue reading

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Significant figures: Katherine Johnson

This year, on 26 August, one of the most memorable and well-known mathematicians, Katherine Coleman Goble Johnson, celebrated her 100th birthday. This is a tribute in honour of her life so far.

Family life and first steps towards mathematics

A postcard of Katherine’s hometown

Katherine was born on 26 August 1918 in White Sulphur Springs, a small town in West Virginia. She was the youngest of four children and was always the smart kid-she finished high school at the age of 14 and earned her Bachelor of Science in mathematics and French from the West Virginia State University at the age of 18. This, in part, was thanks to her father, who moved the family closer to a school to help his children get a better education. She still remembers her family dearly. Especially fondly, Katherine talks about her father’s stepmother, known as granny. They would visit her house to eat some of her delicious pancakes-just as anyone would with their grandmothers! The four children loved their parents very much: they thought of their mother as the prettiest lady in the world and their dad as the most handsome man. Katherine says she was daddy’s girl, but she always remembers her father telling her, “you are as good as anybody in this town, but you’re no better”.

Katherine was always good at mathematics. She has memories from childhood very clearly linking her to the subject: “I counted everything. I counted the steps to the road, the steps up to the church, the number of dishes and silverware I washed… anything that could be counted, I did”. But mathematics wasn’t the only subject for her. Sure, she loved it, but she was just as good in English because it also felt logical to her.

When asked why mathematics became the subject she was most fond of, she says it was because it was the subject you had to work hard for. It was the one subject with a right and a wrong, and once you got it right, it was right-unlike history!

The next push towards maths came at university. In eighth grade she had a maths teacher who happened to teach at the university Katherine went to years later. One day, she happened to meet the teacher again, who told her “if you aren’t in my math class this semester, I’m coming after you!” So Katherine had no choice but to go to maths class and her career in mathematics had begun. Later at university, her maths interests were taken care of by Mr Claytor. He added courses to the university almost exclusively for Katherine because he could see her potential. It was he who steered her towards research mathematics and eventually NASA.

Female mathematician at NASA

Katherine started her career in a rather unusual manner-she became a computer. Back then, these were the people who did the calculations for NASA’s predecessor NACA (the National Advisory Committee for Aeronautics) before the space race began. Katherine was sent to the flight research division.

She counts her character as one of the key things that contributed towards her career as a NASA mathematician. She remembers how her siblings and parents always used to try to shush her because she was always so assertive and stubborn. After she was invited to join the men doing the mathematics behind the calculations she would perform, she started fighting for her own place within the team. She demanded to see all the data, and asked to join the confidential meetings NACA held, slowly gaining respect in the midst of the all-male team.

Especially noted in the recent film Hidden Figures are the times when Johnson overcame and battled racism, as the only female and the only African American in the department. Katherine herself always says, though, it was never anything special: she just did her job and was appreciated for that, not her sex or skin colour.

On 20 February 1962, Friendship 7 was to be launched. Modern technological computers were already running the numbers and everything was being prepared for the mission that would make John Glenn the first American to orbit the earth. The astronaut, however, was feeling uneasy. He made the call to NASA to speak to Katherine Johnson. Would she redo the calculations?, he asked. Because if she got it right, he knew it was right-he would feel safe to go on the mission. Katherine approved the computer’s calculations. She admits the NASA team was much more worried about Glenn never making it back to Earth. If he missed the trajectory by a few degrees or tried entering at a different velocity, he would never return home.

Aerial view of Apollo 11

The mission was successful and Katherine also checked the trajectory for the Freedom 7, Apollo 11 and Apollo 13 missions, further proving her incredible mathematical skills.
As a token from NASA, Katherine received an American flag that flew to the moon.

At this point you may wonder-wait, you’re telling me about all these amazing things she has done, but you aren’t sharing any of the maths. Unfortunately, many of her articles aren’t available to the public. However, the three that are, are described below. I must warn you-all contain sophisticated mathematics and will most certainly take quite a while to wrap your head around.

Skopinski & Johnson, 1960

However, I can give you some insight into what the three papers contain. The first paper (Skopinski & Johnson, 1960) focuses on calculating the azimuth angle when placing a satellite over a predetermined position to ensure safe landing.

The second paper (Westrick & Johnson, 1962) is an analysis of the data from the Echo 1 satellite. It contains a lot of very nice graphs-it’s worth having a look just for the curves!

The third paper (White & Johnson, 1964) probably has the toughest maths, but similar to the first one, it focuses on finding solutions of some variables for the landing of a satellite. Arm yourself with some patience-the papers are worth your time even if they might seem a bit daunting at first!

After NASA

She retired from NASA in 1986, but she still has her hands full. For 50 years she enjoyed singing in a church choir. She loves playing bridge and other mathematical games, and she plays the piano and enjoys spending time with her six grandchildren and 11 great-grandchildren. She has authored or co-authored 26 research papers, and she has worked on the space shuttle and Project Apollo’s lunar lander. For her achievements and lifelong work, she received the Presidential Medal of Freedom in 2015. In 2016, the BBC named Katherine in their 100 Women 2016 as one of the most inspiring women alive. And then there’s the aforementioned film Hidden Figures, which revealed the story behind the three brilliant mathematicians Dorothy Vaughan, Mary Jackson, and of course Katherine, which has since conquered my and many other mathematicians’ hearts.

Katherine Johnson has been an inspiration for mathematicians all around the world, showing how one person can change so much. From gaining respect in one of the most prestigious research facilities in the world in times of unimaginable discrimination, to creating mathematics which helped many astronauts find their way back home; from simply being a wonderful person to being an incredibly talented mathematician-here’s to Katherine Johnson on her 100th birthday!

References

  1. Hidden Figures, directed by Theodor Melfi.
  2. TH Skopinski, Katherine G Johnson, Determination of Azimuth Angle at Burnout for Placing a Satellite Over a Selected Earth Position, September 1960.
  3. Gertrude C Westrick, Katherine G Johnson, Orbital Behavior of the Echo I Satellite and its Rocket Casing During the First 500 Days, June 1962.
  4. Jack A White, Katherine G Johnson, Approximate Solutions for Flight-Path Angle of a Reentry Vehicle in the Upper Atmosphere, July 1964.
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The magnetic pendulum

You have to go a long way to beat the magnetic pendulum for demonstrating the deep and profound nature of physics. One is pictured on the right. On the face of it, a swinging bob seems simple enough to understand. But as is so often the case, simplicity is masking complexity and the motion possesses an almost magical quality. Here, we will take a glimpse into just how unpredictable the predictable can really be.

The magnetic pendulum: it comprises a bob with a small magnet suspended by a string above a base plane that contains similar magnets arranged with opposite poles to ensure attraction.

Pull the pendulum back some distance in any direction, let it go, and prepare to be mesmerised by the way it darts back and forth in an erratic and seemingly random way. The bob is drawn simultaneously to all three base-plane magnets until, finally, it tends towards a precarious state of rest above one of them. Playing like this, it does not take long to become convinced of two immutable facts. Firstly, one can never know beforehand over which magnet the bob will stop. Secondly, and perhaps more subtly, the motion is not reproducible. No matter how hard one tries to replicate the initial displacement, the bob never follows the same winding path twice and the magnet that ultimately `wins’ seems to be governed by chance. How can we possibly find such randomness in a tabletop toy?

Of course, the bob is not moving randomly at all and we are at best starting it off each time from only roughly the same initial conditions. Its motion is prescribed by purely deterministic equations, a combination of classical mechanics and electromagnetics, and at that level there is no randomness. Were we able to release the bob from exactly the same starting position each time, the subsequent motions would all be identical and they would always stop at the same place. No uncertainty. No unpredictability. No endless fascination!

Since our first pull can never be repeated with infinite precision, what we are seeing is sensitive dependence on initial conditions or, more colloquially, the butterfly effect. Any change in input—any, no matter how imperceptibly small—can have a dramatic impact on output. That intriguing phenomenon turns out to be far more widespread and pervasive than one might first imagine. Moreover, it provides our working definition of chaos and crystallises what we mean by saying ”a system is chaotic”. In this article, we will start to explore how the magnetic pendulum embodies chaos in the scientific (rather than the everyday) sense.

A phenomenological model

It is not too difficult to devise a model that exhibits all the key qualitative features of the magnetic pendulum. Our approach is a phenomenological one, meaning that we are aiming to capture the essence of the motion using intuitive physical ideas rather than focusing on all the mathematical minutiae. A relatively easy way forward is to consider looking down on the pendulum from a plan view (see figure below). The bob’s trajectory in the three-dimensional space is projected downwards onto the horizontal base plane, and the origin of the $(x,y)$ coordinates is fixed at the centre of an equilateral triangle. The magnets are subsequently located at vertices $ \mathbf{X_1} $ , $ \mathbf{X_2} $ , and $ \mathbf{X_3} $, all of which lie along the circumference of a circle with a radius taken to be the unit length.

Left: Schematic diagram of the magnetic pendulum. Right: Projecting the position of the bob (white square at position $ \mathbf{x} $ ) onto the horizontal $(x,y)$ plane. The base-plane magnets (red squares 1, 2, and 3) are positioned at the vertices of an equilateral triangle.

The position vector of the bob may be represented by $\mathbf{x}(t)$ at time $t$. To account for gravity, it is sufficient for our purposes to consider a restoring force $\mathbf{F}_\text{grav} \propto -\mathbf{x}$ whose influence, due to the minus sign, always acts to pull the bob towards the origin $\mathbf{x} = \mathbf{0}$ (the constant of proportionality is set equal to 1, just to keep things simple). Dissipation might be introduced by way of the standard velocity-dependent damping force familiar from textbook physics. Here, we use $\mathbf{F}_{\text{losses}} = -b\mathbf{u}$, where $\mathbf{u} \equiv \mathrm{d}\mathbf{x}/\mathrm{d}t$. For some constant $b > 0$, the effect of $\mathbf{F}_{\text{losses}}$ is to drain kinetic energy from the motion through air resistance at low speeds. Finally, we must look to include the attractive forces due to the base-plane magnets. It is tempting to reach immediately for the inverse-square rule familiar from Coulomb’s law of electrostatics and Newton’s law of universal gravitation. We instead opt for a $1/\text{distance}^4$ rule as this form tends to describe the forces exchanged by magnetic dipoles.

Since Newton’s second law of motion equates mass $\times$ acceleration to the combined forces of gravity, magnetism, and damping, we can write down a governing equation of the form
\begin{equation}
\frac{\mathrm{d}^2\mathbf{x}}{\mathrm{d}t^{\hspace{0.5mm}2}} + b\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t} + \mathbf{x} = \sum_{n=1}^3\frac{\mathbf{X}_n – \mathbf{x}}{\left(|\mathbf{X}_n – \mathbf{x}|^2+h^2\right)^{5/2}}.
\end{equation}
Pythagoras’s theorem has been deployed here, and so each contribution in the summation on the right-hand side of (1) corresponds to a `distance/distance$^5 =$ 1/distance$^4$’ type of term. An additional parameter, $h^2$, has also appeared. Its role is to suppress unphysical (that is, infinite!) accelerations that would otherwise result whenever $\mathbf{x}$ approaches $\mathbf{X}_n$. We then interpret $h$ as being related to the average height of the bob above the base plane.

The equilibrium points are defined to be those positions $\mathbf{x}=\mathbf{x}_\text{eq}$ that are unchanging in time. Since the velocity and acceleration of the bob must be zero at those points (hence satisfying $\mathrm{d}\mathbf{x}_\text{eq}/\mathrm{d}t = \mathbf{0}$ and $\mathrm{d}^2\mathbf{x}_\text{eq}/\mathrm{d}t^{\hspace{0.5mm}2} = \mathbf{0}$, respectively), it follows from (1) that
\begin{equation}
\mathbf{x}_{\text{eq}} = \sum_{n=1}^3\frac{\mathbf{X}_n – \mathbf{x}_{\text{eq}}}{\left(|\mathbf{X}_n – \mathbf{x}_{\text{eq}}|^2+h^2\right)^{5/2}}.
\end{equation}
After playing with the pendulum and noting the positions where the bob tends to stop, we might reasonably expect to find maybe three or at most four solutions (with $\mathbf{x}_{\text{eq}} = \mathbf{0}$ being the origin). It is worth mentioning that the nontrivial roots of (2) do not occur at $\mathbf{x}_{\text{eq}} = \mathbf{X}_n$, as one might initially suspect. Instead, they lie at the same angular positions as $\mathbf{X}_n$ (as symmetry demands) but at a radial distance $|\mathbf{x}_\text{eq}|$ that is slightly less than unit length. At these positions, the competing pulls from gravity and magnetism are perfectly balanced.

Equation (1) does a surprisingly good job at mimicking the unpredictability so readily seen in experimental demonstrations; the left-hand side is just the damped harmonic oscillator problem from mechanics while the right-hand side sums over the pairwise magnetic-dipole interactions. On the one hand, any urges to seek analytical solutions should be kept in check. Even this stripped-down toy model confronts us with a formidable mathematical beast living in a four-dimensional realm whose axes are $(x,y,u_x,u_y)$. On the other hand, computing a numerical solution can be relatively quick and easy for given initial conditions, say $\mathbf{x}(0) = \mathbf{x}_0$ and $\mathbf{u}(0) = \mathbf{0}$.

Basins of attraction

Physically, we anticipate that the bob will almost inevitably come to rest at one of the non-trivial equilibrium points $\mathbf{x}_\text{eq}$ [cf. (2)] as $t\rightarrow\infty$. Those special points can be thought of as positions in the $(x,y)$ plane that attract the trajectory, and accordingly they are often referred to as fixed-point attractors. The idea now is to use a computer to carry out a systematic set of simulations, recording which magnet `wins’ (interpreted as the output) as we vary the starting point $\mathbf{x}_0$ (taken to be the input). By associating the outcome of each computation with a colour (eg red for magnet 1, white for magnet 2, and black for magnet 3), we can overlay the output on top of the input $(x_0,y_0)$ plane to produce a kind of abstract map. The set of all initial conditions lying in the red region is the basin of attraction for magnet 1—that is, any $\mathbf{x}_0$ lying on a red point will always end up at magnet 1 (and similar for the other colours and magnets), though the colour itself gives no information about the path taken by the bob to arrive at that point.

Figure 1. Basins of attraction for the magnetic pendulum with $b = 0.1$ and $h^2 = 1/4$. In the first pane, the grey squares denote the position of the base-plane magnets and the dotted grey line is a circle with radius equal to the unit length. The second and third panes show successive magnifications.

Figure 2. An illustration of FSS in the magnetic pendulum. Two initial conditions that are arbitrarily close together can give rise to subsequent trajectories that will start to move away from one another after a finite amount of time. The red path ends at magnet 1, while the black path ends at magnet 3.

Using this recipe, we discover a rather striking pattern (see fig. 1). Regions around the origin appear relatively simple. There are large single-colour lobes which indicate that variations in $\mathbf{x}_0$ tend to have little impact on which magnet wins. Further out beyond the unit circle, there is much greater complexity and all three colours are intertwined in a beautifully complicated way. In those regions, the pendulum is extremely sensitive to changes in $\mathbf{x}_0$. Successive zooming-in suggests the intertwining survives down to smaller and smaller length-scales. That feature—proportional levels of pattern detail persisting under arbitrary magnifications—is a defining characteristic of a fractal. A less obvious but equally fascinating property relates to the nature of the boundary separating two differently-coloured regions. We typically cannot cross from a region of red to an adjacent region of black without touching the white (similar is true for other permutations of colours). This situation is reminiscent of the delightfully strange Wada property from the field of topology.

Final state sensitivity

Figure 3. Variation of the basins of attraction for the magnetic pendulum as damping is increased. The second and third columns show successive magnifications. Other parameters and domains of the $(x_0,y_0)$ plane are the same as those in fig. 1.

The basins of attraction have three-fold rotation symmetry about the origin, which is a consequence of the equilateral-triangle arrangement of magnets and the initial condition $\mathbf{u}(0) = \textbf{0}$. Their details also depend crucially on system parameters. One might consider what happens, for instance, when the level of damping is increased (see figure to the right) through $b=0.125$ (first row), $b=0.150$ (second row), $b=0.175$ (third row), and $b=0.200$ (bottom row). The pattern becomes less complex and, accordingly, the pendulum less sensitive to small fluctuations in $\mathbf{x}_0$. However, their key features remain intact: the persistence of self-similar structure (fractality) and complex boundaries that tend to involve all three colours. The Wada-type property is still present in the right-hand column of the last two rows, but it is not obvious from these figures.

A helpful way to quantify just how strongly the long-term state of a system depends upon small fluctuations at its input is to estimate the fractal dimension $1 < D \leq 2$ of the basin boundaries. One selects a set of $N_\mathit{\Gamma}$ points in a domain $\mathit{\Gamma}$ of the $(x_0,y_0)$ plane and tests each of them in turn for the property of \textit{final-state sensitivity} (FSS) by considering a triplet of initial conditions: say $(x_0,y_0)$, $(x_0 + \epsilon,y_0)$, and $(x_0-\epsilon,y_0)$, where $0 < \epsilon \ll \mathcal{O}(1)$ can be interpreted as an error or as a limit to our experimental precision. When the winning magnet is the same for all three trajectories, then the final state is independent of $\epsilon$ and $(x_0,y_0)$ is, accordingly, free from FSS. Alternatively, think of $\epsilon$ as the radius of a small disc centred on $(x_0,y_0)$, somewhere within which the `true’ initial condition lies. FSS, as demonstrated in fig. 2, appears whenever that disc impinges on a basin boundary and thus overlaps more than one colour.

If the total number of points possessing FSS for a given $\epsilon$ is denoted by $N_\epsilon$, one finds that that $N_\epsilon/N_\mathit{\Gamma} \sim \epsilon^\alpha$. The parameter $\alpha \equiv 2 – D$ is known as the uncertainty exponent and it satisfies the inequality $0 \leq \alpha < 1$. Finally, we obtain the uncertainty dimension D, where
\begin{equation}
D = 2 – \frac{\mathrm{d}\log_{10}(N_\epsilon/N_\mathit{\Gamma})}{\mathrm{d}(\log_{10}\epsilon)}
\end{equation}
and a larger $D$ is indicative of increased susceptibility to initial fluctuations. The patterns shown in the middle panes of figs 1 and 3 turn out to have dimensions in the range $D\approx1.32$ (for $b = 0.1$) to $D\approx1.16$ (for $b = 0.2$). It follows that the basin boundaries for lightly-damped pendula tend to be associated with higher values of $D$. More generally, we now see a connection between the dimension of an abstract fractal pattern (which, crucially, need not be an integer such as 1 or 2) and the physical property of FSS.

Concluding remarks

In this article, we have started to unpick some of the intriguing behaviour exhibited by what is, in reality, a simple toy—one that never fails to capture the imagination of university students in lectures and Ucas applicants (and their parents alike!) at open days. The apparently erratic swinging and unknowable terminus of a bob are not quite so `random’ as one might first suppose from a few naïve observations. The essential ingredient giving rise to all this rich and diverse behaviour is the interplay between the three constituent feedback loops (here, due to gravity, dissipation, and magnetism).

Although we have considered the barest of bare-bones models (from pretending gravity provides a restoring force proportional to $-\mathbf{x}$, to suppressing the fully-vectorial character of magnetic interactions), the beautiful complexity of nature survives and we simply cannot get rid of it. That, it seems to us, is a mind-blowing conclusion!

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Game, set, maths (no more tennis puns)

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 little of the maths behind it.

No More Women

“Simon Singh, no more authors.”
Richard Cooper, CC BY-SA 3.0

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 left 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.

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.

“Courteney Cox, no more actors.”
Alan Light, CC BY 2.0

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 matter 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 written an autobiography count as an author, since they have a published work?

“John Venn, no more logicians.”

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 brunettes” 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 letters”. 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).

This is an excellent game to play when you’re all, for example, sitting 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?

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 categories described by John Venn in 1880, now a staple of set theory and a lovely way to visualise sets of things with certain properties. In this game, by definition, every person falls neatly in or out of a set. For example, you could have “no more authors” which would exclude everyone inside the set ‘authors’ . In terms of Venn diagrams, the nicest way to display this would be to call 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 (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).

A second move (eg “no more brunettes”—agreement will have to be reached whether you count natural brunettes only) could then overlap with this, restricting the next move to those in the intersection of those two circles. You’d also need at least one person to be in the region you’d just excluded (the non-author brunette you just named) and at least one still allowed (one non-author non-brunette you can name if challenged). However, it’s not necessary for there to be a non-brunette author (no offence, JK Rowling), as that category has already been excluded, so we don’t care what happens there.

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 (brunette authors, in the outer region; brunette non-authors; non-brunette authors; and non-brunette non-authors, which is where allowable moves still lie). You’d need to have named a non-brunette non-author politician as your turn (we went with Charles Kennedy, confirmed redhead) and you should also be able to name a non-brunette non-author non-politician (such as Lupita Nyong’o, who was in The Jungle Book, but hasn’t published one yet).

A four-way Venn diagram using only circles is not possible—in the example shown on the left below, not all sets of intersections are present. For example, there would be nowhere to place Steven Tyler (a brunette 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.

 

Two attempts to draw a four-way Venn diagram. In the attempt on the left, you can dream on if you think there’s a space for Steven Tyler.

It is also possible to construct Venn diagrams with five identical pieces—the one on the left, devised by Branko Grünbaum has rotational symmetry. The five-way Venn diagram shown next to it is called an Edwards–Venn diagram, and was 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 left 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 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.

Two five-way diagrams: Grünbaum’s diagram (left) and an Edwards–Venn diagram (right)

Technically any number of segments is possible, although the diagrams get much more complex as you go on. There’s a lovely interactive seven-way Venn diagram by Santiago Ortiz (on his website: 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 cutting the plane in half, with all the allowed persons on one side and the disallowed persons on the other side.

A goat acting out a maths puzzle.

For example, the line $x=2$ describes a vertical line on an $x y$-plane running through the point $2$ on the $x$-axis, and everyone to the left 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 brunettes above and non-brunettes 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).

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 pretty 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:

$$\text{Authors} = \{ \text{Simon Singh}, \text{Stephen King}, \text{JK Rowling}, … \}$$

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:

$$\text{Simon Singh}\in\text{Authors}$$

$$\overline{\text{Authors}} \neq \varnothing$$

Here the $\in$ 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, $\overline{\text{Authors}}$ would be the inside of the ‘no more authors’ circle, and the set $\text{Authors}$ would be everything outside this circle.

We’ve also specified, in the second line, that the set $\overline{\text{Authors}}$—the set of all non-authors—is not equal to $\varnothing$, where this symbol denotes the empty set. This means that set is not empty, because something exists in it.

Play continues:
$$\text{Courteney Cox}\in\text{Brunettes}\cap \overline{\text{Authors}}$$

$$\overline{\text{Brunettes}} \cap \overline{\text{Authors}} \neq \varnothing$$
Here we’ve used the $\cap$ symbol to define an intersection—this is the set of all things that occur in both the given sets—here, brunettes who are also not authors.

Along with the $\cup$ 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 and OR operators in Boolean logic, in a rough sense. On the Venn diagram, the intersection is the part of those two sets which overlaps—here, Brunettes is the outside of one circle, and $\overline{\text{Authors}}$ is the inside of another circle, so this is the portion of the non-Authors circle that doesn’t overlap with the non-Brunettes circle.

Again, the requirement of being able to name an example means that the set of non-brunette non-authors now has to definitely not be empty, and must contain something.

$$\text{Ed Miliband}\in\text{Politicians}\cap \overline{\text{Brunettes}} \cap \overline{\text{Authors}}$$

$$\overline{\text{Politicians}} \cap \overline{\text{Brunettes}} \cap \overline{\text{Authors}} \neq \varnothing$$
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:
$$A \cap B = B \cap A.$$
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 matters (multiplying by three then adding four is different to adding four then multiplying by three, and often 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:
$$A \cap B \cap C = (A \cap B) \cap C = A \cap (B \cap C).$$
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 matter which order you do this in, you’ll get the same result. Someone who’s a brunette 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 afterwards. Non-brunette non-authors are just as hard to think of, and just as numerous, as non-author non-brunettes. 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:

$$\overline{A \cup B} = \overline{A} \cap \overline{B}.$$

This says, the complement of the union of two 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) 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.

In the context of the game, you can consider this to mean that if you’re looking for someone who’s not a brunette AND 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 brunette OR 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

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 getting thrown out, it can get a bit like the coding/drinking/coding-while-drinking game Fizz Buzz.

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.