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 2016-03-15 06:36:52 

The Mathematical Games of Martin Gardner

This article first appeared in issue 03 of Chalkdust. I highly recommend reading the rest of the magazine (and trying to solve the crossnumber I wrote for the issue).
It all began in December 1956, when an article about hexaflexagons was published in Scientific American. A hexaflexagon is a hexagonal paper toy which can be folded and then opened out to reveal hidden faces. If you have never made a hexaflexagon, then you should stop reading and make one right now. Once you've done so, you will understand why the article led to a craze in New York; you will probably even create your own mini-craze because you will just need to show it to everyone you know.
The author of the article was, of course, Martin Gardner.
A Christmas flexagon.
Martin Gardner was born in 1914 and grew up in Tulsa, Oklahoma. He earned a bachelor's degree in philosophy from the University of Chicago and after four years serving in the US Navy during the Second World War, he returned to Chicago and began writing. After a few years working on children's magazines and the occasional article for adults, Gardner was introduced to John Tukey, one of the students who had been involved in the creation of hexaflexagons.
Soon after the impact of the hexaflexagons article became clear, Gardner was asked if he had enough material to maintain a monthly column. This column, Mathematical Games, was written by Gardner every month from January 1956 for 26 years until December 1981. Throughout its run, the column introduced the world to a great number of mathematical ideas, including Penrose tiling, the Game of Life, public key encryption, the art of MC Escher, polyominoes and a matchbox machine learning robot called MENACE.


Gardner regularly received topics for the column directly from their inventors. His collaborators included Roger Penrose, Raymond Smullyan, Douglas Hofstadter, John Conway and many, many others. His closeness to researchers allowed him to write about ideas that the general public were previously unaware of and share newly researched ideas with the world.
In 1970, for example, John Conway invented the Game of Life, often simply referred to as Life. A few weeks later, Conway showed the game to Gardner, allowing him to write the first ever article about the now-popular game.
In Life, cells on a square lattice are either alive (black) or dead (white). The status of the cells in the next generation of the game is given by the following three rules:
For example, here is a starting configuration and its next two generations:
The first three generations of a game of Life.
The collection of blocks on the right of this game is called a glider, as it will glide to the right and upwards as the generations advance. If we start Life with a single glider, then the glider will glide across the board forever, always covering five squares: this starting position will not lead to the sad ending where everything is dead. It is not obvious, however, whether there is a starting configuration that will lead the number of occupied squares to increase without bound.
Gosper's glider gun.
Originally, Conway and Gardner thought that this was impossible, but after the article was published, a reader and mathematician called Bill Gosper discovered the glider gun: a starting arrangement in Life that fires a glider every 30 generations. As each of these gliders will go on to live forever, this starting configuration results in the number of live cells perpetually increasing!
This discovery allowed Conway to prove that any Turing machine can be built within Life: starting arrangements exist that can calculate the digits of pi, solve equations, or do any other calculation a computer is capable of (although very slowly)!

Encrypting with RSA

To encode the message \(809$, we will use the public key:
$$s=19\quad\text{and}\quad r=1769$$
The encoded message is the remainder when the message to the power of \(s\) is divided by \(r$:

Decrypting with RSA

To decode the message, we need the two prime factors of \(r\) (\(29\) and \(61\)). We multiply one less than each of these together:
\begin{align*} a&=(29-1)\times(61-1)\\[-2pt] &=1680. \end{align*}
We now need to find a number \(t\) such that \(st\equiv1\mod a\). Or in other words:
$$19t\equiv1\mod 1680$$
One solution of this equation is \(t=619\) (calculated via the extended Euclidean algorithm).
Then we calculate the remainder when the encoded message to the power of \(t\) is divided by \(r\):


Another concept that made it into Mathematical Games shortly after its discovery was public key cryptography. In mid-1977, mathematicians Ron Rivest, Adi Shamir and Leonard Adleman invented the method of encryption now known as RSA (the initials of their surnames). Here, messages are encoded using two publicly shared numbers, or keys. These numbers and the method used to encrypt messages can be publicly shared as knowing this information does not reveal how to decrypt the message. Rather, decryption of the message requires knowing the prime factors of one of the keys. If this key is the product of two very large prime numbers, then this is a very difficult task.

Something to think about

Gardner had no education in maths beyond high school, and at times had difficulty understanding the material he was writing about. He believed, however, that this was a strength and not a weakness: his struggle to understand led him to write in a way that other non-mathematicians could follow. This goes a long way to explaining the popularity of his column.
After Gardner finished working on the column, it was continued by Douglas Hofstadter and then AK Dewney before being passed down to Ian Stewart.
Gardner died in May 2010, leaving behind hundreds of books and articles. There could be no better way to end than with something for you to go away and think about. These of course all come from Martin Gardner's Mathematical Games:

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 2015-10-21 14:27:38 

How to Kick a Conversion

This post also appeared on the Chalkdust Magazine blog.
If you're like me, then you will be disappointed that all of the home nations have been knocked out of the Rugby World Cup. If you're really like me, doing some maths related to rugby will cheer you up...
The scoring system in rugby awards points in packets of 3, 5 and 7. This leads a number of interesting questions that you can find in my guest puzzle on Alex Bellos's Guardian blog. In this blog post, we will focus on another area of rugby: conversion kicking.

Conversion Kicks

When a try is scored by putting the ball down behind the line, the scoring team gets to take a conversion kick. This kick must be taken in line with where the try was scored but it is up to the kicker how far away the kick should be taken. But how far back should the ball be taken to make the kick easiest?
Too close (red) and too far away (blue) will give small angles to aim at. Somewhere in the middle is needed (green).
One way to answer this question is to look to maximise the angle between the posts which the kicker will have to aim at: if the kick is taken too close to or too far from the goal line there will be a very thin angle to aim at. Somewhere between these extremes there will be a maximum angle to aim at.
When looking to maximise this angle, we can use one of the 'circle theorems' which have tormented many generations of GCSE maths students: 'angles subtended by the same arc at the circumference are equal'. This means that if a circle is drawn going through both posts, then the angle made at any point on this circle will be the same.
The angles made by the red and blue lines are equal because 'angles subtended by the same arc at the circumference are equal'.
A larger circle drawn through the posts will give a smaller angle. If a vertical line is drawn which just touches the right of the circle, then the point at which it touches the circle will be the best place on this line to take a kick. This is because any other point on the line will be on a larger circle and so make a smaller angle.
Using this method for circles of different sizes leads to the following diagram, which shows where the kick should be taken for every position a try could be scored:
The best place to take a kick?
This, however, is not the best place to take the kick.

Taking Account of Height

When a try is scored near the posts, the above method recommends a position from where the ball must be kicked at an impossibly steep angle to go over. To deal with this problem, we are going to have to look at the situation from the side.
When kicked, the ball will travel along a parabola (ignoring air resistance and wind as their effects will be small[citation needed]). Given a distance from the posts, there will be two angles which the ball can be kicked at and just make it over the bar. Kicking at any angle between these two will lead to a successful conversion. Again, we have an angle which we would like to maximise.
The highest (blue) and lowest (red) the ball can be kicked while still going over the bar.
However, the position where this angle is maximised is very unlikely to also maximise the angle we looked at earlier. To find the best place to kick from, we need to find a compromise point where both angles are quite big.
To do this, imagine that the kicker is standing inside a large sphere. For each point on the sphere, kicking the ball at the point will either lead to it going over or missing. We can draw a shape on the sphere so that aiming inside the shape will lead to scoring. Our sensible kicker will aim at the centre of this shape.
But our kicker will not be able to aim perfectly: there will be some random variation. We can predict that this variation will follow a Kent distribution, which is like a normal distribution but on the surface of a sphere. We can use this distribution to calculate the probability that our kicker will score. We would like to maximise this probability.
The Kent distribution can be adjusted to reflect the accuracy of the kicker. Below are the optimal kicking positions for an inaccurate, an average and a very accurate kicker.
The best place to take a kick for a bad kicker (top), an average kicker (middle) and a good kicker (bottom). All the kickers kick the ball at 30m/s.
As you might expect, the less accurate kicker should stand slightly further forwards to make it easier to aim. Perhaps surprisingly, the good kicker should stand further back when between the posts than when in line with the posts.
The model used to create these results could be further refined. Random variation in the speed of the kick could be introduced. Or the kick could be made to have more variation horizontally than vertically: there are parameters in the Kent distribution which allow this to be easily adjusted. In fact, data from players could be used to determine the best position for each player to kick from.
In addition to analysing conversions, this method could be used to determine the probability of scoring 3 points from any point on the pitch. This could be used in conjunction with the probability of scoring a try from a line-out to decide whether kicking a penalty for the posts or into touch is likely to lead to the most points.
Although estimating the probability of scoring from a line-out is a difficult task. Perhaps this will give you something to think about during the remaining matches of the tournament.

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 2015-10-08 04:38:58 

How Much Will I Win on the New National Lottery?

This post also appeared on the Chalkdust Magazine blog. You can read the excellent second issue of Chalkdust here, including the £100 prize crossnumber which I set.
From today, the National Lottery's Lotto draw has 59 balls instead of 49. You may be thinking that this means there is now much less chance of winning. You would be right, except the prizes are also changing.
Camelot, who run the lottery, are saying that you are now "more likely to win a prize" and "more likely to become a millionaire". But what do these changes actually mean?

The Changes

Until yesterday, Lotto had 49 balls. From today, there are 59 balls. Each ticket still has six numbers on it and six numbers, plus a bonus ball, are still chosen by the lottery machine. The old prizes were as follows:
RequirementEstimated Prize
Match all 6 normal balls£2,000,000
Match 5 normal balls and the bonus ball£50,000
Match 5 normal balls£1,000
Match 4 normal balls£100
Match 3 normal balls£25
50 randomly picked tickets£20,000
The prizes have changed to:
RequirementEstimated Prize
Match all 6 normal balls£2,000,000
Match 5 normal balls and the bonus ball£50,000
Match 5 normal balls£1,000
Match 4 normal balls£100
Match 3 normal balls£25
Match 2 normal ballsFree lucky dip entry in next Lotto draw
One randomly picked ticket£1,000,000
20 other randomly picked tickets£20,000

Probability of Winning a Prize

The probability of winning each of these prizes can be calculated. For example, the probability of matching all 6 balls in the new lotto is $$\mathbb{P}(\mathrm{matching\ ball\ 1})\times \mathbb{P}(\mathrm{matching\ ball\ 2})\times...\times\mathbb{P}(\mathrm{matching\ ball\ 6})$$ $$=\frac{6}{59}\times\frac{5}{58}\times\frac{4}{57}\times\frac{3}{56}\times\frac{2}{55}\times\frac{1}{54}$$ $$=\frac{1}{45057474},$$ and the probability of matching 4 balls in the new lotto is $$(\mathrm{number\ of\ different\ ways\ of\ picking\ four\ balls\ out\ of\ six})\times\mathbb{P}(\mathrm{matching\ ball\ 1})\times\\...\times\mathbb{P}(\mathrm{matching\ ball\ 4})\times\mathbb{P}(\mathrm{not\ matching\ ball\ 5})\times\mathbb{P}(\mathrm{not\ matching\ ball\ 6})$$ $$=15\times\frac{6}{59}\times\frac{5}{58}\times\frac{4}{57}\times\frac{3}{56}\times\frac{53}{55}\times\frac{52}{54}$$ $$=\frac{3445}{7509579}.$$ In the second calculation, it is important to include the probabilities of not matching the other balls to prevent double counting the cases when more than 4 balls are matched.
Calculating a probability for every prize and then adding them up gives the probability of winning a prize. In the old draw, the probability of winning a prize was \(0.0186\). In the new draw, it is \(0.1083\). So Camelot are correct in claiming that you are now more likely to win a prize.
But not all prizes are equal: these probabilities do not take into account the values of the prizes. To analyse the actual winnings, we're going to have to look at the expected amount of money you will win. But first, let's look at Camelot's other claim: that under the new rules you are more likely to become a millionaire.

Probability of Winning £1,000,000

In the old draw, the only way to win a million pounds was to match all six balls. The probability of this happening was \(0.00000007151\) or \(7.151\times 10^{-8}\).
In the new lottery, a million pounds can be won either by matching all six balls or by winning the millionaire raffle. This will lead to different probabilities of winning on Wednesdays and Saturdays due to different numbers of people buying tickets. Based on expected sales of 16.5 million tickets on Saturdays and 8.5 million tickets on Wednesdays, the chances of becoming a millionaire on a Wednesday or Saturday are \(0.0000001398\) (\(1.398\times 10^{-7}\)) and \(0.00000008280\) (\(8.280\times 10^{-8}\)) respectively.
These are both higher than the probability of winning a million in the old draw, so again Camelot are correct: you are now more likely to become a millionaire...
But the new chances of becoming a millionaire are actually even higher. The probabilities given above are the chances of winning a million in a given draw. But if two balls are matched, you win a lucky dip: you could win a million in the next draw without buying another ticket. We should include this in the probability calculated above, as you are still becoming a millionaire due to the original ticket you bought.
In order to count this, let \(A_W\) and \(A_S\) be the probabilities of winning a million in a given draw (as given above) on a Wednesday or a Saturday, let \(B_W\) and \(B_S\) be the probabilities of winning a million in this draw or due to future lucky dip tickets on a Wednesday or a Saturday (the values we want to find) and let \(p\) be the probability of matching two balls. We can write $$B_W=A_W+pB_S$$ and $$B_S=A_S+pB_W$$ since the probability of winning a million is the probability of winning in this draw (\(A\)) plus the probability of winning a lucky dip ticket and winning in the next draw (\(pB\)). Substituting and rearranging, we get $$B_W=\frac{A_W+pA_S}{1-p^2}$$ and $$B_W=\frac{A_S+pA_W}{1-p^2}.$$
Using this (and the values of \(A_S\) and \(A_W\) calculated earlier) gives us probabilities of \(0.0000001493\) (\(1.493\times 10^{-7}\)) and \(0.00000009736\) (\(9.736\times 10^{-8}\)) of becoming a millionaire on a Wednesday and a Saturday respectively. These are both significantly higher than the probability of becoming a millionaire in the old draw (\(7.151\times 10^{-8}\)).
Camelot's two claims—that you are more likely to win a prize and you are more likely to become a millionaire—are both correct. It sounds like the new lottery is a great deal, but so far we have not taken into account the size of the prizes you will win and have only shown that a very rare event will become slightly less rare. Probably the best way to measure how good a lottery is is by working out the amount of money you should expect to win, so let's now look at that.

Expected Prize Money

To find the expected prize money, we must multiply the value of each prize by the probability of winning that prize and then add them up, or, in other words,
$$\sum_\mathrm{prizes}\mathrm{value\ of\ prize}\times\mathbb{P}(\mathrm{winning\ prize}).$$
Once this has been calculated, the chance of winning due to a free lucky dip entry must be taken into account as above.
In the old draw, after buying a ticket for 2, you could expect to win 78p or 83p on a Wednesday or Saturday respectively. In the new draw, the expected winnings have changed to 58p and 50p (Wednesday and Saturday respectively). Expressed in this way, it can be seen that although the headline changes look good, the overall value for money of the lottery has significantly decreased.
Looking on the bright side, this does mean that the lottery will make even more money that it can put towards charitable causes: the lottery remains an excellent way to donate your money to worthy charities!

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 2015-03-25 08:30:38 

Optimal Pac-Man

This is an article which I wrote for the first issue of Chalkdust. I highly recommend reading the rest of the magazine (and trying to solve the crossnumber I wrote for the issue).
In the classic arcade game Pac-Man, the player moves the title character through a maze. The aim of the game is to eat all of the pac-dots that are spread throughout the maze while avoiding the ghosts that prowl it.
While playing Pac-Man recently, my concentration drifted from the pac-dots and I began to think about the best route I could take to complete the level.

Seven Bridges of Königsberg

In the 1700s, Swiss mathematician Leonhard Euler studied a related problem. The city of Königsberg had seven bridges, which the residents would try to cross while walking around the town. However, they were unable to find a route crossing every bridge without repeating one of them.
Diagram showing the bridges in Königsberg. If you have not seen this puzzle before, you may like to try to find a route crossing them all exactly once before reading on.
In fact, the city dwellers could not find such a route because it is impossible to do so, as Euler proved in 1735. He first simplified the map of the city, by making the islands into vertices (or nodes) and the bridges into edges.
A graph of the seven bridges problem.
This type of diagram has (slightly confusingly) become known as a graph, the study of which is called graph theory. Euler represented Königsberg in this way as he realised that the shape of the islands is irrelevant to the problem: representing the problem as a graph gets rid of this useless information while keeping the important details of how the islands are connected.
Euler next noticed that if a route crossing all the bridges exactly once was possible then whenever the walker took a bridge onto an island, they must take another bridge off the island. In this way, the ends of the bridges at each island can be paired off. The only bridge ends that do not need a pair are those at the start and end of the circuit.
This means that all of the vertices of the graph except two (the first and last in the route) must have an even number of edges connected to them; otherwise there is no route around the graph travelling along each edge exactly once. In Königsberg, each island is connected to an odd number of bridges. Therefore the route that the residents were looking for did not exist (a route now exists due to two of the bridges being destroyed during World War II).
This same idea can be applied to Pac-Man. By ignoring the parts of the maze without pac-dots the pac-graph can be created, with the paths and the junctions forming the edges and vertices respectively. Once this is done there will be twenty-four vertices, twenty of which will be connected to an odd number of edges, and so it is impossible to eat all of the pac-dots without repeating some edges or travelling along parts of the maze with no pac-dots.
The Pac-graph. The odd nodes are shown in red.
This is a start, but it does not give us the shortest route we can take to eat all of the pac-dots: in order to do this, we are going to have to look at the odd vertices in more detail.

The Chinese Postman Problem

The task of finding the shortest route covering all the edges of a graph has become known as the Chinese postman problem as it is faced by postmen---they need to walk along each street to post letters and want to minimise the time spent walking along roads twice---and it was first studied by Chinese mathematician Kwan Mei-Ko.
As the seven bridges of Königsberg problem demonstrated, when trying to find a route, Pac-Man will get stuck at the odd vertices. To prevent this from happening, all the vertices can be made into even vertices by adding edges to the graph. Adding an edge to the graph corresponds to choosing an edge, or sequence of edges, for Pac-man to repeat or including a part of the maze without pac-dots. In order to complete the level with the shortest distance travelled, Pac-man wants to add the shortest total length of edges to the graph. Therefore, in order to find the best route, Pac-Man must look at different ways to pair off the odd vertices and choose the pairing which will add the least total distance to the graph.
The Chinese postman problem and the Pac-Man problem are slightly different: it is usually assumed that the postman wants to finish where he started so he can return home. Pac-Man however can finish the level wherever he likes but his starting point is fixed. Pac-Man may therefore leave one odd node unpaired but must add an edge to make the starting node odd.
One way to find the required route is to look at all possible ways to pair up the odd vertices. With a low number of odd vertices this method works fine, but as the number of odd vertices increases, the method quickly becomes slower.
With four odd vertices, there are three possible pairings. For the Pac-Man problem there will be over 13 billion (\(1.37\times 10^{10}\)) pairings to check. These pairings can be checked by a laptop running overnight, but for not too many more vertices this method quickly becomes unfeasible.
With 46 odd nodes there will be more than one pairing per atom in the human body (\(2.53\times 10^{28}\)). By 110 odd vertices there will be more pairings (\(3.47\times 10^{88}\)) than there are estimated to be atoms in the universe. Even the greatest supercomputer will be unable to work its way through all these combinations.
Better algorithms are known for this problem that reduce the amount of work on larger graphs. The number of pairings to check in the method above increases like the factorial of the number of vertices. Algorithms are known for which the amount of work to be done increases like a polynomial in the number of vertices. These algorithms will become unfeasible at a much slower rate but will still be unable to deal with very large graphs.

Solution of the Pac-Man Problem

For the Pac-Man problem, the shortest pairing of the odd vertices requires the edges marked in red to be repeated. Any route which repeats these edges will be optimal. For example, the route in green will be optimal.
One important element of the Pac-Man gameplay that I have neglected are the ghosts (Blinky, Pinky, Inky and Clyde), which Pac-Man must avoid. There is a high chance that the ghosts will at some point block the route shown above and ruin Pac-Man's optimality. However, any route repeating the red edges will be optimal: at many junctions Pac-man will have a choice of edges he could continue along. It may be possible for a quick thinking player to utilise this freedom to avoid the ghosts and complete an optimal game.
Additionally, the skilled player may choose when to take the edges that include the power pellets, which allow Pac-Man to reverse the roles and eat the ghosts. Again cleverly timing these may allow the player to complete an optimal route.
Unfortunately, as soon as the optimal route is completed, Pac-Man moves to the next level and the player has to do it all over again ad infinitum.

A Video

Since writing this piece, I have been playing Pac-Man using MAME (Multiple Arcade Machine Emulator). Here is one game I played along with the optimal edges to repeat for reference:

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You're right. In a number of places I could've turned round a few pixels earlier.

There seems to be no world record for just one Pac-Man level (and I don't have time to get good enough to speed run all 255 levels before it crashes!)
This vid was billed as an "optimal" run but around 40 seconds in you eat one "pill" that you don't need to eat. Why don't you just speedrun the first level? This must have been done before. Can you beat the world record?
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