Since it's only two weeks until the IMA Festival of Mathematics and its Applications at the University of Greenwich, which I am helping to organise, it's time I wrote something about it here. This national Festival presents, over two days, about 50 talks, workshops and activities, showcasing the diversity of mathematics and mathematicians. Visitors can learn about applications of mathematics in statistics, cryptography, numbers, fashion, medicine, fire safety engineering, and many more: they can learn how to make boomerangs and they can walk on custard. There is something for everybody!

The Festival has been generously sponsored by the IMA, the University of Greenwich, FMSP, GCHQ, the OR Society, the London South East and Kent & Medway Maths Hubs, and FDM, For information about the sponsors, click on the logos on the Festival website.

I've been delighted that so many top presenters have agreed to take part in the Festival. If you can get to Greenwich on either of these days, you'll have a wonderful time exploring lots of exciting mathematics!

## Monday, 12 June 2017

## Tuesday, 9 May 2017

### A Champions League Mathematical curiosity

(Apologies for any reader who doesn't share my interest in football. You don't have to read on.)

When I'm fed up marking, like so many other middle-aged men, I turn to the computer game Football Manager. And a disaster for my team last night made me aware of a curiosity regarding the tie-break rules in the Champions League.

This is the situation. It's the last minute of the last match of the group stage and my team, Arsenal, are losing 2-1 at home to Olympiakos (we haven;t had much luck and the red card early in the first half for our midfielder Victor Wamyama has cost us). But it's OK - Bayern Munich (who are losing in Lyon) will win the group and we will come second, qualifying for the knock-out stages, while Olympiakos will be out.

But what's this? Bayern have scored a last-minute equaliser against Lyon! They were going to win the group anyway, so presumably we still finish in that all-important second place.

But no - Olympiakos are now above us: we are down to third and we are out! A goal in the other game, which doesn't change the position of the teams in that game, has caused Arsenal and Olympiakos to swap positions, with disastrous results.

It's all to do with tie-break rules. Without Bayern's late goal, all four teams would have had eight points, The tie-break rule looks at the scores in matches involving all the teams which have tied on points. In this case Bayern (who scored six at home to Olympiakos) had the best goal difference, with Arsenal (who had played well until the last match, despite early red cards ion three matches) second best, and so they finish first and second. But Bayern's equaliser means they have nine points, and Arsenal and Olympiakos are tied for second place: and since the two teams drew in Greece and Olympiakos won in London, the Greeks are above Arsenal. That last-minute goal in the other match really has moved Olympiakos above Arsenal. (If you want to check for yourself, I give all the scores below.)

This consequence of the mathematics of the tie-break rules is something I was vaguely aware of, though I didn't realise the peril my team were in until about ten game-minutes before Bayern's goal. But it is slightly counter-intuitive that a goal in match B can cause the teams in match A to change positions. (In fact in my game, before the goal Lyon had been in third place above Olympiakos, so conceding the goal caused them to drop to fourth, but a slightly different set of results could have meant that a game in match B could leave the two teams in that match in the same position while inverting the order of the other two teams.)

Here, for anyone who cares, are the complete results:

Game 1 - Arsenal (2) 2 Bayern (0) 0; Lyon (0) 1 Olympiakos (0) 0

Game 2 - Bayern (0) 0 Lyon (0) 0; Olympiakos (0) 1 Arsenal (0) 1

Game 3 - Lyon (0) 1 Arsenal (1) 3; Olympiakos (0) 1 Bayern (0) 1

Game 4 - Arsenal (0) 0 Lyon (0) 0; Bayern (3) 6 Olympiakos (0) 0

Game 5 - Bayern (0) 1 Arsenal (0) 0; Olympiakos (0) 1 Lyon (0) 0

Game 6 - Arsenal (0) 1 Olympiakos (0) 2; Lyon (3) 3 Bayern (0) 3

When I'm fed up marking, like so many other middle-aged men, I turn to the computer game Football Manager. And a disaster for my team last night made me aware of a curiosity regarding the tie-break rules in the Champions League.

This is the situation. It's the last minute of the last match of the group stage and my team, Arsenal, are losing 2-1 at home to Olympiakos (we haven;t had much luck and the red card early in the first half for our midfielder Victor Wamyama has cost us). But it's OK - Bayern Munich (who are losing in Lyon) will win the group and we will come second, qualifying for the knock-out stages, while Olympiakos will be out.

But what's this? Bayern have scored a last-minute equaliser against Lyon! They were going to win the group anyway, so presumably we still finish in that all-important second place.

But no - Olympiakos are now above us: we are down to third and we are out! A goal in the other game, which doesn't change the position of the teams in that game, has caused Arsenal and Olympiakos to swap positions, with disastrous results.

It's all to do with tie-break rules. Without Bayern's late goal, all four teams would have had eight points, The tie-break rule looks at the scores in matches involving all the teams which have tied on points. In this case Bayern (who scored six at home to Olympiakos) had the best goal difference, with Arsenal (who had played well until the last match, despite early red cards ion three matches) second best, and so they finish first and second. But Bayern's equaliser means they have nine points, and Arsenal and Olympiakos are tied for second place: and since the two teams drew in Greece and Olympiakos won in London, the Greeks are above Arsenal. That last-minute goal in the other match really has moved Olympiakos above Arsenal. (If you want to check for yourself, I give all the scores below.)

This consequence of the mathematics of the tie-break rules is something I was vaguely aware of, though I didn't realise the peril my team were in until about ten game-minutes before Bayern's goal. But it is slightly counter-intuitive that a goal in match B can cause the teams in match A to change positions. (In fact in my game, before the goal Lyon had been in third place above Olympiakos, so conceding the goal caused them to drop to fourth, but a slightly different set of results could have meant that a game in match B could leave the two teams in that match in the same position while inverting the order of the other two teams.)

Here, for anyone who cares, are the complete results:

Game 1 - Arsenal (2) 2 Bayern (0) 0; Lyon (0) 1 Olympiakos (0) 0

Game 2 - Bayern (0) 0 Lyon (0) 0; Olympiakos (0) 1 Arsenal (0) 1

Game 3 - Lyon (0) 1 Arsenal (1) 3; Olympiakos (0) 1 Bayern (0) 1

Game 4 - Arsenal (0) 0 Lyon (0) 0; Bayern (3) 6 Olympiakos (0) 0

Game 5 - Bayern (0) 1 Arsenal (0) 0; Olympiakos (0) 1 Lyon (0) 0

Game 6 - Arsenal (0) 1 Olympiakos (0) 2; Lyon (3) 3 Bayern (0) 3

## Monday, 1 May 2017

### Something I'd forgotten

One of my parents daily rituals was to change the date on a device on top of the bureau which displayed the day, date and month. My equivalent is to change the date on my Rubik's cube-style calendar , which also gives the day, ate and month: the month in three-letter form ("Jan", "Feb", "Mar" etc). It's nice that the names of the months in English make this possible (though www.puzl.co.uk, where I got my calendar cube, have also sold French and German versions).

But I was astonished to find in the cellar yesterday that I had made something similar myself, 35 years ago. I have absolutely no recollection of this at all (but the handwriting is mine), but I am quite impressed by my ingenuity.

In my final year as a student, I played backgammon regularly with my friend Karen. It appears that I constructed this device to keep track of the cumulative score. It's rather the worse for wear, and has lost one of the corner cubies. But I assume that it still records the final state of play when we had played our last game - which shows that, if you want to win at a gambling game, it is best not to play against a future professor of psychology.

But I was astonished to find in the cellar yesterday that I had made something similar myself, 35 years ago. I have absolutely no recollection of this at all (but the handwriting is mine), but I am quite impressed by my ingenuity.

In my final year as a student, I played backgammon regularly with my friend Karen. It appears that I constructed this device to keep track of the cumulative score. It's rather the worse for wear, and has lost one of the corner cubies. But I assume that it still records the final state of play when we had played our last game - which shows that, if you want to win at a gambling game, it is best not to play against a future professor of psychology.

## Saturday, 1 April 2017

### How to use number theory to help bus travellers

Living in London, I use buses a lot. Each day I catch a 180 or 199, ignoring the 188, 286 and others whcih don't take me home. I have a good memory for all the different buses in the parts of central London that I frequent, but wouldn't it be easier if the bus number told me where it was going? If, rather than 180 arbitrarily designating a route between Lewisham and Belvedere, the number itself indicated where the bus goes?

So today I am going to unveil my scheme for taking advantage of the properties of numbers to do exactly that. I will start off with my first idea, and then show enhancements.

The Fundamental Theorem of Arithmetic tells us that any number can be expresssed as the product of prime numbers in an essentially unique way. So my first scheme allocates to every destination a different prime number. For example, we might assign 2 to Waterloo, 3 to Euston and 5 to London Bridge. Then the bus route serving these three places would be number 30. All you need to do, as a passenger, is know the number assigned to your destination. If you want to go to Waterloo, you know that any even-numbered bus will take you there: if you are heading for London Bridge then any bus whose number ends with 0 or 5 will do,

You might object that it is easy to tell when a number is divisible by 2 or 5, but less easy if your destination's number is, say, 17. But there are tricks: to test whether a number is divisible by 17, one simply tests the number obtained by subtracting five times the last digit from the rest (so for 374, we subtract 20 from 37, getting 17: that tells us 17 divides 374).

But this method has a glaring weakness. It doesn't tell us the order of the stops. I want to be able to distinguish the route Euston - Waterloo - London Bridge from Euston - London Bridge - Waterloo since I want to travel direct from Euston to Waterloo, and going via London Bridge would take much longer. So my improved proposal uses Godel numbering. If a bus visits destinations a, b, c, ... in that order, its number is 2^a times 3^b times 5^c times ... And this can be dynamically adjusted during the journey: when the bus has left Euston, the number changes to reflect that the next stop is Londo Bridge. Now, I can find out not only whether the bus takes me to Waterloo, but how many stops there are first. In my example, if I am at Euston, bus number 288 (2^5x3^2) will take me to London Bridge and then Waterloo, while 1944 (2^3*3^5) will go to London Bridge via Waterloo.

But my final scheme is even better. In this one, The bus number is the product of the primes representing the places it visits, each raised to the power of the number of minutes it is expected to take to get there. If a bus doesn't call at destination

You might object that this system assumes bus passengers can carry out mental arithmetic. But of course, there will be apps to do this for those who are not confident. I will just point my smartphone at the front of a bus, and the app will read the number (say 7200) and tell me that it will be at Waterloo (destination 2) in 5 minutes (since the highest power of 2 dividing 7200 is 2^5 = 32).

Yet another way in which mathematics can make our lives easier!

So today I am going to unveil my scheme for taking advantage of the properties of numbers to do exactly that. I will start off with my first idea, and then show enhancements.

The Fundamental Theorem of Arithmetic tells us that any number can be expresssed as the product of prime numbers in an essentially unique way. So my first scheme allocates to every destination a different prime number. For example, we might assign 2 to Waterloo, 3 to Euston and 5 to London Bridge. Then the bus route serving these three places would be number 30. All you need to do, as a passenger, is know the number assigned to your destination. If you want to go to Waterloo, you know that any even-numbered bus will take you there: if you are heading for London Bridge then any bus whose number ends with 0 or 5 will do,

You might object that it is easy to tell when a number is divisible by 2 or 5, but less easy if your destination's number is, say, 17. But there are tricks: to test whether a number is divisible by 17, one simply tests the number obtained by subtracting five times the last digit from the rest (so for 374, we subtract 20 from 37, getting 17: that tells us 17 divides 374).

But this method has a glaring weakness. It doesn't tell us the order of the stops. I want to be able to distinguish the route Euston - Waterloo - London Bridge from Euston - London Bridge - Waterloo since I want to travel direct from Euston to Waterloo, and going via London Bridge would take much longer. So my improved proposal uses Godel numbering. If a bus visits destinations a, b, c, ... in that order, its number is 2^a times 3^b times 5^c times ... And this can be dynamically adjusted during the journey: when the bus has left Euston, the number changes to reflect that the next stop is Londo Bridge. Now, I can find out not only whether the bus takes me to Waterloo, but how many stops there are first. In my example, if I am at Euston, bus number 288 (2^5x3^2) will take me to London Bridge and then Waterloo, while 1944 (2^3*3^5) will go to London Bridge via Waterloo.

But my final scheme is even better. In this one, The bus number is the product of the primes representing the places it visits, each raised to the power of the number of minutes it is expected to take to get there. If a bus doesn't call at destination

*p*, then the number is not a product of*p*. So if I want to get to Waterloo, and bus number*n*arrives, I check to see what is the largest power of 2 which divides*n*: that is how many minutes it will take to get there. Of course, this is adjusted dynamically, in the same way as London's bus stops now tell us how long it will be before the next bus arrives.You might object that this system assumes bus passengers can carry out mental arithmetic. But of course, there will be apps to do this for those who are not confident. I will just point my smartphone at the front of a bus, and the app will read the number (say 7200) and tell me that it will be at Waterloo (destination 2) in 5 minutes (since the highest power of 2 dividing 7200 is 2^5 = 32).

Yet another way in which mathematics can make our lives easier!

## Sunday, 12 March 2017

### Puzzles from my grandmother

I have been reading the autobiography of one of my heroes - the great popular mathematics writer, Martin Gardner, who along with my school maths teachers Jimmy Cowan and Ivan Wells, inspired me with the excitement of mathematics. I have come late to Gardner's autobiography,

One of Gardner's stories reminded me of my own childhood. Gardner recounts his uncle telling him a riddle. "There was one duck with two ducks behind it; one duck with two ducks in front of it; and one duck between two ducks. How many ducks were there?"

SPOILER ALERT: Gardner notes that his uncle began by saying, "There were three ducks", which gave away the answer.

This reminded me of my paternal grandmother giving me two puzzles when I must have been perhaps in the upper levels of primary school. I had to make sense of the following:

First riddle:

11 was a race-horse.

22 was 12.

1111 race.

22112.

Second riddle:

If the B MT put : .

If the B . putting : .

(I have just googled the first of these, and curiously it is described on one website as a tongue-twister, which it isn't!)

SOLUTIONS

The first riddle reads as "One-one was a race-horse. Two-two was one too. One-one won one race. Two-two won one too."

The second is "If the grate be empty, put coal on. If the grate be full, stop putting coal on."

*Undiluted Hocus-Pocus*, which was published a few years ago, partly because of luke-warm reviews, and while I enjoyed many of the anecdotes, I wouldn't regard it as essential reading even for those who, like me, admire Gardner enormously. But I'm glad to have read it.One of Gardner's stories reminded me of my own childhood. Gardner recounts his uncle telling him a riddle. "There was one duck with two ducks behind it; one duck with two ducks in front of it; and one duck between two ducks. How many ducks were there?"

SPOILER ALERT: Gardner notes that his uncle began by saying, "There were three ducks", which gave away the answer.

This reminded me of my paternal grandmother giving me two puzzles when I must have been perhaps in the upper levels of primary school. I had to make sense of the following:

First riddle:

11 was a race-horse.

22 was 12.

1111 race.

22112.

Second riddle:

If the B MT put : .

If the B . putting : .

(I have just googled the first of these, and curiously it is described on one website as a tongue-twister, which it isn't!)

SOLUTIONS

The first riddle reads as "One-one was a race-horse. Two-two was one too. One-one won one race. Two-two won one too."

The second is "If the grate be empty, put coal on. If the grate be full, stop putting coal on."

## Monday, 2 January 2017

### Why (some) mistakes are interesting

As you would expect, there are a number of reasons.

Perhaps a mistake casts interesting light on someone's thought processes, revealing the way a mathematician was approaching a problem or what was in their had as they tackled it.

Perhaps it is simply a cautionary example: seeing how someone else has erred helps one avoid making the same mistake.

Yet another reason is not exactly schadenfreude (though that may be part of it) but that seeing better people than me make errors is encouraging: I make lots of mistakes and it's helpful to realise that most other people do too!

Now, I worked for many years as a software engineer, working on safety-critical systems, and I have taught software engineering.

Errors occur too often in software, and the better we understand how we make errors, the more likely we are to be able to reduce their frequency.

When I was writing software, I felt that there were lessons to be learnt from railway accidents, particularly those where a remarkable combination of circumstances defeated what had seemed to be an infallible system.

You might have had considerable faith in the Tyer electric tablet system, which for many years after its introduction prevented the dreadful collisions on single-track lines which had occurred when two trains travelling in opposite directions entered the same section: but at Abermule in 1921 the combination of many tiny lapses by several individuals subverted what had appeared to be an infallible system.

Equally unlucky was the Hull Paragon accident in 1927, when two apparently independent slips by signalmen interacted in an extremely unlikely way to subvert the signalling system which protected the trains (the photo above comes from www.railwaysarchive.co.uk).

As a software engineer,I felt there was a lot to be learnt from thinking about such system failures: could I be confident that my own system could not fail in some unlikely combination of circumstances, when the accidents at Abermule and Hull Paragon show how even apparently the most secure systems can fail?

Rhese are some reasons I think (some) mistakes are worth our study.

## Thursday, 29 December 2016

### A curious cat and another curious error

The first mathematics book written in English was the snappily-titled, anonymous

The book concludes by presenting lots of interesting problems and their solutions. One is a version of the Josephus problem, in which fifteen out of thirty merchants are to be cast overboard to save an overloaded galley in a storm: the reader is given a mnemonic for arranging the merchants so that the right fifteen (the Christians rather than the Saracens, as one would expect for the time) are saved. The mnemonic is a Latin verse, which seems a little odd for a book whose selling point was that it was in English! Presumably the author was padding his book out with whatever came to hand, not very carefully, as we shall see.

Another example asks about travellers going in opposite directions between London and Paris and when they will meet, or at least I think that is the intention: but since in the book one traveller is going between Paris and London, and the other between Paris and Lyon, they are unlikely to meet unless one of them gets badly lost. It seems that our author was trying to make the problem more relevant to an English readership but didn't carry his intention through.

The most curious of the problems, to my mind, is "The rule and questyon of a Catte". The problem presented (transcribed from a 1546 edition) is,

*An introduction for to lerne to recken with the pen, or with the counters accordyng to the trewe cast of Algorisme, in hole numbers or in broken, newly corrected...*, first published in the 1530s. A few years ago the British Library paid £95,000 for a copy of the first edition. Happily, a facsimile (of the second edition) is now cheaply available in the wonderful series produced by TGR Renascent Books.The book concludes by presenting lots of interesting problems and their solutions. One is a version of the Josephus problem, in which fifteen out of thirty merchants are to be cast overboard to save an overloaded galley in a storm: the reader is given a mnemonic for arranging the merchants so that the right fifteen (the Christians rather than the Saracens, as one would expect for the time) are saved. The mnemonic is a Latin verse, which seems a little odd for a book whose selling point was that it was in English! Presumably the author was padding his book out with whatever came to hand, not very carefully, as we shall see.

Another example asks about travellers going in opposite directions between London and Paris and when they will meet, or at least I think that is the intention: but since in the book one traveller is going between Paris and London, and the other between Paris and Lyon, they are unlikely to meet unless one of them gets badly lost. It seems that our author was trying to make the problem more relevant to an English readership but didn't carry his intention through.

The most curious of the problems, to my mind, is "The rule and questyon of a Catte". The problem presented (transcribed from a 1546 edition) is,

"There is a catte
at the fote of
a tre
the lēght
of 300 fote /
this catte goeth
upwarde
eche
day 17 fote,
and descendeth
the nyghte 12
fote. I demaunde
in howe ōge
tyme
that she be at ŷ
toppe."

Or in today's spelling, "There is a cat at the foot of a tree of height 300 feet. This cat goes upward each day 17 feet, and
descends each night 12 feet. I ask, how
long a time will she take to reach the top?" It's good to see that even as far back as the sixteenth century, authors of maths textbooks were presenting realistic real-life problems to their students.

Luckily our author provides the solution:

"Answere.
Take by and abate the nyghte of
the day, that is 12 of 17 and there remayneth 5, there fore the catte mounteth eche daye 5 fote / deuyde
now 300 by 5 and thereof cometh 60 dayes then she shall be at the toppe. And thus ye maye do of all other semblable."

That is, "Subtract the night from the day, that is 12 from 17: this
gives 5, therefore the cat mounts each day 5 feet. Divide now 300 by 5 and you
get 60 days: then she shall be at the top.
And thus you may do all other similar problems."

Which is very neat and useful. But unfortunately the answer is wrong. After sliding down 12 feet on the 57th night, the cat will be at a height of 285 feet and will reach the top of the tree after 58 days, not 60.

What is curious about this is that the author seems to have missed the point of the puzzle. It is interesting, surely, only because it is a trick question, but the author has fallen for the trick. What has happened?

A historian friend with whom I discussed this had a plausible idea. The author was probably taking his problems from a continental book. This book may have presented the problem of the cat, and first derived the incorrect solution as above, but then went on to say something like, "But in fact this is not the correct answer, because ..." and explained the trick. However the unwary writer of

*An introduction ...*didn't read any further, and reproduced the problem with the incorrect solution. Who knows how many generations of English students were bemused by their cats gaining the top of the tree two days earlier than they had calculated as a result of this carelessness?
Subscribe to:
Posts (Atom)