The three part BBC series The Code about maths begins next Wednesday. Presented by Marcus du Sautoy, and supported by the Open University. We have created some Open University hexagonal shaped cards to accompany the series. They are meant to be drinks mats (think beer mats). I’m worried they will be too thin to be convincing drinks mats. Hope I’m wrong.
There is a treasure hunt accompanying the series. Information on The Code website.
I created this word cloud for a project on visualising uncertainty. The font size relates to certain properties of nuclear reactors in the United States. I won’t state what the data is, because I later discovered it was dodgy, and I don’t want to advertise it. So the word cloud looks good, but doesn’t represent anything meaningful. Here’s a debate between Monbiot and Caldicott on nuclear power.
TED is a group in the USA which organises lectures. Here’s a TED lecture about teaching maths: Dan Meyer.
Dan argues that you should ask practical maths questions in a bare form. For example, you could ask, “You chuck a 1kg rock upwards at 2 metres per second; how high does it go?” Dan suggests instead “How high does a rock travel if you chuck it upwards?” Then the student has to realise what are the important quantities needed to answer the problem. Makes the problem deeper and more attractive.
There’s something in this. Probably doesn’t apply as much to pure maths problems.
E left a book, :59 Seconds – Think a little – Change a lot by Richard Wiseman, lying around which stated that people with surnames such as Short, Little, or Bent are especially likely to suffer feelings of social inferiority. (Wiseman referred to some paper or other for this fact.) I laughed, but the author’s surname… The book also suggested that men with positive initials (such as ACE, HUG, and JOY) live longer, and those with negative initials (PIG, BUM, and DIE) die younger. Sounds unlikely. I read the rest of the book, and it was pretty good. Here are some key points from certain chapters (not in order).
(1) It’s often thought that brainstorming is more productive than thinking alone. Not necessarily true, because of social loafing. Dominant person doing all the thinking. Guess that’s obvious.
(2) In solving a problem, a good strategy is to leave it for a while, and then return. Your subconscious will have a crack. Various experiments in book to justify this. Seems true. He even suggests that you occupy your conscious mind with a distraction from the problem, and this focussing on something else will help with the original problem.
(3) He has four tips for creativity. (i) Prime: think hard about problem, and then distract yourself. (ii) Perspective: how would X think about this problem? (iii) Play: take a break and have fun. (iv) Ask yourself challenging questions. That’s it. Exaggerated inaccurate summary: to solve a problem work on anything but the problem.
(4) Green things are good for solving problems! Plants. Or without plants paint your room green.
(5) More on priming: try looking at inspirational pictures. These can be great works of art (or listen to music), but a simpler inspirational image is as follows. Consider lots of arrows pointing downwards, and amongst them a single arrow pointing up. Look at this image! It helps you think differently. So the referenced experiments indicate.
(1) Rewards don’t necessarily work – rewards make people believe task must be unenjoyable.
(2) In interviews, likeability is often most important factor.
(3) In the TV programme The Weakest Link, the contestants near the middle win more often. How curious.
(4) To make yourself liked: (i) show interest in others, (ii) ask for favours, (iii) show fallibility, (iv) don’t gossip. All sounds believable.
(5) Here’s one I hadn’t thought of: if you want to get someone to agree to something, first ask them some basic questions they are likely to say ‘yes’ too. Then hit them with your request.
(6) In bargaining, humour helps.
(7) Bystander effect – I like this one. Example is of a guy who collapses in a street. When there is only one other person in the street, he or she offers to help the guy. When there are loads of people, nobody helps. Reason is because people look to each other for indications of how to behave. So nobody brings themselves to do anything. Also some kind of social embarrassment or something involved.
(8) Reciprocity: help someone and they help you.
(9) Best way of getting a lost wallet returned: keep a photo of a baby in it.
Well that all seemed devious somehow. I’ll move on to…
(1) Once reach certain modest level of wealth, happiness independent of money. Believable.
(2) Attempting to suppress thoughts makes them come up more regularly.
(3) After traumatic event, writing about it helps (getting things down in an orderly fashion). Talking may not help though: the orderly part is usually missing. So write!
(4) Writing positive things (in all kinds of contexts) helps you out. Be positive! Wiseman is forgiven for the Short thing. I’ll embrace my surname and ridiculous initials.
(5) If you are happy you smile. Converse holds partly too. Smile and you’ll be happier. Have good posture and use positive language too.
(1) The ‘play hard to get’ adage is wrong. Should be ‘play hard to get but at the same time be enthusiastic about the other person’.
(2) Touch a person’s upper arm to be friendly. I won’t be doing that one.
(3) Mimic a person to be agreeable. (What if they catch you?)
(4) Apparently women look for bravery in men and risk taking. Rock climbing number one pursuit for men attracting women. Golf bottom. Aerobics number one pursuit for women attracting men. Body building bottom – ha!
(5) When in love your heart beats faster. Converse holds a bit too. Similar thing with intimate conversations.
(6) Reflected glory: if guy sees girl with guy then guy thinks more of girl than if girl alone.
(1) People in groups take more risks. Their views become polarized.
(2) To persuade people, get your foot in the door. Isn’t this in the wrong chapter?
(3) Don’t spend long on big decisions. Let your subconscious work on them. People regret more if they dwell on decisions for a long time.
(4) To detect liars, ignore tension or nervousness; instead look for slow answers, evasiveness, and lack of detail.
(1) (i) Plan. (ii) Tell others about plan. (iii) Think about good things that come from achieving goal. (iv) Reward yourself for progress. (v) Record progress.
(2) A familiar phenomenon: once you start a task, your mind wants to finish it, and is agitated until it’s done. Once it’s done, mind wipes it and you can’t remember it so well. (Example, waiter who forgets order once it’s paid.)
(3) Employ Orwellian doublethink: optimism and realism simultaneously.
(1) Punching and screaming doesn’t help. Instead search for the positive side of things.
(2) Praying for others helps you. Altruism is good for you.
(3) Laugh more. Go outside in the sun. (To alleviate stress.)
(4) Own a dog. (And pass stress on to your neighbours.)
(1) Apparently its well known that some canny psychologists once went through the dictionary and extracted all the words relating to personality. They then classified those words into five categories, namely openness, conscientiousness, extroversion, agreeableness, and neuroticism. A personality can be well approximated by describing these five attributes.
Didn’t read this one.
Nor this one.
I recently found the book Letters to my grandchildren, by Tony Benn, in my local library. It consists of a series of short letters written by Tony Benn to his grandchildren, on different political themes. I like Benn and I like the book. Here are four small parts.
Letter 4 is about mechanisms by which leaders control their people. Benn mentions (but does not endorse) the quote ‘democracy inevitably leads to Marxism’ from Mein Kamp. The idea behind this quote is that, in a democracy, the wealthy are a minority, and so everyone else would do well to vote for an even distribution of wealth.
Benn reports that he has told students at ceremonies: ‘When you leave this university you will know far less that there is to be known than when you arrived.’ Benn wasn’t referring to the effects of alcohol; he meant that we cannot keep up with the growth of human knowledge.
Of course there is loads on the Labour Party and New Labour. Benn reports that Margaret Thatcher claimed New Labour was her greatest achievement. (In the sense that, with New Labour, traditional Labour Party values were replaced with Conservative policies.)
Benn is against developing Trident. He describes how our nuclear weapons are dependent on the USA (because we use their technology). So we don’t have an independent nuclear deterrent. He questions whether nuclear weapons work as a deterrent for war anyway, and cites the Falklands war and various other wars. He also mentions money spent on Trident, and the disaster of a nuclear war, as you’d expect. He says that a person with good sense would not sanction the use of nuclear weapons, and he refers to some wartime officer or other to back him up. In response to this, a prime minister who supports nuclear weapons would have to reassure everyone that he or she would be prepared to use nuclear weapons. That is, the prime minister would be reassuring us that he or she is prepared to approve the killing of millions of people.
Here is an exercise that can be performed with a group of at least three people. Each person secretly writes down a positive integer. The person with the least positive integer that nobody else has written down wins the game. Everyone is aware of the object of the exercise.
There seems to be game theory in this, and statistics. It’s an interesting exercise with which to begin a class. I’ll give it a go sometime and report back here.
Tom Körner gave a seminar here at the Open University a couple of weeks back about using the Baire Category Theorem to prove a quantitative form of a theorem of Rudin. Tom made some passing remarks on the Central Limit Theorem, which I discuss and develop in this post.
Central Limit Theorem. Let X1, X2,… be identically distributed, independent random variables with mean 0 and variance 1. Then, for a<b,
This theorem is well used in statistics, finance, and so forth, but as stated above it says nothing about the speed of convergence. It may be that convergence is so slow that the theorem is practically useless. Tom mentioned a related theorem (of Bernstein) which gives an idea of rates of convergence of sums of independent random variable.
Theorem. Let X1, X2,… be independent random variables, each with |Xi|≤1 and mean 0. Then, for k>0,
Sketch proof. First, by expanding out exp(tXi), we can check that
Let Then Also,
Choose t=k/(2n) and combine the past two equations for the result. ■
I’ll apply this theorem, with certain simplifying assumptions, to the football Premier League. I have written about the Premier League before, in Football Leagues. Each team in the Premier League plays 38 matches. I assume these matches are all independent, and there is no home bias. I will assume that each match is either won or lost (no draws), and 11/4 points are awarded for a win, and 0 for a loss. (The 11/4 points for a win is less than the usual 3, but this compensates well for ignoring draws.)
Let Y1, Y2, …, Y38 be the points scored by a particular team. Let us suppose this team has a 50% chance of winning, and a 50% chance of losing. Define Xi = (Yi-11/8)×8/11. Then the conditions of the theorem above are satisfied and, roughly, we find that
Choose k=30 and we find that
In other words, there is a less than 0.3% chance that a team from this random league scores more than 94 points. In 2005 Chelsea amassed 95 points, which indicates that teams weren’t all of equal ability in the 2004–2005 season; Chelsea were better than most teams. That is, without knowledge of football, from analysing the distribution of points alone, we see that teams vary in ability.
This quick application of Bernstein’s Theorem was an experiment I carried out through curiosity. There are better ways of analysing football scores.
Some months ago I discussed the following problem with Steve Hewson and Mike Pearson after an NRICH meeting. Two people play a game with a finite graph. They take turns in colouring the uncoloured vertices of the graph, one vertex at a time. One player colours vertices red, and the other player colours vertices green. Two adjacent vertices (vertices joined by an edge) cannot be coloured in the same way. The loser of the game is the first person who is unable to colour a vertex.
The optimal strategy for each player depends on the graph. For a graph with no edges, player 1 wins if there are an odd number of vertices, and player 2 wins if there are an even number of vertices. For a complete graph of order at least 2, player 2 always wins. Complete bipartite graphs are also easy to handle. More interesting are cycles and linear graphs, and I consider these in the next two sections, so don’t read on if you want to think for yourself first.
For a cycle, player 2 can always win. How? Player 2 colours green the vertex U that is one place clockwise from the last vertex V that player 1 coloured red. Is it always possible for player 2 to do this? It is only impossible if U is already occupied. But U cannot be occupied by red; else player 1 wouldn’t have been able to colour the adjacent vertex V red. It cannot be coloured green either if the strategy is followed. So player 2 wins.
For linear graphs, again, player 2 can always win. (Unless the graph has only one vertex.) How? If the graph has an even number of vertices then player 2 can mirror the moves of player 1. I’m sure you see what I mean by this. Here’s a picture to illustrate the effect.
What if there are an odd number of vertices? Player 2 can still win. I have a strategy for player 2 winning, but it’s not as simple as the previous strategies. Generally, he or she should mirror player 1 as before. If player 1 colours the centre vertex, then player 2 colours one of the two vertices adjacent to the centre (which must, necessarily, be empty). There can then arise slight complications moving outwards from the centre as player 1 tries to take advantage of this break in the mirror strategy, but the complications fizzle out. Details omitted; work it out for yourself.
No game theory so far, but when you introduce more players, things get interesting. We introduce a third player, colouring vertices blue. We want to know who loses the game now; who is the first player that is unable to go. Consider the cycle graph with four vertices. The possible outcomes of the game on this graph are shown in Figure 3 (player 1 is red, player 2 is green, and player 3 is blue).
It’s easier (conceptually) using numbers to mark the vertices rather than colours, but the corresponding diagrams look worse. We see from Figure 3 that if green, on his first move, chooses the vertex directly opposite red then red definitely loses. On the other hand, if green chooses one of the two vertices adjacent to red then whether red or green loses is determined completely by blue’s first move. Blue cannot lose. So it makes sense for green to choose the opposite vertex, and red loses.
The five cycle case is better; see Figure 4. I have filtered out various branches from the tree of Figure 4. This missing branches mostly correspond to moves player 3 could make to guarantee he or she loses (when there is a better alternative).
It seems that green is most likely to lose this game. Indeed, suppose each player adopts the strategy that (i) they wish to maximise their chances of not losing, and (ii) from a selection of vertices with an equal chance of losing, they choose randomly. Then there is a 25% chance red loses, a 50% chance green loses, and a 25% chance blue loses.
Suppose instead these three people are sitting round a table. Green can choose to go down either the left or right branch of Figure 5. If he goes left then who loses depends on red’s next move: either green or blue. If he goes right then who loses depends on blue’s next move: either red or green. So, before green’s first move, red may say to green “go left and I will make blue lose.” At the same time blue may say to green “go right and I will make red lose.” Whichever green chooses, he is saved! Rather than losing with 50% probability, green now has the power to determine who loses the game. Unless, of course, red or blue betray him and make him lose anyway.
What next: I don’t know, this is about as far as I got. Seems worth exploring further. Probably someone else has explored this already! Could make a good set of questions for NRICH.
The following problem was suggested by Bernard Murphy at Jenny Piggott’s retirement meeting in Cambridge on Tuesday 13 July.
Consider eight jacket buttons in a column, as shown on the left hand side of the image below.
We can unbutton the eight buttons in various orders; one such order is shown on the right hand side of the above figure. The problem is to choose an ordering so that we have to move up and down the buttons the maximum number of times. For example, if we unbutton in the usual fashion, starting from the top button and finishing with the bottom button, we would make in total seven moves (from first button to second button, from second button to third button, and so forth). In the example above, we travel from button 4 to button 7 (3 moves), then button 7 to button 2 (5 moves), and so forth, totalling 31 moves.
Bernard’s question was What is the maximum number of moves? We add to that question Which unbuttoning strategies yield this maximum?
I’ll state the answers here, and prove them below. The maximum number of moves is 31. One unbuttoning strategy which yields this solution is shown above. What do the others look like? They can all be described as follows. The four odd labels 1, 3, 5, and 7 must all occur in either the top four, or bottom four, positions. Also, labels 1 and 8 must occupy buttons 4 and 5, in either order.
A suggestion made at the time was to regard this as a graph theory problem. Let the buttons represent the vertices of a weighted complete graph, and the weight between two buttons is the number of moves it takes to get from one button to the other. The original question then becomes a special case of a travelling salesman problem. I see this perspective as a significant backwards step; we have moved from a simple linear graph to a complicated complete graph, and I won’t discuss this view any further.
A more obvious graph theoretic model is to consider the buttons to be vertices of the linear graph shown below. If we equip this graph with the usual graph metric d, then the problem is to choose an ordering v1,…,v8 of the vertices such that the quantity d(v1,v2)+d(v2,v3)+…+d(v7,v8) is maximised.
A third form of the problem is to choose an ordering x1,…,x8 of the integers 1,…,8 such that the sum |x1-x2|+|x2-x3|+…+|x7-x8| is maximised. (Here xi is the number of the ith button chosen.) I solve this problem in the next section.
We first solve a more symmetric version of the last problem, which is to choose an ordering x1,…,x8 of the integers 1,…,8 such that the sum |x1-x2|+|x2-x3|+…+|x7-x8|+|x8-x1| is maximised. There is an extra final term in the sum, which corresponds to moving from the finish button back to the start button.
Let S=|x1-x2|+|x2-x3|+…+|x7-x8|+|x8-x1|. The numbers 1,…,8 each occur twice as terms xi in this sum. Consider, for each i, reordering the terms xi and xi+1 from each summand |xi-xi+1|, if necessary, so that the modulus sign can be removed without affecting the value of S. We end up with eight +xi terms, and eight -xi terms. The maximum value of S is attained if and only if the eight plus terms are 5, 6, 7, and 8 (each twice) and the eight minus terms are 1, 2, 3, and 4 (each twice). Thus this maximum is attained if and only if we repeatedly alternate between the top four and bottom four buttons. The maximum has value 2×((5+6+7+8)-(1+2+3+4))=32.
Our simpler, but less symmetric, problem cannot, therefore, have value greater than 31. Since we know that 31 can be achieved, it is the maximum value for the problem. To obtain all the orderings which yield this value 31 we simply select those solutions which have |x8-x1|=1. Now, |x8-x1|=1 if and only if x1 and x8, the start and finish buttons, occupy the middle two button positions.
I haven’t gone into detail above; nor have I taken much care in my explanation. The method generalises to n buttons, and you should find that ½floor(n2)-1 is the maximum number of moves. I haven’t worked out the number of solutions which achieve this maximum, but one could easily do so.
The most natural extension is to choose a graph G equipped with the graph metric d and investigate what ordering of the vertices v1,…,vn of G maximises the sum d(v1,v2)+d(v2,v3)+…+d(vn-1,vn). It’s easy for a cycle, and probably easy for a tree. Later note: I worked out the details for a tree, and may publish the results somewhere.