Here’s the much delayed second part of the Making It Count series which was launched with much fanfare a month ago. The counting technique I will present is simple and powerful:

Counting the complement: Let X be a set whose cardinality is known and Y be a subset of X whose cardinality we must find. We can do this by determining the cardinality of the complement of Y (all the elements which are in X but not in Y) and `subtracting’ it from the cardinality of X.
…A couple of clarifications before I apply this rather innocuous principle to some problems:
If X is finite then the cardinality is just the number of elements in X. But I use a fancy word to include the cases when X is infinite. Note, also that subtracting cardinalities doesn’t always make sense in the infinite case. It depends on the `size’ of the infinity.

Problem 1: There are 128 players in a tennis tournament. Assuming it is a typical knockout tournament, how many matches are played before a winner emerges?
(Lazy solution): This is how most people would solve it. The `shape’ of the number 128 allows them to get away with this solution:
Pair them up. 64 matches are played initially; 64 players progress to play 32 matches and so on.The number of matches played is 64+32+16+8+4+2+1 = 127! Very good, but what about when 125 players start the tournament? One doesn’t have the comfort of repeated division by 2 anymore.
(Proper solution): Use the above counting principle! Being a knockout, a player can lose only once before he is out. So, number of matches played before winner emerges = number of matches lost = number of players who lost. Well, there is 1 winner so 128 -1 =127 players lose and so 127 matches are played. The argument works for any number other than 128. But how can 125 players be paired up in a sensible way? Well they can’t. Either the organisers add 3 players to the draw or they give a whole bunch of players a free passage into the second round, which is what a lot of the smaller tournaments do!

Problem 2: Let p be a prime and n be a non-negative integer. How many non-zero positive integers have no factors (other than 1) in common with ?
Solution: For we can just list these numbers: 1,3,5,7 have no factors in common with 8 so the answer is 4. But this method of listing is rather naive and quickly gets out of hand for even small numbers like or . Instead, we count the number of integers that do have a factor in common with . Ironically, we can list them out every time! They are just p, 2p, 3p, 4p,…,p. There are of them and so, by the principle, exactly non-negative positive integers are relatively prime to it. I haven’t contrived this problem to glorify the virtues of the counting argument. This computation is infact, one of the most fundamental in number theory and leads us to the “Euler phi-function,” denoted for N positive and defined as the number of non-negative integers N and relatively prime to it. The function is multiplicative, which means that if then from the above computation.

For instance,

Mathematics has a well-deserved reputation of being an armchair endeavour. You sit down armed with some concepts and wander off into many wonderful directions by just digging deeper into these concepts. Physics on the other hand has traditionally been a laboratory science. One tried to understand the physical universe with experiments and codified the results with the help of mathematical tools. Einstein however came along and blurred this distinction. He changed the way we view the universe `simply by thinking about it.’ I will, in one of my next posts, show how the postulates of special relativity can be derived in a purely mathematical way in much the same way Einstein might have done. Newtonian physics recognised gravity as one of the fundamental forces that shapes our universe but was unsuccessful in explaining how and why objects attract other objects. Einstein, armchair and all, unraveled this mystery in his theory of general relativity. He visualised space-time as a fabric that stretched and curled through the universe. An object that is plonked on this fabric creates a deformity or depression that sucks in objects around it. The heavier the object, the bigger the depression and greater its powers of attraction. Simple and profound. Einstein was nothing if not that! Armed with this model, it is but natural to conclude that objects which are so plonked onto this fabric create waves that ripple across the fabric. These are the gravitational waves that EInstein’s General Relativity predicts and which are yet to be observed experimentally.

It was under this mathematical framework of general relativity that Stephen’s Hawking made his celebrated conclusion that black holes emit radiation. Whilst being a tour de force in mathematical reasoning, it hasn’t won him a Nobel prize because it hasn’t being verified experimentally! And then along came Edward Witten and others and the traditional barriers between mathematics and physics have all but dissolved.

Stretch you right arm out, palm up. Try to rotate your palm clockwise about the vertical axis while keeping it facing up at all times. You will notice that it takes a continuous 720 degrees rotation rather than a 360 degrees rotation to bring your arm back to its original state. This is an illustration of the remarkable fact that an object rotated through 360 degrees might not always come back to its original state. Paul Dirac assigned a quantity of to the spin of an electron to express the fact that its amplitude changes under a 360 rotation but is restored with a 720 rotation. He justified this assignment with a famous topological insight which has since taken many forms and legends: the Phillipine wine glass trick (which involves sustaining a wineglass placed on the aforementioned palm), the Feynman Plate trick or the Dirac belt trick. Whatever the name, it was probably the first application of algebraic topology to quantum physics. These days it is hard to distinguish one from the other. I will explain this phenomenon in a mathematical framework when I introduce some algebraic topology basics later on in this blog. Here is an excellent visualisation of this trick.

I had presented two betting scenarios in Psyprobs 1: coin-tossing and poker. A friend of mine has mailed in with a fairly comprehensive analysis of the former and a suggestion that the poker problem is exactly the kind of bone a Prospect Theorist would chew on. I have reproduced his mail as is…save for a spelling error that wordpress red-flagged thrice over. And no, he cannot pass it off as a US-UK clash of ideologies.

“When you walked up to the table, you perhaps had some guess of what the p(heads) for the coin being used was. And of course, p(tails) = 1 – p(heads). Using p(heads) = p(tails) = 0.5 seems reasonable, given your trust in the precision of the mint as well as the credentials of bloke doing the tossing. And then, you recognise him – he’s Derren Brown, now you’re not quite sure. You are uncertain of the value of p(heads). We’re looking for a general enough way to solve this problem, so we want to describe everything so that replacing Derren Brown with the patron angel of innocent faces should just be a matter of blindly turning the crankshaft.
The way to do this would be to represent p(heads) with a Beta distribution, given by . Being a distribution, it will have a mean (i.e., the best guess right now) and a variance (i.e., the uncertainty in the best guess). When you were first picked from the audience, you condense your reading of the situation into a Beta distribution that will be the prior belief of p(heads). This will compactly represent how biased you think the coin is, how unsure you are about the parentage of the guy in front of you, etc.
At one end is complete uncertainty, you have no clue about p(heads) for the coin, this is represented by 1. The mean of this distribution is 0.5, but there is maximal doubt about this value, i.e., you think that p(heads) could be any value between 0 and 1 but your best guess is 0.5.
Alternatively, the coin comes with a certificate of authenticity, you know for sure that p(heads)=0.5. This could be represented by having 1000, or some suitably large number. The mean of this distribution is 0.5, and the variance (i.e., your uncertainty) is very low.
Or maybe, this coin can only come down heads, in which case you use ( read ‘a lot bigger than.’) All intermediate values of your prior belief in the fairness of the coin can be got by choosing and appropriately.
Now, the part of incorporating the new information – the guy across the table telling you that the last 100 tosses have been tails. Ignoring this information is wasteful, surely the most efficient thing to do is to use this information somehow. It’s easy when you trust this incoming information. Starting from a prior of , knowing and believing that 100 tails were just observed will lead to your view of the bias in the coin being .
This process we just went through has some nice properties. If you walked in with , it meant that you had no particular opinion, you could be convinced either way. After hearing about the 100 tails, you will now be , that is your best guess of p(heads) for that coin is . On the other hand, if you walked in completely confident that it is a fair coin represented by say , hearing about 100 tails will change this to whose mean is not too far away from the original mean of 0.5. That is, the new evidence has done little to change your mind.
This would be how to deal with uncertainty, but when you bring in money, things get a bit hazy. Given that your belief of p(heads) for the coin is given by , as a rational person, you would be ready to accept a bet where you are asked to pay h£ for a chance to win (h+t)£ if the next count toss came down heads-up. You would of course be ready to accept any bet that gives you better odds – i.e., asking you to pay less than h£ with a chance to win (h+t)£. Goes without saying that you shouldn’t accept any bet that gives you worse odds, but greed and foolishness are funny things.
But we’re all intelligent people. Which is why you refused the bet where Derren Brown asked you for £1.01 for a chance to win £2.00 on a coin-toss with a completely fair coin. But he’s got to put out a show on TV, so he gives you 2£ and you then proceed to enjoy the finer things with life with that money and get used to this rich lifestyle. He arrives at your door the following week and you are forced to return the 2 pounds to him. But he’s a nice guy, he gives you a chance to win the money back. He asks you if you’re in on a £1.01 bet for a chance to get the £2 if the next toss came up heads. The funny thing is, you would probably take the bet now. You see, Derren Brown has been reading about Prospect Theory.”

A counting argument is a method of counting the number of objects with a prescribed property. A good counting argument, apart from being merely clever and elegant, should be able to hold its own in more general mathematical settings. The Making It Count series is a tribute to the art of good counting. I will present nuggets not just from the usual sources like number theory and combinatorics, but also analysis, geometry and physics. It is only fitting to start the series with one of the most natural and widely used counting arguments: the humble and versatile Pigeonhole Principle!

The above problem is a specific instance of what are called Ramsey Numbers, named for Frank Ramsey. Every time I look at stars in the sky I think about Frank Ramsey, his tragic life and his wonderful theory. For two colours, the Ramsey Number R(r,s) is the smallest number of points such that the edges between them, when coloured, contain a monochromatic polygon with either r or s sides. So, the above problem proves that R(3,3) = 6. These numbers are hard to compute for even small values of r and s. For instance, R(10,10) is known to be between 798 and 23556. Establishing bounds for Ramsey Numbers is as important problem in graph theory.

The subtle and celebrated art of psychological warfare has been part of almost every war, battle, skirmish and brawl in history. These days they make bloodhounds bark into the faces of Guantanamo Bay prisoners and market it to kids as ‘PSYOPS.’ I am not going to burn effigies of bloodhounds like the politically aware would, but I will talk about the concept of Psychological Probabilities that I have been forced to explore since I started playing online poker. I have more questions than answers and the reader(s) will probably dismiss Psyprobs as just conditional probabilities in disguise. Consider the following scenario:
You’re engaged in a sophisticated game of strategy and chance where the bloke tosses a fair coin and you win if its heads and cough up the cash if its tails. You shift through your mental gears and conclude ‘50-50′ until the bloke throws a psychological spanner. “I should tell you that the last 100 tosses have all landed tails.” Wocha gonna do? Are you gonna remind yourself that a coin toss is an independent event and you have an even chance on the next toss regardless of prior (history?) Will you conclude that since a ‘balance’ has to maintained in the long run, a heads is just around the corner and double your bet? Or will you just figure it ain’t your day and pack it in? The fact is you’ve been given some extra information. How are you going to use it? Should it bother you that the coin landed 100 consecutive tails? Can (and should) this run of tails be factored into your decision making in a precise mathematical way? Sometimes, what seems to be a Psyprob scenario is just conditional probability in disguise. The classic Monty Hall Problem is an example. I will not go through the setup and solution. This Wikipedia entry does a good job.
I believe that the first example is just a standard coin-toss and the ‘extra’ information should not come in the way of your gambling fortune. The Monty Hall Problem is clear-cut conditional probability in action. Here’s yet another scenario which may require something more than conditional probability. Then again, maybe not. You decide.
So you’re playing poker heads-up. You’re dealt a pair of Kings. The flop (the first 3 of the 5 community cards) is a pair of 7s and a Queen. The suits are such that a flush isn’t a factor. You make a serious bet, your opponent goes all in. You figure maybe he has a pair of As and are ready to fold when he shows you one of his cards: a Queen. Agreed that there are two more cards to follow but, as it stands, can you work out the odds of your having the winning hand?
Well…he has the better hand if his other card is a 7 or a Q. Should this be the sole consideration when you calculate odds or should you think in terms of, “is he showing me the weaker/stronger card?” The former is conditional probability while the latter has Pysprobial undertones. Are both the approaches one and the same? It is tempting to say Yes! If he’s showing the weaker card you calculate the odds and add them to the odds if he’s showing the stronger card and it should tally with the conditional probability calculation. I suspect, however, that each component in this addition should be ‘weighted’ in some way to reflect the psychology of your opponent in showing the weaker/stronger card. Think about it. I’ll post some actual calculations later. All I know is that Stu Ungar wouldn’t have put up with this kinda rubbish.

This blog is an attempt to give the reader(s) a glimpse into The Langlands Program which is an intricate web of results and conjectures linking seemingly disparate areas of mathematics and physics. The content of the posts might show a general bias towards number theory and algebraic geometry not so much because the Program has its origins in these areas but because I tend to be more clueless in other neighbourhoods. The occasional foray into historical and philosophical aspects should promptly be construed as me buying time so I can find something useful to say.There is no linear masterplan to lead the faithful into the heart of the Program. It is more in the spirit of “taking a left and venturing as the crow flies.” Earth-shattering knowledge of mathematics and physics is not a pre-requisite. Dodgy recollection of high school stuff is desirable.