The Mumbling Mathematician

June 23, 2008

Making It Count: The Pigeonhole Principle

Filed under: graph theory, number theory — zetahype @ 5:12 pm

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.

7 Comments »

  1. why the different fonts?

    Comment by Tobias — June 23, 2008 @ 7:14 pm

  2. Well..if you must know. I chose wordpress over blogspot because I heard it allows inline Latex commands which would help me type the maths bits. Turns out it supports latex at a very basic level..so I make pdf files using latex and capture them as images and attach them at the right places in my post. I’m not terribly happy with the outcome but I’m trying different things and will make improvements.

    Comment by zetahype — June 23, 2008 @ 8:03 pm

  3. I can under stand how you use the principle to the first problem but it is not obvious in the second one. It does not appear to a situation where one can use directly the pigeonhole principle that you stated?

    Comment by Anonymous — June 23, 2008 @ 9:49 pm

  4. You are right. I have unintentionally used a generalised version of the pigeonhole principle in the second problem: If n objects are distributed among m boxes then at least one box will have [n/m] objects, where [*] is the smallest integer greater than or equal to *.
    In the problem we want to distribute 5 edges between two colours, so at least one colour will have [5/2] = 3 edges.
    Note that this version of the pigeonhole principle allows us to introduce probabilities. If an object has an even chance of being alloted one of the m boxes (uniform probability 1/m) we ask what the probability of one of the boxes having more than one object is. Ofcourse, if n>m, the probability is 1 but it is not hard to compute the probability even otherwise. The “probabilistic pigeonhole principle” is a useful tool in solving density problems in number theory. I am sure we will encounter it later.

    Comment by zetahype — June 24, 2008 @ 8:03 am

  5. You were talking about Ramsey numbers but didnt introduce the Party Problem? That I would have thought would be the party piece.

    Comment by Meech — June 24, 2008 @ 1:39 pm

  6. umm..i did..the party of 6 problem is the one I solved in the second example.

    Comment by zetahype — June 24, 2008 @ 1:41 pm

  7. [...] under: Uncategorized — zetahype @ 1:21 am 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 [...]

    Pingback by Making It Count: Counting the complement « The Mumbling Mathematician — July 23, 2008 @ 1:21 am


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