Horacio G. Duharthgd20@bath.ac.uk

09/10/2014

How many ways are there to partition a set of size N?

{a} {a,b} {a,b,c}

N=1 N=2 N=3

{a} {a} {b}{a,b}

{a} {b} {c} {d}{a} {b,c,d}{b} {a,c,d}{c} {a,b,d}{d} {a,b,c}

{a,b} {c} {d}{a,c} {b} {d}{a,d} {b} {c}{b,c} {a} {d}{b,d} {a} {c}{c,d} {a} {b}{a,b} {c,d}{a,c} {b,d}{a,d} {b,c}{a,b,c,d}

S=1 S=2

N=4

{a,b,c,d}

{a} {b} {c}{a} {b,c}{b} {a,c}{c} {a,b}{a,b,c}

S=5 S=15

Follow strategies in this book:

Specification

Generating function

Asymptotic estimate

Desired result!

Symbolic method

Complex asymptotics

We start with a simple example:

How many binary words are there of size N?

We first need to define a combinatorial class, namely the alphabet, say {a,b}

We then define a size function from the combinatorial class to the non-negative integers.

In our example, we want this function to represent the length of the word, so each character has size 1, that is |a|=|b|=1.

We need to go through some of the basics:

Combinatorial class: A countable non empty set, finite or infinite.

Size of an element: A function from the combinatorial class to the non-negative integers. The number of elements of any given size is finite.

Counting sequence: The counting sequence of a combinatorial class is the sequence of integers where the n-th term is the number of elements in the combinatorial class that have size n.

Ordinary generating function: This is the formal power series

The set of binary words is another combinatorial class. We can think of it as the set of words of length 1 or 2 or 3, etc… this is what we will call a sequence.

In our example, if we define the combinatorial classes:

Then, we can think of the combinatorial class of the binary words as

These “sequence” operator and addition of combinatorial classes are examples of the possible operators of the symbolic method.

We now want to find a generating function…

Note that the binary words satisfy the following recursion:

Therefore, the OGF is given by:

This is, clearly, a geometric series:

Yes! I said

CLEARLY!

(only say it when talking about the

geometric series… or the exponential

Taylor expansion… or other clear stuff…)

Fortunately, we don’t need to find the generating function of all possible operators on combinatorial classes… someone else has.

Check the book! I got this table from there…

The magic, of course, is to find the coefficient of the n-th power of z in the series expansion of the generating function.

This task is sometimes easy…

But sometimes hard… for the partitions of a set:

So we also define the exponential generating function (EGF) of a counting sequence

For the partition example:

This gif has nothing to do with what we’re talking about… but since the most interesting bit is about to start it is worth start paying attention once more.

In order to find the coefficients of the generating functions we first assume they are defined on the complex numbers.

Basic concepts of complex analysis:

A function f is analytic at if

It is holomorphic at if it is differentiable at

Interestingly, a function is analytic if and only if it is holomorphic.

Finally, a function is meromorphic at if there exists a neighbourhood of where the function can be represented as

If we say that h has a pole of order M at. The coefficient is called the residue.

With this in mind, if the generating function is analytic

Then is meromorphic at 0, and it is a pole of order n+1

And the coefficient of the n-th power of z of the generating function is precisely the residue of this quotient

So we just need to find the residue…

According to the Cauchy Integral Formula, we can calculate the residue at 0 to find the n-th coefficient of the generating function by

Where the contour is a simple closed curved around 0 that has no other pole in its enclosed area.

For the example:

For the curious ones, this is done by partial fractions… and yes, I did it by hand.

How does the magic work? Here we go… (pay attention because I’ll do this only once)

Close to the saddle point: TaylorFar from the saddle point: Negligible

Close to saddle point Far from saddle point

Write the second derivative in polar coordinates:

And do a change of variable:

Approximate by a Gaussian integral:

Putting all together, we get this:

Where the saddle point is the unique solution to:

So… I did this to the binary word problem and found

What about the partition question?

Well… apparently is not that easy.