The shape of a random partition

In this post we will give a heuristic derivation of a result of Vershik, describing the shape of a random partition of a large integer N. (Vershik’s Russian original is available here; English translation is pay-walled.)

By a partition of N, we mean positive integers \lambda_1 \geq \lambda_2 \geq \cdots \geq \lambda_r > 0 with \sum \lambda_i = N. We draw a partition as a collection of boxes: For example, this is 4+2+1+1+1:

smallPart

Suppose we let N \to \infty, select partitions of N uniformly at random and rescale the size of the boxes by 1/\sqrt{N}, so that the diagram of the partition always has area 1. What is the shape of the most likely diagram?

We want to describe, given a particular shape, how many partitions have roughly that shape. To that end, let (x_1, y_1), (x_2, y_2), …, (x_r, y_r) be a sequence of points with x_1<x_2<\cdots<x_r and y_1>y_2>\cdots>x_r. Let's count partitions which fit inside the box \{ (x,y) : x \leq x_r ,\ y \leq y_1 \} and whose boundary passes through the points (x_i, y_i).

To describe such a partition, we simply must describe the portion between (x_1, y_1) and (x_2, y_2), and then describe the portion between (x_2, y_2) and (x_3, y_3), and so forth; each of these portions is independent. Set (\delta x)_i=x_{i+1}-x_i and (\delta y)_i = y_{i+1}-y_{i} (so (\delta y)_i<0). Set (\delta t)_i = (\delta x)_i+(-(\delta y)_i).

So the number of such partitions is

\displaystyle{\prod_i \binom{(\delta t)_i}{(\delta x)_i} }.

From our computations in an earlier post, this is roughly

\displaystyle{\exp \left( \sum_i (\delta t)_i \cdot h\left( \frac{\delta x_i}{\delta t_i} \right) \right)}

where h(p) = - p \log p - (1-p) \log (1-p). Moreover, if we rescale all of the x_i and y_i by \sqrt{N}, this approximation becomes better.
So, if we want there to be a lot of partitions, we should make \sum (\delta t)_i h\left( \frac{\delta x_i}{\delta t_i} \right) large.

We have now described what happens if we require the boundary of the partition to pass through finitely many fixed points. Suppose instead we insist that the partition lie near a fixed curve? Parametrize that curve as (x(t),y(t)), with the normalization that x(t)-y(t) = t. I’m not even going to say what I mean precisely, but we can extrapolate from the preceding formula that the number of partitions whose boundary is near \sqrt{N} (x(t), y(t)) is roughly

\displaystyle{\exp \left( \sqrt{N} \int_t h\left(\frac{dx}{dt} \right) dt \right)}

We thus obtain a problem in the calculus of variations: Consider curves (x(t), y(t)), with the normalizations that x(t)-y(t)=t and the area enclosed by (x(t), y(t)) and the coordinate axes is 1. Subject to these conditions, maximize \int h(x'(t)) dt. The next several paragraphs are solving this variational problem.

Rotating the coordinates 45^{\circ}, we see that the condition on the enclosed area is simply that \int (x(t)+y(t)-|t|) dt = \sqrt{2}.

As always in problems involving the calculus of variations, the first step is to perturb our proposed solution and see how it changes. Replace (x(t), y(t)) by (x(t)+\epsilon u(t), y(t)+\epsilon v(t)). If (x(t), y(t)) was optimal, than \epsilon=0 should be a maximum, so the first order variation in \epsilon should vanish.

This is only valid if our perturbation preserves the conditions on (x(t), y(t)). We want the perturbed functions to obey x(t)-y(t)=t, so we take u(t)=v(t), and we want to preserve the area, so we take \int u(t) dt =0.
For such a perturbation, the first order variation in \epsilon is supposed to vanish. We compute:

0 = \left. \frac{\partial}{\partial \epsilon} \right|_{\epsilon=0} \int_t h(x'(t)+\epsilon u'(t)) dt = \int u'(t) h'(x'(t)) dt.

Integrating by parts:

0 = \int u(t) \frac{d h'(x'(t))}{dt} dt.

We want this to hold for any u obeying \int u(t)=0. This will happen if and only if \frac{d h(x'(t))}{dt} is a constant. So we want h'(x'(t)) = Ct+D for some constants C and D. So x(t) = \int (h')^{-1}(Ct+D).

The formula x(t) = \int (h')^{-1}(Ct+D), in some sense, solves the variational problem. Our next task is to simplify that formula.

From the previous post, we know that \int h^{-1}(u) du = \log(1+e^u). So x(t) = (1/C) \log (1+e^{Ct+D}) + B, where B is another constant of integration.

As t \to - \infty, we want x(t) \to 0. This shows that B=0.

As t \to \infty, we want x(t)-t \to 0. This shows that D=0. So we have x(t) = (1/C) \log(1+e^{Ct}) and y(t) = (1/C) \log(1+e^{-Ct}).

To evaluate the final constant C, we need to use that the area under the curve is 1. It is convenient to eliminate t and relate x and y directly as e^{-C x} + e^{-C y} = 1, or, y = (1/C) \log \frac{1}{1-e^{-C x}}. Now, the area under the curve is \int_{x=0}^{\infty} \frac{dx}{C} \log \frac{1}{1-e^{-Cx}} = \frac{1}{C} \sum_{n=1}^{\infty} \frac{1}{n} \int_{x=0}^{\infty} e^{-nCx} dx = \frac{1}{C^2} \sum_{n=1}^{\infty} \frac{1}{n^2}. We deduce that C = \tfrac{\pi}{\sqrt{6}}.

In other words, our final answer is that the curve is e^{-\pi x/\sqrt{6}} + e^{-\pi y/\sqrt{6}} = 1.

How good is this approximation? See for yourself! Here is a random partition of 1000, and the approximating curve.
rp

The thing which I think is really cool is that we never needed the explicit formula h(p) = -p \log p - (1-p) \log p; we only needed that it was Legendre dual to \log(1+e^x). Suppose that we looked at some other sort of local probability distribution on partitions: For example, maybe we would weight the likehood of a partition by \exp(\mbox{number of outer corners}) or something like that. We could still build a generating function Z_n(x) for partitions fitting in a k \times (n-k) box with this weighting. We could then try to compute z(x) = \lim_{n \to \infty} \frac{1}{n} \log Z_n(x). If we succeeded, the exact same logic as before would show that z(x) was the shape of a random partition of area N chosen from this distribution. Which brings me to the thing I really want to talk about … next time.

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6 thoughts on “The shape of a random partition”

  3. Like this :).

    Without that: Let q(k,n) be the number of partitions of n whose largest part is at most k. We can quickly compute a table of values of q by the recursion q(k,n) = q(k-1,n) + q(k,n-k).

    We then write a recursive function r(k,n) which builds a random partition of n with largest part at most k: If k=1, return (1,1,\ldots,1). Otherwise, choose an integer j between 1 and k with probability (q(j,n)-q(j-1,n))/q(k,n) and return (j, \lambda), where \lambda is the output of r(j,n-j).

    I had this implemented and it ran pretty much instantly up to n=250 or so, when I discovered that Mathematica already had a function to do this.

  6. Pingback: Random three dimensional partitions | Secret Blogging Seminar ~

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