A couple days ago I wrote a post about a probability problem that involved calculating Stirling numbers. There are two kinds of Stirling numbers, creatively called “Stirling numbers of the first kind” and “Stirling numbers of the second kind.” The second kind come up more often in application, and so when authors say “Stirling numbers” without qualification, they mean the second kind. That’s the convention I’ll use here.

The Stirling number S(n, k) can be computed via the following series, though we will take a different approach for reasons explained below.

S(n,k) = frac{1}{k!}sum_{i = 0}^k (-1)^i {k choose i} (k-i)^n

Problem statement

This made me think of the following problem. Suppose you want to calculate Stirling numbers. You want to use integer arithmetic in order to get exact results.

And suppose you can use integers as large as you like, but you have to specify the maximum integer size before you start calculating. Imagine you have to pay $N to work with N-digit integers, so you want to not use a larger value of N than necessary.

Intermediate results

There are a couple problems with using the equation above. Direct implementation of the equation would first calculate k! S(n, k) first, then divide by k! to get S(n, k). This would be costly because it would require calculating a number k! times bigger than the desired final result. (You could refactor the equation so that there is no division by k! at the end, but this gives an expression that includes non-integer terms.)

In the earlier post, I actually needed to calculate k! S(n, k) rather than S(n, k) and so this wasn’t a problem. This brings up the second problem with the equation above. It’s an alternating series, and it’s not clear whether evaluating the series produces intermediate results that are larger than the final result.

Recursion

Stirling numbers can be computed recursively using

S(n+1, k) = k S(n,k) + S(n, k-1)

with the base cases S(n, n) = 1 for n ≥ 0 and S(n, 0) = S(0, n) for n > 0. This computes Stirling numbers via a sum of smaller Stirling numbers, and so the intermediate results are no larger than the final result.

So now we’re left with the question of how big Stirling numbers can be. Before computing S(n, k) you need an upper bound on how many digits it will have.

Bounding Stirling numbers

Stirling numbers can be bound in terms of binomial coefficients.

S(n, k) leq frac{1}{2} binom{n}{k}k^{n-k}

This reduces the problem to finding an upper bound on binomial coefficients. I wrote about a simple bound on binomial coefficients here.

The post Computing Stirling numbers with limited integers first appeared on John D. Cook.

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