Tips for Project Euler Problem #78

Question

This is the problem in question: Problem #78

This is driving me crazy. I've been working on this for a few hours now and I've been able to reduce the complexity of finding the number of ways to stack n coins to O(n/2), but even with those improvements and starting from an n for which p(n) is close to one-million, I still can't reach the answer in under a minute. Not at all, actually.

Are there any hints that could help me with this?

Keep in mind that I don't want a full solution and there shouldn't be any functional solutions posted here, so as not to spoil the problem for other people. This is why I haven't included any code either.

Answer 1

Hi,

here some hints:

1. Divisibility by one million is not the same thing as just being larger than one million. 1 million = 1,000,000 = 10^6 = 2^6 * 5^6.

2. So the question is to find a lowest n so that the factors of p(n) contain six 2's and six 5's.

Answer 2

Wikipedia can help you here. I assume that the solution you already have is a recursion such as the one in the section "intermediate function". This can be used to find the solution to the Euler problem, but isn't fast.

A much better way is to use the recursion based on the pentagonal number theorem in the next section. The proof of this theorem isn't straight forward, so I don't think the authors of the problem expect that you come up with the theorem by yourself. Rather it is one of the problems, where they expect some literature search.

Answer 3

Have you done problems 31 or 76 yet? They form a nice set that is an generalization of the same base problem each time. Doing the easier questions may give you insight into a solution for 78.

Answer 4

This problem is really asking to find the first term in the sequence of integer partitions that’s divisible by 1,000,000.

A partition of an integer, n, is one way of describing how many ways the sum of positive integers, ≤ n, can be added together to equal n, regardless of order. The function p(n) is used to denote the number of partitions for n. Below we show our 5 “coins” as addends to evaluate 7 partitions, that is p(5)=7.

5 = 5 = 4+1 = 3+2 = 3+1+1 = 2+2+1 = 2+1+1+1 = 1+1+1+1+1

We use a generating function to create the series until we find the required n. The generating function requires at most 500 so-called generalized pentagonal numbers, given by n(3n – 1)/2 with 0, ± 1, ± 2, ± 3…, the first few of which are 0, 1, 2, 5, 7, 12, 15, 22, 26, 35, … (Sloane’s A001318).

We have the following generating function which uses our pentagonal numbers as exponents:

1 - q - q^2 + q^5 + q^7 - q^12 - q^15 + q^22 + q^26 + ...

my blog at blog.dreamshire.com has a perl program that solves this in under 10 sec.