## Project Euler 31: Investigating combinations of English currency denominations

#### Project Euler 31 Problem Description

Project Euler 31: In England the currency is made up of pound, £, and pence, p, and there are eight coins in general circulation:

1p, 2p, 5p, 10p, 20p, 50p, £1 (100p) and £2 (200p).

It is possible to make £2 in the following way:

1×£1 + 1×50p + 2×20p + 1×5p + 1×2p + 3×1p

How many different ways can £2 be made using any number of coins?

#### Analysis

This is a typical problem that demonstrates the use of partitions which can be solved by using dynamic programming. To keep this problem simple: order does not matter, there are always enough coins to make a combination and we’re not looking for the optimal way to make change. This a perfect application for tested algorithms where this problem has been used as an example. As the details and nuances of dynamic programming are somewhat involved; we have provided some links below for a better explanation:

Coin Change – Algorithmist

Dynamic Programming Solution to the Coin Changing Problem

#### Project Euler 31 Solution

Runs < 0.001 seconds in Python 2.7.Use this link to get the Project Euler 31 Solution Python 2.7 source.

#### Answer

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#### Afterthoughts

- See also, Project Euler 76 Solution: How many different ways can one hundred be written as a sum of at least two positive integers?
- This problem only wants the number of combinations. It would be a different approach if they wanted a set of combinations.

*Project Euler 31 Solution last updated*

Hi, exactly what does this line mean?

ways[i] += ways[i-coin]

Thanks,

Judy

Meaning, why does it work for this problem?

This code is indeed brilliant, but I have no idea how it works!

This is very elegant code. Thanks for posting. Much prettier than my brute force list of combinations. 🙂

For 2x pounds the number of ways to make change with coins of 1, 2, 5, 10, 20, 50, 100, and 200 pence is:

Ways(2x) = (63 – 16524x + 121100x^2 + 862127x^3 + 1656620x^4 +1395380x^5 + 543200x^6 + 80000x^7) / 63.

I can compute that pretty quickly even for 10^12 pounds!

This solution is brilliant! My code was over 60 lines long. Thanks for the post.