SPOILERS: Using Simple Combinatorics

DISCLAIMER: This is the solution to Project Euler’s problem 15. Please attempt to solve the problem yourself before reading my solution.

Starting in the top left corner of a 2×2 grid, and only being able to move to the right and down, there are exactly 6 routes to the bottom right corner.



How many such routes are there through a 20×20 grid?

I like to use this problem to demonstrate the efficacy of using simple maths to improve code.

Instead of the naive solution….

from itertools import permutations

def unique(iterable):
    seen = set()
    for x in iterable:
        if x in seen:
            continue
        seen.add(x)
        yield x

options = [1]*20 + [0]*20
counter = 0
for a in unique(permutations(options)):
 counter = counter+1

print counter

Use simple combinatorics!

To find the number of unique routes through a 20×20 grid, use our friend: the concept of permutations with repeated elements:

$\frac{\text{number of elements}!}{\text{repetitions of character}!*\text{repetitions of other character}!*…}$

import math
print math.factorial(40)/(math.factorial(20)*math.factorial(20))

Even better – one line in Haskell:

product [1..40] `div` product[1..20]^2

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