Project Euler Problem 15 Solution
Starting in the top left corner of a 2x2 grid, there are 6 routes (without backtracking) to the bottom right corner.
How many routes are there through a 20x20 grid?
The grid can be expressed as Pascal’s Triangle:
1 1 1 1 (2) 1 1 3 3 1 1 4 (6) 4 1 1 5 10 10 5 1 1 6 15 (20) 15 6 1
Note that the solution for a 1x1 grid is 2, a 2x2 grid is 6, and a 3x3 grid is 20.
If we compare these solutions to Pascal’s Triangle, we see that they correspond to the 1st element in the 2nd row, the 2nd element in the 4th row, and the 3rd element in the 6th row, respectively. (Note that Pascal’s Triangle is zero-indexed.)
The binomial coefficient can be used to determine the th element in the th row of Pascal’s Triangle. Thus, we could express the aforementioned solutions as , , and , respectively.
Thus, a general solution for grids of size is
factorial :: Integer -> Integer factorial n = product [1..n] choose :: Integer -> Integer -> Integer choose n k = div (factorial n) $ factorial k * factorial (n - k) main :: IO () main = print $ choose 40 20
$ ghc -O2 -o binom binom.hs $ time ./binom real 0m0.001s user 0m0.000s sys 0m0.000s
#!/usr/bin/env python from gmpy2 import comb print(comb(2 * 20,20))
$ time python3 grid-routes.py real 0m0.079s user 0m0.020s sys 0m0.000s
#!/usr/bin/env ruby class Integer def choose(k) (self-k+1 .. self).inject(1, &:*) / (2 .. k).inject(1, &:*) end end puts 40.choose(20)
$ time ruby pascal.rb real 0m0.051s user 0m0.028s sys 0m0.000s
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