Project Euler Problem 15 Solution

Question

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?

Commentary

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 (nk)\binom {n} {k} can be used to determine the kkth element in the nnth row of Pascal’s Triangle. Thus, we could express the aforementioned solutions as (21)\binom {2} {1}, (42)\binom {4} {2}, and (63)\binom {6} {3}, respectively.

Thus, a general solution for grids of size xx is

routes=(2xx)\displaystyle routes = \binom {2x} {x}.

Haskell

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

Python

#!/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

Ruby

#!/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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