Project Euler Problem 11

Solution

Answer: 70600674

We check every possible group of 4 adjacent numbers in the grid in the four relevant directions:

Horizontal
Vertical
Diagonal down-right
Diagonal down-left

For each starting cell, compute the product of the 4 numbers if the direction stays inside the grid, and keep the maximum product found.

A straightforward brute-force search over the $20 \times 20$ grid is tiny computationally and guarantees the correct answer.

Python code:

grid = [
[8,2,22,97,38,15,0,40,0,75,4,5,7,78,52,12,50,77,91,8],
[49,49,99,40,17,81,18,57,60,87,17,40,98,43,69,48,4,56,62,0],
[81,49,31,73,55,79,14,29,93,71,40,67,53,88,30,3,49,13,36,65],
[52,70,95,23,4,60,11,42,69,24,68,56,1,32,56,71,37,2,36,91],
[22,31,16,71,51,67,63,89,41,92,36,54,22,40,40,28,66,33,13,80],
[24,47,32,60,99,3,45,2,44,75,33,53,78,36,84,20,35,17,12,50],
[32,98,81,28,64,23,67,10,26,38,40,67,59,54,70,66,18,38,64,70],
[67,26,20,68,2,62,12,20,95,63,94,39,63,8,40,91,66,49,94,21],
[24,55,58,5,66,73,99,26,97,17,78,78,96,83,14,88,34,89,63,72],
[21,36,23,9,75,0,76,44,20,45,35,14,0,61,33,97,34,31,33,95],
[78,17,53,28,22,75,31,67,15,94,3,80,4,62,16,14,9,53,56,92],
[16,39,5,42,96,35,31,47,55,58,88,24,0,17,54,24,36,29,85,57],
[86,56,0,48,35,71,89,7,5,44,44,37,44,60,21,58,51,54,17,58],
[19,80,81,68,5,94,47,69,28,73,92,13,86,52,17,77,4,89,55,40],
[4,52,8,83,97,35,99,16,7,97,57,32,16,26,26,79,33,27,98,66],
[88,36,68,87,57,62,20,72,3,46,33,67,46,55,12,32,63,93,53,69],
[4,42,16,73,38,25,39,11,24,94,72,18,8,46,29,32,40,62,76,36],
[20,69,36,41,72,30,23,88,34,62,99,69,82,67,59,85,74,4,36,16],
[20,73,35,29,78,31,90,1,74,31,49,71,48,86,81,16,23,57,5,54],
[1,70,54,71,83,51,54,69,16,92,33,48,61,43,52,1,89,19,67,48]
]

best = 0

for r in range(20):
    for c in range(20):

        # horizontal
        if c + 3 < 20:
            p = grid[r][c] * grid[r][c+1] * grid[r][c+2] * grid[r][c+3]
            best = max(best, p)

        # vertical
        if r + 3 < 20:
            p = grid[r][c] * grid[r+1][c] * grid[r+2][c] * grid[r+3][c]
            best = max(best, p)

        # diagonal down-right
        if r + 3 < 20 and c + 3 < 20:
            p = grid[r][c] * grid[r+1][c+1] * grid[r+2][c+2] * grid[r+3][c+3]
            best = max(best, p)

        # diagonal down-left
        if r + 3 < 20 and c - 3 >= 0:
            p = grid[r][c] * grid[r+1][c-1] * grid[r+2][c-2] * grid[r+3][c-3]
            best = max(best, p)

print(best)

The maximum product found is:

$$89 \times 94 \times 97 \times 87 = 70600674$$

Answer: 70600674