Problem 11: Largest Product in a Grid

Here, we can use the fact that multiplication is both commutative and symmetrical.

You want the maximum product of 4 adjacent numbers in any of the 8 possible directions:

• right
• left
• down
• up
• diagonal down-right
• diagonal up-right
• diagonal down-left
• diagonal up-left

However, thanks to the properties of multiplication mentioned above, we only really need 4 directions:

• right
• down
• down-right
• up-right

Given the grid is only 20x20, we can easily check all possible products of 4 adjacent numbers in these directions.

In terms of python:

x = """08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00
81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65
52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 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 03 45 02 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 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95
78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92
16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57
86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58
19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40
04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66
88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69
04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36
20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16
20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54
01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48"""

grid = [list(map(int, row.split())) for row in x.split('\n')]
n = len(grid)
max_prod = 0
for i in range(n):
    for j in range(n):
        if j + 3 < n:  # right
            max_prod = max(max_prod, grid[i][j]*grid[i][j+1]*grid[i][j+2]*grid[i][j+3])
        if i + 3 < n:  # down
            max_prod = max(max_prod, grid[i][j]*grid[i+1][j]*grid[i+2][j]*grid[i+3][j])
        if i + 3 < n and j + 3 < n:  # diagonal down-right
            max_prod = max(max_prod, grid[i][j]*grid[i+1][j+1]*grid[i+2][j+2]*grid[i+3][j+3])
        if i - 3 >= 0 and j + 3 < n:  # diagonal up-right
            max_prod = max(max_prod, grid[i][j]*grid[i-1][j+1]*grid[i-2][j+2]*grid[i-3][j+3])
print(max_prod)

Explained line by line:

• We start by defining the 20x20 grid as a multi-line string and then convert it into a 2D list of integers called grid.
• We initialize a variable max_prod to keep track of the maximum product found.
• We use nested loops to iterate through each cell in the grid using indices i (row) and j (column).
• For each cell, we check if we can calculate the product of four adjacent numbers in the right direction. If so, we compute the product and update max_prod if this product is greater than the current maximum.
• We repeat this process for the down direction, diagonal down-right direction, and diagonal up-right direction, ensuring we stay within the bounds of the grid.
• Finally, we print the maximum product found.

Running the above code leads to an answer of $70600674$