# Project Euler Problem 11 ("Largest product in a grid") in Python

Can this code in any way be improved or optimized? I'd appreciate any clever tricks or Pythonic expressions that may remove some more brute force elements from my solution.

Also, is there possibly a more intuitive way of approaching this problem that I couldn't see?

Problem statement: Largest product in a grid

In the 20×20 grid below, four numbers along a diagonal line have been marked in red [here bold].

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 102638 40 67 59 54 70 66 18 38 64 70
67 26 20 68 02 62 12 20 956394 39 63 08 40 91 66 49 94 21
24 55 58 05 66 73 99 26 97 177878 96 83 14 88 34 89 63 72
21 36 23 09 75 00 76 44 20 45 351400 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

The product of these numbers is 26 × 63 × 78 × 14 = 1788696.

What is the greatest product of four adjacent numbers in the same direction (up, down, left, right, or diagonally) in the 20×20 grid?

My solution:

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]]

horizontal_max = max(grid[i][j]*grid[i][j+1]*grid[i][j+2]*grid[i][j+3] for i in range(20) for j in range(17))

vertical_max = max(grid[i][j]*grid[i+1][j]*grid[i+2][j]*grid[i+3][j] for i in range(17) for j in range(20))

diagonal1_max = max(grid[i][j]*grid[i+1][j+1]*grid[i+2][j+2]*grid[i+3][j+3] for i in range(17) for j in range(17))

diagonal2_max = max(grid[i][j]*grid[i-1][j+1]*grid[i-2][j+2]*grid[i-3][j+3] for i in range(3,20) for j in range(17))

max_max = max(horizontal_max, vertical_max, diagonal1_max, diagonal2_max)

print(max_max)

If you restricted your lines to 79 columns, as recommended by the Python style guide (PEP8), then we wouldn't have to scroll the code horizontally to read it here.

Anyway, all four loops have the same form:

max(grid[i][j]
* grid[i+i_step][j+j_step]
* grid[i+2*i_step][j+2*j_step]
* grid[i+3*i_step][j+3*j_step]
for i in range(i_min, i_limit)
for j in range(j_limit))

given particular choices of parameters i_step, j_step, i_min, i_limit and j_limit. So we can implement the program as a data-driven loop over those choice of parameters:

print(max(max(grid[i][j]
* grid[i+i_step][j+j_step]
* grid[i+2*i_step][j+2*j_step]
* grid[i+3*i_step][j+3*j_step]
for i in range(i_min, i_limit)
for j in range(j_limit)))
for i_step, j_step, i_min, i_limit, j_limit
in [( 0, 1, 0, 20, 17),  # horizontal
( 1, 0, 0, 17, 20),  # vertical
( 1, 1, 0, 17, 17),  # leading diagonal
(-1, 1, 3, 20, 17)]) # trailing diagonal

If you want to avoid writing out the multiplication, then the best approach would be to define a helper function, for example using functools.reduce and operator.mul:

from functools import reduce
from operator import mul

def product(iterable):
"""Return the product of the numbers in the iterable."""
return reduce(mul, iterable, 1)

and then:

print(max(max(product(grid[i+k*i_step][j+k*j_step] for k in range(4))
for i in range(i_min, i_limit)
for j in range(j_limit)))
for i_step, j_step, i_min, i_limit, j_limit
in [( 0, 1, 0, 20, 17),  # horizontal
( 1, 0, 0, 17, 20),  # vertical
( 1, 1, 0, 17, 17),  # leading diagonal
(-1, 1, 3, 20, 17)]) # trailing diagonal
• Thanks. If we wanted to use the length of the consecutive numbers (here, 4) as our parameter (instead of i_min, etc), e.g. so that there's less mental work involved for the user who isn't interested in understanding the logic of the algorithm, would we be able to avoid writing grid[][] four times? In math, one could use the ∏ (product) notation. Does Python offer a similar functionality? May 25, 2015 at 1:07
• See updated answer. May 25, 2015 at 12:34