Problem

I have recently solved the following Project Euler problem:

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 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`

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?

with the following Python code:

```
import itertools
import numpy as np
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]
]
def diagonals(matrix):
'''
Returns all diagonals from a matrix
'''
# http://stackoverflow.com/questions/6313308/get-all-the-diagonals-in-a-matrix-list-of-lists-in-python
np_array = np.array(matrix)
diags = [np_array[::-1, :].diagonal(i).tolist() for
i in range(-np_array.shape[0]+1, np_array.shape[1])]
diags.extend(np_array.diagonal(i).tolist() for
i in range(np_array.shape[1]-1, -np_array.shape[0], -1))
return diags
def columns(matrix):
'''
Returns all columns from a matrix
'''
np_array = np.array(matrix)
cols = [np_array[:, i].tolist() for i in range(np_array.shape[1])]
return cols
def subsequences(lst, size):
'''
Returns all subsequences of certain size from a list
'''
# http://stackoverflow.com/questions/6670828/find-all-consecutive-sub-sequences-of-length-n-in-a-sequence
return [list(subsequence) for subsequence in zip(*(lst[i:] for i in range(size)))]
def largest_adjacent_product(matrix, subsequence_size):
'''
Returns largest product of a subsequence of a certain size of adjacent
numbers of a matrix. The subsequences can be horizontal, vertical and
diagonal.
'''
diags = diagonals(matrix)
cols = columns(matrix)
horizontal_subsequences = itertools.chain.from_iterable(
[subsequences(row, subsequence_size) for row in matrix])
diagonal_subsequences = itertools.chain.from_iterable(
[subsequences(diag, subsequence_size) for diag in diags])
vertical_subsequences = itertools.chain.from_iterable(
[subsequences(col, subsequence_size) for col in cols])
horizontal_products = [np.prod(subsequence) for
subsequence in horizontal_subsequences]
diagonal_products = [np.prod(subsequence) for
subsequence in diagonal_subsequences]
vertical_products = [np.prod(subsequence) for
subsequence in vertical_subsequences]
max_horizontal_product = max(horizontal_products)
max_diagonal_product = max(diagonal_products)
max_vertical_product = max(vertical_products)
result = max(max_horizontal_product, max_diagonal_product, max_vertical_product)
return result
if __name__ == '__main__':
print(largest_adjacent_product(GRID, 4))
```

So, what are your thoughts?

Solution

If you look at `largest_adjacent_product`

you should see that `{}_subsequences`

, `{}_products`

and `max_{}_product`

are all created the same way.

If you make it its own function, and add `subsequences`

, then you should be able to come to:

```
product = np.prod
chain = itertools.chain.from_iterable
max(product(s) for s in chain(zip(*(row[i:] for i in range(size))) for row in grid))
```

You can actually remove the `chain`

.

This gives you code that is easy to read:

```
product = np.prod
max(product(s)
for row in grid
for s in zip(*(row[i:] for i in range(size))))
```

Alternately I’d use use the second answer for `subsequence`

and use:

```
product = np.prod
max(product(row[i:i + size])
for row in grid
for i in range(len(row) - size + 1))
```

I’d say this is better,

space complexity is O(1)$O(1)$, as opposed to O(kn)$O(kn)$ (or O(kn−−√)$O(k\sqrt{n})$ for the modified ones above).

Time complexity should be the same at O(kn)$O(kn)$,

and the readability of both the above are much better.

k=$k=$ size and n=$n=$ grid.

This would make your code much simpler:

```
def largest_subset(grid, size):
product = np.prod
return max(product(row[i:i + size])
for row in grid
for i in range(len(row) - size + 1))
def largest_adjacent_product(matrix, size):
return max(largest_subset(i, size)
for i in [matrix, columns(matrix), diagonals(matrix)])
```

I don’t like how you change `matrix`

to `np_array`

in both diagonals and columns.

I would use `grid`

rather than both terms, and just overwrite `grid`

when converted to a np array.

Alternately I’d pass both a np array in the first place.