Project Euler 11: Largest product in a grid

Problem 11

Largest product in a grid

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

import numpy as np
initial_max = 0
candidate_max1 = 1
candidate_max2 = 1
candidate_max3 = 1
S = {}
S[1] = "08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08"
S[2] = "49 49 99 40 17 81 18 57 60 87 17 40 98 43 69 48 04 56 62 00"
S[3] = "81 49 31 73 55 79 14 29 93 71 40 67 53 88 30 03 49 13 36 65"
S[4] = "52 70 95 23 04 60 11 42 69 24 68 56 01 32 56 71 37 02 36 91"
S[5] = "22 31 16 71 51 67 63 89 41 92 36 54 22 40 40 28 66 33 13 80"
S[6] = "24 47 32 60 99 03 45 02 44 75 33 53 78 36 84 20 35 17 12 50"
S[7] = "32 98 81 28 64 23 67 10 26 38 40 67 59 54 70 66 18 38 64 70"
S[8] = "67 26 20 68 02 62 12 20 95 63 94 39 63 08 40 91 66 49 94 21"
S[9] = "24 55 58 05 66 73 99 26 97 17 78 78 96 83 14 88 34 89 63 72"
S[10] = "21 36 23 09 75 00 76 44 20 45 35 14 00 61 33 97 34 31 33 95"
S[11] = "78 17 53 28 22 75 31 67 15 94 03 80 04 62 16 14 09 53 56 92"
S[12] = "16 39 05 42 96 35 31 47 55 58 88 24 00 17 54 24 36 29 85 57"
S[13] = "86 56 00 48 35 71 89 07 05 44 44 37 44 60 21 58 51 54 17 58"
S[14] = "19 80 81 68 05 94 47 69 28 73 92 13 86 52 17 77 04 89 55 40"
S[15] = "04 52 08 83 97 35 99 16 07 97 57 32 16 26 26 79 33 27 98 66"
S[16] = "88 36 68 87 57 62 20 72 03 46 33 67 46 55 12 32 63 93 53 69"
S[17] = "04 42 16 73 38 25 39 11 24 94 72 18 08 46 29 32 40 62 76 36"
S[18] = "20 69 36 41 72 30 23 88 34 62 99 69 82 67 59 85 74 04 36 16"
S[19] = "20 73 35 29 78 31 90 01 74 31 49 71 48 86 81 16 23 57 05 54"
S[20] = "01 70 54 71 83 51 54 69 16 92 33 48 61 43 52 01 89 19 67 48"

# for making an input matrix
raw_matrix = np.zeros(shape=(20,20))
for i in range(20):
for j in range(20):
S_list = S[i+1].split(" ")
raw_matrix[i][j] = int(S_list[j])
print raw_matrix[13,5]

for i in range(20):
for j in range(20):
candidate_max1 = raw_matrix[i][j]
candidate_max2 = raw_matrix[i][j]
candidate_max3 = raw_matrix[i][j]
candidate_max4 = raw_matrix[i][j]
try:
for cnt in range(1,4):
candidate_max1 = candidate_max1 * raw_matrix[i][j+cnt] # for horizon
candidate_max2 = candidate_max2 * raw_matrix[i+cnt][j] #for vertical
candidate_max3 = candidate_max3 * raw_matrix[i+cnt][j+cnt] #for diagonal
candidate_max4 = candidate_max4 * raw_matrix[i-cnt][j+cnt] #for reverse diagonal
if initial_max < max(candidate_max1,candidate_max2,candidate_max3,candidate_max4):
initial_max = max(candidate_max1,candidate_max2,candidate_max3,candidate_max4)
except:
continue
print initial_max


• Welcome to Code Review! Please replace that image at the top with simple text. On the front page your post shows up as "Python2.7 Question link: projecteuler.net/problem=..." instead of a nice description. That's not very attractive for reviewers. You'll get more out of Code Review by writing good descriptions as the first paragraph of your post – janos Sep 3 '15 at 15:41

Your approach misses a few products

For instance, replace the last row of your matrix with 99 99 99 99 .... Clearly the maximal product should be 99*99*99*99 = 96059601, but your approach doesn't find this solution.

You can only find this solution, when i = 19. But than you receive an IndexError, because your simultaneously calculate the product of matrix[19,0] - matrix[22,0].

One way of handling this would be to compute each direction separately, each one with each own direction.

But of course you could put it back into a loop. For instance like this:

for i in range(20):
for j in range(20):

for direction in ((0,1), (1,0), (1,1), (-1,-1)):
try:
product = 1
for l in range(4):
product *= raw_matrix[i + direction[0]*l][j + direction[1]*l]
if product > initial_max:
initial_max = product
except IndexError:
pass

print initial_max


No need for numpy-arrays

I would get rid of numpy-arrays. Not really necessary, since you never use any matrix/vector operations. A standard 2D-list should be quite enough. The parsing of the matrix can also a bit simplified using List Comprehensions.

raw_matrix = '''\
08 02 22 97 38 15 00 40 00 75 04 05 07 78 52 12 50 77 91 08
...
01 70 54 71 99 99 99 99 16 92 33 48 61 43 52 01 89 19 67 48'''

matrix = [map(int, line.split()) for line in raw_matrix.split('\n')]


Notice. I import the matrix as a multi-line string. Then I extract each line by splitting at new-lines. Then each line gets split by spaces and mapped to ints.

Variable naming

Well, kinda obvious. What does S stand for, why call the matrix raw_matrix. raw_matrix would be good name instead of S, because the matrix is still in raw form (string). And call the parsed matrix simply matrix.

Others renaming I would prefer: i -> row, j -> column, initial_product -> max_product.

So all in all, here a improved version of your code:

raw_matrix = '''\
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 99 99 99 99 16 92 33 48 61 43 52 01 89 19 67 48'''

matrix = [map(int, line.split()) for line in raw_matrix.split('\n')]

max_product = 0

for row in range(20):
for column in range(20):
for direction in ((0,1), (1,0), (1,1), (-1,-1)):
try:
product = 1
for l in range(4):
product *= matrix[row + direction[0]*l][column + direction[1]*l]
if product > max_product:
max_product = product
except IndexError:
pass

print max_product

• Welcome to code review! Remember to avoid except: it will swallow any exception – Caridorc Sep 3 '15 at 13:29
• @Caridorc Changed it to except IndexError:. I was mostly trying to demonstrate a functional solution and was not really paying attention to style. – Jakube Sep 3 '15 at 13:33
• Sadly this does not produce the right answer according to the Project Euler site. – chicks Nov 10 '15 at 20:22
• @chicks The matrix I used is not exactly the same matrix as in Project Euler. I modified the last row a little bit. Basically I wanted to show off a test case, that the OP's code was unable to find. – Jakube Nov 10 '15 at 20:30

We take take advantage of these features of numpy to reduce the code quite a bit:

1. array reshaping
2. array row, column and diagonal slices
3. np.prod to compute the product of all elements in an array
4. flipping the array to get anti-diagonals

Reshaping

Reshaping allows you to (for instance) reorganize a 1-dimensional array into a 2-dimenstional array. Just the dimension metadata is changed so the operation is very efficient:

a = np.array([1,2,3,4,5,6]).reshape(3,2)
print a

[[1 2]
[3 4]
[5 6]]


Slicing

Here are example of how to take row, column and diagonal slices of arrays:

import numpy as np

a = np.array( [[11,12,13,14,15,16],
[21,22,23,24,25,26],
[31,32,33,34,35,36],
[41,42,43,44,45,46]
] )

print "row",     a[1][3:5]         # horizontal slice, row 1, columns 3..5
print "column:", a[:,4]            # column 4
print "diag:",   np.diagonal(a, 2) # 2nd diagonal


Note that row and column indices start from 0. In the case of diagonals, 0 is the main diagonal, with positive diagonals to the right and negative diagonals to the left of the main.

Products

And here is how to use np.prod:

a = np.array([5,6,7,8])
print np.prod(a)             # prints 1680


Anti-diagonals

We can get anti-diagonals by first flipping the matrix along one of its dimensions:

a = np.array([1,2,3,4,5,6,7,8]).reshape(2,4)
b = a[::-1,]
print b


b has the same shape as a and the diagonals of b are the anti-diagonals of a.

Solution

Putting it all together:

def euler11():
data = """
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
"""
a = np.array([ int(x) for x in data.split() ]).reshape(20,20)
# print a
b = a[::-1,]

maxprod = 0
for i in xrange(20):
for j in xrange(20):
r = np.prod( a[i][j:j+4] )
c = np.prod( a[:,j][i:i+4] )
k = min(i,j)
d1 = np.prod( np.diagonal(a, j-i)[ k:k+4 ] )
d2 = np.prod( np.diagonal(b, j-i)[ k:k+4 ] )
maxprod = max([maxprod,r,c,d1,d2])
print maxprod


Note here is i,j represents the upper left corner of either the row, column or diagonal we are taking the product of. That's why both range from 0 to 16 We just run both i and j from 0 through 19. We examine a few extra products (some with less than 4 terms), but since we are looking for the maximum it doesn't matter.

• This solution using numpy is really clear and clever. Can you recommend any book on learning numpy? or any doc that you consider useful for a better understanding on numpy? – Nick May 7 '16 at 7:59