# Project Euler #18 in Python: Maximum sum down a triangle of numbers

I've solved Project Euler #18 in Python but think my code and algorithm can be improved. Here is the problem:

By starting at the top of the triangle below and moving to adjacent numbers on the row below, the maximum total from top to bottom is 23.

3
7 4
2 4 6
8 5 9 3

That is, 3 + 7 + 4 + 9 = 23.

Find the maximum total from top to bottom of the triangle below:

75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23

NOTE: As there are only 16384 routes, it is possible to solve this problem by trying every route. However, Problem 67, is the same challenge with a triangle containing one-hundred rows; it cannot be solved by brute force, and requires a clever method!

And I want with an bottom-up calculation to find the best path down. Here is my code:

import time

num_pyr = '''75
95 64
17 47 82
18 35 87 10
20 04 82 47 65
19 01 23 75 03 34
88 02 77 73 07 63 67
99 65 04 28 06 16 70 92
41 41 26 56 83 40 80 70 33
41 48 72 33 47 32 37 16 94 29
53 71 44 65 25 43 91 52 97 51 14
70 11 33 28 77 73 17 78 39 68 17 57
91 71 52 38 17 14 91 43 58 50 27 29 48
63 66 04 68 89 53 67 30 73 16 69 87 40 31
04 62 98 27 23 09 70 98 73 93 38 53 60 04 23'''

def break_pyramid(num_pyr):
'''
Breaks the number string into an array
'''
pyramid = []
rows = num_pyr.split('\n')
for row in rows:
pyramid.append([int(i) for i in row.split(' ')])
return pyramid

'''
Takes the maximum of the legs under upper row to each index to add up from bottom
Parameters
----------
upper_row : len n
lower_row : len n + 1

Returns
-------
new_upper_row
'''
greater_vals = [max(lower_row[i:i+2]) for i in range(len(upper_row))]
return [x + y for x, y in zip(upper_row, greater_vals)]

def main():
pyramid = break_pyramid(num_pyr)[::-1]
for i in range(len(pyramid)-1):
pyramid[1] = new_upper
del pyramid[0]
return pyramid[0][0]

if __name__ == '__main__':
t1 = time.time()
best_path = main()
op_time = 1000000 * (time.time() - t1)  # us
print("Calculated {} in {:.1f} us".format(best_path, op_time))

I think improvements can be made in my main() since I am using del. Should del be avoided for any reason?

• "Should del be avoided for any reason?" Let's turn it around: why do you feel you need it in this case?
– Mast
Commented Dec 4, 2017 at 18:23
• it just seemed to simplify the loop since the other option is to have a counter to keep track of what row I'm on. But I'm already doing that with for i in range(len(pyramid)-1) so I can just make the first line new_upper = add_from_under(pyramid[i+1], pyramid[0]). then that would remove the entire del line from the loop. Commented Dec 4, 2017 at 18:47

In break_pyramid you use list comprehension to split a row into numbers – why not use list comprehension as well to split the input into rows?

def break_pyramid(num_pyr):
'''
Breaks the number string into an array
'''
pyramid = [[int(col) for col in row.split(' ')] for row in num_pyr.split('\n')]
return pyramid

greater_vals = [max(lower_row[i:i+2]) for i in range(len(upper_row))]

creates many intermediate slices. The alternative

greater_vals = [max(l, r) for l, r in zip(lower_row, lower_row[1:])]

seems to be slightly faster. You can even combine both zip operations into a single one:

return [upper + max(lower_left, lower_right)
for upper, lower_left, lower_right in zip(upper_row, lower_row, lower_row[1:])]

so that no intermediate list is created.

The main loop applies add_from_under cumulatively to the pyramid rows (in reverse order). That is a "reduce" operation and can be done with functools.reduce:

from functools import reduce

def main():
pyramid = break_pyramid(num_pyr)
return final_row[0]

if we exchange the parameter order in add_from_under to