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py_0018_maximum_path_sum_i.py
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# Solution of;
# Project Euler Problem 18: Maximum path sum I
# https://projecteuler.net/problem=18
#
# 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! ;o)
#
# by lcsm29 http://github.com/lcsm29/project-euler
import timed
def fn_maxsum_from_bottom(n):
def g(num):
return int(0.5 * num * (num + 1))
flat = [int(s) for s in s_triangle.split()]
n_rows = next((i for i in range(int(1e100)) if g(i) >= len(flat)), None)
t = [flat[g(i):g(i + 1)] for i in range(n_rows)]
for i in range(n_rows - 2, -1, -1):
for j in range(len(t[i])):
t[i][j] += max(t[i + 1][j], t[i + 1][j + 1])
return t[0][0]
s_triangle = '''
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
'''
if __name__ == '__main__':
n = None
i = 23_000
prob_id = 18
timed.caller(fn_maxsum_from_bottom, n, i, prob_id)