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)


In [1]:
t4 = [
    [3],
    [7, 4],
    [2, 4, 6],
    [8, 5, 9, 3],]
t4


Out[1]:
[[3], [7, 4], [2, 4, 6], [8, 5, 9, 3]]

In [2]:
t15 = [
    [75],
    [95, 64],
    [17, 47, 82],
    [18, 35, 87, 10],
    [20,  4, 82, 47, 65],
    [19,  1, 23, 75,  3, 34],
    [88,  2, 77, 73,  7, 63, 67],
    [99, 65,  4, 28,  6, 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,  4, 68, 89, 53, 67, 30, 73, 16, 69, 87, 40, 31],
    [04, 62, 98, 27, 23,  9, 70, 98, 73, 93, 38, 53, 60,  4, 23],]
len(t15)


Out[2]:
15

In [3]:
def foo(t):
    for i in range(len(t))[::-1]:
        r = t[i]
        try:
            nr = t[i+1]
        except IndexError:
            for j in range(len(t[i])):
                t[i][j] = (t[i][j], None)
        else:
            for j in range(len(t[i])):
                dir = (t[i+1][j+1][0] > t[i+1][j+0][0])
                t[i][j] = (t[i][j] + t[i+1][j+dir][0], dir)
    return t[0][0][0]

In [4]:
from copy import deepcopy

In [5]:
%timeit foo(deepcopy(t4))
foo(deepcopy(t4))


10000 loops, best of 3: 59.3 us per loop
Out[5]:
23

In [6]:
%timeit foo(deepcopy(t15))
foo(deepcopy(t15))


1000 loops, best of 3: 431 us per loop
Out[6]:
1074