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 in triangle.txt (right click and 'Save Link/Target As...'), a 15K text file containing a triangle with one-hundred rows.

NOTE: This is a much more difficult version of Problem 18. It is not possible to try every route to solve this problem, as there are 2⁹⁹ altogether! If you could check one trillion (10¹²) routes every second it would take over twenty billion years to check them all. There is an efficient algorithm to solve it. ;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: 58.1 us per loop






    Out[5]:





23



In [6]:

    
%timeit foo(deepcopy(t15))
foo(deepcopy(t15))









    



1000 loops, best of 3: 427 us per loop






    Out[6]:





1074



In [7]:

    
t100 = [[int(i) for i in(s.strip().split())] for s in open('triangle.txt').readlines()]
len(t100)









    Out[7]:





100



In [8]:

    
%timeit foo(deepcopy(t100))
foo(deepcopy(t100))









    



100 loops, best of 3: 10.7 ms per loop






    Out[8]:





7273