In my attempt to learn (Iron)Python out in the open, here’s my solution for Project Euler Problem 18.
As always, any feedback is welcome.
# Euler 18
# http://projecteuler.net/index.php?section=problems&id=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)
import time
start = time.time()
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, 9, 70, 98, 73, 93, 38, 53, 60, 04, 23]]
# Loop through each row of the triangle starting at the base.
for a in range(len(triangle) - 1, -1, -1):
for b in range(0, a):
# Get the maximum value for adjacent cells in current row.
# Update the cell which would be one step prior in the path
# with the new total. For example, compare the first two
# elements in row 15. Add the max of 04 and 62 to the first
# position of row 14.This provides the max total from row 14
# to 15 starting at the first position. Continue to work up
# the triangle until the maximum total emerges at the
# triangle's apex.
triangle [a-1][b] += max(triangle [a][b], triangle [a][b+1])
print triangle [0][0]
print "Elapsed Time:", (time.time() - start) * 1000, "millisecs"
a=raw_input('Press return to continue')
