Maximum path sum I
Author: Felix Herrmann
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.
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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)
Source code: prob018-felher.pl
use v6; my $triangle = '75 95 64 17 47 82 18 35 87 10 20 04 82 47 65 19 01 23 75 03 34 88 02 77 73 07 63 67 99 65 04 28 06 16 70 92 41 41 26 56 83 40 80 70 33 41 48 72 33 47 32 37 16 94 29 53 71 44 65 25 43 91 52 97 51 14 70 11 33 28 77 73 17 78 39 68 17 57 91 71 52 38 17 14 91 43 58 50 27 29 48 63 66 04 68 89 53 67 30 73 16 69 87 40 31 04 62 98 27 23 09 70 98 73 93 38 53 60 04 23'; my @lines = string-to-array($triangle).reverse; # reduce the triangle by adding up the lines until only one line with one # element is left; then print it. say "{@lines.reduce: &add-maxima}"; # this function assumes the shorter and longer array to be consecutive lines # in an reversed triangle. It then adds each of the maxima of consecutive fields # of the longer array to their shared diagonal neighbour in the shorter array. sub add-maxima(@longer, @shorter is copy) { for 0 .. @longer - 2 -> $i { @shorter[$i] += max @longer[$i], @longer[$i + 1]; } return @shorter; } sub string-to-array($string) { my @lines = $string.lines; @lines .= map(-> $line { $line.comb(/\d+/).item }); }