Problem
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)
Answer
1 | 1074 |
Python1
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37#!/usr/bin/env python
def find_sum(triangle):
def get_options(row, index):
return triangle[row+1][index], triangle[row+1][index+1]
row = len(triangle) - 2
while True:
try:
for index, node in enumerate(triangle[row]):
best = max([node + option for option in get_options(row, index)])
triangle[row][index] = best
row -= 1
except:
return triangle[0][0]
def main():
triangle_str = '''\
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'''
triangle = [[int(digit) for digit in line.strip().split()] for line in triangle_str.splitlines()]
print(find_sum(triangle))
if __name__ == "__main__":
main()
Ruby1
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28#!/usr/bin/env ruby
triangle_str = <<EOS
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
EOS
triangle = triangle_str.each_line.map { |line| line.split.map(&:to_i) }
(triangle.length - 2).downto(0) do |y|
triangle[y].length.times do |x|
triangle[y][x] += [triangle[y+1][x], triangle[y+1][x+1]].max
end
end
puts triangle[0][0]
Haskell1
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18parse :: String -> [[Integer]]
parse = map (map read . words) . lines
best :: [Integer] -> [Integer]
best row = map maximum choices where
choices = zipWith (\a b -> a : [b]) row (tail row)
maxStep :: [Integer] -> [Integer] -> [Integer]
maxStep current next = zipWith (+) next (best current)
maxPath :: [[Integer]] -> Integer
maxPath [[x]] = x
maxPath (current:next:rest) = maxPath $ (maxStep current next) : rest
main :: IO ()
main = do
str <- readFile "/home/zach/code/euler/018/triangle.txt"
print $ maxPath $ reverse $ parse str
Java1
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49public final class p018 implements EulerSolution {
public static void main(String[] args) {
System.out.println(new p018().run());
}
/*
* We create a new blank triangle with the same dimensions as the original big triangle.
* For each cell of the big triangle, we consider the sub-triangle whose top is at this cell,
* calculate the maximum path sum when starting from this cell, and store the result
* in the corresponding cell of the blank triangle.
*
* If we start at a particular cell, what is the maximum path total? If the cell is at the
* bottom of the big triangle, then it is simply the cell's value. Otherwise the answer is
* the cell's value plus either {the maximum path total of the cell down and to the left}
* or {the maximum path total of the cell down and to the right}, whichever is greater.
* By computing the blank triangle's values from bottom up, the dependent values are always
* computed before they are utilized. This technique is known as dynamic programming.
*/
public String run() {
for (int i = triangle.length - 2; i >= 0; i--) {
for (int j = 0; j < triangle[i].length; j++)
triangle[i][j] += Math.max(triangle[i + 1][j], triangle[i + 1][j + 1]);
}
return Integer.toString(triangle[0][0]);
}
private int[][] triangle = { // Mutable
{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},
{ 4,62,98,27,23, 9,70,98,73,93,38,53,60, 4,23},
};
}
Mathematica1
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28(*
* If we start at a particular cell in the triangle, what is the maximum path total?
* If the cell is in the bottom row, then it is simply the cell's value. Otherwise the answer
* is the cell's value plus either {the maximum path total of the cell down and to the left}
* or {the maximum path total of the cell down and to the right}, whichever is greater.
*)
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}};
MaxPath[i_, j_] := triangle[[i, j]] +
If[i < Length[triangle], Max[MaxPath[i + 1, j], MaxPath[i + 1, j + 1]], 0]
MaxPath[1, 1]
