Project Euler Problem 18

Solution

Answer: 1074

This is a classic dynamic programming problem.

At each position in the triangle, the maximum sum reaching that position depends only on the two possible parents above it. Instead of checking all $2^{14} = 16384$ routes, we propagate the best possible total row by row.

Let $dp[i][j]$ be the maximum total reaching element $j$ in row $i$.

Transition:

$$dp[i][j] = \text{triangle}[i][j] + \max(dp[i-1][j-1], dp[i-1][j])$$

with edge cases on the left and right borders.

A more efficient approach is to work bottom-up:

Start from the second-last row.
Replace each number with:

$$\text{value} + \max(\text{two children below})$$

After collapsing all rows, the top element becomes the maximum total.

Using the triangle from the problem:

triangle = [
    [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]
]

for row in range(len(triangle) - 2, -1, -1):
    for col in range(len(triangle[row])):
        triangle[row][col] += max(
            triangle[row + 1][col],
            triangle[row + 1][col + 1]
        )

print(triangle[0][0])

Tracing the reductions upward yields the final top value:

$$1074$$

Answer: 1074