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.
2 4* 6
8 5 9* 3
3 + 7 + 4 + 9 = 23
Find the maximum total from top to bottom of a given triangle.
This solution together with the solution of James' (or mine) exhibits a nice duality between the "top-down" and "bottom-up" approaches.
Nice! When you go from the bottom, solution automatically goes to a(1).
Thanks. I coded that solution up LONG before movmax became a thing...
02 - Vector Variables 4
Solving Quadratic Equations (Version 1)
Highly divisible triangular number (inspired by Project Euler 12)
Project Euler: Problem 14, Longest Collatz sequence
Project Euler: Problem 11, Largest product in a grid
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