Bottom-up DP minimum-path through a triangle:
2
3 4
6 5 7
4 1 8 3
let min_path_triangle rows =
initialise dp from last row;
for r = n - 2 downto 0 do
for c = 0 to row_len - 1 do
dp.(c) <- row.(c) + min(dp.(c), dp.(c+1))
done
done;
dp.(0)
The optimal path 2 -> 3 -> 5 -> 1 sums to 11.
Tests downto loop, Array.of_list inside loop body, nested arr.(i)
reads + writes, and inline if-then-else for min.
74 baseline programs total.
giles
5b38f4d499