Matrix chain multiplication: Difference between revisions

This is a translation of the Python iterative solution.

implicit none

contains

implicit none

integer :: a(:), n, i, j, k

integer, allocatable :: u(:, :)

integer(8), allocatable :: v(:, :)

n = ubound(a, 1) - 1

allocate (u(n, n), v(n, n))

u(:, 1) = -1

v(:, 1) = 0

do k = 1, j - 1

c = v(i, k) + v(i + k, j - k) + int(a(i), 8) * int(a(i + k), 8) * int(a(i + j), 8)

u(i, j) = k

v(i, j) = c

recursive subroutine aux(i, j)

integer :: i, j, k

k = u(i, j)

if (k < 0) then

end module

implicit none

call optim([1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2])

call optim([1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10])

<pre>

38120 ((((((((1*2)*3)*4)*5)*6)*7)*(8*(9*10)))*(11*12))

1773740 (1*((((((2*3)*4)*(((5*6)*7)*8))*9)*10)*11))