Matrix chain multiplication

From Rosetta Code
Task
Matrix chain multiplication
You are encouraged to solve this task according to the task description, using any language you may know.
Problem

Using the most straightfoward algorithm (which we assume here), computing the product of two matrices of dimensions (n1,n2) and (n2,n3) requires n1*n2*n3 FMA operations. The number of operations required to compute the product of matrices A1, A2... An depends on the order of matrix multiplications, hence on where parens are put. Remember that the matrix product is associative, but not commutative, hence only the parens can be moved.

For instance, with four matrices, one can compute A(B(CD)), A((BC)D), (AB)(CD), (A(BC))D, (AB)C)D. The number of different ways to put the parens is a Catalan number, and grows exponentially with the number of factors.

Here is an example of computation of the total cost, for matrices A(5,6), B(6,3), C(3,1):

  • AB costs 5*6*3=90 and produces a matrix of dimensions (5,3), then (AB)C costs 5*3*1=15. The total cost is 105.
  • BC costs 6*3*1=18 and produces a matrix of dimensions (6,1), then A(BC) costs 5*6*1=30. The total cost is 48.

In this case, computing (AB)C requires more than twice as many operations as A(BC). The difference can be much more dramatic in real cases.

Task

Write a function which, given a list of the successive dimensions of matrices A1, A2... An, of arbitrary length, returns the optimal way to compute the matrix product, and the total cost. Any sensible way to describe the optimal solution is accepted. The input list does not duplicate shared dimensions: for the previous example of matrices A,B,C, one will only pass the list [5,6,3,1] (and not [5,6,6,3,3,1]) to mean the matrix dimensions are respectively (5,6), (6,3) and (3,1). Hence, a product of n matrices is represented by a list of n+1 dimensions.

Try this function on the following two lists:

  • [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
  • [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]

To solve the task, it's possible, but not required, to write a function that enumerates all possible ways to parenthesize the product. This is not optimal because of the many duplicated computations, and this task is a classic application of dynamic programming.

See also Matrix chain multiplication on Wikipedia.

Ada

This example implements the pseudocode in the reference Wiki page. The pseudocode states that the index values for the array to multiply begin at 0 while the cost and order matrices employ index values beginning at 1. Ada supports this pseudocode directly because Ada allows the programmer to define the index range for any array type. This Ada example is implemented using a simple package and a main procedure. The package specification is: <lang ada> package mat_chain is 

  type Vector is array (Natural range <>) of Integer;
  procedure Chain_Multiplication (Dims : Vector);

end mat_chain; </lang> The implementation or body of the package is: <lang ada> with Ada.Text_IO; use Ada.Text_IO; with Ada.Strings.Unbounded; use Ada.Strings.Unbounded;

package body mat_chain is 

  type Result_Matrix is
    array (Positive range <>, Positive range <>) of Integer;

  --------------------------
  -- Chain_Multiplication --
  --------------------------

  procedure Chain_Multiplication (Dims : Vector) is
     n : Natural := Dims'Length - 1;
     S : Result_Matrix (1 .. n, 1 .. n);
     m : Result_Matrix (1 .. n, 1 .. n);
     procedure Print (Item : Vector) is
     begin
        Put ("Array Dimension  = (");
        for I in Item'Range loop
           Put (Item (I)'Image);
           if I < Item'Last then
              Put (",");
           else
              Put (")");
           end if;
        end loop;
        New_Line;
     end Print;

     procedure Chain_Order (Item : Vector) is
        J    : Natural;
        Cost : Natural;
        Temp : Natural;

     begin
        for idx in 1 .. n loop
           m (idx, idx) := 0;
        end loop;

        for Len in 2 .. n loop
           for I in 1 .. n - Len + 1 loop
              J        := I + Len - 1;
              m (I, J) := Integer'Last;
              for K in I .. J - 1 loop
                 Temp := Item (I - 1) * Item (K) * Item (J);
                 Cost := m (I, K) + m (K + 1, J) + Temp;
                 if Cost < m (I, J) then
                    m (I, J) := Cost;
                    S (I, J) := K;
                 end if;
              end loop;
           end loop;
        end loop;
     end Chain_Order;

     function Optimal_Parens return String is
        function Construct
          (S : Result_Matrix; I : Natural; J : Natural)
           return Unbounded_String
        is
           Us         : Unbounded_String := Null_Unbounded_String;
           Char_Order : Character;
        begin
           if I = J then
              Char_Order := Character'Val (I + 64);
              Append (Source => Us, New_Item => Char_Order);
              return Us;
           else
              Append (Source => Us, New_Item => '(');
              Append (Source => Us, New_Item => Construct (S, I, S (I, J)));
              Append (Source => Us, New_Item => '*');
              Append
                (Source => Us, New_Item => Construct (S, S (I, J) + 1, J));
              Append (Source => Us, New_Item => ')');
              return Us;
           end if;
        end Construct;

     begin
        return To_String (Construct (S, 1, n));

     end Optimal_Parens;

  begin
     Chain_Order (Dims);
     Print (Dims);
     Put_Line ("Cost             = " & Integer'Image (m (1, n)));
     Put_Line ("Optimal Multiply = " & Optimal_Parens);
  end Chain_Multiplication;

end mat_chain; </lang> The main procedure is: <lang ada> with Mat_Chain; use Mat_Chain; with Ada.Text_IO; use Ada.Text_IO;

procedure chain_main is 

  V1 : Vector := (5, 6, 3, 1);
  V2 : Vector := (1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2);
  V3 : Vector := (1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10);

begin 

  Chain_Multiplication(V1);
  New_Line;
  Chain_Multiplication(V2);
  New_Line;
  Chain_Multiplication(V3);

end chain_main; </lang>

Output:
Array Dimension  = ( 5, 6, 3, 1)
Cost             =  48
Optimal Multiply = (A*(B*C))

Array Dimension  = ( 1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2)
Cost             =  38120
Optimal Multiply = ((((((((A*B)*C)*D)*E)*F)*G)*(H*(I*J)))*(K*L))

Array Dimension  = ( 1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10)
Cost             =  1773740
Optimal Multiply = (A*((((((B*C)*D)*(((E*F)*G)*H))*I)*J)*K))

C

Translation of: Kotlin

<lang c>#include <stdio.h>

  1. include <limits.h>
  2. include <stdlib.h>

int **m; int **s;

void optimal_matrix_chain_order(int *dims, int n) { 

   int len, i, j, k, temp, cost;
   n--;
   m = (int **)malloc(n * sizeof(int *));
   for (i = 0; i < n; ++i) {
       m[i] = (int *)calloc(n, sizeof(int));
   }

   s = (int **)malloc(n * sizeof(int *));
   for (i = 0; i < n; ++i) {
       s[i] = (int *)calloc(n, sizeof(int));
   }

   for (len = 1; len < n; ++len) {
       for (i = 0; i < n - len; ++i) {
           j = i + len;
           m[i][j] = INT_MAX;
           for (k = i; k < j; ++k) {
               temp = dims[i] * dims[k + 1] * dims[j + 1];
               cost = m[i][k] + m[k + 1][j] + temp;
               if (cost < m[i][j]) {
                   m[i][j] = cost;
                   s[i][j] = k;
               }
           }
       }
   }

}

void print_optimal_chain_order(int i, int j) { 

   if (i == j)
       printf("%c", i + 65);
   else {
       printf("(");
       print_optimal_chain_order(i, s[i][j]);
       print_optimal_chain_order(s[i][j] + 1, j);
       printf(")");
   }

}

int main() { 

   int i, j, n;
   int a1[4]  = {5, 6, 3, 1};
   int a2[13] = {1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2};
   int a3[12] = {1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10};
   int *dims_list[3] = {a1, a2, a3};
   int sizes[3] = {4, 13, 12};
   for (i = 0; i < 3; ++i) {
       printf("Dims  : [");
       n = sizes[i];
       for (j = 0; j < n; ++j) {
           printf("%d", dims_list[i][j]);
           if (j < n - 1) printf(", "); else printf("]\n");
       }        
       optimal_matrix_chain_order(dims_list[i], n);
       printf("Order : ");
       print_optimal_chain_order(0, n - 2);
       printf("\nCost  : %d\n\n", m[0][n - 2]);
       for (j = 0; j <= n - 2; ++j) free(m[j]);
       free(m);
       for (j = 0; j <= n - 2; ++j) free(s[j]);
       free(s);
   }
   return 0;

}</lang>

Output:
Dims  : [5, 6, 3, 1]
Order : (A(BC))
Cost  : 48

Dims  : [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
Order : ((((((((AB)C)D)E)F)G)(H(IJ)))(KL))
Cost  : 38120

Dims  : [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
Order : (A((((((BC)D)(((EF)G)H))I)J)K))
Cost  : 1773740

C#

Translation of: Kotlin

<lang csharp>using System;

class MatrixChainOrderOptimizer { 

   private int[,] m;
   private int[,] s;

   void OptimalMatrixChainOrder(int[] dims) {
       int n = dims.Length - 1;
       m = new int[n, n];
       s = new int[n, n];
       for (int len = 1; len < n; ++len) {
           for (int i = 0; i < n - len; ++i) {
               int j = i + len;
               m[i, j] = Int32.MaxValue;
               for (int k = i; k < j; ++k) {
                   int temp = dims[i] * dims[k + 1] * dims[j + 1];
                   int cost = m[i, k] + m[k + 1, j] + temp;
                   if (cost < m[i, j]) {
                       m[i, j] = cost;
                       s[i, j] = k;
                   }
               }
           }
       }
   }

   void PrintOptimalChainOrder(int i, int j) {
       if (i == j)
           Console.Write((char)(i + 65));
       else {
           Console.Write("(");
           PrintOptimalChainOrder(i, s[i, j]);
           PrintOptimalChainOrder(s[i, j] + 1, j);
           Console.Write(")");
       }
   }

   static void Main() {
       var mcoo = new MatrixChainOrderOptimizer();
       var dimsList = new int[3][];
       dimsList[0] = new int[4] {5, 6, 3, 1};
       dimsList[1] = new int[13] {1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2};
       dimsList[2] = new int[12] {1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10};
       for (int i = 0; i < dimsList.Length; ++i) {
           Console.Write("Dims  : [");
           int n = dimsList[i].Length;
           for (int j = 0; j < n; ++j) {
               Console.Write(dimsList[i][j]);
               if (j < n - 1)
                   Console.Write(", ");
               else
                   Console.WriteLine("]");
           }
           mcoo.OptimalMatrixChainOrder(dimsList[i]);
           Console.Write("Order : ");
           mcoo.PrintOptimalChainOrder(0, n - 2);
           Console.WriteLine("\nCost  : {0}\n",  mcoo.m[0, n - 2]);
       }
   }

}</lang>

Output:
Dims  : [5, 6, 3, 1]
Order : (A(BC))
Cost  : 48

Dims  : [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
Order : ((((((((AB)C)D)E)F)G)(H(IJ)))(KL))
Cost  : 38120

Dims  : [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
Order : (A((((((BC)D)(((EF)G)H))I)J)K))
Cost  : 1773740

Fortran

Translation of: Python

This is a translation of the Python iterative solution.

<lang fortran>module optim_mod 

   implicit none

contains 

   subroutine optim(a)
       implicit none
       integer :: a(:), n, i, j, k
       integer, allocatable :: u(:, :)
       integer(8) :: c
       integer(8), allocatable :: v(:, :)

       n = ubound(a, 1) - 1
       allocate (u(n, n), v(n, n))
       v = huge(v)
       u(:, 1) = -1
       v(:, 1) = 0
       do j = 2, n
           do i = 1, n - j + 1
               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)
                   if (c < v(i, j)) then
                       u(i, j) = k
                       v(i, j) = c
                   end if
               end do
           end do
       end do
       write (*, "(I0,' ')", advance="no") v(1, n)
       call aux(1, n)
       print *
       deallocate (u, v)
   contains
       recursive subroutine aux(i, j)
           integer :: i, j, k

           k = u(i, j)
           if (k < 0) then
               write (*, "(I0)", advance="no") i
           else
               write (*, "('(')", advance="no")
               call aux(i, k)
               write (*, "('*')", advance="no")
               call aux(i + k, j - k)
               write (*, "(')')", advance="no")
           end if
       end subroutine
   end subroutine

end module

program matmulchain 

   use optim_mod
   implicit none

   call optim([5, 6, 3, 1])
   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])

end program</lang>

Output 

48 (1*(2*3))
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))

Go

The first for loop is based on the pseudo and Java code from the Wikipedia article. <lang Go>package main

import "fmt"

// PrintMatrixChainOrder prints the optimal order for chain // multiplying matrices. // Matrix A[i] has dimensions dims[i-1]×dims[i]. func PrintMatrixChainOrder(dims []int) { n := len(dims) - 1 m, s := newSquareMatrices(n)

// m[i,j] will be minimum number of scalar multiplactions // needed to compute the matrix A[i]A[i+1]…A[j] = A[i…j]. // Note, m[i,i] = zero (no cost). // s[i,j] will be the index of the subsequence split that // achieved minimal cost. for lenMinusOne := 1; lenMinusOne < n; lenMinusOne++ { for i := 0; i < n-lenMinusOne; i++ { j := i + lenMinusOne m[i][j] = -1 for k := i; k < j; k++ { cost := m[i][k] + m[k+1][j] + dims[i]*dims[k+1]*dims[j+1] if m[i][j] < 0 || cost < m[i][j] { m[i][j] = cost s[i][j] = k } } } }

// Format and print result. const MatrixNames = "ABCDEFGHIJKLMNOPQRSTUVWXYZ" var subprint func(int, int) subprint = func(i, j int) { if i == j { return } k := s[i][j] subprint(i, k) subprint(k+1, j) fmt.Printf("%*s -> %s × %s%*scost=%d\n", n, MatrixNames[i:j+1], MatrixNames[i:k+1], MatrixNames[k+1:j+1], n+i-j, "", m[i][j], ) } subprint(0, n-1) }

func newSquareMatrices(n int) (m, s [][]int) { // Allocates two n×n matrices as slices of slices but // using only one [2n][]int and one [2n²]int backing array. m = make([][]int, 2*n) m, s = m[:n:n], m[n:] tmp := make([]int, 2*n*n) for i := range m { m[i], tmp = tmp[:n:n], tmp[n:] } for i := range s { s[i], tmp = tmp[:n:n], tmp[n:] } return m, s }

func main() { cases := [...][]int{ {1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2}, {1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10}, } for _, tc := range cases { fmt.Println("Dimensions:", tc) PrintMatrixChainOrder(tc) fmt.Println() } }</lang>

Output:
Dimensions: [1 5 25 30 100 70 2 1 100 250 1 1000 2]
          AB -> A × B           cost=125
         ABC -> AB × C          cost=875
        ABCD -> ABC × D         cost=3875
       ABCDE -> ABCD × E        cost=10875
      ABCDEF -> ABCDE × F       cost=11015
     ABCDEFG -> ABCDEF × G      cost=11017
          IJ -> I × J           cost=25000
         HIJ -> H × IJ          cost=25100
  ABCDEFGHIJ -> ABCDEFG × HIJ   cost=36118
          KL -> K × L           cost=2000
ABCDEFGHIJKL -> ABCDEFGHIJ × KL cost=38120

Dimensions: [1000 1 500 12 1 700 2500 3 2 5 14 10]
         BC -> B × C          cost=6000
        BCD -> BC × D         cost=6012
         EF -> E × F          cost=1750000
        EFG -> EF × G         cost=1757500
       EFGH -> EFG × H        cost=1757506
    BCDEFGH -> BCD × EFGH     cost=1763520
   BCDEFGHI -> BCDEFGH × I    cost=1763530
  BCDEFGHIJ -> BCDEFGHI × J   cost=1763600
 BCDEFGHIJK -> BCDEFGHIJ × K  cost=1763740
ABCDEFGHIJK -> A × BCDEFGHIJK cost=1773740

Haskell

<lang Haskell>import Data.List (elemIndex) import Data.Char (chr, ord) import Data.Maybe (fromJust)

mats :: Int mats = 

 [ [5, 6, 3, 1]
 , [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
 , [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
 ]

cost :: [Int] -> Int -> Int -> (Int, Int) cost a i j 

 | i < j =
   let m =
         [ fst (cost a i k) + fst (cost a (k + 1) j) +
          (a !! i) * (a !! (j + 1)) * (a !! (k + 1))
         | k <- [i .. j - 1] ]
       mm = minimum m
   in (mm, fromJust (elemIndex mm m) + i)
 | otherwise = (0, -1)

optimalOrder :: [Int] -> Int -> Int -> String optimalOrder a i j 

 | i < j =
   let c = cost a i j
   in "(" ++ optimalOrder a i (snd c) ++ optimalOrder a (snd c + 1) j ++ ")"
 | otherwise = [chr ((+ i) $ ord 'a')]

printBlock :: [Int] -> IO () printBlock v = 

 let c = cost v 0 (length v - 2)
 in putStrLn
      ("for " ++
       show v ++
       " we have " ++
       show (fst c) ++
       " possibilities, z.B " ++ optimalOrder v 0 (length v - 2))

main :: IO () main = mapM_ printBlock mats</lang>

Output:
for [5,6,3,1] we have 48 possibilities, z.B (a(bc))
for [1,5,25,30,100,70,2,1,100,250,1,1000,2] we have 38120 possibilities, z.B ((((((((ab)c)d)e)f)g)(h(ij)))(kl))
for [1000,1,500,12,1,700,2500,3,2,5,14,10] we have 1773740 possibilities, z.B (a((((((bc)d)(((ef)g)h))i)j)k)

J

This is no more than a mindless transliteration of the Wikipedia Java code (for moo; for pooc, the author found Go to have the clearest expression for transliteration).

Given J's incredible strengths with arrays and matrices, the author is certain there is a much more succinct and idiomatic approach available, but hasn't spent the time understanding how the Wikipedia algorithm works, so hasn't made an attempt at a more native J solution. Others on RC are welcome and invited to do so.

<lang j>moo =: verb define 

 s =. m =. 0 $~ ,~ n=._1+#y
 for_lmo. 1+i.<:n do.
   for_i. i. n-lmo do.
     j =. i + lmo
     m =. _ (<i;j)} m
     for_k. i+i.j-i do.
       cost =. ((<i;k){m) + ((<(k+1);j){m) + */ y {~ i,(k+1),(j+1)
       if. cost < ((<i;j){m) do.
         m =. cost (<i;j)} m
         s =. k (<i;j)} s
       end.
     end.
   end.
 end.

 m;s

)

poco =: dyad define 

 'i j' =. y
  if. i=j do.
    a. {~ 65 + i      NB. 65 = a.i.'A'
  else.
    k =. x {~ <y      NB. y = i,j
    '(' , (x poco i,k) , (x poco j ,~ 1+k) , ')'
  end.

)

optMM =: verb define 

 'M S' =. moo y
 smoutput 'Cost: ' , ": x: M {~ <0;_1
 smoutput 'Order: ', S poco 0 , <:#M

)</lang>

Output:

<lang j> optMM 5 6 3 1 Cost: 48 Order: (A(BC)) 

  optMM 1 5 25 30 100 70 2 1 100 250 1 1000 2

Cost: 38120 Order: ((((((((AB)C)D)E)F)G)(H(IJ)))(KL)) 

  optMM 1000 1 500 12 1 700 2500 3 2 5 14 10

Cost: 1773740 Order: (A((((((BC)D)(((EF)G)H))I)J)K))</lang>

Java

Thanks to the Wikipedia page for a working Java implementation. <lang java> import java.util.Arrays;

public class MatrixChainMultiplication { 

   public static void main(String[] args) {
       runMatrixChainMultiplication(new int[] {5, 6, 3, 1});
       runMatrixChainMultiplication(new int[] {1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2});
       runMatrixChainMultiplication(new int[] {1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10});
   }
   
   private static void runMatrixChainMultiplication(int[] dims) {
       System.out.printf("Array Dimension  = %s%n", Arrays.toString(dims));
       System.out.printf("Cost             = %d%n", matrixChainOrder(dims));
       System.out.printf("Optimal Multiply = %s%n%n", getOptimalParenthesizations());
   }

   private static int[][]cost;
   private static int[][]order;
   
   public static int matrixChainOrder(int[] dims) {
       int n = dims.length - 1;
       cost = new int[n][n];
       order = new int[n][n];

       for (int lenMinusOne = 1 ; lenMinusOne < n ; lenMinusOne++) {
           for (int i = 0; i < n - lenMinusOne; i++) {
               int j = i + lenMinusOne;
               cost[i][j] = Integer.MAX_VALUE;
               for (int k = i; k < j; k++) {
                   int currentCost = cost[i][k] + cost[k+1][j] + dims[i]*dims[k+1]*dims[j+1];
                   if (currentCost < cost[i][j]) {
                       cost[i][j] = currentCost;
                       order[i][j] = k;
                   }
               }
           }
       }
       return cost[0][n-1];
   }

   private static String getOptimalParenthesizations() {
       return getOptimalParenthesizations(order, 0, order.length - 1);
   }
   
   private static String getOptimalParenthesizations(int[][]s, int i, int j) {
       if (i == j) {
           return String.format("%c", i+65);
       }
       else {
           StringBuilder sb = new StringBuilder();
           sb.append("(");
           sb.append(getOptimalParenthesizations(s, i, s[i][j]));
           sb.append(" * ");
           sb.append(getOptimalParenthesizations(s, s[i][j] + 1, j));
           sb.append(")");
           return sb.toString();
       }
   }

} </lang>

Output:
Array Dimension  = [5, 6, 3, 1]
Cost             = 48
Optimal Multiply = (A * (B * C))

Array Dimension  = [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
Cost             = 38120
Optimal Multiply = ((((((((A * B) * C) * D) * E) * F) * G) * (H * (I * J))) * (K * L))

Array Dimension  = [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
Cost             = 1773740
Optimal Multiply = (A * ((((((B * C) * D) * (((E * F) * G) * H)) * I) * J) * K))

Julia

Works with: Julia version 0.6
Translation of: MATLAB

Module: <lang julia>module MatrixChainMultiplications

using OffsetArrays

function optim(a) 

   n = length(a) - 1
   u = fill!(OffsetArray{Int}(0:n, 0:n), 0)
   v = fill!(OffsetArray{Int}(0:n, 0:n), typemax(Int))
   u[:, 1] .= -1
   v[:, 1] .= 0
   for j in 2:n, i in 1:n-j+1, k in 1:j-1
       c = v[i, k] + v[i+k, j-k] + a[i] * a[i+k] * a[i+j]
       if c < v[i, j]
           u[i, j] = k
           v[i, j] = c
       end
   end
   return v[1, n], aux(u, 1, n)

end

function aux(u, i, j) 

   k = u[i, j]
   if k < 0
       return sprint(print, i)
   else
       return sprint(print, '(', aux(u, i, k), '×', aux(u, i + k, j - k), ")")
   end

end

end # module MatrixChainMultiplications</lang>

Main: <lang julia>println(MatrixChainMultiplications.optim([1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2])) println(MatrixChainMultiplications.optim([1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]))</lang>

Output:
(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))")

Kotlin

This is based on the pseudo-code in the Wikipedia article. <lang scala>// Version 1.2.31

lateinit var m: List<IntArray> lateinit var s: List<IntArray>

fun optimalMatrixChainOrder(dims: IntArray) { 

   val n = dims.size - 1
   m = List(n) { IntArray(n) }
   s = List(n) { IntArray(n) }
   for (len in 1 until n) {
       for (i in 0 until n - len) {
           val j = i + len
           m[i][j] = Int.MAX_VALUE
           for (k in i until j) {
               val temp = dims[i] * dims [k + 1] * dims[j + 1]
               val cost = m[i][k] + m[k + 1][j] + temp
               if (cost < m[i][j]) {
                   m[i][j] = cost
                   s[i][j] = k
               }
           }
       }
   }

}

fun printOptimalChainOrder(i: Int, j: Int) { 

   if (i == j)
       print("${(i + 65).toChar()}")
   else {
       print("(")
       printOptimalChainOrder(i, s[i][j])
       printOptimalChainOrder(s[i][j] + 1, j)
       print(")")
   }

}

fun main(args: Array<String>) { 

   val dimsList = listOf(
       intArrayOf(5, 6, 3, 1),
       intArrayOf(1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2),
       intArrayOf(1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10)
   )
   for (dims in dimsList) {
       println("Dims  : ${dims.asList()}")
       optimalMatrixChainOrder(dims)
       print("Order : ")
       printOptimalChainOrder(0, s.size - 1)
       println("\nCost  : ${m[0][s.size - 1]}\n")
   }

}</lang>

Output:
Dims  : [5, 6, 3, 1]
Order : (A(BC))
Cost  : 48

Dims  : [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
Order : ((((((((AB)C)D)E)F)G)(H(IJ)))(KL))
Cost  : 38120

Dims  : [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
Order : (A((((((BC)D)(((EF)G)H))I)J)K))
Cost  : 1773740

Lua

<lang lua>-- Matrix A[i] has dimension dims[i-1] x dims[i] for i = 1..n local function MatrixChainOrder(dims) 

   local m = {}
   local s = {}
   local n = #dims - 1;
   -- m[i,j] = Minimum number of scalar multiplications (i.e., cost)
   -- needed to compute the matrix A[i]A[i+1]...A[j] = A[i..j]
   -- The cost is zero when multiplying one matrix
   for i = 1,n do
       m[i] = {}
       m[i][i] = 0
       s[i] = {}
   end

   for len = 2,n do -- Subsequence lengths
       for i = 1,(n - len + 1) do
           local j = i + len - 1
           m[i][j] = math.maxinteger
           for k = i,(j - 1) do
               local cost = m[i][k] + m[k+1][j] + dims[i]*dims[k+1]*dims[j+1];
               if (cost < m[i][j]) then
                   m[i][j] = cost;
                   s[i][j] = k; --Index of the subsequence split that achieved minimal cost
               end
           end
       end
   end
   return m,s

end

local function printOptimalChainOrder(s) 

   local function find_path(start,finish)
       local chainOrder = ""
       if (start == finish) then
           chainOrder = chainOrder .."A"..start
       else
           chainOrder = chainOrder .."(" .. 
                        find_path(start,s[start][finish]) ..
                        find_path(s[start][finish]+1,finish) .. ")"
       end
       return chainOrder
   end
   print("Order : "..find_path(1,#s))

end

local dimsList = {{5, 6, 3, 1},{1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2},{1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10}}

for k,dim in ipairs(dimsList) do 

   io.write("Dims  : [")
   for v=1,(#dim-1) do
       io.write(dim[v]..", ")
   end
   print(dim[#dim].."]")
   local m,s = MatrixChainOrder(dim)
   printOptimalChainOrder(s)
   print("Cost  : "..tostring(m[1][#s]).."\n")

end</lang>

Output:
Dims  : [5, 6, 3, 1]
Order : (A1(A2A3))
Cost  : 48

Dims  : [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
Order : ((((((((A1A2)A3)A4)A5)A6)A7)(A8(A9A10)))(A11A12))
Cost  : 38120

Dims  : [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
Order : (A1((((((A2A3)A4)(((A5A6)A7)A8))A9)A10)A11))
Cost  : 1773740

Mathematica / Wolfram Language

Translation of: Fortran

<lang Mathematica>ClearAll[optim, aux] optim[a_List] := Module[{u, v, n, c, r, s}, 

 n = Length[a] - 1;
 u = ConstantArray[0, {n, n}];
 v = ConstantArray[\[Infinity], {n, n}];
 uAll, 1 = -1;
 vAll, 1 = 0;
 Do[
  Do[
   Do[
    c = 
     vi, k + vi + k, j - k + ai ai + k ai + j;
     If[c < vi, j,
      ui, j = k;
      vi, j = c;
     ]
    ,
    {k, 1, j - 1}
    ]
   ,
   {i, 1, n - j + 1}
   ]
  ,
  {j, 2, n}
  ];
 r = v1, n;
 s = aux[u, 1, n];
 {r, s}
 ]

aux[u_, i_, j_] := Module[{k}, 

 k = ui, j;
 If[k < 0,
  i
  ,
  Inactive[Times][aux[u, i, k], aux[u, i + k, j - k]]
  ]
 ]

{r, s} = optim[{1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2}]; r s {r, s} = optim[{1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10}]; r s</lang>

Output:
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)


MATLAB

Translation of: Fortran

<lang matlab>function [r,s] = optim(a) 

   n = length(a)-1;
   u = zeros(n,n);
   v = ones(n,n)*inf;
   u(:,1) = -1;
   v(:,1) = 0;
   for j = 2:n
       for i = 1:n-j+1
           for k = 1:j-1
               c = v(i,k)+v(i+k,j-k)+a(i)*a(i+k)*a(i+j);
               if c<v(i,j)
                   u(i,j) = k;
                   v(i,j) = c;
               end
           end
       end
   end
   r = v(1,n);
   s = aux(u,1,n);

end

function s = aux(u,i,j) 

   k = u(i,j);
   if k<0
       s = sprintf("%d",i);
   else
       s = sprintf("(%s*%s)",aux(u,i,k),aux(u,i+k,j-k));
   end

end</lang>

Output:

<lang matlab>[r,s] = optim([1,5,25,30,100,70,2,1,100,250,1,1000,2])

r = 

      38120


s = 

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


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

r = 

    1773740


s = 

   "(1*((((((2*3)*4)*(((5*6)*7)*8))*9)*10)*11))"</lang>

Nim

Translation of: Kotlin

<lang Nim>import sequtils

type Optimizer = object 

 dims: seq[int]
 m: seq[seq[Natural]]
 s: seq[seq[Natural]]


proc initOptimizer(dims: openArray[int]): Optimizer = 

 ## Create an optimizer for the given dimensions.
 Optimizer(dims: @dims)

proc findMatrixChainOrder(opt: var Optimizer) = 

 ## Find the best order for matrix chain multiplication.

 let n = opt.dims.high
 opt.m = newSeqWith(n, newSeq[Natural](n))
 opt.s = newSeqWith(n, newSeq[Natural](n))

 for lg in 1..<n:
   for i in 0..<(n - lg):
     let j = i + lg
     opt.m[i][j] = Natural.high
     for k in i..<j:
       let cost = opt.m[i][k] + opt.m[k+1][j] + opt.dims[i] * opt.dims[k+1] * opt.dims[j+1]
       if cost < opt.m[i][j]:
         opt.m[i][j] = cost
         opt.s[i][j] = k


proc optimalChainOrder(opt: Optimizer; i, j: Natural): string = 

 ## Return the optimal chain order as a string.
 if i == j:
   result.add chr(i + ord('A'))
 else:
   result.add '('
   result.add opt.optimalChainOrder(i, opt.s[i][j])
   result.add opt.optimalChainOrder(opt.s[i][j] + 1, j)
   result.add ')'


when isMainModule: 

 const
   Dims1 = @[5, 6, 3, 1]
   Dims2 = @[1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
   Dims3 = @[1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]

 for dims in [Dims1, Dims2, Dims3]:
   var opt = initOptimizer(dims)
   opt.findMatrixChainOrder()
   echo "Dims:  ", dims
   echo "Order: ", opt.optimalChainOrder(0, dims.len - 2)
   echo "Cost:  ", opt.m[0][dims.len - 2]
   echo ""</lang>
Output:
Dims:  @[5, 6, 3, 1]
Order: (A(BC))
Cost:  48

Dims:  @[1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
Order: ((((((((AB)C)D)E)F)G)(H(IJ)))(KL))
Cost:  38120

Dims:  @[1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
Order: (A((((((BC)D)(((EF)G)H))I)J)K))
Cost:  1773740

Perl

Translation of: Raku

<lang perl>use strict; use feature 'say';

sub matrix_mult_chaining { 

   my(@dimensions) = @_;
   my(@cp,@path);

   # a matrix never needs to be multiplied with itself, so it has cost 0
   $cp[$_][$_] = 0 for keys @dimensions;

   my $n = $#dimensions;
   for my $chain_length (1..$n) {
       for my $start (0 .. $n - $chain_length - 1) {
           my $end = $start + $chain_length;
           $cp[$end][$start] = 10e10;
           for my $step ($start .. $end - 1) {
               my $new_cost = $cp[$step][$start]
                            + $cp[$end][$step + 1]
                            + $dimensions[$start] * $dimensions[$step+1] * $dimensions[$end+1];
               if ($new_cost < $cp[$end][$start]) {
                   $cp[$end][$start] = $new_cost; # cost
                   $cp[$start][$end] = $step;     # path
               }
           }
      }
   }

   $cp[$n-1][0] . ' ' . find_path(0, $n-1, @cp);

}

sub find_path { 

   my($start,$end,@cp) = @_;
   my $result;

   if ($start == $end) {
       $result .= 'A' . ($start + 1);
   } else {
       $result .= '(' .
                  find_path($start, $cp[$start][$end], @cp) .
                  find_path($cp[$start][$end] + 1, $end, @cp) .
                  ')';
   }
   return $result;

}

say matrix_mult_chaining(<1 5 25 30 100 70 2 1 100 250 1 1000 2>); say matrix_mult_chaining(<1000 1 500 12 1 700 2500 3 2 5 14 10>);</lang>

Output:
38120 ((((((((A1A2)A3)A4)A5)A6)A7)(A8(A9A10)))(A11A12))
1773740 (A1((((((A2A3)A4)(((A5A6)A7)A8))A9)A10)A11))

Phix

As per the wp pseudocode <lang Phix>function optimal_chain_order(int i, int j, sequence s) 

   if i==j then
       return i+'A'-1
   end if
   return "("&optimal_chain_order(i,s[i,j],s)
             &optimal_chain_order(s[i,j]+1,j,s)&")"

end function

function optimal_matrix_chain_order(sequence dims) 

   integer n = length(dims)-1
   sequence {m,s} @= repeat(repeat(0,n),n)
   for len=2 to n do
       for i=1 to n-len+1 do
           integer j = i+len-1
           m[i][j] = -1
           for k=i to j-1 do
               atom cost := m[i][k] + m[k+1][j] + dims[i]*dims[k+1]*dims[j+1]
               if m[i][j]<0
               or cost<m[i][j] then
                   m[i][j] = cost;
                   s[i][j] = k;
               end if
           end for
       end for
   end for
   return {optimal_chain_order(1,n,s),m[1,n]}

end function

constant tests = {{5, 6, 3, 1}, 

                 {1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2},
                 {1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10}}

for i=1 to length(tests) do 

   sequence ti = tests[i]
   printf(1,"Dims  : %s\n",{sprint(ti)})
   printf(1,"Order : %s\nCost  : %d\n",optimal_matrix_chain_order(ti))

end for</lang>

Output:
Dims  : {5,6,3,1}
Order : (A(BC))
Cost  : 48
Dims  : {1,5,25,30,100,70,2,1,100,250,1,1000,2}
Order : ((((((((AB)C)D)E)F)G)(H(IJ)))(KL))
Cost  : 38120
Dims  : {1000,1,500,12,1,700,2500,3,2,5,14,10}
Order : (A((((((BC)D)(((EF)G)H))I)J)K))
Cost  : 1773740

Python

We will solve the task in three steps:

1) Enumerate all ways to parenthesize (using a generator to save space), and for each one compute the cost. Then simply look up the minimal cost.

2) Merge the enumeration and the cost function in a recursive cost optimizing function. The computation is roughly the same, but it's much faster as some steps are removed.

3) The recursive solution has many duplicates computations. Memoize the previous function: this yields a dynamic programming approach.

Enumeration of parenthesizations

<lang python>def parens(n): 

   def aux(n, k):
       if n == 1:
           yield k
       elif n == 2:
           yield [k, k + 1]
       else:
           a = []
           for i in range(1, n):
               for u in aux(i, k):
                   for v in aux(n - i, k + i):
                       yield [u, v]
   yield from aux(n, 0)</lang>

Example (in the same order as in the task description)

<lang python>for u in parens(4): 

   print(u)

[0, [1, [2, 3]]] [0, [[1, 2], 3]] [[0, 1], [2, 3]] [[0, [1, 2]], 3] [[[0, 1], 2], 3]</lang>

And here is the optimization step:

<lang python>def optim1(a): 

   def cost(k):
       if type(k) is int:
           return 0, a[k], a[k + 1]
       else:
           s1, p1, q1 = cost(k[0])
           s2, p2, q2 = cost(k[1])
           assert q1 == p2
           return s1 + s2 + p1 * q1 * q2, p1, q2
   cmin = None
   n = len(a) - 1
   for u in parens(n):
       c, p, q = cost(u)
       if cmin is None or c < cmin:
           cmin = c
           umin = u
   return cmin, umin</lang>

Recursive cost optimization

The previous function optim1 already used recursion, but only to compute the cost of a given parens configuration, whereas another function (a generator actually) provides these configurations. Here we will do both recursively in the same function, avoiding the computation of configurations altogether.

<lang python>def optim2(a): 

   def aux(n, k):
       if n == 1:
           p, q = a[k:k + 2]
           return 0, p, q, k
       elif n == 2:
           p, q, r = a[k:k + 3]
           return p * q * r, p, r, [k, k + 1]
       else:
           m = None
           p = a[k]
           q = a[k + n]
           for i in range(1, n):
               s1, p1, q1, u1 = aux(i, k)
               s2, p2, q2, u2 = aux(n - i, k + i)
               assert q1 == p2
               s = s1 + s2 + p1 * q1 * q2
               if m is None or s < m:
                   m = s
                   u = [u1, u2]
           return m, p, q, u
   s, p, q, u = aux(len(a) - 1, 0)
   return s, u</lang>

Memoized recursive call

The only difference between optim2 and optim3 is the @memoize decorator. Yet the algorithm is way faster with this. According to Wikipedia, the complexity falls from O(2^n) to O(n^3). This is confirmed by plotting log(time) vs log(n) for n up to 580 (this needs changing Python's recursion limit).

<lang python>def memoize(f): 

   h = {}
   def g(*u):
       if u in h:
           return h[u]
       else:
           r = f(*u)
           h[u] = r
           return r
   return g

def optim3(a): 

   @memoize
   def aux(n, k):
       if n == 1:
           p, q = a[k:k + 2]
           return 0, p, q, k
       elif n == 2:
           p, q, r = a[k:k + 3]
           return p * q * r, p, r, [k, k + 1]
       else:
           m = None
           p = a[k]
           q = a[k + n]
           for i in range(1, n):
               s1, p1, q1, u1 = aux(i, k)
               s2, p2, q2, u2 = aux(n - i, k + i)
               assert q1 == p2
               s = s1 + s2 + p1 * q1 * q2
               if m is None or s < m:
                   m = s
                   u = [u1, u2]
           return m, p, q, u
   s, p, q, u = aux(len(a) - 1, 0)
   return s, u</lang>

Putting all together

<lang python>import time

u = [[1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2], 

    [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]]

for a in u: 

   print(a)
   print()
   print("function     time       cost   parens  ")
   print("-" * 90)
   for f in [optim1, optim2, optim3]:
       t1 = time.clock()
       s, u = f(a)
       t2 = time.clock()
       print("%s %10.3f %10d   %s" % (f.__name__, 1000 * (t2 - t1), s, u))
   print()</lang>

Output (timings are in milliseconds) 

[1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]

function     time       cost   parens
------------------------------------------------------------------------------------------
optim1    838.636      38120   [[[[[[[[0, 1], 2], 3], 4], 5], 6], [7, [8, 9]]], [10, 11]]
optim2     80.628      38120   [[[[[[[[0, 1], 2], 3], 4], 5], 6], [7, [8, 9]]], [10, 11]]
optim3      0.373      38120   [[[[[[[[0, 1], 2], 3], 4], 5], 6], [7, [8, 9]]], [10, 11]]

[1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]

function     time       cost   parens
------------------------------------------------------------------------------------------
optim1    223.186    1773740   [0, [[[[[[1, 2], 3], [[[4, 5], 6], 7]], 8], 9], 10]]
optim2     27.660    1773740   [0, [[[[[[1, 2], 3], [[[4, 5], 6], 7]], 8], 9], 10]]
optim3      0.307    1773740   [0, [[[[[[1, 2], 3], [[[4, 5], 6], 7]], 8], 9], 10]]

A mean on 1000 loops to get a better precision on the optim3, yields respectively 0.365 ms and 0.287 ms.

Iterative solution

In the previous solution, memoization is done blindly with a dictionary. However, we need to compute the optimal products for all sublists. A sublist is described by its first index and length (resp. i and j+1 in the following function), hence the set of all sublists can be described by the indices of elements in a triangular array u. We first fill the "solution" (there is no product) for sublists of length 1 (u[0]), then for each successive length we optimize using what when know about smaller sublists. Instead of keeping track of the optimal solutions, the single needed one is computed in the end.

<lang python>def optim4(a): 

   global u
   n = len(a) - 1
   u = [None] * n
   u[0] = None, 0 * n
   for j in range(1, n):
       v = [None] * (n - j)
       for i in range(n - j):
           m = None
           for k in range(j):
               s1, c1 = u[k][i]
               s2, c2 = u[j - k - 1][i + k + 1]
               c = c1 + c2 + a[i] * a[i + k + 1] * a[i + j + 1]
               if m is None or c < m:
                   s = k
                   m = c
           v[i] = [s, m]
       u[j] = v
   def aux(i, j):
       s, c = u[j][i]
       if s is None:
           return i
       else:
           return [aux(i, s), aux(i + s + 1, j - s - 1)]
   return u[n - 1][0][1], aux(0, n - 1)


print(optim4([1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2])) print(optim4([1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]))</lang>

Output 

(38120, [[[[[[[[0, 1], 2], 3], 4], 5], 6], [7, [8, 9]]], [10, 11]])
(1773740, [0, [[[[[[1, 2], 3], [[[4, 5], 6], 7]], 8], 9], 10]])

Racket

Memoization

<lang racket>#lang racket

(define (memoize f) 

 (define table (make-hash))
 (λ args (hash-ref! table args (thunk (apply f args)))))

(struct $ (cost expl)) (define @ vector-ref)

(define (+: #:combine [combine (thunk* #f)] . xs) 

 ($ (apply + (map $-cost xs)) (apply combine (map $-expl xs))))

(define (min: . xs) (argmin $-cost xs))

(define (compute dims) 

 (define loop
   (memoize
    (λ (left right)
      (cond
        [(= 1 (- right left)) ($ 0 left)]
        [else (for/fold ([ans ($ +inf.0 #f)]) ([mid (in-range (add1 left) right)])
                (min: ans (+: (loop left mid) (loop mid right)
                              ($ (* (@ dims left) (@ dims mid) (@ dims right)) #f)
                              #:combine (λ (left-answer right-answer _)
                                          (list left-answer '× right-answer)))))]))))
 (loop 0 (sub1 (vector-length dims))))</lang>

Main

<lang racket>(define-syntax-rule (echo <x> ...) 

 (begin (printf "~a: ~a\n" (~a (quote <x>) #:min-width 12) <x>) ...))

(define (solve input) 

 (match-define-values ((list ($ cost explanation)) _ time _) (time-apply compute (list input)))
 (echo input time cost explanation)
 (newline))

(solve #(1 5 25 30 100 70 2 1 100 250 1 1000 2)) (solve #(1000 1 500 12 1 700 2500 3 2 5 14 10))</lang>

Output (timings are in milliseconds) 

input       : #(1 5 25 30 100 70 2 1 100 250 1 1000 2)
time        : 1
cost        : 38120
explanation : ((((((((0 × 1) × 2) × 3) × 4) × 5) × 6) × (7 × (8 × 9))) × (10 × 11))

input       : #(1000 1 500 12 1 700 2500 3 2 5 14 10)
time        : 0
cost        : 1773740
explanation : (0 × ((((((1 × 2) × 3) × (((4 × 5) × 6) × 7)) × 8) × 9) × 10))

Raku

(formerly Perl 6) This example is based on Moritz Lenz's code, written for Carl Mäsak's Perl 6 Coding Contest, in 2010. Slightly simplified, it fulfills the Rosetta Code task as well. <lang perl6>sub matrix-mult-chaining(@dimensions) { 

   my @cp;
   # @cp has a dual function: 
   # * the upper triangle of the diagonal matrix stores the cost (c) for
   #   multiplying matrices $i and $j in @cp[$j][$i], where $j > $i
   # * the lower triangle stores the path (p) that was used for the lowest cost
   #   multiplication to get from $i to $j.

   # a matrix never needs to be multiplied with itself, so it has cost 0
   @cp[$_][$_] = 0 for @dimensions.keys;
   my @path;

   my $n = @dimensions.end;
   for 1 .. $n -> $chain-length {
       for 0 .. $n - $chain-length - 1 -> $start {
           my $end = $start + $chain-length;
           @cp[$end][$start] = Inf;  # until we find a better connection
           for $start .. $end - 1 -> $step {
               my $new-cost = @cp[$step][$start]
                            + @cp[$end][$step + 1]
                            + [*] @dimensions[$start, $step+1, $end+1];
               if $new-cost < @cp[$end][$start] {
                   @cp[$end][$start] = $new-cost; # cost
                   @cp[$start][$end] = $step;     # path
               }
           }
      }
   }

   sub find-path(Int $start, Int $end) {
       if $start == $end {
           take 'A' ~ ($start + 1);
       } else {
           take '(';
           find-path($start, @cp[$start][$end]);
           find-path(@cp[$start][$end] + 1, $end);
           take ')';
       }
   }

  return @cp[$n-1][0], gather { find-path(0, $n - 1) }.join;

}

say matrix-mult-chaining(<1 5 25 30 100 70 2 1 100 250 1 1000 2>); say matrix-mult-chaining(<1000 1 500 12 1 700 2500 3 2 5 14 10>);</lang>

Output:
(38120 ((((((((A1A2)A3)A4)A5)A6)A7)(A8(A9A10)))(A11A12)))
(1773740 (A1((((((A2A3)A4)(((A5A6)A7)A8))A9)A10)A11)))

Rust

<lang rust>use std::collections::HashMap;

fn main() { 

   println!("{}\n", mcm_display(vec![5, 6, 3, 1]));
   println!(
       "{}\n",
       mcm_display(vec![1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2])
   );
   println!(
       "{}\n",
       mcm_display(vec![1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10])
   );

}

fn mcm_display(dims: Vec<i32>) -> String { 

   let mut costs: HashMap<Vec<i32>, (i32, Vec<usize>)> = HashMap::new();
   let mut line = format!("Dims : {:?}\n", dims);
   let ans = mcm(dims, &mut costs);
   let mut mats = (1..=ans.1.len() + 1)
       .map(|x| x.to_string())
       .collect::<Vec<String>>();
   for i in 0..ans.1.len() {
       let mat_taken = mats[ans.1[i]].clone();
       mats.remove(ans.1[i]);
       mats[ans.1[i]] = "(".to_string() + &mat_taken + "*" + &mats[ans.1[i]] + ")";
   }
   line += &format!("Order: {}\n", mats[0]);
   line += &format!("Cost : {}", ans.0);
   line

}

fn mcm(dims: Vec<i32>, costs: &mut HashMap<Vec<i32>, (i32, Vec<usize>)>) -> (i32, Vec<usize>) { 

   match costs.get(&dims) {
       Some(c) => c.clone(),
       None => {
           let ans = if dims.len() == 3 {
               (dims[0] * dims[1] * dims[2], vec![0])
           } else {
               let mut min_cost = std::i32::MAX;
               let mut min_path = Vec::new();
               for i in 1..dims.len() - 1 {
                   let taken = dims[(i - 1)..(i + 2)].to_vec();
                   let mut rest = dims[..i].to_vec();
                   rest.extend_from_slice(&dims[(i + 1)..]);
                   let a1 = mcm(taken, costs);
                   let a2 = mcm(rest, costs);
                   if a1.0 + a2.0 < min_cost {
                       min_cost = a1.0 + a2.0;
                       min_path = vec![i - 1];
                       min_path.extend_from_slice(&a2.1);
                   }
               }
               (min_cost, min_path)
           };
           costs.insert(dims, ans.clone());
           ans
       }
   }

}</lang>

Output:
Dims : [5, 6, 3, 1]
Order: (1*(2*3))
Cost : 48

Dims : [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
Order: ((((((((1*2)*3)*4)*5)*6)*7)*(8*(9*10)))*(11*12))
Cost : 38120

Dims : [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
Order: (1*((((((2*3)*4)*(((5*6)*7)*8))*9)*10)*11))
Cost : 1773740

Stata

Recursive solution

Translation of: Python

Here is the equivalent of optim3 in Python's solution. Memoization is done with an associative array. Multiple results are returned in a structure. The same effect as optim2 can be achieved by removing the asarray machinery.

<lang stata>mata struct ans { real scalar p,q,s string scalar u }

struct ans scalar function aux(n,k) { external dim,opt struct ans scalar r,r1,r2 real scalar s,i

if (n==1) { r.p = dim[k] r.q = dim[k+1] r.s = 0 r.u = strofreal(k) return(r) } else if (n==2) { r.p = dim[k] r.q = dim[k+2] r.s = r.p*r.q*dim[k+1] r.u = sprintf("(%f*%f)",k,k+1) return(r) } else if (asarray_contains(opt,(n,k))) { return(asarray(opt,(n,k))) } else { r.p = dim[k] r.q = dim[k+n] r.s = . for (i=1; i<n; i++) { r1 = aux(i,k) r2 = aux(n-i,k+i) s = r1.s+r2.s+r1.p*r1.q*r2.q if (s<r.s) { r.s = s r.u = sprintf("(%s*%s)",r1.u,r2.u) } } asarray(opt,(n,k),r) return(r) } }

function optim(a) { external dim,opt struct ans scalar r real scalar t

timer_clear() dim = a opt = asarray_create("real",2) timer_on(1) r = aux(length(a)-1,1) timer_off(1) t = timer_value(1)[1] printf("%10.0f %10.0f %s\n",t*1000,r.s,r.u) }

optim((1,5,25,30,100,70,2,1,100,250,1,1000,2)) optim((1000,1,500,12,1,700,2500,3,2,5,14,10)) end</lang>

Output 

         0      38120 ((((((((1*2)*3)*4)*5)*6)*7)*(8*(9*10)))*(11*12))
        16    1773740 (1*((((((2*3)*4)*(((5*6)*7)*8))*9)*10)*11))

The timing is in milliseconds, but the time resolution is too coarse to get a usable result. A mean on 1000 loops doing the same computation yields respectively 5.772 ms and 4.430 ms for these two cases. For comparison, the computation was made on the same machine as the Python solution.

Iterative solution

Translation of: Fortran

<lang stata>mata function aux(u,i,j) { k = u[i,j] if (k<0) { printf("%f",i) } else { printf("(") aux(u,i,k) printf("*") aux(u,i+k,j-k) printf(")") } }

function optim(a) { n = length(a)-1 u = J(n,n,.) v = J(n,n,.) u[.,1] = J(n,1,-1) v[.,1] = J(n,1,0) for (j=2; j<=n; j++) { for (i=1; i<=n-j+1; i++) { for (k=1; k<j; k++) { c = v[i,k]+v[i+k,j-k]+a[i]*a[i+k]*a[i+j] if (c<v[i,j]) { u[i,j] = k v[i,j] = c } } } } printf("%f ",v[1,n]) aux(u,1,n) printf("\n") }

optim((1,5,25,30,100,70,2,1,100,250,1,1000,2)) optim((1000,1,500,12,1,700,2500,3,2,5,14,10)) end</lang>

Output 

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))

This solution is faster than the recursive one. The 1000 loops run now in 0.234 ms and 0.187 ms per loop on average.

VBA

Translation of: Fortran

<lang vb>Option Explicit Option Base 1 Dim N As Long, U() As Long, V() As Long Sub Optimize(A As Variant) 

   Dim I As Long, J As Long, K As Long, C As Long
   N = UBound(A) - 1
   ReDim U(N, N), V(N, N)
   For I = 1 To N
       U(I, 1) = -1
       V(I, 1) = 0
   Next I
   
   For J = 2 To N
       For I = 1 To N - J + 1
           V(I, J) = &H7FFFFFFF
           For K = 1 To J - 1
               C = V(I, K) + V(I + K, J - K) + A(I) * A(I + K) * A(I + J)
               If C < V(I, J) Then
                   U(I, J) = K
                   V(I, J) = C
               End If
           Next K
       Next I
   Next J

   Debug.Print V(1, N);
   Call Aux(1, N)
   Debug.Print
   Erase U, V

End Sub Sub Aux(I As Long, J As Long) 

   Dim K As Long
   K = U(I, J)
   If K < 0 Then
       Debug.Print CStr(I);
   Else
       Debug.Print "(";
       Call Aux(I, K)
       Debug.Print "*";
       Call Aux(I + K, J - K)
       Debug.Print ")";
   End If

End Sub Sub Test() 

   Call Optimize(Array(5, 6, 3, 1))
   Call Optimize(Array(1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2))
   Call Optimize(Array(1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10))

End Sub</lang>

Output 

48 (1*(2*3))
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))

Wren

Translation of: Kotlin

<lang ecmascript>var m = [] var s = []

var optimalMatrixChainOrder = Fn.new { |dims| 

   var n = dims.count - 1
   m = List.filled(n, null)
   s = List.filled(n, null)
   for (i in 0...n) {
       m[i] = List.filled(n, 0)
       s[i] = List.filled(n, 0)
   }
   for (len in 1...n) {
       for (i in 0...n-len) {
           var j = i + len
           m[i][j] = 1/0
           for (k in i...j) {
               var temp = dims[i] * dims [k + 1] * dims[j + 1]
               var cost = m[i][k] + m[k + 1][j] + temp
               if (cost < m[i][j]) {
                   m[i][j] = cost
                   s[i][j] = k
               }
           }
       }
   }

}

var printOptimalChainOrder printOptimalChainOrder = Fn.new { |i, j| 

   if (i == j) {
       System.write(String.fromByte(i + 65))
   } else {
       System.write("(")
       printOptimalChainOrder.call(i, s[i][j])
       printOptimalChainOrder.call(s[i][j] + 1, j)
       System.write(")")
   }

}

var dimsList = [ 

   [5, 6, 3, 1],
   [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2],
   [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]

] for (dims in dimsList) { 

   System.print("Dims  : %(dims)")
   optimalMatrixChainOrder.call(dims)
   System.write("Order : ")
   printOptimalChainOrder.call(0, s.count - 1)
   System.print("\nCost  : %(m[0][s.count - 1])\n")

}</lang>

Output:
Dims  : [5, 6, 3, 1]
Order : (A(BC))
Cost  : 48

Dims  : [1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2]
Order : ((((((((AB)C)D)E)F)G)(H(IJ)))(KL))
Cost  : 38120

Dims  : [1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10]
Order : (A((((((BC)D)(((EF)G)H))I)J)K))
Cost  : 1773740

zkl

Translation of: Python

<lang zkl>fcn optim3(a){ // list --> (int,list) 

  aux:=fcn(n,k,a){  // (int,int,list) --> (int,int,int,list)
     if(n==1){

p,q := a[k,2]; return(0,p,q,k); 

     }
     if(n==2){

p,q,r := a[k,3]; return(p*q*r, p, r, T(k,k+1)); 

     }
     m,p,q,u := Void, a[k], a[k + n], Void;
     foreach i in ([1..n-1]){
     #if 0	// 0.70 sec for both tests

s1,p1,q1,u1 := self.fcn(i,k,a); s2,p2,q2,u2 := self.fcn(n - i, k + i, a); 

     #else	// 0.33 sec for both tests

s1,p1,q1,u1 := memoize(self.fcn, i,k,a); s2,p2,q2,u2 := memoize(self.fcn, n - i, k + i, a); 

     #endif

_assert_(q1==p2); s:=s1 + s2 + p1*q1*q2; if((Void==m) or (s<m)) m,u = s,T(u1,u2); 

     }
     return(m,p,q,u);
  };

  h=Dictionary();		// reset memoize
  s,_,_,u := aux(a.len() - 1, 0,a);
  return(s,u);

}

var h; // a Dictionary, set/reset in optim3() fcn memoize(f,n,k,a){ 

  key:="%d,%d".fmt(n,k);	// Lists make crappy keys
  if(r:=h.find(key)) return(r);
  r:=f(n,k,a);
  h[key]=r;
  return(r);

}</lang> <lang zkl>fcn pp(u){ // pretty print a list of lists 

  var letters=["A".."Z"].pump(String);
  u.pump(String,
     fcn(n){ if(List.isType(n)) String("(",pp(n),")") else letters[n] })

} fcn prnt(s,u){ "%-9,d %s\n\t-->%s\n".fmt(s,u.toString(*,*),pp(u)).println() }</lang> <lang zkl>s,u := optim3(T(1, 5, 25, 30, 100, 70, 2, 1, 100, 250, 1, 1000, 2)); prnt(s,u);

s,u := optim3(T(1000, 1, 500, 12, 1, 700, 2500, 3, 2, 5, 14, 10)); prnt(s,u);

optim3(T(5,6,3,1)) : prnt(_.xplode());</lang>

Output:
38,120    L(L(L(L(L(L(L(L(0,1),2),3),4),5),6),L(7,L(8,9))),L(10,11))
	-->(((((((AB)C)D)E)F)G)(H(IJ)))(KL)

1,773,740 L(0,L(L(L(L(L(L(1,2),3),L(L(L(4,5),6),7)),8),9),10))
	-->A((((((BC)D)(((EF)G)H))I)J)K)

48        L(0,L(1,2))
	-->A(BC)
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