Matrix chain multiplication

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.

C[edit]

Translation of: Kotlin

#include <stdio.h>
#include <limits.h>
#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;
}

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#[edit]

Translation of: Kotlin

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]);
        }
    }
}

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[edit]

Translation of: Python

This is a translation of the Python iterative solution.

module optim_m
    implicit none
contains
    subroutine optim(a)
        implicit none
        integer :: a(:), n, i, j, k
        integer(8) :: c
        integer, allocatable :: u(:, :)
        integer(8), allocatable :: v(:, :)
 
        n = ubound(a, 1) - 1
        allocate (u(n, n), v(n, n))
        v = -1
        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 (v(i, j) < 0 .or. 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 optim_p
    use optim_m
    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])
end program

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))

Go[edit]

The first for loop is based on the pseudo and Java code from the Wikipedia article.

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()
	}
}

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

Julia[edit]

Works with: Julia version 0.6

Translation of: MATLAB

Module:

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

Main:

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]))

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[edit]

This is based on the pseudo-code in the Wikipedia article.

// 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")
    }
}

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

MATLAB[edit]

Translation of: Fortran

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

Output

[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))"

Perl[edit]

Translation of: Perl 6

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>);

Output:

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

Perl 6[edit]

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.

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>);

Output:

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

Phix[edit]

As per the wp pseudocode

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

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[edit]

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[edit]

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)

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

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]

And here is the optimization step:

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

Recursive cost optimization[edit]

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.

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

Memoized recursive call[edit]

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).

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

Putting all together[edit]

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()

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[edit]

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.

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]))

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]])

Stata[edit]

Recursive solution[edit]

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.

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

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[edit]

Translation of: Fortran

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

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.

zkl[edit]

Translation of: Python

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);
}

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() }

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());

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)

Matrix chain multiplication

Contents

C[edit]

C#[edit]

Fortran[edit]

Go[edit]

Julia[edit]

Kotlin[edit]

MATLAB[edit]

Perl[edit]

Perl 6[edit]

Phix[edit]

Python[edit]

Enumeration of parenthesizations[edit]

Recursive cost optimization[edit]

Memoized recursive call[edit]

Putting all together[edit]

Iterative solution[edit]

Stata[edit]

Recursive solution[edit]

Iterative solution[edit]

zkl[edit]