Matrix Chain Multiplication | DP-8

Given the dimension of a sequence of matrices in an array arr[], where the dimension of the i^th matrix is (arr[i-1] * arr[i]), the task is to find the most efficient way to multiply these matrices together such that the total number of element multiplications is minimum.

Examples:

Input: arr[] = {40, 20, 30, 10, 30}
Output: 26000
Explanation:There are 4 matrices of dimensions 40×20, 20×30, 30×10, 10×30.
Let the input 4 matrices be A, B, C and D.
The minimum number of  multiplications are obtained by 
putting parenthesis in following way (A(BC))D.
The minimum is 20*30*10 + 40*20*10 + 40*10*30

Input: arr[] = {1, 2, 3, 4, 3}
Output: 30
Explanation: There are 4 matrices of dimensions 1×2, 2×3, 3×4, 4×3. 
Let the input 4 matrices be A, B, C and D.  
The minimum number of multiplications are obtained by 
putting parenthesis in following way ((AB)C)D.
The minimum number is 1*2*3 + 1*3*4 + 1*4*3 = 30

Input: arr[] = {10, 20, 30}
Output: 6000
Explanation: There are only two matrices of dimensions 10×20 and 20×30. 
So there  is only one way to multiply the matrices, cost of which is 10*20*30

Matrix Chain Multiplication using Recursion:

We can solve the problem using recursion based on the following facts and observations:

Two matrices of size m*n and n*p when multiplied, they generate a matrix of size m*p and the number of multiplications performed are m*n*p.

Now, for a given chain of N matrices, the first partition can be done in N-1 ways. For example, sequence of matrices A, B, C and D can be grouped as (A)(BCD), (AB)(CD) or (ABC)(D) in these 3 ways. 

So a range [i, j] can be broken into two groups like {[i, i+1], [i+1, j]}, {[i, i+2], [i+2, j]}, . . . , {[i, j-1], [j-1, j]}.

• Each of the groups can be further partitioned into smaller groups and we can find the total required multiplications by solving for each of the groups.
• The minimum number of multiplications among all the first partitions is the required answer.

Follow the steps mentioned below to implement the approach:

• Create a recursive function that takes i and j as parameters that determines the range of a group.
  • Iterate from k = i to j to partition the given range into two groups.
  • Call the recursive function for these groups.
  • Return the minimum value among all the partitions as the required minimum number of multiplications to multiply all the matrices of this group.
• The minimum value returned for the range 0 to N-1 is the required answer.

Below is the implementation of the above approach.

C++

// C++ code to implement the
// matrix chain multiplication using recursion

#include <bits/stdc++.h>
using namespace std;

// Matrix Ai has dimension p[i-1] x p[i]
// for i = 1 . . . n
int MatrixChainOrder(int p[], int i, int j)
{
    if (i == j)
        return 0;
    int k;
    int mini = INT_MAX;
    int count;

    // Place parenthesis at different places
    // between first and last matrix,
    // recursively calculate count of multiplications
    // for each parenthesis placement
    // and return the minimum count
    for (k = i; k < j; k++)
    {
        count = MatrixChainOrder(p, i, k)
                + MatrixChainOrder(p, k + 1, j)
                + p[i - 1] * p[k] * p[j];

        mini = min(count, mini);
    }

    // Return minimum count
    return mini;
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    cout << "Minimum number of multiplications is "
         << MatrixChainOrder(arr, 1, N - 1);
    return 0;
}

// This code is contributed by Shivi_Aggarwal

C

// C code to implement the
// matrix chain multiplication using recursion

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

// Matrix Ai has dimension p[i-1] x p[i]
// for i = 1 . . . n
int MatrixChainOrder(int p[], int i, int j)
{
    if (i == j)
        return 0;
    int k;
    int min = INT_MAX;
    int count;

    // Place parenthesis at different places
    // between first and last matrix,
    // recursively calculate count of multiplications
    // for each parenthesis placement
    // and return the minimum count
    for (k = i; k < j; k++)
    {
        count = MatrixChainOrder(p, i, k)
                + MatrixChainOrder(p, k + 1, j)
                + p[i - 1] * p[k] * p[j];

        if (count < min)
            min = count;
    }

    // Return minimum count
    return min;
}

// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    printf("Minimum number of multiplications is %d ",
           MatrixChainOrder(arr, 1, N - 1));
    getchar();
    return 0;
}

Java

// Java code to implement the
// matrix chain multiplication using recursion

class MatrixChainMultiplication {

    // Matrix Ai has dimension p[i-1] x p[i]
    // for i = 1 . . . n
    static int MatrixChainOrder(int p[], int i, int j)
    {
        if (i == j)
            return 0;

        int min = Integer.MAX_VALUE;

        // Place parenthesis at different places
        // between first and last matrix,
        // recursively calculate count of multiplications
        // for each parenthesis placement
        // and return the minimum count
        for (int k = i; k < j; k++) {
            int count = MatrixChainOrder(p, i, k)
                        + MatrixChainOrder(p, k + 1, j)
                        + p[i - 1] * p[k] * p[j];

            if (count < min)
                min = count;
        }

        // Return minimum count
        return min;
    }

    // Driver code
    public static void main(String args[])
    {
        int arr[] = new int[] { 1, 2, 3, 4, 3 };
        int N = arr.length;

        // Function call
        System.out.println(
            "Minimum number of multiplications is "
            + MatrixChainOrder(arr, 1, N - 1));
    }
}
/* This code is contributed by Rajat Mishra*/

Python3

# Python code to implement the
# matrix chain multiplication using recursion

import sys

# Matrix A[i] has dimension p[i-1] x p[i]
# for i = 1..n
def MatrixChainOrder(p, i, j):
    if i == j:
        return 0

    _min = sys.maxsize

    # Place parenthesis at different places
    # between first and last matrix,
    # recursively calculate count of multiplications
    # for each parenthesis placement
    # and return the minimum count
    for k in range(i, j):

        count = (MatrixChainOrder(p, i, k)
                 + MatrixChainOrder(p, k + 1, j)
                 + p[i-1] * p[k] * p[j])

        if count < _min:
            _min = count

    # Return minimum count
    return _min


# Driver code
if __name__ == '__main__':
    arr = [1, 2, 3, 4, 3]
    N = len(arr)

    # Function call
    print("Minimum number of multiplications is ",
      MatrixChainOrder(arr, 1, N-1))

# This code is contributed by Aryan Garg

C#

// C# code to implement the
// matrix chain multiplication using recursion

using System;

class GFG {

    // Matrix Ai has dimension p[i-1] x p[i]
    // for i = 1..n
    static int MatrixChainOrder(int[] p, int i, int j)
    {

        if (i == j)
            return 0;

        int min = int.MaxValue;

        // Place parenthesis at different places
        // between first and last matrix,
        // recursively calculate count of multiplications
        // for each parenthesis placement
        // and return the minimum count
        for (int k = i; k < j; k++)
        {
            int count = MatrixChainOrder(p, i, k)
                        + MatrixChainOrder(p, k + 1, j)
                        + p[i - 1] * p[k] * p[j];

            if (count < min)
                min = count;
        }

        // Return minimum count
        return min;
    }

    // Driver code
    public static void Main()
    {
        int[] arr = new int[] { 1, 2, 3, 4, 3 };
        int N = arr.Length;

        // Function call
        Console.Write(
            "Minimum number of multiplications is "
            + MatrixChainOrder(arr, 1, N - 1));
    }
}

// This code is contributed by Sam007.

PHP

<?php
// PHP code to implement the
// matrix chain multiplication using recursion

// Matrix Ai has dimension
// p[i-1] x p[i] for i = 1..n
function MatrixChainOrder(&$p, $i, $j)
{
    if($i == $j)
        return 0;
    $min = PHP_INT_MAX;

    // Place parenthesis at different places
    // between first and last matrix,
    // recursively calculate count of multiplications
    // for each parenthesis placement
    // and return the minimum count
    for ($k = $i; $k < $j; $k++)
    {
        $count = MatrixChainOrder($p, $i, $k) +
                 MatrixChainOrder($p, $k + 1, $j) +
                                  $p[$i - 1] *
                                  $p[$k] * $p[$j];

        if ($count < $min)
            $min = $count;
    }

    // Return minimum count
    return $min;
}

// Driver Code
$arr = array(1, 2, 3, 4, 3);
$N = sizeof($arr);

echo "Minimum number of multiplications is " .
      MatrixChainOrder($arr, 1, $N - 1);

// This code is contributed by ita_c
?>

Javascript

<script>

// Javascript code to implement the
// matrix chain multiplication using recursion

// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
function MatrixChainOrder(p , i , j)
{
    if (i == j)
        return 0;

    var min = Number.MAX_VALUE;

    // Place parenthesis at different places
    // between first and last matrix,
    // recursively calculate count of multiplications
    // for each parenthesis placement
    // and return the minimum count
    var k=0;
    for (k = i; k < j; k++)
    {
        var count = MatrixChainOrder(p, i, k)
                    + MatrixChainOrder(p, k + 1, j)
                    + p[i - 1] * p[k] * p[j];

        if (count < min)
            min = count;
    }

    // Return minimum count
    return min;
}

// Driver code
var arr = [ 1, 2, 3, 4, 3 ];
var N = arr.length;

document.write(
    "Minimum number of multiplications is "
    + MatrixChainOrder(arr, 1, N - 1));

// This code contributed by shikhasingrajput

</script>

Output

Minimum number of multiplications is 30

The time complexity of the solution is exponential
Auxiliary Space: O(1)

Dynamic Programming Solution for Matrix Chain Multiplication using Memoization:

Below is the recursion tree for the 2nd example of the above recursive approach:

If observed carefully you can find the following two properties:

1) Optimal Substructure: In the above case, we are breaking the bigger groups into smaller subgroups and solving them to finally find the minimum number of multiplications. Therefore, it can be said that the problem has optimal substructure property.

2) Overlapping Subproblems: We can see in the recursion tree that the same subproblems are called again and again and this problem has the Overlapping Subproblems property. 

So Matrix Chain Multiplication problem has both properties of a dynamic programming problem. So recomputations of same subproblems can be avoided by constructing a temporary array dp[][] in a bottom up manner.

Follow the below steps to solve the problem:

• Build a matrix dp[][] of size N*N for memoization purposes.
• Use the same recursive call as done in the above approach:
  • When we find a range (i, j) for which the value is already calculated, return the minimum value for that range (i.e., dp[i][j]).
  • Otherwise, perform the recursive calls as mentioned earlier.
• The value stored at dp[0][N-1] is the required answer.

Below is the implementation of the above approach 

C++

// C++ program using memoization
#include <bits/stdc++.h>
using namespace std;
int dp[100][100];

// Function for matrix chain multiplication
int matrixChainMemoised(int* p, int i, int j)
{
    if (i == j)
    {
        return 0;
    }
    if (dp[i][j] != -1)
    {
        return dp[i][j];
    }
    dp[i][j] = INT_MAX;
    for (int k = i; k < j; k++)
    {
        dp[i][j] = min(
            dp[i][j], matrixChainMemoised(p, i, k)
                     + matrixChainMemoised(p, k + 1, j)
                       + p[i - 1] * p[k] * p[j]);
    }
    return dp[i][j];
}
int MatrixChainOrder(int* p, int n)
{
    int i = 1, j = n - 1;
    return matrixChainMemoised(p, i, j);
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int n = sizeof(arr) / sizeof(arr[0]);
    memset(dp, -1, sizeof dp);

    cout << "Minimum number of multiplications is "
         << MatrixChainOrder(arr, n);
}

// This code is contributed by Sumit_Yadav

Java

// Java program using memoization
import java.io.*;
import java.util.*;
class GFG
{

  static int[][] dp = new int[100][100];

  // Function for matrix chain multiplication
  static int matrixChainMemoised(int[] p, int i, int j)
  {
    if (i == j)
    {
      return 0;
    }
    if (dp[i][j] != -1)
    {
      return dp[i][j];
    }
    dp[i][j] = Integer.MAX_VALUE;
    for (int k = i; k < j; k++)
    {
      dp[i][j] = Math.min(
        dp[i][j], matrixChainMemoised(p, i, k)
        + matrixChainMemoised(p, k + 1, j)
        + p[i - 1] * p[k] * p[j]);
    }
    return dp[i][j];
  }

  static int MatrixChainOrder(int[] p, int n)
  {
    int i = 1, j = n - 1;
    return matrixChainMemoised(p, i, j);
  }

  // Driver Code
  public static void main (String[] args)
  {

    int arr[] = { 1, 2, 3, 4 };
    int n= arr.length;

    for (int[] row : dp)
      Arrays.fill(row, -1);

    System.out.println("Minimum number of multiplications is " + MatrixChainOrder(arr, n));
  }
}

// This code is contributed by avanitrachhadiya2155

Python3

# Python program using memoization
import sys
dp = [[-1 for i in range(100)] for j in range(100)]

# Function for matrix chain multiplication
def matrixChainMemoised(p, i, j):
    if(i == j):
        return 0

    if(dp[i][j] != -1):
        return dp[i][j]

    dp[i][j] = sys.maxsize

    for k in range(i,j):
        dp[i][j] = min(dp[i][j], matrixChainMemoised(p, i, k) + matrixChainMemoised(p, k + 1, j)+ p[i - 1] * p[k] * p[j])

    return dp[i][j]

def MatrixChainOrder(p,n):
    i = 1
    j = n - 1
    return matrixChainMemoised(p, i, j)

# Driver Code
arr = [1, 2, 3, 4]
n = len(arr)
print("Minimum number of multiplications is",MatrixChainOrder(arr, n))

# This code is contributed by rag2127

C#

// C# program using memoization
using System;
class GFG
{

    static int[,] dp = new int[100, 100];

  // Function for matrix chain multiplication
  static int matrixChainMemoised(int[] p, int i, int j)
  {
    if (i == j)
    {
      return 0;
    }
    if (dp[i, j] != -1)
    {
      return dp[i, j];
    }
    dp[i, j] = Int32.MaxValue;
    for (int k = i; k < j; k++)
    {
      dp[i, j] = Math.Min(
        dp[i, j], matrixChainMemoised(p, i, k)
        + matrixChainMemoised(p, k + 1, j)
        + p[i - 1] * p[k] * p[j]);
    }
    return dp[i,j];
  }

  static int MatrixChainOrder(int[] p, int n)
  {
    int i = 1, j = n - 1;
    return matrixChainMemoised(p, i, j);
  }

  // Driver code
  static void Main()
  {
    int[] arr = { 1, 2, 3, 4 };
    int n = arr.Length;

    for(int i = 0; i < 100; i++)
    {
        for(int j = 0; j < 100; j++)
        {
            dp[i, j] = -1;
        }
    }

    Console.WriteLine("Minimum number of multiplications is " +
                      MatrixChainOrder(arr, n));
  }
}

// This code is contributed by divyeshrabadiya07.

Javascript

<script>

// Javascript program using memoization
let dp = new Array(100);
for(var i = 0; i < dp.length; i++)
{
    dp[i] = new Array(2);
}

// Function for matrix chain multiplication
function matrixChainMemoised(p, i, j)
{
    if (i == j)
    {
        return 0;
    }
    if (dp[i][j] != -1)
    {
        return dp[i][j];
    }

    dp[i][j] = Number.MAX_VALUE;
    for(let k = i; k < j; k++)
    {
        dp[i][j] = Math.min(
            dp[i][j], matrixChainMemoised(p, i, k) +
                      matrixChainMemoised(p, k + 1, j) +
                      p[i - 1] * p[k] * p[j]);
    }
    return dp[i][j];
}

function MatrixChainOrder(p, n)
{
    let i = 1, j = n - 1;
    return matrixChainMemoised(p, i, j);
}

// Driver code
let arr = [ 1, 2, 3, 4 ];
let n = arr.length;

for(var i = 0; i < dp.length; i++)
{
    for(var j = 0; j < dp.length; j++)
    {
        dp[i][j] = -1;
    }
}

document.write("Minimum number of multiplications is " +
               MatrixChainOrder(arr, n));

// This code is contributed by target_2

</script>

Output

Minimum number of multiplications is 18

Time Complexity: O(N³ )
Auxiliary Space: O(N²) ignoring recursion stack space

Dynamic Programming Solution for Matrix Chain Multiplication using Tabulation (Iterative Approach):

In iterative approach, we initially need to find the number of multiplications required to multiply two adjacent matrices. We can use these values to find the minimum multiplication required for matrices in a range of length 3 and further use those values for ranges with higher lengths. 

Build on the answer in this manner till the range becomes [0, N-1].

Follow the steps mentioned below to implement the idea:

• Iterate from l = 2 to N-1 which denotes the length of the range:
  • Iterate from i = 0 to N-1:
    • Find the right end of the range (j) having l matrices.
    • Iterate from k = i+1 to j which denotes the point of partition.
      • Multiply the matrices in range (i, k) and (k, j).
      • This will create two matrices with dimensions arr[i-1]*arr[k] and arr[k]*arr[j].
      • The number of multiplications to be performed to multiply these two matrices (say X) are arr[i-1]*arr[k]*arr[j].
      • The total number of multiplications is dp[i][k]+ dp[k+1][j] + X.
• The value stored at dp[1][N-1] is the required answer.

Below is the implementation of the above approach.

C++

// See the Cormen book for details of the
// following algorithm
#include <bits/stdc++.h>
using namespace std;

// Matrix Ai has dimension p[i-1] x p[i]
// for i = 1..n
int MatrixChainOrder(int p[], int n)
{

    /* For simplicity of the program, one
    extra row and one extra column are
    allocated in m[][]. 0th row and 0th
    column of m[][] are not used */
    int m[n][n];

    int i, j, k, L, q;

    /* m[i, j] = Minimum number of scalar
    multiplications needed to compute the
    matrix A[i]A[i+1]...A[j] = A[i..j] where
    dimension of A[i] is p[i-1] x p[i] */

    // cost is zero when multiplying
    // one matrix.
    for (i = 1; i < n; i++)
        m[i][i] = 0;

    // L is chain length.
    for (L = 2; L < n; L++)
    {
        for (i = 1; i < n - L + 1; i++)
        {
            j = i + L - 1;
            m[i][j] = INT_MAX;
            for (k = i; k <= j - 1; k++)
            {
                // q = cost/scalar multiplications
                q = m[i][k] + m[k + 1][j]
                    + p[i - 1] * p[k] * p[j];
                if (q < m[i][j])
                    m[i][j] = q;
            }
        }
    }

    return m[1][n - 1];
}

// Driver Code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int size = sizeof(arr) / sizeof(arr[0]);

    cout << "Minimum number of multiplications is "
         << MatrixChainOrder(arr, size);

    getchar();
    return 0;
}

// This code is contributed
// by Akanksha Rai

C

// See the Cormen book for details of the following
// algorithm
#include <limits.h>
#include <stdio.h>

// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
int MatrixChainOrder(int p[], int n)
{

    /* For simplicity of the program,
       one extra row and one
       extra column are allocated in m[][].
       0th row and 0th
       column of m[][] are not used */
    int m[n][n];

    int i, j, k, L, q;

    /* m[i, j] = Minimum number of
       scalar multiplications
       needed to compute the matrix
       A[i]A[i+1]...A[j] =
       A[i..j] where dimension of A[i]
       is p[i-1] x p[i] */

    // cost is zero when multiplying one matrix.
    for (i = 1; i < n; i++)
        m[i][i] = 0;

    // L is chain length.
    for (L = 2; L < n; L++) {
        for (i = 1; i < n - L + 1; i++)
        {
            j = i + L - 1;
            m[i][j] = INT_MAX;
            for (k = i; k <= j - 1; k++)
            {
                // q = cost/scalar multiplications
                q = m[i][k] + m[k + 1][j]
                    + p[i - 1] * p[k] * p[j];
                if (q < m[i][j])
                    m[i][j] = q;
            }
        }
    }

    return m[1][n - 1];
}

// Driver  code
int main()
{
    int arr[] = { 1, 2, 3, 4 };
    int size = sizeof(arr) / sizeof(arr[0]);

    printf("Minimum number of multiplications is %d ",
           MatrixChainOrder(arr, size));

    getchar();
    return 0;
}

Java

// Dynamic Programming Java implementation of Matrix
// Chain Multiplication.
// See the Cormen book for details of the following
// algorithm
class MatrixChainMultiplication
{

    // Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
    static int MatrixChainOrder(int p[], int n)
    {
        /* For simplicity of the
        program, one extra row and
        one extra column are allocated in m[][].  0th row
        and 0th column of m[][] are not used */
        int m[][] = new int[n][n];

        int i, j, k, L, q;

        /* m[i, j] = Minimum number of scalar
        multiplications needed to compute the matrix
        A[i]A[i+1]...A[j] = A[i..j] where
        dimension of A[i] is p[i-1] x p[i] */

        // cost is zero when multiplying one matrix.
        for (i = 1; i < n; i++)
            m[i][i] = 0;

        // L is chain length.
        for (L = 2; L < n; L++)
        {
            for (i = 1; i < n - L + 1; i++)
            {
                j = i + L - 1;
                if (j == n)
                    continue;
                m[i][j] = Integer.MAX_VALUE;
                for (k = i; k <= j - 1; k++)
                {
                    // q = cost/scalar multiplications
                    q = m[i][k] + m[k + 1][j]
                        + p[i - 1] * p[k] * p[j];
                    if (q < m[i][j])
                        m[i][j] = q;
                }
            }
        }

        return m[1][n - 1];
    }

    // Driver code
    public static void main(String args[])
    {
        int arr[] = new int[] { 1, 2, 3, 4 };
        int size = arr.length;

        System.out.println(
            "Minimum number of multiplications is "
            + MatrixChainOrder(arr, size));
    }
}
/* This code is contributed by Rajat Mishra*/

Python3

# Dynamic Programming Python implementation of Matrix
# Chain Multiplication. See the Cormen book for details
# of the following algorithm
import sys
maxint=int(1e9+7)
# Matrix Ai has dimension p[i-1] x p[i] for i = 1..n


def MatrixChainOrder(p, n):
    # For simplicity of the program,
    # one extra row and one
    # extra column are allocated in m[][].
    # 0th row and 0th
    # column of m[][] are not used
    m = [[0 for x in range(n)] for x in range(n)]

    # m[i, j] = Minimum number of scalar
    # multiplications needed
    # to compute the matrix A[i]A[i + 1]...A[j] =
    # A[i..j] where
    # dimension of A[i] is p[i-1] x p[i]

    # cost is zero when multiplying one matrix.
    for i in range(1, n):
        m[i][i] = 0

    # L is chain length.
    for L in range(2, n):
        for i in range(1, n-L + 1):
            j = i + L-1
            m[i][j] = maxint
            for k in range(i, j):

                # q = cost / scalar multiplications
                q = m[i][k] + m[k + 1][j] + p[i-1]*p[k]*p[j]
                if q < m[i][j]:
                    m[i][j] = q

    return m[1][n-1]


# Driver code
arr = [1, 2, 3, 4]
size = len(arr)

print("Minimum number of multiplications is " +
      str(MatrixChainOrder(arr, size)))
# This Code is contributed by Bhavya Jain

C#

// Dynamic Programming C# implementation of
// Matrix Chain Multiplication.
// See the Cormen book for details of the
// following algorithm
using System;

class GFG
{

    // Matrix Ai has dimension p[i-1] x p[i]
    // for i = 1..n
    static int MatrixChainOrder(int[] p, int n)
    {

        /* For simplicity of the program, one
        extra row and one extra column are
        allocated in m[][]. 0th row and 0th
        column of m[][] are not used */
        int[, ] m = new int[n, n];

        int i, j, k, L, q;

        /* m[i, j] = Minimum number of scalar
        multiplications needed
        to compute the matrix A[i]A[i+1]...A[j]
        = A[i..j] where dimension of A[i] is
        p[i-1] x p[i] */

        // cost is zero when multiplying
        // one matrix.
        for (i = 1; i < n; i++)
            m[i, i] = 0;

        // L is chain length.
        for (L = 2; L < n; L++)
        {
            for (i = 1; i < n - L + 1; i++)
            {
                j = i + L - 1;
                if (j == n)
                    continue;
                m[i, j] = int.MaxValue;
                for (k = i; k <= j - 1; k++)
                {
                    // q = cost/scalar multiplications
                    q = m[i, k] + m[k + 1, j]
                        + p[i - 1] * p[k] * p[j];
                    if (q < m[i, j])
                        m[i, j] = q;
                }
            }
        }

        return m[1, n - 1];
    }

    // Driver code
    public static void Main()
    {
        int[] arr = new int[] { 1, 2, 3, 4 };
        int size = arr.Length;

        Console.Write("Minimum number of "
                      + "multiplications is "
                      + MatrixChainOrder(arr, size));
    }
}

// This code is contributed by Sam007

PHP

<?php
// Dynamic Programming Python implementation
// of Matrix Chain Multiplication.

// See the Cormen book for details of the
// following algorithm Matrix Ai has
// dimension p[i-1] x p[i] for i = 1..n
function MatrixChainOrder($p, $n)
{
    /* For simplicity of the program, one
    extra row and one extra column are
    allocated in m[][]. 0th row and 0th
    column of m[][] are not used */
    $m[][] = array($n, $n);

    /* m[i, j] = Minimum number of scalar
    multiplications needed to compute the
    matrix A[i]A[i+1]...A[j] = A[i..j] where
    dimension of A[i] is p[i-1] x p[i] */

    // cost is zero when multiplying one matrix.
    for ($i = 1; $i < $n; $i++)
        $m[$i][$i] = 0;

    // L is chain length.
    for ($L = 2; $L < $n; $L++)
    {
        for ($i = 1; $i < $n - $L + 1; $i++)
        {
            $j = $i + $L - 1;
            if($j == $n)
                continue;
            $m[$i][$j] = PHP_INT_MAX;
            for ($k = $i; $k <= $j - 1; $k++)
            {
                // q = cost/scalar multiplications
                $q = $m[$i][$k] + $m[$k + 1][$j] +
                     $p[$i - 1] * $p[$k] * $p[$j];
                if ($q < $m[$i][$j])
                    $m[$i][$j] = $q;
            }
        }
    }

    return $m[1][$n-1];
}

// Driver Code
$arr = array(1, 2, 3, 4);
$size = sizeof($arr);

echo"Minimum number of multiplications is ".
              MatrixChainOrder($arr, $size);

// This code is contributed by Mukul Singh
?>

Javascript

<script>

// Dynamic Programming javascript implementation of Matrix
// Chain Multiplication.
// See the Cormen book for details of the following
// algorithm


// Matrix Ai has dimension p[i-1] x p[i] for i = 1..n
function MatrixChainOrder(p , n)
{
    /* For simplicity of the
    program, one extra row and
    one extra column are allocated in m.  0th row
    and 0th column of m are not used */
    var m = Array(n).fill(0).map(x => Array(n).fill(0));

    var i, j, k, L, q;

    /* m[i, j] = Minimum number of scalar
    multiplications needed to compute the matrix
    A[i]A[i+1]...A[j] = A[i..j] where
    dimension of A[i] is p[i-1] x p[i] */

    // cost is zero when multiplying one matrix.
    for (i = 1; i < n; i++)
        m[i][i] = 0;

    // L is chain length.
    for (L = 2; L < n; L++)
    {
        for (i = 1; i < n - L + 1; i++)
        {
            j = i + L - 1;
            if (j == n)
                continue;
            m[i][j] = Number.MAX_VALUE;
            for (k = i; k <= j - 1; k++)
            {
                // q = cost/scalar multiplications
                q = m[i][k] + m[k + 1][j]
                    + p[i - 1] * p[k] * p[j];
                if (q < m[i][j])
                    m[i][j] = q;
            }
        }
    }

    return m[1][n - 1];
}

// Driver code
var arr = [ 1, 2, 3, 4 ];
var size = arr.length;

document.write(
    "Minimum number of multiplications is "
    + MatrixChainOrder(arr, size));

// This code contributed by Princi Singh

</script>

Output

Minimum number of multiplications is 18

Time Complexity: O(N³ )
Auxiliary Space: O(N²)

Matrix Chain Multiplication (A O(N^2) Solution)
Printing brackets in Matrix Chain Multiplication Problem
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Applications:
Minimum and Maximum values of an expression with * and +