Matrix Chain Multiplication (A O(N^2) Solution)

24 May, 2021
Given a sequence of matrices, find the most efficient way to multiply these matrices together. The problem is not actually to perform the multiplications, but merely to decide in which order to perform the multiplications.We have many options to multiply a chain of matrices because matrix multiplication is associative. In other words, no matter how we parenthesize the product, the result will be the same. For example, if we had four matrices A, B, C, and D, we would have:  
    (ABC)D = (AB)(CD) = A(BCD) = ....
However, the order in which we parenthesize the product affects the number of simple arithmetic operations needed to compute the product or the efficiency. For example, suppose A is a 10 × 30 matrix, B is a 30 × 5 matrix, and C is a 5 × 60 matrices. Then,  
    (AB)C = (10×30×5) + (10×5×60) = 1500 + 3000 = 4500 operations
    A(BC) = (30×5×60) + (10×30×60) = 9000 + 18000 = 27000 operations.
Clearly, the first parenthesization requires less number of operations.Given an array p[] which represents the chain of matrices such that the ith matrix Ai is of dimension p[i-1] x p[i]. We need to write a function MatrixChainOrder() that should return the minimum number of multiplications needed to multiply the chain.  
  Input: p[] = {40, 20, 30, 10, 30}   
  Output: 26000  
  There are 4 matrices of dimensions 40x20, 20x30, 30x10 and 10x30.
  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 --> 20*30*10 + 40*20*10 + 40*10*30

  Input: p[] = {10, 20, 30, 40, 30} 
  Output: 30000 
  There are 4 matrices of dimensions 10x20, 20x30, 30x40 and 40x30. 
  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 --> 10*20*30 + 10*30*40 + 10*40*30

  Input: p[] = {10, 20, 30}  
  Output: 6000  
  There are only two matrices of dimensions 10x20 and 20x30. So there 
  is only one way to multiply the matrices, cost of which is 10*20*30
 
We have discussed a O(n^3) solution for Matrix Chain Multiplication Problem. Note: Below solution does not work for many cases. For example: for input {2, 40, 2, 40, 5}, the correct answer is 580 but this method returns 720 There is another similar approach to solving this problem. More intuitive and recursive approach. 
Assume there are following available method minCost(M1, M2) -> returns min cost of multiplying matrices M1 and M2Then, for any chained product of matrices like, M1.M2.M3.M4...Mn min cost of chain = min(minCost(M1, M2.M3...Mn), minCost(M1.M2..Mn-1, Mn)) Now we have two subchains (sub problems) : M2.M3...Mn M1.M2..Mn-1 
 
C++
Java
Python3
C#
PHP
Javascript
// CPP program to implement optimized matrix chain multiplication problem.#include<iostream>using namespace std; // Matrix Mi has dimension p[i-1] x p[i] for i = 1..nint MatrixChainOrder(int p[], int n){     /* For simplicity of the program, one extra row and one extra column are allocated in    dp[][]. 0th row and 0th column of dp[][] are not used */    int dp[n][n];     /* dp[i, j] = Minimum number of scalar multiplications needed to compute the matrix M[i]M[i+1]...M[j]                = M[i..j] where dimension of M[i] is p[i-1] x p[i] */                     // cost is zero when multiplying one matrix.    for (int i=1; i<n; i++)        dp[i][i] = 0;     // Simply following above recursive formula.    for (int L=1; L<n-1; L++)    for (int i=1; i<n-L; i++)            dp[i][i+L] = min(dp[i+1][i+L] + p[i-1]*p[i]*p[i+L],                    dp[i][i+L-1] + p[i-1]*p[i+L-1]*p[i+L]);             return dp[1][n-1];} // Driver codeint main(){    int arr[] = {10, 20, 30, 40, 30} ;    int size = sizeof(arr)/sizeof(arr[0]);     printf("Minimum number of multiplications is %d ",                    MatrixChainOrder(arr, size));     return 0;}
// Java program to implement optimized matrix chain multiplication problem.import java.util.*;import java.lang.*;import java.io.*;  class GFG{// Matrix Mi has dimension p[i-1] x p[i] for i = 1..nstatic int MatrixChainOrder(int p[], int n){     /* For simplicity of the program, one extra row and one extra column are     allocated in dp[][]. 0th row and 0th column of dp[][] are not used */    int [][]dp=new int[n][n];     /* dp[i, j] = Minimum number of scalar multiplications needed to compute the matrix M[i]M[i+1]...M[j]                = M[i..j] where dimension of M[i] is p[i-1] x p[i] */                     // cost is zero when multiplying one matrix.    for (int i=1; i<n; i++)        dp[i][i] = 0;     // Simply following above recursive formula.    for (int L=1; L<n-1; L++)    for (int i=1; i<n-L; i++)            dp[i][i+L] = Math.min(dp[i+1][i+L] + p[i-1]*p[i]*p[i+L],                    dp[i][i+L-1] + p[i-1]*p[i+L-1]*p[i+L]);             return dp[1][n-1];} // Driver codepublic static void main(String args[]){    int arr[] = {10, 20, 30, 40, 30} ;    int size = arr.length;     System.out.print("Minimum number of multiplications is "+                    MatrixChainOrder(arr, size)); }}
# Python3 program to implement optimized# matrix chain multiplication problem. # Matrix Mi has dimension# p[i-1] x p[i] for i = 1..ndef MatrixChainOrder(p, n):         # For simplicity of the program, one    # extra row and one extra column are    # allocated in dp[][]. 0th row and    # 0th column of dp[][] are not used    dp = [[0 for i in range(n)]             for i in range(n)]     # dp[i, j] = Minimum number of scalar    # multiplications needed to compute    # the matrix M[i]M[i+1]...M[j] = M[i..j]    # where dimension of M[i] is p[i-1] x p[i]                     # cost is zero when multiplying one matrix.    for i in range(1, n):        dp[i][i] = 0     # Simply following above recursive formula.    for L in range(1, n - 1):        for i in range(n - L):            dp[i][i + L] = min(dp[i + 1][i + L] +                                p[i - 1] * p[i] * p[i + L],                               dp[i][i + L - 1] +                                p[i - 1] * p[i + L - 1] * p[i + L])         return dp[1][n - 1] # Driver codearr = [10, 20, 30, 40, 30]size = len(arr)print("Minimum number of multiplications is",                 MatrixChainOrder(arr, size)) # This code is contributed# by sahishelangia
// C# program to implement optimized// matrix chain multiplication problem.using System; class GFG{// Matrix Mi has dimension// p[i-1] x p[i] for i = 1..nstatic int MatrixChainOrder(int []p, int n){     /* For simplicity of the program,       one extra row and one extra       column are allocated in dp[][].       0th row and 0th column of dp[][]       are not used */    int [,]dp = new int[n, n];     /* dp[i, j] = Minimum number of scalar       multiplications needed to compute       the matrix M[i]M[i+1]...M[j] = M[i..j]       where dimension of M[i] is p[i-1] x p[i] */                     // cost is zero when multiplying    // one matrix.    for (int i = 1; i < n; i++)        dp[i, i] = 0;     // Simply following above    // recursive formula.    for (int L = 1; L < n - 1; L++)    for (int i = 1; i < n - L; i++)            dp[i, i + L] = Math.Min(dp[i + 1, i + L] +                                 p[i - 1] * p[i] * p[i + L],                                dp[i, i + L - 1] + p[i - 1] *                                 p[i + L - 1] * p[i + L]);         return dp[1, n - 1];} // Driver codepublic static void Main(){    int []arr = {10, 20, 30, 40, 30} ;    int size = arr.Length;     Console.WriteLine("Minimum number of multiplications is " +                                  MatrixChainOrder(arr, size));}} // This code is contributed by anuj_67
<?php// PHP program to implement optimized// matrix chain multiplication problem. // Matrix Mi has dimension// p[i-1] x p[i] for i = 1..nfunction MatrixChainOrder($p, $n){     /* For simplicity of the program,    one extra row and one extra column    are allocated in dp[][]. 0th row    and 0th column of dp[][] are not used */    $dp = array();     /* dp[i, j] = Minimum number of scalar                  multiplications needed to                  compute the matrix M[i]M[i+1]...M[j]                = M[i..j] where dimension of M[i]                  is p[i-1] x p[i] */                     // cost is zero when multiplying    // one matrix.    for ($i=1; $i<$n; $i++)        $dp[$i][$i] = 0;     // Simply following above    // recursive formula.    for ($L = 1; $L < $n - 1; $L++)    for ($i = 1; $i < $n - $L; $i++)        $dp[$i][$i + $L] = min($dp[$i + 1][$i + $L] + $p[$i - 1] *                               $p[$i] * $p[$i + $L],                               $dp[$i][$i + $L - 1] + $p[$i - 1] *                               $p[$i + $L - 1] * $p[$i + $L]);         return $dp[1][$n - 1];} // Driver code$arr = array(10, 20, 30, 40, 30) ;$size = sizeof($arr); echo "Minimum number of multiplications is ".               MatrixChainOrder($arr, $size); // This code is contributed// by Akanksha Rai(Abby_akku)?>
<script>  // Javascript program to implement optimized// matrix chain multiplication problem. // Matrix Mi has dimension p[i-1] x p[i] for i = 1..nfunction MatrixChainOrder(p, n){     /* For simplicity of the program, one extra row    and one extra column are allocated in    dp[][]. 0th row and 0th column of dp[][] are not used */    var dp = Array.from(Array(n), ()=> Array(n));     /* dp[i, j] = Minimum number of scalar multiplications    needed to compute the matrix M[i]M[i+1]...M[j]                = M[i..j] where dimension                of M[i] is p[i-1] x p[i] */                     // cost is zero when multiplying one matrix.    for (var i=1; i<n; i++)        dp[i][i] = 0;     // Simply following above recursive formula.    for (var L=1; L<n-1; L++)    for (var i=1; i<n-L; i++)            dp[i][i+L] = Math.min(dp[i+1][i+L] +        p[i-1]*p[i]*p[i+L],        dp[i][i+L-1] + p[i-1]*p[i+L-1]*p[i+L]);             return dp[1][n-1];} // Driver codevar arr = [10, 20, 30, 40, 30];var size = arr.length;document.write("Minimum number of multiplications is  "+                MatrixChainOrder(arr, size));  </script>
Minimum number of multiplications is 30000
 
Thanks to Rishi_Lazy for providing above optimized solution.Time Complexity : O(n2) Printing Brackets in Matrix Chain Multiplication 
tufan_gupta2000
vt_m
Akanksha_Rai
sahilshelangia
2019ucp1394
rutvik_56
matrix-chain-multiplication
Dynamic Programming
Matrix
Dynamic Programming
Matrix
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