Given the dimension of a sequence of matrices in an array arr[], where the dimension of the ith 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*30Input: 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 = 30Input: 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 . . . nint 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 Codeint 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 . . . nint 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 codeint 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 recursionimport java.io.*;import java.util.*;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 recursionimport sys# Matrix A[i] has dimension p[i-1] x p[i]# for i = 1..ndef 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 codeif __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 recursionusing 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..nfunction 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..nfunction 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 codevar 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 multiplicationint 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 Codeint 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 sysdp = [[-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 Codearr = [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 memoizationusing 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 codelet 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.
- Iterate from i = 0 to N-1:
- 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..nint 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 Codeint 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..nint 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 codeint 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// algorithmimport java.util.*;import java.io.*;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 algorithmimport sysmaxint=int(1e9+7)# Matrix Ai 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 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 codearr = [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 algorithmusing 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..nfunction 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..nfunction 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 codevar 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 +
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
