Optimal Matrix Chain Multiplication

Matrix Chain Multiplication Problem

Problem Statement:

You are given an array arr[] of length n, where arr[i] represents the dimensions of matrix Ai. The matrix multiplication of matrices A1, A2, ..., An is a chain multiplication problem, where the matrices are multiplied in sequence.

The goal is to find the minimum number of scalar multiplications required to multiply the entire chain of matrices.

Matrix Multiplication Rules:

For matrices to be multiplied, the number of columns of the first matrix must be equal to the number of rows of the second matrix.

  • Matrix Ai has dimensions arr[i-1] x arr[i].

  • For example, if arr[] = [10, 20, 30, 40], then matrix A1 is 10x20, matrix A2 is 20x30, and matrix A3 is 30x40.

Problem Explanation:

The problem asks us to find the optimal parenthesization of the matrix chain so that the total number of scalar multiplications is minimized.

Example 1:

Input:

arr = [10, 20, 30, 40, 50]

Output:

38000

Explanation: We have 4 matrices:

  • Matrix A1 has dimensions 10x20.

  • Matrix A2 has dimensions 20x30.

  • Matrix A3 has dimensions 30x40.

  • Matrix A4 has dimensions 40x50.

The optimal way to parenthesize the matrices is:

((A1 * A2) * A3) * A4

The number of scalar multiplications would be:

(10 * 20 * 30) + (10 * 30 * 40) + (10 * 40 * 50) = 6000 + 12000 + 20000 = 38000.

Best Technique – Dynamic Programming

This problem is best solved using Dynamic Programming (DP) because of its overlapping subproblems and optimal substructure properties. The key idea is to compute the minimum cost for each possible sub-chain of matrices and build up the solution for the entire chain.

Different Approaches:

Approach 1: Brute Force (Recursive)

In this approach, we recursively try all possible ways to parenthesize the matrix chain. However, it’s inefficient because it repeatedly calculates the same subproblems, leading to an exponential time complexity.

public class MatrixChainMultiplication {

    // Recursive function for matrix chain multiplication
    public static int matrixChainOrder(int[] arr, int i, int j) {
        if (i == j) {
            return 0;
        }
        
        int minCost = Integer.MAX_VALUE;
        
        // Try all possible positions for splitting the chain
        for (int k = i; k < j; k++) {
            int cost = matrixChainOrder(arr, i, k) + matrixChainOrder(arr, k + 1, j) 
                    + arr[i - 1] * arr[k] * arr[j];
            minCost = Math.min(minCost, cost);
        }
        
        return minCost;
    }
    
    public static void main(String[] args) {
        int[] arr = {10, 20, 30, 40, 50};
        int n = arr.length;
        System.out.println("Minimum number of multiplications: " + matrixChainOrder(arr, 1, n - 1));
    }
}

Approach 2: Memoization (Top-Down DP)

In this approach, we store the results of subproblems in a table so that we don’t recompute them. It improves upon the brute force approach by using a top-down approach with memoization.

import java.util.*;

public class MatrixChainMultiplication {

    // Memoization approach
    public static int matrixChainOrder(int[] arr, int i, int j, int[][] dp) {
        if (i == j) {
            return 0;
        }
        
        if (dp[i][j] != -1) {
            return dp[i][j];
        }
        
        int minCost = Integer.MAX_VALUE;
        
        // Try all possible positions for splitting the chain
        for (int k = i; k < j; k++) {
            int cost = matrixChainOrder(arr, i, k, dp) + matrixChainOrder(arr, k + 1, j, dp) 
                    + arr[i - 1] * arr[k] * arr[j];
            minCost = Math.min(minCost, cost);
        }
        
        dp[i][j] = minCost;
        return dp[i][j];
    }
    
    public static void main(String[] args) {
        int[] arr = {10, 20, 30, 40, 50};
        int n = arr.length;
        int[][] dp = new int[n][n];
        
        // Initialize dp array with -1
        for (int[] row : dp) {
            Arrays.fill(row, -1);
        }
        
        System.out.println("Minimum number of multiplications: " + matrixChainOrder(arr, 1, n - 1, dp));
    }
}

Approach 3: Optimized DP (Bottom-Up DP)

In this approach, we use a bottom-up DP approach, iterating over increasing lengths of subproblems and solving them iteratively. This approach eliminates the overhead of recursion and the need for memoization.

public class MatrixChainMultiplication {

    // Bottom-up DP approach
    public static int matrixChainMultiplication(int[] arr) {
        int n = arr.length;
        int[][] dp = new int[n][n];
        
        // Fill dp[][] for chains of length 2 to n
        for (int len = 2; len < n; len++) {
            for (int i = 1; i < n - len + 1; i++) {
                int j = i + len - 1;
                dp[i][j] = Integer.MAX_VALUE;
                for (int k = i; k < j; k++) {
                    int cost = dp[i][k] + dp[k + 1][j] + arr[i - 1] * arr[k] * arr[j];
                    dp[i][j] = Math.min(dp[i][j], cost);
                }
            }
        }
        
        return dp[1][n - 1];
    }
    
    public static void main(String[] args) {
        int[] arr = {10, 20, 30, 40, 50};
        System.out.println("Minimum number of multiplications: " + matrixChainMultiplication(arr));
    }
}

⚠️Common Pitfalls:

  • Recursion Depth: In recursive solutions, you may run into issues with deep recursion for large values of n. Memoization or bottom-up DP can help alleviate this.

  • Off-by-One Errors: Ensure that you handle array indices properly, especially when dealing with 1-based indexing for matrix dimensions.

📊 Time and Space Complexity:

ApproachTime Complexity    Space Complexity
Brute Force    O(2^n)    O(n)
Memoization (Top-Down)    O(n^3)    O(n^2)
Optimized DP (Bottom-Up)    O(n^3)    O(n^2)

🌍 Real-World Applications:

  • Computer Graphics: Efficient rendering algorithms where matrix multiplication is a core part of the calculation.

  • AI and Machine Learning: Optimization problems that require multiplying large matrices efficiently.

  • Scientific Computing: Numerical simulations often require matrix chain multiplication for large systems of equations.


I hope you found this article insightful! Keep exploring, keep learning, and don’t forget to check back daily for more exciting problem breakdowns. If you know someone who would benefit from this, share it with them and help them grow too! That’s it for today—see you in the next post!

                                                                                                                                         Signing off!! 

                                                                                                Master DSA One Problem at a Time :)

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