Matrix Chain Multiplication in DSA Java

Mental Model

We want to multiply a chain of matrices with the least number of scalar multiplications by choosing the best order to multiply them.

Analogy: Imagine you have several long ropes to tie together. The order in which you tie pairs affects how much effort you spend. Choosing the best order saves effort.

Matrices: A1[10x30] -> A2[30x5] -> A3[5x60]
Goal: Find best way to multiply A1xA2xA3 with minimum cost.

Dry Run Walkthrough

Input: matrices dimensions: [10, 30, 5, 60] (means A1:10x30, A2:30x5, A3:5x60)

Goal: Find the minimum number of scalar multiplications needed to multiply A1xA2xA3

Step 1: Calculate cost of multiplying A1 and A2 first, then multiply result with A3

Cost for (A1xA2)xA3 = 10x30x5 + 10x5x60 = 1500 + 3000 = 4500

Why: We multiply first two matrices, then multiply the result with the third

Step 2: Calculate cost of multiplying A2 and A3 first, then multiply A1 with the result

Cost for A1x(A2xA3) = 30x5x60 + 10x30x60 = 9000 + 18000 = 27000

Why: We multiply last two matrices first, then multiply the first matrix with the result

Step 3: Compare both costs and choose the minimum

Minimum cost = 4500 (from step 1)

Why: Choosing the order with fewer multiplications saves time

Result:

Minimum multiplication cost = 4500

Annotated Code

DSA Java

public class MatrixChainMultiplication {

    // Function to find minimum multiplication cost
    public static int matrixChainOrder(int[] dims) {
        int n = dims.length;
        int[][] dp = new int[n][n];

        // dp[i][j] stores minimum cost to multiply matrices from i to j
        for (int length = 2; length < n; length++) {
            for (int i = 1; i < n - length + 1; i++) {
                int j = i + length - 1;
                dp[i][j] = Integer.MAX_VALUE;
                for (int k = i; k < j; k++) {
                    int cost = dp[i][k] + dp[k + 1][j] + dims[i - 1] * dims[k] * dims[j];
                    if (cost < dp[i][j]) {
                        dp[i][j] = cost;
                    }
                }
            }
        }
        return dp[1][n - 1];
    }

    public static void main(String[] args) {
        int[] dims = {10, 30, 5, 60};
        int result = matrixChainOrder(dims);
        System.out.println("Minimum multiplication cost = " + result);
    }
}

for (int length = 2; length < n; length++) {

increase chain length from 2 to n-1 matrices

for (int i = 1; i < n - length + 1; i++) {

set start index of matrix chain

int j = i + length - 1;

set end index of matrix chain

dp[i][j] = Integer.MAX_VALUE;

initialize minimum cost for this chain to max

for (int k = i; k < j; k++) {

try all possible split points to find minimum cost

int cost = dp[i][k] + dp[k + 1][j] + dims[i - 1] * dims[k] * dims[j];

calculate cost of splitting at k plus cost of multiplying resulting matrices

if (cost < dp[i][j]) { dp[i][j] = cost; }

update minimum cost if current split is cheaper

return dp[1][n - 1];

return minimum cost to multiply full chain

Output

Minimum multiplication cost = 4500

Complexity Analysis

Time: O(n^3) because we check all pairs and all split points for chains of length up to n

Space: O(n^2) because we store minimum costs for all subchains in a 2D table

vs Alternative: Naive approach tries all parenthesizations recursively with exponential time; DP reduces it to polynomial time

Edge Cases

Only one matrix (dims length = 2)

No multiplication needed, cost is zero

DSA Java

for (int length = 2; length < n; length++) {

Two matrices (dims length = 3)

Direct multiplication cost calculated without splits

DSA Java

for (int k = i; k < j; k++) {

When to Use This Pattern

When asked to find the best order to multiply matrices to minimize operations, use Matrix Chain Multiplication dynamic programming to avoid exponential checks.

Common Mistakes

Mistake: Using the wrong indices for dimensions when calculating cost
Fix: Use dims[i-1] * dims[k] * dims[j] to correctly get matrix dimensions for multiplication cost

Mistake: Not initializing dp[i][j] to max before finding minimum
Fix: Set dp[i][j] = Integer.MAX_VALUE before checking splits to ensure correct minimum calculation

Summary

Finds the minimum number of scalar multiplications needed to multiply a chain of matrices.

Use when you have multiple matrices to multiply and want to minimize computation cost.

The key is to try all split points and store minimum costs for subchains to avoid repeated work.

Practice

1. What is the main goal of the Matrix Chain Multiplication problem?

easy

A. To find the minimum number of scalar multiplications needed to multiply a chain of matrices

B. To multiply all matrices in the chain directly without optimization

C. To find the maximum number of matrices that can be multiplied

D. To count the total number of matrices in the chain

Solution

1. Step 1: Understand the problem objective

   The Matrix Chain Multiplication problem aims to find the best way to multiply matrices to minimize cost.

2. Step 2: Identify the cost metric

   The cost is measured by the number of scalar multiplications needed, not just multiplying matrices directly.

3. Final Answer:

   To find the minimum number of scalar multiplications needed to multiply a chain of matrices -> Option A

4. Quick Check:

   Goal = minimum scalar multiplications [OK]

Hint: Focus on minimizing multiplication cost, not counting matrices [OK]

Common Mistakes:

• Thinking the goal is to multiply matrices directly without optimization
• Confusing number of matrices with multiplication cost
• Assuming maximum matrices multiplied is the goal

2. Which of the following is the correct way to declare the dimensions array for 4 matrices with dimensions 10x20, 20x30, 30x40, and 40x50 in Java?

easy

A. int[] dims = {10, 20, 20, 30, 30, 40, 40, 50};

B. int[] dims = {10, 20, 30, 40};

C. int[] dims = {10, 20, 30, 40, 50, 60};

D. int[] dims = {10, 20, 30, 40, 50};

Solution

1. Step 1: Understand the dimensions array format

   For n matrices, the dimensions array has length n+1, storing rows of first matrix and columns of each matrix.

2. Step 2: Apply to given matrices

   Given matrices: 10x20, 20x30, 30x40, 40x50, so dims = {10, 20, 30, 40, 50}.

3. Final Answer:

   int[] dims = {10, 20, 30, 40, 50}; -> Option D

4. Quick Check:

   Length = 5 for 4 matrices [OK]

Hint: Dimensions array length = number of matrices + 1 [OK]

Common Mistakes:

• Using length equal to number of matrices instead of number+1
• Repeating dimensions unnecessarily
• Including extra unrelated dimensions

3. What will be the output of the following Java code snippet for matrix chain multiplication cost calculation?

int[] dims = {5, 10, 3};
int n = dims.length;
int[][] dp = new int[n][n];
for (int l = 2; l < n; l++) {
  for (int i = 1; i < n - l + 1; i++) {
    int j = i + l - 1;
    dp[i][j] = Integer.MAX_VALUE;
    for (int k = i; k < j; k++) {
      int cost = dp[i][k] + dp[k+1][j] + dims[i-1]*dims[k]*dims[j];
      if (cost < dp[i][j]) dp[i][j] = cost;
    }
  }
}
System.out.println(dp[1][n-1]);

medium

A. 0

B. 75

C. 150

D. Syntax Error

Solution

1. Step 1: Understand the input matrices

   Dimensions: 5x10 and 10x3 matrices. Only one way to multiply.

2. Step 2: Calculate multiplication cost

   Cost = 5*10*3 = 150 scalar multiplications.

3. Final Answer:

   150 -> Option C

4. Quick Check:

   5*10*3 = 150 [OK]

Hint: Multiply dims[i-1]*dims[k]*dims[j] for cost [OK]

Common Mistakes:

• Confusing indices in cost calculation
• Assuming zero cost for single multiplication
• Syntax errors from incorrect loops

4. Identify the error in the following Java code snippet for matrix chain multiplication:

int[] dims = {2, 3, 4};
int n = dims.length;
int[][] dp = new int[n][n];
for (int l = 2; l <= n; l++) {
  for (int i = 1; i <= n - l + 1; i++) {
    int j = i + l - 1;
    dp[i][j] = Integer.MAX_VALUE;
    for (int k = i; k < j; k++) {
      int cost = dp[i][k] + dp[k+1][j] + dims[i-1]*dims[k]*dims[j];
      if (cost < dp[i][j]) dp[i][j] = cost;
    }
  }
}
System.out.println(dp[1][n-1]);

medium

A. Loop condition l <= n causes ArrayIndexOutOfBoundsException

B. Incorrect cost calculation formula

C. dp array should be one-dimensional

D. Missing initialization of dp array

Solution

1. Step 1: Analyze loop bounds

   Outer loop runs l from 2 to n inclusive, but dp indices go up to n-1.

2. Step 2: Identify index error

   When l = n, j = i + l - 1 can exceed dp array size causing ArrayIndexOutOfBoundsException.

3. Final Answer:

   Loop condition l <= n causes ArrayIndexOutOfBoundsException -> Option A

4. Quick Check:

   Loop bound off by one causes index error [OK]

Hint: Check loop bounds carefully to avoid index errors [OK]

Common Mistakes:

• Using <= n instead of < n for loop limit
• Ignoring array size limits
• Misunderstanding dp array dimensions

5. Given matrices with dimensions {40, 20, 30, 10, 30}, what is the minimum number of scalar multiplications needed to multiply the entire chain?

hard

A. 30000

B. 26000

C. 18000

D. 35000

Solution

1. Step 1: Understand matrix dimensions

   Matrices: 40x20, 20x30, 30x10, 10x30.

2. Step 2: Calculate minimum cost using dynamic programming

   Possible parenthesizations and costs:
   - ((40x20)x30)x(10x30):
   Cost1 = 40*20*30=24000 + 40*30*10=12000 + 0 = 36000 (too high)
   - (40x(20x30))x(10x30):
   Cost2 = 20*30*10=6000 + 40*20*10=8000 + 40*10*30=12000 = 26000 (minimum)

3. Final Answer:

   26000 -> Option B

4. Quick Check:

   Minimum scalar multiplications = 26000 [OK]

Hint: Try all splits and pick minimum cost [OK]

Common Mistakes:

• Calculating cost for only one parenthesization
• Mixing up matrix dimensions order
• Ignoring intermediate multiplication costs