Matrix Chain Multiplication in DSA Javascript

Mental Model

We want to find the best way to multiply a chain of matrices to use the least number of multiplications.

Analogy: Imagine you have several boxes to stack, and the order you stack them changes how easy it is to move them. Finding the best stacking order saves effort.

Matrices: A1(10x30) -> A2(30x5) -> A3(5x60)
Goal: Find best way to multiply A1 x A2 x A3

Chain: [10, 30, 5, 60]
Each number is a matrix dimension

Initial:
A1 -> A2 -> A3
Dimensions:
10x30 -> 30x5 -> 5x60

Dry Run Walkthrough

Input: matrices dimensions: [10, 30, 5, 60]

Goal: Find the minimum number of scalar multiplications to multiply A1(10x30), A2(30x5), A3(5x60)

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

Cost for (A1xA2) = 10x30x5 = 1500
Then multiply result (10x5) with A3 (5x60): 10x5x60 = 3000
Total cost = 1500 + 3000 = 4500

Why: We try one way of parenthesizing to find total cost

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

Cost for (A2xA3) = 30x5x60 = 9000
Then multiply A1 (10x30) with result (30x60): 10x30x60 = 18000
Total cost = 9000 + 18000 = 27000

Why: Try the other way of parenthesizing to compare costs

Step 3: Compare both costs and choose minimum

Minimum cost = 4500 (from step 1)

Why: We want the least multiplication cost to optimize performance

Result:

Minimum multiplication cost = 4500

Annotated Code

DSA Javascript

class MatrixChainMultiplication {
  constructor(dimensions) {
    this.dimensions = dimensions;
    this.n = dimensions.length - 1;
    this.dp = Array.from({ length: this.n }, () => Array(this.n).fill(0));
  }

  computeMinCost() {
    for (let length = 2; length <= this.n; length++) {
      for (let i = 0; i <= this.n - length; i++) {
        let j = i + length - 1;
        this.dp[i][j] = Infinity;
        for (let k = i; k < j; k++) {
          const cost = this.dp[i][k] + this.dp[k + 1][j] + this.dimensions[i] * this.dimensions[k + 1] * this.dimensions[j + 1];
          if (cost < this.dp[i][j]) {
            this.dp[i][j] = cost;
          }
        }
      }
    }
    return this.dp[0][this.n - 1];
  }
}

// Driver code
const dimensions = [10, 30, 5, 60];
const mcm = new MatrixChainMultiplication(dimensions);
console.log(mcm.computeMinCost());

for (let length = 2; length <= this.n; length++) {

increase chain length from 2 to n to consider all subproblems

for (let i = 0; i <= this.n - length; i++) {

set start index of matrix chain for current length

let j = i + length - 1;

calculate end index of matrix chain

this.dp[i][j] = Infinity;

initialize minimum cost for this subchain

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

try all possible splits between i and j

const cost = this.dp[i][k] + this.dp[k + 1][j] + this.dimensions[i] * this.dimensions[k + 1] * this.dimensions[j + 1];

calculate cost of splitting at k plus multiplication cost

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

update minimum cost if current split is cheaper

return this.dp[0][this.n - 1];

return minimum cost for full chain multiplication

Output

4500

Complexity Analysis

Time: O(n^3) because we check all subchains and all split points for each subchain

Space: O(n^2) for the dp table storing minimum costs for all subchains

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

Edge Cases

Only one matrix (dimensions length = 2)

No multiplication needed, cost is zero

DSA Javascript

for (let length = 2; length <= this.n; length++) {

Two matrices (dimensions length = 3)

Single multiplication cost calculated directly

DSA Javascript

for (let length = 2; length <= this.n; length++) {

Matrices with same dimensions (e.g., all 10x10)

Algorithm still finds minimum cost, which may be equal for all splits

DSA Javascript

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

When to Use This Pattern

When you see a problem asking for the best order to multiply matrices or combine operations to minimize cost, use Matrix Chain Multiplication dynamic programming.

Common Mistakes

Mistake: Confusing matrix dimensions order when calculating multiplication cost
Fix: Use dimensions[i] * dimensions[k+1] * dimensions[j+1] consistently for cost calculation

Mistake: Not initializing dp[i][j] to Infinity before checking splits
Fix: Set dp[i][j] = Infinity before inner loop to find minimum cost

Summary

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

Use when you want to optimize the order of matrix multiplication to reduce computation.

The key insight is to break the problem into smaller subproblems and combine their solutions using dynamic programming.

Practice

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

easy

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

B. To multiply all matrices in the chain in the given order without optimization

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

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

Solution

1. Step 1: Understand the problem objective

   The Matrix Chain Multiplication problem aims to multiply a sequence of matrices with minimum cost.

2. Step 2: Identify what is minimized

   The cost minimized is the number of scalar multiplications, not just multiplying in order or counting matrices.

3. Final Answer:

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

4. Quick Check:

   Goal = minimum scalar multiplications [OK]

Hint: Remember: minimize multiplication cost, not count matrices [OK]

Common Mistakes:

• Thinking the problem is about counting matrices
• Assuming multiplication order is fixed without optimization
• Confusing maximum with minimum cost

2. Which of the following is the correct way to declare the dimensions array p for matrices A1 (10x30), A2 (30x5), A3 (5x60) in JavaScript?

easy

A. const p = [10, 30, 5, 60];

B. const p = [[10, 30], [30, 5], [5, 60]];

C. const p = [10, 30, 5];

D. const p = [30, 5, 60];

Solution

1. Step 1: Understand the dimensions array format

   The array p stores dimensions such that matrix Ai has dimensions p[i-1] x p[i].

2. Step 2: Apply to given matrices

   For A1 (10x30), A2 (30x5), A3 (5x60), p = [10, 30, 5, 60].

3. Final Answer:

   const p = [10, 30, 5, 60]; -> Option A

4. Quick Check:

   Dimensions array = [10,30,5,60] [OK]

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

Common Mistakes:

• Using nested arrays instead of flat array
• Missing last dimension in array
• Confusing matrix dimensions order

3. What is the output of the following JavaScript code snippet for matrix dimensions p = [40, 20, 30, 10, 30]?

function matrixChainOrder(p) {
  const n = p.length - 1;
  const m = Array.from({ length: n }, () => Array(n).fill(0));
  for (let L = 2; L <= n; L++) {
    for (let i = 0; i <= n - L; i++) {
      let j = i + L - 1;
      m[i][j] = Infinity;
      for (let k = i; k < j; k++) {
        let q = m[i][k] + m[k + 1][j] + p[i] * p[k + 1] * p[j + 1];
        if (q < m[i][j]) m[i][j] = q;
      }
    }
  }
  return m[0][n - 1];
}
console.log(matrixChainOrder(p));

medium

A. 45000

B. 30000

C. 26000

D. None of the above

Solution

1. Step 1: Understand the code logic

   The code implements the dynamic programming solution for matrix chain multiplication to find minimum cost.

2. Step 2: Calculate minimum multiplication cost for p = [40,20,30,10,30]

   Minimum cost is 26000 scalar multiplications after checking all splits.

3. Final Answer:

   26000 -> Option C

4. Quick Check:

   Minimum cost for given p = 26000 [OK]

Hint: Use DP table to find minimum cost, result is m[0][n-1] [OK]

Common Mistakes:

• Confusing indices i, j, k in loops
• Incorrectly calculating multiplication cost
• Returning wrong DP table value

4. Identify the error in the following JavaScript code snippet for matrix chain multiplication and select the fix:

function matrixChainOrder(p) {
  const n = p.length - 1;
  const m = Array.from({ length: n }, () => Array(n).fill(0));
  for (let L = 2; L <= n; L++) {
    for (let i = 0; i < n - L; i++) {
      let j = i + L - 1;
      m[i][j] = Infinity;
      for (let k = i; k < j; k++) {
        let q = m[i][k] + m[k + 1][j] + p[i] * p[k + 1] * p[j + 1];
        if (q < m[i][j]) m[i][j] = q;
      }
    }
  }
  return m[0][n - 1];
}

medium

A. Initialize m with Infinity instead of 0

B. Change inner loop condition to i <= n - L

C. Use for (let k = i; k <= j; k++) instead of k < j

D. Return m[0][n] instead of m[0][n - 1]

Solution

1. Step 1: Analyze the outer loop range for i

   The loop for (let i = 0; i < n - L; i++) misses the case when i equals n - L, causing incomplete computation.

2. Step 2: Fix loop condition to include i = n - L

   Changing to i <= n - L ensures all subproblems are covered correctly.

3. Final Answer:

   Change inner loop condition to i <= n - L -> Option B

4. Quick Check:

   Loop range fix = i <= n - L [OK]

Hint: Check loop boundaries carefully for full coverage [OK]

Common Mistakes:

• Off-by-one error in loop ranges
• Incorrect DP table initialization
• Wrong return index from DP table

5. Given matrices with dimensions p = [5, 10, 3, 12, 5, 50, 6], what is the minimum number of scalar multiplications needed to multiply the entire chain? Use the matrix chain multiplication approach.

hard

A. 1510

B. 2010

C. 4050

D. None of the above

Solution

1. Step 1: Understand the problem and dimensions

   We have 6 matrices with dimensions from p array: A1(5x10), A2(10x3), A3(3x12), A4(12x5), A5(5x50), A6(50x6).

2. Step 2: Apply matrix chain multiplication DP to find minimum cost

   Calculating minimum scalar multiplications yields 1510 as the optimal cost.

3. Final Answer:

   1510 -> Option A

4. Quick Check:

   Minimum scalar multiplications = 1510 [OK]

Hint: Use DP table and check all splits for minimum cost [OK]

Common Mistakes:

• Calculating cost for wrong matrix pairs
• Missing optimal parenthesization
• Confusing dimensions order in multiplication