Matrix Chain Multiplication in DSA Javascript - Build from Scratch

Matrix Chain Multiplication

📖 Scenario: Imagine you have a series of matrices that you want to multiply together. The order in which you multiply them can change how many calculations you need. This project helps you find the best order to multiply matrices to do the least work.

🎯 Goal: You will write a program that takes the sizes of matrices and finds the minimum number of multiplications needed to multiply them all together.

📋 What You'll Learn

Create an array called dimensions with the exact values [40, 20, 30, 10, 30]

Create a variable called n that stores the length of dimensions

Write a function called matrixChainOrder that takes dimensions and n as inputs and returns the minimum multiplication cost

Print the minimum multiplication cost using console.log

💡 Why This Matters

🌍 Real World

Matrix chain multiplication is used in computer graphics, scientific computing, and optimizing database queries where multiple matrix multiplications happen.

💼 Career

Understanding dynamic programming and optimization problems like matrix chain multiplication is important for software engineers working on performance-critical applications.

1

Create the dimensions array

Create an array called dimensions with these exact values: [40, 20, 30, 10, 30]

DSA Javascript

# Your code here

Hint

Use const to create the array with the exact numbers given.

2

Create the variable for array length

Create a variable called n that stores the length of the dimensions array

DSA Javascript

const dimensions = [40, 20, 30, 10, 30];
# Your code here

Hint

Use dimensions.length to get the number of elements.

3

Write the matrixChainOrder function

Write a function called matrixChainOrder that takes dimensions and n as inputs and returns the minimum number of multiplications needed to multiply the chain of matrices. Use dynamic programming with a 2D array called dp to store results. Initialize dp[i][i] to 0 for all i. Use nested loops with variables length, i, j, and k to fill dp. Calculate cost as dp[i][k] + dp[k+1][j] + dimensions[i-1] * dimensions[k] * dimensions[j] and update dp[i][j] with the minimum cost.

DSA Javascript

const dimensions = [40, 20, 30, 10, 30];
const n = dimensions.length;

# Your code here

Hint

Use a 2D array dp to store minimum costs. Start with chains of length 2 and increase. Calculate cost for each split and keep the minimum.

4

Print the minimum multiplication cost

Call the function matrixChainOrder with dimensions and n and print the result using console.log

DSA Javascript

const dimensions = [40, 20, 30, 10, 30];
const n = dimensions.length;

function matrixChainOrder(dimensions, n) {
    const dp = Array.from({ length: n }, () => Array(n).fill(0));

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

# Your code here

Hint

Call matrixChainOrder(dimensions, n) and print the result with console.log.

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