Matrix Chain Multiplication in DSA Javascript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Matrix Chain Multiplication Cost Calculation

What is the minimum number of scalar multiplications needed to multiply the chain of matrices with dimensions [10, 20, 30, 40] using the Matrix Chain Multiplication algorithm?

DSA Javascript

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([10, 20, 30, 40]));

A 18000

B 30000

C 24000

D 36000

💡 Hint

Think about the cost of multiplying matrices in different orders and choose the minimum.

The minimum cost is calculated by trying all possible splits and choosing the one with the least scalar multiplications. For matrices with dimensions 10x20, 20x30, and 30x40, the minimum cost is 18000.

🧠 Conceptual

intermediate

Understanding Matrix Chain Multiplication Dimensions

Given matrices A (5x10), B (10x3), C (3x12), and D (12x5), what is the dimension of the resulting matrix after multiplying A * B * C * D?

A 5 x 5

B 5 x 12

C 10 x 5

D 12 x 5

💡 Hint

The resulting matrix dimension is the first matrix's row count and the last matrix's column count.

When multiplying matrices, the resulting matrix has the number of rows of the first matrix and the number of columns of the last matrix. Here, A is 5x10 and D is 12x5, so the result is 5x5.

🔧 Debug

advanced

Identify the Bug in Matrix Chain Multiplication Implementation

What error will the following code produce when executed?

DSA Javascript

function matrixChainOrder(p) {
  const n = p.length - 1;
  const m = Array(n).fill(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([10, 20, 30, 40]));

A Output: 18000

B TypeError: Cannot assign to read only property '1' of object '[object Array]'

C ReferenceError: m is not defined

D SyntaxError: Unexpected token

💡 Hint

Check how the 2D array 'm' is initialized and whether each row is a separate array.

The code uses Array(n).fill(Array(n).fill(0)) which fills the outer array with references to the same inner array. Modifying one row modifies all rows, causing assignment errors.

❓ Predict Output

advanced

Output of Matrix Chain Multiplication Parenthesization

What is the optimal parenthesization for matrices with dimensions [40, 20, 30, 10, 30] that minimizes multiplication cost?

DSA Javascript

function printOptimalParens(s, i, j) {
  if (i === j) return `A${i + 1}`;
  return `(${printOptimalParens(s, i, s[i][j])} x ${printOptimalParens(s, s[i][j] + 1, j)})`;
}

function matrixChainOrder(p) {
  const n = p.length - 1;
  const m = Array.from({ length: n }, () => Array(n).fill(0));
  const s = 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;
          s[i][j] = k;
        }
      }
    }
  }
  return printOptimalParens(s, 0, n - 1);
}

console.log(matrixChainOrder([40, 20, 30, 10, 30]));

A ((A1 x (A2 x A3)) x A4)

B ((A1 x A2) x (A3 x A4))

C (A1 x (A2 x (A3 x A4)))

D (((A1 x A2) x A3) x A4)

💡 Hint

The optimal parenthesization minimizes the total scalar multiplications by choosing the best split points.

The optimal order is to multiply A3 and A4 first, then multiply A2 with that result, and finally multiply A1 with the result, represented as (A1 x (A2 x (A3 x A4))).

🧠 Conceptual

expert

Number of Scalar Multiplications for Large Chain

Given a chain of 5 matrices with dimensions [5, 10, 3, 12, 5, 50], what is the minimum number of scalar multiplications required to multiply the entire chain?

A 2015

B 2100

C 2070

D 2010

💡 Hint

Use dynamic programming to find the minimum cost by checking all splits.

The minimum cost is found by trying all possible splits and choosing the minimum. For these dimensions, the minimum scalar multiplications is 2010.

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