Matrix Chain Multiplication in DSA Typescript - 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] using the matrix chain multiplication algorithm?

DSA Typescript

function matrixChainOrder(p: number[]): number {
  const n = p.length - 1;
  const m: number[][] = 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] = Number.MAX_SAFE_INTEGER;
      for (let k = i; k < j; k++) {
        const 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]));

A 9000

B 6000

C 18000

D 15000

💡 Hint

Think about the cost formula for multiplying two matrices and how the algorithm tries all splits.

The minimum cost is calculated by trying all possible splits. For matrices with dimensions 10x20 and 20x30, the cost is 10*20*30 = 6000 scalar multiplications.

🧠 Conceptual

intermediate

Understanding the Role of the 'k' Variable in Matrix Chain Multiplication

In the matrix chain multiplication dynamic programming solution, what does the variable 'k' represent inside the nested loops?

A 'k' represents the index where the matrix chain is split to compute the cost of multiplication.

B 'k' is the total number of matrices in the chain.

C 'k' is the dimension of the resulting matrix after multiplication.

D 'k' is the minimum cost found so far for multiplying the matrices.

💡 Hint

Think about how the algorithm tries different places to split the chain.

The variable 'k' is used to try all possible places to split the matrix chain between i and j to find the minimum multiplication cost.

🔧 Debug

advanced

Identify the Error in Matrix Chain Multiplication Implementation

What error will the following TypeScript code produce when executed?

DSA Typescript

function matrixChainOrder(p: number[]): number {
  const n = p.length - 1;
  const m: number[][] = 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] = Number.MAX_SAFE_INTEGER;
      for (let k = i; k < j; k++) {
        const 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]));

A TypeError: Cannot read property 'j+1' of undefined

B IndexError: Array index out of bounds

C The function returns 0 instead of the correct minimum cost

D No error, outputs 6000

💡 Hint

Check the loop boundaries carefully, especially the inner loop conditions.

The original code had 'i < n - L' which caused skipping the last valid index. Changing it to 'i <= n - L' fixes the issue and the function outputs the correct minimum cost 6000.

❓ Predict Output

advanced

Output of Matrix Chain Multiplication with Multiple Matrices

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

DSA Typescript

function matrixChainOrder(p: number[]): number {
  const n = p.length - 1;
  const m: number[][] = 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] = Number.MAX_SAFE_INTEGER;
      for (let k = i; k < j; k++) {
        const 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([40, 20, 30, 10, 30]));

A 26000

B 30000

C 45000

D 35000

💡 Hint

Try to find the optimal split points to minimize multiplications.

The minimum cost is 26000 scalar multiplications by splitting the chain optimally.

🚀 Application

expert

Number of Optimal Parenthesizations for Matrix Chain Multiplication

Given the matrix dimensions [10, 20, 30, 40], how many different ways are there to parenthesize the matrix chain multiplication that achieve the minimum number of scalar multiplications?

A 3

B 2

C 1

D 4

💡 Hint

Consider how many ways the chain can be split to achieve the same minimum cost.

There is exactly 1 way to parenthesize the chain [10,20,30,40] that yields the minimum cost of 18000: ((A1 A2) A3). The other parenthesization (A1 (A2 A3)) costs 32000.

Practice

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

easy

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

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

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

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 find the best way to multiply matrices to minimize cost.

2. Step 2: Identify what is minimized

   The cost minimized is the number of scalar multiplications, not the number of matrices or direct multiplication.

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:

• Confusing multiplication count with matrix count
• Thinking the problem multiplies matrices directly
• Assuming maximum matrices multiplication is the goal

2. Which of the following is the correct TypeScript function signature for a recursive Matrix Chain Multiplication function that takes an array of matrix dimensions?

easy

A. function matrixChainOrder(dimensions: string): number

B. function matrixChainOrder(dimensions: number): number[]

C. function matrixChainOrder(dimensions: number[]): number

D. function matrixChainOrder(dimensions: number[][]): void

Solution

1. Step 1: Identify input type for matrix dimensions

   Matrix dimensions are given as an array of numbers representing sizes, so number[] is correct.

2. Step 2: Identify return type for minimum cost

   The function returns the minimum number of scalar multiplications, which is a number.

3. Final Answer:

   function matrixChainOrder(dimensions: number[]): number -> Option C

4. Quick Check:

   Input = number[], Output = number [OK]

Hint: Input is array of numbers, output is a single number [OK]

Common Mistakes:

• Using string instead of number array for dimensions
• Returning an array instead of a number
• Using 2D array or void return incorrectly

3. What is the output of the following TypeScript code snippet?

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

medium

A. 18000

B. 24000

C. 30000

D. 60000

Solution

1. Step 1: Understand the input and matrix dimensions

   The input array [10, 20, 30, 40] represents matrices of dimensions 10x20, 20x30, and 30x40.

2. Step 2: Calculate minimum multiplication cost

   Possible parenthesizations:
   - (A1 x (A2 x A3)) = 20*30*40 + 10*20*40 = 24000 + 8000 = 32000
   - ((A1 x A2) x A3) = 10*20*30 + 10*30*40 = 6000 + 12000 = 18000 (minimum)

3. Final Answer:

   18000 -> Option A

4. Quick Check:

   Minimum multiplications = 18000 [OK]

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

Common Mistakes:

• Calculating cost for only one parenthesization
• Mixing matrix dimensions order
• Forgetting to add multiplication costs correctly

4. Identify the error in the following TypeScript code snippet for Matrix Chain Multiplication and select the fix:

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

medium

A. Change p[i] * p[k] * p[j] to p[i] * p[k + 1] * p[j + 1]

B. Initialize m with Infinity instead of 0

C. Change loop for (let k = i; k <= j; k++) to k < j

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

Solution

1. Step 1: Identify the multiplication cost formula error

   The cost formula uses incorrect indices: p[i] * p[k] * p[j] should use p[k + 1] and p[j + 1] to represent matrix dimensions correctly.

2. Step 2: Confirm other parts are correct

   Initialization with 0 is correct for diagonal, loop ranges are correct, and return index is correct.

3. Final Answer:

   Change p[i] * p[k] * p[j] to p[i] * p[k + 1] * p[j + 1] -> Option A

4. Quick Check:

   Correct cost formula indices fix error [OK]

Hint: Check matrix dimension indices carefully in cost formula [OK]

Common Mistakes:

• Using wrong indices for matrix dimensions in cost calculation
• Changing loop bounds unnecessarily
• Returning wrong matrix cell

5. Given matrices with dimensions [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. 3500

Solution

1. Step 1: Understand matrix dimensions and chain length

   The array represents matrices: 5x10, 10x3, 3x12, 12x5, 5x50, 50x6 (6 matrices).

2. Step 2: Use dynamic programming to find minimum cost

   By calculating all splits and costs:
   - Minimum cost found is 2010 scalar multiplications.

3. Final Answer:

   2010 -> Option B

4. Quick Check:

   Minimum scalar multiplications = 2010 [OK]

Hint: Use DP table to check all splits and pick minimum [OK]

Common Mistakes:

• Calculating only one split option
• Mixing up matrix dimensions order
• Forgetting to add multiplication costs for subproblems