Matrix Chain Multiplication in DSA C# - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the dp table with zeros.

DSA C#

int n = dims.Length - 1;
int[,] dp = new int[n, n];
for (int i = 0; i < n; i++) {
    dp[i, i] = [1];
}

Drag options to blanks, or click blank then click option'

A-1

B1

C0

Dint.MaxValue

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[i, i] to 1 or -1 instead of 0.

Using int.MaxValue which is for initialization of minimum cost comparisons.

The diagonal elements represent single matrices, so their multiplication cost is zero.

2fill in blank

medium

Complete the code to calculate the cost of multiplying two matrices.

DSA C#

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

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Adims[j + 1]

Bdims[j]

Cdims[k]

Ddims[i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using dims[j] incorrectly in the formula.

Using dims[k] or dims[i] incorrectly in the formula.

The cost formula uses dims[i] * dims[k+1] * dims[j+1] for matrix multiplication cost.

3fill in blank

hard

Fix the error in the loop condition to correctly iterate over chain lengths.

DSA C#

for (int length = 2; length <= [1]; length++) {
    for (int i = 0; i <= n - length; i++) {
        int j = i + length - 1;
        // calculation code
    }
}

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An - 2

Bn

Cn - 1

Dn + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using n - 1 which misses the last chain length.

Using n + 1 which causes out of range errors.

The chain length goes up to n, which is the number of matrices.

4fill in blank

hard

Fill both blanks to correctly update the minimum cost in dp table.

DSA C#

for (int k = i; k < j; k++) {
    int cost = dp[i, k] + dp[k + 1, j] + dims[i] * dims[k + 1] * dims[j + 1];
    if (cost < dp[i, j]) {
        dp[i, j] = [1];
        split[i, j] = [2];
    }
}

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Acost

Bk

Cdp[i, k]

Ddims[j]

Attempts:

3 left

💡 Hint

Common Mistakes

Assigning dp[i, j] to dp[i, k] instead of cost.

Storing cost instead of k in split array.

We update dp[i, j] with the new minimum cost and record the split position k.

5fill in blank

hard

Fill all three blanks to complete the recursive function to print optimal parenthesization.

DSA C#

void PrintOptimalParens(int[,] split, int i, int j) {
    if (i == j) {
        Console.Write("A" + [1]);
        return;
    }
    Console.Write("(");
    PrintOptimalParens(split, i, [2]);
    PrintOptimalParens(split, [3] + 1, j);
    Console.Write(")");
}

Drag options to blanks, or click blank then click option'

Ai

Bj

Csplit[i, j]

Dsplit[j, i]

Attempts:

3 left

💡 Hint

Common Mistakes

Using j instead of i when printing matrix name.

Using split[j, i] which is incorrect indexing.

When i == j, print matrix name with index i. Recursively print left and right parts using split[i, j].

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 computation.

2. Step 2: Identify the optimization goal

   The goal is to minimize the total scalar multiplications, not just multiply matrices or count them.

3. Final Answer:

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

4. Quick Check:

   Minimum scalar multiplications = To find the minimum number of scalar multiplications needed to multiply a chain of matrices [OK]

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

Common Mistakes:

• Thinking the problem is about counting matrices
• Assuming direct multiplication without optimization
• Confusing maximum with minimum operations

2. Which of the following is the correct way to declare the dimensions array for matrices A1(10x20), A2(20x30), A3(30x40) in C# for Matrix Chain Multiplication?

easy

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

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

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

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

Solution

1. Step 1: Understand the dimensions array format

   For n matrices, dimensions array has length n+1, where dims[i] and dims[i+1] represent matrix i's rows and columns.

2. Step 2: Apply to given matrices

   Given A1(10x20), A2(20x30), A3(30x40), dims = {10, 20, 30, 40} correctly represents all matrices.

3. Final Answer:

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

4. Quick Check:

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

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

Common Mistakes:

• Using array length equal to number of matrices
• Skipping first or last dimension
• Mixing rows and columns order

3. What is the minimum number of scalar multiplications needed to multiply matrices with dimensions {40, 20, 30, 10, 30} using Matrix Chain Multiplication?

medium

A. 18000

B. 26000

C. 24000

D. 30000

Solution

1. Step 1: Understand the matrices and dimensions

   Dimensions array {40, 20, 30, 10, 30} means matrices: A1(40x20), A2(20x30), A3(30x10), A4(10x30).

2. Step 2: Calculate minimum multiplications using DP or manual check

   Optimal parenthesization ((A1 x (A2 x A3)) x A4):
   A2xA3: 20x30x10 = 6000 (result 20x10)
   A1x(20x10): 40x20x10 = 8000 (result 40x10)
   (40x10)xA4(10x30): 40x10x30 = 12000
   Total: 6000 + 8000 + 12000 = 26000.
   Other parenthesizations (e.g., (A1xA2)x(A3xA4) = 69000, ((A1xA2)xA3)xA4 = 48000) are higher.

3. Final Answer:

   26000 -> Option B

4. Quick Check:

   Minimum scalar multiplications = 26000 [OK]

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

Common Mistakes:

• Calculating cost for wrong parenthesization
• Mixing matrix dimensions order
• Ignoring multiplication order impact

4. Identify the error in the following C# code snippet for Matrix Chain Multiplication:

int n = dims.Length;
int[,] dp = new int[n, n];
for (int i = 1; i < n; i++) dp[i, i] = 0;
for (int len = 2; len <= n; len++) {
  for (int i = 1; i <= n - len + 1; i++) {
    int j = i + len - 1;
    dp[i, j] = int.MaxValue;
    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;
    }
  }
}

medium

A. The cost calculation uses wrong indices; it should be 'dims[i] * dims[k+1] * dims[j+1]' instead

B. The initialization 'dp[i, i] = 0;' should start from i=0 instead of i=1

C. The outer loop should run till 'len < n' instead of 'len <= n' to avoid index out of range

D. The dp array should be declared as 'int[,] dp = new int[n-1, n-1];' to match matrix count

Solution

1. Step 1: Check loop bounds for index safety

   n = dims.Length (e.g., 5 for 4 matrices). Outer loop len <= n allows len = n, i=1, j = n, causing dims[j] (dims[n]) index out of range (valid 0 to n-1).

2. Step 2: Other components are correct

   dp initialization i=1 to n-1 covers single matrices 1 to n-1. Cost formula uses correct indices dims[i-1] * dims[k] * dims[j]. dp size n x n sufficient for indices up to n-1.

3. Final Answer:

   The outer loop should run till 'len < n' instead of 'len <= n' to avoid index out of range -> Option C

4. Quick Check:

   len max = dims.Length - 1 to match num matrices [OK]

Hint: Outer loop len < dims.Length to avoid dims[n] access [OK]

Common Mistakes:

• len loop exceeding number of matrices
• Incorrect dp array size
• Misusing dimension indices in cost

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 using Matrix Chain Multiplication?

hard

A. 2010

B. 2015

C. 2011

D. 2500

Solution

1. Step 1: Understand the matrices and dimensions

   Dimensions array {5, 10, 3, 12, 5, 50, 6} means matrices:
   A1(5x10), A2(10x3), A3(3x12), A4(12x5), A5(5x50), A6(50x6).

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

   Calculate minimum cost for all subchains:
   - For example, cost(A1xA2) = 5*10*3 = 150
   - Cost for longer chains computed by checking all splits and picking minimum.
   After full DP computation, minimum cost is 2010 scalar multiplications.

3. Final Answer:

   2010 -> Option A

4. Quick Check:

   Minimum scalar multiplications = 2010 [OK]

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

Common Mistakes:

• Calculating cost for only one split
• Mixing matrix dimensions order
• Ignoring multiplication order impact