Matrix Chain Multiplication in DSA C++ - Cheat Sheet & Quick Revision

Recall & Review

beginner

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

To find the most efficient way to multiply a chain of matrices by minimizing the total number of scalar multiplications.

beginner

What does the dimension array represent in Matrix Chain Multiplication?

An array where each element represents the number of rows or columns of matrices in the chain. For n matrices, the array has n+1 elements representing dimensions.

intermediate

Explain the role of the 'm' table in the dynamic programming solution.

The 'm' table stores the minimum number of scalar multiplications needed to multiply matrices from index i to j, helping avoid repeated calculations.

intermediate

What is the time complexity of the Matrix Chain Multiplication dynamic programming solution?

O(n^3), where n is the number of matrices in the chain.

beginner

Why is Matrix Chain Multiplication not solved by simply multiplying matrices in the given order?

Because the order of multiplication affects the total number of scalar multiplications, and some orders are much more efficient than others.

What does the Matrix Chain Multiplication problem optimize?

A The order of matrix multiplication to minimize scalar multiplications

B The size of matrices

C The number of matrices in the chain

D The addition of matrices

The problem focuses on finding the multiplication order that results in the least scalar multiplications.

If you have matrices with dimensions 10x30, 30x5, and 5x60, what is the dimension array?

A [10, 5, 30, 60]

B [10, 30, 60]

C [10, 30, 5, 60]

D [30, 5, 60]

The dimension array lists the row count of the first matrix and the column counts of each matrix in order.

What does the entry m[i][j] represent in the DP table?

A Dimensions of matrix i

B Minimum cost to multiply matrices from i to j

C Number of matrices between i and j

D Maximum cost to multiply matrices from i to j

m[i][j] stores the minimum scalar multiplications needed to multiply matrices from i to j.

What is the base case cost for multiplying one matrix (i == j)?

A 0

B 1

C Dimension of matrix

D Undefined

Multiplying one matrix alone requires no scalar multiplications, so cost is zero.

Which programming technique is used to solve Matrix Chain Multiplication efficiently?

A Divide and Conquer without memoization

B Greedy Algorithm

C Brute Force only

D Dynamic Programming

Dynamic Programming stores intermediate results to avoid repeated calculations.

Describe how the dynamic programming table is filled in the Matrix Chain Multiplication problem.

Think about starting with single matrices and building up to the full chain.

Explain why the order of multiplying matrices affects the total number of scalar multiplications.

Consider multiplying a small matrix with a large one versus the opposite order.

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

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 correct goal

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

3. Final Answer:

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

4. Quick Check:

   Goal = Minimum multiplications [OK]

Hint: Remember: minimize multiplication cost, not just multiply [OK]

Common Mistakes:

• Thinking the goal is to multiply matrices directly without optimization
• Confusing counting matrices with minimizing operations
• Assuming maximum multiplications is the goal

2. Which of the following is the correct way to declare the dimensions array for 4 matrices with dimensions 10x20, 20x30, 30x40, and 40x50 in C++?

easy

A. int dims[] = {10, 20, 30, 40, 50};

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

C. int dims[] = {10, 20, 30, 40, 50, 60};

D. int dims[] = {20, 30, 40, 50};

Solution

1. Step 1: Understand the dimensions array format

   For n matrices, the dimensions array has n+1 elements representing rows and columns of each matrix.

2. Step 2: Apply to given matrices

   For matrices 10x20, 20x30, 30x40, 40x50, dims = {10, 20, 30, 40, 50}.

3. Final Answer:

   int dims[] = {10, 20, 30, 40, 50}; -> Option A

4. Quick Check:

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

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

Common Mistakes:

• Using fewer elements than needed in the dimensions array
• Including extra unrelated numbers
• Confusing matrix dimensions order

3. What is the output of the following C++ code snippet for Matrix Chain Multiplication cost calculation?

int dims[] = {5, 10, 3, 12, 5};
int n = 4;
int m[5][5] = {0};
for (int L = 2; L <= n; L++) {
  for (int i = 1; i <= n - L + 1; i++) {
    int j = i + L - 1;
    m[i][j] = 1000000;
    for (int k = i; k < j; k++) {
      int cost = m[i][k] + m[k+1][j] + dims[i-1]*dims[k]*dims[j];
      if (cost < m[i][j]) m[i][j] = cost;
    }
  }
}
std::cout << m[1][n] << std::endl;

medium

A. 270

B. 150

C. 201

D. 405

Solution

1. Step 1: Understand matrix dimensions and indices

   Dimensions: 5x10, 10x3, 3x12, 12x5; n=4 matrices; m[i][j] stores min cost from i to j.

2. Step 2: Calculate minimum multiplication cost

   Length 2: m[1][2]=150, m[2][3]=360, m[3][4]=180
   Length 3: m[1][3]=min(0+360+600,150+0+180)=330; m[2][4]=min(0+180+150,360+0+600)=330
   Length 4: m[1][4]=min(0+330+250,150+180+75,330+0+300)=405

3. Final Answer:

   405 -> Option D

4. Quick Check:

   Minimum cost = 405 [OK]

Hint: Trace splits and pick minimum cost [OK]

Common Mistakes:

• Mixing up indices i, j, k
• Incorrect cost formula application
• Forgetting to initialize m[i][j] to max

4. Identify the error in the following C++ code snippet for Matrix Chain Multiplication and choose the correct fix:

int dims[] = {40, 20, 30, 10, 30};
int n = 4;
int m[5][5];
for (int i = 1; i <= n; i++)
  m[i][i] = 0;
for (int L = 2; L <= n; L++) {
  for (int i = 1; i <= n - L + 1; i++) {
    int j = i + L - 1;
    m[i][j] = 0;
    for (int k = i; k < j; k++) {
      int cost = m[i][k] + m[k+1][j] + dims[i-1]*dims[k]*dims[j];
      if (cost < m[i][j])
        m[i][j] = cost;
    }
  }
}

medium

A. Change dims indexing to dims[i-1]*dims[k]*dims[j]

B. Initialize m[i][j] to INT_MAX instead of 0 before inner loop

C. Start i loop from 0 instead of 1

D. Remove the if condition inside the inner loop

Solution

1. Step 1: Identify initialization mistake

   m[i][j] is set to 0 before checking for minimum cost, which causes incorrect comparisons.

2. Step 2: Correct initialization

   m[i][j] should be initialized to a very large number (INT_MAX) to find minimum cost correctly.

3. Final Answer:

   Initialize m[i][j] to INT_MAX instead of 0 before inner loop -> Option B

4. Quick Check:

   Use INT_MAX to find minimum cost [OK]

Hint: Initialize costs to max before minimizing [OK]

Common Mistakes:

• Setting m[i][j] to 0 instead of INT_MAX
• Incorrect dims indexing causing wrong cost
• Ignoring initialization leading to wrong min

5. Given matrices with dimensions 10x15, 15x20, 20x25, and 25x30, what is the minimum number of scalar multiplications needed to multiply the entire chain? Use the Matrix Chain Multiplication approach.

hard

A. 15500

B. 15000

C. 13000

D. 11250

Solution

1. Step 1: List matrix dimensions and possible splits

   Matrices: A1(10x15), A2(15x20), A3(20x25), A4(25x30). dims=[10,15,20,25,30]. Calculate costs for chain length 2 to 4.

2. Step 2: Calculate minimum costs for each subchain

   Length 2: m[1][2]=10*15*20=3000; m[2][3]=15*20*25=7500; m[3][4]=20*25*30=15000
   Length 3: m[1][3]=min(0+7500+10*15*25=11250, 3000+0+10*20*25=8000)=8000;
   m[2][4]=min(0+15000+15*20*30=24000, 7500+0+15*25*30=18750)=18750
   Length 4: m[1][4]=min(0+18750+10*15*30=23250, 3000+15000+10*20*30=24000, 8000+0+10*25*30=15500)=15500

3. Final Answer:

   15500 -> Option A

4. Quick Check:

   Minimum scalar multiplications = 15500 [OK]

Hint: Calculate costs for smaller chains first, then combine [OK]

Common Mistakes:

• Incorrectly calculating subproblems costs
• Mixing up matrix dimensions in cost formula
• Not checking all split points k