Matrix Chain Multiplication in DSA C - Build from Scratch

Matrix Chain Multiplication

📖 Scenario: You work in a software company that optimizes mathematical computations. One common task is to multiply a chain of matrices in the most efficient way to save time and resources.Matrix Chain Multiplication helps find the best order to multiply matrices so that the total number of scalar multiplications is minimized.

🎯 Goal: You will write a program in C that uses dynamic programming to find the minimum number of multiplications needed to multiply a chain of matrices.The program will take the dimensions of matrices and output the minimum cost.

📋 What You'll Learn

Create an integer array dims with the exact values {40, 20, 30, 10, 30}

Create an integer variable n that stores the number of matrices (length of dims - 1)

Create a 2D integer array m of size n x n to store minimum multiplication costs

Implement the matrix chain multiplication dynamic programming logic using nested loops

Print the minimum multiplication cost stored in m[0][n-1]

💡 Why This Matters

🌍 Real World

Matrix chain multiplication is used in computer graphics, scientific computing, and database query optimization to speed up matrix operations.

💼 Career

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

1

Create the dimensions array

Create an integer array called dims with the exact values {40, 20, 30, 10, 30}.

DSA C

# Your code here

Hint

The array dims holds the dimensions of 4 matrices: A1 is 40x20, A2 is 20x30, A3 is 30x10, A4 is 10x30.

2

Set the number of matrices and declare the cost matrix

Create an integer variable called n that stores the number of matrices (which is the length of dims minus 1). Then declare a 2D integer array called m of size n by n to store minimum multiplication costs.

DSA C

int dims[] = {40, 20, 30, 10, 30};
# Your code here

Hint

Use sizeof(dims) / sizeof(dims[0]) - 1 to get the number of matrices.

Declare m as a 2D array to hold costs.

3

Implement the matrix chain multiplication logic

Initialize the diagonal of m with zeros using a for loop with variable i. Then use nested for loops with variables L, i, and k to compute minimum multiplication costs. Use variables j and q inside the loops. Update m[i][j] with the minimum cost found.

DSA C

int dims[] = {40, 20, 30, 10, 30};
int n = sizeof(dims) / sizeof(dims[0]) - 1;
int m[n][n];
// Your code here

Hint

Set the diagonal of m to zero because multiplying one matrix costs nothing.

Use the chain length L from 2 to n.

Calculate cost q for each split and update minimum.

4

Print the minimum multiplication cost

Print the minimum multiplication cost stored in m[0][n-1] using printf with the format string "Minimum number of multiplications is %d\n".

DSA C

int dims[] = {40, 20, 30, 10, 30};
int n = sizeof(dims) / sizeof(dims[0]) - 1;
int m[n][n];

for (int i = 0; i < n; i++) {
    m[i][i] = 0;
}

for (int L = 2; L <= n; L++) {
    for (int i = 0; i <= n - L; i++) {
        int j = i + L - 1;
        m[i][j] = 2147483647; // large number for infinity
        for (int k = i; k < j; k++) {
            int q = m[i][k] + m[k + 1][j] + dims[i] * dims[k + 1] * dims[j + 1];
            if (q < m[i][j]) {
                m[i][j] = q;
            }
        }
    }
}
// Your code here

Hint

Print the value m[0][n-1] which holds the minimum multiplication cost for the full chain.

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 so that the total number of scalar multiplications is minimized.

2. Step 2: Identify the correct goal

   Among the options, only To find the minimum number of scalar multiplications needed to multiply a chain of matrices describes minimizing scalar multiplications, which is the main goal.

3. Final Answer:

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

4. Quick Check:

   Goal = minimize multiplications [OK]

Hint: Remember: minimize multiplication cost, not count matrices [OK]

Common Mistakes:

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

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

easy

A. int p[5] = {10, 20, 30, 40};

B. int p[4] = {10, 20, 30, 40};

C. int p[4] = {10, 20, 30, 40, 50};

D. int p[5] = {10, 20, 30, 40, 50};

Solution

1. Step 1: Understand the dimensions array size

   For n matrices, the dimensions array size is n+1. Here, 4 matrices need an array of size 5.

2. Step 2: Check the array values

   The array should list the row count of the first matrix and then the column counts of each matrix in order: {10, 20, 30, 40, 50}.

3. Final Answer:

   int p[5] = {10, 20, 30, 40, 50}; -> Option D

4. Quick Check:

   Array size = matrices + 1, values match dimensions [OK]

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

Common Mistakes:

• Using array size equal to number of matrices instead of number + 1
• Missing the last dimension in the array
• Including extra or fewer elements than needed

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

int p[] = {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 q = m[i][k] + m[k+1][j] + p[i-1]*p[k]*p[j];
            if (q < m[i][j])
                m[i][j] = q;
        }
    }
}
printf("%d", m[1][n]);

medium

A. 201

B. 405

C. 270

D. 150

Solution

1. Step 1: Understand the matrix dimensions and initialization

   p = {5,10,3,12,5} means matrices: 5x10, 10x3, 3x12, 12x5. The code calculates minimum multiplication cost using dynamic programming.

2. Step 2: Calculate minimum cost m[1][4]

   Calculations:
   - m[1][2] = 5*10*3 = 150
   - m[2][3] = 10*3*12 = 360
   - m[3][4] = 3*12*5 = 180
   Then:
   - m[1][3] = min( m[1][1]+m[2][3]+5*10*12=0+360+600=960, m[1][2]+m[3][3]+5*3*12=150+0+180=330 ) = 330
   - m[2][4] = min( m[2][2]+m[3][4]+10*3*5=0+180+150=330, m[2][3]+m[4][4]+10*12*5=360+0+600=960 ) = 330
   Finally:
   - m[1][4] = min( m[1][1]+m[2][4]+5*10*5=0+330+250=580, m[1][2]+m[3][4]+5*3*5=150+180+75=405, m[1][3]+m[4][4]+5*12*5=330+0+300=630 ) = 405

3. Final Answer:

   405 -> Option B

4. Quick Check:

   Minimum cost = 405 [OK]

Hint: Trace inner loops carefully to find minimum cost [OK]

Common Mistakes:

• Misindexing arrays starting from 0 instead of 1
• Incorrectly calculating multiplication costs
• Ignoring initialization of m[i][j]

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

int p[] = {2, 3, 4};
int n = 2;
int m[3][3] = {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] = INT_MAX;
        for (int k = i; k < j; k++) {
            int q = m[i][k] + m[k+1][j] + p[i]*p[k]*p[j];
            if (q < m[i][j])
                m[i][j] = q;
        }
    }
}
printf("%d", m[1][n]);

medium

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

B. Initialize m array with -1 instead of 0

C. Change loop to start i from 0 instead of 1

D. Use float instead of int for m array

Solution

1. Step 1: Identify the indexing error in cost calculation

   The multiplication cost formula uses p[i]*p[k]*p[j], but correct indexing for dimensions array is p[i-1]*p[k]*p[j] because p has length n+1 and matrices are from i to j.

2. Step 2: Fix the formula to use correct indices

   Changing to p[i-1]*p[k]*p[j] fixes the dimension mismatch and ensures correct cost calculation.

3. Final Answer:

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

4. Quick Check:

   Dimension indexing = p[i-1]*p[k]*p[j] [OK]

Hint: Remember dimension array indexing starts at i-1 for cost [OK]

Common Mistakes:

• Using wrong indices in multiplication cost formula
• Not initializing m array properly
• Incorrect loop ranges causing out-of-bound errors

5. Given matrices with dimensions p = {40, 20, 30, 10, 30}, what is the minimum number of scalar multiplications needed to multiply the chain? Use the Matrix Chain Multiplication approach.

hard

A. 18000

B. 30000

C. 26000

D. 24000

Solution

1. Step 1: Understand matrix dimensions and problem size

   p = {40, 20, 30, 10, 30} means matrices: 40x20, 20x30, 30x10, 10x30. We need minimum multiplication cost for 4 matrices.

2. Step 2: Calculate minimum multiplication cost using DP

   Calculate costs:
   - m[1][2] = 40*20*30 = 24000
   - m[2][3] = 20*30*10 = 6000
   - m[3][4] = 30*10*30 = 9000
   Calculate m[1][3]:
   - (m[1][1]+m[2][3]+40*20*10) = 0 + 6000 + 8000 = 14000
   - (m[1][2]+m[3][3]+40*30*10) = 24000 + 0 + 12000 = 36000
   Minimum = 14000
   Calculate m[2][4]:
   - (m[2][2]+m[3][4]+20*30*30) = 0 + 9000 + 18000 = 27000
   - (m[2][3]+m[4][4]+20*10*30) = 6000 + 0 + 6000 = 12000
   Minimum = 12000
   Calculate m[1][4]:
   - (m[1][1]+m[2][4]+40*20*30) = 0 + 12000 + 24000 = 36000
   - (m[1][2]+m[3][4]+40*30*30) = 24000 + 9000 + 36000 = 69000
   - (m[1][3]+m[4][4]+40*10*30) = 14000 + 0 + 12000 = 26000
   Minimum = 26000

3. Final Answer:

   26000 -> Option C

4. Quick Check:

   Minimum scalar multiplications = 26000 [OK]

Hint: Calculate costs for all splits, pick minimum [OK]

Common Mistakes:

• Calculating only one split instead of all
• Mixing up matrix dimensions in multiplication
• Forgetting to add multiplication costs properly