Matrix Chain Multiplication in DSA C#

Mental Model

We want to find the best way to multiply a chain of matrices so that the total number of multiplications is as small as possible.

Analogy: Imagine you have several long ropes to tie together. The order in which you tie them affects how much effort you spend. Matrix Chain Multiplication is like finding the order to tie ropes that uses the least effort.

Matrices: A1 [10x30] -> A2 [30x5] -> A3 [5x60]
Goal: Find best order to multiply A1xA2xA3

Initial chain:
A1(10x30) -> A2(30x5) -> A3(5x60)

Dry Run Walkthrough

Input: Matrices dimensions: [10, 30, 5, 60] representing matrices A1(10x30), A2(30x5), A3(5x60)

Goal: Find the minimum number of scalar multiplications needed to multiply A1xA2xA3

Step 1: Calculate cost of multiplying A1 and A2 first, then multiply result with A3

Cost for (A1xA2) = 10x30x5 = 1500 multiplications
Then multiply (A1xA2) with A3: 10x5x60 = 3000
Total cost = 1500 + 3000 = 4500

Why: We try one way of parenthesizing to find its cost

Step 2: Calculate cost of multiplying A2 and A3 first, then multiply A1 with result

Cost for (A2xA3) = 30x5x60 = 9000 multiplications
Then multiply A1 with (A2xA3): 10x30x60 = 18000
Total cost = 9000 + 18000 = 27000

Why: We try the other way of parenthesizing to compare costs

Step 3: Compare both costs and choose the minimum

Costs: (A1xA2)xA3 = 4500 vs A1x(A2xA3) = 27000
Minimum cost = 4500

Why: We want the least multiplication cost for efficiency

Result:

Minimum multiplication cost = 4500

Annotated Code

DSA C#

using System;

class MatrixChainMultiplication
{
    // Function to find minimum multiplication cost
    public static int MatrixChainOrder(int[] dims)
    {
        int n = dims.Length - 1;
        int[,] dp = new int[n, n];

        // dp[i,j] stores minimum cost to multiply matrices from i to j
        for (int length = 2; length <= n; length++)
        {
            for (int i = 0; i <= n - length; i++)
            {
                int j = i + length - 1;
                dp[i, j] = int.MaxValue;
                for (int k = i; k < j; k++)
                {
                    // Cost = cost of left + cost of right + cost of multiplying two results
                    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] = cost;
                    }
                }
            }
        }
        return dp[0, n - 1];
    }

    static void Main()
    {
        int[] dims = { 10, 30, 5, 60 };
        int minCost = MatrixChainOrder(dims);
        Console.WriteLine(minCost);
    }
}

for (int length = 2; length <= n; length++)

increase chain length from 2 to n to consider all subproblems

for (int i = 0; i <= n - length; i++)

set start index of matrix chain

int j = i + length - 1;

set end index of matrix chain

dp[i, j] = int.MaxValue;

initialize minimum cost for this subchain

for (int k = i; k < j; k++)

try all possible split points to find minimum cost

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

calculate cost of splitting at k

if (cost < dp[i, j]) dp[i, j] = cost;

update minimum cost if current split is cheaper

return dp[0, n - 1];

return minimum cost for full chain

Output

4500

Complexity Analysis

Time: O(n^3) because we check all subchains and all split points for each subchain

Space: O(n^2) because we store minimum costs for all subchains in a 2D table

vs Alternative: Naive approach tries all parenthesizations recursively with exponential time; DP reduces it to polynomial time by storing intermediate results

Edge Cases

Single matrix (dims length = 2)

No multiplication needed, cost is zero

DSA C#

for (int length = 2; length <= n; length++)

Two matrices (dims length = 3)

Only one way to multiply, cost is straightforward

DSA C#

for (int k = i; k < j; k++)

Matrices with same dimensions

Algorithm still finds minimum cost correctly

DSA C#

if (cost < dp[i, j]) dp[i, j] = cost;

When to Use This Pattern

When you see a problem asking for the best way to multiply a sequence of matrices or combine operations with different costs, reach for Matrix Chain Multiplication dynamic programming to minimize total cost.

Common Mistakes

Mistake: Confusing matrix dimensions order when calculating cost
Fix: Remember cost formula uses dims[i] * dims[k+1] * dims[j+1] to match matrix multiplication rules

Summary

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

Use when you want to optimize the order of matrix multiplications to reduce computation.

The key insight is to break the problem into smaller subproblems and store their solutions to avoid repeated work.

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