Matrix Chain Multiplication in DSA C# - Execution Trace

Concept Flow - Matrix Chain Multiplication

Start with matrix dimensions array

↓

Initialize DP table with zeros on diagonal

↓

For chain length = 2 to n

↓

For each start index i

↓

Calculate end index j = i + chain length - 1

↓

Try all split points k between i and j

↓

Calculate cost = cost(i,k) + cost(k+1,j) + multiplication cost

↓

Update DP[i

↓

Repeat until full chain cost is computed

↓

Result is DP[0

This flow shows how the algorithm fills a table to find the cheapest way to multiply a chain of matrices by trying all split points and storing minimum costs.

Execution Sample

DSA C#

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

This code calculates the minimum multiplication cost for matrix chain using dynamic programming.

Execution Table

Step	Operation	Indices (i,j)	Split k	Cost Calculated	DP Table Update	DP Table Visual State
1	Initialize diagonal	(0,0)	-	0	dp[0,0] = 0	[0, -, -, -] [-, -, -, -] [-, -, -, -] [-, -, -, -]
2	Initialize diagonal	(1,1)	-	0	dp[1,1] = 0	[0, -, -, -] [-, 0, -, -] [-, -, -, -] [-, -, -, -]
3	Initialize diagonal	(2,2)	-	0	dp[2,2] = 0	[0, -, -, -] [-, 0, -, -] [-, -, 0, -] [-, -, -, -]
4	Initialize diagonal	(3,3)	-	0	dp[3,3] = 0	[0, -, -, -] [-, 0, -, -] [-, -, 0, -] [-, -, -, 0]
5	Chain length = 2	(0,1)	k=0	dims[0]*dims[1]*dims[2] = 10*20*30=6000	dp[0,1] = 6000	[0, 6000, -, -] [-, 0, -, -] [-, -, 0, -] [-, -, -, 0]
6	Chain length = 2	(1,2)	k=1	20*30*40=24000	dp[1,2] = 24000	[0, 6000, -, -] [-, 0, 24000, -] [-, -, 0, -] [-, -, -, 0]
7	Chain length = 2	(2,3)	k=2	30*40*30=36000	dp[2,3] = 36000	[0, 6000, -, -] [-, 0, 24000, -] [-, -, 0, 36000] [-, -, -, 0]
8	Chain length = 3	(0,2)	k=0	dp[0,0]+dp[1,2]+10*20*40=0+24000+8000=32000	dp[0,2] = 32000 (tentative)	[0, 6000, 32000, -] [-, 0, 24000, -] [-, -, 0, 36000] [-, -, -, 0]
9	Chain length = 3	(0,2)	k=1	dp[0,1]+dp[2,2]+10*30*40=6000+0+12000=18000	dp[0,2] = 18000 (updated)	[0, 6000, 18000, -] [-, 0, 24000, -] [-, -, 0, 36000] [-, -, -, 0]
10	Chain length = 3	(1,3)	k=1	dp[1,1]+dp[2,3]+20*30*30=0+36000+18000=54000	dp[1,3] = 54000 (tentative)	[0, 6000, 18000, -] [-, 0, 24000, 54000] [-, -, 0, 36000] [-, -, -, 0]
11	Chain length = 3	(1,3)	k=2	dp[1,2]+dp[3,3]+20*40*30=24000+0+24000=48000	dp[1,3] = 48000 (updated)	[0, 6000, 18000, -] [-, 0, 24000, 48000] [-, -, 0, 36000] [-, -, -, 0]
12	Chain length = 4	(0,3)	k=0	dp[0,0]+dp[1,3]+10*20*30=0+48000+6000=54000	dp[0,3] = 54000 (tentative)	[0, 6000, 18000, 54000] [-, 0, 24000, 48000] [-, -, 0, 36000] [-, -, -, 0]
13	Chain length = 4	(0,3)	k=1	dp[0,1]+dp[2,3]+10*30*30=6000+36000+9000=51000	dp[0,3] = 51000 (updated)	[0, 6000, 18000, 51000] [-, 0, 24000, 48000] [-, -, 0, 36000] [-, -, -, 0]
14	Chain length = 4	(0,3)	k=2	dp[0,2]+dp[3,3]+10*40*30=18000+0+12000=30000	dp[0,3] = 30000 (updated)	[0, 6000, 18000, 30000] [-, 0, 24000, 48000] [-, -, 0, 36000] [-, -, -, 0]
15	End	-	-	-	-	Final minimum cost dp[0,3] = 30000

💡 All chain lengths processed; dp[0,3] holds minimum multiplication cost.

Variable Tracker

Variable	Start	After Step 5	After Step 9	After Step 11	After Step 14	Final
dp[0,0]	0	0	0	0	0	0
dp[0,1]	-	6000	6000	6000	6000	6000
dp[0,2]	-	-	18000	18000	18000	18000
dp[0,3]	-	-	-	-	30000	30000
dp[1,1]	0	0	0	0	0	0
dp[1,2]	-	-	-	24000	24000	24000
dp[1,3]	-	-	-	48000	48000	48000
dp[2,2]	0	0	0	0	0	0
dp[2,3]	-	-	-	-	36000	36000
dp[3,3]	0	0	0	0	0	0

Key Moments - 3 Insights

Why do we initialize the diagonal of the DP table with zeros?

Because multiplying one matrix alone costs nothing, so dp[i][i] = 0. This is shown in steps 1-4 where dp[i,i] is set to 0.

Why do we try all split points k between i and j?

Because the chain can be split in different ways, and we want the minimum cost. Steps 8-14 show trying different k and updating dp[i,j] with the smallest cost.

Why does the cost formula use dims[i] * dims[k+1] * dims[j+1]?

Because multiplying two matrices of dimensions p x q and q x r costs p*q*r scalar multiplications. dims array stores these dimensions, so this formula calculates the cost of multiplying the two resulting matrices after splitting.

Visual Quiz - 3 Questions

Look at the execution table at step 9, what is the updated dp[0,2] value?

A 32000

B 18000

C 6000

D 24000

💡 Hint

Check the 'DP Table Update' and 'Cost Calculated' columns at step 9 in the execution_table.

At which step does dp[0,3] get its final minimum cost?

A Step 14

B Step 12

C Step 13

D Step 15

💡 Hint

Look at the 'DP Table Update' column for dp[0,3] updates and the final value in the DP Table Visual State.

If the dimension dims[1] changed from 20 to 25, which cost calculation would be affected?

A Cost at step 7 (dp[2,3])

B Cost at step 6 (dp[1,2])

C Cost at step 5 (dp[0,1])

D Cost at step 14 (dp[0,3])

💡 Hint

Check the cost formulas in the execution_table rows and see which use dims[1].

Concept Snapshot

Matrix Chain Multiplication:
- Input: array dims of length n+1 for n matrices
- dp[i,j] = min cost to multiply matrices i to j
- Initialize dp[i,i] = 0
- For chain length l=2 to n:
  For i=0 to n-l:
    j = i+l-1
    dp[i,j] = min over k in [i,j-1] of dp[i,k] + dp[k+1,j] + dims[i]*dims[k+1]*dims[j+1]
- Result: dp[0,n-1]

Full Transcript

Matrix Chain Multiplication finds the cheapest way to multiply a sequence of matrices. We use a table dp where dp[i,j] stores the minimum cost to multiply matrices from i to j. We start by setting dp[i,i] to zero because a single matrix needs no multiplication. Then, for increasing chain lengths, we try all possible split points k between i and j. For each split, we calculate the cost as the sum of costs to multiply the left and right parts plus the cost to multiply the resulting two matrices. We update dp[i,j] with the minimum cost found. After filling the table, dp[0,n-1] gives the minimum multiplication cost for the whole chain. This process is shown step-by-step in the execution table with costs and dp updates.

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