Matrix Chain Multiplication in DSA Javascript - Execution Trace

Concept Flow - Matrix Chain Multiplication

Start with matrix dimensions array

↓

Initialize DP table for costs

↓

Set cost of single matrices to 0

↓

For chain lengths from 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-1

↓

Calculate cost = cost[i

↓

Update cost[i

↓

Repeat until full chain cost computed

↓

Result is cost[1

The flow shows how we fill a table with minimum multiplication costs for matrix chains of increasing length, trying all splits.

Execution Sample

DSA Javascript

const dims = [10, 20, 30, 40];
const n = dims.length;
const cost = Array.from({length: n}, () => Array(n).fill(0));
for (let length = 2; length < n; length++) {
  for (let i = 1; i < n - length + 1; i++) {
    let j = i + length - 1;
    cost[i][j] = Infinity;
    for (let k = i; k < j; k++) {
      let q = cost[i][k] + cost[k+1][j] + dims[i-1]*dims[k]*dims[j];
      if (q < cost[i][j]) cost[i][j] = q;
    }
  }
}
console.log(cost[1][n-1]);

This code calculates the minimum number of scalar multiplications needed to multiply a chain of matrices with given dimensions.

Execution Table

Step	Operation	Indices (i,j)	Split k	Cost Computed	Cost Table State (partial)
1	Initialize cost[i][i] = 0	i=1..3, j=i	-	0	[[0,0,0,0],[0,0,0,0],[0,0,0,0],[0,0,0,0]]
2	Chain length = 2	i=1, j=2	k=1	cost[1][1]+cost[2][2]+10*20*30=0+0+6000=6000	[[0,0,0,0],[0,0,6000,0],[0,0,0,0],[0,0,0,0]]
3	Chain length = 2	i=2, j=3	k=2	cost[2][2]+cost[3][3]+20*30*40=0+0+24000=24000	[[0,0,0,0],[0,0,6000,0],[0,0,0,24000],[0,0,0,0]]
4	Chain length = 3	i=1, j=3	k=1	cost[1][1]+cost[2][3]+10*20*40=0+24000+8000=32000	[[0,0,0,0],[0,0,6000,0],[0,0,0,24000],[0,0,0,0]]
5	Chain length = 3	i=1, j=3	k=2	cost[1][2]+cost[3][3]+10*30*40=6000+0+12000=18000	[[0,0,0,0],[0,0,6000,18000],[0,0,0,24000],[0,0,0,0]]
6	Minimum cost for full chain	i=1, j=3	-	18000	[[0,0,0,0],[0,0,6000,18000],[0,0,0,24000],[0,0,0,0]]

💡 All chain lengths processed; minimum multiplication cost is cost[1][3] = 18000

Variable Tracker

Variable	Start	After Step 2	After Step 3	After Step 5	Final
cost[1][1]	0	0	0	0	0
cost[2][2]	0	0	0	0	0
cost[3][3]	0	0	0	0	0
cost[1][2]	Infinity	6000	6000	6000	6000
cost[2][3]	Infinity	Infinity	24000	24000	24000
cost[1][3]	Infinity	Infinity	Infinity	18000	18000

Key Moments - 3 Insights

Why do we set cost[i][i] = 0 at the start?

Because multiplying one matrix alone requires no multiplication, so cost is zero (see Step 1 in execution_table).

Why do we try all split points k between i and j-1?

Because the chain can be split at any matrix between i and j to find the minimal cost (see Steps 4 and 5 where k=1 and k=2 are tried).

Why is the final answer cost[1][n-1]?

Because cost[1][n-1] stores the minimal cost to multiply the entire chain from first to last matrix (see Step 6).

Visual Quiz - 3 Questions

Look at the execution_table at Step 2, what is the cost computed for multiplying matrices 1 and 2?

A 18000

B 24000

C 6000

D 0

💡 Hint

Check the 'Cost Computed' column at Step 2 in execution_table.

At which step does the algorithm find the minimum cost for multiplying matrices 1 to 3?

A Step 4

B Step 5

C Step 3

D Step 1

💡 Hint

Look at the 'Minimum cost for full chain' operation in execution_table.

If the dimension array dims changed to [10, 20, 10, 40], how would the cost at Step 5 change?

A It would decrease

B It would stay the same

C It would increase

D It would be zero

💡 Hint

Check how cost is calculated using dims[i-1]*dims[k]*dims[j] in Step 5.

Concept Snapshot

Matrix Chain Multiplication:
- Input: array of matrix dimensions dims
- Use DP table cost[i][j] for min multiplications from i to j
- cost[i][i] = 0 (single matrix)
- For chain lengths 2 to n, try all splits k
- cost[i][j] = min(cost[i][k] + cost[k+1][j] + dims[i-1]*dims[k]*dims[j])
- Result: cost[1][n-1]

Full Transcript

Matrix Chain Multiplication finds the minimum number of scalar multiplications needed to multiply a chain of matrices. We use a dynamic programming table to store costs for multiplying subchains. Initially, cost for single matrices is zero. Then, for increasing chain lengths, we try all possible split points to find the minimal cost. The final answer is stored in cost[1][n-1]. This process reduces the exponential complexity of naive multiplication to polynomial time.

Practice

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

easy

A. To count the total number of matrices in the chain

B. To multiply all matrices in the chain in the given order without optimization

C. To find the maximum number of matrices that can be multiplied

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

Solution

1. Step 1: Understand the problem objective

   The Matrix Chain Multiplication problem aims to multiply a sequence of matrices with minimum cost.

2. Step 2: Identify what is minimized

   The cost minimized is the number of scalar multiplications, not just multiplying in order or counting matrices.

3. Final Answer:

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

4. Quick Check:

   Goal = minimum scalar multiplications [OK]

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

Common Mistakes:

• Thinking the problem is about counting matrices
• Assuming multiplication order is fixed without optimization
• Confusing maximum with minimum cost

2. Which of the following is the correct way to declare the dimensions array p for matrices A1 (10x30), A2 (30x5), A3 (5x60) in JavaScript?

easy

A. const p = [10, 30, 5, 60];

B. const p = [[10, 30], [30, 5], [5, 60]];

C. const p = [10, 30, 5];

D. const p = [30, 5, 60];

Solution

1. Step 1: Understand the dimensions array format

   The array p stores dimensions such that matrix Ai has dimensions p[i-1] x p[i].

2. Step 2: Apply to given matrices

   For A1 (10x30), A2 (30x5), A3 (5x60), p = [10, 30, 5, 60].

3. Final Answer:

   const p = [10, 30, 5, 60]; -> Option A

4. Quick Check:

   Dimensions array = [10,30,5,60] [OK]

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

Common Mistakes:

• Using nested arrays instead of flat array
• Missing last dimension in array
• Confusing matrix dimensions order

3. What is the output of the following JavaScript code snippet for matrix dimensions p = [40, 20, 30, 10, 30]?

function matrixChainOrder(p) {
  const n = p.length - 1;
  const m = Array.from({ length: n }, () => Array(n).fill(0));
  for (let L = 2; L <= n; L++) {
    for (let i = 0; i <= n - L; i++) {
      let j = i + L - 1;
      m[i][j] = Infinity;
      for (let k = i; k < j; k++) {
        let q = m[i][k] + m[k + 1][j] + p[i] * p[k + 1] * p[j + 1];
        if (q < m[i][j]) m[i][j] = q;
      }
    }
  }
  return m[0][n - 1];
}
console.log(matrixChainOrder(p));

medium

A. 45000

B. 30000

C. 26000

D. None of the above

Solution

1. Step 1: Understand the code logic

   The code implements the dynamic programming solution for matrix chain multiplication to find minimum cost.

2. Step 2: Calculate minimum multiplication cost for p = [40,20,30,10,30]

   Minimum cost is 26000 scalar multiplications after checking all splits.

3. Final Answer:

   26000 -> Option C

4. Quick Check:

   Minimum cost for given p = 26000 [OK]

Hint: Use DP table to find minimum cost, result is m[0][n-1] [OK]

Common Mistakes:

• Confusing indices i, j, k in loops
• Incorrectly calculating multiplication cost
• Returning wrong DP table value

4. Identify the error in the following JavaScript code snippet for matrix chain multiplication and select the fix:

function matrixChainOrder(p) {
  const n = p.length - 1;
  const m = Array.from({ length: n }, () => Array(n).fill(0));
  for (let L = 2; L <= n; L++) {
    for (let i = 0; i < n - L; i++) {
      let j = i + L - 1;
      m[i][j] = Infinity;
      for (let k = i; k < j; k++) {
        let q = m[i][k] + m[k + 1][j] + p[i] * p[k + 1] * p[j + 1];
        if (q < m[i][j]) m[i][j] = q;
      }
    }
  }
  return m[0][n - 1];
}

medium

A. Initialize m with Infinity instead of 0

B. Change inner loop condition to i <= n - L

C. Use for (let k = i; k <= j; k++) instead of k < j

D. Return m[0][n] instead of m[0][n - 1]

Solution

1. Step 1: Analyze the outer loop range for i

   The loop for (let i = 0; i < n - L; i++) misses the case when i equals n - L, causing incomplete computation.

2. Step 2: Fix loop condition to include i = n - L

   Changing to i <= n - L ensures all subproblems are covered correctly.

3. Final Answer:

   Change inner loop condition to i <= n - L -> Option B

4. Quick Check:

   Loop range fix = i <= n - L [OK]

Hint: Check loop boundaries carefully for full coverage [OK]

Common Mistakes:

• Off-by-one error in loop ranges
• Incorrect DP table initialization
• Wrong return index from DP table

5. Given matrices with dimensions p = [5, 10, 3, 12, 5, 50, 6], what is the minimum number of scalar multiplications needed to multiply the entire chain? Use the matrix chain multiplication approach.

hard

A. 1510

B. 2010

C. 4050

D. None of the above

Solution

1. Step 1: Understand the problem and dimensions

   We have 6 matrices with dimensions from p array: A1(5x10), A2(10x3), A3(3x12), A4(12x5), A5(5x50), A6(50x6).

2. Step 2: Apply matrix chain multiplication DP to find minimum cost

   Calculating minimum scalar multiplications yields 1510 as the optimal cost.

3. Final Answer:

   1510 -> Option A

4. Quick Check:

   Minimum scalar multiplications = 1510 [OK]

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

Common Mistakes:

• Calculating cost for wrong matrix pairs
• Missing optimal parenthesization
• Confusing dimensions order in multiplication