Matrix Chain Multiplication in DSA Rust - Execution Trace

Concept Flow - Matrix Chain Multiplication

Start with matrix dimensions array p

↓

Initialize DP table m with zeros

↓

Set chain length l = 2 to n

↓

For each chain length l

↓

For each i from 1 to n-l+1

↓

Set j = i + l - 1

↓

Set m[i

↓

Repeat until full table filled

↓

Result is m[1

We fill a table step-by-step to find the cheapest way to multiply a chain of matrices by trying all split points.

Execution Sample

DSA Rust

let p = [10, 20, 30, 40];
let n = p.len() - 1;
let mut m = vec![vec![0; n]; n];
// Fill m with minimal costs
// Result in m[0][n-1]

This code sets up the dimensions and a table to store minimal multiplication costs for matrix chains.

Execution Table

Step	Chain Length (l)	i	j	k (split)	Cost Calculation	m[i][j] Updated	DP Table State
1	2	1	2	1	m[1][1] + m[2][2] + p[0]*p[1]*p[2] = 0 + 0 + 10*20*30 = 6000	6000	[[0, 6000, 0], [0, 0, 0], [0, 0, 0]]
2	2	2	3	2	m[2][2] + m[3][3] + p[1]*p[2]*p[3] = 0 + 0 + 20*30*40 = 24000	24000	[[0, 6000, 0], [0, 0, 24000], [0, 0, 0]]
3	3	1	3	1	m[1][1] + m[2][3] + p[0]*p[1]*p[3] = 0 + 24000 + 10*20*40 = 32000	32000	[[0, 6000, 32000], [0, 0, 24000], [0, 0, 0]]
4	3	1	3	2	m[1][2] + m[3][3] + p[0]*p[2]*p[3] = 6000 + 0 + 10*30*40 = 18000	18000	[[0, 6000, 18000], [0, 0, 24000], [0, 0, 0]]
Exit	-	-	-	-	-	-	All subproblems computed, minimal cost is m[1][3] = 18000

💡 All subproblems computed, minimal cost is m[1][3] = 18000

Variable Tracker

Variable	Start	After Step 1	After Step 2	After Step 3	After Step 4	Final
m	[[0,0,0],[0,0,0],[0,0,0]]	[[0,6000,0],[0,0,0],[0,0,0]]	[[0,6000,0],[0,0,24000],[0,0,0]]	[[0,6000,32000],[0,0,24000],[0,0,0]]	[[0,6000,18000],[0,0,24000],[0,0,0]]	[[0,6000,18000],[0,0,24000],[0,0,0]]
l	undefined	2	2	3	3	3
i	undefined	1	2	1	1	1
j	undefined	2	3	3	3	3
k	undefined	1	2	1	2	2

Key Moments - 3 Insights

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

Because the minimal cost depends on where we split the chain; checking all k ensures we find the cheapest multiplication order (see execution_table steps 3 and 4).

Why is m[i][i] always zero?

Because multiplying one matrix alone costs nothing, so the diagonal entries m[i][i] are zero (initial state in variable_tracker for m).

Why do we fill the table by increasing chain length l?

Because smaller chains must be solved first to use their results for longer chains, as shown by the order of steps in execution_table.

Visual Quiz - 3 Questions

Look at the execution_table at step 4, what is the updated value of m[1][3]?

A 32000

B 6000

C 18000

D 24000

💡 Hint

Check the 'm[i][j] Updated' column at step 4 in execution_table.

At which step does the chain length l become 3 for the first time?

A Step 3

B Step 2

C Step 1

D Step 4

💡 Hint

Look at the 'Chain Length (l)' column in execution_table rows.

If the dimension array p was changed to [10, 20, 10, 40], how would the cost at step 1 change?

A It would increase

B It would decrease

C It would stay the same

D It would become zero

💡 Hint

Compare the cost formula in step 1: p[0]*p[1]*p[2], changing p[2] from 30 to 10 reduces cost.

Concept Snapshot

Matrix Chain Multiplication:
- Input: array p of matrix dimensions (length n+1 for n matrices)
- Goal: find minimal scalar multiplications to multiply chain
- Use DP table m[n][n], m[i][i] = 0
- Fill m by increasing chain length l from 2 to n
- m[i][j] = min over k of (m[i][k] + m[k+1][j] + p[i-1]*p[k]*p[j])
- Result in m[1][n]

Full Transcript

Matrix Chain Multiplication finds the cheapest way to multiply a sequence of matrices. We use a table m where m[i][j] stores the minimal cost to multiply matrices from i to j. Initially, m[i][i] is zero because one matrix alone needs no multiplication. We fill the table by increasing chain length l, trying all split points k between i and j to find the minimal cost. The final answer is in m[1][n]. The execution table shows step-by-step how costs are computed and updated. Variable tracker shows how m and indices change. Key moments clarify why we try all splits, why diagonal is zero, and why we fill by chain length. The quiz tests understanding of table updates and cost calculations.

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 goal

   The Matrix Chain Multiplication problem aims to find the best way to multiply matrices to minimize cost.

2. Step 2: Identify the cost metric

   The cost is measured by the number of scalar multiplications needed, not just multiplying matrices directly.

3. Final Answer:

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

4. Quick Check:

   Goal = minimize scalar multiplications [OK]

Hint: Focus on minimizing multiplication cost, not counting matrices [OK]

Common Mistakes:

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

2. Which of the following Rust function signatures correctly represents a function to compute the minimum multiplication cost for matrix chain multiplication?

easy

A. fn matrix_chain_order(dimensions: usize) -> usize

B. fn matrix_chain_order(dimensions: Vec<usize>) -> Vec<usize>

C. fn matrix_chain_order(dimensions: &mut [usize]) -> bool

D. fn matrix_chain_order(dimensions: &[usize]) -> usize

Solution

1. Step 1: Analyze parameter type

   The function should take a slice of dimensions because matrices are represented by their dimensions in order.

2. Step 2: Analyze return type

   The function returns the minimum cost as a single number (usize), not a vector or boolean.

3. Final Answer:

   fn matrix_chain_order(dimensions: &[usize]) -> usize -> Option D

4. Quick Check:

   Input slice, output cost number [OK]

Hint: Use slice input and usize output for cost [OK]

Common Mistakes:

• Using Vec instead of slice for input unnecessarily
• Returning a vector instead of a single cost
• Using incorrect parameter types like usize alone

3. Given the dimensions array [10, 20, 30, 40], what is the minimum number of scalar multiplications needed to multiply the chain of matrices?

medium

A. 18000

B. 24000

C. 30000

D. 36000

Solution

1. Step 1: Identify matrices and dimensions

   There are 3 matrices: A1 (10x20), A2 (20x30), A3 (30x40).

2. Step 2: Calculate costs for parenthesizations

   Option 1: (A1 x A2) x A3 = cost1 + cost2 = (10*20*30) + (10*30*40) = 6000 + 12000 = 18000
   Option 2: A1 x (A2 x A3) = cost1 + cost2 = (20*30*40) + (10*20*40) = 24000 + 8000 = 32000

3. Final Answer:

   18000 -> Option A

4. Quick Check:

   Minimum cost = 18000 [OK]

Hint: Try both parenthesizations, pick minimum cost [OK]

Common Mistakes:

• Calculating only one parenthesization cost
• Mixing up matrix dimensions order
• Adding dimensions instead of multiplying

4. Consider this Rust code snippet for matrix chain multiplication cost calculation:

fn matrix_chain_order(dimensions: &[usize]) -> usize {
    let n = dimensions.len();
    let mut dp = vec![vec![0; n]; n];
    for length in 2..n {
        for i in 0..n - length {
            let j = i + length;
            dp[i][j] = usize::MAX;
            for k in i..j {
                let cost = dp[i][k] + dp[k+1][j] + dimensions[i] * dimensions[k+1] * dimensions[j+1];
                if cost < dp[i][j] {
                    dp[i][j] = cost;
                }
            }
        }
    }
    dp[0][n-1]
}

What is the main error causing a panic or wrong result?

medium

A. Index out of bounds due to accessing dimensions[j+1]

B. Incorrect initialization of dp matrix size

C. Loop ranges do not cover all matrix pairs

D. Using usize::MAX causes overflow in cost calculation

Solution

1. Step 1: Check dimensions indexing

   dimensions has length n, valid indices 0 to n-1. Accessing dimensions[j+1] when j = n-1 causes out of bounds.

2. Step 2: Confirm loop ranges and indexing

   Loop runs j up to n-1, so j+1 equals n, which is invalid index for dimensions.

3. Final Answer:

   Index out of bounds due to accessing dimensions[j+1] -> Option A

4. Quick Check:

   dimensions[j+1] out of range [OK]

Hint: Check array indices carefully, especially j+1 [OK]

Common Mistakes:

• Ignoring array bounds on dimensions
• Assuming dp size matches dimensions length
• Not verifying loop end conditions

5. You have matrices with dimensions [5, 10, 3, 12, 5, 50, 6]. Using matrix chain multiplication, what is the minimum number of scalar multiplications needed to multiply the entire chain?

hard

A. 1510

B. 2010

C. 4050

D. 3500

Solution

1. Step 1: Identify matrices and dimensions

   There are 6 matrices: A1(5x10), A2(10x3), A3(3x12), A4(12x5), A5(5x50), A6(50x6).

2. Step 2: Use dynamic programming or known solution

   Using DP or known results, minimum cost is 2010 scalar multiplications for this chain.

3. Final Answer:

   2010 -> Option B

4. Quick Check:

   Minimum cost = 2010 [OK]

Hint: Use DP table or trusted calculator for complex chains [OK]

Common Mistakes:

• Calculating only partial parenthesizations
• Mixing matrix dimensions order
• Forgetting to multiply correct dimension triples