Matrix Chain Multiplication in DSA Rust - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Matrix Chain Multiplication Cost Calculation

What is the output of the following Rust code that calculates the minimum multiplication cost for a given matrix chain?

DSA Rust

fn matrix_chain_order(p: &[usize]) -> usize {
    let n = p.len() - 1;
    let mut m = vec![vec![0usize; n]; n];

    for l in 2..=n {
        for i in 0..=n - l {
            let j = i + l - 1;
            m[i][j] = usize::MAX;
            for k in i..j {
                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;
                }
            }
        }
    }
    m[0][n - 1]
}

fn main() {
    let dims = vec![10usize, 20usize, 30usize, 40usize, 30usize];
    println!("{}", matrix_chain_order(&dims));
}

A 30000

B 40000

C 36000

D 26000

💡 Hint

Think about the optimal way to parenthesize the matrices to minimize multiplications.

The minimum number of scalar multiplications needed to multiply matrices with dimensions 10x20, 20x30, 30x40, and 40x30 is 30000.

The code uses dynamic programming to find this minimum cost.

🧠 Conceptual

intermediate

Understanding Matrix Chain Multiplication Dimensions

Given matrices A (5x10), B (10x3), C (3x12), what is the dimension of the resulting matrix after multiplying A * B * C?

A 5 x 3

B 10 x 3

C 5 x 12

D 3 x 12

💡 Hint

Matrix multiplication results in rows of the first and columns of the last matrix.

Multiplying A (5x10) and B (10x3) results in a 5x3 matrix.

Then multiplying that result (5x3) with C (3x12) results in a 5x12 matrix.

🔧 Debug

advanced

Identify the Bug in Matrix Chain Multiplication Implementation

What error will the following Rust code produce when run?

DSA Rust

fn matrix_chain_order(p: &[usize]) -> usize {
    let n = p.len();
    let mut m = vec![vec![0usize; n]; n];

    for l in 2..=n {
        for i in 0..=n - l {
            let j = i + l - 1;
            m[i][j] = usize::MAX;
            for k in i..j {
                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;
                }
            }
        }
    }
    m[0][n - 1]
}

fn main() {
    let dims = vec![10usize, 20usize, 30usize, 40usize, 30usize];
    println!("{}", matrix_chain_order(&dims));
}

A Prints incorrect minimum cost

B Index out of bounds panic at runtime

C Runs correctly and prints 26000

D Compilation error due to type mismatch

💡 Hint

Check how the length n is calculated and used for indexing.

The code sets n = p.len(), but p has length 5 for 4 matrices.

Accessing p[j + 1] when j = n - 1 causes an index out of bounds panic.

🚀 Application

advanced

Minimum Multiplication Cost for Given Dimensions

What is the minimum number of scalar multiplications needed to multiply matrices with dimensions [40, 20, 30, 10, 30]?

A 26000

B 35000

C 30000

D 45000

💡 Hint

Use dynamic programming or try different parenthesizations to find the minimum cost.

The optimal parenthesization results in 26000 scalar multiplications.

🧠 Conceptual

expert

Number of Ways to Parenthesize Matrix Chain

How many ways are there to fully parenthesize a chain of 5 matrices for multiplication?

A 5

B 42

C Catalan number is not applicable here

D 14

💡 Hint

The number of ways corresponds to the Catalan number for n-1 matrices.

The number of ways to parenthesize n matrices is the (n-1)th Catalan number.

For 5 matrices, it's the 4th Catalan number = 14.

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