Matrix Chain Multiplication in DSA Rust - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the DP table with infinity.

DSA Rust

let n = dims.len();
let mut dp = vec![vec![[1]; n]; n];

Drag options to blanks, or click blank then click option'

A0

B1

Cusize::MAX

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Initializing with 0 causes no updates since costs are positive and greater than 0.

Using 1 or -1 is incorrect for infinity.

The DP table should be initialized with usize::MAX to represent infinity for uncomputed minimum costs. The diagonal will be overridden to 0.

2fill in blank

medium

Complete the code to set the initial cost of multiplying one matrix to zero.

DSA Rust

for i in 0..n {
    dp[i][i] = [1];
}

Drag options to blanks, or click blank then click option'

A0

Busize::MAX

C1

D-1

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[i][i] to 1 or -1 instead of 0.

Leaving it as usize::MAX would make single matrix cost infinity.

The cost to multiply a single matrix is zero, so dp[i][i] should be set to 0.

3fill in blank

hard

Fix the error in the cost calculation inside the nested loops.

DSA Rust

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

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Adims[j + 1]

Bdims[k]

Cdims[j]

Ddims[i + 1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using dims[j] instead of dims[j+1] uses wrong dimension.

Using dims[k] or dims[i+1] causes wrong cost calculation.

The cost formula is dp[i][k] + dp[k+1][j] + dims[i] * dims[k+1] * dims[j+1], where dims[j+1] is the number of columns of the last matrix in the chain.

4fill in blank

hard

Fill both blanks to correctly update the minimum cost in the DP table.

DSA Rust

if cost < dp[[1]][[2]] {
    dp[i][j] = cost;
}

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Ai

Bj

Ck

Dk + 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using k or k+1 as indices instead of i and j.

Updating dp[k][j] or dp[i][k] instead of dp[i][j].

We compare the new cost with dp[i][j] to update the minimum cost for multiplying matrices from i to j.

5fill in blank

hard

Fill all three blanks to complete the nested loops for the matrix chain multiplication algorithm.

DSA Rust

for length in 2..=[1] {
    for i in 0..=[2] - length {
        let j = i + length - [3];
        // compute dp[i][j]
    }
}

Drag options to blanks, or click blank then click option'

An

B1

C2

Dn - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using n instead of n-1 for loop limits.

Incorrect calculation of j index.

The outer loop runs length from 2 to n-1, inner loop i runs from 0 to n-1-length, and j is calculated as i + length - 1.

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