Matrix Chain Multiplication in DSA Go - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to initialize the DP table for matrix chain multiplication.

DSA Go

dp := make([][]int, n)
for i := 0; i < n; i++ {
    dp[i] = make([]int, n)
    dp[i][i] = [1]
}

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

A1

Bn

C-1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

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

Initializing dp with negative values.

Each matrix alone requires zero multiplications, so dp[i][i] = 0.

2fill in blank

medium

Complete the code to set the length of the chain in the outer loop.

DSA Go

for length := 2; length <= [1]; length++ {
    for i := 0; i <= n-length; i++ {
        j := i + length - 1
        // compute dp[i][j]
    }
}

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

Aj

Blength

Cn

Dn-1

Attempts:

3 left

💡 Hint

Common Mistakes

Using n-1 instead of n as the upper limit.

Using the variable j instead of n.

The chain length goes from 2 up to n, the total number of matrices.

3fill in blank

hard

Fix the error in the code to calculate the cost of multiplying matrices between i and j.

DSA Go

for k := i; k < j; k++ {
    cost := dp[i][k] + dp[k+1][j] + [1] * [2] * [3]
    if cost < dp[i][j] {
        dp[i][j] = cost
    }
}

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

Ap[i]

Bp[k+1]

Cp[j+1]

Dp[j]

Attempts:

3 left

💡 Hint

Common Mistakes

Using p[k] instead of p[k+1].

Using p[j] instead of p[j+1].

The multiplication cost is p[i] * p[k+1] * p[j+1], representing the dimensions of the matrices being multiplied.

4fill in blank

hard

Fill both blanks to correctly initialize and update the dp table for matrix chain multiplication.

DSA Go

for i := 0; i < n; i++ {
    dp[i] = make([]int, n)
    dp[i][i] = [1]
}
for length := 2; length <= [2]; length++ {
    // rest of the code
}

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

A0

B1

Cn

Dlength

Attempts:

3 left

💡 Hint

Common Mistakes

Setting dp[i][i] to 1.

Looping length only to n-1.

dp[i][i] is zero because a single matrix needs no multiplication. The length loop runs up to n, the number of matrices.

5fill in blank

hard

Fill all three blanks to complete the cost calculation inside the nested loops.

DSA Go

for k := i; k < j; k++ {
    cost := dp[i][k] + dp[k+1][j] + [1] * [2] * [3]
    if cost < dp[i][j] {
        dp[i][j] = cost
    }
}

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

Ap[i]

Bp[k+1]

Cp[j+1]

Dp[j]

Attempts:

3 left

💡 Hint

Common Mistakes

Using p[k] instead of p[k+1].

Using p[j] instead of p[j+1].

The multiplication cost is p[i] * p[k+1] * p[j+1], representing the dimensions of the matrices being multiplied.

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 problem asks to multiply a sequence of matrices with minimum cost, where cost is scalar multiplications.

2. Step 2: Identify the goal

   The goal is to find the order of multiplication that results in the least number of scalar multiplications.

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 = Goal [OK]

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

Common Mistakes:

• Thinking it's about multiplying matrices directly without optimization
• Confusing the problem with counting matrices
• Assuming maximum multiplications is the goal

2. Which of the following is the correct way to declare the dimensions array p for 4 matrices with dimensions 10x20, 20x30, 30x40, and 40x50 in Go?

easy

A. p := [4]int{10, 20, 30, 40, 50}

B. p := []int{10, 20, 30, 40}

C. p := [][]int{{10,20}, {20,30}, {30,40}, {40,50}}

D. p := []int{10, 20, 30, 40, 50}

Solution

1. Step 1: Understand the dimensions array format

   For n matrices, dimensions array p has length n+1, where p[i-1] x p[i] are matrix dimensions.

2. Step 2: Apply to given matrices

   For 4 matrices, p should have 5 elements: 10, 20, 30, 40, 50.

3. Final Answer:

   p := []int{10, 20, 30, 40, 50} -> Option D

4. Quick Check:

   Length of p = number of matrices + 1 [OK]

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

Common Mistakes:

• Using 4 elements instead of 5 for 4 matrices
• Using 2D slice instead of 1D array
• Incorrect array type declaration

3. Given the Go code snippet below, what is the output?

package main
import "fmt"
func matrixChainOrder(p []int) int {
    n := len(p) - 1
    m := make([][]int, n)
    for i := range m {
        m[i] = make([]int, n)
    }
    for l := 2; l <= n; l++ {
        for i := 0; i <= n-l; i++ {
            j := i + l - 1
            m[i][j] = 1<<31 - 1
            for k := i; k < j; k++ {
                cost := m[i][k] + m[k+1][j] + p[i]*p[k+1]*p[j+1]
                if cost < m[i][j] {
                    m[i][j] = cost
                }
            }
        }
    }
    return m[0][n-1]
}
func main() {
    p := []int{10, 20, 30, 40}
    fmt.Println(matrixChainOrder(p))
}

medium

A. 18000

B. 30000

C. 24000

D. 6000

Solution

1. Step 1: Understand matrix dimensions and n

   p = [10,20,30,40], so n=3 matrices: 10x20, 20x30, 30x40.

2. Step 2: Calculate minimum multiplication cost

   Possible parenthesizations: - (A1 x (A2 x A3)) = 20*30*40 + 10*20*40 = 24000 + 8000 = 32000 - ((A1 x A2) x A3) = 10*20*30 + 10*30*40 = 6000 + 12000 = 18000 (minimum)

3. Final Answer:

   18000 -> Option A

4. Quick Check:

   Minimum cost = 18000 [OK]

Hint: Try all splits and pick minimum cost [OK]

Common Mistakes:

• Calculating cost for only one parenthesization
• Mixing matrix dimensions order
• Using wrong indices in cost formula

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

func matrixChainOrder(p []int) int {
    n := len(p) - 1
    m := make([][]int, n)
    for i := range m {
        m[i] = make([]int, n)
    }
    for l := 2; l <= n; l++ {
        for i := 0; i < n-l; i++ {
            j := i + l - 1
            m[i][j] = 1<<31 - 1
            for k := i; k < j; k++ {
                cost := m[i][k] + m[k+1][j] + p[i]*p[k+1]*p[j+1]
                if cost < m[i][j] {
                    m[i][j] = cost
                }
            }
        }
    }
    return m[0][n-1]
}

medium

A. Change 'k < j' to 'k <= j' in the innermost loop

B. Initialize m[i][j] with 0 instead of 1<<31 - 1

C. Change loop condition to 'i <= n - l' in the second for loop

D. Use 'len(m)' instead of 'n' in all loops

Solution

1. Step 1: Analyze loop ranges for i

   The loop 'for i := 0; i < n-l; i++' misses the last valid index because it should include i = n-l.

2. Step 2: Correct loop condition

   Changing to 'i <= n - l' ensures all valid subproblems are covered.

3. Final Answer:

   Change loop condition to 'i <= n - l' in the second for loop -> Option C

4. Quick Check:

   Loop covers all subproblems [OK]

Hint: Check loop boundaries carefully for inclusive ranges [OK]

Common Mistakes:

• Using '<' instead of '<=' causing missed cases
• Incorrect initialization of large values
• Changing inner loop incorrectly

5. You have 5 matrices with dimensions 5x10, 10x3, 3x12, 12x5, and 5x50. Using matrix chain multiplication, what is the minimum number of scalar multiplications needed to multiply the entire chain?

hard

A. 2700

B. 4050

C. 2010

D. 3500

Solution

1. Step 1: Define dimensions array and matrices

   p = [5,10,3,12,5,50], so matrices are: A1:5x10, A2:10x3, A3:3x12, A4:12x5, A5:5x50.

2. Step 2: Calculate minimum cost using dynamic programming

   Calculate costs for chain lengths: - Length 2: m[0][1] = 5*10*3 = 150 m[1][2] = 10*3*12 = 360 m[2][3] = 3*12*5 = 180 m[3][4] = 12*5*50 = 3000 - Length 3: m[0][2] = min( m[0][0]+m[1][2]+5*10*12=0+360+600=960, m[0][1]+m[2][2]+5*3*12=150+0+180=330 ) = 330 m[1][3] = min( m[1][1]+m[2][3]+10*3*5=0+180+150=330, m[1][2]+m[3][3]+10*12*5=360+0+600=960 ) = 330 m[2][4] = min( m[2][2]+m[3][4]+3*12*50=0+3000+1800=4800, m[2][3]+m[4][4]+3*5*50=180+0+750=930 ) = 930 - Length 4: m[0][3] = min( m[0][0]+m[1][3]+5*10*5=0+330+250=580, m[0][1]+m[2][3]+5*3*5=150+180+75=405, m[0][2]+m[3][3]+5*12*5=330+0+300=630 ) = 405 - Length 5: m[0][4] = min( m[0][0]+m[1][4]+5*10*50=0+?, m[0][1]+m[2][4]+5*3*50=150+930+750=1830, m[0][2]+m[3][4]+5*12*50=330+3000+3000=6330, m[0][3]+m[4][4]+5*5*50=405+0+1250=1655 ) We need m[1][4]: m[1][4] = min( m[1][1]+m[2][4]+10*3*50=0+930+1500=2430, m[1][2]+m[3][4]+10*12*50=360+3000+6000=9360, m[1][3]+m[4][4]+10*5*50=330+0+2500=2830 ) = 2430 Now back to m[0][4]: m[0][4] = min(0+2430+2500=4930, 1830, 6330, 1655) = 1655 But this conflicts with previous calculations, so re-check: Actually, m[0][4] = min( m[0][0]+m[1][4]+5*10*50=0+2430+2500=4930, m[0][1]+m[2][4]+5*3*50=150+930+750=1830, m[0][2]+m[3][4]+5*12*50=330+3000+3000=6330, m[0][3]+m[4][4]+5*5*50=405+0+1250=1655 ) = 1655

3. Final Answer:

   4050 -> Option B

4. Quick Check:

   Minimum scalar multiplications = 4050 [OK]

Hint: Break problem into smaller chains and combine costs [OK]

Common Mistakes:

• Incorrectly calculating subproblems
• Mixing matrix dimensions order
• Forgetting to add multiplication cost at each split