Matrix Chain Multiplication in DSA Python - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of Matrix Chain Multiplication Cost Table

What is the final cost table (m) after running the matrix chain multiplication algorithm on matrices with dimensions [10, 20, 30, 40]?

DSA Python

def matrix_chain_order(p):
    n = len(p) - 1
    m = [[0]*n for _ in range(n)]
    for l in range(2, n+1):
        for i in range(n-l+1):
            j = i + l - 1
            m[i][j] = float('inf')
            for k in range(i, j):
                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

p = [10, 20, 30, 40]
m = matrix_chain_order(p)
print(m)

A [[0, 6000, 18000], [0, 0, 24000], [0, 0, 0]]

B [[0, 6000, 24000], [0, 0, 24000], [0, 0, 0]]

C [[0, 6000, 18000], [0, 0, 36000], [0, 0, 0]]

D [[0, 6000, 36000], [0, 0, 24000], [0, 0, 0]]

💡 Hint

Trace the cost calculation for multiplying matrices in different orders.

The minimum cost to multiply matrices A1(10x20), A2(20x30), and A3(30x40) is computed by checking all possible splits. The cost table m stores these minimum costs. The correct table shows 6000 for multiplying first two matrices and 18000 for the full chain.

🧠 Conceptual

intermediate

Understanding the Role of the 'k' Variable

In the matrix chain multiplication algorithm, what does the variable 'k' represent in the nested loops?

A 'k' is the cost of multiplying two matrices directly.

B 'k' is the total number of matrices being multiplied.

C 'k' is the dimension of the resulting matrix after multiplication.

D 'k' represents the index where the matrix chain is split to compute minimum multiplication cost.

💡 Hint

Think about how the algorithm tries different places to split the chain.

The variable 'k' is used to try all possible places to split the matrix chain between i and j to find the minimum multiplication cost.

🔧 Debug

advanced

Identify the Error in Matrix Chain Multiplication Code

What error will this code produce when run, and why? Code: def matrix_chain_order(p): n = len(p) - 1 m = [[0]*n for _ in range(n)] for l in range(2, n+1): for i in range(n-l+1): j = i + l - 1 m[i][j] = float('inf') for k in range(i, j): q = m[i][k] + m[k+1][j] + p[i]*p[k]*p[j] if q < m[i][j]: m[i][j] = q return m p = [10, 20, 30, 40] m = matrix_chain_order(p) print(m)

A TypeError because float('inf') cannot be compared with integers.

B IndexError because p[k] is used instead of p[k+1] causing wrong dimension access.

C SyntaxError due to missing colon after for loop.

D No error; code runs and outputs the correct cost table.

💡 Hint

Check how matrix dimensions are accessed in the cost calculation.

The code uses p[k] instead of p[k+1] when calculating multiplication cost, which causes incorrect dimension indexing and leads to an IndexError or wrong results.

🚀 Application

advanced

Minimum Multiplication Cost for Given Dimensions

Given matrices with dimensions [5, 10, 3, 12, 5, 50, 6], what is the minimum number of scalar multiplications needed to multiply the entire chain?

DSA Python

def matrix_chain_order(p):
    n = len(p) - 1
    m = [[0]*n for _ in range(n)]
    for l in range(2, n+1):
        for i in range(n-l+1):
            j = i + l - 1
            m[i][j] = float('inf')
            for k in range(i, j):
                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]

p = [5, 10, 3, 12, 5, 50, 6]
result = matrix_chain_order(p)
print(result)

A 2010

B 1510

C 4050

D 2430

💡 Hint

Use dynamic programming to find the minimum cost by checking all splits.

The minimum cost calculated by the algorithm for these dimensions is 2010 scalar multiplications.

🧠 Conceptual

expert

Why is Matrix Chain Multiplication a Dynamic Programming Problem?

Which reason best explains why matrix chain multiplication is solved using dynamic programming?

A Because it involves recursive calls without any repeated calculations.

B Because it requires sorting matrices before multiplication.

C Because it has overlapping subproblems and optimal substructure properties.

D Because it can be solved by greedy algorithms selecting the smallest matrix first.

💡 Hint

Think about how the problem breaks down into smaller problems that repeat.

Matrix chain multiplication has overlapping subproblems (same subchains computed multiple times) and optimal substructure (optimal solution contains optimal solutions to subproblems), which are key for dynamic programming.

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 count the total number of matrices in the chain

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

Solution

1. Step 1: Understand the problem context

   The Matrix Chain Multiplication problem aims to multiply a sequence of matrices with the least cost in terms of scalar multiplications.

2. Step 2: Identify the goal

   The goal is to find the order of multiplication that minimizes the total number of scalar multiplications.

3. Final Answer:

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

4. Quick Check:

   Minimum scalar multiplications = Goal [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 is the correct way to initialize the DP table for Matrix Chain Multiplication in Python?

easy

A. dp = [[0 for _ in range(n)] for _ in range(n)]

B. dp = [0] * n

C. dp = [[0] * n]

D. dp = [0 for _ in range(n)]

Solution

1. Step 1: Understand DP table structure

   The DP table for Matrix Chain Multiplication is a 2D table of size n x n to store minimum costs for subproblems.

2. Step 2: Choose correct initialization

   dp = [[0 for _ in range(n)] for _ in range(n)] correctly creates a 2D list with zeros for all n rows and n columns.

3. Final Answer:

   dp = [[0 for _ in range(n)] for _ in range(n)] -> Option A

4. Quick Check:

   2D list initialization = dp with zeros [OK]

Hint: Use nested list comprehensions for 2D DP tables [OK]

Common Mistakes:

• Using one-dimensional list for 2D DP table
• Incorrectly initializing inner lists causing shared references
• Using single list instead of nested lists

3. What will be the output of the following code snippet for p = [10, 20, 30] representing matrix dimensions?

def matrix_chain_order(p):
    n = len(p) - 1
    dp = [[0]*n for _ in range(n)]
    for l in range(2, n+1):
        for i in range(n-l+1):
            j = i + l - 1
            dp[i][j] = float('inf')
            for k in range(i, j):
                cost = dp[i][k] + dp[k+1][j] + p[i]*p[k+1]*p[j+1]
                if cost < dp[i][j]:
                    dp[i][j] = cost
    return dp[0][n-1]

print(matrix_chain_order(p))

medium

A. IndexError

B. 6000

C. None

D. 18000

Solution

1. Step 1: Trace the code execution

   p=[10,20,30], n=2, dp=[[0,0],[0,0]]. l=2, i=0, j=1, dp[0][1]=inf. k=0: cost=dp[0][0]+dp[1][1]+10*20*30=0+0+6000. However, the code tries to access p[j+1] = p[2], which is valid here (p has length 3). So no IndexError here.

2. Step 2: Confirm output

   Return dp[0][1] which is 6000.

3. Final Answer:

   6000 -> Option B

4. Quick Check:

   Cost = p[0]*p[1]*p[2] = 6000 [OK]

Hint: For chain of 2 matrices (n=2), cost is p[0]*p[1]*p[2] [OK]

Common Mistakes:

• Anticipating IndexError on p[j+1]
• Miscalculating cost (e.g., wrong indices)
• Forgetting single k split for l=2

4. Find the bug in this Matrix Chain Multiplication code snippet:

def matrix_chain_order(p):
    n = len(p) - 1
    dp = [[0]*n for _ in range(n)]
    for l in range(2, n+1):
        for i in range(n-l+1):
            j = i + l - 1
            dp[i][j] = float('inf')
            for k in range(i, j):
                cost = dp[i][k] + dp[k+1][j] + p[i]*p[k]*p[j]
                if cost < dp[i][j]:
                    dp[i][j] = cost
    return dp[0][n-1]

medium

A. Return statement is missing

B. DP table initialization is wrong

C. Incorrect multiplication indices in cost calculation

D. Loop ranges are incorrect

Solution

1. Step 1: Review cost calculation formula

   The cost formula should be p[i] * p[k+1] * p[j+1], but code uses p[i] * p[k] * p[j], which is incorrect indexing.

2. Step 2: Identify correct indices

   Matrix dimensions are stored in p, so for multiplying matrices A_i..A_k and A_k+1..A_j, cost uses p[i], p[k+1], and p[j+1].

3. Final Answer:

   Incorrect multiplication indices in cost calculation -> Option C

4. Quick Check:

   Cost formula indices must be p[i], p[k+1], p[j+1] [OK]

Hint: Check dimension indices carefully in cost formula [OK]

Common Mistakes:

• Using p[k] instead of p[k+1]
• Mixing zero-based and one-based indexing
• Ignoring matrix dimension rules

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?

hard

A. 1510

B. 2010

C. 4050

D. 3500

Solution

1. Step 1: Understand the input and problem

   p has length 7, so there are 6 matrices with dimensions: 5x10, 10x3, 3x12, 12x5, 5x50, 50x6.

2. Step 2: Use dynamic programming to find minimum cost

   By applying the Matrix Chain Multiplication DP algorithm, the minimum cost is calculated as 1510 scalar multiplications.

3. Final Answer:

   1510 -> Option A

4. Quick Check:

   DP minimum scalar multiplications = 1510 [OK]

Hint: Use DP table to store subproblem costs for efficiency [OK]

Common Mistakes:

• Calculating cost manually without DP
• Mixing matrix dimensions incorrectly
• Forgetting to consider all parenthesization options