Matrix Chain Multiplication in DSA Python - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Python

n = len(p) - 1
m = [[[1] for _ in range(n)] for _ in range(n)]

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

A0

B-1

CNone

D1

Attempts:

3 left

💡 Hint

Common Mistakes

Using 1 or None instead of 0 for initialization.

We initialize the DP table with zeros because initially, the cost of multiplying one matrix is zero.

2fill in blank

medium

Complete the code to set the chain length for the outer loop in matrix chain multiplication.

DSA Python

for L in range(2, [1] + 1):

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

An

Blen(p)

Cn - 1

Dlen(p) - 1

Attempts:

3 left

💡 Hint

Common Mistakes

Using len(p) instead of n.

The chain length L goes from 2 to n (number of matrices).

3fill in blank

hard

Fix the error in the code to calculate the minimum cost for multiplying matrices from i to j.

DSA Python

for k in range(i, j):
    cost = m[i][k] + m[k+1][j] + p[i]*p[k+1]*p[j+1]
    m[i][j] = min(m[i][j], [1])

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

Ap[i]*p[k]*p[j]

Bm[i][k]

Ccost

Dm[k+1][j]

Attempts:

3 left

💡 Hint

Common Mistakes

Using wrong variables instead of cost in min function.

We update m[i][j] with the minimum of its current value and the newly calculated cost.

4fill in blank

hard

Fill both blanks to correctly set the indices for the subproblems in matrix chain multiplication.

DSA Python

for i in range(0, n - [1] + 1):
    j = i + [2] - 1

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

AL

B1

D0

Attempts:

3 left

💡 Hint

Common Mistakes

Mixing up indices or using wrong values for i and j.

i loops from 0 to n-L, and j is set to i+L-1 to cover the chain length L.

5fill in blank

hard

Fill all three blanks to complete the matrix chain multiplication function that returns the minimum multiplication cost.

DSA Python

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

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

Acost

Bm[0][n-1]

Cm[i][j]

Dcost + m[i][j]

Attempts:

3 left

💡 Hint

Common Mistakes

Returning wrong value or updating m[i][j] incorrectly.

Update m[i][j] with the current cost if it's less, and finally return m[0][n-1] which holds the minimum cost for the full chain.

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