Matrix Chain Multiplication in DSA Python - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main goal of the Matrix Chain Multiplication problem?

To find the most efficient way to multiply a chain of matrices by minimizing the total number of scalar multiplications.

beginner

Why do we not actually multiply the matrices in the Matrix Chain Multiplication problem?

Because the problem focuses on finding the order of multiplication that minimizes cost, not the actual multiplication itself.

beginner

What does the 'cost' represent in Matrix Chain Multiplication?

The number of scalar multiplications needed to multiply the entire chain of matrices in a specific order.

intermediate

What is the role of dynamic programming in solving Matrix Chain Multiplication?

It stores results of smaller subproblems to avoid repeated calculations and efficiently find the minimum multiplication cost for the entire chain.

intermediate

What is the time complexity of the Matrix Chain Multiplication dynamic programming solution?

O(n^3), where n is the number of matrices in the chain.

What does the input array represent in Matrix Chain Multiplication?

A Number of matrices only

B Values inside the matrices

C Dimensions of matrices in the chain

D Order of matrix multiplication

The input array contains the dimensions of matrices, where the i-th matrix has dimensions p[i-1] x p[i].

Which technique is commonly used to solve Matrix Chain Multiplication efficiently?

A Dynamic programming

B Brute force only

C Divide and conquer without memoization

D Greedy algorithm

Dynamic programming is used to store intermediate results and avoid repeated calculations.

What does the matrix m[i][j] represent in the DP solution?

A Minimum cost to multiply matrices from i to j

B Maximum cost to multiply matrices from i to j

C Number of matrices between i and j

D Dimensions of matrix i

m[i][j] stores the minimum number of scalar multiplications needed to multiply matrices from i to j.

If you have 4 matrices with dimensions 10x20, 20x30, 30x40, and 40x30, what is the minimum number of multiplications?

A 18000

B 30000

C 24000

D 36000

The optimal order results in 30000 scalar multiplications.

What is the base case cost for multiplying a single matrix in the DP table?

A 1

B Infinity

C Dimension of matrix

D 0

Multiplying one matrix requires no multiplications, so cost is zero.

Explain how dynamic programming helps solve the Matrix Chain Multiplication problem.

Think about how storing results of smaller problems avoids repeating work.

Describe the meaning of the input array and how it relates to the matrices in the chain.

Each matrix's size is defined by two consecutive numbers in the array.

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