Time Complexity of Loops - Problem Analysis

Overview

• Discussing time complexity of a given C code segment.
• Focus on understanding Tn, the frequency count of the for loop based on input n.
• Tn will be represented using Big O and Big Omega notation.

Problem Description

• Need to analyze a C code segment with a for loop.
• The goal is to determine the correct asymptotic representation of Tn.

Code Segment Explanation

• For Loop Initialization:
  • i is initialized to 2.
  • Loop continues while i is less than or equal to sqrt(n).
• Loop Execution:
  • If condition n mod i == 0 is true, it indicates n is not a prime number and returns 0, terminating the loop early.
  • If n is prime, the loop continues up to sqrt(n).

Analyzing Tn

• Frequency Count: Tn represents the number of times the loop is executed.
• For a non-prime n, the loop can terminate early, leading to fewer iterations.
• When n is prime, the loop will execute approximately sqrt(n) - 1 times.

Simplified Analysis

• For a simplified loop without the if condition:
  • The loop will run from 2 to sqrt(n).
  • Total iterations: sqrt(n) - 1, leading to Tn = theta(sqrt(n)).
  • Both best and worst cases yield the same result:
    • Tn = O(sqrt(n)) and Tn = Omega(sqrt(n)).

Best and Worst Case Analysis

• Best Case: Occurs when n is not prime.
  • Example: If n = 16, the loop can terminate within 1 iteration since n mod 2 == 0.
  • Resulting time complexity: Omega(1).
• Worst Case: Occurs when n is prime.
  • In this case, the loop runs sqrt(n) - 1 times.
  • Resulting time complexity: O(sqrt(n)).

Conclusion

• Final Asymptotic Representation:
  • Tn = O(sqrt(n)) (worst case)
  • Tn = Omega(1) (best case)
• Correct option confirmed as: Option B - Tn is O(sqrt(n)) and Tn is Omega(1).

Closing Statement

• Presentation concludes with gratitude for attendance and an invitation to the next session.