Solution

The correct answer is f(n) is 0 (g(n))

Key Points

The notation Ω(n) is the formal way to express the lower bound of an algorithm's running time. It measures the best case time complexity or the best amount of time an algorithm can possibly take to complete.

The notation O(n) is the formal way to express the upper bound of an algorithm's running time. It measures the worstcase time complexity of the algorithms.

 

Let's analyze the functions and their growth rates to determine which of the statements is true.

Given Functions:
1. f(n) = \(3n^{\sqrt{n}}\)
2. g(n) = \(2^{\Gamma n} \log_2 n\)
3. h(n) = n!

Analyzing the Growth Rates:

1. f(n) = \(3n^{\sqrt{n}}\)
This function grows very quickly due to the \(n^{\sqrt{n}}\) term, which is larger than any polynomial but smaller than exponential functions like \(n^n\) .

2. g(n) = \(2^{\Gamma n} \log_2 n\)
This function involves the Gamma function \(\Gamma(n)\) , which grows faster than factorial functions but depends on the context. Given that \(\Gamma(n)\) is closely related to (n 1)! , this function also grows extremely fast, much faster than polynomial functions.

3. h(n) = n!
Factorial growth is very rapid and typically outpaces both polynomial and exponential functions for large n .

Comparing the Growth Rates:

• h(n) compared to f(n) :
  •   h(n) = n! grows faster than f(n) = \(3n^{\sqrt{n}}\) , so h(n) is not O(f(n)) . This eliminates Option 1.
• h(n) compared to g(n) :
  •   While h(n) = n! grows quickly, g(n) = \(2^{\Gamma n} \log_2 n\) with \(\Gamma n \) growing similarly to factorial but involving an extra exponential factor \(2^{\Gamma n}\) . This suggests that g(n) might grow faster, making h(n) not O(g(n)) . This eliminates Option 2.
• g(n) compared to f(n) :
  •   g(n) involves an exponential term \(2^{\Gamma n}\) which grows much faster than \(f(n) = 3n^{\sqrt{n}}\) , implying that g(n) is not O(f(n)) . So, Option 3 is also incorrect.
• f(n) compared to g(n) :
  •   Since g(n) grows faster than f(n) , f(n) is O(g(n)) . This supports Option 4.

Conclusion:
The correct answer is:

3) g(n) is not O(f(n))

This indicates that g(n) grows faster than f(n) , meaning f(n) is not an upper bound on g(n) .

Here's a list of functions in asymptotic notation that we often encounter when analyzing algorithms, ordered by slowest to fastest growing:

• \(\Theta(1)\)
• \(\Theta(\log_2 n)\)
• \(\Theta(n)\)
• \(\Theta(n \log_2 n)\)
• \( \Theta(n^2)\)
• \(\Theta(n^2 \log_2 n)\)
• \(\Theta(n^3)\)
• \(\Theta(2^n)\)
• \(\Theta(n!)\)