GATE CS 2022 Q.11: Asymptotic Analysis and Time/Space Complexity

Which one of the following statements is TRUE for all positive functions $f(n)$ ?

$f(n^2) = \Theta(f(n)^2)$ , when $f(n)$ is a polynomial

$f(n^2) = o(f(n)^2)$

$f(n^2) = O(f(n)^2)$ , when $f(n)$ is an exponential function

$f(n^2) = \Omega(f(n)^2)$

Official answer and solution

Official answer: A
From: Official GATE 2022 answer key

Solution

Analyze the asymptotic growth of $f(n^2)$ versus $(f(n))^2$ for polynomials and exponentials.

Consider $f(n) = c n^k$ where $c > 0$ and $k > 0$ (a polynomial):

f(n^2) = c (n^2)^k = c n^{2k}

(f(n))^2 = (c n^k)^2 = c^2 n^{2k}

Taking the ratio as $n \to \infty$ :

\lim_{n \to \infty} \frac{f(n^2)}{(f(n))^2} = \lim_{n \to \infty} \frac{c n^{2k}}{c^2 n^{2k}} = \frac{1}{c} > 0

Since the limit is a non-zero positive constant, $f(n^2) = \Theta((f(n))^2)$ for any polynomial function. This confirms Option (A) is TRUE.

For an exponential function like $f(n) = 2^n$ :

f(n^2) = 2^{n^2}, \quad (f(n))^2 = 2^{2n}

Here $2^{n^2}$ grows strictly faster than $2^{2n}$ , so $f(n^2) \ne O((f(n))^2)$ , showing Option (C) is FALSE. Similarly, if $f(n) = n$ , $f(n^2) = n^2$ and $(f(n))^2 = n^2$ , so $f(n^2) \ne o((f(n))^2)$ , ruling out Option (B).

Final answer

Option (A)

Common trap

Assuming polynomial properties apply to exponential functions or confusing little-o with big-Theta.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2022 Computer Science and Information Technology (CS)
Question: 11 of 65
Section: Computer Science and Information Technology

Report an error in this question

GATE questions are the work of the organising institutes. This site is independent and is not affiliated with them.

Question

Solution

Source