GATE CS 2017 Set 2 Q.30 (CS Q.30): Asymptotic Analysis and Time/Space Complexity

Consider the recurrence function

T(n) = \begin{cases} 2T(\sqrt{n}) + 1, & n > 2 \\ 2, & 0 < n \le 2 \end{cases}

Then $T(n)$ in terms of $\Theta$ notation is

$\Theta(\log \log n)$

$\Theta(\log n)$

$\Theta(\sqrt{n})$

$\Theta(n)$

Official answer and solution

Official answer: B
From: Official GATE 2017 answer key

Solution

Apply the substitution method ( $n = 2^m$ ) to transform the square-root recurrence into a standard divide-and-conquer recurrence solvable by the Master Theorem.

Given the recurrence $T(n) = 2T(\sqrt{n}) + 1$ for $n > 2$ .
Let $n = 2^m$ , so $m = \log_2 n$ and $\sqrt{n} = (2^m)^{1/2} = 2^{m/2}$ .
Define $S(m) = T(2^m) = T(n)$ . Then:

S(m) = 2 S(m/2) + 1

Solve $S(m) = 2S(m/2) + 1$ using the Master Theorem:

$a = 2, b = 2, f(m) = 1 = m^0$
Critical exponent: $\log_b a = \log_2 2 = 1$
Since $f(m) = O(m^{1 - \epsilon})$ for $\epsilon = 1$ , Case 1 of the Master Theorem applies:

S(m) = \Theta(m^1) = \Theta(m)

Substitute back $m = \log_2 n$ :

T(n) = S(m) = \Theta(\log n)

This matches option B.

Final answer

Common trap

Assuming $T(n) = \Theta(\log \log n)$ by confusing this recurrence with $T(n) = T(\sqrt{n}) + 1$ .

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

Source

Paper: GATE 2017 Computer Science and Information Technology (CS) Session 2
Question: 30 of 65, printed as CS Q.30
Section: Computer Science and Information Technology

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