For constants a ≥q 1 and b > 1, consider the following recurrence defined on the non-negative integers: T(n) = a · T≤ft(n/b) + f(n) Which one of the following options is correct about the recurrence T(n)?

If f(n) is O(n^_b(a) - ) for some > 0, then T(n) is (n^_b(a)).

If f(n) is n _2(n), then T(n) is (n _2(n)).

If f(n) is n_2(n), then T(n) is (_2(n)).

If f(n) is O(n^_b(a) - ) for some > 0, then T(n) is (n^_b(a)).

If f(n) is (n^_b(a)), then T(n) is (n^_b(a)).

GATE CSE 2021 Set 2 Q49: Master theorem cases for T(n) = aT(n/b) + f(n)

Algorithms
2 marks
MCQ
−2/3 if wrong

For constants $a \geq 1$ and $b > 1$ , consider the following recurrence defined on the non-negative integers:

$T(n) = a \cdot T\left(\frac{n}{b}\right) + f(n)$

Which one of the following options is correct about the recurrence $T(n)$ ?

Try it: multiple choice: exactly one correct option

Wrong answer costs 2/3 mark in GATE.

Show answer and explanation

Answer: C (official GATE 2021 final answer key)

Concept used: Master theorem: compare $f(n)$ with $n^{\log_b a}$ .

First step: Recall case 1 (f polynomially smaller gives $\Theta(n^{\log_b a})$ ) and case 2 (equal growth adds a $\log n$ factor); A and B ignore $a$ and $b$ .

Full step-by-step solution

Detailed working, common mistakes and linked mock tests will be available with a free GATEiit account. Log-in is coming soon.

Share this question

WhatsAppDownload image

Source: official GATE 2021 Set 2 CS question paper and final answer key, organised by IIT Bombay. Question paper · Answer key

PracticePractise Algorithms questions →Full paperAll GATE CSE 2021 Set 2 questions →

GATE CSE 2021 Set 2 Q49: Master theorem cases for T(n) = aT(n/b) + f(n)

Question

Full step-by-step solution

More Algorithms PYQs