GATE CS 2022 Q.51: Combinatorics, Counting, Recurrence Relations and Generating Functions

Consider the following recurrence:

f(1) = 1;

f(2n) = 2f(n) - 1, \quad \text{for } n \ge 1;

f(2n + 1) = 2f(n) + 1, \quad \text{for } n \ge 1.

Then, which of the following statements is/are TRUE?

$f(2^n - 1) = 2^n - 1$

$f(2^n) = 1$

$f(5 \cdot 2^n) = 2^{n+1} + 1$

$f(2^n + 1) = 2^n + 1$

Official answer and solution

Solution

Relate the recurrence to the binary representation of integers (the Josephus problem).

This recurrence is the classic Josephus recurrence with step size 2:

f(1) = 1, \quad f(2n) = 2f(n) - 1, \quad f(2n + 1) = 2f(n) + 1

Writing any integer $N$ as $N = 2^m + l$ where $0 \le l < 2^m$ , the closed-form solution is $f(N) = 2l + 1$ .

Evaluating each option:

(A) For $N = 2^n - 1$ : We have $N = 2^{n-1} + (2^{n-1} - 1)$ , so $m = n-1$ and $l = 2^{n-1} - 1$ .
$f(2^n - 1) = 2(2^{n-1} - 1) + 1 = 2^n - 2 + 1 = 2^n - 1$ (TRUE)
(B) For $N = 2^n$ : Here $l = 0$ .
$f(2^n) = 2(0) + 1 = 1$ (TRUE)
(C) For $N = 5 \cdot 2^n = (4 + 1) \cdot 2^n = 2^{n+2} + 2^n$ : Here $m = n+2$ and $l = 2^n < 2^{n+2}$ .
$f(5 \cdot 2^n) = 2(2^n) + 1 = 2^{n+1} + 1$ (TRUE)
(D) For $N = 2^n + 1$ : Here $m = n$ and $l = 1$ .
$f(2^n + 1) = 2(1) + 1 = 3 \ne 2^n + 1$ for $n \ge 2$ . (FALSE)

Therefore, (A), (B), and (C) are TRUE.

Final answer

Options (A), (B), (C)

Common trap

Assuming $f(2^n + 1) = 2^n + 1$ by extrapolating from $n=1$ where $f(3) = 3$ .

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