ITPEC FE Subject B April 2024 Question 7

Fibonacci sequence implemented via recursion with base cases and recursive call.

The sequence starts at position 1: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34.

Base cases: fibo(1) = 0 and fibo(2) = 1. Since the function returns n - 1 in the base branch, both n = 1 (returns 0) and n = 2 (returns 1) must be caught by condition A:

→ A = (n = 1) or (n = 2)

Recursive case: every other value is the sum of the two preceding values:

→ B = fibo(n-1) + fibo(n-2)

Verification: fibo(9) = fibo(8) + fibo(7) = 13 + 8 = 21 ✓

Why not others:
- (a), (b), (c) — condition n = 1 misses n = 2, so fibo(2) would recurse into fibo(1) + fibo(0), and fibo(0) has no base case → infinite recursion
- (d), (e), (f) — condition n > 1 puts recursion in the if branch and the base case in else, which is inverted logic; only n = 1 returns a base value, missing n = 2
- (g) — correct condition but fibo(n-1) + 1 is not Fibonacci addition
- (h) — correct condition but fibo(n-1) + n adds the position instead of the previous term

Key rule: Fibonacci recursion needs two base cases (n = 1 and n = 2) because the recursive formula references two prior positions.

AI-generated — may contain errors