ITPEC FE Subject B April 2026 Question 12

A is s2[i] ≠ s2[j] and B is s2[i] = s2[j].

The double loop compares every pair of positions and looks for the two ways a mapping can break:

• •Blank A — the same character in s1 at both positions must produce the same character in s2. If s1[i] = s1[j] but s2[i] ≠ s2[j], one source character maps to two different targets
• •Blank B — different characters in s1 must map to different characters in s2. If s1[i] ≠ s1[j] but s2[i] = s2[j], two source characters collapse onto one target

The table shows both failures: {"f", "o", "o"} against {"b", "a", "r"} breaks the first rule, and {"a", "b"} against {"c", "c"} breaks the second.

Answer (d)

Why not others:
- (a) and (b) — compare s1[i] with s2[j], matching a character in one array against a character in the other; isomorphism says nothing about whether the two alphabets overlap
- (c) — inverts both tests, rejecting exactly the pairs that are consistent and accepting the ones that are not

Key rule: A one-to-one mapping needs both directions checked: equal sources must give equal targets, and different sources must give different targets. Testing only one direction accepts {"a", "b"} → {"c", "c"}.

AI-generated — may contain errors