ITPEC FE Morning April 2023 Question 26

Candidate Key from Functional Dependencies — find the minimal set of attributes that determines all others.

Given R = (M, N, O, P, S, T) with FDs:
- O → T
- S → M
- OS → P
- M → N

Step 1: Find attributes that never appear on the right side of any FD.

Right-side attributes: T, M, P, N
Never on right side: `O` and `S` → must be part of every key.

Step 2: Compute closure of `{O, S}`:
- Start: {O, S}
- O → T → {O, S, T}
- S → M → {O, S, T, M}
- OS → P → {O, S, T, M, P}
- M → N → {O, S, T, M, P, N} = all attributes ✓

OS is a superkey, and neither O alone nor S alone determines all attributes → OS is minimal → it is a candidate key.

Why not others:
- (a) MO — cannot derive S, so cannot derive P via OS → P
- (b) MS — cannot derive O, so cannot derive T via O → T
- (c) OP — cannot derive S, so cannot derive M via S → M

Key rule: Attributes absent from the right side of all FDs must be in every candidate key.

AI-generated — may contain errors