ITPEC FE Morning October 2020 Question 25
ITPEC FE Morning October 2020 — Question 25 of 80
Candidate Key from Functional Dependencies — find the minimal set of attributes that determines all others.
Given R(A, B, C, P, Q, T) with FDs:
- A → B
- A → C
- CP → Q
- CP → T
Step 1: Find attributes that never appear on the right side of any FD.
- Right-side attributes: B, C, Q, T
- Never on right: A, P → these must be in every candidate key
Step 2: Compute closure of {A, P}.
- A → B → add B
- A → C → add C
- CP → Q → add Q
- CP → T → add T
- Closure: {A, P, B, C, Q, T} = all attributes ✓
Answer: `(c) A, P`
Why not others:
- (a) A — closure is {A, B, C}, missing P, Q, T
- (b) A, B — closure is {A, B, C}, still no P to trigger CP → Q/T
- (d) C, P — closure is {C, P, Q, T}, missing A, B
Key rule: Attributes that never appear on the right side of any FD must be part of every candidate key.
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