Project ForgeDatabase Management Systems
Normalization
GATE 2014
Expert
Consider the relation scheme R=(E,F,G,H,I,J,K,L,M,N) and the set of functional dependencies {{E,F}→{G}, {F}→{I,J}, {E,H}→{K,L}, {K}→{M}, {L}→{N}} on R. What is the key for R?
- A{E,F}
- {E,F,H}
- C{E,F,H,K,L}
- D{E}
Explanation
Closure of {E,F,H}: EF→G gives G; F→IJ gives I,J; EH→KL gives K,L; K→M gives M; L→N gives N. So {E,F,H}⁺ = all 10 attributes — it's a superkey. Checking minimality: {E,F}⁺={E,F,G,I,J} (missing H,K,L,M,N) — not a key without H. {E,H}⁺={E,H,K,L,M,N} (missing F,G,I,J) — not a key without F. {F,H}⁺={F,H,I,J} — not a key without E. So all three of E, F, H are needed, and {E,F,H} is the minimal key.
Source: GATE CS 2014 (Session 1) (2014)
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