Consider a relation R (A, B, C, D, E, F, G, H), where each attribute is…
2017
Consider a relation R (A, B, C, D, E, F, G, H), where each attribute is atomic, and following functional dependencies exist.
CH → G
A → BC
B → CFH
E → A
F → EG
The relation R is __________ .
- A.
in 1NF but not in 2NF
- B.
in 2NF but not in 3NF
- C.
in 3NF but not in BCNF
- D.
in BCNF
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Correct answer: A
Answer: the relation is in 1NF but not in 2NF.
Step 1 — Verify 1NF: All attributes are atomic, so the relation is in 1NF.
Step 2 — Find candidate keys: No FD determines D, so D must be in every key. Compute closures to find full keys: for example, with E: E → A, A → B,C, B → F,H, and F → E,G; therefore {D,E}+ contains all attributes, so {D,E} is a candidate key. By similar closure chains, {D,A}, {D,B}, and {D,F} are also candidate keys.
Step 3 — Identify prime and non-prime attributes: Prime attributes (appear in some candidate key): D, A, B, E, F. Non-prime attributes: C, G, H.
Step 4 — Check for partial dependencies (2NF): A → C is an FD where A is part of a candidate key (for example {D,A}) and C is non-prime. This is a partial dependency on a part of a candidate key, so 2NF is violated. Similarly, B → H is another partial dependency (B is part of {D,B} and H is non-prime).
Conclusion: Because the relation is in 1NF but has partial dependencies from part of candidate keys to non-prime attributes, it is not in 2NF. Consequently it cannot be in 3NF or BCNF.
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