Relation \(R\) has eight attributes \(ABCDEFGH\). Fields of \(R\) contain only…
2013
Relation \(R\) has eight attributes \(ABCDEFGH\). Fields of \(R\) contain only atomic values.
\(F={CH→G, A→BC, B→CFH, E→A, F→EG}\) is a set of functional dependencies (FDs) so that \(F^+\) is exactly the set of FDs that hold for \(R\).
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: R is in 1NF but not in 2NF.
1NF: Satisfied because every field contains only atomic values.
Find candidate keys: D does not appear on the right side of any functional dependency, so every candidate key must include D.
Closure of B,D: Start with {B,D}. B → C,F,H adds C,F,H. F → E,G adds E,G. E → A adds A. A → B,C adds B,C. Closure yields all attributes A,B,C,D,E,F,G,H, so {B,D} is a candidate key.
Closure of A,D: A → B,C adds B,C. B → C,F,H adds F,H. F → E,G adds E,G. All attributes are obtained, so {A,D} is a candidate key.
Similarly, {E,D} and {F,D} are candidate keys because E → A and F → E,G let them derive all attributes when combined with D.
Prime and non-prime attributes: Attributes that appear in some candidate key are prime: A, B, E, F, D. Non-prime attributes are C, G, H.
Check 2NF: There is a functional dependency A → B,C. Attribute C is non-prime and is functionally dependent on A, which is a proper subset of the candidate key {A,D}. This is a partial dependency, so the relation violates 2NF.
Conclusion: Since 1NF holds but 2NF is violated, the relation is in 1NF but not in 2NF. Consequently it is not in 3NF or BCNF.