Consider relation with the set of functional dependency FDs ABCDE,
Duration: 4 min
Consider a relation R = (A, B, C, D, E) with the set of functional dependency FDs
{A -> ABCDE, B -> C }. Which of the following statement is true?
- A.
R1 = (A, C, D, E) and R2 = (B, C) are both in BCNF and preserve lossless-join.
- B.
R1 = (A, B, D, E) and R2 = (B, C) are both in BCNF and preserve lossless-join.
- C.
both (a) and (b)
- D.
None of the above.
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Correct answer: B
Answer: Decompose into R1 = (A, B, D, E) and R2 = (B, C). This decomposition is in BCNF, is lossless, and preserves dependencies.
Key observation: A -> ABCDE means A is a candidate key for the original relation.
BCNF violation: B -> C violates BCNF on the original relation because B is not a key of the original relation.
Decompose on B -> C to get R1 = (A, B, D, E) and R2 = (B, C).
Check BCNF: In R1, A -> B,D,E (from A -> ABCDE) so A is a key for R1. In R2, B -> C so B is a key for R2. Thus both relations are in BCNF.
Check lossless-join: The intersection R1 ∩ R2 = {B}. Since B -> C is in the FDs, the intersection functionally determines R2 (B -> B,C), so the decomposition is lossless.
Check dependency preservation: Projected dependencies include A -> B,D,E (in R1) and B -> C (in R2). From these we can infer A -> C via A -> B and B -> C, so the original dependencies are preserved.
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