You are given the following set of functional dependencies for a relation…
You are given the following set of functional dependencies for a relation R(A,B,C,D,E,F),
F = {AB → C, DC → AE, E → F }.
i. Is this relation in BCNF?
ii. Is the decomposition (A,B,C,D) (B,C,D,E,F) a dependency preserving decomposition?
Answer: D. i: No, ii: Yes — Given relation R(A, B, C, D, E, F) with functional dependencies: AB → C, DC → AE, E → F. Part i: Is R in BCNF? BCNF requires that for every FD X → Y, X is a…
- A.
i: Yes, ii: Yes
- B.
i: Yes, ii: No
- C.
i: No, ii: No
- D.
i: No, ii: Yes
Attempted by 20 students.
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Correct answer: D
Given relation R(A, B, C, D, E, F) with functional dependencies: AB → C, DC → AE, E → F.
Part i: Is R in BCNF?
BCNF requires that for every FD X → Y, X is a superkey. Check the closures:
AB → C: AB+ = {A, B, C}. Cannot derive D, E, or F, so AB is not a superkey.
DC → AE: From DC → AE and E → F, DC+ = {D, C, A, E, F}. B is not obtained, so DC is not a superkey.
E → F: E+ = {E, F}, which is not all attributes, so E is not a superkey.
Because none of the FD left-hand sides are superkeys, R is not in BCNF.
Part ii: Is the decomposition (A,B,C,D) and (B,C,D,E,F) dependency preserving?
Check where each original FD appears in the decomposed relations:
AB → C: All attributes A, B, C are in relation (A,B,C,D), so this FD is preserved in that relation.
DC → AE: This FD splits across the decomposed relations: DC → A is preserved in (A,B,C,D) and DC → E is preserved in (B,C,D,E,F). Together they preserve DC → AE.
E → F: Both attributes E and F appear in (B,C,D,E,F), so E → F is preserved there.
Since every original functional dependency is present in one of the decomposed relations (or can be reconstructed from FDs present in the components), the decomposition is dependency preserving.
Final Answer: i: No (R is not in BCNF). ii: Yes (the decomposition is dependency preserving).