Given relation R(ABCD) and FD’s under this relation is {A → B, B → C} the…
Given relation R(ABCD) and FD’s under this relation is {A → B, B → C} the decomposition of the above relation R into BCNF is
I. Lossless decomposition
II. Dependency preserving
Answer: C. Both I & II — Step 1: Find the candidate key: Attributes that do not appear on the right-hand side of any FD are A and D, so any key must include A and D. Compute closure:…
- A.
Only I
- B.
Only II
- C.
Both I & II
- D.
None of the above
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Correct answer: C
Step 1: Find the candidate key: Attributes that do not appear on the right-hand side of any FD are A and D, so any key must include A and D. Compute closure: AD+ contains A; from A → B add B; from B → C add C; therefore AD+ = ABCD. Hence AD is a candidate key (and the only one).
Step 2: Identify BCNF violations: Both A → B and B → C have left sides that are not superkeys of R, so they violate BCNF and we must decompose.
Decompose using B → C: produce relation BC and relation ABD.
In ABD, A → B still violates BCNF (A is not a key for ABD), so decompose ABD into AB and AD.
Final BCNF relations: BC, AB, and AD.
Dependency preservation: Both original FDs are preserved in the decomposition: A → B is preserved within AB, and B → C is preserved within BC.
Lossless decomposition: One of the resulting relations (AD) contains the candidate key of the original relation. Because a decomposition that includes a relation containing a candidate key is lossless, the decomposition is lossless.
Conclusion: The BCNF decomposition of R yields relations BC, AB, and AD, which is both lossless and dependency preserving. Therefore both statements are true.