Let the set of functional dependencies F = {QR → S, R → P, S → Q} hold on a…
2019
Let the set of functional dependencies F = {QR → S, R → P, S → Q} hold on a relation schema X = (PQRS). X is not in BCNF. Suppose X is decomposed into two schemas Y and Z, where Y = (PR) and Z = (QRS).
Consider the two statements given below.
I. Both Y and Z are in BCNF
II. Decomposition of X into Y and Z is dependency preserving and lossless
Which of the above statements is/are correct?
- A.
I only
- B.
Neither I nor II
- C.
II only
- D.
Both I and II
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Correct answer: C
Answer: II only
Check Y = (P,R): Relevant FD is R->P. Since R determines P (and trivially R), R is a key for Y, so every nontrivial FD on Y has a superkey on the left. Therefore Y is in BCNF.
Check Z = (Q,R,S): Relevant FDs are QR->S and S->Q. The FD S->Q holds but S does not determine R, so S is not a key of Z. Thus S->Q violates BCNF and Z is not in BCNF.
Lossless check: Intersection Y ∩ Z = {R}. Since R->P is in F, R->PR holds, so R is a key for Y and the decomposition is lossless.
Dependency preservation: Projecting F onto Y gives R->P; projecting onto Z gives QR->S and S->Q. These cover all original FDs, so the decomposition is dependency-preserving.
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