Consider the following relational schema (P,Q, R, S, T) with FD set (P->QR,…
Consider the following relational schema (P,Q, R, S, T) with FD set (P->QR, RS->T, Q->S, T->P) if the relation decomposed into R1(P,Q,R) and R2(P, S, T), which of the following is true for given decomposition?
Answer: B. Lossless but not dependency preserving — Final answer: Lossless but not dependency preserving. Lossless-join test: R1 ∩ R2 = {P}. Given P -> Q R, P functionally determines all attributes of R1, so…
- A.
Lossless join and dependency preserving
- B.
Lossless but not dependency preserving
- C.
Lossy and dependency preserving
- D.
Lossy but not dependency preserving
Attempted by 126 students.
Show answer & explanation
Correct answer: B
Final answer: Lossless but not dependency preserving.
Lossless-join test:
R1 ∩ R2 = {P}. Given P -> Q R, P functionally determines all attributes of R1, so the decomposition into R1(P,Q,R) and R2(P,S,T) is lossless.
Dependency preservation:
Projection on R1(P,Q,R): P -> Q,R (so P -> QR) is preserved in R1.
Projection on R2(P,S,T): T -> P is preserved in R2.
Original dependencies Q -> S and RS -> T are not contained in either projection and are not implied by the union {P -> QR, T -> P}. For example, Q+ under the projected set is just {Q}, so Q -> S is not implied; RS+ under the projected set does not include T, so RS -> T is not implied.
Conclusion: The decomposition is lossless but not dependency preserving.