Consider a schema R(W,X,Y,Z) and functional dependencies W->X and Y->Z. Then…
Consider a schema R(W,X,Y,Z) and functional dependencies W->X and Y->Z.
Then the decomposition of R into R1(WX) and R2(YZ) is
Answer: C. dependency preserving but not lossless join — Answer: The decomposition is dependency preserving but not lossless. Why it is dependency preserving: Each functional dependency is contained entirely in one…
- A.
dependency preserving and lossless join
- B.
lossless join but not dependency preserving
- C.
dependency preserving but not lossless join
- D.
not dependency preserving and not lossless join
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Correct answer: C
Answer: The decomposition is dependency preserving but not lossless.
Why it is dependency preserving:
Each functional dependency is contained entirely in one of the decomposed relations: W→X is preserved in R1(WX), and Y→Z is preserved in R2(YZ). Therefore the set of original FDs is preserved by the decomposition.
Why it is not lossless:
R1 and R2 have no attributes in common (R1∩R2 is empty). The standard lossless-join condition requires the intersection to functionally determine at least one of the relations. Since there is no nonempty common attribute set and there is no FD from the empty set to any attributes, the condition fails. Hence the decomposition is not lossless.