Consider the following relation R(ABCDE) with their functional dependencies:
F = {A → BC, CD → E, B → D, E → A}
Choose the correct option
(A) Decomposition R1 = (A, B, C), R2 = (A, D, E) of R is lossy join decomposition.
(B) Decomposition R3 = (A, B, C), R4 = (C, D, E) of R is lossless join decomposition.
(C) Decomposition R1 = (A, B, C), R2 = (A, D, E) of R is preserving dependency.
(D) None of these
Answer: (D)
Explanation: Decomposition R1 = (A, B, C), R2 = (A, D, E) of R is lossless join and dependency preserving decomposition.
R1 ∩ R2 = A; (A → BC) hence (A → ABC) hence (R1 ∩ R2 → R1), so this is a lossless-join decomposition.
And F1 = {A → BC}, F2 = {E → A}, (F1 ∪ F2)+ ≠ F+, some FDs are not covered, so not dependency preserving decomposition.
Also, decomposition R3 and R4 are not lossless join, because common attribute (i.e., C) is not superkey of either sub-relatins.
Option (D) is correct.
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