Which of the following relations can not be decomposed in to BCNF with a…

Answer: The relation R(V, W, X, Y) with FDs {VW→X, X→VY} cannot be decomposed into BCNF while keeping both a lossless join and dependency preservation.

1. Step 1: Identify keys and BCNF violations.

   Compute VW+:

   VW+ = {V, W} → X (from VW→X), then V and Y (from X→VY).

   So VW+ = {V, W, X, Y}, hence VW is a key and VW→X does not violate BCNF.

   Compute X+: X+ = {X, V, Y}. X+ does not include W, so X is not a key; therefore X→VY violates BCNF.

2. Step 2: Perform BCNF decomposition on the violating FD X→VY.

   Decompose R into:

   • R1(X, V, Y) (attributes determined by X)

   • R2(X, W) (remaining attributes plus the common attribute X)

   This decomposition is lossless because the intersection R1 ∩ R2 = {X} and X→VY holds, so X determines the attributes of R1.

3. Step 3: Check dependency preservation.

   Original FD VW→X involves attributes in both decomposed relations (V in R1, W in R2, X in both). Projecting FDs onto R1 and R2 yields X→VY in R1 and no FD that yields X from V and W in R2. Therefore VW→X cannot be inferred from the FDs that hold within R1 and R2 alone, so the FD VW→X is not preserved.

4. Conclusion:

   You can decompose R into BCNF relations with a lossless join, but doing so breaks dependency preservation (VW→X is lost). Hence it is impossible to obtain a decomposition that is simultaneously in BCNF, lossless, and dependency-preserving for this relation.