Which of the following relations can not be decomposed in to BCNF with a lossless join and dependency-preserving decomposition ?
R (V, W, X, Y) {VW→X, X→VY}
A relation is in BCNF when the left side of every non-trivial functional dependency is a superkey. BCNF decomposition is always possible losslessly, but dependency preservation can fail when an attribute on the right of one dependency is part of the left side of a key dependency, so splitting on the violating dependency separates the attributes the key dependency needs.
Step 1 — R(V, W, X, Y) with {VW → X, X → VY}: keys are VW (VW → X → Y) and WX (X → V, so WX → V, Y). X → VY violates BCNF because X alone is not a superkey.
Step 2 — To remove the violation, X and its dependents V, Y must go into a table (X, V, Y), leaving (X, W). Then VW → X spans both tables: V is in one, W in the other and no table holds V, W, X together, so VW → X is lost. Any table that keeps V, W and X together also contains X → V with X not a superkey, so it is not in BCNF. Hence this relation cannot be put into BCNF losslessly while preserving dependencies.
Step 3 — R(V, W, X, Y, Z) with {VW → X, WX → V, VX → W, W → Y, Y → Z}: decompose into (W, Y), (Y, Z) and (V, W, X). In (V, W, X) the keys are VW, WX and VX, so all three dependencies there have superkey left sides. Every dependency lives in one table, so the decomposition is lossless, preserving and in BCNF.
Step 4 — R(V, W, X, Y) with {VW → W, X → Y}: VW → W is trivial; X → Y violates BCNF. Decomposing into (X, Y) and (V, W, X) is lossless (X is a key of (X, Y)), both tables are in BCNF and X → Y is kept.
Step 5 — R(V, W, X, Y, Z) with {VW → X, X → Y, Y → Z}: the key is VW. Decompose into (Y, Z), (X, Y) and (V, W, X); each table's dependency has a key on its left, and no dependency links X back to V or W, so (V, W, X) with key VW is in BCNF. All three dependencies are preserved and the joins are lossless.
R (V, W, X, Y) {VW→X, X→VY} option is correct.
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