Normalization

A functional dependency is cheap to enforce when all its attributes sit in one relation of the decomposition: each insert or update is checked against that relation alone. In a decomposition that is not dependency-preserving, some dependency \(X \to Y\) cannot be derived from the dependencies holding within the individual relations, because its attributes are spread over several of them. To check it after every change, those relations must first be joined to bring \(X\) and \(Y\) back together, so joins are executed far more often. Selection, projection and set union cannot reunite attributes stored in different relations.

Answer: C

Find the keys with attribute closures.

• \(\mathrm{A}^{+}\): A \(\rightarrow\) BCEF, then E \(\rightarrow\) DG, so \(\mathrm{A}^{+} = \mathrm{ABCDEFG}\) and A is a key.
• \(\mathrm{BC}^{+}\): BC \(\rightarrow\) A, and everything follows as above. \(\mathrm{B}^{+} = \mathrm{B}\) and \(\mathrm{C}^{+} = \mathrm{C}\), so BC is a candidate key.
• \(\mathrm{E}^{+} = \mathrm{DEG}\), so E is not a key.

A appears on a right side only in BC \(\rightarrow\) A, and B and C are determined only by A, so every key contains A or both B and C. The candidate keys are exactly A and BC: B holds, while A and C fail.

In E \(\rightarrow\) DG the left side E is not a superkey, so R is not in BCNF and D holds.

Answer: B, D

Compute attribute closures.

• \(\{V, W\}^{+}\): \(V \to X\) adds \(X\), and then \(WX \to Y\) and \(WX \to Z\) add \(Y\) and \(Z\). So \(\{V, W\}^{+} = \{V, W, X, Y, Z\}\).
• \(\{W, X\}^{+} = \{W, X, Y, Z\}\), straight from the two given dependencies.

Hence \(VW \to YZ\) (A) and \(VW \to Y\) (D) hold, and \(WX \to YZ\) (B) is the union rule applied to \(WX \to Y\) and \(WX \to Z\). But \(U\) is not in \(\{V, W\}^{+}\): no dependency has \(U\) on its right-hand side, so only sets containing \(U\) determine it, and C cannot be derived.

Answer: A, B, D