Consider a relational schema 𝑆 = (π‘ˆ,𝑉,π‘Š,𝑋,π‘Œ,𝑍) on which the following…

2020

Consider a relational schema 𝑆 = (π‘ˆ,𝑉,π‘Š,𝑋,π‘Œ,𝑍)Β on which the following functional dependencies hold:

(π‘ˆβ†’π‘‰,π‘‰π‘Šβ†’π‘‹,π‘Œβ†’π‘Š,π‘‹β†’π‘ˆ)

Which are the candidate keys among following options?

  1. A.

    π‘ˆπ‘Œ,π‘‰π‘Œ

  2. B.

    π‘ˆπ‘Œ,π‘‰π‘Œ,π‘‹π‘Œ

  3. C.

    π‘ˆπ‘Œπ‘,π‘‰π‘Œπ‘,π‘‰π‘Šπ‘

  4. D.

    π‘ˆπ‘Œπ‘,π‘‰π‘Œπ‘,π‘‹π‘Œπ‘

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Correct answer: D

Answer: The candidate keys are U Y Z, V Y Z, and X Y Z.

Key idea: Use attribute closure and the given dependencies: U β†’ V, VW β†’ X, Y β†’ W, X β†’ U.

  • For X Y Z: compute closure. X β†’ U, then U β†’ V, and Y β†’ W. With V and W we have VW β†’ X (already present). Thus XYZ+ = {U, V, W, X, Y, Z} (all attributes). Removing any attribute breaks this: X Y lacks Z, X Z lacks Y, and Y Z lacks X, so minimality holds.

  • For U Y Z: compute closure. U β†’ V and Y β†’ W. With V and W we get VW β†’ X, and X β†’ U gives no new attribute. So UYZ+ = {U, V, W, X, Y, Z}. Removing any attribute prevents deriving the missing one (for example, UY lacks Z), so U Y Z is minimal.

  • For V Y Z: compute closure. Y β†’ W, and with V and W we get VW β†’ X, then X β†’ U and U β†’ V. So VYZ+ = {U, V, W, X, Y, Z}. Removing any attribute prevents deriving the missing one (for example, VY lacks Z), so V Y Z is minimal.

Why other sets fail: Any set that does not include Z cannot produce Z because no functional dependency yields Z. Some sets without Z can generate many other attributes (e.g., X Y or U Y produce U, V, W, X, Y), but they remain incomplete without Z.

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