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The "cover" of a Set of Functional Dependencies

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• Cover set

  • Cover set

    Let ℉1 and ℉2 be two different sets of functional dependencies ℉1 is cover set of ℉2 if: every functional dependency in ℉2 can be inferred from the functional dependency in ℉1

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  • Example:

    Relation R = (A, B, C, D, E, F) ℉1 = { A → BC D → DF } ℉2 = { A → B }

    ℉₁ covers ℉₂ because

    ℉2 = { A → B } 1. A → B can be inferred from ℉1 as follows: A → A (Reflexivity rule) A → BC (A → A and A → BC) A → B (Subset rule)

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• How to determine if ℉₁ covers ℉₂

  • Let ℉₁ and ℉₂ be two sets of functional dependencies

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  • Algorithm to determine if ℉₁ covers ℉₂ :

    For each functional dependency X → Y ∈ ℉2 do: { compute X+ using the FDs in ℉1 if ( Y ⊄ X+ ) return(NO); } return(YES);

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• Example of functional dependency cover
  • Consider the following two sets of functional dependencies:

    Relation R = (A, B, C, D, E, F) ℉1 = { A → C AC → D E → AD E → F } ℉2 = { A → CD E → DF }

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  • Question:

    Does ℉1 covers ℉2 ???

    Check each functional dependency in ℉₂:

    A → CD   ∈   ℉2 : Compute A+ with respect to ℉1: ℉1 = { A → C AC → D E → AD E → F } A+ = A (Initialization) = AC (A → C) = ACD (AC → D) stop... Check: CD ⊆ ACD ?? Yes. ---> Check next functional dependency in ℉2.... E → DF   ∈   ℉2 : Compute E+ with respect to ℉1: ℉1 = { A → C AC → D E → AD E → F } E+ = E (Initialization) = ADE (E → AD) = ADEF (E → F) = ACDEF (A → C) Done Check: DF ⊆ ACDEF ?? Yes. Continue (no more functional dependency - DONE)

    Conclusion: ℉₁ covers ℉₂

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  • Does ℉₂ covers ℉₁ ???

    Check each functional dependency in ℉₁:

    A → C   ∈   ℉1 : Compute A+ with respect to ℉2: ℉2 = { A → CD E → DF } A+ = A (Initialization) = ACD A → CD Done Check: C ⊆ ACD ?? Yes. Continue (next functional dependency in ℉1) AC → D   ∈   ℉1 : Compute AC+ with respect to ℉2: ℉2 = { A → CD E → DF } AC+ = AC (Initialization) = ACD (A → CD) Done Check: D ⊆ ACD ?? Yes. Continue (next functional dependency in ℉1) E → AD   ∈   ℉1 : Compute E+ with respect to ℉2: ℉2 = { A → CD E → DF } E+ = E (Initialization) = DEF (E → DF) Done Check: AD ⊆ DEF ?? NO !!!

    Conclusion: ℉₂ do not cover ℉₁