Dependency preserving decomposition - solved exercises 2

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Dependency preserving decomposition

Consider a relation R (A, B, C, D) with the following set of functional dependencies;

F = {A → B, B → C, and C → D}.

Is the decomposition of R (A, B, C, D) into R1 (A, C, D) and R2 (B, C) a dependency preserving decomposition?

Solution:

The above said decomposition of R into R1 and R2 is a dependency preserving decomposition if (F₁ U F₂)⁺ = F⁺, where F₁ is set of FDs hold by R1, F₂ is set of FDs hold by R2, and F is the set of FDs hold by R. (F₁ U F₂)⁺ is the closure of (F₁ U F₂), and F⁺ is the closure of F.

Step 1: for R1, the derivable non-trivial functional dependency is, C → D. Hence, F₁ = {C → D}

Step 2: for R2, the derivable non-trivial functional dependency is, B → C. Hence, F₂ = {B → C}

(F₁ U F₂) = ({C → D} U {B → C}) = {C → D, B → C} ≠ F.

(F₁ U F₂)⁺ = {C → D, B → C, B → D}

F⁺ = {A → B, B → C, C → D, A → C, A → D, B → D}

Hence, (F₁ U F₂)⁺ ≠ F⁺

The FD A → B is not supported by (F₁ U F₂)⁺. Hence, the decomposition of R (A, B, C, D) into R1 (A, C, D) and R2 (B, C) is not a dependency preserving decomposition.

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