Dependency preserving decomposition - solved exercises 3

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Dependency preserving decomposition

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

F = {A → B, B → C}.

Is the decomposition of R (A, B, C) into R1 (A, C) 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 non-trivial functional dependency is A → C. Hence F1 = {A → C}.

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

Step 3: find F1 U F2 and the closure of F1 U F2.

F1 U F2 = {A → C, B → C} ≠ F.

(F1 U F2)⁺ = {A → C, B → C}

F⁺ = {A → B, B → C, A → C}

And, (F1 U F2)⁺ ≠ F.

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

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