## Dependency preserving decomposition - Dependency preserving decomposition solved exercises - How to verify that a decomposition is dependency preserving? - Steps to find dependency preserving decomposition - Dependency preserving decomposition examples

**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 (F

_{1}U F_{2})^{+}= F^{+}, where F_{1}is set of FDs hold by R1, F_{2}is set of FDs hold by R2, and F is the set of FDs hold by R. (F_{1}U F_{2})^{+ }is**the closure**of (F_{1}U F_{2}), and F^{+}is**the closure**of F.*: for R1, the derivable non-trivial functional dependency is, C → D. Hence, F*

**Step 1**_{1}= {C → D}

*: for R2, the derivable non-trivial functional dependency is, B → C. Hence, F*

**Step 2**_{2}= {B → C}

(

**F**) = ({C → D} U {B → C}) = {C → D, B → C} ≠_{1}U F_{2}**F**.
(F

_{1}U F_{2})^{+}= {C → D, B → C, B → D}
F

^{+}= {A → B, B → C, C → D, A → C, A → D, B → D}
Hence, (F

_{1}U F_{2})^{+}≠ F^{+}
The FD A → B is not supported by (F

_{1}U F_{2})^{+}. Hence, t*.***he 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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