## 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) 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 (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 non-trivial functional dependency is A → C. Hence F1 = {A → C}.*

**Step 1***: for R2, the non-trivial functional dependency is B → C. Hence, F2 = {B → C}.*

**Step 2***: find F1 U F2 and the closure of F1 U F2.*

**Step 3****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 (F**. Hence, the decomposition of R (A, B, C) into R1 (A, C) and R2 (B, C) is not a dependency preserving decomposition._{1}U F_{2})^{+}