Wednesday, 9 March 2016

Dependency preserving decomposition - solved exercises 3

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 (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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