Sunday, 14 February 2016

Dependency preserving decomposition - solved exercises 2

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 (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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