Advanced Database Management System - Tutorials and Notes: Find minimal cover of set of functional dependencies

Thursday, 12 May 2016

Find minimal cover of set of functional dependencies

Find minimal cover of set of functional dependencies example, Solved exercise - how to find minimal cover of F? Easy steps to find minimal cover of FDs, What is minimal cover?


Question:

5. Find the minimal cover of the set of functional dependencies given; {A → BC, B → C, AB → D}


Solution:
Minimal cover:
Definition 1:
A minimal cover of a set of FDs F is a minimal set of functional dependencies Fmin that is equivalent to F. There can be many such minimal covers for a set of functional dependencies F.
Definition 2:
A set of FDs F is minimum if F has as few FDs as any equivalent set of FDs.


Simple properties/steps of minimal cover:
1. Right Hand Side (RHS) of all FDs should be single attribute.
2. Remove extraneous attributes. [What is extraneous attribute? Refer here].
3. Eliminate redundant functional dependencies.

Let us apply these properties to F = {A → BC, B → C, AB → D}
1. Right Hand Side (RHS) of all FDs should be single attribute. So we write F as F1, as follows;
F1 = {A → B, A → C, B → C, AB → D}
2. Remove extraneous attributes.
Extraneous attribute is a redundant attribute on the LHS of the functional dependency. In the set of FDs, on AB → D has more than one attribute in the LHS. Hence, we check one of A and B is extraneous or not.
First we check whether A is extraneous or not. To do that, we need to find the closure of the remaining attribute B with respect to F1.
B+ = BC.
This does not include D, so A is not extraneous.
Now we check whether B is extraneous or not. To do that, we need to find the closure of the remaining attribute A with respect to F1.
A+ = ABCD.
This includes D, so B is extraneous, ie., we can identify D without B on the LHS.
Now, we can write the new set of FDs, F2 as follows;
F2 = {A → B, A → C, B → C, A → D}
3. Eliminate redundant functional dependency.
If A → B, and B → C, then A → C is true (according to transitive rule). Hence, the FD A → C is redundant. We can eliminate this and we get final set of FDs F3 as follows;
F3 = {A → B, B → C, A → D}

The set of FDs F3 is the minimal cover of F.


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Similar topics

How to find closure of set of functional dependencies?

How to find closure of attributes?

How to find canonical cover for a set of functional dependencies?

How to find extraneous attribute?




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