Find minimal cover/canonical cover in a set of FDs

Find minimal cover/canonical cover in a set of given functional dependencies

Question:

For a relation schema R = (A, B, C, D, E), consider the following set of functional dependencies;

F = {A → BC, CD → E, B → D, E → A}

Using the functional dependencies compute the canonical cover F_c.

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. [How to find minimal cover? – Refer here].

Follow these steps to find the canonical (minimal) cover;

1. Decompose each functional dependency in F to have singleton (single) attribute on the Right Hand Side (RHS).

2. Remove extraneous (redundant) attributes from Left Hand Side of each functional dependency.
[What is extraneous attribute? Refer here].

3. Find and eliminate redundant functional dependencies.

Let us apply the above said properties to F;

F = {A → BC, CD → E, B → D, E → A}

1. RHS of all functional dependencies should be singleton. To achieve this we decompose each functional dependency that has two or more attributes in their RHS. We get the following set of functional dependencies F1;

F1 = {A → B, A → C, CD → E, B → D, E → A}

2. Remove extraneous (redundant) attributes. This is applicable only for those functional dependencies that have two or more attributes on its LHS.

We need to check whether or not an attribute/set of attributes of LHS of a functional dependency is redundant or not. For checking this, we have to find the closure of each individual LHS attribute of each FD that has two or more LHS attributes.

With this property we have only one FD, CD → E. We need to check whether C alone or E alone can find E uniquely.

C+ = C by reflexivity rule

D+ = D by reflexivity rule

From this we found that either C or D alone cannot determine the value of E. Hence, the FD CD → E does not have any extraneous attributes.

Hence, our F1 is as follows;

F1 = {A → B, A → C, CD → E, B → D, E → A}

3. Eliminate redundant functional dependencies.

To find redundant FDs, we have to find whether the RHS of a FD can be determined using other FDs.

(i) Let us check first for FD, A → B. To proceed, we forget , A → B from F, and use only the remaining FDs A → C, CD → E, B → D, E → A of F. That is our F becomes F1 = {A → C, CD → E, B → D, E → A}

Now find A+ using F1. A+ = {A,C} which does not include B. Hence, A → B is not redundant.

(ii) Now check for A → C. can we find C using following FDs?

F1 = {A → B, CD → E, B → D, E → A}

A+ = ABD

B+ = BD

We are not able to determine C using F1. Hence, A → C is not redundant.

If we repeat this for each FD in F, we found that none of them are redundant.

Hence our Fc is as follows;

Fc = {A → BC, CD → E, B → D, E → A} which is same as F in the question.

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Similar topics

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How to find closure of attributes?

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How to find extraneous attribute?

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