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:
6. Find the minimal cover of the set
of functional dependencies given; {A → C, AB → C, C → DI, CD → I, EC → AB, EI →
C}
Solution:
Minimal cover:
Definition 1:
A minimal cover
of a set of FDs F is a minimal set of functional dependencies F_{min}
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
→ C, AB → C, C → DI, CD → I, EC → AB, EI → C}
1. Right
Hand Side (RHS) of all FDs should be single attribute. So we write F as
F1, as follows;
F1
= {A → C, AB → C, C → D, C → I, CD → I, EC → A, EC → B, EI → C}
2. Remove
extraneous attributes.
Extraneous attribute is a redundant
attribute on the LHS of the functional dependency. In the set of FDs, AB → C, CD
→ I, EC → A, EC → B, and EI → C have more than one attribute in the LHS. Hence,
we check one of these LHS attributes are extraneous or not.
To check, we need to find the closure
of each attribute on the LHS; [apply the closure finding algorithm – refer here]
(i) A+ = ACDI
(ii) B+ = B
(iii) C+ = CDI
(iv) D+ = D
(v) E+ = E
(vi) I+ = I
From (i), the closure of A included the
attribute C. So, B is extraneous in AB → C, and B can be removed.
From (iii), the closure of C included
the attribute I. So, D is extraneous in CD → I, and D can be removed.
No more extraneous attributes are
found. Hence, we write F1 as F2 after removing extraneous attributes from F1 as
follows;
F2
= {A → C, C → D, C → I, EC → A, EC → B, EI → C}
3. Eliminate
redundant functional dependency.
None of the functional dependencies in F2 are redundant (Please refer here). Hence,
the minimal cover Fc
= {A → C, C → D, C → I, EC → A, EC → B, EI → C}.
Hence, set of functional dependencies Fc is the
minimal cover for the set 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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Thx for the article
ReplyDeleteThank u
ReplyDeleteIf we first find redundant fds first, then C>I misses out
ReplyDeleteMore than one minimal covers are possible
DeleteWhat about F3 = {A → C, C → D, C → I, EI → A, EI → B} ? It has only five FD's and it is equivalent to F2!
ReplyDeleteYes. A set F can have several canonical cover. It depends on which FD or where do we start. Thanks for your comment.
DeleteI made a mistake. Either A>C or EI > C is redundant. Hence, the final Fc would be {A → C, C → D, C → I, EC → A, EC → B}
Thank You
ReplyDeleteIf you remove the redundancy EI > C, how do you infer this FD back using the remaining FDs? In other words, won't we lose this dependency information if we remove EI > C?
ReplyDeleteThis not about inferring the FDs. It is about determining RHS attributes. We can determine C without the help of EI.
DeleteBut only those FDs are considered redundant if the LHS can determine the RHS with its closure (not considered the FD on which the rule is applied) and in this case EI > C does not have C in its closure EI = {E,I,} so EI>C is not redundant
DeleteCorrected. Thanks
DeleteThis is really helpful! Thank you for this article.
ReplyDelete