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+1 vote

Two minimal covers are possible : A->B, B->C, C->A     AND  A->C, C->B , B->A

answered by Active (1.3k points)
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Can you please provide the procedure ?
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1) A->C, B->C,C->AB

2) A->C,B->A,C->AB

aren't these are minimal covers?

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In ypour 2nd Minimal cover , A->C,B->A,C->AB It will be A->C,B->A,C->B as from C->B if we find closure of C we will get {C,B,A}.

Plus we can have more minimal covers:

1) A->BC, B->A,C->A

2) A->B,B->AC,C->B
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While identifying the minimal cover three steps should be followed

1: Split the given FDs such that RHS contain only single attribute.

2: Identify the extraneous attribute in LHS and remove it.

3: Eliminate the redundant FD.
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@Shiv Gaur, yes procedure is correct...

but my question is

A->C, B->C,C->AB

A->BC, B->A,C->A ( As Iqra Islam said )

A->B,B->AC,C->B

isn't these are minimal?

i didn't find any redundant term or FD in that.

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I don't know the exact reason but I guess , when we talk about minimal cover of a given  FD F,  It is a minimal set of FDs that is equivalent to the given FD F.

Here  {A->C, B->C,C->AB} covers the given F but it could be further split to {A->C, B->C,C->A, C->B} ie 4 FDs and we already have A->B, B->C, C->A  and A->C, C->B , B->A  ie 3 FDs   that's why we split the RHS  in first step

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Those are also SHOULD BE minimal covers,

note that minimal covers are not unique.

compare this in DIGITAL LOGIC

all those are irreducible, but which you mentioned are minimal expression

but in DBMS we don't have two terminologies... therefore all are minimal covers only

+1 vote

Two minimal covers are possible.

answered by (241 points)
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@vikas999, yes procedure is correct...

but my question is

A->C, B->C,C->AB

A->BC, B->A,C->A

A->B,B->AC,C->B

isn't these are minimal?

i didn't find any redundant term or FD in that.