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+1 vote

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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

Plus we can have more minimal covers:

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

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

0

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.

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.

0

@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.

0

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

+1 vote

0

@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.

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