Test Series

Question

Answer 1

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

Comments

@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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@Shiv Gaur,

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

Answer 2

Two minimal covers are possible.

Comments

@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.