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7 votes
7 votes

Consider the following FD set {A → BC,B → AC, C → AB}. The number of different minimal covers possible for the above FD set __________ .

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2 Comments

What is the proper method to find the minimal cover
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{A->C, B->C, C->AB}
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7 Answers

14 votes
14 votes
here every key is a primary key..
numbber of minimal covers = 5 i think
1)A-->B, B-->C, C-->A
2)A-->C, C-->B, B-->A
3)B-->AC, A-->B ,C-->B
4)A-->BC , B-->A, C-->A
5)C-->AB ,A-->C, B-->C
did i miss any?

4 Comments

Insufficient answer without any detailed approach.
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@srestha

Maam. the answer should be 4 right ?

A-> C, B-> C, C-> A, C-> B.

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no

selected ans is correct
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3 votes
3 votes

The minimal cover is A->C,B->C,C->A,C->B

Another two are A->B,B->C.C->A and A->C,B->A,C->B .

Thus total different minimal covers are 3.

edited by

4 Comments

We have the find out the number of different minimal cover possible
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Is it 3?

Other two are

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

And A->C,B->A,C->B
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Can someone suggest the proper method to find all minimal covers
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@Arjun

@srestha

@Mk Utkarsh

Is this the correct answer?

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1 vote
1 vote

4 IS THE CORRECT ANS

0 votes
0 votes
A--->B, A--->C, B-->A, C-->A

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

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

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

 A--->C, B-->C, C-->A

total 5

2 Comments

How did you find?
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I don't have any proper method

I just some intution in selecting the set and then see if it is minimal or not
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