write down a minimal cover

Question

Consider the attribute set R = ABCDEGH (where Functional Dependencies are
F= {AB·--> C,AC --> B: AD ---> E, B -----> D, BE --> A, B -> G}.

Answer 1

Minimal cover means the smallest set of FDs which can derive all other FDs. The given FDs are

$AB \to C$
$AC \to B$
$AD \to E$
$B \to D$
$BE \to A$
$B \to G$

Take each FD and see if any attribute on the LHS can be avoided. There is no such reductions possible. Also, the RHS of the FDs are unique. So, I suppose the given FDs are already a minimal cover. 

The candidate keys can be found as follows. If any attribute is not on the right side of any FD, it must be in any candidate key. So, here $H$ must be in any candidate key. 

Trial and error with $AB$ shows it can determine, $C, D,E$. So, $ABH$ must be a candidate key. Since $AC \to B$, $ACH$ is another candidate key. Since $BE \to A$, $BEH$ is another candidate key. So, totally 3 candidate keys $\left\{ABH, ACH, BEH\right\}$.  

Comments

no u cant remove B in AB->C

bcz if yu remove B yu cant reach C as u need both AB to reach C

here we can apply union rule on    B->D and B->G   as  B->DG

so we have 5 min fd's

 

comment plz!!!

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closure(A) is no where giving B and viceversa also. so how can u remove.???????

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yes, we cannot remove.

Answer 2

I think here we should check the LHS.  At the LHS reduction is possible. Check it yourself.If anything wrong or you can't understand ,then comment..Thank you