super key / candidate key

Question

R (A, B, C ,D ,E )

F={ A->B, BC-> D, D->BC, B->E, B->A }

Check whether BD is  C.Key or S.key . Also find the candidate key.

Answer 1

BD⁺ ={BDEAC}

D⁺={DBCEA} 

Since D alone is Ck implies BD is super key

Candidate Keys

D⁺={ABCDE}                 By replacing D with BC

BC⁺={ABCDE}               By replacing B with A

AC⁺={ABCDE}

These are candidate keys D,AC,BC

Comments

ohh..okk...f9..

So every time when we will have more than 2 C.Key..  we will subtract all the possible combination of repetetion then add 1 S.K  which is the mixture of all the attributes.

EX-- R (A,B,C,D,E,F)

Suppose AC, DE, BF are the Candidate keys..

then ,

    SK= n(AC)+ n(DE)  +n(BF) - n( AC ∩ DE )  - n( DE ∩ BF )  - n( AC ∩ BF ) + n( ABCDEF )

Am i right...??

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yes,Right. It's kind of Set theory Problem.

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okk...thank u @LeenSharma .. :D :D

Answer 2

Ck with single attribute :

D+ =ABCDE            // This is the only Ck with single attribute

Ck with 2 attributes :

AC+ =ABCDE

BC+ =ABCDE          // These are the only 2 Ck's with 2 attributes

There is no Ck with 3 attributes :

There is no Ck with 4 attributes :

Now your question whether BD is a Ck Or Sk

you can see that D is already proven to be Ck So BD will be a Sk Here not a Ck .

Comments

CEA ,CEB are super keys right

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chauhan when...AC is candidate key how CEA can b candidate key

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my mistake sorry