R5(A,B,C,D,E)
{A→ CE, D→ CE}

Please can anyone help if this is in BCNF or 3NF. Also what are the candidate keys? Thanks

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arjunsinghra's picture

minimal candidate key is {ABD

FD set -

A--->CE {prime attribute----->non prime attribute } here is partial dependencies so 1nf

D---->CE {prime attribute----->non prime attribute } here is partial dependencies so 1nf

so relation in 1nf

Thank you

amit17's picture

Steps are simple first find candidate key , look in RHS , those are not there will be in C.K(candidate key.)

now START FRM ELIMINATION IF THESE EXITS MEANS THEY ARE NOT IN THAT NORMAL FORM

 P/N.P-->P(BCNF)    ,  N.P-->N.P(3NF) TRANISTIVITY  ,   P-->N.P(2NF) PARTIAL DEPENDENCY .

NOW QUESTION

   CK WOULD BE ---> ABD     P(A,B,D),

NOW A-->CE A IS NOT A SUPER KEY NO BCNF

A-->CE A IS A PRIME ATTRITUBE AND CE IS NON PRIME SO NOT IN 2NF ALSO.

SO THIS RELATION WILL BE IN 1NF. BY DEFAULT WE ASSUME ATOMICITY IS THERE.

 

R(A, B, C, D) and F = {C ->A, D ->B, AD->C, BC->D}.

I want this 3rd normal form to convert in boyce codd normal form(bcnf) ?

please help me

sahastranshu's picture

In BCNF, for given R, if x->y (FD), then, x will be a super key i.e. x will have to define the whole relation R.
But here none of the FDs are defining R(ABCD), thus, we will divide it like R1(CA), R2(DB), R3(ADC), R4(BCD).
Now, every relations R1,R2,R3,R4 are in BCNF.

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