##### BCNF OR 3NF?

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

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

ABD is the candidate key

it is not even in 2NF..only in 1NF

Thanks for your reply. Could you explain how ABD is the candidate key?

basic method is to see RHS of every FD...here we have CE..means definitly ABD has to be present in our candidate key else how we will derive them..

now minimal CK will be ABD...u cant reduce it and no need to expand also else it will become Superkey

so ABD suffices our condition of minimal CK here

minimal candidate key is {

}ABDFD 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

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

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.