##### 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

Shobhit sudsho 29 Nov 2016 08:55 pm

ABD is the candidate key
it is not even in 2NF..only in 1NF

Emmanuel Tejuosho emmanueltejuosh 29 Nov 2016 08:56 pm

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

Shobhit sudsho 29 Nov 2016 08:58 pm

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

Arjun arjunsinghra 29 Nov 2016 08:57 pm

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

Emmanuel Tejuosho emmanueltejuosh 29 Nov 2016 09:07 pm

Thank you

AMIT CHAUDHARY amit17 2 Dec 2016 05:08 pm

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.

Berism mesojava 16 Dec 2016 04:13 am

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) ?