general normalisation

Question

Can anyone explain me the conditions of 2NF, 3NF and BCNF.....in short.....and how to find normalisation in fastest way possible......

Answer 1

condition for cheaking normal forms:

BCNF: if determinat side every attribute is SK or CD then it's in BCNF(X--->Y,X is determinant Y is determiner)

3NF: if determinant is SK or determiner is prime attribute then it;s 3NF

2NF:  for 2nf avoid PD conditioin. if proper subset of CD belongs to non prime attribute then PD exist so not in 2NF

1NF: free form multivalued attribute..

Comments

consider this eg:

R(ABCD)

F{AB---->C,BC----->D}

here the relation is not in BCNF becoz AB is CD but BC is not

so CD is AB so prime attribut is A,B

non prime attribute is C,D

it's not in 3nf becoz BC is not CD & D not belongs to prime attribute

it's in 2nf becoz proper subset of CD that is clouser of A & B belongs to prime attribute so here is no PD exist so given relation is in 2NF.

Answer 2

2NF::

Find Ck(Candiadate Key by closure property)

IF partial keys exist then it will not be in 2NF. Otherwise YES.

3NF::

After finding out Ck ,checkout any TD(transitive dependency).If exist then it will not be in 3NF but still in 2NF

BCNF::

Every left hand side of each and every superkays should be superkeys then it must be in BCNF.

Answer 3

All relational databases are in 1NF. (have atomic values) 

Here, NP: Non-Prime and P: Prime 

For 2NF 

No partial dependency should be there. (We don't care about transitive dependency here) 

P -> NP (Not P or NP -> P or NP -> NP ) is the only thing we have to check for.

P shouldn't be a proper subset of Candidate Key. 

For 3 NF 

No transitive dependency allowed (NP -> NP ) and must be in 2NF. 
  

For BCNF 

Suppose we have A -> B 

A should be super key. 

All FD should satisfy these conditions. 
This is all you need to check for which normal form is satisfied.

Comments

Partial dependencies exists if

Proper subset of CK ->Non prime attributes.

Then 2NF is not possible.