data base system

Question

Find highest normal form?

R(ABCDE)

FD={AB-->C,  C-->A,  C-->D,  BD-->E}

here I have doubt that BD-->E is partial dependency or not?

BD-->E is 1nf or 2nf or 3nf?

Answer 1

candidate keys are : AB , BC 

NOT IN BCNF 

NOT IN 3NF 

NOT IN 2 nf why ??? 

 because C------>D is partial dependency so not in 2nf and also BD---->E is not partial dependency because BD is not proper subset of candidate key .

so it is in 1nf (answer) 

Comments

I have one contradiction example for above example:

R(ABCD)

FD={AB-->C,  C-->B,  BC-->D}

here I have doubt that BC-->D is partial dependency or not?

if partial dependency then why?

Answer 2

Ur doubt was particular out of given 4FDs BD->E falls under which NF ??

Ans is 3NF. Follow the below Tour...

Here Candidate Keys are:- AB,BC

BD->E  (Lets Check)

1.Partial Dependency:-

PartOfKey->NonKey

Here if B(part of key) is alone in LHS of FD then only it will Partial Dependency So here it is clearly no partial Dependency.

2.Transitive Dependency:-

non-key -> non-key

BD->E (Here LHS not satisfy non-key cz B is prime attribute)

So there is no transitive dependency So this particular FD (BD->E) out of 4 FDs as u asked is in 3NF.

and also note that this is not BCNF bcz BD is not a superkey.

Comments

@arjun sir Here BD->E violates BCNF as BD is not a Superkey.

So can't I tell like that BD->E satisfy Upto 3NF ??

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Not exactly. "Normal forms" are for detabase schema- not FDs. Lets say we call a person rich if he has a car. But we never call the car rich. Likewise presence of certain FDs make the schema not in certain Normal Forms.

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@Rajesh

bd->e should be in 2NF bcoz bd is not a superkey and e is not prime attribute. So why u have mentioed it is in 3NF?