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

But all total the relation is 1NF, because C-> D partial dependency present. rt?

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Yes U r Absolutely Correct @srestha

Bcz Here BC is candidate key and C is part of the key which identifies non-key D. means

Partial Dependency:-

PartOfKey->NonKey

C-> D 

holds here so the whole relation is not in 2NF hence in 1NF.

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Part of key means proper subset of key, right ?

Is BD a proper subset of BC ?

Straight from wiki : 

 Specifically: a table is in 2NF if it is in 1NF and no non-prime attribute is dependent on any proper subset of any candidate key of the table

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We are talking about

BC-> ABCDE

as, BC is a key.

C-> D is partial dependency

C is proper subset of BC

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Yes, PartOfKey(informally) I mean proper subset of candidate key @kapil .

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

A FD cannot be in a normal form. Rather presence of a FD can violate certain normal forms. We should never say a FD is in some normal form. This is very basic and all basic stuffs are important for GATE.

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@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?