What is the Normal Form?


#1

Consider the following set of functional dependencies
{
(Eid, pno) --­­> hours,
Eid ­­–> Ename,
Pno ­­–> (Pname, Plocation),
Pname --­­> Plocation
}

in a relation R (Eid, Pno, hours, Ename, Pname, Plocation). After decomposition, the relation R1 (Pno, Pname, Plocation) is in __( Marks: 0.00 )

  1. BCNF
  2. 3NF but not in BCNF
  3. 1NF

Please avoid sharing links… if possible give your own solution.


#2

I think it is 3 NF but not in BCNF ! as no transitive dependency is there! As BCNF is more stricter than 3NF


#3

In given functional dependencies candidate key is (Eid,Pno).
After decomposition into R1(Pno,Pname,Plocation) . Here functional dependencies are

Pno ­­–> (Pname, Plocation),
Pname --­­> Plocation

Candidate key is Pno and Pno is the only prime attribute here. So no partial dependencies ( If proper subset of candidate key determines non-prime attribute, it is called partial dependency.
) are there. So it must be in 2NF.

A relation is in 3NF iff at least one of the following condition holds in every non-trivial function dependency X –> Y
1.X is a super key.
2.Y is a prime attribute (each element of Y is part of some candidate key).

Pname --­­> Plocation
is not follow any of the above rules so we can say it is not in 3NF


#4

but yhe answer is 1NF.


#5

It is asked about the relation R1.

R1(Pno,Pname,Plocation) has the following dependencies after decomposition.
Pno->(Pname,Plocation)
Pname->Plocation

Clearly the candidate key is {Pno}

PrimeAttributes = {Pno}
NonPrimeAttributes={Pname,Plocation}

So there is a nonkey-nonkey dependency.{Pname->Plocation}.
So R1 is not in 3NF. But there is no partial dependency, so it is in 2NF.

The best option that matches is 3 as a table in 2NF is already in 1NF.