What is the Normal Form?


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.


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


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


but yhe answer is 1NF.


It is asked about the relation R1.

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

Clearly the candidate key is {Pno}

PrimeAttributes = {Pno}

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.