Test Series

Question

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 __

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

Comments

It will be in 2NF.

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@ Ashwin Kulkarni  Reason for 2NF..?

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R(Eid, Pno, hours, Ename, Pname, Plocation)
Eid, pno --­­> hours, 
Eid ­­--> Ename, 
Pno ­­--> Pname, Plocation
Pname --­­> Plocation 

{Eid, Pno}+ = Eid, Ename, Pno, Hours, pname, plocation, 
CK = {Eid, Pno}

R1(Pno, Pname, Plocation)
Pno ­­--> Pname, Plocation
Pname --­­> Plocation ( Non Prime Attribute -> Non Prime Attribute)
CK={Pno}
R1 is in 2NF.

Note - Any relation with simple candidate key is by default in 2NF.

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@sahil

Pname --­­> Plocation (Non PA -> Non PA) which is violating 3NF condition.

Answer 1

the given relation wll be in 2 NF as there is single key for candidate key.

2NF says : If there is  a single attribute in the key then the relation will be in 2 NF.