normalization

Question

Consider the relation R(ABCD) with FD's set F={ AB->CD,C->.A,D->B}

which of the following is false ?

1.C->A is a partial Dependency

2.C->A is a transitive Dependency

3.D->B IS A Partial dependency

4. All of these

Answer 1

A,B,C,D all are prime attribute so no partial as well as no transitive dependencies..

Comments

@Arjun sir : i think CD ---> A satisfying 3NF condition so  2NFcondition trivially satisfied..

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got u sir.. thank u :-)

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@Arjun Sir I did not get your point... did you mean to say that if CD->A was present then C->A would satisfy partial dependency as C is a proper subset of CD... you meant something like this sir? @Arjun sir pls clarify pls..

Answer 2

Here AB , BC , CD , AD are candidate keys.

So , A , B , C , D all are part of prime attributes.

So , the relation is in 3NF .

 

So , ans is (d) : all are false

Comments

Relation is in 3NF only..
if all attribute are prime then 3NF relation.
proper subset of candidate key -----> prime not allowed in BCNF.
C ---->A, D----> B
neither C nor D are super key.

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let primary key A and a non-key domain B and C where C is more dependent on B than A and B is directly dependent on A, then C can be considered transitively dependant on A.
if A---->B , B----->C then A----->C is transitive dependency.

Transitive dependencies not allowed in 3NF.

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okk.. sorry... my bad.. thanks a lot for clarifying :)

Answer 3

Candidate keys={AB,AD,BC,CD}

Prime attributes are A,B,C,D

1.C->A (P->P) Not PD(P->NP)

2.C->A(P->P) Not TD(NP->NP)

3.D->B(P>P) Not PD

All are false hence

Answer 4

CK-AB,C,D so

1-C->A No partial dependency

2-No NPA so no transitive dependency

3-this is also wrong

4-All are false

So option 4 is right option