ACE Volume 1

Question

consider the relation R(ABCD) with dependencies F:{A-->B,B-->C,C-->D,D-->A}.if decomposed into R1(ABC) and R2(BD),then the relation R1(ABC) is in ___________Normal form.

a)1NF       b)2NF

c)3NF       d)BCNF

here , the ans is option(d) but i have doubt that if above relation is decomposed into R1(ABC) then the dependencies A-->B,

B--->C comes under R1.and in that Dependency B--->C not satisfied for NF3 bcz non-key--->non-key so,how this relation i.e R1 in BCNF ,plz explain me.

Comments

Bro... Who says B is Non-Key ? C is Non Key ?

 

C->D->A ==> C->A right?

B->C->A ==> B->A right ?

 

therefore all are keys here.... if every attribute is key, then the relation is in BCNF.

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i know that in relation R each attributes are key.but after decomposition R1(ABC) contain only two dependencies

i.e A--->B and B--->C(so,here A is key for R1 and B, C are non-key) the say me how this relation R1 in BCNF.

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why only 2 dependencies are there in the relation R1 ?

C->A and C->B and B->A also there in the relation R1

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@meethunjadhav

It's not that what are given are the only FDs we are having.

We have to consider all the FDs(you can find closure or apply axioms), both in the original or in the decomposed relation, and then check for normal forms.

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got it bro..

After decomposition we can write any number of dependency that satisfy the relation

here, in the above case R1(ABC) contain {A-->B  , B-->C} i.e {A-->BC} and this is non trivial dependency with 'A' is super key that is why R1 in BCNF.

aise hi hai na...