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Question

Consider the following relation R (A B C D E) WITH FD F= (A ⇒BC, CD ⇒E, B⇒D, E ⇒A ).

How many minimum relations required to decompose into BCNF with Dependency preserving and Lossless join decomposition?

Answer 1

Here A,E,BC,CD are candidate keys.

B-->D is violation of bcnf ...

Made two tables R1={A,B,C,E} and R2{B,D}..

Here it is lossless but not dependency preserving...

CD-->E is lost...so for it we will create one more relation...

So total 3 relations are required...

Comments

we can take E as common attribute to make lossless join decomposition.

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@ARJUN SIR
PLEASE LOOK

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how  A---> BC in ABCE relation is in BCNF?
since A is not superkey for this relation

Answer 2

Given relation is either BCNF or Dependency Preserving.

FD F= (A ->BC, CD -> E, B -> D, E  -> A )

Here CD -> E , E ->C Because of this relation is either BCNF or Dependency Preserving.

Comments

since taking cde will not result in dependency preserving decomposition ,
we can take E  as a common attribute and e is a candidate key..

so by taking E as a common attribute it will be in BCNF and lossless join decomposition.

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this should be correct answer @arjun sir please varify it

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If B → D and CD-->E then due to PseudoTransitive Rule CB -->E .

Now perform closure of  FD’s from R(ABCE) and R(BD).

So minimum two relations required R(ABCE) and R(BD) to be in BCNF.