made easy mock 3 DBMS

Question

Consider a relation R(ABCD) with FD’s {A → D, B → D, D → BC}. What is the minimum number of decomposition required to make above relation BCNF?

A) 1

B) 2

C) 3

D) 4

Comments

very sorry iita did a major silly mistake..reposting it..

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am9 B won't be a candidate key ..as B+ =BCD it will not cover attribute A

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@iita

C.K= A

as A+= ABCD

now the relation is in 2nf.

but it is not in 3nf.as transitivity exists in A->D and D->BC.

so we break relation into AD(A->D is preserved and A is superkey),BCD(D->BC is preserved and D is superkey) and BD(B->D and b is a superkey).so for decomposition into bcnf minimum three relations are needed..not two..

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but in R(BCD) B->D is also preserved since D is the SK and B⁺ (BCD) so B is also SK hence 2 are enough