decomposition of relation

Question

if you are given table like R(A,B,C,D,E,F)= {AB$\rightarrow$C, BC $\rightarrow$ A, AC$\rightarrow$ B , B $\rightarrow$ D , D$\rightarrow$ E }

How many tables are required in each normal form

A- 2NF

B- 3 NF

C- BCNF

please illustrate 2NF only

Comments

@Utkarsh Joshi

We can decompose R as R1(A,B,C) and R2(B,D,E)

where is F ?

if you add F in R1 then it will lead to PFD, isn't it?

for 2NF, it requires 3 tables i.e., one of the decomposition is R₁(A,B,C) , R₂(B,D,E) and R₃(A,C,F)

for 3NF/BCNF, it requires 4 tables i.e., one of the decomposition is R₁(A,B,C) , R₂(B,D) R₄(D,E) and R₃(A,C,F)

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Shaik Masthan the thing is i completely forgot to take F in consideration i thought we are having just ABCDE as F was not part of any FD.

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i expected brother...

Answer 1

https://gateoverflow.in/?qa=blob&qa_blobid=11466423765690194514

Maybe for bcnf.

Comments

Clearly F should be a part of the key, as F is not on the RHS of any functional dependency. So clearly either ABF, ACF, BCF should be separated which will not have any functional dependency. Thus the division as done by hitendra is perfect.

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as given first three FD's are made of A,B,C

In every AB→C, BC → A, AC→ B we will get a subset of C.K {ABF,ACF,BCF} LHS of the FD's are subset these will violate 2NF condition .so we have to split

I hope you get it ,

revert back if u didn't.

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thanks @hitendra singh @tejasv