BCNF DECOMPOSTION

Question

Can the given relation can be decomposed into BCNF with dependency preserving and lossless join decomposition

               R(ABCDEFG) : {AB → C, BC → A, AC → B, B → D, D → E}.

Comments

Yes.

Relations are (ABFG, ABC, BD, DE)

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BCFG ABC BD DE

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@  Tesla!

Can you show the steps how you decomposed this relation?

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Here ABFG, BCFG, ABFG are keys, we can include anyone in 1st relation.

Answer 1

The correct way to solve such question is to start from lowest level of Normalization and go upwards

so we have R(ABCDEFG) with  AB → C, BC → A, AC → B, B → D, D → E as FD's

our current relation is not in 2NF due to partial dependency as our candidate key are ABFG, ACFG, BCFG so all FD's violate 2NF 

step 1 break it to fulfill 2NF condition. here i am taking ABFG as key

R1(ABFG) with FD A->A, B->B, F->F, G->G

R2(ABCDE) with FD AB → C, BC → A, AC → B, B → D, D → E.

R1 is in 2NF, 3NF and BCNF but R2 violate 2 NF because of B-> partial dependency

R11(ABC) R12(BDE) with there respective FD's.

R11 is in 2NF, 3NF and BCNF 

R12 is not in 3NF due to D->E (non prime -> non prime)

break R12 as R121(BD) R122(DE)

Now each relation is BCNF dependency preserving lossless decomposition 

which are R1(ABFG) R2(ABC) R3(BD) R4(DE) 

If you first try to fix lower normal form higher will be in place here i first converted it into 2 NF and the 3 NF it turned out it automatically became BCNF, there is no algorithm for my method its just aptitude based 

Comments

D->E violate so 

R(A,B,E,F,G)-(E-D)= R(A,B,F,G) and R(ED)

Here, as we are subtracting (E-D) from (A,B,E,F,G), won't the D will be added to (A,B,F,G)?

If not then what is the actual significance of subtraction?

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First perform B-A then perform R - result of B-A

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Please see the answer now as i have changed it @srivivek95