BCNF Decomposition

Question

 

Comments

Brother look Carefully Will ya !

R1(BCD) intersection R2(ABC) = BC Which is candidate key in R1 because of BC -> D.

R1(AB) R2(AC) R3(BCD):-

R1 intersection R2 = A which is key in both so Merging R1 and R2 in R = (ABC)

Now R(ABC) intersection R3 (BCD) = BC which is key in R3 so Relation is lossless.

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well well i dnt knw what i was reading yes both lossy..thnks for correction:)

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For this particular question both A and D are correct irrespective of minimum number of relations.

Answer 1

Ans is A & D.

option A> BCD F:{BC->D} , ABC {A->B,A->C}   loss less And dependency preserving

option D> AB{A->B},AC{A->C},BCD{BC->D} lossless And dependency preserving