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BCNF Decomposition
Na462
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Jul 14, 2018
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database-normalization
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Na462
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Na462
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Jul 14, 2018
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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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Shubham Shukla 6
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Jul 14, 2018
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well well i dnt knw what i was reading yes both lossy..thnks for correction:)
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Mk Utkarsh
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Jul 25, 2018
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For this particular question both A and D are correct irrespective of minimum number of relations.
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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
manav kothari
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Oct 5, 2018
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