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+18 votes
1k views
State True or False with reason

There is always a decomposition into Boyce-Codd normal form (BCNF) that is lossless and dependency preserving.
asked in Databases by Veteran (59.8k points) | 1k views

4 Answers

+29 votes
Best answer
False

BCNF decomposition can always be lossless, but it may not be always possible to get a dependency preserving BCNF decomposition.
answered by Active (3.6k points)
selected by
+6 votes
bcnf decomposition guarantees lossless and d.p may not is correct one
answered by Junior (633 points)
+5 votes
FALSE, as BCNF guarantee only lossless property.
answered by Loyal (7.8k points)
+1 vote

false not every relation can decompose into BCNF with dependency preserving and lossless.

you can see here: relation R is not possible to decompose into BCNF with dependency preserving and lossless.

R(ABC)   F={AB->C , C->A } 

answered by Active (4.7k points)
0
Bt example is wrong !!!
+1
An example for it may be R(ABCDE) FD:(AB->CDE, C->A, D->E ).It is lossless but not fd preserving

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