The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+17 votes
888 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 (59.6k points) | 888 views

4 Answers

+27 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 (3.5k points)
selected by
+6 votes
bcnf decomposition guarantees lossless and d.p may not is correct one
answered by Junior (607 points)
+4 votes
FALSE, as BCNF guarantee only lossless property.
answered by Loyal (7.5k 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 (4.5k 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

Related questions



Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

42,598 questions
48,599 answers
155,642 comments
63,709 users