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5 votes
5 votes
1) dependency preservation
2)lossless join

a)If a relation is in 3NF , which of the above points is guaranteed.
b)If a relation is in BCNF , which of the above points is guaranteed

[ I am confused  right now, can I say , if a relation is in BCNF, then is it  by default  lossless/dependency preserving straight away , or is it the fact that,if I derive BCNF decomposed relations using a particular algorithm, then only  I can say that the decomposed relations is lossless/dependency preserving same goes for 3NF].

Please help!

4 Answers

Best answer
18 votes
18 votes
The question is not making sense as it is. So, making correction.

Let $R$ be a relation. Now if we decompose it and make the decomposition $3NF$, then it is possible to satisfy both lossless and dependency preserving. i.e., for any relation there always exist a decomposition to 3NF, which is guaranteed to satisfy both the given properties.

For BCNF, the decomposition is not guaranteed to satisfy both the properties. Sometimes when lossless property is satisfied, dependency preserving is no longer possible.
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10 votes
10 votes

If a relation is in 3NF , which of the above points is guaranteed ==> both are neccessary  

If a relation is in BCNF , which of the above points is guaranteed ==> lossless join is guaranteed but dependency preserving may or may not be guaranteed

https://www.iitg.ernet.in/awekar/teaching/cs344fall11/lecturenotes/september%2006.pdf

1 votes
1 votes

R[A,B,C,D,E,F] 
A->BCDEF
BC->ADEF
D->E
D->B
B->F
Decompose this into 3NF and BCNF and find the diff ;)

0 votes
0 votes
i think if a relation is in 3NF then both lossless and dependency preserving are guaranteed but in case of BCNF only lossless is guaranteed.
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