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69 votes
69 votes

Consider the schema $R=(S,T, U, V)$ and the dependencies $S \rightarrow T, T \rightarrow U, U \rightarrow V$ and $V \rightarrow S$. Let $R = (R1\text{ and } R2)$ be a decomposition such that $R1 \cap R2 \neq \phi$. The decomposition is

  1. not in $\textsf{2NF}$
  2. in $\textsf{2NF}$ but not $\textsf{3NF}$
  3. in $\textsf{3NF}$ but not in $\textsf{2NF}$
  4. in both $\textsf{2NF}$ and $\textsf{3NF}$
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11 Answers

5 votes
5 votes
If all the attributes are candidate keys, then any subset of attributes will be a super key

Whatever FD you can form, your LHS will have a subset of attributes

From above stataments, we can say that any FD that can be formed will have LHS as a super-key

This is good enough to say that the relation is in BCNF ...

This is also applicable even after the decomposition of relation irrespective of what common attributes we have (even if there is no common attributes, though lossy decompositions are not recommended)
4 votes
4 votes
since no partial dependency and non key to non key

every determinant is deriving all atributes so in bcnf

 by default 1nf

so it is all 1nf,2nf,3nf, and bcnf
0 votes
0 votes
a option is right because r1 intersection r2 is phi and it shows that it is lossy decomposition which is not supported by any form i.e. (none of the 2, 3,and BCNF).
Answer:

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