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Consider the relation $R(ABCDE)$ and functional dependency 

$F=\{AB \rightarrow C, C \rightarrow D,D \rightarrow E, E \rightarrow A \}.$

If we convert given relation in Boyce Codd Normal Form ( BCNF ) then

  1. In BCNF it is lossless decomposition and dependency preserving
  2. In BCNF it is lossless decomposition but dependency is not preserved
  3. In BCNF it is lossy decomposition but dependency is preserved
  4. In BCNF it is lossy decomposition but dependency not preserved
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According to the given FD's , the attribute B do not appear on the right hand side of any FD, hence it is an essential attribute and must be included in the candidate key.

To find the Candidate key,we will pair up B with other attributes and see which pair's closure gives us the relation R.

Some homework will yield that AB,EB are the candidate keys of the above relation.

Now let us observe the given FD's one by one.

The first FD $\left ( AB\rightarrow C \right )$ has a super-key on LHS and hence satisfy BCNF.

Therefore Option A seems correct.

For all other FD's , the LHS is not a super-key. So we need decomposition.

Clearly a decomposition such as $R_{1}\left ( ABC \right ), R_{2}\left ( CD \right ), R_{3}\left ( DE \right ), R_{4}\left ( EA \right )$ puts all sub-relations in BCNF.

$R_{1}$ and $R_{2}$ have C as common attribute and C is also a key on $R_{2}$ , so their join is lossless.

Similarly,we can join these sub-relations based on common attributes which are also key in atleast one of the sub-relations.

Hence,the BCNF decomposition is lossless join.

Now it is trivial to see that the dependencies covered by these sub-relations and those given originally are equivalent.

Hence, the BCNF decomposition is also dependency preserving.

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Shefali asked Jul 23, 2015
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The answer given is YES. But how is the dependency C - DE preserved in this decomposition?