retagged by
631 views
3 votes
3 votes

Decompose the following table into BCNF:
$R(ABCD)$
$A \rightarrow C$
$C \rightarrow A$
$AB \rightarrow D$

The result is:

  1. (AC) (ABD)
  2. (AC) (CBD)
  3. (AB)(BAD)
  4. Both (A) & (B)
retagged by

1 Answer

Best answer
6 votes
6 votes

The question is asking for decomposition into BCNF. It isn't asking if the decomposition is dependency preserving or not.

$R(ABCD) $
$A \rightarrow C$
$C \rightarrow A$ 
$AB \rightarrow D$ 

Here the candidate keys are $AB$ and $BC$.

For decomposing a relation into BCNF, we first find a functional dependency which is not trivial and violates the BCNF property.

We get $A \rightarrow C$, here $A$ is not a superkey, and hence causes a violation of BCNF.

So, decomposing it into $(A, C)$ and $(A, B, D)$. This is now in BCNF. Option (A).

Now if in the first step of "finding a functional dependency which is not trivial and violates the BCNF property", we chose $C \rightarrow A$, we get the following decomposition: 

$(C, A)$ and $(C, B, D)$. This is also in BCNF(but does not preserve dependencies). Option (B).

So, both (A) and (B) are correct.

selected by
Answer:

Related questions

3 votes
3 votes
1 answer
2
1 votes
1 votes
2 answers
4
Bikram asked Aug 26, 2017
316 views
The relation R (ABCDE) with FD set $\{AB \rightarrow CDE, \ A \rightarrow C, \ C \rightarrow D\}$ is in which of the following normal forms? 1NF but not in 2NF in both 2...