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GO Classes DPP | GATE CS, DA | DBMS | Lossy Decomposition

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2 2 votes

Consider $R(A,B,C,D)$ with $F=\{B\to C,\ D\to A\}$.

The relation is decomposed into $R_1(B,C)$ and $R_2(A,D)$.

Which of the following statements are correct?

  1. The decomposition is lossy.
     
  2. The decomposition is dependency preserving.
     
  3. $R_1$ is in BCNF.
     
  4. $R_2$ is in BCNF.

1 Answer

1 1 vote

First, $R_1\cap R_2=\varnothing$.

For a binary lossless decomposition, the intersection must determine one of the component relations.

Here the intersection is empty, and no FD allows $\varnothing\to R_1$ or $\varnothing\to R_2$.

Therefore, the decomposition is lossy.

So A is correct.

Now consider dependency preservation.

$B\to C$ is entirely contained in $R_1$.

$D\to A$ is entirely contained in $R_2$.

Therefore, both original FDs can be enforced without joining the relations.

Hence, the decomposition is dependency preserving.

So B is correct.

For $R_1(B,C)$, the non-trivial FD is $B\to C$.

Therefore, $B$ is a key of $R_1$.

Hence $R_1$ is in BCNF.

So C is correct.

Similarly, in $R_2(A,D)$, $D\to A$, so $D$ is a key of $R_2$.

Thus $R_2$ is also in BCNF.

So D is correct.

Therefore, all four statements are correct.

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