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2 2 votes

Consider $X(A,B,C,D,E,F)$ with $F=\{AB\to CD,\ CD\to EF\}$.

The relation is decomposed into $X_1(A,B,C,D)$ and $X_2(C,D,E,F)$.

Which statement is correct?

  1. The decomposition is lossless but not dependency preserving.
     
  2. The decomposition is dependency preserving but lossy.
     
  3. The decomposition is both lossless and dependency preserving.
     
  4. The decomposition is neither lossless nor dependency preserving.

1 Answer

1 1 vote

First check the binary lossless condition.

$X_1\cap X_2=\{C,D\}$.

For a binary decomposition to be lossless, the common attributes must functionally determine all attributes of at least one component.

We have $CD\to EF$.

Therefore, $CD\to CDEF$.

But, $CDEF=X_2$.

Hence,

$(X_1\cap X_2)\to X_2$.

Therefore, the decomposition is lossless.

Now check dependency preservation.

The FD $AB\to CD$ can be enforced entirely inside $X_1(A,B,C,D)$.

The FD $CD\to EF$ can be enforced entirely inside $X_2(C,D,E,F)$.

Thus both original dependencies are preserved locally.

Therefore, the decomposition is also dependency preserving.

Hence, the correct answer is C.

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