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GO Classes DPP | GATE CS, DA | DBMS | Non-Binary Decomposition

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1 1 vote

Consider $R(S,N,C,P,X,Y,Q)$ with $F=\{S\to NC,\ P\to XY,\ SP\to Q\}$.

The relation is decomposed into $R_1(S,N,C),$ $R_2(P,X,Y), $ $R_3(S,P,Q)$.

Which statement is correct?

  1. The decomposition is lossy because no single component contains all attributes of $R$.
     
  2. The decomposition is lossless, and the Chase eventually produces a row containing all distinguished values.
     
  3. The decomposition is lossless only if $NC\to S$ is additionally given.
     
  4. Chase cannot be applied because the decomposition contains three relations.

1 Answer

1 1 vote

One way to see this is through the Chase.

For $R_3(S,P,Q)$, initially the distinguished values are available for $S$, $P$, and $Q$.

Using $S\to NC$, the values of $N$ and $C$ can be forced to their distinguished values in the row corresponding to $R_3$.

Using $P\to XY$, the values of $X$ and $Y$ can also be forced to their distinguished values.

Thus the $R_3$ row eventually contains distinguished values for $S,N,C,P,X,Y,Q$.

Therefore, the Chase contains a completely distinguished row.

Hence, the decomposition is lossless.

We can also prove it using successive combining:

$R_1\cap R_3={S}$.

Since

$S\to SNC$,

$R_1$ and $R_3$ combine losslessly.

Their combined schema is $SNCPQ$.

Now its intersection with $R_2$ is $\{P\}$.

Since

$P\to PXY$, the second combination is also lossless.

Therefore, all three relations reconstruct $R$ losslessly.

Answer : B

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