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

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1 1 vote

Consider the relation $R(T,I,G,E,R,S)$ with $F=\{G\to ES,\ RT\to I,\ IR\to E\}$.

It is decomposed into $R_1(T,E),$ $R_2(I,G),$ $R_3(I,E,R,S)$.

What does the Chase Test conclude?

  1. Lossless, because $E$ occurs in both $R_1$ and $R_3$.
     
  2. Lossless, because $I$ occurs in both $R_2$ and $R_3$.
     
  3. Lossy, because the Chase reaches a fixed point without producing an all-distinguished row.
     
  4. This is not a valid decomposition because no attribute occurs in all three relations.

1 Answer

1 1 vote

Construct the initial Chase tableau.

Using $t,i,g,e,r,s$ as the distinguished values:

\[
\begin{array}{|c|c|c|c|c|c|c|}
\hline
\text{Relation} & \text{T} & \text{I} & \text{G} & \text{E} & \text{R} & \text{S} \\
\hline
\text{R}_1 & t & i_1 & g_1 & e & r_1 & s_1 \\
\hline
\text{R}_2 & t_2 & i & g & e_2 & r_2 & s_2 \\
\hline
\text{R}_3 & t_3 & i & g_3 & e & r & s \\
\hline
\end{array}
\]

Now apply the FDs.

For $G\to ES$, no two rows have the same $G$ value.

For $RT\to I$, no two rows agree on both $R$ and $T$.

For $IR\to E$, no two rows agree on both $I$ and $R$.

Therefore, no Chase substitution can be made.

The table has reached a fixed point.

No row contains $t,i,g,e,r,s$ simultaneously.

Therefore, the decomposition is lossy.

Answer : C

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