GATE 2025 DA Q.16: solution and answer

Official answer: CSource: Official answer key, GATE 2025 DA (IIT Roorkee)

1. A decomposition of $R$ into $R_1,\dots ,R_n$ is dependency-preserving if every functional dependency of $F$ can be enforced inside a single decomposed relation, i.e. $(F_1\cup F_2\cup \cdots \cup F_n{)}^{+}=F^{+}$, where $F_i$ is the projection of $F$ onto $R_i$.

2. If the decomposition is not dependency-preserving, at least one dependency $X\to Y$ has its attributes spread over two or more decomposed relations, so no single table can verify it.

3. To check such a dependency after an insert or update, the attributes of $X$ and $Y$ must first be brought side by side in one relation. The operator that combines tuples of different relations on common attributes is the (natural) join: $R_1\bowtie R_2\bowtie \cdots \bowtie R_n$. The dependency is then tested on the joined result.

4. Selection ($\sigma$) and projection ($\pi$) act on one relation and cannot bring attributes of different relations together. Union ($\cup$) stacks tuples of relations that have the same schema; it does not place attributes of different relations side by side. So none of these helps check a dependency that spans relations.

5. Hence the join is the operator executed more frequently to maintain the dependencies. Answer: (C).