GATE CS 2005 Q30 — Relational Model: Relational Algebra

For any relation $r$ with schema $R$ decomposed into projections $r_1 = \pi_{X}(r)$ and $r_2 = \pi_{Y}(r)$ such that $X \cup Y = R$, the natural join of the projections always contains at least all the tuples of the original relation. That is:

Let $s = r_1 \bowtie r_2$. Thus, we always have $r \subseteq s$.

• Lossless Join: If the decomposition is lossless, then $s = r$.
• Lossy Join: If the decomposition is lossy, the join generates additional spurious tuples, meaning $s$ contains more tuples than $r$ ($r \neq s$).

Since the problem states that the decomposition is lossy, we have:

Let's evaluate the options:

• Option A ($s \subset r$): False, because $s$ contains more tuples than $r$.
• Option B ($r \cup s = r$): False. Since $r \subset s$, we have $r \cup s = s$.
• Option C ($r \subset s$): True, as derived above.
• Option D ($r * s = s$): False. Since both $r$ and $s$ share the same schema $(A, B, C, D)$, their natural join is equivalent to their intersection: $r \bowtie s = r \cap s$. Since $r \subset s$, we get $r \cap s = r \neq s$.