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GO Classes DPP | GATE CS, DA | DBMS | Relational Algebra

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Let $R$ and $S$ be union-compatible relations.

Which expression computes $R\cap S$ using only union and set difference?

  1. $(R\cup S)-((R-S)\cup(S-R))$
     
  2. $(R\cup S)-(R-S)$
     
  3. $(R-S)\cup(S-R)$
     
  4. $(R\cup S)-(R\cap S)$

1 Answer

1 1 vote

Start with $R\cup S$.

This contains every tuple appearing in either relation.

Now, $R-S$ contains tuples appearing only in $R$, while $S-R$ contains tuples appearing only in $S$.

Therefore, $(R-S)\cup(S-R)$ contains all tuples belonging exclusively to one of the two relations.

Remove these from the union $:$

$(R\cup S)-((R-S)\cup(S-R))$.

Only tuples appearing in both relations remain.

$\therefore (R\cup S)-((R-S)\cup(S-R))=R\cap S$

Why the others fail:

B does not isolate the intersection. It retains all of $S$.

C gives the symmetric difference, meaning tuples present in exactly one relation.

D already uses $\cap$, so it does not express intersection using only the requested operators.

Hence, the correct answer is A.

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