GATE 2026 DA Q.42: solution and answer

GATE 2026 · DAQ.42 · 2 marks · MCQ

Consider the given relations $X$ , $Y$ and $Z$ . The relation $X$ has three columns $P$ , $Q$ and $R$ . The relation $Y$ has three columns $P$ , $Q$ and $S$ . The relation $Z$ has two columns $P$ and $T$ .

Relation $X (P, Q, R)$ : (P1, Q1, R1), (P2, Q2, R2), (P3, Q3, R2).

Relation $Y (P, Q, S)$ : (P1, Q1, 2), (P1, Q2, 5), (P2, Q1, 6), (P3, Q3, 1).

Relation $Z (P, T)$ : (P1, T1), (P3, T2), (P4, T3), (P4, NULL).

Consider the relational algebra expression

π_{P R S} [(σ_{(Q = Q 3 \lor R = R 2)} [X ⋈ Y]) ⋈ (σ_{(S > 1)} [Y ⋈ Z])]

where $⋈$ denotes natural join operation.

Which of the following options is the correct output for the given expression?

GATE 2026 DA question 42: Three small tables side by side. — Figure from the GATE 2026 DA paper. It is part of the question.

(A)
Two rows (P1, R1, 2) and (P1, R1, 5)
(B)
Three rows (P1, R1, 2), (P1, R1, 5) and (P2, R2, 6)
(C)
One row (P1, R1, 2)
(D)
Zero rows

Answer and solution

ANSWER

The expression returns zero rows, which is option (D).

Official answer: DSource: Official answer key, GATE 2026 DA (IIT Guwahati)

Left part: $X ⋈ Y$ is a natural join on the common attributes $P, Q$ . The pairs $(P, Q)$ in $X$ are $(P 1, Q 1)$ , $(P 2, Q 2)$ , $(P 3, Q 3)$ . The pairs in $Y$ are $(P 1, Q 1)$ , $(P 1, Q 2)$ , $(P 2, Q 1)$ , $(P 3, Q 3)$ . Only $(P 1, Q 1)$ and $(P 3, Q 3)$ occur in both, so the join has the rows $(P 1, Q 1, R 1, 2)$ and $(P 3, Q 3, R 2, 1)$ over $(P, Q, R, S)$ .
Apply $σ_{Q = Q 3 \lor R = R 2}$ . The row $(P 1, Q 1, R 1, 2)$ has $Q = Q 1$ and $R = R 1$ , so both conditions fail and it is removed. The row $(P 3, Q 3, R 2, 1)$ satisfies both and stays. Left result: $L = {(P 3, Q 3, R 2, 1)}$ .
Right part: $Y ⋈ Z$ is a natural join on $P$ . $Z$ contains $P 1, P 3, P 4$ . $Y$ rows with $P 1$ give $(P 1, Q 1, 2, T 1)$ and $(P 1, Q 2, 5, T 1)$ . The $Y$ row with $P 3$ gives $(P 3, Q 3, 1, T 2)$ . The $Y$ row with $P 2$ has no partner in $Z$ and is dropped. $P 4$ appears in no $Y$ row.
Apply $σ_{S > 1}$ . The row with $S = 1$ is removed, leaving $R = {(P 1, Q 1, 2, T 1), (P 1, Q 2, 5, T 1)}$ over $(P, Q, S, T)$ .
Final natural join $L ⋈ R$ is on the common attributes $P, Q, S$ . Every tuple of $L$ has $P = P 3$ , and every tuple of $R$ has $P = P 1$ , so no pair of tuples agrees on $P$ . The join is empty.
The projection $π_{P R S}$ of an empty relation is empty, so the output has zero rows. This is option (D).

KEY IDEA

A natural join keeps only tuples that agree on every common attribute. Evaluate each side of the final join separately and compare them on the shared attributes $P, Q, S$ .

WATCH OUT FOR

Reading the final join as being on $P$ alone, or forgetting that the selection with $Q = Q 3 \lor R = R 2$ removes the $P 1$ row from the left side. Another slip is to treat $Y ⋈ Z$ as the only source of rows and read off $(P 1, R 1, 2)$ and $(P 1, R 1, 5)$ , ignoring that the left side contains only $P 3$ .

From the GATE 2026 DA paper. Every question of this paper · Download the full paper · All 5 Relational algebra questions, with notes

GATE 2026 DA Q.42: solution and answer

Answer and solution

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