GATE DA 2026 | Question: 32

Question

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 $T$.

 

X

P	Q	R
P1	Q1	R1
P2	Q2	R2
P3	Q3	R2

 

Y

P	Q	S
P1	Q1	2
P1	Q2	5
P2	Q1	6
P3	Q3	1

 

Z

P	T
P1	T1
P3	T2
P4	T3
P4	NULL

Consider the relational algebra expression

\[ \pi_{P,R,S} \Big[ (\sigma_{(Q=Q3 \vee R=R2)}(X \bowtie Y)) \ \bowtie\ (\sigma_{(S>1)}(Y \bowtie Z)) \Big] \]

where $\bowtie$ denotes natural join operation.

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

• A. Two rows $\text{(P1, R1, 2)}$ and $\text{(P1, R1, 5)}$
• B. Three rows $\text{(P1, R1, 2), (P1, R1, 5)}$ and $\text{(P2, R2, 6)}$
• C. One row $\text{(P1, R1, 2)}$
• D. Zero rows

Answer 1

\[
\Pi_{PRS}
\Big(
(\sigma_{Q=Q3 \vee R=R2}(X \bowtie Y))
\bowtie
(\sigma_{S>1}(Y \bowtie Z))
\Big)
\]

Step 1:  Compute $X \bowtie $ Y

Natural join is performed on common attributes $P$ and $Q$.

The resulting relation is:

\[
\begin{array}{|c|c|c|c|}
\hline
P & Q & R & S \\ \hline
P1 & Q1 & R1 & 2 \\
P3 & Q3 & R2 & 1 \\ \hline
\end{array}
\]

Now apply the selection condition $Q=Q3 \vee R=R2$.

Only the second row satisfies the condition.

\[
\begin{array}{|c|c|c|c|}
\hline
P & Q & R & S \\ \hline
P3 & Q3 & R2 & 1 \\ \hline
\end{array}
\]

Step 2:  Compute $Y \bowtie $ Z

Natural join is performed on attribute $P$.

\[
\begin{array}{|c|c|c|c|}
\hline
P & Q & S & T \\ \hline
P1 & Q1 & 2 & T1 \\
P1 & Q2 & 5 & T1 \\
P3 & Q3 & 1 & T2 \\ \hline
\end{array}
\]

Now apply the selection condition $S > 1$.

\[
\begin{array}{|c|c|c|c|}
\hline
P & Q & S & T \\ \hline
P1 & Q1 & 2 & T1 \\
P1 & Q2 & 5 & T1 \\ \hline
\end{array}
\]

Step 3: Final Natural Join

The first intermediate relation contains only $P3$.

The second intermediate relation contains only $P1$.

Since there is no matching value of $P$, the natural join results in an empty relation.

 

Step 4: Projection

Applying $\Pi_{PRS}$ on an empty relation still gives an empty relation.

\[
\boxed{\text{Final Output: Zero rows}}
\]

Correct Option: (D)

Comments

for X $\bowtie$ Y, the second row should be $<P_3, Q_3, R_3, 1>$

 

---

@Kushagr_Sharma

For \(X \bowtie Y\), the natural join is performed on the common attributes \(P\) and \(Q\).

The matching rows are:

\[
(P_1,Q_1,R_1,2)
\]

and

\[
(P_3,Q_3,R_2,1)
\]

So the second row contains \(R_2\), not \(R_3\).

Therefore,

\[
\langle P_3,Q_3,R_3,1\rangle
\]

is incorrect. The correct row is:

\[
\boxed{\langle P_3,Q_3,R_2,1\rangle}
\]

---

@Aditya Jyotishi ohh my bad, I misread the table values. Yes yes, then the given row is correct. Thanks!

Answer 2

we can also do by option matching
First need to be compute (σ(Q=Q3∨R=R2)​(X⋈Y))
and then we get only 

so option D will satisfied.

Comments

@Pranay_VG  Yes, this option-matching approach is correct. After the selections, the two relations have different P values, so their natural join produces zero rows.