Question

Let there be two relations \(X\) and \(Y\) as shown. \(X\) has three columns \(P\), \(Q\) and \(R\). \(Y\) has two columns \(P\) and \(S\).

Relation X:

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

Relation Y:

P	S
P1	10
P1	15
P2	20
P3	1

Consider that the following tuple relational calculus expression is evaluated.
\[ \{t \mid t \in X \wedge \exists z \in X (t[P] = z[P]) \wedge \exists m \in Y (m[P] = t[P] \wedge m[S] > 1)\} \]
The number of tuples that will be returned is _______. (Answer in integer)

Hint:

The self-reference on X is always true; the real filter checks whether each tuple's P value appears in Y with an S value strictly greater than 1.

Answer 1

Correct Answer: 2

A clean way to read this tuple relational calculus expression is to translate it into an equivalent SQL query, since that makes the logic more familiar. The condition $\exists z \in X(t[P] = z[P])$ is a self-reference that is always satisfied (just pick $z = t$), so it contributes nothing to the filtering and can be dropped. What remains is: keep a tuple $t$ from $X$ only if there is some row in $Y$ with the same $P$ value and $S > 1$. In SQL terms, this is:

SELECT * FROM X WHERE P IN (SELECT P FROM Y WHERE S > 1)

1. Find which P values in Y have S > 1. Y's rows are (P1, 10), (P1, 15), (P2, 20), (P3, 1). Filtering for S > 1 keeps (P1, 10), (P1, 15), and (P2, 20), while (P3, 1) is dropped since 1 is not greater than 1. So the surviving P values are P1 and P2.
2. Match these P values against X. X's rows are (P1, Q1, R1), (P2, Q2, R2), (P3, Q3, R2). The row with P1 matches, the row with P2 matches, but the row with P3 does not, since P3 was excluded from the surviving P values in step 1.

So exactly two rows of X, the ones with P1 and P2, satisfy the original expression, while the P3 row is excluded because its only matching row in Y has S = 1, which fails the strict S > 1 test.

$$\boxed{2}$$