Tuple Calculus Practice Questions

The query aims to find the names of players who belong to the team named 'MI'.

1. The outermost part of the query, $\{ p.pname \mid \dots \}$, specifies that we want to retrieve the player names.
2. The condition $p \in players$ indicates that $p$ is a tuple from the 'players' relation, ensuring we are selecting player attributes.
3. The existential quantifier $\exists t$ introduces a tuple $t$ from the 'teams' relation.
4. The condition $t \in teams$ confirms that $t$ is a tuple from the 'teams' relation.
5. To link players to teams, we need to join the 'players' and 'teams' relations using their common attribute, tid. This is represented by $p.tid = t.tid$.
6. Finally, to filter for teams named 'MI', we add the condition $t.tname = 'MI'$.

Combining these, the correct tuple relational calculus query is: $\\{ p.pname \\mid p \\in players \\wedge \\exists t ( t \\in teams \\wedge p.tid = t.tid \\wedge t.tname = 'MI' ) \\}$

The expression to evaluate is $\sigma_{B < C}(R \bowtie_{R.A=S.A} S)$.
First, perform the natural join $R \bowtie_{R.A=S.A} S$ on attribute $A$. This joins tuples where $R.A = S.A$.
Let's list the tuples from R and S and find matches on A:
R:
(10, 20)
(20, 30)
(30, 40)
(30, 50)
(50, 95)

S:
(10, 90)
(30, 45)
(40, 80)

Join results:

1. $R.A=10$: (10, 20) from R and (10, 90) from S. Result: (10, 20, 10, 90) (attributes R.A, R.B, S.A, S.C)
2. $R.A=30$: (30, 40) from R and (30, 45) from S. Result: (30, 40, 30, 45)
3. $R.A=30$: (30, 50) from R and (30, 45) from S. Result: (30, 50, 30, 45)

The joined table (with common A attribute removed for simplicity in standard natural join representation, but here we keep distinct A's to show R.A, S.A as per solution's example) for $R \bowtie_{R.A=S.A} S$ is:
(R.A, R.B, S.A, S.C)
(10, 20, 10, 90)
(30, 40, 30, 45)
(30, 50, 30, 45)

Next, apply the selection condition $\sigma_{B < C}$. We compare $R.B$ and $S.C$ for each tuple in the joined result.

1. (10, 20, 10, 90): Is $20 < 90$? Yes. This tuple is selected.
2. (30, 40, 30, 45): Is $40 < 45$? Yes. This tuple is selected.
3. (30, 50, 30, 45): Is $50 < 45$? No. This tuple is NOT selected.

The resulting tuples after selection are:
(10, 20, 10, 90)
(30, 40, 30, 45)

The total number of tuples obtained is 2.

The probability of $A > 10$ is calculated as the number of values greater than 10 (11 to 20, which is 10 values) divided by the total range of $A$ (6 to 20, which is 15 values), so $P(A > 10) = \frac{10}{15} = \frac{2}{3}$.
The probability of $B = 18$ is 1 value out of the total range of $B$ (1 to 20, which is 20 values), so $P(B = 18) = \frac{1}{20}$.
Since $A$ and $B$ are independent, $P((A > 10) \wedge (B = 18)) = P(A > 10) \times P(B = 18) = \frac{2}{3} \times \frac{1}{20} = \frac{1}{30}$.
Using the principle of inclusion-exclusion, $P((A > 10) \vee (B = 18)) = P(A > 10) + P(B = 18) - P((A > 10) \wedge (B = 18))$.
This gives $P((A > 10) \vee (B = 18)) = \frac{2}{3} + \frac{1}{20} - \frac{1}{30} = \frac{40+3-2}{60} = \frac{41}{60}$.
The estimated number of tuples is the total number of tuples multiplied by this probability: $1200 \times \frac{41}{60} = 20 \times 41 = 820$.

A query in tuple relational calculus is considered "safe" if it is guaranteed to produce a finite set of tuples as its result. Unsafe queries can potentially generate infinite results, often due to negations over infinite domains or lack of sufficient constraints.

In this problem, all three queries are constrained by the tuples that exist within the EMP and DEPT relations. Since these base relations are finite, any result derived from them will also be finite. Therefore, all three queries are safe.

Let $R-S = \{a,b\}$.

1. Query I Analysis: The expression $I.\pi _{R-S}(r)-\pi_{R-S}(\pi_{R-S}(r) \times S -\pi_{R-S,S}(r))$ is the standard relational algebra definition for the division operator $r \div s$. This query returns tuples $(a,b)$ from $r$ such that for every tuple $u$ in $s$, there exists a tuple $v$ in $r$ where $v[\{a,b\}]$ is the current $(a,b)$ tuple and $v[c] = u[c]$.

2. Query II Analysis: The expression $II.\{t|t\in \pi _{R-S}(r)\wedge \forall u \in s (\exists v \in r(u=v[s]\wedge t=v[R-S]))\}$ is the formal definition of relational division in tuple relational calculus. It selects an $(a,b)$ tuple $t$ from $\pi_{R-S}(r)$ if for every $c$-value $u$ in $s$, there exists a tuple $v$ in $r$ such that $v[c]$ matches $u[c]$ and $v[\{a,b\}]$ matches $t$. This is precisely what relational division computes.

3. Query III Analysis: The expression $III.\{t|t\in \pi _{R-S}(r)\wedge \forall v\in r(\exists u\in s(u=v[s]\wedge t=v[R-S]))\}$ has a crucial difference in quantification. It requires that for every tuple $v$ in $r$, there exists a $c$-value $u$ in $s$ such that $v[c]=u[c]$ AND $t=v[\{a,b\}]$. This means all tuples in $r$ must project to the same $(a,b)$ tuple $t$, which is not what relational division implies.

4. Query IV Analysis: The SQL query $IV. Select \ R.a, R.b \ From \ R,S \ Where \ R.c=S.c$ performs an inner join between $R$ and $S$ on attribute 'c', followed by a projection onto attributes $a$ and $b$. This returns all $(a,b)$ pairs from $r$ that are associated with at least one $c$-value present in $s$. This is not relational division, which requires association with all $c$-values in $s$.

Based on the analysis, Query I and Query II both represent the relational division operation.

The final answer is $\boxed{A}$

The universal quantification $\forall t \in r(P(t))$ states that the predicate $P(t)$ is true for every element $t$ in set $r$. According to the laws of predicate logic, this is logically equivalent to asserting that there does not exist an element $t$ in $r$ for which $P(t)$ is false. Mathematically, this transformation is represented as $\neg \exists t \in r(\neg P(t))$, which corresponds exactly to Statement III. Statement I, $\neg \exists t \in r(P(t))$, is equivalent to $\forall t \in r(\neg P(t))$, which is the opposite of the original intent. Statement IV, $\exists t \in r(\neg P(t))$, represents the negation of the universal statement. Thus, only Statement III is correct.

The query selects e.name for an employee e based on the given condition.
The condition is employee(e) ^ (∀x) [¬employee(x) v x.supervisorName ≠ e.name v x.sex = "male"].
Let's focus on the universally quantified part:
(∀x) [¬employee(x) v x.supervisorName ≠ e.name v x.sex = "male"]

This can be rewritten using the implication A -> B which is equivalent to ¬A v B:
(∀x) [employee(x) -> (x.supervisorName ≠ e.name v x.sex = "male")]

This means, "For all employees x, if x is an employee, then either x's supervisor is not e.name OR x's sex is male."
The disjunction (x.supervisorName ≠ e.name v x.sex = "male") is equivalent to ¬(x.supervisorName = e.name ^ x.sex ≠ "male").
Assuming sex is either "male" or "female", x.sex ≠ "male" means x.sex = "female".

So, the condition becomes:
(∀x) \$[employee(x) -> ¬(x.supervisorName = e.name ^x.sex = "female")]\$

This universal statement is logically equivalent to the negation of an existential statement:
¬(∃x) \$[employee(x) ^ (x.supervisorName = e.name ^x.sex = "female")]\$

In plain language, this means "There does not exist an employee x such that x is an employee, x is supervised by e.name, AND x is female."
Therefore, e is an employee who has no immediate female subordinates.

The final answer is $\boxed{C}$

Codd's Theorem establishes that relational algebra, safe tuple relational calculus (TRC), and safe domain relational calculus (DRC) are computationally equivalent in expressive power. Relational algebra operations such as selection $\sigma$, projection $\pi$, union $\cup$, set difference $-$, and Cartesian product $\times$ can be mapped directly to logical predicates within safe TRC and DRC. Conversely, any safe query expressed in TRC or DRC can be translated into an equivalent expression in relational algebra. Since these formalisms define the same class of "relational complete" queries, all three languages possess identical expressive power. Therefore, I, II, and III are all equivalent.

A relational calculus expression is considered safe if it produces a finite result for any database instance. The expression in option (c), $\{t \mid \neg (t \in R_1)\}$, selects all possible tuples $t$ that are not in the relation $R_1$. Because the domain of all possible tuples is infinite, the set of tuples satisfying $\neg (t \in R_1)$ is also infinite. Since the resulting relation is not finite, the expression violates the safety condition. In contrast, options (a), (b), and (d) are bounded by values existing within $R_1$ or $R_2$, ensuring their results are always finite.

The tuple calculus expression $\{t \mid t \in r \land (t[A] = 10 \land t[B]=20)\}$ selects all tuples $t$ in relation $r$ that satisfy both conditions $t[A]=10$ and $t[B]=20$. In relational algebra, this is equivalent to $\sigma_{(A=10 \land B=20)}(r)$. According to set-theoretic properties of relational algebra, the intersection of two selections on the same relation, $\sigma_{P_1}(r) \cap \sigma_{P_2}(r)$, results in a set of tuples that satisfy both predicates $P_1$ and $P_2$. Thus, $\sigma_{(A=10 \land B=20)}(r)$ is identical to the intersection of the two separate selections $\sigma_{(A=10)}(r) \cap \sigma_{(B=20)}(r)$.