Relational Algebra Practice Questions

We want to find the $eIds$ of employees who own all brands. This is a relational division problem.
Given relations:
Employee(eId, Name)
Brand(bId, bName)
Own(eId, bId) (links employees to brands they own)

We need to select eId from Employee such that for every bId in Brand, there exists a tuple (eId, bId) in Own.
Let's analyze the given options for relational division:
The division operator $R \div S$ (where $S$ has attributes $B$ and $R$ has attributes $A, B$) is equivalent to $\pi_A (R) - \pi_A ( (\pi_A (R) \times S) - R' )$.
In our case, $R = Own(eId, bId)$ and $S = \pi_{bId}(Brand)$. We want to divide Own by $\pi_{bId}(Brand)$ on attribute bId.
So, the expression should be equivalent to: $\pi_{eId}(Own) \div \pi_{bId}(Brand)$.

Let's check Option (A): $\Pi _{eId}(\Pi _{eId,bId}(Own)/\Pi _{bId}(Brand))$
This directly uses the division operator. $\Pi _{eId,bId}(Own)$ is simply Own. So it simplifies to $\Pi _{eId}(Own / \Pi _{bId}(Brand))$. This correctly expresses employees who own all brands. So, statement A is TRUE.

Let's check Option (B): $\Pi _{eId}(Own)-\Pi _{eId}\left ((\Pi _{eId}(Own) \times \Pi _{bId}(Brand) )-\Pi _{eId,bId}(Own) \right )$
This is the standard set-theoretic equivalent definition for relational division:

1. $\pi_{eId}(Own)$: All employee IDs that own at least one brand.
2. $\pi_{bId}(Brand)$: All brand IDs.
3. $(\pi_{eId}(Own) \times \pi_{bId}(Brand))$: All possible (eId, bId) pairs.
4. $(\pi_{eId}(Own) \times \pi_{bId}(Brand) ) - \pi_{eId,bId}(Own))$: This identifies (eId, bId) pairs where an employee $eId$ does not own brand $bId$.
5. $\Pi _{eId}\left ((\pi _{eId}(Own) \times \pi _{bId}(Brand) )-\pi _{eId,bId}(Own) \right )$: This projects the result from step 4 onto eId, giving the IDs of employees who do not own at least one of the brands.
6. $\Pi _{eId}(Own) - (\text{result from step 5})$: Subtracting these eIds from the set of all eIds who own at least one brand yields the eIds of employees who own all brands. So, statement B is TRUE.

Both options (A) and (B) are correct ways to express the query. The question asks for the expressions that return the set.
Therefore, the correct options are (A) and (B).

The relational algebra expression performs a self-join on the empAge relation. The ρ_{empNo1,age1}(empAge) operation renames the attributes of the second instance of empAge to empNo1 and age1.

The join condition (age > age1) selects tuples where an employee's age is strictly greater than another employee's age.
Projecting Π_{empNo} then retrieves the employee numbers of all employees who have an age greater than at least one other employee's age. This set precisely corresponds to all employees whose age is not the minimum.

The query performs a set difference between the results of two subqueries.

1. First subquery: $\Pi_X(\sigma_{P.Y=R.Y \land R.V=V2}(P \times R))$. We join P and R on attribute Y, but only for tuples in R where V is 'V2'. The R tuples are (Y3, V2) and (Y2, V2). Only Y2 exists in P's Y column, corresponding to the tuple (X2, Y2, Z2). The projection on X gives the set {X2}.
2. Second subquery: $\Pi_X(\sigma_{Q.Y=R.Y \land Q.T>2}(Q \times R))$. We join Q and R on Y, for tuples in Q where T > 2. The Q tuples are (X1, Y2, 5) and (X1, Y1, 6). Both Y1 and Y2 exist in R. In both cases, the X value from Q is X1. The projection on X gives the set {X1}.
   The final result is the set difference: {X2} - {X1} = {X2}. This result contains one tuple.

The original query, $Q:r \Join (\sigma _{B\lt 5}(s))$, first selects rows from table s where B is less than 5, and then performs a natural join with table r. This is an inner join, so any row in r whose B value is 5 or greater will be discarded because it won't find a match in the filtered s.

Now consider option C: $r \text{ LOJ}(\sigma _{B\lt 5}(s))$. This also filters s first. However, it then performs a left outer join. This means if a row in r has a B value $\ge 5$, it will still be included in the final result, with NULL values for the columns from s. Since this query preserves rows from r that the original query would discard, it is not equivalent.

The first query, $T1\leftarrow \pi _{CourseName}(\sigma _{StudentName='SA'}(CR))$, selects the set of courses taken by student 'SA'. From the provided table, these courses are {CA, CB, CC}.

The second query, $T2\leftarrow CR\div T1$, performs relational division. It finds all students who have taken every course listed in T1. We must check the table for students who are enrolled in all three courses: CA, CB, and CC.

The students who satisfy this condition are SA, SC, SD, and SF. Therefore, the resulting relation T2 will contain 4 rows.

The join of a relation $R$ with $m$ tuples and a relation $S$ with $n$ tuples combines their attributes based on a join condition.
The maximum size occurs in the case of a Cartesian product, where every tuple in $R$ pairs with every tuple in $S$, resulting in $m \times n$ tuples.
The minimum size occurs when the join condition is never satisfied, meaning there are no matching values between the joining attributes, resulting in $0$ tuples.
Therefore, the maximum number of tuples is $mn$ and the minimum is $0$.

Basic relational algebra operations ($\sigma, \pi, \times , \Join, \cup, \cap, -$) are designed to manipulate sets of tuples by filtering, projecting, or joining relations. These operators are inherently unable to perform aggregate calculations, such as summing values across rows, because standard relational algebra is defined to produce relations, not scalar values. Therefore, calculating "The sum of all employees' salaries" requires an extended aggregation operator (often denoted as $\gamma$ or $\mathcal{G}$), which is not part of the basic set. Options A, B, and D involve only standard filtering and joining, both of which are fully supported by these basic operators.

The full natural outer join of $R_1(A, B)$ and $R_2(A, C)$ on attribute $A$ combines matching tuples and retains all unmatched tuples padded with $null$ values.
Matching $A=1$: tuples $(1, 5) \in R_1$ and $(1, 7) \in R_2$ combine to form tuple $e = (1, 5, 7)$.
Unmatched $A=3$: tuple $(3, 7) \in R_1$ produces tuple $f = (3, 7, \text{null})$ since no match exists in $R_2$.
Unmatched $A=4$: tuple $(4, 9) \in R_2$ produces tuple $g = (4, \text{null}, 9)$ since no match exists in $R_1$.
The resulting set $R$ consists exactly of $\{e, f, g\}$, as no other combinations or padding result in the provided options $a, b, c,$ or $d$.
Therefore, $R$ contains $e, f, g$ but not $a, b$.

To minimize disk I/O accesses in a nested loop join, the smaller relation should be in the outer loop.
Given that size(r(R)) < size(s(S)), relation r(R) is smaller than s(S).
Therefore, placing r(R) in the outer loop will result in fewer disk block accesses. This is because the outer loop relation is scanned once, while the inner loop relation is scanned multiple times for each block of the outer relation.
By iterating fewer blocks in the outer loop, the total number of scans of the larger inner loop relation is reduced, thus minimizing I/O.

The given relational algebra expression is $\pi _{A1}(\pi _{A2}(\sigma _{F1}(\sigma_{F2}(r))))$.

1. Combine Selection Operations: According to relational algebra equivalences, a sequence of selection operations $\sigma_{F1}(\sigma_{F2}(r))$ can be combined into a single selection using the conjunction of the conditions: $\sigma_{(F1 \wedge F2)}(r)$.
   So, the expression becomes $\pi _{A1}(\pi _{A2}(\sigma_{(F1 \wedge F2)}(r)))$.

2. Optimize Projection Operations: We have $\pi _{A1}(\pi _{A2}(...))$. Since $A_1 \subset A_2$, projecting on $A_2$ first and then on $A_1$ is redundant. The outer projection $\pi_{A1}$ means we only need the attributes in $A_1$. If $A_1 \subset A_2$, projecting on $A_2$ first does not remove any attributes necessary for the final projection on $A_1$. Therefore, $\pi_{A1}(\pi_{A2}(X))$ is equivalent to $\pi_{A1}(X)$ if $A_1 \subset A_2$.
   The expression simplifies to $\pi_{A1}(\sigma_{(F1 \wedge F2)}(r))$.

This is the optimized version of the given relational algebra expression.

The given relational algebra query is:
$\Pi_{empId} (employee) - \Pi_{empId} (employee \bowtie_{(empId=eID) \land (empAge \leq depAge)} dependent)$
This query can be broken down into two parts:

1. The right-hand side subquery: $\Pi_{empId} (employee \bowtie_{(empId=eID) \land (empAge \leq depAge)} dependent)$. This part selects empIds of employees who have at least one dependent whose depAge is greater than or equal to the employee's empAge.

2. The main query: The outer query subtracts the empIds obtained from the subquery (those employees with at least one dependent older than or equal to them) from the set of all empIds in the employee relation.

Therefore, the result will be the empIds of employees whose age is strictly greater than all of their dependents' ages. This means their age is greater than that of all of his/her dependents.

The given relational algebra expression involves a union, a join, and a selection condition.
First, compute the union $A \cup B$. This combines all unique tuples from relations A and B.
Next, perform a join operation between $(A \cup B)$ and relation C. The join condition is $A.Id > 40 \vee C.Id < 15$.
This means a tuple from $(A \cup B)$ will be combined with a tuple from C if either the $Id$ from $(A \cup B)$ is greater than 40 OR the $Id$ from C is less than 15.
After evaluating the join condition for each possible pair, count the number of resulting tuples. The final table after all operations, as shown in the solution, contains 7 tuples.

Here's a step-by-step explanation:

Referential integrity constraint implies that for every tuple in $R1$, the value of its foreign key attribute $B$ must exist as a value in the primary key attribute $C$ of $R2$. This means that the set of all $B$ values in $r1$ must be a subset of all $C$ values in $r2$.

Mathematically, this relationship can be expressed as:
$\Pi_{B}(r_{1}) \subseteq \Pi_{C}(r_{2})$

If a set $X$ is a subset of set $Y$ (i.e., $X \subseteq Y$), then the set difference $X - Y$ must be an empty set.
Applying this principle, the difference between the projection of $B$ from $r1$ and the projection of $C$ from $r2$ must be empty.
$\Pi_{B}(r_{1}) - \Pi_{C}(r_{2}) = \phi$

This is always true when referential integrity is satisfied.

Explanation:
Hashing is most efficient for exact match lookups, as it directly maps the key to the data location. Query Q1, $\sigma _{A_{p}=c}(r)$, involves an exact match on attribute $A_p$ with a constant $c$. Therefore, hashing on $A_p$ will be very effective for Q1.

Ordered indexing is more suitable for range queries because it stores data in a sorted order, allowing for efficient retrieval of data within a specified range. Query Q2, $\sigma _{c_{1}\leq A_{p}\leq c_{2}}(r)$, is a range query on attribute $A_p$. For Q2, ordered indexing on $A_p$ will generally outperform hashing, as hashing would require probing multiple buckets within the range.

Thus, hashing will outperform ordered indexing on Q1, but not on Q2.

The maximum number of tuples in the natural join R ⋈ S is determined by the values of the common attribute, which is 'B'. Let's analyze the given functional dependencies (FDs) for relation R(A, B, C): B → A and A → C.

1. Find the key for R: Using the property of transitivity on the given FDs (since B → A and A → C), we can infer that B → C. This means B determines all other attributes in relation R (B → AC). Therefore, B is a candidate key for relation R.

2. Implication of the key: Since B is a key for R, every tuple in R must have a unique value for attribute B. The relation R has 200 tuples, so there are 200 distinct B values in R.

3. Analyze the join: The natural join R ⋈ S pairs tuples from R and S that have the same value for B. Consider a tuple from relation S. For it to be part of the join result, its B value must match a B value in R. Because B is a key in R, any given B value appears in at most one tuple in R.

4. Calculate maximum tuples: Each of the 100 tuples in S can match with at most one tuple in R. To maximize the join result, we assume that every B value present in S is also present in R. In this best-case scenario, each of the 100 tuples from S will successfully join with exactly one tuple from R. Therefore, the maximum number of tuples in the natural join R ⋈ S is 100.

1. Let $A = \Pi_{P,Q}(R)$ and $B = \Pi_{P,Q}(S)$. Since $\{P, Q\}$ is the key for both schemas, the natural join $R \Join S$ is equivalent to $A \cap B$ regarding the attributes $\{P, Q\}$.
2. Query I, $\Pi_{P}(R \Join S)$, is therefore equivalent to $\Pi_{P}(A \cap B)$.
3. Query III, $\Pi_{P}(A \cap B)$, is identical to the simplified Query I.
4. Query IV uses the set identity $A - (A - B) = A \cap B$, which simplifies to $\Pi_{P}(A \cap B)$, making it equivalent to I and III.
5. Query II calculates $\Pi_{P}(R) \cap \Pi_{P}(S)$, which finds $P$ values present in both relations regardless of whether they share the same $Q$ value. This is not equivalent to the join operation where $P$ and $Q$ must both match.

The join operation, denoted as $R \bowtie_{\theta} S$, is a fundamental relational algebra operation used to combine tuples from two relations, $R$ and $S$, based on a join condition $\theta$. The process mathematically begins by performing a Cartesian product ($R \times S$), which generates all possible pairings of tuples from both relations. Subsequently, a selection operator ($\sigma_{\theta}$) is applied to this result to filter out the pairings that do not satisfy the specific join predicate $\theta$. Thus, the formal definition is expressed as $R \bowtie_{\theta} S = \sigma_{\theta}(R \times S)$, combining a Cartesian product followed by a selection.

The query calculates the pass percentage for each school $t$. The denominator $|\{x \mid x \in \text{Enrolment} \land x.\text{school-id}=t\}|$ counts the total number of exam enrolments for the school. The numerator $|\{x \mid x \in \text{Enrolment} \land x.\text{school-id}=t \land (\exists B \in \text{ExamResult} \dots B.\text{marks}>35)\}|$ counts the total number of exam enrolments for that school where the student passed (marks $> 35$). The fraction represents the ratio of passed exams to total exams taken by the school. Thus, the condition $> 35$ filters for schools with an overall pass percentage greater than 35% across all exams combined.

In relational algebra, the operation used to isolate specific attributes (columns) from a relation $R$ is known as Projection. The Projection operator is mathematically denoted as $\pi_{a_1, a_2, ..., a_n}(R)$, where $a_1, a_2, ..., a_n$ represent the specific columns to be extracted from the dataset. While the Selection operator ($\sigma$) filters rows based on a condition, Projection narrows the table's structure to only the requested columns. Other options like Join combine multiple relations, while "Extract" and "Substitute" are not standard relational algebra operators for this purpose. Therefore, the Project operation is the correct choice.