Consider the following three relations
Employee (eid, eName), Comp(cid, cName), Own(eid, cid). Which of the following relational algebra expression return the set of eids who own all brands:
Understanding the Relational Algebra Problem
The question asks us to identify the specific relational algebra expression that correctly retrieves the set of employee IDs ($eid$) for employees who own *every* brand ($cid$) available in the $Comp$ table. We are given three relations:
- $Employee(eid, eName)$: Stores employee information.
- $Comp(cid, cName)$: Stores company/brand information.
- $Own(eid, cid)$: Represents the ownership relationship between employees and brands.
The core challenge here is finding employees associated with *all* items in a set, which typically involves the relational algebra division operator ($/$).
Relational Division Explained
The division operator, denoted by $/$, is used when you want to find tuples in one relation that are related to *every* tuple in another relation based on certain attributes. If relation $R$ has schema $(X, Y)$ and relation $S$ has schema $(Y)$, then $R / S$ yields a relation with schema $(X)$ containing tuples $t$ from $R$ such that for *all* tuples $u$ in $S$, the tuple $tu$ (concatenation) exists in $R$.
In our case:
- We want to find $eid$s (the 'X' attribute).
- The condition is owning *all* $cid$s (the 'Y' attribute).
- The relation containing the pairs is $Own(eid, cid)$.
- The set of all $cid$s we need to check against is found in the $Comp$ relation, specifically $\Pi_{cid}(Comp)$.
Therefore, the general form should be $\Pi_{eid}(Own) / \Pi_{cid}(Comp)$, but we need to consider the exact structure from the options.
Analyzing the Options
Option 1: $\Pi_{eid}(\Pi_{eid,cid} (\text{Own})/\Pi_{cid} (\text{Comp}))$
Let's break this down:
- $\Pi_{cid} (\text{Comp})$: This projection retrieves all unique brand IDs ($cid$) from the $Comp$ relation. This represents the set of *all* brands that an employee must own. Let's call this set $AllBrands$.
- $\Pi_{eid,cid} (\text{Own})$: This projection retrieves all pairs of (employee ID, brand ID) from the $Own$ relation. Since the schema of $Own$ is already $(eid, cid)$, this expression simply returns the entire $Own$ relation. Let's call this $OwnPairs$.
- $OwnPairs / AllBrands$ (i.e., $\Pi_{eid,cid} (\text{Own})/\Pi_{cid} (\text{Comp})$): This is the relational division. It finds all the $eid$s from $OwnPairs$ that are associated with *every* $cid$ present in $AllBrands$. This correctly identifies employees who own all brands.
- $\Pi_{eid}(...)$: The final projection takes the result of the division (which would be a relation containing the desired $eid$s) and ensures the output is only the set of $eid$s.
This expression correctly uses the division operator to find employees who own all brands.
Option 2: $\Pi_{eid}(\Pi_{eid} (\text{Own}) \times \Pi_{cid} (\text{Comp}))$
This option calculates the Cartesian product ($\times$) of all employee IDs and all brand IDs. This results in a list pairing every employee with every brand, which doesn't tell us which employees own *all* brands. It simply lists all possible combinations.
Option 3: $\Pi_{eid} (\Pi_{eid,cid} (\text{Own}) \times \Pi_{cid} (\text{Comp}))$
This calculates the Cartesian product of the $Own$ relation with the set of all brand IDs. For each actual ownership record ($eid1, cid1$), it pairs it with every brand ID $cid2$ from $Comp$, producing tuples like $(eid1, cid1, cid2)$. This does not isolate employees owning *all* brands.
Option 4: $\Pi_{eid} (\Pi_{eid} (\text{Own}) \times (\Pi_{cid, cName} (\text{Own})/ \Pi_{cid} (\text{Comp})))$
This option is structurally flawed. Firstly, $\Pi_{cid, cName} (\text{Own})$ is invalid because the $Own$ relation does not contain the $cName$ attribute. Even if we assume a corrected structure for the division part, the overall logic involves multiplying the set of all $eid$s with the result of a division, which doesn't fit the requirement of finding employees who own all brands.
Conclusion
Based on the analysis, the relational division operator is the correct tool for this task. Option 1 correctly applies the division operator $/$ to the $Own$ relation and the set of all brand IDs derived from $Comp$ to find the employees who own every brand.
Important Questions from Relational Algebra
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 $T$.
Table X
P Q R P1 Q1 R1 P2 Q2 R2 P3 Q3 R2 Table Y
P Q S P1 Q1 2 P1 Q2 5 P2 Q1 6 P3 Q3 1 Table Z
P T P1 T1 P3 T2 P4 T3 P4 NULL Consider the relational algebra expression $$ \Pi_{P, R, S} \left[ \left( \sigma_{(Q = Q3 \lor R = R2)} [X \bowtie Y] \right) \bowtie \left( \sigma_{(S > 1)} [Y \bowtie Z] \right) \right] $$ where $\bowtie$ denotes natural join operation.
Which of the following options is the correct output for the given expression?
- Consider a relational database schema with two relations $R(P, Q)$ and $S(X, Y)$.
Let $E = \{\langle u \rangle \mid \exists v\ \exists w\ \langle u, v \rangle \in R\ \land\ \langle v, w \rangle \in S\}$ be a tuple relational calculus expression.
Which one of the following relational algebraic expressions is equivalent to $E$ ? Consider a database that includes the following relations:
Defender($name, rating, side, goals$)
Forward($name, rating, assists, goals$)
Team($name, club, price$)
Which ONE of the following relational algebra expressions checks that every name occurring in Team appears in either Defender or Forward, where $\phi$ denotes the empty set?
Consider the following three relations:
Car (model, year, serial, color)
Make (maker, model)
Own (owner, serial)
A tuple in Car represents a specific car of a given model, made in a given year, with a serial number and a color. A tuple in Make specifies that a maker company makes cars of a certain model. A tuple in Own specifies that an owner owns the car with a given serial number. Keys are underlined; (owner, serial) together form key for Own. ($\bowtie$ denotes natural join)
$ \pi_{\text{owner}} (\text{Own} \bowtie (\sigma_{\text{color}=\text{"red"}} (\text{Car} \bowtie (\sigma_{\text{maker}=\text{“ABC”}} \text{Make})))) $Consider the following tables, Loan and Borrower of a bank.
Loan Borrower loan_num branch_name amount customer_name loan_num L11 Banjara Hills 90000 Anand L11 L14 Kondapur 50000 Karteek L11 L15 SR Nagar 40000 Ankita L15 L22 SR Nagar 25000 Gopal L19 L23 Balanagar 80000 Karteek L22 L25 Kondapur 70000 Karteek L23 L19 SR Nagar 65000 Sunil L23 Sunil L25
Query: $\pi_{\text{branch\_name}, \text{customer\_name}}(\mathbf{Loan} \bowtie \mathbf{Borrower}) \div \pi_{\text{branch\_name}}(\mathbf{Loan})$
where $\bowtie$ denotes natural join.
The number of tuples returned by the above relational algebra query is _______
(Answer in integer)
