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})))) $
Database Query Analysis: Owners of Red Cars by ABC
This solution breaks down the relational algebra query to identify specific car owners based on car attributes.
Understanding the Relational Algebra Query
The query is:
$ \pi_{\text{owner}} (\text{Own} \bowtie (\sigma_{\text{color}=\text{"red"}} (\text{Car} \bowtie (\sigma_{\text{maker}=\text{“ABC”}} \text{Make})))) $
We will analyze this expression step-by-step, starting from the innermost operations.
Step 1: Identify Cars Made by "ABC"
The operation $ \sigma_{\text{maker}=\text{“ABC”}} \text{Make} $ filters the $Make$ relation to find all models produced by the maker "ABC".
Then, the natural join $ \text{Car} \bowtie (\dots) $ combines this result with the $Car$ relation based on the $model$ attribute. This yields tuples representing cars that are made by "ABC".
Step 2: Filter for Red Cars Made by "ABC"
The selection $ \sigma_{\text{color}=\text{"red"}} (\dots) $ is applied to the result from Step 1. This filters the set of cars made by "ABC" to include only those whose $color$ is "red".
The result at this stage represents red cars made by "ABC".
Step 3: Link Cars to Owners
The natural join $ \text{Own} \bowtie (\dots) $ connects the red cars made by "ABC" (from Step 2) with the $Own$ relation using the $serial$ attribute. This creates tuples linking owners to specific red cars made by "ABC".
Step 4: Extract Owner Information
Finally, the projection $ \pi_{\text{owner}} (\dots) $ selects only the $owner$ attribute from the joined result. This gives the final list of unique owners who own at least one red car made by "ABC".
Conclusion
The query effectively finds all owners associated with cars that satisfy both conditions: being red and being made by "ABC".
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 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)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:
