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 Table Y Table Z 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?P Q R P1 Q1 R1 P2 Q2 R2 P3 Q3 R2 P Q S P1 Q1 2 P1 Q2 5 P2 Q1 6 P3 Q3 1 P T P1 T1 P3 T2 P4 T3 P4 NULL
Relational Schema Definition
The given relations are defined as:
- Relation X: Columns (P, Q, R)
- Relation Y: Columns (P, Q, S)
- Relation Z: Columns (P, T)
Instance data for each relation:
| 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 |
Expression Step-by-Step Evaluation
The relational algebra expression to evaluate is:
$ \Pi_{P, R, S} \left[ \left( \sigma_{(Q = Q3 \lor R = R2)} [X \bowtie Y] \right) \bowtie \left( \sigma_{(S \gt 1)} [Y \bowtie Z] \right) \right] $
Left Join and Selection Evaluation
First, compute the natural join of X and Y, then apply the selection.
- $X \bowtie Y$ (on columns P, Q):
P Q R S P1 Q1 R1 2 P3 Q3 R2 1
- Selection $\sigma_{(Q = Q3 \lor R = R2)}$ on the result:
- (P1, Q1, R1, 2): Condition ($Q = Q3 \lor R = R2$) is false.
- (P3, Q3, R2, 1): Condition ($Q = Q3 \lor R = R2$) is true.
P Q R S P3 Q3 R2 1
Right Join and Selection Evaluation
Next, compute the natural join of Y and Z, then apply the selection.
- $Y \bowtie Z$ (on column P):
P Q S T P1 Q1 2 T1 P1 Q2 5 T1 P3 Q3 1 T2
- Selection $\sigma_{(S \gt 1)}$ on the result:
- (P1, Q1, 2, T1): Condition ($S \gt 1$) is true.
- (P1, Q2, 5, T1): Condition ($S \gt 1$) is true.
- (P3, Q3, 1, T2): Condition ($S \gt 1$) is false.
P Q S T P1 Q1 2 T1 P1 Q2 5 T1
Final Join and Projection
Now, compute the natural join of the LHS and RHS relations, then project the required columns.
- LHS $\bowtie$ RHS (on common columns P, Q, S):
- LHS: (P3, Q3, R2, 1)
- RHS: (P1, Q1, 2, T1) and (P1, Q2, 5, T1)
- Projection $\Pi_{P, R, S}$ on the empty set.
Result and Conclusion
The final result of the relational algebra expression is an empty set, meaning zero rows.
Important Questions from Relational Algebra
- 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)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:
