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?
The question requires finding a relational algebra expression that verifies if every player name listed in the $Team$ relation is also present in either the $Defender$ relation or the $Forward$ relation. The condition is that the set of names from $Team$ that do *not* appear in either $Defender$ or $Forward$ must be the empty set ($\phi$).
Relational Algebra Expression Derivation
To solve this, we need to perform the following steps:
- Project the $name$ attribute from the $Team$ relation.
- Project the $name$ attribute from the $Defender$ relation.
- Project the $name$ attribute from the $Forward$ relation.
- Combine the names from $Defender$ and $Forward$ using the union operator ($\cup$). This gives a set of all names that are either defenders or forwards.
- Find the names that are in $Team$ but *not* in the combined set of defenders and forwards. This is done using the set difference operator ($\setminus$).
- The expression should evaluate to the empty set ($\phi$) if all names in $Team$ are indeed present in the union of $Defender$ and $Forward$ names.
Evaluating the Options
Let's represent the projections:
- Names in Team: $\Pi_{name} (Team)$
- Names in Defender: $\Pi_{name} (Defender)$
- Names in Forward: $\Pi_{name} (Forward)$
- Names in Defender or Forward: $\Pi_{name} (Defender) \cup \Pi_{name} (Forward)$
The expression that checks if names in $Team$ are *not* in the combined list of $Defender$ or $Forward$ names is:
$ \Pi_{name} (Team) \setminus (\Pi_{name} (Defender) \cup \Pi_{name} (Forward)) $
For the condition "every name occurring in Team appears in either Defender or Forward" to be true, this resulting set difference must be empty ($\phi$).
Therefore, the correct expression is:
$ \Pi_{name} (Team) \setminus (\Pi_{name} (Defender) \cup \Pi_{name} (Forward)) = \phi $
This matches Option 3.
Final Answer Check
- Option 1 checks for names in Team not common to both Defender and Forward.
- Option 2 checks for names common to Defender and Forward but not in Team.
- Option 3 correctly checks if names in Team are absent from the union of Defender and Forward names, requiring this difference to be empty.
- Option 4 checks if names in Defender or Forward are absent from Team.
Option 3 accurately represents the required condition.
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 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 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 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:
