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 $E$ ?
Tuple Relational Calculus Analysis
The given tuple relational calculus (TRC) expression is:
$E = \{\langle u \rangle \mid \exists v\ \exists w\ \langle u, v \rangle \in R\ \land\ \langle v, w \rangle \in S\}$
With relations $R(P, Q)$ and $S(X, Y)$.
- The condition $\langle u, v \rangle \in R$ means that $u$ corresponds to attribute $P$ and $v$ corresponds to attribute $Q$ in relation $R$. Thus, $u = R.P$ and $v = R.Q$.
- The condition $\langle v, w \rangle \in S$ means that $v$ corresponds to attribute $X$ and $w$ corresponds to attribute $Y$ in relation $S$. Thus, $v = S.X$ and $w = S.Y$.
- The existential quantifiers ($\exists v, \exists w$) indicate that we need to find matching tuples from $R$ and $S$, implying a join operation.
- The common variable $v$ links the two relations. From $R$, $v = R.Q$. From $S$, $v = S.X$. This establishes the join condition: $R.Q = S.X$.
- The expression $E$ selects $\langle u \rangle$, where $u$ is derived from $R.P$. Therefore, a projection on attribute $P$ is required from the result of the join.
Relational Algebra Equivalence
Combining the join condition ($R.Q = S.X$) and the projection on $P$, the equivalent relational algebra expression is:
$\Pi_{P} (R \bowtie_{R.Q = S.X} S)$
Join and Projection Match
Comparing this derived expression with the provided options:
- Option 1: $\Pi_P (R \bowtie_{R.P = S.X} S)$ - The join condition $R.P = S.X$ is incorrect.
- Option 2: $\Pi_P (S \bowtie_{S.X = R.Q} R)$ - The join condition $S.X = R.Q$ is equivalent to $R.Q = S.X$. The projection is on $P$. This expression matches the derived equivalent.
- Option 3: $\Pi_P (R \bowtie_{R.P = S.Y} S)$ - The join condition $R.P = S.Y$ is incorrect.
- Option 4: $\Pi_P (S \bowtie_{S.Y = R.Q} R)$ - The join condition $S.Y = R.Q$ is incorrect.
Option 2 correctly represents the equivalent relational algebraic expression.
Important Questions from Relational Algebra
Which of the following statements is/are correct regarding the Finance Commission of India?
A. The Finance Commission consist of a Chairman and four other members.
B. The recommendations made by the Finance Commission are binding on the government and the government needs to grant funds according to the advice of the Commission,
C. Article 280 of the Indian Constitution talks about the recommendations of the Finance Commission.
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
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: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})))) $
