The relation R contains 200 tuples and the relation S contains 100 tuples. What is the maximum number of tuples possible in the natural join R $$\Join$$ S?
R (P, Q, R1, R2, R3)
S (P, Q, S1, S2)
Where {P, Q} is the key for both schemas. Which of the following queries are equivalent?I. $$\Pi_P \left(R \bowtie S\right)$$
II. $$\Pi_P \left(R\right) \bowtie \Pi_P\left(S\right)$$
III. $$\Pi_P \left(\Pi_{P, Q} \left(R\right) \cap \Pi_{P,Q} \left(S\right) \right)$$
IV. $$\Pi_P \left(\Pi_{P, Q} \left(R\right) - \left(\Pi_{P,Q} \left(R\right) - \Pi_{P,Q} \left(S\right)\right)\right)$$
b-Schema = (b-name, b-city, assets)
a-Schema = (a-num, b-name, bal)
d-Schema = (c-name, a-number)
Let branch, account and depositor be respectively instances of the above schemas. Assume that account and depositor relations are much bigger than the branch relation.
Consider the following query:Пc-name (σb-city = “Agra” ⋀ bal < 0 (branch $$ \Join $$ (account $$ \Join $$ depositor))
Which one of the following queries is the most efficient version of the above query ?
Consider the relation employee(name, sex, supervisorName) with name as the key, supervisorName gives the name of the supervisor of the employee under consideration. What does the following Tuple Relational Calculus query produce?
$$\eqalign{ & \{ e.name\,|\,employee(e) \wedge \cr & (\forall x)[\neg employee(x) \vee \cr & x.\sup ervisorName \ne e.name\, \vee \cr & x.sex = 'male']\} \cr} $$