Division operation is ideally suited to handle queries of the type :
customers who have an account at all branches in Delhi.
For relations r(X, Y) and s(Y), the division \(r \div s\) returns those X values that are paired in r with every Y value present in s.
So division expresses universal (“for all”) queries. With r = depositor(customer, branch) and s = the set of branches located in Delhi, \(r \div s\) gives the customers related to every Delhi branch.
A query about having an account in at least one Delhi branch is existential and needs only a join with projection; having no account needs set difference; neither needs division.
Hence, the answer is customers who have an account at all branches in Delhi..
Consider the following statements :
I. Re-construction operation used in mixed fragmentation satisfies commutative rule.
II. Re-construction operation used in vertical fragmentation satisfies commutative rule
Which of the following is correct ?
Which of the following is true ?
I. Implementation of self-join is possible in SQL with table alias.
II. Outer-join operation is basic operation in relational algebra.
III. Natural join and outer join operations are equivalent.
Consider the following three tables R, S and T. In this question, all the join operations are natural joins (\(\bowtie\)). (\(\pi\)) is the projection operation of a relation :
| R | S | T | |||
| A | B | B | C | A | C |
| 1 | 2 | 6 | 2 | 7 | 1 |
| 3 | 2 | 2 | 4 | 1 | 2 |
| 5 | 6 | 8 | 1 | 9 | 3 |
| 7 | 8 | 8 | 3 | 5 | 4 |
| 9 | 8 | 2 | 5 | 3 | 5 |
Possible answer tables for this question are also given as below :
(a)
| A | B | C |
| 1 | 2 | 4 |
| 1 | 2 | 5 |
| 3 | 2 | 4 |
| 3 | 2 | 5 |
| 5 | 6 | 2 |
| 7 | 8 | 1 |
| 7 | 8 | 3 |
| 9 | 8 | 1 |
| 9 | 8 | 3 |
(b)
| A | B | C |
| 1 | 2 | 2 |
| 3 | 2 | 5 |
| 5 | 6 | 4 |
| 7 | 8 | 1 |
| 9 | 8 | 3 |
(c)
| A | B | C |
| 1 | 6 | 2 |
| 3 | 2 | 5 |
| 5 | 2 | 4 |
| 7 | 8 | 1 |
| 9 | 8 | 3 |
(d)
| A | B | C |
| 3 | 2 | 5 |
| 7 | 8 | 1 |
| 9 | 8 | 3 |
What is the resulting table of \(\pi_{A,B}(R \bowtie T) \bowtie \pi_{B,C}(S \bowtie T)\) ?
A ________ operation combines the rows of the two relations and outputs a new relation that has both input relations’ rows in it.
Given below are the two statements:
Statement I : Any retrieval request that is specified in the basic relational algebra can also be specified in relational calculus.
Statement II : Any retrieval request that is specified in the relational calculus can also be specified in basic relational algebra.
In the light of the above statements, choose the correct answer from the options given below:
Consider the relations R(A, B) and S(B, C) and the following four relational algebra queries over R and S :
I. ΠA, B (R ⋈ S)
II. R ⋈ ΠB(S)
III. R ∩ (ΠA(R) × ΠB(S))
IV. ΠA,R.B (R × S) where R⋅B refers to the column B in table R.
One can determine that :
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 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)