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.
Only I is correct.
Statement I: a self-join joins a table with itself, which SQL does by giving the table two aliases, e.g. FROM emp e1, emp e2 WHERE e1.mgr = e2.id. Statement I is true.
Statement II: the basic (fundamental) operations of relational algebra are selection, projection, union, set difference, Cartesian product and rename. Outer join is a derived (extended) operation built from join, difference and union, so Statement II is false.
Statement III: a natural join keeps only matching tuples, while an outer join also keeps the unmatched tuples of one or both relations padded with nulls, so the results differ in general. Statement III is false.
Only Statement I is true.
Hence, the answer is Only I is correct..
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 ?
Division operation is ideally suited to handle queries of the type :
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)