Relational-Algebra
Question 1 |
Consider the following relations P(X,Y,Z), Q(X,Y,T) and R(Y,V).
How many tuples will be returned by the following relational algebra query?
∏x(σ(P.Y=R.Y ∧ R.V=V2)(P × R)) - ∏x(σ(Q.Y=R.Y ∧ Q.T>2)(Q × R))
0 | |
1 | |
2 | |
3 |

∏x(σ(P.Y = R.Y ∧ R.V = V2)(P × R))

σ(Q.Y = R.Y ∧ Q.T>2)(Q × R)


∏x(σ(P.Y = R.Y ∧ R.V = V2)(P × R)) - ∏x(σ(Q.Y = R.Y ∧ Q.T>2)(Q × R))

Question 2 |
Consider the relations r(A, B) and s(B, C), where s.B is a primary key and r.B is a foreign key referencing s.B. Consider the query
-
Q: r⋈(σB<5(s))
Let LOJ denote the natural left outer-join operation. Assume that r and s contain no null values.
Which one of the following is NOT equivalent to Q?
σB<5 (r ⨝ s) | |
σB<5 (r LOJ s)
| |
r LOJ (σB<5(s))
| |
σB<5(r) LOJ s |
Consider the following tables without NULL values.

Q: r⨝(σB<5(S))
The result of σB<5(S) is

The result of σB<5(S) is

Option (A):
The result of r⨝S is

The result of σB<5(r⨝S) is

Option (B):
The result of r LOJ S is

The result of σB<5(r LOJ S) is

Option (C):
The result of σB<5(S) is

Now, the result of r LOJ(σB<5(S))

Option (D):
The result of σB<5(r) is

Now, the result of σB<5(r) LOJ S is

Therefore, from the output of above four options, the results of options, the results of options (A), (B) and (D) are equivalent to Q.
Question 3 |
Consider a database that has the relation schemas EMP(EmpId, EmpName, DeptId), and DEPT(DeptName, DeptId). Note that the DeptId can be permitted to a NULL in the relation EMP. Consider the following queries on the database expressed in tuple relational calculus.
-
(I) {t│∃u ∈ EMP(t[EmpName] = u[EmpName] ∧ ∀v ∈ DEPT(t[DeptId] ≠ v[DeptId]))}
(II) {t│∃u ∈ EMP(t[EmpName] = u[EmpName] ∧ ∃v ∈ DEPT(t[DeptId] ≠ v[DeptId]))}
(III) {t│∃u ∈ EMP(t[EmpName] = u[EmpName] ∧ ∃v ∈ DEPT(t[DeptId] = v[DeptId]))}
Which of the above queries are safe?
(I) and (II) only | |
(I) and (III) only | |
(II) and (III) only | |
(I), (II) and (III) |
(I) Gives EmpNames who do not belong to any Department. So, it is going to be a finite number of tuples as a result.
(II) Gives EmpNames who do not belong to some Department. This is also going to have finite number of tuples.
(III) Gives EmpNames who do not belong to same Department. This one will also give finite number of tuples.
All the expressions I, II and III are giving finite number of tuples. So, all are safe.
Question 4 |
Consider a database that has the relation schema CR(StudentName, CourseName). An instance of the schema CR is as given below.

The following query is made on the database.
T1 ← πCourseName(σStudentName='SA'(CR)) T2 ← CR ÷ T1
The number of rows in T2 is ____________.
4 | |
5 | |
6 | |
7 |
The σStudentName = 'SA'(CR) will produce the following

⇾ The result of T1 ← πCourseName(σStudentName='SA'(CR)) is

(2) T2 ← CR÷T1
⇾ We see that SA is enrolled for CA, CB and CC.
⇾ T2 will give the StudentNames those who have enrolled for all the CA, C, CC courses. So, the following Students are enrolled for the given 3 courses.

⇾ So, the output of T2 will have 4 rows.
Question 5 |
Consider two relations R1(A,B) with the tuples (1,5), (3,7) and R2(A,C) = (1,7), (4,9). Assume that R(A,B,C) is the full natural outer join of R1 and R2. Consider the following tuples of the form (A,B,C): a = (1.5,null), b = (1,null,7), c = (3,null,9), d = (4,7,null), e = (1,5,7), f = (3,7,null), g = (4,null,9). Which one of the following statements is correct?
R contains a,b,e,f,g but not c, d.
| |
R contains all of a,b,c,d,e,f,g | |
R contains e,f,g but not a,b | |
R contains e but not f,g |

⋆ So, from the above resultant table, R contains e, f, g only but not a, b.
Question 6 |
Suppose R1(A, B) and R2(C, D) are two relation schemas. Let r1 and r2 be the corresponding relation instances. B is a foreign key that refers to C in R2. If data in r1 and r2 satisfy referential integrity constraints, which of the following is ALWAYS TRUE?
∏B (r1) - ∏C (r2) = ∅ | |
∏C (r2) - ∏B (r1) = ∅ | |
∏B (r1) = ∏C (r2) | |
∏B (r1) - ∏C (r2) ≠ ∅ |
So we can say that r2(C) is the superset of r1(B).
So (subset - superset) is always empty.
Question 7 |
Consider the following relations A, B, C.

How many tuples does the result of the following relational algebra expression contain? Assume that the schema of AUB is the same as that of A.
(AUB)⋈A.Id>40∨C.Id<15 C
7 | |
4 | |
5 | |
9 |


Performs the cross product and selects the tuples whose A∙Id is either greater than 40 or C∙Id is less than 15. It yields:

Question 8 |
Which of the following functional dependencies hold for relations R(A, B, C) and S(B, D, E):
B -> A A -> C
The relation R contains 200 tuples and the rel ation S contains 100 tuples. What is the maximum number of tuples possible in the natural join of R and S (R natural join S)
100 | |
200 | |
300 | |
2000 |
R(A, B, C) – 200 tuples
S(B, D, E) – 100 tuples
FD’s:
B → A
A → C
― ‘B’ is primary key for R and foreign key of S from the given FDs.
― Maximum tuples in natural join of R and S is min(200, 100) = 100.
Question 9 |
Let R and S be two relations with the following schema
- 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. ΠP(R ⨝ S)
II. ΠP(R) ⨝ ΠP(S)
III. ΠP(ΠP,Q(R) ∩ ΠP,Q(S))
IV. ΠP(ΠP,Q(R) - (ΠP,Q(R) - ΠP,Q(S)))
Only I and II | |
Only I and III | |
Only I, II and III | |
Only I, III and IV |
We have two common columns in 'R' and 'S' which are 'P' and 'Q'.
(I) Both P and Q are used while doing the join, i.e., both P and Q are used to filter.
(II) Q is not used here for filtering. Natural join is done on all P's from R and all P's from S. So different from option (I).
(III) Through venn diagram it can be proved that A∩B = A - (A-B).
So through above formula we can say that (III) and (IV) are equivalent.
So, finally (I), (III) and (IV) are equivalent.
Question 10 |
Information about a collection of students is given by the relation studinfo(studId, name, sex). The relation enroll(studId, courseId) gives which student has enrolled for (or taken) that course(s). Assume that every course is taken by at least one male and at least one female student. What does the following relational algebra expression represent?
ΠcourseId((ΠstudId(σsex="female"(studInfo))×ΠcourseId(enroll))-enroll)Courses in which all the female students are enrolled. | |
Courses in which a proper subset of female students are enrolled. | |
Courses in which only male students are enrolled.
| |
None of the above |
Option B: Yes, True. It selects the proper subset of female students which are enrolled because in the expression we are performing the Cartesian product.
Option C: False. It doesn’t shows (or) display the males students who are enrolled.
Question 11 |
Let r be a relation instance with schema R = (A, B, C, D). We define r1 = ΠA,B,C(R) and r2 = ΠA,D(R). Let s = r1*r2 where * denotes natural join. Given that the decomposition of r into r1 and r2 is lossy, which one of the following is TRUE?
s ⊂ r | |
r ∪ s = r | |
r ⊂ s | |
r * s = s
|
Table r: R(A, B, C, D)

Table r1: ΠA,B,C(R)

Table r2: ΠA,D(R)

S = r1 * r2 (* denotes natural join)
Table S:

Table r ⊂ Table S
⇒ r ⊂ S
Question 12 |
Let R1(A,B,C) and R2(D,E) be two relation schema, where the primary keys are shown underlined, and let C be a foreign key in R1 referring to R2. Suppose there is no violation of the above referential integrity constraint in the corresponding relation instances r1 and r2. Which one of the following relational algebra expressions would necessarily produce an empty relation?
ΠD(r2) - ΠC(r1) | |
ΠC(r1) - ΠD(r2) | |
ΠD(r1⨝C≠Dr2) | |
ΠC(r1⨝C=Dr2) |
→ Based on referral integrity C is subset of values in R2 then,
ΠC(r1) - ΠD(r2) results empty relation.
Question 13 |
Consider the following relation schema pertaining to a students database:
Student (rollno, name, address) Enroll (rollno, courseno, coursename)
Where the primary keys are shown underlined. The number of tuples in the Student and Enroll tables are 120 and 8 respectively. What are the maximum and minimum number of tuples that can be present in (Student * Enroll), where '*' denotes natural join ?
8, 8 | |
120, 8 | |
960, 8 | |
960, 120 |
→ In the question only enroll Id's are same with the student table.
→ The no. of minimum and maximum tuples is same i.e., 8, 8.
Question 14 |
Consider the relation Student (name, sex, marks), where the primary key is shown underlined, pertaining to students in a class that has at least one boy and one girl. What does the following relational algebra expression produce? (Note: p is the rename operator).

The condition in join is "(sex = female ^ x = male ^ marks ≤ m)"
names of girl students with the highest marks
| |
names of girl students with more marks than some boy student | |
names of girl students with marks not less than some boy students
| |
names of girl students with more marks than all the boy students |
Question 15 |
With regard to the expressive power of the formal relational query languages, which of the following statements is true?
Relational algebra is more powerful than relational calculus. | |
Relational algebra has the same power as relational calculus. | |
Relational algebra has the same power as safe relational calculus. | |
None of the above. |
A query can be formulated in safe Relational Calculus if and only if it can be formulated in Relational Algebra.
Question 16 |
Suppose the adjacency relation of vertices in a graph is represented in a table Adj(X,Y). Which of the following queries cannot be expressed by a relational algebra expression of constant length?
List of all vertices adjacent to a given vertex | |
List all vertices which have self loops | |
List all vertices which belong to cycles of less than three vertices | |
List all vertices reachable from a given vertex |
(b) Finding a self loop is also simple (Oop(X,X))
(c) If a → b, b → c then c!=a, finding this is also simple.
(d) List all the elements reachable from a given vertex is too difficult in Relational Algebra.
Question 17 |
Let r and s be two relations over the relation schemes R and S respectively, and let A be an attribute in R. Then the relational algebra expression σA=a(r⋈s) is always equal to
σA=a (r) | |
r | |
σA=a (r)⨝s | |
None of the above |
(b) Display table
(c) A=a for all Tables r and s
Question 18 |
Given the relations
employee (name, salary, deptno) and department (deptno, deptname, address)
Which of the following queries cannot be expressed using the basic relational algebra operations (σ, π, ×, ⋈, ∪, ∩, -)?
Department address of every employee | |
Employees whose name is the same as their department name | |
The sum of all employees’ salaries | |
All employees of a given department |
Question 19 |
Consider the join of a relation R with a relation S. If R has m tuples and S has n tuples then the maximum and minimum sizes of the join respectively are
m + n and 0 | |
mn and 0 | |
m + n and |m – n| | |
mn and m + n |
Suppose there is no common attribute in R and S due to which natural join will act as cross product. So then in cross product total no. of tuples will be mn.
For minimum:
Suppose there is common attribute in R and S, but none of the row of R matches with rows of S then minimum no. of tuples will be 0.
Question 20 |
The relational algebra expression equivalent to the following tuple calculus expression:
{t| t ∈ r ∧(t[A] = 10 ∧ t[B] = 20)} isσ(A=10∨B=20) (r) | |
σ(A=10) (r) ∪ σ(B=20) (r) | |
σ(A=10) (r) ∩ σ(B=20) (r) | |
σ(A=10) (r) - σ(B=20) (r) |
σ(A=10) (r) ∩ σ(B=20) (r)
Question 21 |
Given two union compatible relations R1(A,B) and R2(C,D), what is the result of the operation R1A = CAB = DR2?
R1 ∪ R2 | |
R1 × R2 | |
R1 - R2 | |
R1 ∩ R2 |
Question 22 |
Which of the following query transformations (i.e. replacing the l.h.s. expression by the r.h.s. expression) is incorrect? R1 and R2 are relations, C1, C2 are selection conditions and A1, A2 are attributes of R1?
σC1(σC1(R1)) → σC2(σC2(R1)) | |
σC1(σA1(R1)) → σA1(σC1(R1)) | |
σC1(R1 ∪ R2) → σC1(R1) ∪ σC1 | |
πA1(σC1(R1)) → σC1(σA1(R1)) |
Question 23 |
Give a relational algebra expression using only the minimum number of operators from (∪, −) which is equivalent to R ∩ S.
Out of syllabus (For explanation see below) |
→ No need of using Union operation here. → In question they gave (∪, −) but we don't use both.
→ And also they are saying that only the minimum number of operators from (∪, −) which is equivalent to R ∩ S.
So, the expression is minimal.
Question 24 |
Consider a selection of the form σA≤100(r), where r is a relation with 1000 tuples. Assume that the attribute values for A among the tuples are uniformly distributed in the interval [0, 500]. Which one of the following options is the best estimate of the number of tuples returned by the given selection query ?
50 | |
100 | |
150 | |
200 |
Values for A among the tuples are uniformly distributed in the interval [0, 500]. This can be split to 5 mutually exclusive and exhaustive intervals of same width of 100 ([0-100], [101-200], [201-300], [301-400], [401-500], 0 makes the first interval larger - this must be type in this question) and we can assume all of them have same number of values due to uniform distribution. So no. of tuples with A value in first interval should be,
Total no. of tuples/5 = 1000/5 = 200
Question 25 |
Consider the following relation schemas:
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 ⋈ (account ⋈ depositor)
Which one of the following queries is the most efficient version of the above query?
Пc-name (σbal < 0 (σb-city = “Agra” branch ⋈ account) ⋈ depositor) | |
Пc-name (σb-city = “Agra”branch ⋈ (σbal < 0 account ⋈ depositor)) | |
Пc-name (σb-city = “Agra” branch ⋈ σb-city = “Agra” ⋀ bal < 0 account) ⋈ depositor) | |
Пc-name (σb-city = “Agra” ⋀ bal < 0 account ⋈ depositor)) |
Options (C) and (D) are invalid as there is no b-city column in a-schema.
Question 26 |
Consider the relations r1(P, Q, R) and r2(R, S, T) with primary keys P and R respectively. The relation r1 contains 2000 tuples and r2 contains 2500 tuples. The maximum size of the join r1⋈ r2 is :
2000 | |
2500 | |
4500 | |
5000 |
Question 27 |
Project | |
Join | |
Extract | |
Substitute |
Projection is used to project required column data from a relation. By Default projection removes duplicate data.
Example :
R(A B C)
----------
1 2 4
2 2 3
3 2 3
4 3 4
π (BC)
B C
-----
2 4
2 3
3 4
Question 28 |
a cartesian product of two relations followed by a selection | |
a cartesian product of two relations | |
a union of two relations followed by cartesian product of the two relations | |
a union of two relations |
→ The join operation can be defined as a cartesian product of two relations followed by a selection.
→ A SQL JOIN clause is used to combine rows from two or more tables, based on a related column between them.
Different Types of SQL JOINs
1. INNER JOIN: Returns records that have matching values in both tables

2. LEFT (OUTER) JOIN: Return all records from the left table, and the matched records from the right table

3. RIGHT (OUTER) JOIN: Return all records from the right table, and the matched records from the left table

4. FULL (OUTER) JOIN: Return all records when there is a match in either left or right table

Question 29 |
m + n & 0 | |
mn & 0 | |
m + n & | m – n | | |
mn & m + n |
If there is common attribute in R and S, and every row of R match with every row of S then total no. of tuples will be mn.
For minimum:
If there is no common attribute between R and S or if there is common attribute but none of the row of R matches with rows of S then output tuples will be 0.
Question 30 |
Department address of every employee | |
Employees whose name is the same as their department name | |
The sum of all employees’ salaries | |
All employees of a given department |
Question 31 |
Set intersection | |
Assignment | |
Natural Join | |
None of the above |
1.Select
2.Project
3.Cartesian Product
4.Rename
5.Union
6.Set Difference
Question 32 |
Set intersection | |
Natural Join | |
Assignment | |
None of the above |
2.Project
3.Cartesian Product
4.Rename
5.Union
6.Set Difference
Question 33 |
Consider the schema R = (A, B, C, D, E, F) on which the following functional dependencies hold :
A ➝ B B, C ➝ D E ➝ C D ➝ A
What are the candidate keys of R ?
AEF, BEF and DEF
| |
AEF, BEF and BCF | |
AE and BE | |
AE, BE and DE |
EFB+ = {EFABCD}
EFC+ = {EFC}
EFD+ = {EFDCAB}
So, EFA, EFB, EFD are the keys for the given relation R = (A, B, C, D, E, F).
Question 34 |
Record | |
Field | |
File | |
Database |
Equivalent Database Concepts
Relation → Table
Tuple → Row or record
Attribute → Column or field
Cardinality → Number of rows
Degree → Number of columns
Primary key → Unique identifier
Domain → Pool of legal values
Question 35 |
|R|<=|dom(A1)X dom(A2)..dom(An)| | |
|R|>=|dom(A1)X dom(A2)..dom(An)| | |
|R|=max(|dom(A1)|,|dom(A2)|,..|dom(An)|) | |
|R|=min(|dom(A1)|,|dom(A2)|,..|dom(An)|) |
● Cardinality refers to a number. It gives the number of unique values that appear in the table for a particular column.
● For eg: you have a table called Person with column Gender. Gender column can have values either 'Male' or 'Female''.
● Then the cardinality of Gender column is 2, since there are only two unique values that could possibly appear in that column – Male and Female.
Question 36 |
join | |
self join | |
outer join | |
Equi join |
● To join a table itself means that each row of the table is combined with itself and with every other row of the table.
Question 37 |
mn | |
m+n | |
(m+n)/2 | |
2(m+n) |
If there is common attribute in R and S, and every row of R match with every row of S then total no. of tuples will be mn.
Question 38 |
Aggregate Computation | |
Multiplication | |
Finding transitive closure | |
All of the above |
● Multiplication means cartesian product
●Transitive closure:
Given a domain D, let binary relation R be a subset of D×D. The transitive closure R + of R is the smallest subset of D×D that contains R and satisfies the following condition:
∀ x∀y∀z((x, y ) ∈ R + ⋀ ( y, z ) ∈ R + ⇒ ( x, z ) ∈ R + )
Question 39 |
Natural join, outer join | |
Outer join, natural join | |
Left outer join, right outer join | |
Left outer join, natural join |
→ A NATURAL JOIN can be an INNER join, a LEFT OUTER join, or a RIGHT OUTER join. The default is INNER join.
→ The SQL OUTER JOIN returns all rows from both the participating tables which satisfy the join condition along with rows which do not satisfy the join condition. The SQL OUTER JOIN operator (+) is used only on one side of the join condition only.
Question 40 |
6 | |
2 | |
3 | |
5 |
Consider following two tables

→ Result of natural join R * S (If domain of attribute C in the two tables are same )
→ You can see both R and S contain the attribute C whose value is 2 in each and every tuple. Table R contains 3 tuples, Table S contains 2 tuples, where Result table contains 3*2=6 tuples.
Note: While performing a natural join, if there were no common attributes between the two relations, Natural join will behave as Cartesian Product.
Question 41 |
cartesian product always | |
combination of union and filtered cartesian product | |
combination of selection and filtered cartesian product | |
combination of projection and filtered cartesian product |
→ Example: If the tables R and S contains common attributes and value of that attribute in each tuple in both tables are same, then the natural join will result n*m tuples as it will return all combinations of tuples.
Question 42 |
Inner Join | |
Outer Join | |
Semi Join | |
Anti Join |
Question 43 |
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:
I, III and IV are the same query. | |
II, III and IV are the same query. | |
I, II and IV are the same query. | |
I, II and III are the same query. |




Question 44 |
Only the joining attributes are sent from one site to another and then all of the rows are returned. | |
All of the attributes are sent from one site to another and then only the required rows are returned. | |
Only the joining attributes are sent from one site to another and then only the required rows are returned. | |
All of the attributes are sent from one site to another and then only the required rows are returned. |
Question 45 |

The number of tuples in the resulting table of RA1, RA2 and RA3 are given by:
2, 4, 2 respectively | |
2, 3, 2 respectively | |
3, 3, 1 respectively | |
3, 4, 1 respectively |


Question 46 |

(a) | |
(b) | |
(c) | |
(d) |

Question 47 |
(a) join
(b) Intersection
(c) Cartisian product
(d) Project
(a) and (b) | |
(b) and (c) | |
(c) and (d) | |
(b) and (d) |
Question 48 |
(a) Relational algebra and Domain relational calculus
(b) Relational algebra and Tuple relational calculus
(c) Relational algebra and Domain relational calculus restricted to safe expression
(d) Relational algebra and Tuple relational calculus restricted to safe expression
(a) and (b) only | |
(c) and (d) only | |
(a) and (c) only | |
(b) and (d) only |
Question 49 |
In a system for a restaurant, the main scenario for placing order is given below:
(a) Customer reads menu
(b) Customer places an order
(c) Order is sent to kitchen for preparation
(d) Ordered items are served
(e) Customer requests for a bill for the order
(f) Bill is prepared for this order
(g) Customer is given the bill
(h) Customer pays the bill
A sequence diagram for the scenario will have at least how many objects among whom the messages will be exchanged.
3 | |
4 | |
5 | |
6 |
Question 50 |
30 | |
20 | |
50 | |
1500 |
NATURAL JOIN requires that the two join attributes (or each pair of join attributes) have the same name in both relations.
Let us take a small example where we are having two relations named Employee(EID, Ename) and Department(EID, DID)

Since relation R2 is having 30 tuples, so in best case natural join of R1 and R2 will 30 tuples.
Question 51 |
Considering the relation schemas R (A, B, C, D) and S (C. D. E. F), what will be the degree of the resultant relation of the following Relational Algebra expression. Where “*” represents the”natural join" operation?

3 | |
4 | |
6 | |
5 |
Question 52 |
Consider the following relation schema R and S along with their tuple sets.
R(A, B) = {
S(A) = {a1, a2, a3}
What is the value of TR / S. where "/" represents the Relational Algebra “division" operation?
T(B) = {b1, b3} | |
T(B) = {b1, b2, b4} | |
T(B) = {b1, b4} | |
T(B) = {b1, b3, b4} |

Question 53 |
Relations and fields can be renamed in relational algebra using the renaming operator
p
| |
σ | |
Θ | |
⟕
|
Question 54 |
The ____ defines a set of operations on relations, paralleling the usual algebraic operations such as addition, subtraction or multiplication, which operates on numbers.
Relational calculus | |
Referential Integrity
| |
Relational Algebra | |
Relations
|
Question 55 |
Which of the following is a unary operation?
Intersection | |
Projection
| |
Join | |
Cartesian Product
|
Question 56 |
The number of tuples in the result of a left outer join operation will always be
greater than the number of tuples in the result of the corresponding join operation.
| |
at least equal to the number of tuples in the result of the corresponding join operation. | |
less than the number of tuples in the result of the corresponding join operation.
| |
greater than the number of tuples in the result of the corresponding right outer join operation.
|
Question 57 |
R1∪R2 | |
R1×R2 | |
R1-R2 | |
R1∩R2 |
Question 58 |
m+n and O | |
mn and O
| |
m+n and m-n | |
mn and m+n |
If every row of r matches with each row of s, i.e., it means, the join attribute has the same value in all the rows of both r and s, then maximum mn tuples possible.
For minimum no. of tuples, if condition of join is not satisfied for any tuple then, minimum 0 tuples will be there.
Question 59 |
Outer join | |
Inner join | |
Natural join | |
Self join |
Question 60 |
π, ∞, σ | |
π, σ, ∞ | |
σ, π, ∞ | |
σ, ∞, π |
SELECT
FROM
WHERE
in a sequence.
So it is equivalent to
SELECT-π
FROM- ∞
WHERE-σ