Relational-Algebra
Question 1 |
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
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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)
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σB<5 (r LOJ s)
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r LOJ (σB<5(s))
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σB<5(r) LOJ s |
Question 1 Explanation:
Consider the following 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 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.
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 2 |
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.
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(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) |
Question 2 Explanation:
An expression is said to be safe, if it yield a finite number of tuples, otherwise unsafe.
(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.
(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 3 |
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 |
Question 3 Explanation:
(1) T1 ← πCourseName(σStudentName = 'SA'(CR))
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.
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 4 |
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.
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R contains all of a,b,c,d,e,f,g
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R contains e,f,g but not a,b
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R contains e but not f,g
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Question 4 Explanation:
⋆ So, from the above resultant table, R contains e, f, g only but not a, b.
Question 5 |
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) ≠ ∅ |
Question 5 Explanation:
Referential integrity means, all the values in foreign key should be present in primary key.
So we can say that r2(C) is the superset of r1(B).
So (subset - superset) is always empty.
So we can say that r2(C) is the superset of r1(B).
So (subset - superset) is always empty.
Question 6 |
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 A U B is the same as that of
7 | |
4 | |
5 | |
9 |
Question 6 Explanation:
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 7 |
Consider a relational table r with sufficient number of records, having attributes A1, A2,…, An and let 1 <= p <= n. Two queries Q1 and Q2 are given below.
The database can be configured to do ordered indexing on Ap or hashing on Ap. Which of the following statements is TRUE?
The database can be configured to do ordered indexing on Ap or hashing on Ap. Which of the following statements is TRUE?
Ordered indexing will always outperform hashing for both queries | |
Hashing will always outperform ordered indexing for both queries | |
Hashing will outperform ordered indexing on Q1, but not on Q2 | |
Hashing will outperform ordered indexing on Q2, but not on Q1. |
Question 7 Explanation:
Hashing works well on the "equal" queries, while ordered indexing works well better on range queries.
For example, consider B+ tree, once you have searched a key in B+; you can find range of values via the block pointers pointing to another block of values on the leaf node level.
For example, consider B+ tree, once you have searched a key in B+; you can find range of values via the block pointers pointing to another block of values on the leaf node level.
Question 8 |
Which of the following functional dependencies hold for relations R(A, B, C) and S(B, D, E):
B -> A A -> CThe 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 |
Question 8 Explanation:
Given
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.
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
Only I and II
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Only I and III
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Only I, II and III | |
Only I, III and IV |
Question 9 Explanation:
Natural join is based on the common columns of the two tables.
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.
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?
Courses in which all the female students are enrolled.
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Courses in which a proper subset of female students are enrolled.
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Courses in which only male students are enrolled.
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None of the above
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Question 10 Explanation:
Option A: It is not result the all the female students who are enrolled. So option A is false.
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.
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 |
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 |
Question 11 Explanation:
σA≤100(r), r has 1000 tuples.
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 typ0 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
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 typ0 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 12 |
Consider the following relation schemas :
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- b-Schema = (b-name, b-city, assets)
- a-Schema = (a-num, b-name, bal)
- d-Schema = (c-name, a-number)
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)) |
Question 12 Explanation:
Answer should be (A) and not (B), because we are doing a join between two massive tables whereas in (A) we are doing join between relatively smaller table and larger one and the output that this inner table gives (which is smaller in comparision to join that we are doing in (B)) is used for join with depositor table with the selection condition.
Options (C) and (D) are invalid as there is no b-city column in a-schema.
Options (C) and (D) are invalid as there is no b-city column in a-schema.
Question 13 |
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
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r ∪ s = r
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r ⊂ s | |
r * s = s
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Question 13 Explanation:
Let us consider an example:
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
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 14 |
Let R1 (A, B, (D)) 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)
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ΠC(r1) - ΠD(r2)
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ΠD(r1⨝C≠Dr2)
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ΠC(r1⨝C=Dr2)
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Question 14 Explanation:
C be a foreign key in R1 referring to R2.
→ Based on referral integrity C is subset of values in R2 then,
ΠC(r1) - ΠD(r2) results empty relation.
→ Based on referral integrity C is subset of values in R2 then,
ΠC(r1) - ΠD(r2) results empty relation.
Question 15 |
8, 8 | |
120, 8
| |
960, 8
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960, 120 |
Question 15 Explanation:
Natural join is a JOIN operation that creates an implicit join cause for you based on common columns in the two tables being joined.
→ 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.
→ 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 16 |
names of girl students with the highest marks
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names of girl students with more marks than some boy student
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names of girl students with marks not less than some boy students
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names of girl students with more marks than all the boy students
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Question 16 Explanation:
In the operator '-' between the two subexpression gives the names of all female students whose marks are greater than the all the male students of the class.
Question 17 |
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. |
Question 17 Explanation:
Relational algebra has the same power as safe relational calculus.
A query can be formulated in safe Relational Calculus if and only if it can be formulated in Relational Algebra.
A query can be formulated in safe Relational Calculus if and only if it can be formulated in Relational Algebra.
Question 18 |
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 18 Explanation:
The sum of all employee’s salaries can’t be represented by using the given six basic algebra operation. If we want to represent sum of salaries then we need to use aggregation operator.
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 |
Question 19 Explanation:
For maximum:
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.
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 |
σ(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) |
Question 20 Explanation:
The given relational algebra expression represents tuples having A=10 and B=20 which is equivalent to
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 21 Explanation:
The join here will be selecting only those tuples where A=C and B=D, meaning it is the intersection.
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(R2) | |
πA1(σC1(R1)) → σC1(σA1(R1)) |
Question 22 Explanation:
If the selection condition is on attribute A2, then we cannot replace it by RHS as there will not be any attribute A2 due to projection of A1 only.
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) |
Question 23 Explanation:
R - (R - S)
→ 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.
→ 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 |
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 |
Question 24 Explanation:
(a) Finding a adjaceny vertex for the given vertex is too simple i.e., Adj(X,Y).
(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.
(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 25 |
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 |
Question 25 Explanation:
(a) A=a for all r
(b) Display table
(c) A=a for all Tables r and s
(b) Display table
(c) A=a for all Tables r and s
There are 25 questions to complete.