## Functional-Dependency

 Question 1
The following functional dependencies hold true for the relational schema {V, W, X, Y, Z} :
V → W
VW → X
Y → VX
Y → Z
Which of the following is irreducible equivalent for this set of functional dependencies?
 A V→WV→XY→VY→Z B V→WW→XY→VY→Z C V→WV→XY→VY→XY→Z D V→WW→XY→VY→XY→Z
Database-Management-System       Functional-Dependency       GATE 2017 [Set-1]       Video-Explanation
Question 1 Explanation:
Step 1:
V → W, VW → X, Y → V, Y → X, Y→ Z
Step 2:
V → W, VW → X, Y → V, Y → X, Y→ Z
(V)+ = V ×
(VW)+ = VW ×
(Y)+ = YXZ
(Y)+ = YVW ×
(Y)+ = YVWX
Without Y → X, the closure of Y is deriving ‘X’ from the remaining attributes.
So, we can remove Y → X as its redundant.
Step 3:
V → W, VW → X, Y → V, Y → Z
(V)+ = VW, the closure of V is deriving W from the remaining FD’s.
So, W is redundant. We can remove it.
So, the final canonical form is
V→W, V→X, Y→V, Y→Z
⇾ So, option (A) is correct.
 Question 2

Consider the relation X(P, Q, R, S, T, U) with the following set of functional dependencies

```F = {
{P, R} → {S,T},
{P, S, U} → {Q, R}
}```

Which of the following is the trivial functional dependency in F+ is closure of F?

 A {P,R}→{S,T} B {P,R}→{R,T} C {P,S}→{S} D {P,S,U}→{Q}
Database-Management-System       Functional-Dependency       GATE 2015 [Set-3]
Question 2 Explanation:
X→Y is trivial if Y⊆ X
 Question 3

Consider the relation scheme R = (E, F, G, H, I, J, K, L, M, N) and the set of functional dependencies  {{E,F} → {G}, {F} → {I,J}, {E,H} → {K,L}, {K} → {M}, {L} → {N}} on R. What is the key for R?

 A {E,F} B {E,F,H} C {E,F,H,K,L} D {E}
Database-Management-System       Functional-Dependency       GATE 2014 [Set-1]
Question 3 Explanation:
A) (EF)+ = EFGIJ. So, EF is not a key.
B) (EFH)+ = EFHGIJKLMN, EFH is deriving all the attributes of R. So, EFH is a key.
C) If EFH is a key, then EFHKL will become the super key.
D) (E)+ = E
 Question 4

The following functional dependencies are given:

`  AB → CD, AF → D, DE → F, C → G, F → E, G → A.  `

Which one of the following options is false?

 A {CF}+ = {ACDEFG} B {BG}+ = {ABCDG} C {AF}+ = {ACDEFG} D {AB}+ = {ABCDFG}
Database-Management-System       Functional-Dependency       GATE 2006
Question 4 Explanation:
AB → CD
AF → D
DE → F
C → G
F → E
G → A
CF+ = {G,E,A,D,C,F} = {A,C,D,E,F,G} (✔️)
BG+ = {B,G,A,C,D} = {A,B,C,D,G} (✔️)
AF+ = {D,E,A,F} = {A,D,E,F} (❌)
AB+ = {A,B,C,D,G} (❌)
 Question 5

Consider a relation scheme R = (A, B, C, D, E, H) on which the following functional dependencies hold: {A → B, BC → D, E → C, D → A}. What are the candidate keys of R?

 A AE, BE B AE, BE, DE C AEH, BEH, BCH D AEH, BEH, DEH
Database-Management-System       Functional-Dependency       GATE 2005
Question 5 Explanation:
Using the given functional dependencies and looking at the dependent attributes, E and H are not dependent on any. So, they must be a part of any candidate key.
So only option (D) contains E and H as part of candidate key.
 Question 6

From the following instance of a relation scheme R(A,B,C), we can conclude that: A A functionally determines B and B functionally determines C B A functionally determines B and B does not functionally determines C C B does not functionally determines C D A does not functionally determines B and B does not functionally determines C
Database-Management-System       Functional-Dependency       GATE 2002
Question 6 Explanation:
From the given instance of relation it can be seen that A functionally determines B, but we cannot conclude this for the entire relation.
But for the given instance it can be seen that B does not functionally determines C, and it can be concluded for entire relation.
 Question 7

Consider a schema R(A,B,C,D) and functional dependencies A → B and C → D. Then the decomposition of R into R1(AB) and R2(CD) is

 A dependency preserving and lossless join B lossless join but not dependency preserving C dependency preserving but not lossless join D not dependency preserving and not lossless join
Database-Management-System       Functional-Dependency       GATE 2001
Question 7 Explanation:
If the given relations are to be lossless then
R1∩R2 ≠ 0
Given R1(A,B), R2
R1∩R2 = 0
Not lossless.
The given relation decomposed into R1(A,B) and R2(C,D) and there are only two functional dependencies A→B and C→D. So the given decomposition is dependency preserving.
 Question 8

Given the following relation instance.

```          x  y  z
1  4  2
1  5  3
1  6  3
3  2  2  ```

Which of the following functional dependencies are satisfied by the instance?

 A XY → Z and Z → Y B YZ → X and Y → Z C YZ → X and X → Z D XZ → Y and Y → X
Database-Management-System       Functional-Dependency       GATE 2000
Question 8 Explanation:
A functional dependency A→B is said to hold if for two tuples t1 and t2.
If for t1[A] = t2[A] then t1[Y] = t2[Y].
 Question 9

Let R = (A, B, C, D, E, F) be a relation scheme with the following dependencies: C→F, E→A, EC→D, A→B. Which of the following is a key of R?

 A CD B EC C AE D AC
Database-Management-System       Functional-Dependency       GATE 1999
Question 9 Explanation:
Let's check closure for each option,
A) (CD)+ = cdf
Not a key.
B) (EC)+ = ecdabf
Yes, it is a key.
C) (AE)+ = aeb
Not a key. D) (AC)+ = abcf
Not a key.
 Question 10

Let R (A, B, C, D) be a relational schema with the following functional dependencies: A → B, B → C, C → D and D → B.

The decomposition of R into (A, B), (B, C), (B, D)

 A gives a lossless join, and is dependency preserving B gives a lossless join, but is not dependency preserving C does not give a lossless join, but is dependency preserving D does not give a lossless join and is not dependency preserving
Database-Management-System       Functional-Dependency       GATE 2008-IT
Question 10 Explanation:
(A, B) (B, C) - common attribute is (B) and due to B→C, B is a key for (B, C) and hence ABC can be losslessly decomposed into (A, B)and (B, C).
(A, B, C) (B, D) - common attributes is B and B→D is a FD (via B→C, C→D), and hence, B is a key for (B, D). So, decomposition of (A, B, C, D) into (A, B, C) (B, D) is lossless.
Thus the given decomposition is lossless.
The given decomposition is also dependency preserving as the dependencies A→B is present in (A, B), B→C is present in (B, C), D→B is present in (B, D) and C→D is indirectly present via C→B in (B, C) and B→D in (B, D).
 Question 11

Consider a relation R with five attributes V, W, X, Y, and Z. The following functional dependencies hold: VY → W, WX → Z, and ZY → V.
Which of the following is a candidate key for R?

 A VXZ B VXY C VWXY D VWXYZ
Database-Management-System       Functional-Dependency       GATE 2006-IT
Question 11 Explanation:
As we can see that attribute XY do not appear in RHS of an FD, they need to be a part of key.
Candidate keys are
VXY, WXY, ZXY
 Question 12
Suppose the following functional dependencies hold on a relation U with attributes P, Q, R, S, and T:
P⟶ QR
RS⟶ T
Which of the following functional dependencies can be inferred from the above functional dependencies?
 A PS ⟶ Q B PS ⟶ T C R ⟶ T D P ⟶ R
Database-Management-System       Functional-Dependency       GATE 2021 CS-Set-2
Question 12 Explanation: Question 13
Consider a relation R ( A , B , C , D , E ) with the following three functional dependencies.
AB → C; BC → D; C → E;
The number of superkeys in the relation R is _____________.
 A 8
Database-Management-System       Functional-Dependency       GATE 2022       Video-Explanation
Question 13 Explanation:
R(A, B, C, D, E)
AB → C
BC → D
C → E
¯ The attributes A, B are not there in the right hand side of any of the given FDs.
¯ So, AB can be a candidate key.
(AB)^+ = ABCDE
¯ (AB)^+ is deriving all the attributes of R Hence, AB is a candidate key.
¯ The number of super keys possible for R with “AB” candidate kay is:
2^5 – 2 = 2^3
= 8
 Question 14
The set of attributes X will be fully functionally dependent on the set of attributes Y if the following conditions are satisfied.
 A X is functionally dependent on Y B X is not functionally dependent on any subset of Y C Both (a) and (b) D None of these
Database-Management-System       Functional-Dependency       ISRO-2018       Video-Explanation
Question 14 Explanation:
→ A functional dependency is a constraint between two sets of attributes in a relation from a database. In other words, functional dependency is a constraint that describes the relationship between attributes in a relation.
→ An attribute is fully functional dependent on another attribute, if it is Functionally Dependent on that attribute and not on any of its proper subset.
For example, an attribute X is fully functional dependent on another attribute Y, if it is Functionally Dependent on X and not on any of the proper subset of Y.

Must satisfied conditions:
i) X is functionally dependent on Y and
ii) X is not functionally dependent on any subset of Y.
 Question 15
In functional dependency Armstrong inference rules refers to
 A Reflexive, Augmentation and Decomposition B Transitive, Augmentation and Reflexive C Augmentation, Transitive, Reflexive and Decomposition D Reflexive, Transitive and Decomposition
Database-Management-System       Functional-dependency       ISRO CS 2011
Question 15 Explanation:
Armstrong's axioms are a set of axioms (or, more precisely, inference rules) used to infer all the functional dependencies on a relational database.
The axioms are sound in generating only functional dependencies in the closure of a set of functional dependencies when applied to that set
Axiom of Reflexivity:
→If X is a set of attributes and Y is a subset of X, then X holds Y. Hereby, X holds Y ( X -> Y) means that X functionally determines Y.
Axiom of Augmentation:
→If X holds Y and Z is a set of attributes, then XZ holds YZ. It means that attribute in dependencies does not change the basic dependencies.
Axiom of Transitivity:
The axiom of transitivity says if X holds Y, and Y holds Z, then X must also hold Z.
 Question 16
Consider the schema R(A,B,C,D) and the functional dependencies A→ B and C→ D. If the decomposition is made as R1(A,B) and R2(C,D), then which of the following is TRUE?
 A Preserves dependency but cannot perform lossless join B Preserves dependency and performs lossless join C Does not perform dependency and cannot perform lossless join D Does not preserve dependency but perform lossless join
Database-Management-System       Functional-Dependency       ISRO CS 2014
Question 16 Explanation:
Given Data,
Schema R(A, B, C, D)
Functional dependencies are A→ B and C→ D
Decomposed Schema is R1(A,B) and R2(C,D)
Dependency preservation decomposition:
It is another property of decomposed relational database schema D in which each functional dependency X → Y specified in F either appeared directly in one of the relation schemas Ri in the decomposed D or could be inferred from the dependencies that appear in some Ri.
Decomposition D={ R1 , R2, R3,,.., ,Rm} of R is said to be dependency-preserving with respect to F if the union of the projections of F on each Ri , in D is equivalent to F.
In other words, R ⊂ join of R1, R1 over X. The dependencies are preserved because each dependency in F represents a constraint on the database. If decomposition is not dependency-preserving, some dependency is lost in the decomposition.
→ R1(A,B) is covered A→ B
→ R2(C,D) is covered C→ D
It is Functional Dependency preserving because both the functional dependencies are covered.
Lossless join:
The decomposition is a lossless-join decomposition of R if at least one of the following functional dependencies are in F+ (where F+ stands for the closure for every attribute or attribute sets in F):
R1 ∩ R2 → R1
R1 ∩ R2 → R2
According to functional dependency R1(A,B) ∩ R2(C,D) = null. There is no common key in both the tables. So, it is not lossless join.
 Question 17
Every time the attribute A appears, it is matched with the same value of attribute B but not the same value of attribute C. Which of the following is true?
 A A→ (B,C) B A → B, A→→ C C A→ B, C→→A D A→→B, B→ C
Database-Management-System       Functional-Dependency       ISRO CS 2014
Question 17 Explanation:
→ represents functional dependency and
→→ is multivalued dependency.
Functional Dependency:
A → B means that the values of B are determined by the values of A. Two tuples sharing the same values of A will necessarily have the same values of B.
Multivalued dependency: It is a special case of a join dependency, with only two sets of values involved, i.e. it is a binary join dependency.
A multivalued dependency exists when there are at least three attributes (like A,B and C) in a relation and for a value of A there is a well defined set of values of B and a well defined set of values of C. However, the set of values of B is independent of set C and vice versa.
 Question 18
In functional dependency between two sets of attribute A and B then set of attributes A of database is classified as:
 A Top right side B Down left side C Left hand side D Right hand side
Database-Management-System       Functional-Dependency       Nielit Scientist-B IT 4-12-2016
Question 18 Explanation:
A functional dependency (FD) is a relationship between two attributes, typically between the PK and other non-key attributes within a table. For any relation R, attribute Y is functionally dependent on attribute X (usually the PK), if for every valid instance of X, that value of X uniquely determines the value of Y.
This relationship is indicated by the representation below :
X ———–> Y
The left side of the above FD diagram is called the determinant, and the right side is the dependent.
 Question 19
Every Boyce-Codd Normal Form(BCNF) decomposition is
 A dependency preserving B not dependency preserving C need not be dependency preserving D none of these
Database-Management-System       Functional-Dependency       Nielit Scientist-C 2016 march
Question 19 Explanation:
Normalization requires definitely loss less decomposition but not dependency preserving. But BCNF need to be functional dependency.
 Question 20
A functional dependency of the form x → y is trivial if
 A y ⊆ x B y ⊂ x C x ⊆ y D x ⊂ y and y ⊂ x
Database-Management-System       Functional-Dependency       Nielit Scientist-C 2016 march
Question 20 Explanation:
Trivial − If a functional dependency (FD) X → Y holds, where Y is a subset of X, then it is called a trivial FD. Trivial FDs always hold.
 Question 21
Given following relation instance: Which of the following functional dependencies are satisfied by the instance?
 A XY→ Z and Z → Y B YZ →X and Y → Z C YZ → X and X→ Z D XZ→ Y and Y → X
Theory-of-ComputationDatabase-Management-System       Functional-Dependency       Nielit Scientist-C 2016 march
Question 21 Explanation:
Functional dependency is represented by an arrow sign (→) that is, X→Y, where X functionally determines Y.
The left-hand side attributes determine the values of attributes on the right-hand side.
 Question 22
Consider the schema R=(S,T,U,V) and the dependencies S→ T, T→ U, U→ V and V→ S. If R=(R1 and R2) be a decomposition such that R1 ∩R2=​ ∅ ​ , then decomposition is
 A not in 2 NF B in 2 NF but not in 3 NF C in 3 NF not in 2 NF D in both 2NF and 3 NF
Database-Management-System       Functional-Dependency       Nielit Scientist-C 2016 march
Question 22 Explanation:
● R1 ∩ R2 = ∅. This makes the decomposition lossless join, as all the attributes are keys, R1 ∩ R2 will be a key of the decomposed relations (lossless condition says the common attribute must be a key in at least one of the decomposed relation).
● Now, even the original relation R is in 3NF (even BCNF) as all the attributes are prime attributes (in fact each attribute is a candidate key). Hence, any decomposition will also be in 3NF (even BCNF).
 Question 23
Let R=(A,B,C,D,E,F) be a relation scheme with the following dependencies: C→ F, E→ A, EC→ D, A→ B Which of the following is a key for R?
 A CD B EC C AE D AC
Database-Management-System       Functional-Dependency       Nielit Scientist-B CS 2016 march
Question 23 Explanation:
Here, simple way to solve this question is,
1. First we have to find the right hand side values of a dependencies. Check whether it cover all variables in a relation .
2. In right hand side, CE are missing. It means definitely CE should be there in candidate key.
(CE)​ +​ =(A,B,C,D,E,F).
Remaining all are not satisfying candidate key properties.
 Question 24
For a database relation R(a,b,c,d) where the domains of a,b,c,d include only atomic values, only the following functional dependencies and those that can be inferred from them hold: a→ c b→ d the relation is in
 A first normal form but not in second normal form B Second normal form but not in third normal form C Third normal form D None of these
Database-Management-System       Functional-Dependency       Nielit Scientist-B CS 2016 march
Question 24 Explanation:
● First normal form (1NF) is a property of a relation in a relational database. A relation is in first normal form if and only if the domain of each attribute contains only atomic (indivisible) values, and the value of each attribute contains only a single value from that domain.
● The Second Normal Form states that it should meet all the rules for 1NF and there must be no partial dependencies of any of the columns on the primary key.
● The questions states that relation consists of automatic values but there no information about prime key.So we don’t about partial dependency.
 Question 25
Rule which states that addition of same attributes to right side and left side will result in other valid dependency is classified as:
 A referential rule B inferential rule C augmentation rule D reflexive rule
Database-Management-System       Functional-Dependency       Nielit Scientist-B CS 4-12-2016
Question 25 Explanation:  Question 26
If every functional dependency in set E is also in closure of F then this is classified as:
 A FD is covered by E B E is covered by F C F is covered by E D F plus is covered by E
Database-Management-System       Functional-Dependency       Nielit Scientist-B CS 4-12-2016
Question 26 Explanation:
A covers B, if every FD in B can be inferred from A.
A covers B if A​ + ​ ​ ⊆ B​ + ​ (+ means closure ).
Every set of functional dependencies has a canonical cover.
 Question 27
Considering relational database, functional dependency between two attributes A and B is denoted by:
 A A--> B B B<-- A C AB--> R D R<-- AB
Database-Management-System       Functional-Dependency       Nielit Scientist-B CS 4-12-2016
Question 27 Explanation:
Given a relation R, a set of attributes X in R is said to functionally determine another set of attributes Y, also in R, (written X → Y) if, and only if, each X value in R is associated with precisely one Y value in R; R is then said to satisfy the functional dependency X → Y.
 Question 28
If there is more than one key for relation schema in DBMS then each key in relation schema is classified as:
 A Primary key B Super key C candidate key D Primary key
Database-Management-System       Functional-Dependency       Nielit Scientist-B CS 4-12-2016
Question 28 Explanation:
If there is more than one key for relation schema in DBMS then each key in relation schema is classified as candidate key.
→ In the relational model of databases, a candidate key of a relation is a minimal superkey for that relation; that is, a set of attributes such that:
1.The relation does not have two distinct tuples (i.e. rows or records in common database language) with the same values for these attributes (which means that the set of attributes is a superkey)
2.There is no proper subset of these attributes for which (1) holds (which means that the set is minimal).
→ Candidate keys are also variously referred to as primary keys, secondary keys or alternate keys. The constituent attributes are called prime attributes. Conversely, an attribute that does not occur in ANY candidate key is called a non-prime attribute.
→ Since a relation contains no duplicate tuples, the set of all its attributes is a superkey if NULL values are not used. It follows that every relation will have at least one candidate key. The candidate keys of a relation tell us all the possible ways we can identify its tuples. As such they are an important concept for the design of database schema.
 Question 29
Let R = (A, B, C, D, E, F) be a relation schema with the following dependencies C→ F, E→ A, EC→ D, A→ B. Which of the following is a key of R?
 A CD B EC C AE D AC
Database-Management-System       Functional-Dependency       ISRO CS 2015
Question 29 Explanation:
Here, simple way to solve this question is,
First we have to find the right hand side values of a dependencies. Check whether it cover all variables in a relation .
In right hand side, CE are missing. It means definitely CE should be there in candidate key.
(CE)+=(A,B,C,D,E,F).
Remaining all are not satisfying candidate key properties.
 Question 30
Given a relation schema R(ABCDEFGH) in first normal form. For the set of dependencies F={ A→ B, A→ C, CG→ H, B→ H, G→ F}, which dependency is logically implied?
 A AC→ H B C→ H C G→ H D A→ H
Database-Management-System       Functional-Dependency       KVS 22-12-2018 Part-B
Question 30 Explanation:
From the dependencies A→ B and B→ H, we can imply A→ H by using transitive dependency
 Question 31
Given an instance of the relation R(ABCD). Which of the following functional dependencies hold?
 A {AB} → D and D → A B {AB → D and B → D C {AB} → C and B → C D {AB} → D and A → D
Database-Management-System       Functional-Dependency       KVS 22-12-2018 Part-B
Question 31 Explanation:
Rules for determining functional dependency hold or not : → If LHS of a functional dependency is not repeating any value then you. you can say that LHS of a functional dependency is determining RHS of a functional dependency.
→ If LHS of a functional dependency is repeating then see the value of RHS. If values of RHS are same for repeated value of LHS then you can say that LHS of a functional dependency is determining RHS of a functional dependency else not determining.
Now if you see the given options then only option(C) is satisfying above two rules.
 Question 32

If R(A,B,C,D) is a relation schema, which is decomposed into R1(A,B,C) and R2(C,D), which of the following ensures that the given decomposition is non additive(or lossless)?

 A R1 → R2 B R2 → R1 C R1 ∩ R2 → R1 or R1 ∩ R2 → R2 D R1 U R2 → R1 R1 U R2 → R2
Database-Management-System       Functional-Dependency       JT(IT) 2018 PART-B Computer Science
Question 32 Explanation:
Option (C) is the correct option because a decomposition can be lossless if and only if there exists a common attribute between the decomposed relations which is either a candidate key or a key attribute in any of the decomposed tables.
For lossless decomposition:
R1 ∩ R2 → R1 (OR) R1 ∩ R2 → R1 - R2
R1 ∩ R2 → R2 (OR) R1 ∩ R2 → R2 - R1
 Question 33
Which of the statements are true about functional dependency X→ Y in relation R? a.Whenever two tuples t1 and t2 in R have t1(X)=t2(X) then t1(Y)=t2(Y) b.Whenever two tuples t1 and t2 in R have t1(Y)=t2(Y) then t1(X)=t2(X)
 A a and b B Only a C Only b D Neither a nor b
Database-Management-System       Functional-Dependency       KVS 30-12-2018 Part B
Question 33 Explanation:
→Functional dependency is a relationship that exists when one attribute uniquely determines another attribute.
→If R is a relation with attributes X and Y, a functional dependency between the attributes is represented as X→Y, which specifies Y is functionally dependent on X. Here X is a determinant set and Y is a dependent attribute. Each value of X is associated with precisely one Y value.
 Question 34
Consider a schema R(MNPQ) and functional dependencies M → N, P → Q. Then the decomposition of R into R​ 1​ (MN) and R​ 2​ (PQ) is________.
 A Dependency preserving but not lossless join B Dependency preserving and lossless join C Lossless join but not dependency preserving D Neither dependency preserving nor lossless join.
Database-Management-System       Functional-Dependency       UGC NET CS 2017 Jan -paper-2
Question 34 Explanation:
Definition of lossless decomposition:​ Let R be the relational schema decomposed into R​ 1
and R​ 2​ . Given decomposition is lossless only if
1. R​ 1​ U R​ 2​ =R
2. R​ 1​ U R​ 2​ =φ
3. R​ 1 ∩ R​ 2​ → R​ 1 ​ (or) R​ 1 ∩ R​ 2​ → R​ 2
→ In this schema, there is no common key attribute between R​ 1​ and R​ 2​ . So, this relation is lossy relation.
Definition of Functional dependency preserving:
→ Let R be the relational schema with functional dependency set F is decomposed into R​ 1​ , R​ 2​ ,..,R​ n ​ with functional dependency sets F​ 1​ ,F​ 2​ ,...,F​ n​ respectively. In general F​ 1​ ,F​ 2​ ,...,F​ n ​ can be ⊆ F.
So, dependencies are preserved in the given decomposition.
 Question 35
A relation R = {A, B, C, D, E, F,G} is given with following set of functional dependencies: F = {AD → E, BE → F, B → C, AF → G} Which of the following is a candidate key ?
 A A B AB C ABC D ABD
Database-Management-System       Functional-Dependency       UGC NET CS 2015 Dec- paper-2
Question 35 Explanation:
Since no functional dependency is having ABD in their right hand side, so every key should include ABD as a part of them to become a candidate key.
Now check whether ABD is itself a candidate key or not by finding the closure.
(ABD)​ +​ = { ABDCEFG }
since ABD can identify each attribute of the given relation uniquely.
So, ABD is the candidate key.
 Question 36
A relation R={A, B, C, D, E, F} is given with following set of functional dependencies : F =, {A →B,AD →C,B→F,A →E}
Which of the following is candidate key?
 A A B AC C AD D None of these
Database-Management-System       Functional-Dependency       UGC NET CS 2006 June-Paper-2
Question 36 Explanation:
Here, Attribute A and D are not present in the right hand side of any production so every key should include AD to be a candidate key.
Step-1: Check whether AD could be a candidate key or not.
(AD)​ +​ = {A,D,C,B,F,E}
Step-2: AD can uniquely identifies each attribute. So, AD is the candidate key for given relation.
 Question 37
Identify the minimal key for relational scheme R(A, B, C, D, E) with functional dependencies F = {A → B, B → C, AC → D}
 A A B AE C BE D CE
Database-Management-System       Functional-dependency       UGC NET CS 2014 Dec-Paper-2
Question 37 Explanation:
Given functional dependencies are F={A → B, B → C, AC → D}
Step-1: Attribute A and E are not present in the right hand side of any production. So, every key should include AE to be a candidate key.
Step-2: To check whether AE could be a candidate key or not.
(AE)​ +​ = {A,B,C,D,E}
Since AE can uniquely identify each attribute and we can say this is the candidate key for given relation.
 Question 38
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 ?
 A AEF, BEF and DEF B AEF, BEF and BCF C AE and BE D AE, BE and DE
Database-Management-System       Functional-Dependency       UGC NET CS 2018-DEC Paper-2
Question 38 Explanation:
EFA​ +​ = {EFABCD}
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 39
Let R ={A, B, C, D, E, F} be a relation schema with the following dependencies C → F, E → A, EC → D, A → B Which of the following is a key for R ?
 A CD B EC C AE D AC
Database-Management-System       Functional-Dependency       UGC NET CS 2014 June-paper-2
Question 39 Explanation:
Simple way to solve this question is,
→First we have to find the right hand side values of a dependencies. Check whether it cover all variables in a relation .
→In right hand side, CE are missing. It means definitely EC should be there in candidate key.
(EC)+=(A,B,C,D,E,F).
Remaining all are not satisfying candidate key properties.
 Question 40
For a database relation R(a, b, c, d) where the domains of a, b, c and d include only atomic values, and only the following functional dependencies and those that can be inferred from them hold :
a → c
b → d
The relation is in .
 A First normal form but not in second normal form B Second normal form but not in third normal form C Third normal form D BCNF
Database-Management-System       Functional-Dependency       UGC NET CS 2018 JUNE Paper-2
Question 40 Explanation:
Primary key of given relation is “ab” And there is a partial dependency exist in given FD’s so the given relation is in 1NF but not in second normal form.
 Question 41
A many-to-one relationship exists between entity sets r1 and r2. How will it be represented using functional dependencies if Pk(r) denotes the primary key attribute of relation r?
 A Pk(r1 ) → Pk(r2 ) B Pk(r2 ) → Pk(r1 ) C Pk(r2 ) → Pk(r1 ) and Pk(r1 ) → Pk(r2 ) D Pk(r2 ) → Pk(r1 ) or Pk(r1 ) → Pk(r2 )
Database-Management-System       Functional-Dependency       UGC NET CS 2018 JUNE Paper-2
Question 41 Explanation: Here we have a many to one relationship between between Set(r1) and set(r2).
→ Elements of set(r2) can’t identify elements of sert(r1) because one value element in set(r2) is pointing to more than one element of set(r1).
→ So we can’t say Pk(r2 ) → Pk(r1) but elements of set(r1) are pointing to exactly one element of set(r2) so we can say that Pk(r2 ) → Pk(r1 ) because r1 is uniquely identifying r2.
 Question 42
In airline reservation system, the entities are date, flight number, place of departure, destination, type of plane and seats available. The primary key is
 A flight number B flight number + place of departure C flight number + date D flight number + destination
Database-Management-System       Functional-Dependency       NIELIT Junior Teachnical Assistant_2016_march
 Question 43
The dependency preservation decomposition is a property to decompose database schema D, in which each functional dependency X → Y specified in F,
 A appeared directly in one of the relation schemas Ri in the decomposed D. B could be inferred from dependencies that appear in some Ri. C both (A) and (B) D None of these
Database-Management-System       Functional-Dependency       UGC NET CS 2010 Dec-Paper-2
Question 43 Explanation:
The dependency preservation decomposition is a property to decompose database schema D, in which each functional dependency X → Y specified in F
Appeared directly in one of the relation schemas Ri in the decomposed D.
Could be inferred from dependencies that appear in some Ri.
 Question 44
Match the following : A 1 – d, 2 – b, 3 – a, 4 – c B 2 – d, 3 – a, 1 – b, 4 – c C 4 – a, 3 – b, 2 – c, 1 – d D 3 – a, 4 – b, 1 – c, 2 – d
Database-Management-System       Functional-Dependency       UGC NET CS 2009 Dec-Paper-2
Question 44 Explanation:
Determinants→ Group of attributes on the left hand side of arrow of function dependency.
Candidate key→ Uniquely identified a row. Candidate key is minimal super key.
Non-redundancy→ No attribute can be added
Functional dependency→ A constraint between two attribute
 Question 45
A function that has no partial functional dependencies is in __________ form.
 A 3 NF B 2 NF C 4 NF D BCNF
Database-Management-System       Functional-Dependency       UGC NET CS 2009 Dec-Paper-2
Question 45 Explanation:
To qualify for second normal form a relation must:
(1) be in first normal form (1NF)
(2) not have any non-prime attribute that is dependent on any proper subset of any candidate key of the relation. A non-prime attribute of a relation is an attribute that is not a part of any candidate key of the relation.
 Question 46
Suppose R is a relation schema and F is a set of functional dependencies on R. Further, suppose R1 and R2 form a decomposition of R. Then the decomposition is a lossless join decomposition of R provided that :
 A R1 ∩ R2 → R1 is in F+ B R1 ∩ R2 → R2 is in F+ C both R1 ∩ R2 → R1 and R1 ∩ R2 → R2 functional dependencies are in F+ D at least one from R1 ∩ R2 → R1 and R1 ∩ R2 → R2 is in F+
Database-Management-System       Functional-Dependency       UGC NET CS 2008 Dec-Paper-2
Question 46 Explanation:
Suppose R is a relation schema and F is a set of functional dependencies on R. Further, suppose R1 and R2 form a decomposition of R. Then the decomposition is a lossless join decomposition of R provided that at least one from
R1 ∩ R2 → R1 and R1 ∩ R2 → R2 is in F+
Lossless join:
The decomposition is a lossless-join decomposition of R if at least one of the following functional dependencies are in F+ (where F+ stands for the closure for every attribute or attribute sets in F):
R1 ∩ R2 → R1
R1 ∩ R2 → R2
 Question 47
A Relation R = {A,B,C,D,E,F} is given with following set of functional dependencies
F={A →B, AD→C, B→F, A→ E}.
Which of the following is Candidate Key ?
 A A B AC C AD D None of these
Database-Management-System       Functional-Dependency       UGC NET CS 2007 June-Paper-2
Question 47 Explanation: Question 48
Consider a schema R(A, B, C, D) and following functional dependencies.
A → B
B → C
C → D
D → B
Then decomposition of R into R1(A, B), R2(B, C) and R3(B, D) is __________ .
 A Dependency preserving and lossless join. B Lossless join but not dependency preserving. C Dependency preserving but not lossless join. D Not dependency preserving and not lossless join.
Database-Management-System       Functional-Dependency       UGC NET CS 2017 Nov- paper-3
Question 48 Explanation:
(A, B) (B, C) - common attribute is (B) and due to B→C, B is a key for (B, C) and hence ABC can be losslessly decomposed into (A, B)and (B, C).
(A, B, C) (B, D) - common attributes is B and B→D is a FD (via B→C, C→D), and hence, B is a key for (B, D). So, decomposition of (A, B, C, D) into (A, B, C) (B, D) is lossless.
Thus the given decomposition is lossless.
The given decomposition is also dependency preserving as the dependencies A→B is present in (A, B), B→C is present in (B, C), D→B is present in (B, D) and C→D is indirectly present via C→B in (B, C) and B→D in (B, D).
 Question 49
Which of the following statements is TRUE?

D1 : The decomposition of the schema R(A, B, C) into R1(A, B) and R2 (A, C) is always lossless.

D2 : The decomposition of the schema R(A, B, C, D, E) having AD → B, C → DE, B → AE and AE → C, into R1 (A, B, D) and R2 (A, C, D, E) is lossless.

 A Both D1 and D2 B Neither D1 nor D2 C Only D1 D Only D2
Database-Management-System       Functional-Dependency       UGC NET CS 2016 July- paper-3
Question 49 Explanation:
Statement(1) is incorrect because functional dependencies of relation R are not given, so we can't say whether the decomposition of R into R1 and R2 is lossless or not. Question 50
Consider the table R with attributes A, B and C. The functional dependencies that hold on R are : A → B, C → AB. Which of the following statements is/are True ?
I. The decomposition of R into R1(C, A) and R2(A, B) is lossless.
II. The decomposition of R into R1(A, B) and R2(B, C) is lossy.
 A Only I B Only II C Both I and II D Neither I nor II
Database-Management-System       Functional-Dependency       UGC NET CS 2016 Aug- paper-3
Question 50 Explanation: Here ‘B’ is common attribute but ‘B’ is not the primary key in any of the decomposed relations (R1 or R2). Hence this decomposition is not a lossless decomposition i.e., it is a lossy decomposition.
Hence, option (C) is correct.
 Question 51
The relation schemas R​ 1​ and R​ 2​ form a Lossless join decomposition of R if and only if:
R​ 1​ ∩ R​ 2​ ↠ (R​ 1​ - R​ 2​ )
R​ 1​ → R​ 2
R​ 1​ ∩ R​ 2​ ↠ (R​ 2​ - R​ 1​ )
(R​ 2​ → R​ 1​ ) ∩ R​ 2
 A (a) and (b) happens B (a) and (d) happens C (a) and (c) happens D (b) and (c) happens
Database-Management-System       Functional-Dependency       UGC NET CS 2015 June Paper-3
Question 51 Explanation:
The decomposition of a relation is said to be lossless if and only if there exists a common attribute which is a primary key in any of the two decomposed relations. In this case statement (a) and statement (c) are satisfying this criteria. Hence option (C) is correct.
 Question 52

Comprehension:

Answer question (96-100) based on the problem statement given below:

An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functionally dependency set F is given as : F: {A→ BC, D → E, BC → D, A →D}. Consider a universal relation R(A, B, C, D, E) with functional dependency set F. Also all attributes are simple and take atomic values only.

:→ Minimal cover F’ of functional dependency set F is
 A F’ = {A → B, A → C, BC → D, D →E} B F’ = {A → BC, B → D, D → E} C F’ = {A → B, A → C, A → D, D → E} D F’ = {A → B, A → C, B → D, C → D.D → E}
Database-Management-System       Functional-Dependency       UGC-NET DEC-2019 Part-2
Question 52 Explanation:
Steps to find minimal cover:
Step1: Write all FDs in such a way that the RHS of each FD contain only one attribute.
A→ B
A→ C
D → E
BC → D
A →D
Step2: Then for each FD see whether that RHS attribute can be driven by the LHS attribute using remaining FDs, if yes then remove that FD otherwise keep it. So step 1 results in following FDs:
A→ B
A→ C
D → E
BC → D
Step3: Now see the FD which is having 2 or more attributes in its LHS.Then find the closure of LHS attributes and then eliminate the attributes from LHS which are common in clsure. Above BC are two attributes in LHS.
B+ = {B}
C+ = {C}
Since nothing is common in closure so keep both attributes in LHS.
Hence minimal cover is
A→ B
A→ C
D → E
BC → D
 Question 53

Comprehension:

Answer question (96-100) based on the problem statement given below:

An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functionally dependency set F is given as : F: {A→ BC, D → E, BC → D, A →D}. Consider a universal relation R(A, B, C, D, E) with functional dependency set F. Also all attributes are simple and take atomic values only.

→ Assume that given table R is decomposed in two tables Which of the following option is true w.r.t. given decomposition?
 A Dependency preservation property is followed B R1 and R2 are both in 2 NF C R2 is in 2 NF and R3 is in 3 NF D R1 is in 3 NF and R2 is in 2 NF
Database-Management-System       Functional-Dependency       UGC-NET DEC-2019 Part-2
Question 53 Explanation:
Since In R1 and R2 BC can’t determine BC → D of relation “R”. Hence R1 and R2 are not following the Dependency preservation property.
Candidate key of R1 is “A”. And since KHS of R1 contains only “A” so R1 is in 3NF.
Candidate key of R2 is “A” , But Since D→E neither have Super key in its LHS nor have a prime key attribute in its RHS, So R2 is in 2NF but not in 3NF.
 Question 54

Comprehension:

Answer question (96-100) based on the problem statement given below:

An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functionally dependency set F is given as : F: {A→ BC, D → E, BC → D, A →D}. Consider a universal relation R(A, B, C, D, E) with functional dependency set F. Also all attributes are simple and take atomic values only.

→ Identify the redundant functional dependency in F
 A BC→D B D→E C A→D D A→BC
Database-Management-System       Functional-Dependency       UGC-NET DEC-2019 Part-2
Question 54 Explanation:
A→D is redundant because A can determine D using FD’s A→ BC and BC → D.
 Question 55

Comprehension:

Answer question (96-100) based on the problem statement given below:

An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functionally dependency set F is given as : F: {A→ BC, D → E, BC → D, A →D}. Consider a universal relation R(A, B, C, D, E) with functional dependency set F. Also all attributes are simple and take atomic values only.

→ Identify primary key of table R with functional dependency set F
 A BC B AD C A D AB
Database-Management-System       Functional-Dependency       UGC-NET DEC-2019 Part-2
Question 55 Explanation:
Since “A” is not in RHS of any FD so “A” is the key of relation R. NOw to see whether “A” is the primary key of “R” or not find its closure.
A+ = { A,B,C,D,E}. Hence A is the primary key of relation R
 Question 56

Comprehension:

Answer question (96-100) based on the problem statement given below:

An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functionally dependency set F is given as : F: {A→ BC, D → E, BC → D, A →D}. Consider a universal relation R(A, B, C, D, E) with functional dependency set F. Also all attributes are simple and take atomic values only.

→ Identify the normal form in which relation R belong to
 A 1 NF B 2 NF C 3 NF D BCNF
Database-Management-System       Functional-Dependency       UGC-NET DEC-2019 Part-2
Question 56 Explanation:
Since “A” is the primary key or “R” and there is no partial dependency So “R” is in 2NF.
Since, D → E, BC → D neither have a super key in their LHS nor a prime key attribute in their RHS so “R” is not in 3NF.
Since “R” is not in 3NF it can’t be in BCNF.
Hence option(B) is correct
 Question 57
If D={R1,R2} is a decomposition of R, and F is the set of functional dependencies on R, then which of the following ensures that the decomposition D is lossless (nonadditive)?
 A ((R1∩ R2) →(R1-R2)) ∈ F+ B ((R1 ∩R2) → (R1 - R2)) ∉ F+ C ((R1 ∪ R2) →(R1 - R2)) ∈ F+ D ((R1∩R2) →R1 ) ∈ F+
Database-Management-System       Functional-Dependency       CIL Part - B
Question 57 Explanation: Question 58
The 9’s compliment of the decimal number 139452 is:
 A 971658 B 971659 C 860547 D 860548
Database-Management-System       Functional-Dependency       CIL Part - B
Question 58 Explanation:
→To obtain the 9’s complement of any number we have to subtract the number with (10n – 1) where n = number of digits in the number, or in a simpler manner we have to divide each digit of the given decimal number with 9
→Subtract each and every digit of given number by digit “9”.
999999
139452
________
860547
 Question 59
Consider the relation R shown in the below Figure.
 X Y Z X1 Y1 Z1 X1 Y1 Z2 X2 Y1 Z1 X2 Y1 Z3
List all the functional dependencies that this relation instance satisfies.
 A R: Z → Y, X → Y, and X → Z B R: Z →Y, X → Y, and XZ → Y C R: X →Y, X → Z, and XZ → Y D R: Z →Y, X → Z, and XZ → Y
Database-Management-System       Functional-Dependency       APPSC-2016-DL-CS
Question 59 Explanation:
option 1 is wrong bacause X → Z does not satisfy, since X1 is mapped to two values Z1 and Z2.
Option 3 is wrong because of X → Z.
Option 4 is wrong because of X → Z
 Question 60

Let the functional dependencies, (A → B, B → C) hold for relation P(A,B,C). If it is decomposed in R(A,B) and S(B,C), then the decomposition will be

 A Lossy B Lossless C Non-Lossless D Strongly
Database-Management-System       Functional-Dependency       APPSC-2012-DL CA
Question 60 Explanation:
The decomposition is lossless because B is common attribute and is a key in relation S ,so the decomposition is lossless.
 Question 61
Consider a schema R(A, B, C, D) and functional dependencies A → B and C → D. Then the decomposition of R into R1 (A B) and R2(C D) is
 A Dependency preserving and lossless Join B Lossless Joint but not dependency preserving C Dependency preserving but not lossless Join D Not dependency preserving and not lossless Join
Database-Management-System       Functional-Dependency       APPSC-2012-DL-CS
Question 61 Explanation:
There is no common attribute in the R1 and R2. So the given decomposition is not lossless join. But they are dependency preserving because A → B is covered by R1 and C → D is covered by R2.
 Question 62
Functional dependencies are a generalization of
 A Key dependencies B Relation dependencies C Database dependencies D None of these
Database-Management-System       Functional-Dependency       TNPSC-2012-Polytechnic-CS
Question 62 Explanation:
A functional dependency is a generalization of the notion of a key. It requires the value for a certain set of attributes determines uniquely the value for another set of attributes.
 Question 63
Let R = (A, B, C, D, E, F) be a relation schema with the following dependencies:
`C → F,E → A, EC → D, A → B `
Which of the following is a key for R?
 A CD B EC C AE D AC
Database-Management-System       Functional-Dependency       TNPSC-2012-Polytechnic-CS
Question 63 Explanation:
Let's find closure of each options,
(CD)+ = CDF X
(EC)+ = ECFDAB ✓
(AE)+ = AEB X
(AC)+ = ACBF X
 Question 64
A functional dependency is a relationship between:
 A Tables B Attributes C Rows D Relations
Database-Management-System       Functional-Dependency       TNPSC-2017-Polytechnic-CS
Question 64 Explanation:
A functional dependency is a relationship between attributes.
 Question 65
Consider the following dependencies :
`AB → CD, AF → D, DE → F, C → G, F → E, G → A `
Which one of the following options is false ?
 A BG+={ABCDG} B CF+={ACDEFG} C AB+={ABCDG} D AF+={ACDEFG}
Database-Management-System       Functional-Dependency       TNPSC-2017-Polytechnic-CS
Question 65 Explanation:
(BG)+ = BGACD
(AB)+ = ABCDG
(AF)+ = AFDE
Hence option (D) is False.
 Question 66
Let R = ABCDE is a relational scheme with functional dependency set F = {A → B, B → C, AC → D}. The attribute closures of A and E are
 A ABCD,ɸ B ABCD, E C ABC, E
Database-Management-System       Functional-Dependency       UGC NET CS 2014 Dec - paper-3
Question 66 Explanation:
A+ = { A, B, C, D}
E+ = {E}
 Question 67
Consider a Relational schema R(A, B, C, D) and functional dependencies A → B and C → D. Then the decomposition of R into R1 (A, B) and R2(C, D) is
 A dependency preserving and lossless join B lossless join but not dependency preserving C dependency preserving but not lossless join D neither dependency·preserving nor lossless join
Database-Management-System       Functional-Dependency       HCU PHD CS MAY 2019
Question 67 Explanation:
R1(A,B) contains FD A→B
R2(C,D) contains FD C→D
So, yes dependency preserving.
But there is no common attribute between R1 and R2, hence not lossless join.
 Question 68

From the following instance of a relation scheme R(X,Y,Z), we can conclude that:

X Y Z

1 1 1

1 1 0

2 3 2

2 3 2
 A X functionally determines Y and Y functionally determines Z B Y does not functionally determine Z C X does not functionally determine Y and Y does not functionally determine Z D None of the above
Database-Management-System       Functional-Dependency       HCU PHD CS 2018 December
Question 68 Explanation:
From the given instanceof relation it can be seen that X functionally determines Y, but we cannot conclude this for the entire relation.
But for the given instance it can be seen that Y does not functionally determines Z, and it can be concluded for the entire relation
 Question 69
Consider a schema R = {A, B, C, D} and functional dependencies A → B and C → D. Then the decomposition R1 = {A, B} and R2 = {C, D} is
 A Dependency preserving but not lossless join B Dependency preserving and lossless join C Lossless join but not dependency preserving D Neither lossless join nor dependency preserving
Database-Management-System       Functional-Dependency       HCU PHD CS MAY 2017
Question 69 Explanation:
The given decomposition does not have any attribute in common, so not lossless decomposition.
The relation R1 covers FD A → B and relation R2 covers FD C → D.Hence the decomposition is functional dependency preserving.
 Question 70
Consider a schema R(A, B, C, D) and functional dependencies A→ B and C→ D. Then the decomposition of R into R1 (A, B) and R2(C, D) is
 A lossless join but not dependency preserving B dependency preserving but not lossless join C not dependency preserving and not lossless join D part dependency preserving and lossless join
Database-Management-System       Functional-Dependency       HCU PHD CS MAY 2014
Question 70 Explanation:
The given decomposition is not lossless join because there is no common attributes between R1 and R2. But the given decomposition is dependency preserving, A→ B is covered by R1 and C→ D is covered by R2.
 Question 71
A minimal super key (i.e, one of the super keys for which no proper subset is a super key) is called:
 A Super Key B Candidate Key C Primary Key D Candidate and Primary Key E B and D
Database-Management-System       Functional-Dependency       NIC-NIELIT STA 2020
Question 71 Explanation:
A minimal superkey is a candidate key but as per the options it suited as Candidate and primary key.
 Question 72
Let R = (A, B, C, D, E) having following FD's. F = {A → BC, CD → E, B → D, E → A} Which of the following is not a Candidate key?
 A A B B C E D BC
Database-Management-System       Functional-Dependency       NIC-NIELIT STA 2020
Question 72 Explanation:

Closure(A)+ = { A, B, C, D, E }
Closure(B)+ = { B, D }
Closure(E)+ = { A, B, C, D, E }
Closure(BC)+ = { A, B, C, D, E }
 Question 73
Domain constraints, functional dependency and referential integrity are special forms of ______.
 A Foreign key B Primary key C Assertion D Referential constraint
Database-Management-System       Functional-Dependency       NIC-NIELIT STA 2020
Question 73 Explanation:
Domain constraints, functional dependency and referential integrity are special forms of assertion.
There are 73 questions to complete.