Functional-Dependency

Question 1

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-SystemFunctional-DependencyGATE 1999
Question 1 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 2

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-SystemFunctional-DependencyGATE 2000
Question 2 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 3

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-SystemFunctional-DependencyGATE 2001
Question 3 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 4

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-SystemFunctional-DependencyGATE 2002
Question 4 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 5
 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-SystemFunctional-DependencyGATE 2021 CS-Set-2Video-Explanation
Question 5 Explanation: 
Question 6

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-SystemFunctional-DependencyGATE 2006-IT
Question 6 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 7

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-SystemFunctional-DependencyGATE 2008-IT
Question 7 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 8
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-SystemFunctional-DependencyGATE 2022Video-Explanation
Question 8 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 9
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→W
V→X
Y→V
Y→Z
B
V→W
W→X
Y→V
Y→Z
C
V→W
V→X
Y→V
Y→X
Y→Z
D
V→W
W→X
Y→V
Y→X
Y→Z
Database-Management-SystemFunctional-DependencyGATE 2017 [Set-1]Video-Explanation
Question 9 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 10

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-SystemFunctional-DependencyGATE 2015 [Set-3]Video-Explanation
Question 10 Explanation: 
X→Y is trivial if Y⊆ X
Question 11

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-SystemFunctional-DependencyGATE 2014 [Set-1]Video-Explanation
Question 11 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 12

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-SystemFunctional-DependencyGATE 2005
Question 12 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 13

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-SystemFunctional-DependencyGATE 2006
Question 13 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 14
A functional dependency is a relationship between:
A
Tables
B
Attributes
C
Rows
D
Relations
Database-Management-SystemFunctional-DependencyTNPSC-2017-Polytechnic-CS
Question 14 Explanation: 
A functional dependency is a relationship between attributes.
Question 15
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-SystemFunctional-DependencyTNPSC-2017-Polytechnic-CS
Question 15 Explanation: 
(BG)+ = BGACD
(CF)+ = CFGEAD
(AB)+ = ABCDG
(AF)+ = AFDE
Hence option (D) is False.
Question 16
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-SystemFunctional-DependencyCIL Part - B
Question 16 Explanation: 
Question 17
The 9’s compliment of the decimal number 139452 is:
A
971658
B
971659
C
860547
D
860548
Database-Management-SystemFunctional-DependencyCIL Part - B
Question 17 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 18
Functional dependencies are a generalization of
A
Key dependencies
B
Relation dependencies
C
Database dependencies
D
None of these
Database-Management-SystemFunctional-DependencyTNPSC-2012-Polytechnic-CS
Question 18 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.
There are 18 questions to complete.

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