FunctionalDependency
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?
CD  
EC  
AE  
AC 
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?
XY → Z and Z → Y  
YZ → X and Y → Z  
YZ → X and X → Z  
XZ → Y and Y → X 
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 R_{1}(AB) and R_{2}(CD) is
dependency preserving and lossless join  
lossless join but not dependency preserving  
dependency preserving but not lossless join  
not dependency preserving and not lossless join 
R_{1}∩R_{2} ≠ 0
Given R_{1}(A,B), R_{2}
R_{1}∩R_{2} = 0
Not lossless.
The given relation decomposed into R_{1}(A,B) and R_{2}(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 functionally determines B and B functionally determines C  
A functionally determines B and B does not functionally determines C  
B does not functionally determines C  
A does not functionally determines B and B does not functionally determines C 
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 
P⟶ QR
RS⟶ T
Which of the following functional dependencies can be inferred from the above functional dependencies?
PS ⟶ Q  
PS ⟶ T  
R ⟶ T  
P ⟶ R 
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?
VXZ  
VXY  
VWXY  
VWXYZ 
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)
gives a lossless join, and is dependency preserving  
gives a lossless join, but is not dependency preserving  
does not give a lossless join, but is dependency preserving  
does not give a lossless join and is not dependency preserving

(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 
AB → C; BC → D; C → E;
The number of superkeys in the relation R is _____________.
8 
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 
V → W
VW → X
Y → VX
Y → Z
Which of the following is irreducible equivalent for this set of functional dependencies?
V→W V→X Y→V Y→Z  
V→W W→X Y→V Y→Z  
V→W V→X Y→V Y→X Y→Z  
V→W W→X Y→V Y→X Y→Z 
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?
{P,R}→{S,T}  
{P,R}→{R,T}  
{P,S}→{S}  
{P,S,U}→{Q} 
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?
{E,F}  
{E,F,H}  
{E,F,H,K,L}  
{E} 
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?
AE, BE  
AE, BE, DE  
AEH, BEH, BCH  
AEH, BEH, DEH 
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?
{CF}^{+} = {ACDEFG}  
{BG}^{+} = {ABCDG}
 
{AF}^{+} = {ACDEFG}  
{AB}^{+} = {ABCDFG} 
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 
Tables  
Attributes  
Rows  
Relations 
Question 15 
AB → CD, AF → D, DE → F, C → G, F → E, G → AWhich one of the following options is false ?
BG^{+}={ABCDG}  
CF^{+}={ACDEFG}  
AB^{+}={ABCDG}  
AF^{+}={ACDEFG} 
(CF)^{+} = CFGEAD
(AB)^{+} = ABCDG
(AF)^{+} = AFDE
Hence option (D) is False.
Question 16 
((R1∩ R2) →(R1R2)) ∈ F^{+}  
((R1 ∩R2) → (R1  R2)) ∉ F^{+}  
((R1 ∪ R2) →(R1  R2)) ∈ F^{+}  
((R1∩R2) →R1 ) ∈ F^{+} 
Question 17 
971658  
971659  
860547  
860548 
→Subtract each and every digit of given number by digit “9”.
999999
139452
________
860547
Question 18 
Key dependencies  
Relation dependencies
 
Database dependencies  
None of these 