FunctionalDependency
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?
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 
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.
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 
{P,R}→{S,T}  
{P,R}→{R,T}  
{P,S}→{S}  
{P,S,U}→{Q} 
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?
{E,F}  
{E,F,H}  
{E,F,H,K,L}  
{E} 
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
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 
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

Question 4 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 decomposedinto (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).
(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 5 
{CF}^{+} = {ACDEFG}  
{BG}^{+} = {ABCDG}
 
{AF}^{+} = {ACDEFG}  
{AB}^{+} = {ABCDFG}

Question 5 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} (❌)
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 6 
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 
Question 6 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.
So only option (D) contains E and H as part of candidate key.
Question 7 
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 
Question 7 Explanation:
From the given instanceof 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
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 8 
XY → Z and Z → Y  
YZ → X and Y → Z  
YZ → X and X → Z  
XZ → Y and Y → X 
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].
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 for R?
CD  
EC  
AE  
AC 
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.
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 
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 
Question 10 Explanation:
If the given realations are to be lossless then
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.
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.
There are 10 questions to complete.