## 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-01
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
 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-03)
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-01)
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

 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 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).
 Question 5

 A {CF}+ = {ACDEFG} B {BG}+ = {ABCDG} C {AF}+ = {ACDEFG} D {AB}+ = {ABCDFG}
Database-Management-System       Functional-Dependency       Gate-2006
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} (❌)
 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?
 A AE, BE B AE, BE, DE C AEH, BEH, BCH D AEH, BEH, DEH
Database-Management-System       Functional-Dependency       Gate-2005
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.
 Question 7

 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 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
 Question 8

 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 for 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
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 10 Explanation:
If the given realations 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.
There are 10 questions to complete.