NVS PGT CS 2019 Part-B
October 5, 2023NIC-NIELIT Scientist-B 2020
October 5, 2023Functional-Dependency
| 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?
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 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.
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.
Correct Answer: A
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.
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.
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