UGC NET CSE | August 2024 | Part 2 | Question: 29

Question

Consider a relation schema $\mathrm{R}=(\mathrm{U}, \mathrm{V}, \mathrm{W}, \mathrm{X}, \mathrm{Y}, \mathrm{Z})$, on which the following functional dependencies hold:

$$\{\mathrm{U} \rightarrow \mathrm{~V}, \mathrm{VW} \rightarrow \mathrm{X}, \mathrm{Y} \rightarrow \mathrm{~W} ; \mathrm{X} \rightarrow \mathrm{U}\} $$

The candidate keys of $\mathrm{R}$ are:

• A. $\mathrm{UY, ~VY}$
• B. $\mathrm{UY, ~VY, ~XY}$
• C. $\mathrm{UYZ, ~VYZ, ~VWZ}$
• D. $\mathrm{UYZ, ~VYZ, ~XYZ}$

Answer 1

Ans is : D
 

YZ attributes are not present on the RHS of any functional dependency, so YZ has to be the part of candidate keys.
we can eliminate the options A and B , because no YZ are present. and Option C is eliminated because CKs, UYZ,VYZ Contains YZ attributes, but VWZ does not contains Y, so we can eliminate the VWZ and hence ans is D.
 

Answer 2

Just solve for closure to get your first candidate key. 
$$ U\cancel{V}\cancel{W}\cancel{X}YZ^* = \left \{ U, V, W, X, Y, Z \right \} $$ 

1. check if any subset of candidate key determines all the attribute: No
2. check if any candidate key attributes on RHS: Yes

Check and solve again, you will get option (d) as correct answer.

Answer 3

The correct answer is D. The reason is that the attributes Y and Z do not appear on the right-hand side of any given functional dependency. Therefore, they cannot be derived from any other attributes and must be included in every candidate key.

Since Y and Z can only determine themselves ( {Y, Z} → {Y, Z}), excluding them from a key would mean that not all attributes can be derived. In that case, the set would fail to qualify even as a superkey.

Because a candidate key is defined as a minimal superkey, Y and Z must be part of every candidate key. None of the other given options except D include both Y and Z, so they can be eliminated.