GATE IT 2006 | Question: 60

Question

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 $R?$

• A. $VXZ$
• B. $VXY$
• C. $VWXY$
• D. $VWXYZ$

Comments

Can i Say by transitive property answer is B?

If we check XYV it cannot derive Z but it can derive XYVW.

XYVW can derive XYVWZ hence we can say that XYV  --> XYVW  and XYVW --> XYVWZ hence XYV --> XYVWZ?

---

Candidate key ... Minimal Super key .....

---

No need to check larger sets or supersets if a proper subset has already been determined to be a candidate key.

Answer 1

As we can see attributes  $X$ and $Y$ do not appear in the RHS of any FD and so they need to be part of any super/candidate key. So, candidate keys are: $VXY, WXY, ZXY$ as these three can determine any other attribute where as a proper subset of any of them cannot determine all other attributes.

$VXZ$ is not a super key as $Y$ is not there where as $VWXY$ and $VWXYZ$ are super keys but since their proper subsets are also super keys they are not candidate keys.

The answer is B.

Comments

V is on rhs :)

---

Edited:)

---

VWXY and VWXYZ are super keys.

Answer 2

As we can see that w z and v can be generated from the given functional dependencies and there is no way we can generate X and Y therefore X and Y must be the part of candidate key: so, xy+ = xy & vxy + = vxywz So it turns out vxy is the key, ie. option (b) and so option c and d are super keys.

Answer 3

Given Relation Schema R(V, W, X, Y, Z)

VY---->W

WX---->Z

ZY---->V

Short Trick= The attributes which are not present on RHS of a relation are part of Candidate Keys 

here WZV present on R.H.S of a Relation but X,Y not present on R.H.S So XY Must be part of CK.

Find Closure of XY = XY     

Since From XY we cannot find all attributes so check remaining attributes with XY 

•  Closure of XYV = XYVWZ           All Attributes came from XYV So it is CK

• Closure of XYW = XYWZV          All Attributes came from XYW So it is CK

• Closure of XYZ =XYZVW                           All Attributes came from XYZ So it is CK

Dont check with 4 attributes bec 4 atributes are superset of above CK So they are not CK

Now check option

a-  VXZ    WRONG        XY not present here 

b-   VXY   RIGHT 

c-  VWXY   WRONG                 it is SK not CK

d-   VWXYZ     WRONG               it is SK not CK