Question

(DATABASE QUESTION)

Which of the following is in 3NF?

(A) R(VWXYZ) FD's: V → WX; Z → VY; W → Z

(B) R(VWXYZ) FD's: V → WZ; X → V; W → XY; Z → Y

(C) R(VWXYZ) FD's: XY → Z; W → XY; Z → VW; X → V

(D) R(VWXY) FD's: XY → W; XY → V; V → X

Answer 1

Option (A) R(VWXYZ) FD's: V → WX; Z → VY; W → Z is in 3NF

A relation schema R is in 3NF if, whenever a FD X -> A holds in R, either
(a) X is a superkey of R, or
(b) A is a prime attribute of R.

A Superkey is the union of all candidate keys. All members of superkey are prime attributes

To find 3NF:

1. Find the closure of RHS of all functional dependencies
2. Determine the candidate keys
3. For all FD's, check either RHS is a subset of superkey or LHS is a prime attribute

Now consider R(VWXYZ) FD's: V → WX; Z → VY; W → Z

1. Find the closure of RHS:

V+ = {V, W, X, Y, Z}

Z+ = {V, W, X, Y, Z}

W+ = {V, W, X, Y, Z}

2. Determine the candidate keys:

candidate keys = {V,Z,W}

superkey = VZW

prime attributes = {V, Z, W}

3. For all FD's, check either RHS is a subset of superkey or LHS is a prime attribute:

• V → WX: V belongs to superkey
• Z → VY: Z belongs to superkey
• W → Z: W belongs to superkey

Since all the FD's satisfies the 3NF condition, the relation is in 3NF

Alternatively for all other options, some FD's does not satisfy the 3NF condition