Question

2. Given R(x,y, z, w, k, t). There are two keys: (x,y) and z. Given the following functional dependency: F = { {x,y}  {z,w,k,t}, z  {x,y,w,k,t }, yt}. Is R in 2nd normal form? Justify your answer.

3. Given R(x,y, z, w, k, t). There are two keys: (x,y) and z. Given the following functional dependency: F = { fd1:{x,y}  {z,w,k,t}, fd2: z  {x,y,w,k,t }, fd3:k x}. Is R in 3rd normal form? Justify your answer.

4. Given R(x,y, z, w, k, t). There are two keys: (x,y) and z. Given the following functional dependency: F = { fd1:{x,y}  {z,w,k,t}, fd2: z  {x,y,w,k,t }, fd3: w y}. Is R in BCNF? Justify your answer.

Answer 1

2. R (x,y, z, w, k, t).

Given FDs

{x,y} -> {z,w,k,t}

y->t

A Relation R is in second normal form if following conditions hold:

1.The relation R is in first normal form.

2.No non-prime attribute is dependent on proper subset of any candidate key. (Partial dependency)

Since there is a partial dependency, (ie. in y->t, y is a prime attribute.) , it violates 2NF.

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(3) Given fd1:   {x,y} -> {z,w,k,t},

            fd2:   z -> {x,y,w,k,t },

                fd3:   k-> x

A Relation R is in third normal form if following conditions hold:

1.The relation R is in second normal form.

2.Every non-prime attribute of R is non-transitively dependent on every key of R.

Since there is no partial dependency, it is in 2NF and every non prime attributes ie. w, k and t are non transitively depend on the keys of R, the relation is in 3 NF.

4.   Given fd1:{x,y} -> {z,w,k,t},

                 fd2: z -> {x,y,w,k,t },

                 fd3: w -> y

BCNF is a strict form of 3NF.

For a table to satisfy the Boyce-Codd Normal Form, it should satisfy the following two conditions:

1. It should be in the Third Normal Form.
2. And, for any dependency A → B, A should be a super key.

Here xy and z are super keys but w is not a super key. so the relation is not in BCNF.