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Normalization Examples

Before seeing the examples please follow previous chapters: 

Find Super and Candidate Key from Attribute Closer,  2NF, 3NF, BCNF.

For more details about partial dependency and transitive dependency follow: 

2NF, 3NF

Example 4:

R (ABCDE)

AB → C, C → D, D → E, E → A, D → B}. 

Find the highest normal form.

Solution:

Now we can drive candidate key, prime attribute, and non-prime attribute.

Candidate Key (CK) = {AB, C, D, BE}, Prime Attribute (PA) = {A, B, C, D, E}

Here all attributes are prime attributes.

So, one relation consists of all attributes as prime attributes then we can say it is in 3NF.

But not in BCNF because E → A, E is not Super Key. 

So, this relation is in 3NF but not in BCNF.

Example 5:

R (ABCDEF)

{AB → C, C → DE, E → F, F → A}. 

Find the highest normal form. 

Solution:

Candidate Key = {AB, BF, BE, BC}, Non-Prime Attribute = {D}.

Answer: R (ABCDEF) is in 1NF but in 2NF.

Example 6:

R (ABCDEFGH) 

Candidate Key= {A, H, F, C}. 

Find the highest normal form.

Solution:

This relation R at least in 2NF because all the Super Key is atomic Candidate Key = {A, H, F, C} i.e. a subset of the candidate key is not possible.

For more details follow: 2NF

We can check for further 3NF.

Check for 3NF:

Candidate Key = {A, H, F, C} so, prime attributes = {A, H, F, C} and non-prime attributes = {B, D, E, G}

AB → CD is in 3NF because ‘AB’ is super key (if A is candidate key then AB must be super key).

D → EG is not in 3NF because D is not super key and EG is non-prime attributes.

So, it is not required to check other FDs.

So, this relation is in 2NF.