Normalization - solved exercises - Normal forms 1

Set of solved exercises in Normalization / Normalization Solved Examples / How to find candidate keys, and primary keys in database? / Sets of examples to find the keys of a tables / Process of Key finding in a database - Examples / Normalization to 1NF, 2NF, 3NF

Let us assume a table User_Personal as given below;

UserID	U_email	Fname	Lname	City	State	Zip
MA12	Mani@ymail.com	MANISH	JAIN	BILASPUR	CHATISGARH	458991
PO45	Pooja.g@gmail.co	POOJA	MAGG	KACCH	GUJRAT	832212
LA33	Lavle98@jj.com	LAVLEEN	DHALLA	RAIPUR	CHATISGARH	853578
CH99	Cheki9j@ih.com	CHIMAL	BEDI	TRICHY	TAMIL NADU	632011
DA74	Danu58@g.com	DANY	JAMES	TRICHY	TAMIL NADU	645018

• Is this table in First Normal Form?

Yes. All the attributes contain only atomic values.

• Is this table in Second Normal Form?

To verify this property, we need to find all the functional dependencies which are holding in User_Personal table, and have to identify a Primary key.

Let us do that by using the sample data. This leads to the following set of FDs;

F = { UserID →U_email Fname Lname City State Zip, 
Zip → City State, 
City →Zip State }

As UserID attribute can uniquely determine all the other attributes, we can have UserID as the Primary key for User_Personal table.

The next step is to check for the 2NF properties;

Property 1 – The table should be in 1NF.

Property 2 – There should not be any partial key dependencies.

Our table is in 1NF, hence property 1 is holding.

Primary key of our table is UserID and UserID is single simple attribute. As the key is not composite, there is no chance for partial key dependency to hold. Hence property 2 is also holding.

User_Personal table is in 2NF.

• Is User_Personal in 3NF?

To verify this we need to check the 3NF properties;

Property 1 – Table should be in 2NF.

Property 2 – There should not be any Transitive Dependencies in the table.

Table User_Personal is in 2NF, hence property 1 is satisfied.

User_Personal table holds the following Transitive dependency;

UserID →Zip, Zip →City State

Hence, property 2 is not satisfied and the table is not in 3NF.

Solution:

Decompose User_Personal. For this, we can use the functional dependencies Zip →City State and UserID →U_email Fname Lname City State Zip.

As a result, we can have the following tables (primary keys are underlined);

User_Personal (UserID, U_email, Fname, Lname, Zip)

City (Zip, City, State)

UserID	U_email	Fname	Lname	Zip
MA12	Mani@ymail.com	MANISH	JAIN	458991
PO45	Pooja.g@gmail.co	POOJA	MAGG	832212
LA33	Lavle98@jj.com	LAVLEEN	DHALLA	853578
CH99	Cheki9j@ih.com	CHIMAL	BEDI	632011
DA74	Danu58@g.com	DANY	JAMES	645018

Table - User_Personal

Zip	City	State
458991	BILASPUR	CHATISGARH
832212	KACCH	GUJRAT
853578	RAIPUR	CHATISGARH
632011	TRICHY	TAMIL NADU
645018	TRICHY	TAMIL NADU

Table – City

Both tables are in 3NF.

Hence, tables are normalized to Third Normal Form.