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- 1. Database Chapter 3: Normalization Dr. Fatma M. Talaat
- 2. Chapter 3 Content: 1. Why Normalize? 2. Normalization 3. Dependencies 4. Normalization Example 2 Chapter 3: Normalization
- 3. 1. Why Normalize? • If your database is not normalized, it can be inaccurate, slow, inefficient. • It might not produce the data you expect, or want (update and delete anomalies). • It makes creating queries, forms, and reports are much easier to design.
- 4. 4 • Normalization: sequence of steps by which RDB is both created and improved. Advantages of Normalization: 1. Get relations in more desirable form. 2. Make database more accurate and efficient. 3. Make creating queries easier. 4. Remove duplication. 2. Normalization
- 5. 5 • Normalization: sequence of steps by which RDB is both created and improved. 2. Normalization (Normalization Flow)
- 6. • A relation is said to be in first normal form if the data is held in a two- dimensional table with each attribute represented by an atomic value. • The intersection of a row and a column must contain an indivisible value. • Each row and column position in the table there exists one value, never a set of values. • All attributes are atomic - any single attribute must not be composed of multiple attributes. 6 First Normal Form (1NF)
- 7. The data we would want to store could be expressed as: Project No Project Name Employee No Employee Name Rate category Rate 1203 Madagascar travel site 11 Jessica Brookes A £90 12 Andy Evans B £80 16 Max Fat C £70 1506 Online estate agency 11 Jessica Brookes A £90 17 Alex Branton B £80 7 First Normal Form (1NF)
- 8. • The intersection of a row and a column must contain an indivisible value Project No. Project Name Employee No. Employee Name Rate category Rate 1203 Madagascar travel site 11 Jessica Brookes A £90 1203 Madagascar travel site 12 Andy Evans B £80 1203 Madagascat travel site 16 Max Fat C £70 1506 Online estate agency 11 Jessica Brookes A £90 1506 Online estate agency 17 Alex Branton B £70 8 Project No. Project Name Employee No. Solution
- 9. Three problems become apparent with our current model: Tables in a RDBMS use a simple grid structure • All tables in an RDBMS need a key • Data entry should be kept to a minimum • Redundant data 9 First Normal Form (1NF)
- 10. 10 1NF: • Relation is said to be in 1NF if data is held in a table with each attribute is represented by atomic value. 2NF: • Relation is said to be in 2NF if: 1) It is in 1NF. 2) Remove partial dependency. Normalization (Normalization Flow)
- 11. 11 3NF: • Relation is said to be in 3NF if: 1) It is in 2NF. 2) Remove transitive dependency. Normalization (Normalization Flow)
- 12. 12 Partial dependency: when non-key attribute is determined by a part, but not all composite P.K. Transitive dependency: when non-key attribute determines another non-key attribute. 3. Dependencies
- 13. • A relation is said to be in second normal form if the relation (1) is in 1NF and (2) all attributes that are not part of the primary key are completely functionally dependent on the primary key. (Partial dependencies must be removed) • Second Normal Form (2NF) the relation must be in 1NF and each non key attribute must be fully dependent on the whole key (not a subset of the key). 13 Second normal form 2NF
- 14. Third normal form 3NF • A relation is said to be in 3NF if it (1) is in 2NF and (2) no attributes that are not part of the primary key are transitively dependent on the primary key. • The key then to move 2NF relations into 3NF is removing any transitive dependencies that may exist in the relations. 14
- 15. Dependencies: Definitions • Partial Dependency – when an non-key attribute is determined by a part, but not the whole, of a COMPOSITE primary key. 15 CUSTOMER Cust_ID Name Order_ID 101 AT&T 1234 101 AT&T 156 125 Cisco 1250 Partial Dependency
- 16. • Transitive Dependency – when a non-key attribute determines another non-key attribute. 16 EMPLOYEE Emp_ID F_Name L_Name Dept_ID Dept_Name 111 Mary Jones 1 Acct 122 Sarah Smith 2 Mktg Transitive Dependency Dependencies: Definitions
- 17. Example 1 17 • Is the table in the 0NF
- 20. 3 NF 20 . Service Place Service Type Service Date . 12/2002 3/2000 11/99 Service Type Service Place Service Type Transitive Dependency
- 21. 21 • What is meant by third normal form (3NF)? Examine the following table to check if it is in 3NF. If yes, explain your answer. Otherwise convert the table into 3NF. 4. Normalization Example Client_no CName PropertyNo Address rent_start rent_end rent ownerNo oName CR76 John kay PG4 PG16 6 st.G 5 Novar 1-Jul-00 1-Sep-02 31-Aug-01 1-Sep-02 350 450 C040 C093 Tina Tony CR56 Aline Set PG4 PG36 PG16 6 st.G 2 Manor 5 Novar 1-Sep-99 10-oct-00 1-Nov-02 10-Jun-00 1-Dec-01 1-Aug-03 350 370 450 C040 C093 C093 Tina Tony Tony
- 22. 22 Solution: 3NF: Relation is said to be in 3NF if: 1) It is in 2NF. 2) Remove transitive dependency. • It is not in 3NF. 4. Normalization Example (Ex.1): Solution
- 23. 23 1) 1NF: make each attribute is represented by atomic value. 4. Normalization Example (Ex.1): Solution Client_no CName PropertyNo Address rent_start rent_end rent ownerNo oName CR76 CR76 John kay John kay PG4 PG16 6 st.G 5 Novar 1-Jul-00 1-Sep-02 31-Aug-01 1-Sep-02 350 450 C040 C093 Tina Tony CR56 CR56 CR56 Aline Set Aline Set Aline Set PG4 PG36 PG16 6 st.G 2 Manor 5 Novar 1-Sep-99 10-oct-00 1-Nov-02 10-Jun-00 1-Dec-01 1-Aug-03 350 370 450 C040 C093 C093 Tina Tony Tony
- 24. 24 Choose Client_no + PropertyNo composite primary key 4. Normalization Example (Ex.1): Solution Client_no PropertyNo CName Address rent_start rent_end rent ownerNo oName CR76 CR76 PG4 PG16 John kay John kay 6 st.G 5 Novar 1-Jul-00 1-Sep-02 31-Aug-01 1-Sep-02 350 450 C040 C093 Tina Tony CR56 CR56 CR56 PG4 PG36 PG16 Aline Set Aline Set Aline Set 6 st.G 2 Manor 5 Novar 1-Sep-99 10-oct-00 1-Nov-02 10-Jun-00 1-Dec-01 1-Aug-03 350 370 450 C040 C093 C093 Tina Tony Tony
- 25. 25 2) 2NF: remove partial dependency Client_no CName PropertyNo Address, rent, ownerNo, oName 4. Normalization Example (Ex.1): Solution Client_no PropertyNo rent_start rent_end CR76 CR76 PG4 PG16 1-Jul-00 1-Sep-02 31-Aug-01 1-Sep-02 CR56 CR56 CR56 PG4 PG36 PG16 1-Sep-99 10-oct-00 1-Nov-02 10-Jun-00 1-Dec-01 1-Aug-03 Client_no CName CR76 John kay CR56 Aline Set
- 26. 26 3) 3NF: remove transitive dependency ownerNo oName 4. Normalization Example (Ex.1): Solution Client_no PropertyNo rent_start rent_end CR76 CR76 PG4 PG16 1-Jul-00 1-Sep-02 31-Aug-01 1-Sep-02 CR56 CR56 CR56 PG4 PG36 PG16 1-Sep-99 10-oct-00 1-Nov-02 10-Jun-00 1-Dec-01 1-Aug-03 Client_no CName CR76 John kay CR56 Aline Set PropertyNo Address rent ownerNo PG4 6 st.G 350 C040 PG16 5 Novar 450 C093 PG36 2 Manor 370 C093 ownerNo oName C040 Tina C093 Tony
- 27. 27 • Apply the various normalization steps to convert the following table into a normal form. 4. Normalization Example (Ex.2) Invoice No. Date Cust. No. Cust. Name Cust. Address Cust. City Cust. State ItemID Item Description Item.Q ty Item Price Item Total Order total price 125 9/13/2002 56 Foo, Inc. 23 Main St,thorpleb urg thorpleb urg TX 563 56"Blue Fre 4 3.50 $ 14.00 $ 82.00 $ 851 Spline End I 32 0.25 $ 8.00 $ 82.00 $ 652 3' Red Fre 5 12.00 $ 60.00 $ 82.00 $ 126 9/14/2002 2 Freens R Us 1600 Pennsylva nia Washing ton DC 563 56"Blue Fre 500 3.50 $ 1750.00 $ 10750.0 0 $ 652 3' Red Fre 750 12.00 $ 9000.00 $ 10750.0 0 $
- 28. 28 Solution: • To be in 1NF 1) 1NF: make each attribute is represented by atomic value. 4. Normalization Example (Ex.2): Solution Invoic e No. Date Cust.No . Cust. Name Cust. Address Cust. City Cust. State ItemID Item Description Item.Qt y Item Price Item Total Order total price 125 9/13/2002 56 Foo, Inc. 23 Main St,thorplebur g thorplebur g TX 563 56"Blue Fre 4 3.50 $ 14.00 $ 82.00 $ 125 9/13/2002 56 Foo, Inc. 23 Main St,thorplebur g thorplebur g TX 851 Spline End I 32 0.25 $ 8.00 $ 82.00 $ 125 9/13/2002 56 Foo, Inc. 23 Main St,thorplebur g thorplebur g TX 652 3' Red Fre 5 12.00 $ 60.00 $ 82.00 $ 126 9/14/2002 2 Freens R Us 1600 Pennsylvani a Washingto n DC 563 56"Blue Fre 500 3.50 $ 1750.00 $ 10750.00 $ 126 9/14/2002 2 Freens R Us 1600 Pennsylvani a Washingto n DC 652 3' Red Fre 750 12.00 $ 9000.00 $ 10750.00 $
- 29. 29 Choose Invoice No. + ItemID composite primary key 4. Normalization Example (Ex.2): Solution Invoice No. ItemID Date Cust.No. Cust. Name Cust. Address Cust. City Cust. State Item Description Item.Qt y Item Price Item Total Order total price 125 563 9/13/2002 56 Foo, Inc. 23 Main St,thorplebur g thorplebu rg TX 56"Blue Fre 4 3.50 $ 14.00 $ 82.00 $ 125 851 9/13/2002 56 Foo, Inc. 23 Main St,thorplebur g thorplebu rg TX Spline End I 32 0.25 $ 8.00 $ 82.00 $ 125 652 9/13/2002 56 Foo, Inc. 23 Main St,thorplebur g thorplebu rg TX 3' Red Fre 5 12.00 $ 60.00 $ 82.00 $ 126 563 9/14/2002 2 Freens R Us 1600 Pennsylvania Washingt on DC 56"Blue Fre 500 3.50 $ 1750.0 0 $ 10750.00 $ 126 652 9/14/2002 2 Freens R Us 1600 Pennsylvania Washingt on DC 3' Red Fre 750 12.00 $ 9000.0 0 $ 10750.00 $
- 30. 30 2) 2NF: remove partial dependency Invoice No. Date, Cust.No., Cust. Name, Cust. Address, Cust. City, Cust. State, Order total price ItemID Item Description, Item Price Invoice No. + ItemID Item.Qty, Item Total 4. Normalization Example (Ex.2): Solution Invoice No. ItemID Item.Qty Item Total 125 563 4 14.00 $ 125 851 32 8.00 $ 125 652 5 60.00 $ 126 563 500 1750.00 $ 126 652 750 9000.00 $ Invoice No. Date Cust.N o. Cust. Name Cust. Address Cust. City Cust. State Order total price 125 9/13/2002 56 Foo, Inc. 23 Main St,thorple burg thorple burg TX 82.00 $ 126 9/14/2002 2 Freens R Us 1600 Pennsylv ania Washi ngton DC 10750.00 $ ItemID Item Description Item Price 563 56"Blue Fre 3.50 $ 851 Spline End I 0.25 $ 652 3' Red Fre 12.00 $
- 31. 31 3) 3NF: Tables are in 3 NF as there is no transitive dependency 4. Normalization Example (Ex.2): Solution
- 32. Thanks for your attention!