Database Normalization 1NF, 2NF, 3NF, BCNF, 4NF, 5NF
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The document discusses normalization in database design. Normalization is the process of organizing data to avoid redundancy and dependency. It involves splitting tables and restructuring relationships between tables. The document outlines various normal forms including 1NF, 2NF, 3NF, BCNF, 4NF and 5NF and provides examples to illustrate how to normalize tables to conform to each form.
Database Normalization 1NF, 2NF, 3NF, BCNF, 4NF, 5NF
- 1. Normalization Oum Saokosal Master's Degree in Information Systems, South Korea 012-252-752 / 010-878-992 oumsaokosal@gmail.com 1
- 2. Normalization The biggest problem needed to be solved in database is data redundancy. Why data redundancy is the problem? Because it causes: Insert Anomaly Update Anomaly Delete Anomaly 2 Teacher Subject Teacher Degree Tel Sok San Database Master's 012666777 Van Sokhen Database Bachelor's 017678678 Sok San E-Commerce Master's 012666777
- 3. Normalization (Cont.) Normalization is the process of removing redundant data from your tables to improve storage efficiency, data integrity, and scalability. Normalization generally involves splitting existing tables into multiple ones, which must be re-joined or linked each time a query is issued. Why normalization? The relation derived from the user view or data store will most likely be unnormalized. The problem usually happens when an existing system uses unstructured file, e.g. in MS Excel. 3
- 4. Steps of Normalization First Normal Form (1NF) Second Normal Form (2NF) Third Normal Form (3NF) Boyce-Codd Normal Form (BCNF) Fourth Normal Form (4NF) Fifth Normal Form (5NF) In practice, 1NF, 2NF, and 3NF are enough for database. 4
- 5. First Normal Form (1NF) The official qualifications for 1NF are: 1. Each attribute name must be unique. 2. Each attribute value must be single. 3. Each row must be unique. 4. There is no repeating groups. Additional: Choose a primary key. Reminder: A primary key is unique, not null, unchanged. A primary key can be either an attribute or combined attributes. 5
- 6. First Normal Form (1NF) (Cont.) Example of a table not in 1NF : It violates the 1NF because: Attribute values are not single. Repeating groups exists. 6 Group Topic Student Score Group A Intro MongoDB Sok San 18 marks Sao Ry 17 marks Group B Intro MySQL Chan Tina 19 marks Tith Sophea 16 marks
- 7. First Normal Form (1NF) (Cont.) After eliminating: Now it is in 1NF. However, it might still violate 2NF and so on. 7 Group Topic Family Name Given Name Score A Intro MongoDB Sok San 18 A Intro MongoDB Sao Ry 17 B Intro MySQL Chan Tina 19 B Intro MySQL Tith Sophea 16
- 8. Functional Dependencies 8 We say an attribute, B, has a functional dependency on another attribute, A, if for any two records, which have the same value for A, then the values for B in these two records must be the same. We illustrate this as: A B (read as: A determines B or B depends on A) employee name email address employee name project email address Sok San POS Mart Sys soksan@yahoo.com Sao Ry Univ Mgt Sys sao@yahoo.com Sok San Web Redesign soksan@yahoo.com Chan Sokna POS Mart Sys chan@gmail.com Sao Ry DB Design sao@yahoo.com
- 9. Functional Dependencies (cont.) 9 EmpNum EmpEmail EmpFname EmpLname 123 jdoe@abc.com John Doe 456 psmith@abc.com Peter Smith 555 alee1@abc.com Alan Lee 633 pdoe@abc.com Peter Doe 787 alee2@abc.com Alan Lee If EmpNum is the PK then the FDs: EmpNum EmpEmail, EmpFname, EmpLname must exist.
- 10. Functional Dependencies (cont.) 10 EmpNum EmpEmail, EmpFname, EmpLname EmpNum EmpEmail EmpFname EmpLname EmpNum EmpEmail EmpFname EmpLname 3 different ways you might see FDs depicted
- 11. Determinant 11 Functional Dependency EmpNum EmpEmail Attribute on the left hand side is known as the determinant • EmpNum is a determinant of EmpEmail
- 12. Second Normal Form (2NF) The official qualifications for 2NF are: 1. A table is already in 1NF. 2. All nonkey attributes are fully dependent on the primary key. All partial dependencies are removed to place in another table. 12
- 13. CourseID SemesterID Num Student Course Name IT101 201301 25 Database IT101 201302 25 Database IT102 201301 30 Web Prog IT102 201302 35 Web Prog IT103 201401 20 Networking Example of a table not in 2NF: Primary Key The Course Name depends on only CourseID, a part of the primary key not the whole primary {CourseID, SemesterID}.It’s called partial dependency. Solution: Remove CourseID and Course Name together to create a new table. 13
- 14. CourseID Course Name IT101 Database IT101 Database IT102 Web Prog IT102 Web Prog IT103 Networking Semester Done? Oh no, it is still not in 1NF yet. Remove the repeating groups too. Finally, connect the relationship. CourseID Course Name IT101 Database IT102 Web Prog IT103 Networking CourseID SemesterID Num Student IT101 201301 25 IT101 201302 25 IT102 201301 30 IT102 201302 35 IT103 201401 20 14
- 15. Third Normal Form (3NF) The official qualifications for 3NF are: 1. A table is already in 2NF. 2. Nonprimary key attributes do not depend on other nonprimary key attributes (i.e. no transitive dependencies) All transitive dependencies are removed to place in another table. 15
- 16. StudyID Course Name Teacher Name Teacher Tel 1 Database Sok Piseth 012 123 456 2 Database Sao Kanha 0977 322 111 3 Web Prog Chan Veasna 012 412 333 4 Web Prog Chan Veasna 012 412 333 5 Networking Pou Sambath 077 545 221 Example of a Table not in 3NF: Primary Key The Teacher Tel is a nonkey attribute, and the Teacher Name is also a nonkey atttribute. But Teacher Tel depends on Teacher Name. It is called transitive dependency. Solution: Remove Teacher Name and Teacher Tel together to create a new table. 16
- 17. Teacher Name Teacher Tel Sok Piseth 012 123 456 Sao Kanha 0977 322 111 Chan Veasna 012 412 333 Chan Veasna 012 412 333 Pou Sambath 077 545 221 Done? Oh no, it is still not in 1NF yet. Remove Repeating row. Teacher Name Teacher Tel Sok Piseth 012 123 456 Sao Kanha 0977 322 111 Chan Veasna 012 412 333 Pou Sambath 077 545 221 Note about primary key: -In theory, you can choose Teacher Name to be a primary key. -But in practice, you should add Teacher ID as the primary key. ID Teacher Name Teacher Tel T1 Sok Piseth 012 123 456 T2 Sao Kanha 0977 322 111 T3 Chan Veasna 012 412 333 T4 Pou Sambath 077 545 221 StudyID Course Name T.ID 1 Database T1 2 Database T2 3 Web Prog T3 4 Web Prog T3 5 Networking T4 17
- 18. Boyce Codd Normal Form (BCNF) – 3.5NF The official qualifications for BCNF are: 1. A table is already in 3NF. 2. All determinants must be superkeys. All determinants that are not superkeys are removed to place in another table. 18
- 19. Boyce Codd Normal Form (BCNF) (Cont.) Example of a table not in BCNF: Key: {Student, Course} Functional Dependency: {Student, Course} Teacher Teacher Course Problem: Teacher is not a superkey but determines Course. 19 Student Course Teacher Sok DB John Sao DB William Chan E-Commerce Todd Sok E-Commerce Todd Chan DB William
- 20. 20 Student Course Sok DB Sao DB Chan E-Commerce Sok E-Commerce Chan DB Course Teacher DB John DB William E-Commerce Todd Course DB E-Commerce Solution: Decouple a table contains Teacher and Course from from original table (Student, Course). Finally, connect the new and old table to third table contains Course.
- 21. Forth Normal Form (4NF) The official qualifications for 4NF are: 1. A table is already in BCNF. 2. A table contains no multi-valued dependencies. Multi-valued dependency: MVDs occur when two or more independent multi valued facts about the same attribute occur within the same table. A B (B multi-valued depends on A) 21
- 22. Forth Normal Form (4NF) (Cont.) Example of a table not in 4NF: Key: {Student, Major, Hobby} MVD: Student Major, Hobby 22 Student Major Hobby Sok IT Football Sok IT Volleyball Sao IT Football Sao Med Football Chan IT NULL Puth NULL Football Tith NULL NULL
- 23. 23 Student Major Sok IT Sao IT Sao Med Chan IT Puth NULL Tith NULL Student Hobby Sok Football Sok Volleyball Sao Football Chan NULL Puth Football Tith NULL Student Sok Sao Chan Puth Tith Solution: Decouple to each table contains MVD. Finally, connect each to a third table contains Student.
- 24. Fifth Normal Form (5NF) The official qualifications for 5NF are: 1. A table is already in 4NF. 2. The attributes of multi-valued dependencies are related. 24
- 25. Fifth Normal Form (5NF) (Cont.) Example of a table not in 5NF: Key: {Seller, Company, Product} MVD: Seller Company, Product Product is related to Company. 25 Seller Company Product Sok MIAF Trading Zenya Sao Coca-Cola Corp Coke Sao Coca-Cola Corp Fanta Sao Coca-Cola Corp Sprite Chan Angkor Brewery Angkor Beer Chan Cambodia Brewery Cambodia Beer
- 26. 26 Seller Company Sok MIAF Trading Sao Coca-Cola Corp Chan Angkor Brewery Chan Cambodia Brewery Seller Product Sok Zenya Sao Coke Sao Fanta Sao Sprite Chan Angkor Beer Chan Cambodia Beer Company Product MIAF Trading Zenya Coca-Cola Corp Coke Coca-Cola Corp Fanta Coca-Cola Corp Sprite Angkor Brewery Angkor Beer Cambodia Brewery Cambodia Beer Seller Sok Sao Chan Company MIAF Trading Coca-Cola Corp Angkor Brewery Cambodia Brewery Product Zenya Coke Fanta Sprite Angkor Beer Cambodia Beer 1 1 1 1 1 1 M M M M M