1. 1 CMPT 275 Phase: Design

2. Janice Regan, 2008 2 Map of design phase DESIGN HIGH LEVEL DESIGN Modularization User Interface Module Interfaces Data Persistance Subsystem User Manual architecture LOW LEVEL DESIGN Classes Class Interfaces Interaction Diagrams Implementation

3. 3 Implementation issues related to Data Persistence NORMALIZATION

4. Janice Regan, 2008 4 Relational DB Design  We will structure our relational database table(s) using Normalization  process of assigning attributes to tables  series of stages called Normal Forms  1 st normal form: fixed length records  2 nd normal form: remove partial dependencies  3 rd normal form: remove transitive dependencies

5. Janice Regan, 2008 5 Relational DB Design  We will structure our relational database table(s) using Normalization  Advantage:  assures equal length records  reduces data redundancies hence helps eliminate problems that result from redundancies  Disadvantage:  decrease performance as we normalize to higher forms, higher forms require more tables

6. Janice Regan, 2008 6 An example…  To illustrate how to normalize, we shall use the example of a Student Registration System. Here are some requirements: 1. For each student, we need to remember: student-id, name, address, phone, courses 2. For each of the course taken, must remember: credit, semester, grade, room, instructor’s office, instructor.

7. Janice Regan, 2008 7 An example…  To illustrate how to normalize, we shall use the example of a Student Registration System. Here are some requirements: 3. Students can repeat same course in a later semester 4. There is only one offering a a given course in a semester 5. For each of the course attempted, must remember: semester, grade, room, instructor, instructor's office

8. Janice Regan, 2008 8 OO Classes: Class diagram StudentCourse offering Receives a grade for takes Student ID name address phone Course name semester room Instructor Instructor’s office List of courses List of students 0..* Course Name credit 0..* 1 List of grades

9. Janice Regan, 2008 9 Our First Table  From our requirements, we could create the following database table: (horizontal lines separate records, representing single student objects) Instructor’s office Std-id Std-nameStd-address Std-phone CoursenameSem Grade Credit Room Instructor 15438 25636 47352 21544 Paul K. Will B. Kim L. Xiao T. Brook St., Bby Elf Ave., Van. Mer Cr., Poco Alpha St., Bby 294.2563 256.2453 939.2766 295.9976 Cmpt 101 Cmpt 150 Bus 152 Engl 102 Biol 234 Cmpt 354 03-2 03-1 03-2 03-1 03-2 03-1 03-2 C- B A- A+ B+ D A- 43323334332333 AQ2 AQ1 ASB WM EDC AQ1 AQ2 ASB985 ASB352 WM543 AQ834 EDC243 ASB985 ASB111 Dr. Klaus M. Nole V. Karu W. Loti Dr. Quel Dr. Klaus Dr. Yu Each record is uniquely identified by the student number (Std-id). The primary key for the table is therefore Std-id. This table is unnormalized (contains records of varying length)

10. Janice Regan, 2008 10  So far, attributes are in an unnormalized form.  Objects, transformed into DB records, will not all be of same length. Each record contains all information about one student. student-id name address phone course credit semester grade room instructor instructor’s office Group of attributes repeated each time a particular course is attempted by 1 student Group of attributes repeated for each course taken by 1 student Our First Table: Is there a problem? NOT ALL RECORDS ARE OF THE SAME LENGTH !!

11. Janice Regan, 2008 11 What is the problem?  Problem: attributes that are lists (multiple courses per student, multiple attempts per course) do not produce fixed length records  Solution: remove lists by adding additional rows one to hold each attribute in the list  Consider the example: for a student, have 1 complete row per course attempted/taken  This results in a table in First Normal Form

12. Janice Regan, 2008 12 First Normal Form  Definition of First Normal Form (1NF):  Tables do not have repeating groups, i.e., each row/column intersection can contain one and only one value, not a set of values.  All the key attributes are defined, no blank (null) values of keys are permitted

13. Janice Regan, 2008 13 Our example Defining primary key attributes  Which attributes are needed to assure each record is uniquely identified  Std-Id is not enough  a student can take multiple courses  Std-id and course name is not enough  a student can take the same course more than once if they wish  Std-id, course name and semester is enough  Each time the student takes a course it is uniquely identified as a single record in the table

14. Janice Regan, 2008 14 Std-phone First Normal Form of Table (1NF) Std-id Std-name Std-address Course-nameSemester Grade Credit Room Instructor Instructor’s office 15438 25636 47352 21544 Paul K. Will B. Kim L.Xiao T. Brook St., Bby Elf Ave., Van. Merry Cr., Poco Alpha St., BBY Alpha St., Bby 294.2563 256.2453 939.2766 295.9976 295-9976 Cmpt 101 Cmpt 150 Bus 152 Engl 102 Biol 234 Cmpt 354 03-2 03-1 03-2 03-1 03-2 03-1 03-2 C- B A- A+ B+ D A- 43323334332333 AQ2 AQ1 ASB WM EDC AQ1 AQ2 ASB985 ASB352 WM543 AQ834 EDC243 ASB 985 ASB111 Dr. Klaus M. Nole V. Karu W. Loti Dr. Quel Dr. Klaus Dr. Yu  Result: Single table with compound (multi-attribute) primary key.  Primary Key: each row uniquely identified by one single attribute  Compound Primary Key: each row uniquely identify by a or group of attributes  The compound primary key for the above table is: Std-id, Course-name, Semester

15. Janice Regan, 2008 15 Is there still a problem? Y es!  Our table in First Normal Form could still contain data redundancies due to partial dependencies.  Partial dependencies are based only on a part of the compound primary key.  Consider an attribute A, that is dependent on the compound primary key K  If A is dependent on all components of the compound primary key the A is fully dependent on K  If A is dependent on some but not all of the components of the primary key then A is partially dependent on K

16. Janice Regan, 2008 16 Redundancy: Examples + problems  Examples of redundancy and partial dependence:  For each course a student takes the student’s name, address and phone number are repeated. A student’s name and address are dependent on the student’s id but not on the course name or semester  For each course a student repeats the course credit is repeated. The course credit is dependent on the course name but not the student’s id or the semester

17. Janice Regan, 2008 17 Problems related to Redundancy  Redundancy  Insert anomalies: e.g. Each time a student takes a course the student information must be entered, this adds to the potential for error  Delete anomalies: if delete the row where info about Std-id 47352 is stored will also delete info that cannot be found anywhere else in DB table namely that Dr. Quel’s office is EDC243  Update problems: because of redundant data, if a student moves, need to change student's address in all rows corresponding to every instance of every course the student had ever taken. Problems occur if one occurrence is missed or an error is made in one occurrence

18. Janice Regan, 2008 18 Partial Dependencies  Definition: non-key attribute(s) dependent on only some of primary key(s)  Examples:  Phone #, Std-name, and address depend only on Std- id (not course name or semester)  Credit depends only on course name (not on Std-id or semester)  Instructor, Instructor’s office, room, and grade depend on course and semester (not Std-id)

19. Janice Regan, 2008 19 Partial Dependencies  Definition: non-key attribute(s) dependent on only some of primary key(s)  When an attribute is only partially dependent on the primary keys of the table there may be redundant occurrences of that attribute in the table  Therefore, To remove redundancies we should remove partial dependencies

20. Janice Regan, 2008 20 Problem with 1NF  non-key attribute(s) may depend on some but not all of the primary key(s)  e.g.: primary keys are Std_id, Course- name and semester address depends only on Std-id

21. Janice Regan, 2008 21 From 1NF to 2NF  Solution: Remove partial dependencies  Determine if there are any partial dependencies.  If so, divide 1NF table into several tables such that in each table each non-primary key attribute is dependent on only the primary key (or compound primary key) of that table.  Note that if primary key of 1NF table is not a compound primary key, there cannot be partial dependencies and hence the 1NF table is already in 2NF.

22. Janice Regan, 2008 22 Second Normal Form - Example Transform our 1NF table into a 2NF table  STEP 1: We determine dependencies on single primary key:  Std-id Phone #, Std-name, Std-address  Course-name credit  Semester none dependent only on semester

23. Janice Regan, 2008 23 2NF Example - Step 1  DB tables look like: Std-id Std-name Std-address Std-phone 15438 Paul K. Brook St. Bby 294.2563 21544 Xiao T. Alpha St., Bby 295.9976 25636 Will B. Elf Ave., Van. 256.2453 Student Table 47352 Kim L. Merry Cr., Poco 939.2766 Course Table Course-name credit Cmpt 101 4 Cmpt 354 3 Engl 102 2 Bus 152 3 Cmpt 150 3 Biol 234 3

24. Janice Regan, 2008 24 Second Normal Form - Example  STEP 2: We determine dependencies on pairs of primary keys:  Course-name + Semester room, instructor, instructor’s office  Course-name + Std-id none  Semester + Std-id none

25. Janice Regan, 2008 25 Course-name Semester Room Instructor Instructor’s office 2NF Example - Step 2 Cmpt 10103-2 AQ2 Dr. Klaus ASB985 Cmpt 150 03-1 AQ1 M. Nole ASB352 Bus 152 03-2 ASB V. Karu WM543 Engl 102 03-1 WM W. Loti AQ834 Course Offering Table Biol 234 03-2 EDC Dr. Quel EDC243 Cmpt 354 03-1 AQ1 Dr. Klaus ASB985 Cmpt 354 03-2 AQ2 Dr. Yu ASB111

26. Janice Regan, 2008 26 Second Normal Form - Example Step 3  We determine dependencies on whole compound primary key:  Course-name + Semester + Std-id grade

27. Janice Regan, 2008 27 2NF Example - Step 3 Student Registration Table Std-id Course-name Semester Grade 15438 25636 47352 21544 Cmpt 101 Cmpt 150 Bus 152 Engl 102 Biol 234 Cmpt 354 03-2 03-1 03-2 03-1 03-2 03-1 03-2 C- B A- A+ B+ D A-

28. Janice Regan, 2008 28 2NF Example, alternate Step 3-1 Introducing Association  Looking at the data model (class diagram), we can recognize the “many-to-many” multiplicity relationship between Student, grade and CourseOffering StudentCourse offering Receives a grade fo r takes Student ID name address phone Course name semester List of grades room Instructor Instructor’s office List of courses List of students Course Name credit 0..* 1  These 3 attributes are used to implement “many-to-many” multiplicity relationships

29. Janice Regan, 2008 29 Association Class  However, this data model does not lead to DB tables with records of fixed length because these 3 attributes are of varying size for each object of Student and Course Offering class types, so…  … we introduce yet another “class” that associates 1 student to many attempts at (registrations to) one course and 1 course to many attempts (registrations) per 1 student. For each of these attempts there is one grade

30. Janice Regan, 2008 30 Association Class  An association class takes a many-many relation and breaks it into two 1-many relationships  Student and Student Registration have a “1-to-many” multiplicity relationship  Course Offering and Student Registration have a “1-to- many” multiplicity relationship  The association class will contain the attributes that are lists (that cause the many to may relationship)  The association class wil contain the attributes that depend upon all the variables (lists) in the association class.

31. Janice Regan, 2008 31 2NF Example, alternate Step 3 - 2  This relationship can be broken down into 2 “1-to-many” multiplicity relationships by creating an association class Student-Registration StudentCourse offering Student ID name address phone Course name semester room Instructor Instructor’s office Course Name credit 0..* 1 Student Registration Course name semester grade 0..* Student ID 1 1 0..*

32. Janice Regan, 2008 32 2NF Example, alternate Step 3 - 3  We can therefore store the attributes that depend on this association into a Student Registration table. The compound primary key of this table is the union of the primary keys of the Student and the Course Offering tables: C- B A- A+ B+ D A- Student Registration Table Std-id Course-name Semester Grade 15438 25636 47352 21544 Cmpt 101 Cmpt 150 Bus 152 Engl 102 Biol 234 Cmpt 354 03-2 03-1 03-2 03-1 03-2 03-1 03-2 Std-id Std-name Std-address Std-phone 21544 Xiao T. Alpha St., Bby 295.9976 25636 Will B. Elf Ave., Van. 256.2453 Student Table 47352 Kim L. Merry Cr., Poco 939.2766 Course-name Semester Room Instructor Instructor’s office Cmpt 10103-2 AQ2 Dr. Klaus ASB985 Cmpt 150 03-1 AQ1 M. Nole ASB352 Bus 152 03-2 ASB V. Karu WM543 Engl 102 03-1 WM W. Loti AQ834 Course Offering Table Biol 234 03-2 EDC Dr. Quel EDC243 Cmpt 354 03-1 AQ1 Dr. Klaus ASB985 Cmpt 354 03-2 AQ2 Dr. Yu ASB111 Course Table Course-name credit Cmpt 101 4 Cmpt 354 3 Engl 102 2 Bus 152 3 Cmpt 150 3 Biol 234 3

33. Janice Regan, 2008 33 Second Normal Form  To get Student Registration System in 2NF we need 4 tables (files)  Multiplicities come from our Requirement Analysis phase  With this 2NF DB, students do not have to register to a course to be admitted to an institution  Room and instructor for a course offering can be entered even if there are no students registered yet  Less redundancy: Most update problems have been eliminated, but we can still have multiple occurrences of instructor and instructor’s office

34. Janice Regan, 2008 34 Second Normal Form  Definition of 2NF:  The table is in 1NF  The table includes no partial dependencies

35. Janice Regan, 2008 35 Is there still a problem? Yes!  Our tables in 2NF could still contains data redundancies due to transitive dependencies.  When one non-primary key attribute is dependent on another non-primary key attribute, the second non-primary key attribute is transitively dependent on the first non-primary key attribute.

36. Janice Regan, 2008 36 Transitive Dependencies: example  instructor’s office (non-primary key attribute) is transitively dependent on instructor (another non- primary key attribute) but not on any of the primary key attributes for that particular table (course and/or semester)  Solution: Conversion from 2NF to 3NF  Determine the transitive dependencies.  Split 2NF table containing the transitive dependency such that the dependency is represented by its own table.

37. Janice Regan, 2008 37 3NF Example C- B A- A+ B+ D A- Student Registration Table Std-id Course-name Semester Grade 15438 25636 47352 21544 Cmpt 101 Cmpt 150 Bus 152 Engl 102 Biol 234 Cmpt 354 03-2 03-1 03-2 03-1 03-2 03-1 03-2 Std-id Std-name Std-address Std-phone 21544 Xiao T. Alpha St., Bby 295.9976 25636 Will B. Elf Ave., Van. 256.2453 Student Table 47352 Kim L. Merry Cr., Poco 939.2766 Course-name Semester Room Instructor Cmpt 10103-2 AQ2 Dr. Klaus Cmpt 150 03-1 AQ1 M. Nole Bus 152 03-2 ASB V. Karu Engl 102 03-1 WM W. Loti Course Offering Table Biol 234 03-2 EDC Dr. Quel Cmpt 354 03-1 AQ1 Dr. Klaus Cmpt 354 03-2 AQ2 Dr. Yu Course Table Course-name credit Cmpt 101 4 Cmpt 354 3 Engl 102 2 Bus 152 3 Cmpt 150 3 Biol 234 3 Dr. Klaus ASB985 W. Loti AQ834 Dr. Quel EDC243 Dr. Yu ASB111 M. Nole ASB352 V. Karu WM543 Instructor Instructor’s office Instructor Table

38. Janice Regan, 2008 38 Third Normal Form  Definition:  Every table is in 2NF.  There are no transitive dependencies.

39. Janice Regan, 2008 39 Normalization Summary  When normalizing, we seek to make sure that attributes depend  on the key (1NF)  on the whole key (2NF)  on nothing but the key (3NF)  When normalized:  records have fixed length  no insert/delete/update anomalies  minimize redundancy