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CMPT 275 Phase: Design

CMPT 275 Phase: Design. Map of design phase. DESIGN. LOW LEVEL DESIGN. HIGH LEVEL DESIGN. Data Persistance Subsystem. Classes Class Interfaces Interaction Diagrams. Module Interfaces. Modularization. architecture. User Interface. User Manual. Implementation.

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CMPT 275 Phase: Design

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  1. CMPT 275 Phase: Design

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

  3. Implementation issues related to Data Persistence NORMALIZATION

  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 • 1st normal form: fixed length records • 2nd normal form: remove partial dependencies • 3rd normal form: remove transitive dependencies

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

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

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

  9. Our First Table • From our requirements, we could create the following database table: (horizontal lines separate records, representing single student objects) Instructor Std-name Std-address Room Std-phone Std-id Coursename Sem Grade Credit Instructor’s office Paul K. Will B. Kim L. Xiao T. Dr. Klaus M. Nole V. Karu W. Loti Dr. Quel Dr. Klaus Dr. Yu 15438 25636 47352 21544 Brook St., Bby Elf Ave., Van. Mer Cr., Poco Alpha St., Bby AQ2 AQ1 ASB WM EDC AQ1 AQ2 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 4 3 3 2 3 3 3 C- B A- A+ B+ D A- ASB985 ASB352 WM543 AQ834 EDC243 ASB985 ASB111 294.2563 256.2453 939.2766 295.9976 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. student-idnameaddressphonecoursecreditsemestergraderoominstructorinstructor’s office Group of attributes repeated for each course takenby 1 student Group of attributes repeated each time a particular course is attempted by 1 student Our First Table: Is there a problem? • 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. NOT ALL RECORDS ARE OF THE SAME LENGTH !!

  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. 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. 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. First Normal Form of Table (1NF) Instructor’s office Instructor Std-address Room Std-phone Std-id Course-name Semester Grade Credit Std-name • 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 Paul K. Paul K. Paul K. Will B. Kim L. Xiao T.Xiao T. Dr. Klaus M. Nole V. Karu W. Loti Dr. Quel Dr. Klaus Dr. Yu 15438 15438 15438 25636 47352 21544 21544 Brook St., Bby Brook St., Bby Brook St., Bby Elf Ave., Van. Merry Cr., Poco Alpha St., BBY Alpha St., Bby AQ2 AQ1 ASB WM EDC AQ1 AQ2 Cmpt 101 Cmpt 150 Bus 152 Engl 102 Biol 234 Cmpt 354 Cmpt 354 03-2 03-1 03-2 03-1 03-2 03-1 03-2 4 3 3 2 3 3 3 C- B A- A+ B+ D A- ASB985 ASB352 WM543 AQ834 EDC243 ASB 985 ASB111 294.2563 294.2563 294.2563 256.2453 939.2766 295.9976 295-9976

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

  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. 2NF Example - Step 2 Course Offering Table Course-nameSemester Room Instructor Instructor’s office Cmpt 101 03-2 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 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. Second Normal Form - Example Step 3 • We determine dependencies on whole compound primary key: • Course-name + Semester + Std-id grade

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

  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 Student Course offering Course Student ID Course name Course Name Receives a grade for takes name semester 0..* 1 0..* 0..* List of grades address credit room phone Instructor List of courses Instructor’s office List of students • These 3 attributes are used to implement “many-to-many” multiplicity relationships

  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. 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. 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 Student Course offering Course 1 Course name Student ID Course Name name semester 0..* address room credit Student Registration phone Instructor Instructor’s office Student ID 1 Course name 1 0..* semester 0..* grade

  32. 2NF Example, alternate Step 3 - 3 Course Offering Table Course-name Semester Room Instructor Instructor’s office • 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: Cmpt 101 03-2 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 Biol 234 03-2 EDC Dr. Quel EDC243 Cmpt 354 03-1 AQ1 Dr. Klaus ASB985 Cmpt 354 03-2 AQ2 Dr. Yu ASB111 Student Registration Table Std-id Course-name Grade Semester Course Table Student Table Course-name credit Std-id Std-name Std-address Std-phone 03-2 03-1 03-2 03-1 03-2 03-1 03-2 Cmpt 101 Cmpt 150 Bus 152 Engl 102 Biol 234 Cmpt 354 Cmpt 354 15438 15438 15438 25636 47352 21544 21544 C- B A- A+ B+ D A- Cmpt 101 4 25636 Will B. Elf Ave., Van. 256.2453 Cmpt 150 3 47352 Kim L. Merry Cr., Poco 939.2766 Bus 152 3 21544 Xiao T. Alpha St., Bby 295.9976 Engl 102 2 Biol 234 3 Cmpt 354 3

  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. Second Normal Form • Definition of 2NF: • The table is in 1NF • The table includes no partial dependencies

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

  38. Third Normal Form • Definition: • Every table is in 2NF. • There are no transitive dependencies.

  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

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