Relation Normalization

Transcription

1 Relation Normalization Why Normalization? Functional Dependencies. First, Second, and Third Normal Forms. Boyce/Codd Normal Form. No loss Decomposition. Summary Relation Normalization 1 Why Normalization? An ill-structured relation contains redundant data Data redundancy causes modification anomalies: Insertion anomalies -- Suppose we want to enter SCUBA as an activity that costs $100, we can t until a student signs up for it Update anomalies -- If we change the price of swimming for student 150, there is no guarantee that student 200 will pay the new price Deletion anomalies -- If we delete Student 100, we lose not only the fact that he/she is a skier, but also the fact that skiing costs $200 Normalization is the process used to remove modification anomalies ACTIVITY Relation Normalization 2 SID Activity Fee 100 Skiing Swimming Squash Swimming 50 How can this table be changed to fix these problems??? Dave McDonald,, GSU 10-1

2 Why Normalization... Course SID s1 s1 s1 s2 s2 s3 s3 s3 Name Joseph Joseph Joseph Alice Alice Tom Tom Tom Grade A B A A A B B A Course# Text b1 b2 b5 b1 b5 b1 b5 b1 Major CS CS Dept MCS MCS Is there any redundant data? Insertion anomalies? Update anomalies? Deletion anomalies? Relation Normalization 3 Functional Dependencies Given two attributes, X and Y, of a relation R, Y is functionally dependent on X iff each X value must always occur with the same Y value in R. R.X --> R.Y or X --> Y List all FDs in the Course relation: Relation Normalization 4 Dave McDonald,, GSU 10-2

3 Functional Dependencies... X is called the determinant of Y. X and Y may be composite. Dependency relationships change with attribute semantics. X and Y could be mutually dependent on each other. Husband --> Wife, Wife --> Husband, Husband <--> Wife X may or may not be the key attribute of R. AY value can occur in more than one tuple in R. Course# --> Text Relation Normalization 5 Fully Functional Dependencies A fully functional dependence ( FFD ) exists between attributes X and Y if Y is not functional dependent on any proper subset of X. ( SID, Course# ) --> Name? ( SID, Course# ) --> Grade? ( SID, Name ) --> Major? ( SID, Name ) --> SID? Note that if X is not composite, then X --> Y is always a FFD. By default, the term FD refers to FFD Relation Normalization 6 Dave McDonald,, GSU 10-3

4 Transitively Functional Dependencies Given attributes X, Y, and Z of a relation R, Z is transitively dependent on X iff X --> Y and Y --> Z. Given SID --> Dept and Dept --> College SID -->? Given SID --> Major and Major --> Dept, SID --> >? Relation Normalization 7 Graphical Representation Course (SID, Name, Grade, Course#, Text, Major, Dept) Major Dept Primary Key SID Course# Name Grade Text Relation Normalization 8 Dave McDonald,, GSU 10-4

5 First Normal Form (1NF) A relation R is in 1NF iff all attribute domains contain atomic values only. A relation in 1NF has modification anomalies Part# QTY WAddress WHouse# Relation Normalization 9 INVENTORY (Part#, WHouse#, WAddress, QTY) Second Normal Form (2NF) A relation is in 2NF iff R is in 1NF and every non key attribute is fully dependent on the primary key (i.e. has no partial functional dependencies). The term, non key attribute, refers to any attribute that does not belong to any candidate key. Part# QTY WAddress WHouse# Relation Normalization 10 INVENTORY (Part#, WHouse#, WAddress, QTY) Dave McDonald,, GSU 10-5

6 Modification Anomalies in 2NF 2NF relations have modification anomalies: Redundant Information? Update anomalies? Insertion anomalies? Deletion anomalies? Which FD causes the redundant data? INVENTORY Part# WHouse# WAddress QTY Relation Normalization Atlanta Birmingham Columbus Oakland Denver 2 Third Normal Form (3NF) A relation R is in 3NF iff R is in 2NF and every non key attribute is non transitively dependent on the primary key. Student (SID, Name, Major, Dept) Discussion: If a relation does not have any non-key attribute, would it automatically be in 3NF? Relation Normalization 12 Dave McDonald,, GSU 10-6

7 Modification Anomalies in 3NF LOCATION (Employee, Department, Location) Redundant Information? Update anomalies? Insertion anomalies? Deletion anomalies? All determinants? Employee Department Location Relation Normalization 13 Boyce/Codd Normal Forms (BCNF) A relation R is in BCNF iff every determinant is a candidate key. BCNF is applied to a relation R if 1. Those candidate keys are composite, and 2. The candidate keys are overlapped, ADVISE (Student, Major, Advisor) STUDENT ADVISOR MAJOR Relation Normalization 14 Dave McDonald,, GSU 10-7

8 BCNF Example Student Course Instructor Narayan Database Mark Smith Database Jeffries Smith Operating Ammar Systems Smith Theory Schulman Wallace Database Mark Wallace Operating Ahamad Systems Wong Database Omiecinski Zelaya Database Jeffries Narayan Operating Ammar Systems Relation Normalization 15 Teach(Student, Course, Instructor) BCNF Example (cont d) Student Course Instructor There are three possible decompositions (Student, Instructor) (Course, Instructor) and and (Student, Course) (Course, Student,) (Instructor, Course) and (Instructor, Student) Which of the three will not generate spurious tuples after a join? Relation Normalization 16 Dave McDonald,, GSU 10-8

9 Rules for No Loss Decomposition Independent Projection ( Rissanen ) -- Two projections R1 and R2 of a relation R are independent iff Every FD in R can be logically deduced from those in R1 and R2, and The common attributes of R1 and R2 form a candidate key for at least one of the pair. An independent projection guarantees no loss decomposition Each decomposed relation can be maintained independently. The contents of the original relation can be restored correctly by joining the decomposed relations on common attributes. Find out the loss decompositions in the examples Relation Normalization 17 Summary 4NF and 5NF deal with multi-value dependences Relations mapped from E-R schema need no 4NF and 5NF Limitationsi i Can not detect redundancy between relations Results fragmented relations which may slow down data retrieval Denormalization is needed if Relations in higher normal form cause the performance problem, Majority of queries are data retrieval, Denormalization can speed up the data retrieval, and Denormalization does not introduce severe update anomalies. Relation Normalization 18 Dave McDonald,, GSU 10-9

10 Exercise Given relation, Lab_Usage, has been defined as follows. where Lab_usage( SID, Class#, Course, SName, Account#, Lab_Hours ) SID is a unique student ID, Class# is a unique class number, SName is a student name, Lab_Hours is the maximum laboratory hours assigned to each student in a class, Account# is a unique computer account. A student is assigned an account for each class he/she takes. Assume that no student takes the same course twice. Relation Normalization 19 Exercise Determine all candidate keys and select a primary key. 2. List all FFDs. 3. Discuss update anomalies found in the relation. 4. Decompose the relation into 2NF relations. 5. Discuss update anomalies found in the 2NF relations. 6. Decompose the relation into 3NF relations. 7. Discuss update anomalies found in the 3NF relations if any. Relation Normalization 20 Dave McDonald,, GSU 10-10