Kingdom of Saudi Arabia Ministry of Higher Education

Kingdom of Saudi Arabia Ministry of Higher Education Al-Imam Muhammad Ibn Saud Islamic University College of Computer and Information Sciences Normalization IS 320: Introduction to Database Hatoon Al. Sagri Introduction to Databases

Informal Design Guidelines for Relational Databases • Relational database design: The grouping of attributes to form "good" relation schemas • Two levels of relation schemas: • The logical "user view" level • The storage "base relation" level • Design is concerned mainly with base relations Introduction to Databases 2

Informal Design Guidelines for Relational Databases Four informal measures of quality for relation schema design: 1. Semantics of the Relation Attributes 2. Reducing the redundant information in tuples 3. Reducing Null values in tuples 4. Disallowing the possibility of one generating spurious tuples. Introduction to Databases 3

1 - Semantics of the Relation Attributes Each tuple in a relation should represent one entity or relationship instance Guideline #1: Design a schema that can be explained easily relation by relation. The semantics of attributes should be easy to interpret. Introduction to Databases 4

2 - Redundant Information in Tuples and Update Anomalies • Mixing attributes of multiple entities may cause problems: • Information is stored redundantly wasting storage • Problems with update anomalies: • Insertion anomalies • Deletion anomalies • Modification anomalies Guideline #2: Design a schema that does not suffer from the insertion, deletion and update anomalies. If there any present, then note them so that applications can be made to take them into account Introduction to Databases 5

Base Relations EMP_PROJ with redundant information Introduction to Databases 6

3 - Null Values in Tuples Reasons for nulls: a. attribute not applicable or invalid b. attribute value unknown (may exist) c. value known to exist, but unavailable Guideline #3: Relations should be designed such that their tuples will have as few NULL values as possible Attributes that are NULL frequently could be placed in separate relations (with the primary key) Introduction to Databases 7

4 - Spurious Tuples • Bad designs for a relational database may result in erroneous results for certain JOIN operations Guideline #4: The relations should be designed to satisfy the lossless join condition. No spurious tuples should be generated by doing join of any relations. Introduction to Databases 8

Functional Dependencies • Functional dependencies (FDs) are used to specify formal measures of the "goodness" of relational designs • FDs and keys are used to define normal forms for relations • FDs are constraints that are derived from the meaning and interrelationships of the data attributes Introduction to Databases 9

Examples of FD constraints • Social security number determines employee name SSN -> ENAME • Project number determines project name and location PNUMBER -> {PNAME, PLOCATION} • Employee ssn and project number determines the hours per week that the employee works on the project {SSN, PNUMBER} -> HOURS Introduction to Databases 10

Functional Dependencies Describes the relationship between attributes in a relation. If A and B are attributes of relation R, B is functionally dependent on A, denoted by A B, if each value of A is associated with exactly one value of B. B may have several values of A. A Determinant B is functionally dependent on A Introduction to Databases B Dependent 11

Functional Dependencies Example 1: 1 or M: 1 relationship Staff. No Position is functionally position between dependent on Staffno attributes in a SL 21 Manager relation 1: M relationship position Staff. No is NOT functionally Staff. No dependent on position between attributes in a Manager SL 21 relation SG 5 Introduction to Databases 12

Trivial Functional Dependencies A B is trivial if B A Staff. No, Sname SName Staff. No, SName Staff. No We are not interested in trivial functional dependencies as it provides no genuine integrity constraints on the value held by these attributes. Introduction to Databases 13

Question Find FDs of the relation shown below that lists dentist/patient appointment data; known that: • A patient is given an appointment at a specific time and date with a dentist located at a particular surgery. • On each day of patient appointments, a dentist is allocated to a specific surgery for that day. Dentist-patient (staff. No, a. Date, a. Time, dentist. Name, pat. No, pat. Name, surgery. No) Introduction to Databases 14

Question Dentist-patient (staff. No, a. Date, a. Time, dentist. Name, pat. No, pat. Name, surgery. No) FDs list FD 1: staff. No, a. Date, a. Time pat. No, pat. Name FD 2: staff. No dentist. Name FD 3: pat. No pat. Name, surgery. No FD 4: staff. No, a. Date surgery. No FD 5: a. Date, a. Time, pat. No dentist. Name, staff. No Introduction to Databases 15

Introduction to Normalization • Normalization: Process of decomposing unsatisfactory "bad" relations by breaking up their attributes into smaller relations • Normal form: Condition using keys and FDs of a relation to certify whether a relation schema is in a particular normal form Introduction to Databases 16

Normalization into 1 NF Introduction to Databases 17

Examples First Normal Form • EMP_PROJ (Ssn, Ename, {Phone#}) EMP_PROJ 1 (Ssn, Ename) { } Mulitvalue attribute EMP_PROJ 2 (Ssn, Phone#) • EMP_PROJ (Ssn, Ename (Fname, Lname)) ( ) composite attribute EMP_PROJ (Ssn, Fname, Lname) • EMP_PROJ (Ssn, Ename, {PROJS (Pnamber, Hours)}) EMP_PROJ 1 (Ssn, Ename) EMP_PROJ 2 (Ssn, Pnamber, Hours) Introduction to Databases 18

Second Normal Form • Uses the concepts of FDs, primary key • Definitions: • Prime attribute - attribute that is member of the primary key K • Full functional dependency - a FD Y Z where removal of any attribute from Y means the FD does not hold any more Introduction to Databases 19

Full Functional Dependency If A and B are attributes of a relation. B is fully functionally dependent on A if B is functionally dependent on A, but not on any proper subset of A. B is partial functional dependent on A if some attributes can be removed from A & the dependency still holds. Staff. No, Sname Branch. No Partial dependency Client. No, Property. No Rent. Date Full dependency Introduction to Databases 20

1 NF 2 NF 1. Start with 1 NF relation. 2. Find the FDs of a relation. 3. Test the FDs whose determinant attribute is part of the PK. Introduction to Databases 21

Examples Second Normal Form • {SSN, PNUMBER} HOURS is a full FD since neither SSN HOURS nor PNUMBER HOURS hold • {SSN, PNUMBER} ENAME is not a full FD (it is called a partial dependency ) since SSN ENAME also holds • A relation schema R is in second normal form (2 NF) if every non- prime attribute A in R is fully functionally dependent on the primary key • R can be decomposed into 2 NF relations via the process of 2 NF normalization Introduction to Databases 22

Introduction to Databases Examples Second Normal Form 23

Second Normal Form Note: The test for 2 NF involves testing for functional dependencies whose left-hand side attributes are part of the primary key. If the primary key contains a single attribute, the test need not be applied at all. Introduction to Databases 24

Third Normal Form • Definition • Transitive functional dependency – a FD X Y in R is a transitive dependency if there is a set of attributes Z that are neither a primary or candidate key and both X Z and Z Y holds. • Examples: • SSN DMGRSSN is a transitive FD since SSN DNUMBER and DNUMBER DMGRSSN hold • SSN ENAME is non-transitive since there is no set of attributes X where SSN X and X ENAME Introduction to Databases 25

3 rd Normal Form A relation schema R is in third normal form (3 NF) if it is in 2 NF and no non-prime attribute A in R is transitively dependent on the primary key Introduction to Databases 26

Introduction to Databases Examples Third Normal Form 27

SUMMARY OF NORMAL FORMS based on Primary Keys Introduction to Databases 28

BCNF (Boyce-Codd Normal Form) • A relation schema R is in Boyce-Codd Normal Form (BCNF) if whenever an FD X A holds in R, then X is a superkey of R • Each normal form is strictly stronger than the previous one: • Every 2 NF relation is in 1 NF • Every 3 NF relation is in 2 NF • Every BCNF relation is in 3 NF • There exist relations that are in 3 NF but not in BCNF • The goal is to have each relation in BCNF (or 3 NF) Introduction to Databases 29

Introduction to Databases BCNF R 1(A, C) R 2(C, B) 30

Introduction to Databases 31

BCNF FDs: • {Student, course} Instructor • Instructor Course It is in 3 NF not in BCNF • Decomposing into 2 schemas {Student, Instructor} {Instructor, Course} Introduction to Databases 32

Introduction to Databases Examples BCNF n R ( Client#, Problem, Consultant _name) R 1 (Client#, Consultant _name) R 2 (Consultant _name, Problem) ■ R (Stud#, Class#, Instructor, Grade) R 1 (Stud#, Instructor, Grade) R 2 (Instructor, Class#) 33

Example Consider the following relation for published books: BOOK (Book_title, Author_name, Book_type, Listprice, Author_affil, Publisher) - Author_affil referes to the affiliation of the author. Suppose thefollowing dependencies exist: Book_title -> Publisher, Book_type -> Listprice Author_name -> Author-affil (a) What normal form is the relation in? Explain your answer. (b) Apply normalization until you cannot decompose the relations further. State the reasons behind each decomposition. Introduction to Databases 34

Answer BOOK (Book_title, Authorname, Book_type, Listprice, Author_affil, Publisher) (a) The key for this relation is (Book_title, Authorname). This relation is in 1 NF and not in 2 NF as no attributes are Full FD on the key. It is also not in 3 NF. (b) 2 NF decomposition: Book 0(Book_title, Authorname) Book 1(Book_title, Publisher, Book_type, Listprice) Book 2(Authorname, Author_affil) This decomposition eliminates the partial dependencies. 3 NF decomposition: Book 0(Book_title, Authorname) Book 1 -1(Book_title, Publisher, Book_type) Book 1 -2(Book_type, Listprice) Book 2(Authorname, Author_affil) This decomposition eliminates the transitive dependency of Listprice Introduction to Databases 35

Example Given the relation schema Car_Sale (Car#, Salesman#, Date_sold, Commission%, Discount_amt) with the functional dependencies Date_sold -> Discount_amt Salesman# -> Commission% Car# -> Date_sold This relation satisfies 1 NF but not 2 NF (Car# -> Date_sold and Salesman# -> Commission%) so these two attributes are not Full FD on the primary key and not 3 NF Introduction to Databases 36

Answer To normalize, 2 NF: Car_Sale 1 (Car#, Salesman#) Car_Sale 2 (Car#, Date_sold, Discount_amt) Car_Sale 3 (Salesman#, Commission%) 3 NF: Car_Sale 1(Car#, Salesman#) Car_Sale 2 -1(Car#, Date_sold) Car_Sale 2 -2(Date_sold, Discount_amt) Car_Sale 3(Salesman#, Commission%) Introduction to Databases 37

Question Given the following Dentist-patient database schema: Dentist-patient (staff. No, a. Date, a. Time, dentist. Name, pat. No, pat. Name, surgery. No) Normalize the above relation, showing appropriate dependency diagrams to justify decomposition. FDs List: FD 1: staff. No, a. Date, a. Time pat. No, pat. Name FD 2: staff. No dentist. Name FD 3: pat. No pat. Name, surgery. No FD 4: staff. No, a. Date surgery. No FD 5: a. Date, a. Time, pat. No dentist. Name, staff. No Introduction to Databases 38

Answer 1 NF Dentist-patient (staff. No, a. Date, a. Time, dentist. Name, pat. No, pat. Name, surgery. No) 2 NF (fd 2 and fd 4 violates 2 NF) Dentist-patient (staff. No, a. Date, a. Time, pat. No, pat. Name) Surgery (staff. No, a. Date, surgery. No) Dentist (staff. No, dentist. Name) 3 NF (Fd 3’ violates 3 NF) Dentist-patient (staff. No, a. Date, a. Time, pat. No) Surgery (staff. No, a. Date, surgery. No) Dentist (staff. No, dentist. Name) Patient (pat. No, pat. Name) Introduction to Databases 39

Answer BCNF (No violation) Dentist-patient (staff. No, a. Date, a. Time, pat. No) Surgery (staff. No, a. Date, surgery. No) Dentist (staff. No, dentist. Name) Patient (pat. No, pat. Name) Introduction to Databases 40