IS 230Lecture 8Slide 1 Normalization Lecture 9. IS 230Lecture 8Slide 2 Lecture 8: Normalization 1. Normalization 2. Data redundancy and anomalies 3. Spurious

IS 230 Lecture 8 Slide 1

Normalization

Lecture 9

IS 230 Lecture 8 Slide 2IS 230 Lecture 8 Slide 2

Lecture 8: Normalization1. Normalization2. Data redundancy and anomalies3. Spurious information4. Functional dependencies5. Normalization

first normal form second normal form third normal form BCNF normal form

6. Normalization Methodology: Example


1. Normalization A technique for producing a set of

relations with desirable properties, given the data requirements of the applications


2. Data redundancy and anomalies

Emp-Dept relation

NI# Name DateOfBirth Dept# Dname Manager 21 AA - 5 CS 91 22 BB - 5 CS 91 23 CC - 6 TS 93 24 DD - 7 PSV 94 25 EE - 7 PSV 94

Insertion

How do we insert a new department with no employees yet? (keys?) Entering employees is difficult as department information must be

entered correctly.

Deletion What happens when we delete CC's data - do we lose department 6?

Modification If we change the manager of department 5, we must change it for

tuples with Dept# = 5.


3. Spurious information Avoid breaking up relations in such a

way that spurious information is createdNI# Name ProjName ProjLocation 123 XX Accounts London 123 XX Analysis Paris 124 YY PI London

may be broken into:

Joining them back together, we get NEW TUPLES!

NI# Name ProjName 123 XX Accounts 123 XX Analysis 124 YY PI

NI# Name ProjName ProjLocation 123 XX Accounts Paris 123 XX Analysis London

NI# ProjLocation 123 London 123 Paris 124 London


4. Functional dependencies Formal concepts that may be used to

exhibit “goodness” and “badness” of individual relational schemas, and describe relationships between attributes

Examples of Functional Dependency

Name, DateOfBirth, Dept all depend on NI

Dname & Manager depend on Dept

ProjLocation depends on ProjName.


Functional Dependence An attribute, X, of a relation is functionally

dependent on attributes A, B, ..., N if the same values of A, ..., N are always associated with the same value of X

{A,…,N} X

A, ..., N is called the determinant of the functional dependency


Full Functional Dependency X fully depends on A, …, N if it is

not dependent on any subset of A, ..., N; otherwise we talk of partial dependency

e.g. age is dependent on NI and Name but only fully dependent on NI.


5. Normalization Process

taking a set of relations and decomposing them into more relations satisfying some criteria.

Decomposition essentially a series of projections so that the

original data can be reconstituted using joins.

Normal form form of the relations which satisfy the

criteria


5.1. First Normal Form (1)

A relation is in first normal form (1NF) if all values are atomic, i.e. single values - small strings and numbers.

Identify and remove repeating groups (multi-valued attributes)


First Normal Form (2) DEPARTMENT

Dnumber Dname Locations 5 C.S. { Paris, London }

Two ways of normalising this:

Have a tuple for each location of each department:

Have a separate relation for (Dnumber, Locations) pairs:

The latter is better as it avoids redundancy.

Dnumber Dname Locations 5 C.S. Paris 5 C.S. London

Dnumber Dname 5 C.S

Dnumber Locations 5 Paris 5 London


First Normal Form (3) Example of composite attribute:

Room is not atomic Room includes Deptname and

Room#

TeacherID Course Room 23568 230 CS234 23669 240 IS225 25644 352 IS232 26338 455 CE255

TeacherID Course Room# 23568 230 234 23669 240 225 25644 352 232 26338 455 255

Room# Deptname 234 CS 225 IS 232 IS 255 CE


5.2. Second Normal Form By the definition of the primary key, every other attribute

is functionally dependent on it.

If all the other attributes are fully functionally dependent then the relation is in Second Normal Form (2NF).

Clearly, any relation with a single primary key will be 2NF.

If there are two primary key attributes, A & B, then each other attribute is either

dependent on A alone; dependent on B alone; or dependent on both.

2NF Normal consists of creating a separate relation for each of the three cases.


Example of 2NF decomposition

EMP_PROJ(SSN, Pnumber, Hours, Ename, Pname, Ploc)

SSN Pnumber

Hours

Ename Pname Ploc

Decomposition into three 2NF relations:

Work(SSN,Pnumber,Hours)EMP(SSN,Ename)Project(Pnumber,Pname,Ploc)

SSN EnamePnumber Pname,PlocSSN, Pnumber Hours


5.3. Third Normal Form Third Normal Form eliminates transitive dependencies - i.e.

those dependencies which hold only because of some intermediary. An attribute is transitively dependent on the primary key if there is

some other attribute which is dependent on and which is, in turn, dependent on the key.

NI# Dnumber Dname 1234 5 C.S. 1235 5 C.S.

Dname is dependent on NI, but only because it is dependent on Dnumber which is, in turn, dependent on NI.

Non-3NF relations are likely to hold redundant information. A relation is in 3NF if for any pair of attribute A & B such that A

B, there is no attribute such that A X and X B.

NI# Dnumber 1234 5 1235 5

Dnumber Dname 5 C.S.

Normalizing this would create:


Example of 3NF decomposition EMP_DEPT(SSN, Ename, Bdate, Address, Dnum, Dname, Dman)

Ename Bdate Address Dnum

Dname Dman

SSN

Decomposition into two 3NF relations:

EMPLOYEE(SSN, Ename, Bdate, Address, Dnum)DEPT(Dnum, Dname, Dman)

SSN Ename, Bdate, Address, DnumDnum Dname, Dman


General Definitions of Normal Forms


5.4. Boyce-Codd Normal Form Every relation in BCNF is also in 3NF

Relation in 3NF is not necessarily in BCNF

Nontrivial FD means not trivial FD A trivial functional dependency X Y

is one in which Y is a subset of X Example: A, B B is a trivial FD

Most relation schemas that are in 3NF are also in BCNF


Example of not BCNF The following is not in BCNF:

bor_loan = (customer_id, loan_number, amount)

The following is a functional dependency that may hold:

loan_numberamount but loan_number is not a superkey of bor_loan

We decompose into two relations: R1=(customer_id, loan_number) R2=(loan_number, amount) R1 and R2 are in BCNF


R=(A, B, C, D, E, F) A, B D B E (not in BCNF) D F (not in 3NF)


Normalization Methodology: Example

Consider the following description of a company: The company is divided into departments. Each department is identified by its department number. A department has a name and a manager (an employee). The employees of the company are identified by their National Insurance number. An employee has a name, an address, an age, and work in one department only. An employee is supervised by several supervisors (employees), and a supervisor can supervise several employees (supervisees). An employee can have dependents, where each dependent is described by his/her name and age. The company has a number of running projects. A project is identified by its project number. A project has also a name and a description. Several employees can work on a project, and an employee can work on several projects, each a fixed number of hours.


A B1 41 53 7

1. Functional dependencies An attribute, X, of a relation is

functionally dependent on attributes A, B, ..., N if the same values of A, ..., N are always associated with the same value of X

{A,…,N} X

Example

A B does NOT hold, B A does hold


1. Functional dependencies (Cont.)

The company is divided into departments. Each department is identified by its department number. A department has a name and a manager (an employee).

Dnumber Dname, Manager

The employees of the company are identified by their National Insurance number. An employee has a name, an address, an age, and work in one department only.

NI Ename, Address, Eage, Dnumber



An employee is supervised by several supervisors (employees), and a supervisor can supervise several employees (supervisees).

NI Supervisor (is not a valid functional dependency since an employee can have several supervisors)

An employee can have dependents, where each dependent is described by his/her name and age.

NI, DepName DepAge



The company has a number of running projects. A project is identified by its project number. A project has also a name and a description.

Pno Pname, Description

Several employees can work on a project, and an employee can work on several projects, each a fixed number of hours.

Pno, NI Hours


1. All Functional dependencies

Dnumber Dname, ManagerNI Ename, Address, Eage, DnumberNI, DepName DepAgePno Pname, Description Pno, NI Hours

The Universal Relation:U(Dnumber, Dname, Manager, NI, Ename, Address, Eage, Pno, Pname, Description, DepName, DepAge, Hours, Supervisor)


2. The Primary key

What is the primary key?(Dnumber, NI, Pno, DepName)?

Supervisor must be part of the primary key because it is a multivalued attribute

DepName must be part of the primary key since there can be several dependents to an employee

Dnumber is not part of the primary key since it can be derived from NI

Thus the primary key is (NI, Pno, DepName, Supervisor)


3. Is U in 1NF?

A relation is in first normal form (1NF) if all values are atomic, i.e. single values

Supervisor is a multivalued attribute, hence U is not in 1NF

To remove the multivalued attribute, the relation U is decomposed as follows

U1(Dnumber, Dname, Manager, NI, Ename, Address, Eage, Pno, Pname, Description, DepName, DepAge, Hours)

Supervise(NI, Supervisor)


4. Is U in 2NF? A relation is in Second Normal Form (2NF) if all the attributes

are fully functionally dependent on the primary key. A relation with a single primary key is in 2NF.

Supervise is in 2NF U1 is not in 2NF because some attributes are partially

functionally dependent on the primary key (e.g. DepAge, Ename, etc).

We decompose U1 into the following 2NF relations:

Dept_Emp(Dnumber, Dname, Manager, NI, Ename, Address, Eage)

Dependent(NI, DepName, DepAge) Project(Pno, Pname, Description) Work(NI, Pno, Hours)


5. Is the previous decomposition in 3NF? A relation is in 3NF if for any pair of attributes A

& B such that A B, there is no attribute such that A X and X B

Supervise is in 3NF Among the above relations, only Dept_Emp is

not in 3NF, since it has transitive dependencies (e.g. NI Dnumber Dname).

We therefore decompose the relation into the following two relations, which are in 3NF:

Department(Dnumber, Dname, Manager)

Employee(NI, Ename, Address, Eage, Dnumber)


6. Complete Schema in 3NF

Department(Dnumber, Dname, Manager) Employee(NI, Ename, Address, Eage,Dnumber) Dependent(NI, DepName, DepAge) Project(Pno, Pname, Description) Work(NI, Pno, Hours) Supervise(NI, Supervisor)

IS 230 Lecture 8 Slide 32

End of Chapter

Documents

IS 230Lecture 8Slide 1 Normalization Lecture 9. IS 230Lecture 8Slide 2 Lecture 8: Normalization 1. Normalization 2. Data redundancy and anomalies 3. Spurious