The Relational Model of Data Prof. Yin-Fu Huang CSIE, NYUST Chapter 2.

The Relational Model of DataThe Relational Model of Data

Prof. Yin-Fu HuangProf. Yin-Fu HuangCSIE, NYUST CSIE, NYUST

Chapter 2Chapter 2

Database Systems Yin-Fu Huang

2.12.1 An Overview of Data ModelsAn Overview of Data Models2.1.1 What is a Data model2.1.1 What is a Data model

Data model: a notation for describing data or information Three parts:

1) Structure of the dataa. Conceptual model

2) Operations on the data a. Queries and modificationsb. By limiting operations, it is possible for programmers to describe database operations at

a very high level, yet have the DBMS implement the operations efficiently.

3) Constraints on the data

Two important data models:1) The relational model, including object-relational extensions2) The semistructured-data model, including XML and related standards

2.1.22.1.2 Important Data ModelsImportant Data Models

(See Fig. 2.1)

This physical implementation is only one possible way the table could be implemented in physical data structures.

2.1.32.1.3 The Relational Model in BriefThe Relational Model in Brief

(See Fig. 2.2) XML: a way to represent data by hierarchically nested tagged

elements The operations usually involve following paths in the implied

tree from an element to one or more of its nested subelements, then to subelements nested within those, and so on.

The constraints often involve the data type of values associated with a tag.

2.1.42.1.4 The The SemistructuredSemistructured Model in Brief Model in Brief

Object-relational model1) Values can have structure, rather than being elementary

types such as integer or strings.2) Relations can have associated methods.

Object-oriented model Hierarchical model Network model

2.1.52.1.5 Other Data ModelsOther Data Models

It appear that semistructured models have more flexibility than relations.

However, 1) SQL enables the programmer to express their wishes

at a very high level.

2) The short SQL programs can be optimized to run as fast, or faster than the code written in alternative languages.

2.1.62.1.6 Comparison of Modeling ApproachesComparison of Modeling Approaches

2.22.2 Basics of the Relational ModelBasics of the Relational Model

(See Fig. 2.3)

2.2.1 Attributes

2.2.2 Schemas Movies(title, year, length, genre) Database schema

2.2.3 Tuples (Gone With the Wind, 1939, 231, drama)

2.2.4 Domains Each component of each tuple should be atomic.

2.2.5 Equivalent Representations of a Relation Relations are sets of tuples, not lists of tuples (See Fig. 2.4)

2.2.6 Relation Instances A relation about movies is not static; rather, relations change

over time. It is less common for the schema of a relation to change. The current instance

2.2.7 Keys of Relations A set of attributes forms a key for a relation if we do not allow

two tuples in a relation instance to have the same values in all the attributes of the key.

2.2.8 An Example Database Schema (See Fig. 2.5)

2.32.3 Defining a Relation Schema in SQL Defining a Relation Schema in SQL

Current standard for SQL: SQL-991) Data-definition sublanguage2) Data-manipulation sublanguage

2.3.1 Relations in SQL Stored relations called tables Views defined by a computation Temporary tables

2.3.2 Data Types CHAR(n), VARCHAR(n) BIT(n), BIT VARYING(n) BOOLEAN INT or INTEGER, SHORTINT FLOAT or REAL, DOUBLE PRECISION, DECIMAL(n,d) DATE, TIME

2.3.3 Simple Table Declarations CREATE TABLE

(See Fig. 2.7 and Fig. 2.8)

2.3.4 Modifying Relation Schemas DROP TABLE ALTER TABLE

1) ADD followed by an attribute name and its data typeALTER TABLE MovieStar ADD phone CHAR(16);

2) DROP followed by an attribute nameALTER TABLE MovieStar DROP birthdate;

2.3.5 Default Values gender CHAR(1) DEFAULT ‘?’,

birthdate DATE DEFAULT DATE ‘0000-00-00’ ALTER TABLE MovieStar ADD phone CHAR(16)

DEFAULT ‘unlisted’;

2.3.6 Declaring Keys We may declare one attribute to be a key when that attribute is

listed in the relation schema. We may add to the list of items declared in the schema an

additional declaration that says a particular attribute or set of attributes forms the key.

Two declarations:1) PRIMARY KEY2) UNIQUE

Two tuples in R cannot agree on all of the attributes in set S (i.e., key), unless one of them is NULL.

If PRIMARY KEY is used, then attributes in S are not allowed to have NULL as a value for their components.(See Fig. 2.9, Fig. 2.10, and Fig. 2.11)

2.42.4 An Algebraic Query LanguageAn Algebraic Query Language

Relational algebra

2.4.1 Why Do We Need a Special Query Language? Relational algebra is useful because it is less powerful than C

or Java. Two huge rewards:

1) Ease of programming2) Producing highly optimized code

2.4.2 What Is an Algebra? Atomic operands:

1) Variables that stand for relations2) Constants, which are finite relations

2.4.3 Overview of Relational Algebra Operations: Four broad classes

1) The usual set operations – union, intersection, and difference2) Operations that remove parts of a relation – selection and projection

3) Operations that combine the tuples of two relations – Cartesian product and join

4) An operation called renaming We generally shall refer to expressions of relational algebra as queries.

2.4.4 Set Operations on Relations R S, R∩S, R-S∪ Some conditions on R and S:

1) Schemas with identical sets of attributes, and the types (domains) for each attribute must be the same

2) The order of attributes is the same for both relations. Different names renaming operator⇒ (See Fig. 2.12)

2.4.5 Projection πA1, A2, …, An(R) (See Fig. 2.13)

πtitle, year, length(Movies), πgenre(Movies)

2.4.6 Selection σC(R)

σlength 100≧ (Movies), σlength 100 AND studioName=’Fox’≧ (Movies)

2.4.7 Cartesian Product R×S The components from R precede the components from S in the

attribute order for the result. If R and S should happen to have some attributes in common,

then we need to invent new names for at least one of each pair of identical attributes.

(See Fig. 2.14)

2.4.8 Natural Joins R∞S (See Fig. 2.15)

Joined tuple vs. dangling tuple (See Fig. 2.16)

2.4.9 Theta-Joins R∞CS Constructed as follows:

1) Take the product of R and S.2) Select from the product only those tuples that satisfy the condition C.

U∞A<DV (See Fig. 2.17) U∞A<D AND U.B≠V.BV

2.4.10 Combining Operations to Form Queries One can construct expressions of relational algebra by

applying operators to subexpressions, using parentheses when necessary to indicate grouping of operands.

“What are the titles and years of movies made by Fox that are at least 100 minutes long?”

Four steps:(See Fig. 2.18)

πtitle, year(σlength 100≧ (Movies)∩σstudioName=’Fox’(Movies)) πtitle, year(σlength 100 AND studioName=’Fox’≧ (Movies))

2.4.11 Naming and Renaming ρS(A1, A2, …, An)(R) R×ρS(X, C, D)(S), ρRS(A, B, X, C, D)(R×S)

(See Fig. 2.19)

2.4.12 Relationships among Operations Some of the operations can be expressed in terms of other

relational-algebra operations. R∩S=R-(R-S) R∞CS=σC(R×S)

R∞S=πL(σC(R×S))1) C: R.A1=S.A1 AND R.A2=S.A2 AND … AND R.An=S.An2) L: Attributes of R followed by those attributes in S that are not in R.

U∞V ⇒ πA, U.B, U.C, D(σU.B=V.B AND U.C=V.C(U×V)) U∞A<D AND U.B≠V.BV ⇒ σA<D AND U.B≠V.B(U×V) The six remaining operations – union, difference, selection,

projection, product, and renaming – form an independent set.

2.4.13 A Linear Notation for Algebraic Expressions The notation:

1) A relation name and parenthesized list of attributes for that relation

2) The assignment symbol :=3) Any algebraic expression on the right

R(t, y, l, i, s, p):=σlength 100≧ (Movies)S(t, y, l, i, s, p):=σstudioName=’Fox’(Movies)T(t, y, l, i, s, p):=R∩SAnswer(title, year):=πt, y(T)

2.52.5 Constraints on RelationsConstraints on Relations

Many kinds of constraints can be expressed in relational algebra.

2.5.1 Relational Algebra as a Constraint Language Two ways:

1) Equal-to-the-emptyset; e.g., R = 2) Set-containment; e.g., R S⊆

2.5.2 Referential Integrity Constraints Movies(title, year, length, genre, studioName, producerC#)

MovieExec(name, address, cert#, netWorth)πproducerC#(Movies) ⊆ πcertC#(MovieExec)

StarsIn(movieTitle, movieYear, starName)Movies(title, year, length, genre, studioName, producerC#)

πmovieTitle, movieYear(StarsIn) ⊆ πtitle, year(Movies)

2.5.3 Key Constraints If two tuples agree on name, then they must also agree on

address.σ MS1.name=MS2.name AND MS1.address≠MS2.address (MS1×MS2) =

2.5.4 Additional Constraint Examples Domain constraint

σgender≠’F’ AND gender≠’M’(MovieStar) =

MovieExec(name, address, cert#, netWorth)Studio(name, address, presC#)

σnetWorth<10000000(Studio∞presC#=cert#MovieExec) =

πpresC#(Studio) ⊆ πcert#(σnetWorth 10000000≧ (MovieExec))

The End.

The Relational Model of Data Prof. Yin-Fu Huang CSIE, NYUST Chapter 2.

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