cloused

raghad_mm

IT403_Wk03_ch21.ppt

Home >Computer Science homework help >cloused

Database System Concepts, 6th Ed. ©Silberschatz, Korth and Sudarshan See www.db-book.com for conditions on re-use

Chapter 2: Intro to Relational Model

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Example of a Relation

attributes

(or columns)

tuples

(or rows)

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Attribute Types

The set of allowed values for each attribute is called the domain of the attribute
Attribute values are (normally) required to be atomic; that is, indivisible
The special value null is a member of every domain
The null value causes complications in the definition of many operations

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Relation Schema and Instance

A1, A2, …, An are attributes

R = (A1, A2, …, An ) is a relation schema

Example:

instructor = (ID, name, dept_name, salary)

Formally, given sets D1, D2, …. Dn a relation r is a subset of
D1 x D2 x … x Dn
Thus, a relation is a set of n-tuples (a1, a2, …, an) where each ai  Di

The current values (relation instance) of a relation are specified by a table
An element t of r is a tuple, represented by a row in a table

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Relations are Unordered

Order of tuples is irrelevant (tuples may be stored in an arbitrary order)
Example: instructor relation with unordered tuples

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Database

A database consists of multiple relations
Information about an enterprise is broken up into parts

instructor
student
advisor

Bad design:
univ (instructor -ID, name, dept_name, salary, student_Id, ..)
results in
repetition of information (e.g., two students have the same instructor)
the need for null values (e.g., represent an student with no advisor)
Normalization theory (Chapter 7) deals with how to design “good” relational schemas

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Keys

Let K  R
K is a superkey of R if values for K are sufficient to identify a unique tuple of each possible relation r(R)
Example: {ID} and {ID,name} are both superkeys of instructor.
Superkey K is a candidate key if K is minimal
Example: {ID} is a candidate key for Instructor
One of the candidate keys is selected to be the primary key.
which one?
Foreign key constraint: Value in one relation must appear in another
Referencing relation
Referenced relation

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Schema Diagram for University Database

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Relational Query Languages

Procedural vs.non-procedural, or declarative
“Pure” languages:
Relational algebra
Tuple relational calculus
Domain relational calculus
Relational operators

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Selection of tuples

Relation r

Select tuples with A=B and D > 5
σ A=B and D > 5 (r)

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Selection of Columns (Attributes)

Relation r:

Select A and C
Projection
Π A, C (r)

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Joining two relations – Cartesian Product

Relations r, s:

r x s:

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Union of two relations

Relations r, s:

r  s:

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Set difference of two relations

Relations r, s:

r – s:

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Set Intersection of two relations

Relation r, s:

r  s

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Joining two relations – Natural Join

Let r and s be relations on schemas R and S respectively.
Then, the “natural join” of relations R and S is a relation on schema R  S obtained as follows:
Consider each pair of tuples tr from r and ts from s.
If tr and ts have the same value on each of the attributes in R  S, add a tuple t to the result, where
t has the same value as tr on r
t has the same value as ts on s

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Natural Join Example

Relations r, s:

Natural Join
r s

©Silberschatz, Korth and Sudarshan

2.*

Database System Concepts - 6th Edition

Figure in-2.1

End of Chapter 2

2.*

Database System Concepts - 6th Edition

Figure 2.01

2.*

Database System Concepts - 6th Edition

Figure 2.02

2.*

Database System Concepts - 6th Edition

Figure 2.03

2.*

Database System Concepts - 6th Edition

Figure 2.04

2.*

Database System Concepts - 6th Edition

Figure 2.05

2.*

Database System Concepts - 6th Edition

Figure 2.06

2.*

Database System Concepts - 6th Edition

Figure 2.07

2.*

Database System Concepts - 6th Edition

Figure 2.10

2.*

Database System Concepts - 6th Edition

Figure 2.11

2.*

Database System Concepts - 6th Edition

Figure 2.12

2.*

Database System Concepts - 6th Edition

Figure 2.13