Chap 3,6 6e & 7 5e : Relational Model Parts 1 & 2
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Prof. Steven A. Demurjian, Sr.
Computer Science & Engineering Department
The University of Connecticut
191 Auditorium Road, Box U-155
Storrs, CT 06269-3155
steve@engr.uconn.edu
http://www.engr.uconn.edu/~steve
(860) 486 - 4818


A large portion of these slides are being used with the permission of Dr. Ling
Lui, Associate Professor, College of Computing, Georgia Tech.
The remainder of these slides have been adapted from the AWL web site for
the textbook.
Chapter 3-1
Combining Chapters 3 and 6 6ed
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What is a Relational Data Model?
 Schema, Tables, Attributes/Columns, Tuples
 Characteristics and Examples
Referential Integrity
 Superkeys, Candidate Primary and Foreign Keys
 Referential Integrity Constrains
Relational Algebra
 Selection, Project, Join, Union, Intersection
 Advanced Concepts
Operations on Relations
Chapter 3-2
Essentials of Relational Approach
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Relational Model of Data is Based on the Concept of
Relations
 A Relation is a Mathematical Concept Based on Sets
Theory of Relations Provides a Formal Foundation for
the Relational Data Model
The Model Was First Proposed by Dr. E.F. Codd
(IBM) in 1970 in the Paper, Entitled "A Relational
Model for Large Shared Data Banks," Communications
of the ACM, June 1970
Chapter 3-3
Relational Data Model: Data Structure
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Relational Data Model
 Structures a Database as a Set of Relations.
A Relation
 Set of Tuples and Typically Shown as a Table With
Columns and Rows.
 Column (Field) Represents an Attribute
 Row (Tuple) Represents an Entity or a Relationship
Attributes
relation name
t1
t2
A1
v11
v21
tm
vm1
R
Tuples
.
.
.
A2
v12
v22
vm2
......
An
v1n
v2n
t1[An]
vmn
Chapter 3-4
Two Versions of a Student Relation
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Chapter 3-5
Basic Concepts - Relation Schema
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
A Schema of a Relation
 Denoted as R(A :D , A :D , ..., A :D )
1 1
2 2
n n
 Set of Attributes That Describe a Relation Denoted
by {A1:D1, A2:D2 , ..., An:Dn}, where Ai (i=1, …, n)
is Attribute Name and Di is Domain Over Which Ai
is Defined
Domain
 The Set of Values From which the Values of an
Attribute Aj are Drawn, Denoted by Domain(Aj)
Example
STUDENT (s#, sname, email, dept)
Domain(s#): Number(9)
Domain(sname): Char(30)
Domain(email): Char(20)
Domain(dept): Char(15)
Chapter 3-6
Basic Concepts
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Relation Scheme - Definition of a Relation
 Set of Attributes that Describe a Relation
e.g., R( A1, A2 , ..., An)
Domain - Set of Values from which the Values of an
Attribute Are Drawn
 Denoted by Domain(aj)
Relation (Relation Instance)
 Subset of the Cartesian Product of Domains that
Defines its Schema
 Occurrence of a Relation Scheme
 R(r) = {T1, T2, ..., Tm}.
Cardinality is the Number of Tuples
Chapter 3-7
Basic Concepts
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Tuple
 A Row in a Relational Table - ti = {vi1,vi2, ...,vin}
Attribute
 A Column in a Relational Table
 Projection of an Attribute Aj is {v1j,v2j, ...,vmj}, a
Subset of Domain(Aj.)
 Several Attributes may be Defined on the same
Domain (e.g., date of purchase, date of order, etc.)
Null Value
 Special Value Meaning “not known” or “not
applicable” …
 Must be a Value - Even if it is Null
Degree - the Number of Attributes
Chapter 3-8
What is an Example?
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R(A, B) is a Relation Schema Defined over A and B
Let domain(A) = {a1, a2} and domain(B) = {0, 1, 2}
Tuples are:
 <a1, 0>, <a1, 2>, <a2, 2> etc.
 How Many Possible Tuples are there?
Entire Relation is a set:
{<a1,
0>, <a1, 2>, <a2, 2> etc. }
Chapter 3-9
Relation Schemes
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Example
 EMP(ENO, ENAME, TITLE, SAL)
 PROJ (PNO, PNAME, BUDGET)
 WORKS(ENO, PNO, RESP, DUR)
Underlined Attributes are Relation Keys which
Uniquely Distinguish Among Tuples (Rows)
Tabular Form
EMP
ENO
ENAME
TITLE
PROJ
PNO
PNAME
BUDGET
WORKS
ENO
PNO
RESP
SAL
DUR
Chapter 3-10
Relation Instances
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EMP
WORKS
ENO
ENAME
TITLE
E1
E2
E3
E4
E5
E6
E7
E8
J. Doe
M. Smith
A. Lee
J. Miller
B. Casey
L. Chu
R. Davis
J. Jones
Elect. Eng.
Syst. Anal.
Mech. Eng.
Programmer
Syst. Anal.
Elect. Eng.
Mech. Eng.
Syst. Anal.
ENO
PNO
E1
E2
E2
E3
E3
E4
E5
E6
E7
E7
E8
P1
P1
P2
P3
P4
P2
P2
P4
P3
P5
P3
RESP
DUR
Manager
Analyst
Analyst
Consultant
Engineer
Programmer
Manager
Manager
Engineer
Engineer
Manager
12
24
6
10
48
18
24
48
36
23
40
PROJ
PNO
PNAME
BUDGET
P1
P2
P3
P4
P5
Instrumentation
Database Develop.
CAD/CAM
Maintenance
CAD/CAM
150000
135000
250000
310000
500000
PROJ[PNO]
P1
P2
P3
P4
P5
EMP[TITLE]
Elect.Eng
Syst. Anal
Mech. Eng
Programmer
Chapter 3-11
Examples (cont.)
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Quiz:
 R(A, B) is a Relation Schema Defined over A and B
 Let domain(A) = {a1, a2} and domain(B) = {0, 1, 2}
 Which of the Following are Relations of R?
{(a1,
1), (a1, 2), (a2, 0)}
{(a1, 0), (a1, 1), (a1, 2)}
{(a1, 1), (a2, 2}, (a0, 0)}
{(a1, 1), (a2, a2}, (a0, a0)}
{(a1, 1, c1), (a2, 2)}

What if Attribute A is a Key?
Chapter 3-12
Characteristics of Attributes
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Attribute Name

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Attribute Value - Must have a Value

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An Attribute Name Refers to a Position in a Tuple by Name
Rather than Position
An Attribute Name Indicates the Role of a Domain in a
Relation
Attribute Names must be Unique Within Relations
By Using Attribute Names we can Disregard the Ordering of
Field Values in Tuples
Must Be an Atomic Value
Can Be a Null Value Meaning “Not Known”, “Not
Applicable” ...
Not Possible to have Undefined Value
Chapter 3-13
Characteristics of Relations
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No Duplicate Tuples
 It is a Set!
 The Primary Key Always Exists
No Explicit or Implicit Ordering of Tuples
No Ordering of Attributes (If They Are Referred to by
Their Names)
All Attribute Values Are Atomic
 A Special Null Value is Used to Represent Values
that are Unknown or Inapplicable to Certain Tuples
 Thus - If “No” Value is Desired, “Null” is Used
Chapter 3-14
Other Examples
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Which of Following are Relations in a Relational
Model?
R1 A
B
C
D
a2 {b1, b2} c1 d5
a2 b7
c9 d5
a2 b23
c22 d1
…...
Employee
E# Ename
AGE
E2 Diamond 45
E1 Smith
30
E3 Evan
R2
A
a2
a2
a2
B
…...
b2
b7
b7
C
D
c6 d1
c9 d5
c9 d5
ADDRESS
1888 Buford Hyw.
3302 Peachtree Rd., Atlanta, GA
Baker Ct. Atlanta
Chapter 3-15
Data Structure: Summary
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Relational Schema R( A1:D1, A2 :D2 , ..., An :Dn)
Relation R(r) With
 Tuples of n Columns
Denoted as Ti = {vi1,vi2, ...,vin}
 Attributes Aj (I=1,…,m) and R[aj] = {v1j,v2j, ...,vmj},
 Domain(aj.) is a Subset of D , and Several Attributes
1
may be Defined on the Same Domain
 Degree N: Number of Attributes
 Cardinality M: Number of Tuples
Chapter 3-16
Quiz
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R(A, B) is a Relation Schema Defined Over A and B
Let Domain(A) = {a1, a2} and Domain(B) = {0, 1, 2}
 Is R(A, B) Equivalent to R(B, A)? Yes
 How May Possible Tuples?
2×3=6
 Is the Set {(a1, 1), (a2, 2}, (a0, 0)} a Relation of
Schema R? No
 What is the Degree of a Relation of Schema R? 2
 What is the Cardinality of the Following Relation
3
{(a1, 1), (a1, 2), (a2, 0)} of Schema R?
Chapter 3-17
Summary of Model: Common Terms
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Informal
 Table
 Column
 Row (Instance)
 Table Definition
 Populated Table

Formal
 Relation
 Attribute
 Tuple
 Schema of Relation
 Extension
Chapter 3-18
Summary of Model: Theoretical Foundation
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Notion of Relation and Tuple is Modeled as in Set
Theory
Changes From Set Theory
 Existence of Null Value in the Tuples
 Most Implementation Allow Duplicate Tuples in
Result Sets (such as Projection)
Interpretation of Relations:
Interpretation
Linguistic
Logical
Schema
declaration
assertion
Tuple
fact
instance
of assertion
Logical
predicate
values of
satisfying
predicate
Chapter 3-19
Summary of Model: Features
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Simple and Mathematically Elegant
 Simple, Uniform Data Structure
 Solid Theoretical Foundation
Advantage of the Relational Model: Simplicity
 Separation Between Data and Data Access
 Easier to Define Data and Data Structure
 Easier to Write Queries (Specify What Not How)
 Relational DBMS can do More for You
PC-Based Systems have Brought DB to Masses
 MS Access - Easy to Use
 Integration with Office Tools (Word, Excel)
Chapter 3-20
Relational Integrity Constraints
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Integrity Constraints (ICs): Conditions that Must Hold
on All Valid Relation Instances at Any Given DB State
Why are Integrity Constraints Needed?
Multiple Flights, Customers, and Cust/Flight/Date
What Happens when we try to Delete a Flight?
FLT-SCHEDULE
FLT#
DepT
CUSTOMER
Dest
ArrT
CUST#
CUST-NAME
RESERVATION
FLT#
DATE
CUST#
Chapter 3-21
Relational Integrity Constraints Classification
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There are Three Main Types of Constraints:
 Key Constraints
 Entity Integrity Constraints
 Referential Integrity Constraints
Other Types of Semantic Constraints:
 Domain Constraints
 Transition Constraints
 Set Constraints
DBMSs Handle Some But Not All Constraints
Think About Programming Related Constraints
Throughout Our Upcoming Discussion
Chapter 3-22
Key Constraints
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Superkey (SK):
 Any Subset of Attributes Whose Values are
Guaranteed to Distinguish Among Tuples
 Could be All Attributes of Entire Relation
Candidate Key (CK):
 A Superkey with a Minimal Set of Attributes (No
Attribute Can Be Removed Without Destroying the
Uniqueness -- Minimal Identity)
 A Value of an Attribute or a Set of Attributes in a
Relation That Uniquely Identifies a Tuple
 There may be Multiple Candidate Keys
Chapter 3-23
Key Constraints
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Primary Key (PK):
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Choose One From Candidate Keys
The Primary Key Attributed are Underlined
Foreign Key (FK):



An Attribute or a Combination of Attributes (Say A) of
Relation R1 Which Occurs as the Primary Key of another
Relation R2 (Defined on the Same Domain)
Allows Linkages Between Relations that are Tracked and
Establish Dependencies
What are Foreign Keys in:
Chapter 3-24
Superkeys and Candidate Keys: Examples
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Example:
 The CAR relation schema:
CAR(State, Reg#, SerialNo, Make, Model, Year)
 Its primary key is {State, Reg#}
 It has two candidate keys

Key1 = {State, Reg#}
 Key2 = {SerialNo}
 which are also superkeys


{SerialNo, Make} is a Superkey but not a Key
Why?
If Remove SerialNo, Make is not a Primary Key
Chapter 3-25
Another Schema with Key
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CAR(License#, EngineSerialNumber, Make, Model, Year)
What are Typically Used as Keys for Cars?
Chapter 3-26
A Complete Schema with Keys ...
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Keys Allow us to
Establish Links
Between Relations
Chapter 3-27
…and Corresponding DB Tables
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Which Represent Tuples/Instances of Each Relation
A
S
C
null
W
B
null
null
1
4
5
5
Chapter 3-28
…with Remaining DB Tables
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Chapter 3-29
Another View: What Do Arrows Represent?
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Chapter 3-30
Entity Integrity Constraint
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Relational Database Schema:
 A Set S of Relation Schemas (R1, R2, ..., Rn) That
Belong to the Same Database
 S is the Name of the Database
 S = {R1, R2, ..., Rn}
Entity Integrity:
 For Any Ri in S, Pki is the Primary Key of R
 Attributes in Pki Cannot Have Null Values in any
Tuple of R(ri)
 T[pki] < > Null for Any Tuple T in R(r)
Chapter 3-31
Referential Integrity Constraints
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A Constraint Involving Two Relations Used to Specify
a Relationship Among Tuples in


Referencing Relation and Referenced Relation
Definition: R1and R2 have a Referential Integrity
Constraint If


Tuples in the Referencing Relation R1 have a Set of
Foreign Key (FK) Attributes That Reference the
Primary Key PK of the Referenced Relation R2
A Tuple T1 in R1( A1, A2 , ..., An) is Said to
Reference a Tuple T2 in R2 if $ FK {A1, A2 , ...,
An} such that T1[fk] = T2[pk]
Chapter 3-32
Examples
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WORKS
EMP
ENO
ENAME
TITLE
ENO
PNO
E1
E2
E3
E4
E5
E6
E7
E8
J. Doe
M. Smith
A. Lee
J. Miller
B. Casey
L. Chu
R. Davis
J. Jones
Elect. Eng.
Syst. Anal.
Mech. Eng.
Programmer
Syst. Anal.
Elect. Eng.
Mech. Eng.
Syst. Anal.
E1
E2
E2
E3
E3
E4
E5
E6
E7
E7
E8
P1
P1
P2
P3
P4
P2
P2
P4
P3
P5
P3
PROJ
RESP
Manager
Analyst
Analyst
Consultant
Engineer
Programmer
Manager
Manager
Engineer
Engineer
Manager
DUR
12
24
6
10
48
18
24
48
36
23
40
Can we Add this Tuple to WORKS?
PNO
PNAME
BUDGET
P1
P2
P3
P4
P5
Instrumentation
Database Develop.
CAD/CAM
Maintenance
CAD/CAM
150000
135000
250000
310000
500000
E9
P3
Engineer
30
Chapter 3-33
Referential Integrity Constraints
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A Referential Integrity Constraint Can Be Displayed in
a Relational Database Schema as a Directed Arc From
R1.FK to R2.PK
EMP
PROJ
ENO ENAME TITLE
WORK
ENO PNO
PNO PNAME BUDGET
RESP
DUR
WORK[ENO] is a subset of EMP[ENO]
WORK[PNO] is a subset of PROJ[PNO]
Chapter 3-34
Integrity Constraints Summary
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Relational Database: Set of Relations Satisfying the
Integrity Constraints
Integrity Constraints (ICs): Conditions that Must Hold
on All Valid Relation Instances
 Key Constraints - Uniqueness of Keys
 Entity ICs - No Primary Key Value is Null
 Referential ICs Between Two Relations, Cross
References Must Point to Existing Tuples
 Domain ICs are Limits on the Value of Particular
Attribute
 Transition ICs Indicate the Way Values Changes
Due to Database Update
Chapter 3-35
What is Relational Algebra?
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Relational Algebra is a Procedural Paradigm
You Need to Tell What/How to Construct the Result
Consists of a Set of Operators Which, When Applied
to Relations, Yield Relations (Closed Algebra)
Basic Relational Operations:

Unary Operations
 SELECT s


or P.
Binary Operations
 Set operations:
 UNION 
 INTERSECTION 
 DIFFERENCE –
 CARTESIAN PRODUCT 
 JOIN operations

 PROJECT
Chapter 3-36
Relational Algebra
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RS
RS
R-S
RS
union
intersection
set difference
Cartesian product
A1, A2, ..., An (R)
projection
sF (R)
selection
R S
natural join
R S
theta-join
RS
division
 [A1 B1,.., An Bn]rename
Chapter 3-37
Selection
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Selects the Tuples (Rows) From a Relation, Which
Satisfy a Selection Condition
Selection Produces a Horizontal Subset of the Operand
Relation
General Form sF (R)
 R is a Relation
 F is a Boolean Expression on the Attributes of R
 Resulting Relation Has the Same Schema as R
Select Finds and Retrieves All Relevant Rows (Tuples)
of Table/Relation R which Includes ALL of its
Columns (Attributes)
Chapter 3-38
Selection Example
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EMP
ENO
ENAME
TITLE
E1
E2
E3
J. Doe
M. Smith
A. Lee
Elect. Eng.
Syst. Anal.
Mech. Eng.
E4
E5
E6
J. Miller
B. Casey
L. Chu
Programmer
Syst. Anal.
Elect. Eng.
E7
E8
R. Davis
J. Jones
Mech. Eng.
Syst. Anal.
s TITLE='Elect. Eng.'(EMP)
ENO
E1
E6
ENAME
J. Doe
L. Chu
TITLE
Elect. Eng
Elect. Eng.
s TITLE='Elect. Eng.’ OR TITLE=‘Mech.Eng’(EMP)
Chapter 3-39
Another Selection Example
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A
S
C
null
W
B
null
null
Chapter 3-40
Selection Condition
A SELECT Condition is a Boolean Expression
 Form F1 Y F2 Y ..., Y Fq (Q>=1), Where
 Fi (I=1,…,q) are Atomic Boolean Expressions of the Form
ac or ab,
 a, b are Attributes of R and c is a Constant.
 The Operator Q is one of the Arithmetic Comparison
Operators: <, >, =, <>, >=, <=

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The Operator Y is one of the Logical Operators: , , ¬
 Nesting: ( )
Conditions are Essentially “Conditional Expressions” in PLs


Chapter 3-41
Projection
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


Extract Only Certain Columns (Attributes) Specified in
an Attribute List X From a Relation R
Produces a New Relation, which is a Vertical Subset of
the Operand Relation R
The Schema (Columns) of the Resulting Relation is X
General Form X(R)
 R is a Relation
 X is a Subset of the Attributes of R Over Which the
Projection is Performed
Project Retrieves Specified Columns of Table/Relation
R which Includes ALL of its Rows (Tuples)
Chapter 3-42
Projection Example
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PROJ
PNO
PNAME
BUDGET
P1
Instrumentation
150000
P2
Database Develop.
135000
P3
CAD/CAM
250000
P4
P5
Maintenance
CAD/CAM
310000
500000
 PNO,BUDGET(PROJ)
PNO
BUDGET
P1
150000
P2
135000
P3
P4
P5
250000
310000
500000
Chapter 3-43
Other Projection Examples
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Chapter 3-44
Characteristics of Projection

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The PROJECT Operation Eliminates Duplicate Tuples
in the Resulting Relation
 Why?
 Projection Must Maintain a Mathematical Set (No
Duplicate Elements)
EMP
 TITLE(PROJ)
ENO
ENAME
TITLE
TITLE
E1
E2
E3
J. Doe
M. Smith
A. Lee
Elect. Eng.
Syst. Anal.
Mech. Eng.
E4
E5
E6
E7
E8
J. Miller
B. Casey
L. Chu
R. Davis
J. Jones
Programmer
Syst. Anal.
Elect. Eng.
Mech. Eng.
Syst. Anal.
Elect.Eng
Syst.Anal
Mec.Eng
Programmer
Chapter 3-45
Relational Algebra Expression

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Several Operations can be Combined to form a
Relational Algebra Expression (query)
Example: Retrieve all Customers over age 60?
 Method 1:
 CNAME, ADDRESS, AGE (s AGE>60(CUSTOMER) )
 Method 2:
Senior-CUST(C#, Addr, Age)

=  CNAME, ADDRESS, AGE (s AGE>60(CUSTOMER) )
Method 3:
 CNAME, ADDRESS, AGE (C) where
C = s AGE>60(CUSTOMER)
Chapter 3-46
Selection with Projection Example
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Chapter 3-47
Union

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General Form
RS
where R, S are Relations
 Result contains Tuples from both R and S
 Duplications are Removed
 The two Operands R, S should be union-compatible
(type-compatible w.r.t Columns/Attributes)
Example:
“find students registered for course C1 or C3”
s#(sCNO=‘C1’ (S-C))  s#(sCNO=‘C3’ (S-C))
Chapter 3-48
Union Compatibility

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
Two Relations R1(A1, A2, ..., An) and R2(B1, B2, ..., Bn)
are said Union-compatible If and Only If They Have
 The Same Number of Attributes
 The Domains of Corresponding Attributes are
Compatible, i.e., Dom(Ai)=dom(Bi) for I=1, 2, ..., N
 Names Do Not Have to be Same!
For Relational Union and Difference Operations, the
Operand Relations Must Be Union Compatible
The Resulting Relation for Relational Set Operations
 Has the Same Attribute Names as the First Operand
Relation R1 (by Convention)
Chapter 3-49
Set Difference

General Form
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R–S
where R and S are Relations


Result Contains all tuples that are in R, but not in S.

R – S <> S – R

Again, there Must be Compatibility
Example
“Find the students who registered course C1 but not C3”
s#(sCNO=‘C1’ (S-C)) – s#(sCNO=‘C3’ (S-C))
Chapter 3-50
Set Intersection

General Form
CSE
4701
RS
where R and S are Relations




Result Contains all Tuples that are in R and S.
R  S = R – (R – S)
Again, there Must be Compatibility
Example
“find the students who registered for both C1 and C3”
s#(sCNO=‘C1’ (S-C))  s#(sCNO=‘C3’ (S-C))
Chapter 3-51
Union, Difference, Intersection Examples
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4701
What are these Other
Three Result Tables?
Chapter 3-52
Cartesian Product

CSE
4701



Given Relations
 R of Degree k1 and Cardinality card1
 S of Degree k2 and Cardinality card2
Cartesian Product
RS
is a Relation of Degree (k1+ k2) and Consists of Tuples
of Degree (k1+ k2) where each Tuple is a Concatenation
of one Tuple of R with one Tuple of S
Cardinality of the Result of the Cartesian Product
R  S is card1 * card2
What is One Problem with Cartesian Product w.r.t. the
Result Set?
Chapter 3-53
Cartesian Product: Example
CSE
4701
R
A
B
C
a1
a2
a3
b1
b1
b4
c3
c5
c7
R S
S
A
a1
a1
a2
a2
a3
a3
B
b1
b1
b1
b1
b4
b4
C
c3
c3
c5
c5
c7
c7
E
F
e1
e2
f1
f5
E
e1
e2
e1
e2
e1
e2
F
f1
f5
f1
f5
f1
f5
Chapter 3-54
Cartesian Product
CSE
4701


Given R(A1, …,An) and S(B1,…,Bm), the result of a
Cartesian product R  S is a relation of schema
R’(A1, …, An, B1, …, Bm).
Example
“Get a list containing (S#, C#) for all students who live in Storrs
but are not registered for the database course”
(S#(scity=‘Storrs’(STUDENT)) 
C# (sCNAME=‘Database’(COURSE))) – S#, C#(S-C)
Chapter 3-55
Cartesian Product Example
Generates Lots of “Data” that Doesn’t Make Sense
CSE
4701 EMP
ENO
EMP  SAL
ENAME
TITLE
E1
E2
J. Doe
M. Smith
Elect. Eng
Syst. Anal.
E3
A. Lee
Mech. Eng.
E4
E5
E6
J. Miller
B. Casey
L. Chu
Programmer
Syst. Anal.
Elect. Eng.
E7
E8
R. Davis
J. Jones
Mech. Eng.
Syst. Anal.
SAL
TITLE
Elect. Eng.
Syst. Anal.
Mech. Eng.
Programmer
SAL
40000
34000
27000
24000
ENO
ENAME
EMP.TITLE
SAL.TITLE
SAL
E1
E1
E1
J. Doe
J. Doe
J. Doe
Elect. Eng.
Elect. Eng.
Elect. Eng.
Elect. Eng.
Syst. Anal.
Mech. Eng.
40000
34000
27000
E1
J. Doe
Elect. Eng.
Programmer
24000
E2
E2
E2
M. Smith
M. Smith
M. Smith
Syst. Anal.
Syst. Anal.
Syst. Anal.
Elect. Eng.
Syst. Anal.
Mech. Eng.
40000
34000
27000
E2
E3
E3
M. Smith
A. Lee
A. Lee
Syst. Anal.
Mech. Eng.
Mech. Eng.
Programmer
Elect. Eng.
Syst. Anal.
24000
40000
34000
E3
E3
A. Lee
A. Lee
Mech. Eng.
Mech. Eng.
Mech. Eng.
Programmer
27000
24000
E8
E8
E8
J. Jones
J. Jones
J. Jones
Syst. Anal.
Syst. Anal.
Syst. Anal.
Elect. Eng.
Syst. Anal.
Mech. Eng.
40000
34000
27000
E8
J. Jones
Syst. Anal.
Programmer
24000
Chapter 3-56
Homework 2 Spring 15 Problems
CSE
4701

Problem 6.18, parts a, b, c, and d – Using Cartesian
Product
Chapter 3-57
The Library Schema (Figure 6.14)
CSE
4701
Chapter 3-58
Parts a and b
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4701
a. How many copies of the book titled The Lost Tribe are owned by the
library branch whose name is ‘Sharpstown’? (BaB – books at Branches)
BaB= (BOOKCOPIES × (sTitle=‘The Lost Tribe’ (BOOK))) )
Ans = No_Of_Copies( (sBranchName=‘Sharpstown’ (LIBRARY-BRANCH)) × BaB)
b. How many copies of the book titled The Lost Tribe are owned by
each library branch? (CaB- Copies at Branches)
CaB = BOOKCOPIES × LIBRARY_BRANCH)
Ans = BranchName, No_Of_Copies( (sTitle=‘The Lost Tribe’ (BOOK)) × CaB)
Chapter 3-59
Reminder – Discuss Homework 1
CSE
4701
Chapter 3-60
Theta Join (-Join)

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4701
General Form
R



S
where
 R, S are Relations,
 F is a Boolean Expression, called a Join Condition.
A Derivative of Cartesian Product
 R
 S = s (R  S)
R(A1, A2, ..., Am, B1, B2, ..., Bn) is the Resulting
Schema of a -Join over R1 and R2:
R1(A1, A2, ..., Am)
 R2 (B1, B2, ..., Bn)
Chapter 3-61
-Join Condition

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4701
A -Join Condition is a Boolean Expression of the
form F1 y1 F2 y2 ..., yn-1 Fq (q>=1), where
 Fi (i=1,…,q) are Atomic Boolean Expressions of
the form
Ai  Bj,
 Ai, Bj are Attributes of R1 and R2 Respectively

 is one of the Algorithmic Comparison Operators
=, <>, >, <. >=, <=


The Operator yi (i=1,…,n-1) is Either a Logical
AND operator  or a logical OR operator 
Again – a Conditional Expression
Chapter 3-62
-Join Example
CSE
4701
EMP
Notice “Absence” of Nonsensical Data from Cart Product.
ENO
E1
E2
E3
E4
E5
E6
E7
E8
ENAME
J. Doe
M. Smith
A. Lee
J. Miller
B. Casey
L. Chu
R. Davis
J. Jones
TITLE
Elect. Eng
Syst. Anal.
Mech. Eng.
Programmer
Syst. Anal.
Elect. Eng.
Mech. Eng.
Syst. Anal.
SAL
TITLE
Elect. Eng.
Syst. Anal.
Mech. Eng.
Programmer
SAL
40000
34000
27000
24000
EMP
E.TITLE=SAL.TITLE
SAL
TITLE
SAL.TITLE
SAL
J. Doe
M. Smith
Elect. Eng.
Analyst
Elect. Eng.
Analyst
40000
34000
E3
A. Lee
Mech. Eng.
Mech. Eng.
27000
E4
J. Miller
Programmer
E5
B. Casey
Syst. Anal.
Programmer 24000
Syst. Anal. 34000
E6
L. Chu
Elect. Eng.
Elect. Eng.
40000
E7
E8
R. Davis
J. Jones
Mech. Eng.
Syst. Anal.
Mech. Eng.
Syst. Anal.
27000
34000
ENO
ENAME
E1
E2
Chapter 3-63
Other Types of Join

CSE
4701

Equi-join (EQUIJOIN)
 The  Expression only Contains one or more Equality
Comparisons Involving Attributes from R1 and R2
 User Specifies Equality Conditions
Natural Join
 Denoted as R
S
 Special Equi-join of Two Relations R and S Over a Set of
Attributes Common to both R and S
 By Common, it means that each Join Attribute in A has not
only Compatible Domains but also the Same Name in both
Relations R and S
 System Automatically Picks the Equality Conditions
Chapter 3-64
Examples
CSE
4701
R
A
B
C
a1
a2
a3
b1
b1
b4
c3
c5
c7
S
EQUIJOIN
R
R.B=S.B S
A
R.B S.B
a1 b1 b1
a2 b1 b1
B
E
b1
b5
e1
e2
Natural Join
R
S
C
E
A R.B
C
c3
c5
e1
e1
a1
a2
c3 e1
c5 e1
b1
b1
E
Chapter 3-65
Homework 2 Spring 15 Problems
CSE
4701

Problem 6.18, parts a, b, c, and d – with Join
Chapter 3-66
The Library Schema (Figure 6.14)
CSE
4701
Chapter 3-67
Parts a and b
CSE
4701
a. How many copies of the book titled The Lost Tribe are owned by the
library branch whose name is ‘Sharpstown’? (BaB – books at Branches)
BaB= (BOOKCOPIES * (sTitle=‘The Lost Tribe’ (BOOK))) )
BookId
Ans = No_Of_Copies( (sBranchName=‘Sharpstown’ (LIBRARY-BRANCH)) * BaB)
BranchID
b. How many copies of the book titled The Lost Tribe are owned by
each library branch? (CaB- Copies at Branches)
CaB = BOOKCOPIES * LIBRARY_BRANCH)
BranchId
Ans = BranchName, No_Of_Copies( (sTitle=‘The Lost Tribe’ (BOOK)) * CaB)
BookId
Chapter 3-68
Parts c and d
CSE
4701
c. Retrieve the names of all borrowers who do not have any books checked out.
NO_CHECKOUT_B =CardNo(BORROWER) - CardNo(BOOK_LOANS)
Ans = Name(BORROWER * NO_CHECKOUT_B)
d. For each book that is loaned out from the Sharpstown branch and whose Due_date is
today, retrieve the book title, the borrower’s name, and the borrower’s address.
S =BranchID(s BranchName=‘Sharpstwon’ (LIBRARY-BRANCH))
B_FROM_S = BookID, CardNo ( (s DueDate=‘Today’ (BOOKLOANS)) * S )
BranchId
Ans = Title, Name, Address ( BOOK * BORROWER * B_FROM_S )
BookId
CardNo
Chapter 3-69
Natural Join

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4701

Natural Join Combines Relations on Attributes with the
Same Names
 STUDENT(S#, SN, CITY, Email)
 S-C(S#, C#, G)
Example Query 1:
“list of students with complete course grade info”
STUDENT
S-C


All Natural Joins can be Expressed by a Combination of
Primitive Operators
Example Query 2:
“print all students info (courses taken and grades)”
S#, SN, CITY, Email, C#, G (sSTUDENT.S# = S-C.S# (STUDENT  S-C))
Chapter 3-70
Natural Join Example
CSE
4701
EMP
ENO
ENAME
TITLE
EMP
SAL
E1
E2
J. Doe
M. Smith
Elect. Eng
Syst. Anal.
E3
A. Lee
Mech. Eng.
ENO
ENAME
E.TITLE
E4
E5
E6
J. Miller
B. Casey
L. Chu
Programmer
Syst. Anal.
Elect. Eng.
E1
J. Doe
Elect. Eng.
70000
E2
M. Smith
Syst. Anal.
80000
E7
E8
R. Davis
J. Jones
Mech. Eng.
Syst. Anal.
E3
E4
A. Lee
J. Miller
Mech. Eng.
Programmer
56000
60000
SAL
TITLE
Elect. Eng.
Syst. Anal.
Mech. Eng.
Programmer
SAL
70000
80000
56000
60000
SAL
E5
B.Casey
Syst.Anal
80000
E6
L. Chu
Elect.Eng
70000
E7
R.Davis
Mech.Eng
56000
E8
J. Jones
Syst. Anal.
80000
Chapter 3-71
Another Natural Join Example
CSE
4701
Chapter 3-72
Yet Another Natural Join Example
CSE
4701
1
4
5
5
Chapter 3-73
Quotient (Division)
CSE
4701




Given Relations
 R(T,U) of degree r
 S(U) of degree s
The Division of R by S,
R÷S
Results is a Relation of Degree (rs)
Consists of all (rs)-tuples t such that for all s-tuples u
in S, the tuple tu is in R.
Chapter 3-74
Division Example
CSE
4701
R
ENO PNO
E1
E2
E2
E3
E3
E4
E5
E6
E7
E8
P1
P1
P2
P1
P4
P2
P2
P4
P3
P3
PNAME
BUDGET
Instrumentation
Instrumentation
Database Develop.
Instrumentation
Maintenance
Instrumentation
Instrumentation
Maintenance
CAD/CAM
CAD/CAM
150000
150000
135000
150000
310000
150000
150000
310000
250000
250000
S
Find the
employees who
work for both
project P1 and
project P4?
R÷S
PNO
PNAME
P1
P4
Instrumentation
Maintenance
BUDGET
150000
310000
ENO
E3
Chapter 3-75
Relational Algebra
CSE
4701







Selection
Projection
Union
Difference
Cartesian Product
Intersection
Join, Equi-join, Natural Join
Derivable from the
fundamental operators
Fundamental Operators
Chapter 3-76
All Relational Algebra Operations

CSE
4701

A Set of Relational Algebra Operations Is Called a
Complete Set, If and Only If
 Any Relational Algebra Operator in the Set Cannot
be Derived in Terms of a Sequence of Others in Set
 Any Relational Algebra Operator Not in the Set Can
Be Derived in Terms of a Sequence of Only the
Operators in the Set
Important Concepts:
1. The Set of Algebra Operations {, s, , –, } is a
Complete Set of Relational Algebra Operations
2. Any Query Language Equivalent to These Five
Operations is Called Relationally Complete
Chapter 3-77
Relational Algebra: Summary
CSE
4701

Fundamental Operators
 Selection
 Projection
 Union
 Set Difference
 Cartesian Product


Additional Operators
 Join
 Intersection
 Quotient (Division)
Union Compatibility
 Same Degree
 Corresponding
Attributes Defined Over
the Same Domain
Form:
<Operator><Operand(s)>  <Result>


Relation (s)
Relation
Chapter 3-78
Semi-Join

CSE
4701



General form: R
FS
R is Target Relation, S is the Source Relation
Semi-join
 A Form of Join where the Result Contains ONLY
those Tuples of the Target Relation, which
Participate in the Join with the Source Relation
Benefit: Decreases the Number of Tuples that need to
be Handled in the Join
R
F S = P(R
FS)
= P(R) FP(S) = R FP(S)
 where A is a set of Attributes of R, B is a set of
attributes of S
Chapter 3-79
Semi-Join Example
CSE
4701
R
B
A
a1
a2
a3
a4
b1
b1
b3
b4
R
SAL
=R
S
B
b2
b2
b3
b4
b5
b6
b7
b8
b9
b10
C
D
c1
c1
c1
c2
c2
c2
c3
c3
c4
c4
d9
d8
d7
d9
d8
d7
d5
d4
d4
d9
A
a3
a4
(P B S)
B
b3
b4
Chapter 3-80
Aggregate Functions

CSE
4701
Commonly Used Aggregate Functions:
 SUM, COUNT, AVERAGE, MIN, MAX
 They Often Applied to sets of Values or sets of
Tuples in Database Applications
[<grouping attributes> ] f <function list> (R)
 The Grouping Attributes are Optional
Chapter 3-81
Examples of Aggregate Functions
CSE
4701

Example 1:

Retrieve the average salary of all employees (no grouping):
R(AVG_SAL) = s

AVG(SALARY) (EMPLOYEE)
Example 2:

For each department, retrieve the department number, the
number of employees, and the average salary (in the
department):
R(D#,Num_E,AVG_SAL)
= GROUPDNO( s DNO, COUNT(ENO), AVG(SALARY)(EMPLOYEE))

DNO is called the grouping attribute in this example
Chapter 3-82
Operations on Relations
CSE
4701





A DBMS Operates via User Queries to Read and
Change Data in a Database
Changes Can be Inserting, Deleting, or Updating
(Equivalent to a Delete followed by Insert)
One Critical Issue in DB Operations is Integrity
Constraints Maintenance in the Presence of
 INSERTING a Tuple
 DELETING a Tuple
 UPDATING/MODIFYING a Tuple.
We’ll discuss Each case in Turn
What is Constraint Maintenance Similar to in PL?
Chapter 3-83
Problem Statements

CSE
4701


Integrity Constraints (ICs) Should Not Be Violated by
Update Operations
To Maintain ICs, Updates may Need to be Propagated
and Cause Other Updates Automatically
 Common Method: Group Several Update Operations
Together As a Single Transaction
If Integrity Violation, Several Actions Can Be Taken:
 Cancel Operation that Caused Violation (REJECT)
 Perform the Operation but Inform User of Violation
 Trigger Additional Updates So the Violation is
Corrected (CASCADE Option, SET NULL Option)
 Execute a User-specified Error-Correction Routine
(Similar to What in a PL Like Java?)
Chapter 3-84
Insertion Operations on Relations

CSE
4701



Insert a Duplicate Key Violates Key Integrity:
 Check If Duplicates Occur
Insert a Null Key Violates Entity Integrity:
 Check If Null is in Any Key
Insert a Tuple Whose Foreign Key Attribute Pointing
to an Non-existent Tuple Violates Referential Integrity:
 Check the Existence of Referred Tuple
Correction Actions:
 Reject the Update
 Correct the Violation - Change Null, Duplicate, Etc.
 Cascade the Access - Insert a New Tuple That Did
Not Exist/Delete Tuples that are being Referenced
Chapter 3-85
Examples
CSE
4701
EMP
?
?
WORKS
ENO
ENAME
TITLE
E1
E2
E3
E4
E5
J. Doe
M. Smith
A. Lee
J. Miller
B. Casey
Elect. Eng.
Syst. Anal.
Mech. Eng.
Programmer
Syst. Anal.
E6
L. Chu
E3
R. Davis
ENO
PNO
E1
E2
E2
E3
E3
E4
E5
P1
P1
P2
P3
P4
P2
P2
RESP
Manager
Analyst
Analyst
Consultant
Engineer
Programmer
Manager
DUR
12
24
6
10
48
18
24
Mech. Eng.
E1
PROJ
PNO
PNAME
BUDGET
P1
P2
P3
P4
P5
Instrumentation
Database Develop.
CAD/CAM
Maintenance
CAD/CAM
150000
135000
250000
310000
500000
Engineer
E1
P5
Engineer
E8
P3
Manager
36
40
?
?
?
Chapter 3-86
Deletion Operations on Relations

CSE
4701


Deleting a Tuple Referred to by Other Tuples in
Database (via FKs) would Violate Referential Integrity
Action:
 Check for Incoming Pointers of the Deleted Tuple.
 Group the Deletion and the Post-processing of the
Referencing Pointers in a Single Transaction
Three Options If Deletion Causes a Violation
 Reject the Deletion
 Attempt to Cascade (Propagate) the Deletion by
Deleting the Tuples which Reference the Tuple
being or to be Deleted
 Modify the Referencing Attribute Values that Cause
the Violation; Each Values is Set to Null or Changed
to Reference to Another Valid Tuple
Chapter 3-87
Example
CSE
4701
EMP
WORKS
ENO
ENAME
TITLE
E1
E2
E3
E4
E5
E6
J. Doe
M. Smith
A. Lee
J. Miller
B. Casey
L. Chu
Elect. Eng.
Syst. Anal.
Mech. Eng.
Programmer
Syst. Anal.
Elect. Eng.
ENO
PNO
E1
E2
E2
E3
E3
E4
E5
E6
P1
P1
P2
P3
P5
P2
P2
P4
RESP
Manager
Analyst
Analyst
Consultant
Engineer
Programmer
Manager
Manager
DUR
12
24
6
10
48
18
24
48
1. Cascading
Deleting
this tuple?
PROJ
PNO
PNAME
BUDGET
P1
P2
P3
P4
P5
Instrumentation
Database Develop.
CAD/CAM
Maintenance
CAD/CAM
150000
135000
250000
310000
500000
E5
2. reference revision?
Chapter 3-88
Modify Operations on Relations
CSE
4701



Modify Operation Changes Values of One or More
Attributes in a Tuple (or Tuples) of a Given Relation R
Maintaining ICs Requires to Check If the Modifying
Attributes Are Primary Key or Foreign Keys.
Integrity Check Actions:
 Case 1:
If
the Attributes to be Modified are Neither a Primary Key
nor a Foreign Key, Modify Causes No Problems
Must Check and Confirm that the New Value is of
Correct Data Type and Domain

Case 2:
Modifying
a Primary Key Value Similar to Deleting One
Tuple and Insert Another in its Place
Chapter 3-89
Constraints and Update Operations

CSE
4701



Three Types of Update Operations:
 INSERT, DELETE, MODIFY
Constraint Maintenance During Updates
The Types of Constraints
That Most DBMSs
Maintain are
 Key Constraints
 Entity Constraints
 Referential Integrity
Constraints

Other Semantic Constraints Need
to Be Maintained by Application
Developers/programmers
 Transition Constraints
 Domain Constraints
 Etc.
Some DB Do Maintain Domain Constraints via
Enumeration and Value-Range Data Types
Chapter 3-90
Concluding Remarks

CSE
4701

What have we Seen in Chapters 3 & 6?
 Basic Concepts of Relational Model Including
Relation/Table, Tuple/Row, Attribute/Column,
Domain/Attribute Value
 Concept of SK, CK, PK, and FK for Identification
and Referential Integrity
 Integrity Constraints as they Relate to Referential
Dependencies Check for Modification Operations
Overall, Relational Theory is Basis for SQL, Normal
Forms, ER-Relational Translation, etc.
Chapter 3-91