CSE 4701 Chapter 6-1 Chapter 6 6e & 7 5e : Relational Algebra – Part 2 Prof. Steven A. Demurjian, Sr. Computer Science & Engineering Department The University

CSE 4701

Chapter 6-1

Chapter 6 6e & 7 5e : Relational Algebra – Part 2

Prof. Steven A. Demurjian, Sr. Computer Science & Engineering Department

The University of Connecticut191 Auditorium Road, Box U-155

Storrs, CT [email protected]

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.

CSE 4701

Chapter 6-2

Basic Relational Operations: Unary Operations

SELECT s PROJECT or P.

Binary Operations Set operations:

UNION INTERSECTION DIFFERENCE –

CARTESIAN PRODUCT JOIN operations

What is Relational Algebra?

Relational Algebra is a Procedural ParadigmYou Need to Tell What/How to Construct the Result

Consists of a Set of Operators Which, When Applied to Relations, Yield Relations (Closed Algebra)

CSE 4701

Chapter 6-3

Relational Algebra

R S unionR S intersectionR \ S set differenceR S Cartesian

product

A1, A2, ..., An (R) projection

F (R) selection

R S natural join

R S theta-joinRS division[A1 B1,.., An Bn] rename

CSE 4701

Chapter 6-4

Selection

Selects the Tuples (Rows) From a Relation, Which Satisfy a Selection Condition

Selection Produces a Horizontal Subset of the Operand Relation

General Form F (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)

CSE 4701

Chapter 6-5

ENO ENAME TITLE

E1 J. Doe Elect. EngE6 L. Chu Elect. Eng.

TITLE='Elect. Eng.'(EMP)ENO ENAME TITLE

E1 J. Doe Elect. Eng.

E2 M. Smith Syst. Anal.

E3 A. Lee Mech. Eng.

E4 J. Miller Programmer

E5 B. Casey Syst. Anal.

E6 L. Chu Elect. Eng.

E7 R. Davis Mech. Eng.

E8 J. Jones Syst. Anal.

EMP

TITLE='Elect. Eng.’ OR TITLE=‘Mech.Eng’(EMP)

Selection Example

CSE 4701

Chapter 6-6

Another Selection Example

ASCnullWBnullnull

CSE 4701

Chapter 6-7

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

aqc or aqb, a, b are Attributes of R and c is a Constant. The Operator Q is one of the Arithmetic

Comparison Operators: <, >, =, <>, >=, <= The Operator Y is one of the Logical Operators: , , ¬

Nesting: ( )

Selection Condition

CSE 4701

Chapter 6-8

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)

Projection

CSE 4701

Chapter 6-9

Projection Example

PNO,BUDGET(PROJ)

PNO BUDGET

P1 150000

P2 135000

P3 250000

P4 310000

P5 500000

PROJ

PNO BUDGET

P2 135000

P3 250000

P4 310000P5 500000

PNAME

P1 150000Instrumentation

Database Develop.

CAD/CAM

MaintenanceCAD/CAM

CSE 4701

Chapter 6-10

Other Projection Examples

CSE 4701

Chapter 6-11

Relational Algebra Expression 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)

CSE 4701

Chapter 6-12

ENO ENAME TITLE









EMP TITLE(PROJ)

Elect.EngSyst.AnalMec.EngProgrammer

TITLE

Characteristics of Projection The PROJECT Operation Eliminates Duplicate Tuples

in the Resulting Relation Why? Projection Must Maintain a Mathematical Set (No

Duplicate Elements)

CSE 4701

Chapter 6-13

Selection with Projection Example

CSE 4701

Chapter 6-14

General FormR 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#(CNO=‘C1’ (S-C))

s#(

CNO=‘C3’ (S-C))

Union

CSE 4701

Chapter 6-15

Union Compatibility 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)

CSE 4701

Chapter 6-16

General Form

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#(CNO=‘C1’ (S-C)) –

s#(

CNO=‘C3’ (S-C))

Set Difference

CSE 4701

Chapter 6-17

General Form

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#(CNO=‘C1’ (S-C))

s#(

CNO=‘C3’ (S-C))

Set Intersection

CSE 4701

Chapter 6-18

Union, Difference, Intersection Examples

What are these OtherThree Result Tables?

CSE 4701

Chapter 6-19

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?

Cartesian Product

CSE 4701

Chapter 6-20

Cartesian Product: Example

A B C

a1a2a3

b1b1b4

c3c5c7

FE

f1f5

e1e2

A B C E

a1a1a2a2a3a3

b1b1b1b1b4b4

c3c3c5c5c7c7

e1e2e1e2e1e2

R S

R S F

f1f5f1f5f1f5

CSE 4701

Chapter 6-21

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”

Cartesian Product

(S#(

city=‘Storrs’(STUDENT))

C#

(CNAME=‘Database’

(COURSE))) – S#,

C#(S-C)

CSE 4701

Chapter 6-22

Cartesian Product Example

ENO ENAME EMP.TITLE SAL.TITLE SAL





Elect. Eng. 40000

Syst. Anal. 34000

Mech. Eng. 27000

Programmer 24000





Elect. Eng. 40000

Syst. Anal. 34000

Mech. Eng. 27000

Programmer 24000

Elect. Eng. 40000

Syst. Anal. 34000

Mech. Eng. 27000

Programmer 24000

Elect. Eng. 40000

Syst. Anal. 34000

Mech. Eng. 27000

Programmer 24000









EMP SAL

ENO ENAME TITLE

E1 J. Doe Elect. Eng








EMP

TITLE SAL

SAL

Elect. Eng. 40000

Syst. Anal. 34000

Mech. Eng. 27000

Programmer 24000

CSE 4701

Chapter 6-23

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 = (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)

Theta Join (-Join)

CSE 4701

Chapter 6-24

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 q 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

-Join Condition

CSE 4701

Chapter 6-25

-Join Example

ENO ENAME TITLE









EMP

TITLE SAL

SAL

Elect. Eng. 40000

Syst. Anal. 34000

Mech. Eng. 27000

Programmer 24000

ENO ENAME

E1 J. Doe

M. SmithE2

E3 A. Lee

E4 J. Miller

E5 B. Casey

E6 L. Chu

E7 R. DavisE8 J. Jones

TITLE

Elect. Eng.

Analyst

Mech. Eng.

Programmer

Syst. Anal.

Elect. Eng.

Mech. Eng.Syst. Anal.

SAL

40000

34000

27000

24000

34000

40000

2700034000

EMP E.TITLE=SAL.TITLE SAL

SAL.TITLE

Elect. Eng.

Analyst

Mech. Eng.

Programmer

Syst. Anal.

Elect. Eng.

Mech. Eng.Syst. Anal.

CSE 4701

Chapter 6-26

Other Types of Join Equi-join (EQUIJOIN)

The Expression only Contains one or more Equality Comparisons Involving Attributes from R1 and R2

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

CSE 4701

Chapter 6-27

Examples

A B C

a1a2a3

b1b1b4

c3c5c7

B E

b1b5

e1e2

R S

R R.B=S.B S

A R.B

a1a2

b1b1

C E

c3c5

e1e1

S.B

b1b1

EQUIJOIN

A R.B C E

a1a2

b1b1

c3c5

e1e1

R SNatural Join

CSE 4701

Chapter 6-28

Natural Join 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 (

STUDENT.S# = S-C.S# (STUDENT S-C))

CSE 4701

Chapter 6-29

Natural Join Example

ENO ENAME TITLE









EMP

TITLE SAL

SAL

Elect. Eng. 70000

Syst. Anal. 80000

Mech. Eng. 56000

Programmer 60000

ENO ENAME E.TITLE SAL

E1 J. Doe Elect. Eng. 70000

E2 M. Smith Syst. Anal. 80000

56000

80000



EMP SAL

60000E4 J. Miller Programmer

80000E5 B.Casey Syst.Anal

70000E6 L. Chu Elect.Eng

56000E7 R.Davis Mech.Eng

CSE 4701

Chapter 6-30

Another Natural Join Example

CSE 4701

Chapter 6-31

Yet Another Natural Join Example

1455

CSE 4701

Chapter 6-32

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 (rs) Consists of all (rs)-tuples t such that for all s-tuples u in S, the tuple tu is in R.

Quotient (Division)

CSE 4701

Chapter 6-33

Division Example

ENO

E3

R ÷ S

ENO PNO PNAME

E1 P1 Instrumentation 150000

BUDGET

E2 P1 Instrumentation 150000E2 P2 Database Develop. 135000E3 P1 InstrumentationE3 P4 MaintenanceE4 P2 InstrumentationE5 P2 InstrumentationE6 P4E7 P3 CAD/CAME8 P3 CAD/CAM

310000150000150000310000250000250000

R

Maintenance

150000

S

PNO PNAME BUDGET

P1 Instrumentation 150000P4 Maintenance 310000

Find the employees who work for both project P1 and project P4?

CSE 4701

Chapter 6-34

Division: Another Example “list the S# of students that have taken all of the

courses listed in S-C”

S#

(S-C)C#

(S-C)

“list the S# of students who have taken all of the courses taught by instructor Smith”

S#

(S-CC#

(Instructor=‘Smith’(C)))

CSE 4701

Chapter 6-35

Relational Algebra

Fundamental OperatorsDerivable from the fundamental operators

Selection Projection Union Difference Cartesian Product

Intersection Join, Equi-join, Natural Join

CSE 4701

Chapter 6-36

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:The Set of Algebra Operations {S ,P , , –, } is a

Complete Set of Relational Algebra Operations Any Query Language Equivalent to These Five

Operations is Called Relationally Complete

All Relational Algebra Operations

CSE 4701

Chapter 6-37

Fundamental Operators Selection Projection Union Set Difference Cartesian Product

Relational Algebra: Summary

Form:

<Operator><Operand(s)> Result>

Relation (s) Relation

Additional Operators Join Intersection Quotient (Division)

Union Compatibility Same Degree Corresponding

Attributes Defined Over the Same Domain

CSE 4701

Chapter 6-38

Semi-Join General form: R F S

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 = (R FS)

= (R) F(S) = R F(S) where A is a set of Attributes of R, B is a set of

attributes of S

CSE 4701

Chapter 6-39

Semi-Join Example

B C D

b2b2b3b4b5b6b7b8b9b10

S

A B

R

b1b1b3b4

a1a2a3a4. R

BSALR ( P B S)=

c1c1c1c2c2c2c3c3c4c4

d9d8d7d9d8d7d5d4d4d9

A B

b3b4

a3a4.

CSE 4701

Chapter 6-40

Additional DB Operations

Additional Operations, Needed for Database Applications, but were Not Part of the Original Relational Algebra, Include: OUTER JOIN OUTER UNION Aggregate Functions and Grouping

CSE 4701

Chapter 6-41

Definition LEFT OUTER JOIN: R1 F R2 lets every Tuple in

R1 Appear in the result. RIGHT OUTER JOIN: R1 F R2 lets every Tuple

in R2 Appear in the result. FULL OUTER JOIN: R1 F R2 lets every Tuple in

R1 or R2 Appear in the result Examples - See Textbook Chapter 7

Outer Joins

CSE 4701

Chapter 6-42

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

Aggregate Functions

CSE 4701

Chapter 6-43

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

Examples of Aggregate Functions

CSE 4701

Chapter 6-44

Concluding Remarks

What have we Seen in Chapter 7? 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

Relational Algebra Operations that Allow Querying via Selection, Project, Join, and Other Set Operations

Overall, Relational Theory is Basis for SQL, Normal Forms, ER-Relational Translation, etc.

Documents

CSE 4701 Chapter 6-1 Chapter 6 6e & 7 5e : Relational Algebra – Part 2 Prof. Steven A. Demurjian, Sr. Computer Science & Engineering Department The University