Algebra relacional

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Relational AlgebraRelational Algebra

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database Modulesdatabase Modules
Module 1: Database SystemsModule 1: Database Systems
Module 2: Relational ModelModule 2: Relational Model
Module 3: Entity Relationship ModelModule 3: Entity Relationship Model
Module 4: ER to Relational MappingModule 4: ER to Relational Mapping
Module 5: FDs and NormalizationModule 5: FDs and Normalization
Module 6: Relational AlgebraModule 6: Relational Algebra
Module 7: SQLModule 7: SQL
Module 8: Database SystemsModule 8: Database Systems
ArchitectureArchitecture
FundamentalFundamental
ss
√√
DataData
ModelingModeling√√
DataData
DesignDesign√√
DataData
AccessAccess
ArchitectureArchitecture

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MotivationMotivation
The relational data modelThe relational data model
provides a means ofprovides a means of
defining the databasedefining the database
structure and constraintsstructure and constraints
A data model must alsoA data model must also
provide a set ofprovide a set of
operations to manipulateoperations to manipulate
the datathe data
NAME SALARY ADDRESS DEPT
Smith 50k St. Lucia Printing
Dilbert 40k Taringa Printing
Jones 60k Kenmore Printing
Trump 65k Auchenflower Head Office
Harrison 78k St. Lucia Head Office
Find the names and departments of allFind the names and departments of all
employees who earn more than 55Kemployees who earn more than 55K
Increment the salary of all employeesIncrement the salary of all employees
in the printing department by 10%in the printing department by 10%
What is the address of employeeWhat is the address of employee
JonesJones
The basic set of relational model operationsThe basic set of relational model operations
constitute the Relational Algebraconstitute the Relational Algebra

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ContentsContents
• What is a Relational QueryWhat is a Relational Query
• Relational Query LanguagesRelational Query Languages
• Relational Algebra OperationsRelational Algebra Operations
• Query Formulation in Relational AlgebraQuery Formulation in Relational Algebra
• Exercises in Relational AlgebraExercises in Relational Algebra

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What is a Relational QueryWhat is a Relational Query
• Data in a relational database can beData in a relational database can be
manipulated in the following ways:manipulated in the following ways:
–– INSERTINSERT : New tuples may be inserted: New tuples may be inserted
–– DELETEDELETE : Existing tuples may be deleted: Existing tuples may be deleted
–– UPDATEUPDATE : Values of attributes in existing: Values of attributes in existing
tuplestuples
may be changedmay be changed
–– RETRIEVERETRIEVE: Attributes of specific tuples,: Attributes of specific tuples,
entireentire
tuples, or even entire relations may betuples, or even entire relations may be
retrievedretrieved
• Relational Query Languages should provide allRelational Query Languages should provide all
of the aboveof the above

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Relational Query LanguagesRelational Query Languages
Relational Queries are formulated in RelationalRelational Queries are formulated in Relational
Query LanguagesQuery Languages
• Relational Algebra (RA)Relational Algebra (RA)
–– Formal query language for a relational databaseFormal query language for a relational database
• Structured Query Language (SQL)Structured Query Language (SQL)
–– Comprehensive, commercial query language withComprehensive, commercial query language with
widely accepted international standardwidely accepted international standard
• Query by Example (QBE)Query by Example (QBE)
–– Commercial, graphical query language withCommercial, graphical query language with
minimum syntaxminimum syntax

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SQL and Relational AlgebraSQL and Relational Algebra
SQLSQL
• Declarative languageDeclarative language
–– Users specify what theUsers specify what the
result of the query shouldresult of the query should
be, DBMS decidesbe, DBMS decides
operations and order ofoperations and order of
executionexecution
• OperationsOperations
–– Provides commands toProvides commands to
create and modifycreate and modify
database structure anddatabase structure and
constraints (DDL)constraints (DDL)
–– Provides commands toProvides commands to
insert, delete, update andinsert, delete, update and
retrieve (DML)retrieve (DML)
RARA
• Procedural languageProcedural language
–– Algebraic expressionsAlgebraic expressions
specify an order ofspecify an order of
operations ie. How theoperations ie. How the
query will be processedquery will be processed
• OperationsOperations
–– Provides operators, thatProvides operators, that
enable a user to specifyenable a user to specify
retrieval requests onlyretrieval requests only

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ContentsContents

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OperationsOperations
• Relational algebra operations are applied onRelational algebra operations are applied on
relationsrelations
• Result of relational algebra operations are alsoResult of relational algebra operations are also
relations, i.e the algebra operations producerelations, i.e the algebra operations produce
new relations from oldnew relations from old
• A sequence of relational algebra operationsA sequence of relational algebra operations
forms a relational algebra expression, whoseforms a relational algebra expression, whose
result will also be a relationresult will also be a relation

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Types of RA OperationsTypes of RA Operations
• Set operations from mathematical set theorySet operations from mathematical set theory
(Applicable because each relation is also a(Applicable because each relation is also a setset
of tuples)of tuples)
–– UNIONUNION
–– INTERSECTIONINTERSECTION
–– DIFFERENCEDIFFERENCE
–– CARTESIAN PRODUCTCARTESIAN PRODUCT
• Operations developed specifically for RDBsOperations developed specifically for RDBs
–– SELECTSELECT
–– PROJECTPROJECT
–– JOINJOIN
–– DIVISIONDIVISION

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Operators and NotationOperators and Notation
Traditional SetTraditional Set
OperatorsOperators
• IntersectionIntersection ∩∩
• UnionUnion ∪∪
• DifferenceDifference ——
• Cartesian ProductCartesian Product ××
Specific DatabaseSpecific Database
OperatorsOperators
• SelectSelect σσ
• ProjectProject ΠΠ
• JoinJoin ⋈⋈
• DivisionDivision ÷÷

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Understanding RAUnderstanding RA
• SELECTSELECT
• PROJECTPROJECT
• Assignment and NamingAssignment and Naming
• UNIONUNION
• INTERSECTIONINTERSECTION
• DIFFERENCEDIFFERENCE
• Properties of operatorsProperties of operators
• CARTESIAN PRODUCTCARTESIAN PRODUCT
• JOINJOIN
• DIVISIONDIVISION
Discuss FirstDiscuss First
Discuss SecondDiscuss Second
Discuss ThirdDiscuss Third

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SelectSelect
σσ << selection condition >selection condition > ( < relation name > )( < relation name > )
Select those rows which satisfy a given condition ThisSelect those rows which satisfy a given condition This
operation is also called “restriction”operation is also called “restriction”
NAME SALAR
Y
ADDRESS DEPT
Trump 65k Auchenflowe
r
Head Office
Harriso
n
78k St. Lucia Head Office
SelectedSelected
TuplesTuples

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Select ExampleSelect Example
1. List all details of employees working in1. List all details of employees working in
department 4?department 4?
EMPLOYEEEMPLOYEE [Ename,[Ename, SSNSSN, DOB, Address, Sex,, DOB, Address, Sex,
Salary, SuperSSN, Dno]Salary, SuperSSN, Dno]
σσ Dno = 4Dno = 4 (EMPLOYEE)(EMPLOYEE)

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2. List all details of employees earning more than2. List all details of employees earning more than
$30000?$30000?
σσ Salary > 30000Salary > 30000 (EMPLOYEE)(EMPLOYEE)

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3. List all details about employees who work in3. List all details about employees who work in
department 4 and earn over $25000, or workdepartment 4 and earn over $25000, or work
in department 5 and earn over $30000?in department 5 and earn over $30000?
σσ (Dno = 4(Dno = 4 ∧∧ Salary > 25000)Salary > 25000) ∨∨
(Dno = 5 ∧ Salary > 30000)(Dno = 5 ∧ Salary > 30000) (EMPLOYEE)(EMPLOYEE)

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ProjectProject
ΠΠ< atribute column name´s< atribute column name´s >> ( < relation name > )( < relation name > )
Select those columns specified in the listSelect those columns specified in the list
NAME SALAR
Y
ADDRESS DEPT
Trump 65k Auchenflowe
r
Head Office
Harriso
n
78k St. Lucia Head Office
SelectedSelected
attributesattributes

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Project ExampleProject Example
4. For each employee, list their name, date of birth4. For each employee, list their name, date of birth
and salary.and salary.
EMPLOYEEEMPLOYEE [Ename,[Ename, SSNSSN, Bdate, Address, Sex,, Bdate, Address, Sex,
ΠΠ Ename,Bdate,SalaryEname,Bdate,Salary (EMPLOYEE)(EMPLOYEE)

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Project ExampleProject Example
5. List the salaries paid to employees in each5. List the salaries paid to employees in each
department and the department number.department and the department number.
ΠΠ Dno, SalaryDno, Salary (EMPLOYEE)(EMPLOYEE)

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Handling Complex QueriesHandling Complex Queries
Formulation of complex queries may requireFormulation of complex queries may require
several relational algebra operations one afterseveral relational algebra operations one after
the otherthe other
–– Operations can be written as a single relationalOperations can be written as a single relational
algebraalgebra expressionexpression by nesting the operationsby nesting the operations
–– Operations can be applied one at a time byOperations can be applied one at a time by
creatingcreating intermediateintermediate result relationsresult relations
Intermediate Results have to beIntermediate Results have to be “assigned”“assigned” toto
temporary relations which must betemporary relations which must be “named”“named”

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Relation Assignment and NamingRelation Assignment and Naming
• Relation AssignmentRelation Assignment
–– Result RelationResult Relation ←← Relational ExpressionRelational Expression
• Relation NamingRelation Naming
–– TEMPTEMP ←←
• Attribute (re)NamingAttribute (re)Naming
–– TEMP (dept, emp-salary)TEMP (dept, emp-salary) ←←

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Assignment ExampleAssignment Example
6. Create a new relation named RESULT,6. Create a new relation named RESULT,
containing each employee and their date of birth.containing each employee and their date of birth.
Label the resulting columns with ‘Employee’ andLabel the resulting columns with ‘Employee’ and
‘DOB’.‘DOB’.
EMPLOYEEEMPLOYEE [Ename,[Ename, SSNSSN, Bdate, Address, Sex,, Bdate, Address, Sex,
RESULT(Employee,DOB )RESULT(Employee,DOB ) ←← ΠΠ Ename,BdateEname,Bdate (EMPLOYEE)(EMPLOYEE)

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Assignment ExampleAssignment Example
7. List the names and salaries of all employees who7. List the names and salaries of all employees who
work for department 5work for department 5
EMPLOYEEEMPLOYEE [Ename,[Ename, SSNSSN, Bdate, Address,, Bdate, Address,
Sex,Sex,
ΠΠ Ename,SalaryEname,Salary (( σσ Dno = 5Dno = 5 (EMPLOYEE )(EMPLOYEE ) ))
EMPS - DEP5EMPS - DEP5 ←← σσ Dno = 5Dno = 5 (EMPLOYEE )(EMPLOYEE )
RESULTRESULT ←← ΠΠ Ename,SalaryEname,Salary (EMPS_ DEP5(EMPS_ DEP5))
Query withQuery with
ExpressionExpression
Query withQuery with
IntermediateIntermediate
RelationsRelations

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• SELECTSELECT
• PROJECTPROJECT
• UNIONUNION
• JOINJOIN

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Basic Set OperatorsBasic Set Operators
• Relation is aRelation is a setset ofof
tuples (no duplicates)tuples (no duplicates)
• Set theory, andSet theory, and
hence elementary sethence elementary set
operators also applyoperators also apply
to relationsto relations
–– UNIONUNION
–– INTERSECTIONINTERSECTION
–– DIFFERENCEDIFFERENCE
–– CARTESIANCARTESIAN
PRODUCTPRODUCT
Union
A∪B
Intersection
A∩B
Difference
A - B
A B
A B
A B

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Union Compatibility in RelationsUnion Compatibility in Relations
Two relations R(A1, A2, ..., An) and S(B1, B2,Two relations R(A1, A2, ..., An) and S(B1, B2,
..., Bn) are..., Bn) are union compatibleunion compatible iffiff
–– They have theThey have the same degree nsame degree n, (number of, (number of
columns)columns)
–– Their columnsTheir columns have corresponding domainshave corresponding domains, i.e, i.e
dom(Ai) = dom(Bi) for 1dom(Ai) = dom(Bi) for 1 ≤≤ ii ≤≤ nn
Applies to union, intersection and differenceApplies to union, intersection and difference

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…… Union CompatibilityUnion Compatibility
Although domains need to correspond they doAlthough domains need to correspond they do
not have to have the same namenot have to have the same name
WORKS_ONWORKS_ON [ESSN, Pno, Hours][ESSN, Pno, Hours]
WORKED_ONWORKED_ON [Employee, Project, Duration][Employee, Project, Duration]
wherewhere dom (ESSN) = dom (Employee)dom (ESSN) = dom (Employee)
dom (Pno) = dom (project)dom (Pno) = dom (project)
dom (Hours) = dom (Duration)dom (Hours) = dom (Duration)

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UnionUnion
R1R1 ∪∪ R2R2
Produces a relation that includes all tuplesProduces a relation that includes all tuples
that appear only in R1, or only in R2, or inthat appear only in R1, or only in R2, or in
both R1 and R2both R1 and R2
Duplicate Tuples are eliminatedDuplicate Tuples are eliminated
R1 and R2 must be union compatibleR1 and R2 must be union compatible

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Union ExampleUnion Example
8.Identify the employees who either work on8.Identify the employees who either work on
projects or have dependents.projects or have dependents.
WORKS_ONWORKS_ON [[ESSN, PNoESSN, PNo, Hours], Hours]
DEPENDENTDEPENDENT [[ESSN, Dep_NameESSN, Dep_Name, Sex, DOB,, Sex, DOB,
Relationship]Relationship]
WORKS_ONWORKS_ON ∪∪ DEPENDENTDEPENDENT
The relations are not UNION compatible !The relations are not UNION compatible !

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Union ExampleUnion Example
9.List the ESSN’s of employees who either have9.List the ESSN’s of employees who either have
dependents or work on projects.dependents or work on projects.
ΠΠ ESSNESSN (( DEPENDENT )DEPENDENT ) ∪∪ ΠΠ ESSNESSN (WORKS_ON )(WORKS_ON )

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IntersectionIntersection
R1R1 ∩∩ R2R2
Produces a relation that includes the tuplesProduces a relation that includes the tuples
that appearthat appear
in both R1 and R2.in both R1 and R2.
R1 and R2 must be union compatible.R1 and R2 must be union compatible.

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Intersection ExampleIntersection Example
11. List the ESSN’s of employees who have11. List the ESSN’s of employees who have
dependents and are managersdependents and are managers..
DEPARTMENTDEPARTMENT [DName,[DName, DNumberDNumber, MgrSSN, MgrStart], MgrSSN, MgrStart]
ΠΠ ESSNESSN (( DEPENDENT )DEPENDENT ) ∩∩ ΠΠ MGRSSNMGRSSN (DEPARTMENT )(DEPARTMENT )

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DifferenceDifference
R1R1 -- R2R2
Produces a relation that includes all theProduces a relation that includes all the
tuples thattuples that
appear in R1, but do not appear in R2.appear in R1, but do not appear in R2.
R1 and R2 must be union compatible.R1 and R2 must be union compatible.

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Difference ExampleDifference Example
11. List the ESSN’s of employees who have11. List the ESSN’s of employees who have
dependents but do not work on projectsdependents but do not work on projects..
DEPENDENTDEPENDENT [[ESSN, Dep_NameESSN, Dep_Name, Sex,, Sex,
DOB,DOB,
ΠΠ ESSNESSN (( DEPENDENT )DEPENDENT ) —— ΠΠ ESSNESSN (WORKS_ON )(WORKS_ON )

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Properties of OperatorsProperties of Operators
• Commutative and Associative OperatorsCommutative and Associative Operators
• Precedence among operators in relationalPrecedence among operators in relational
algebra expressionsalgebra expressions
• De Morgan’s LawsDe Morgan’s Laws

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Commutative and AssociativeCommutative and Associative
A ∪ B Commutative A ∪ B = B ∪ A
associative (A ∪ B) ∪ C = A ∪ ( B ∪ C )
A∩ B commutative A ∩ B = B ∩ A
associative (A ∩ B) ∩ C = A ∩ ( B ∩ C )
A —
B
not commutative A — B ≠ B — A
not commutative (A — B) — C ≠ A — (B —
C )

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Operator PrecedenceOperator Precedence
==, ≠,, ≠, <, >, ≤, ≥<, >, ≤, ≥
notnot
andand
oror
σσ,, ΠΠ
∩∩, ∪, —, ×, +, ⋈, ∪, —, ×, +, ⋈
• Operators performed left to right in theOperators performed left to right in the
expressionexpression
• ( ) can be used to alter operator precedence,( ) can be used to alter operator precedence,
that is operations in ( ) will be performed beforethat is operations in ( ) will be performed before
even if they have a lower precedence ordereven if they have a lower precedence order
HigherHigher
LowerLower

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Precedence ExamplePrecedence Example
12.12. List all employees who are male, and eitherList all employees who are male, and either
earn less than $40000 or work for deptment 5.earn less than $40000 or work for deptment 5.
Salary, SuperSSN, Dnum]Salary, SuperSSN, Dnum]
ΠΠ EnameEname ((σσ (Sex = ‘M’ and (Salary < 40000 or Dnum = 5))(Sex = ‘M’ and (Salary < 40000 or Dnum = 5)) (EMPLOYEE )(EMPLOYEE ) ))
How does the above solution differ from the following?How does the above solution differ from the following?
ΠΠ EnameEname ((σσ (Sex = ‘M’ and Salary < 40000 or Dnum = 5)(Sex = ‘M’ and Salary < 40000 or Dnum = 5) (EMPLOYEE )(EMPLOYEE ) ))

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De Morgan’s LawsDe Morgan’s Laws
¬¬ ( p( p ∧∧ q )q ) ≡ ¬≡ ¬ pp ∨∨ ¬¬ qq
¬¬ ( p( p ∨∨ q )q ) ≡ ¬≡ ¬ pp ∧∧ ¬¬ qq
where p and q are predicates, e.g Age>20,where p and q are predicates, e.g Age>20,
Dept=Research, e.gDept=Research, e.g
¬¬ (Salary(Salary >> 40000)40000) ∧∧ (Dept(Dept == Research) )Research) )
≡≡
¬¬ (Salary(Salary >> 40000)40000) ∨∨ ¬¬ (Dept(Dept == Research)Research)

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DeMorgan’s Law ExampleDeMorgan’s Law Example
13. List all projects which are neither located in13. List all projects which are neither located in
Brisbane, nor controlled by department 4.Brisbane, nor controlled by department 4.
PROJECTPROJECT [PName,[PName, PNoPNo, Plocation, Dnum], Plocation, Dnum]
ΠΠ PnamePname ((σσ not (Plocation = ‘Brisbane’) and not (Dnum=4)not (Plocation = ‘Brisbane’) and not (Dnum=4) (PROJECT)(PROJECT)))
≡≡
ΠΠ PnamePname ((σσ not (Plocation = ‘Brisbane’ or Dnum=4)not (Plocation = ‘Brisbane’ or Dnum=4) (PROJECT)(PROJECT) ))
≡≡
ΠΠ PnamePname ((σσ not (Plocation <> ‘Brisbane’ and Dnum <> 4)not (Plocation <> ‘Brisbane’ and Dnum <> 4) (PROJECT)(PROJECT) ))

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Cartesian ProductCartesian Product
R1R1 ×× R2R2
• Also known as a cross-product or cross-join R1Also known as a cross-product or cross-join R1
and R2 needand R2 need NOT be union compatibleNOT be union compatible
• The result of R1 (A1, A2, … An) x R2 (B1, B2, …The result of R1 (A1, A2, … An) x R2 (B1, B2, …
Bm) is a relation Q withBm) is a relation Q with n + m attributesn + m attributes QQ
(A1, A2, … An, B1, B2, … Bm) in that order(A1, A2, … An, B1, B2, … Bm) in that order
• Q has one tuple for each combination of tuplesQ has one tuple for each combination of tuples
from R1 and R2, thus if R1 has r tuples and R2from R1 and R2, thus if R1 has r tuples and R2
has t tuples, thenhas t tuples, then Q will haveQ will have r * t tuplesr * t tuples

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Cartesian Product ExampleCartesian Product Example
Subject
CS114
CS115
CS180
Subject Student Degree
CS114 Anna BIT
CS114 Fred BSc
CS115 Anna BIT
CS115 Fred BSc
CS180 Anna BIT
CS180 Fred BSc
Student Degree
Anna BIT
Fred BSc
=×

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Cartesian Product ExampleCartesian Product Example
14. For each female employee, list the names14. For each female employee, list the names
of all of her dependents.of all of her dependents.
EMPLOYEEEMPLOYEE [Ename,[Ename,SSNSSN,DOB,Address,Sex,Salary,SuperSSN, Dno],DOB,Address,Sex,Salary,SuperSSN, Dno]
DEPENDENTDEPENDENT [[ESSN, DepNameESSN, DepName, Sex, DOB, Relationship], Sex, DOB, Relationship]
FEMALE_EMPSFEMALE_EMPS ←← σσ Sex = ‘F’Sex = ‘F’ (EMPLOYEE)(EMPLOYEE)
EMP_NAMESEMP_NAMES ←← ΠΠ Ename, SSNEname, SSN (FEMALE_EMPS)(FEMALE_EMPS)
EMP_DEPENDEMP_DEPEND ←← EMP_NAMESEMP_NAMES ×× DEPENDENTDEPENDENT
ACTUAL_DEPENDACTUAL_DEPEND ←← σσ SSN = ESSNSSN = ESSN (EMP_DEPEND)(EMP_DEPEND)
RESULTRESULT ←← ΠΠ Ename, DepNameEname, DepName (ACTUAL_DEPEND)(ACTUAL_DEPEND)

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• SELECTSELECT
• PROJECTPROJECT
• UNIONUNION
• JOINJOIN

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Join OperationsJoin Operations
• A Join is similar to Cartesian Product, but onlyA Join is similar to Cartesian Product, but only
selected pairs of tuples appear in the resultselected pairs of tuples appear in the result
• It is used to combine related tuples from twoIt is used to combine related tuples from two
relations into a single tuple in a new relation.relations into a single tuple in a new relation.
This is needed when information is contained inThis is needed when information is contained in
more than one relationmore than one relation
• There are three types of Join Operations:There are three types of Join Operations:
–– Thieta-JoinThieta-Join
–– Equi-JoinEqui-Join
–– Natural JoinNatural Join

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Thieta-JoinThieta-Join
R1R1 ⋈⋈ < join condition>< join condition> R2R2
A join condition(s) is of the form AA join condition(s) is of the form A θθ B, where AB, where A
∈∈ R1 and BR1 and B ∈∈ R2,R2,
andand θθ is one of {=, <, ≤, >, ≥}is one of {=, <, ≤, >, ≥}

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Thieta-Join ExampleThieta-Join Example
15. For each employee, list all the employees who15. For each employee, list all the employees who
earn more (than the first employee).earn more (than the first employee).
EMPLOYEEEMPLOYEE [Ename,[Ename, SSNSSN, DOB, Address, Sex, Salary,, DOB, Address, Sex, Salary,
SuperSSN, Dno]SuperSSN, Dno]
DEPARTMENTDEPARTMENT [DName,[DName, DNumberDNumber, MgrSSN, MgrStart], MgrSSN, MgrStart]
AA ←← EMPLOYEEEMPLOYEE
BB ←← EMPLOYEEEMPLOYEE
RESULTRESULT ←← ΠΠ A.Ename, B.EnameA.Ename, B.Ename (A(A⋈⋈ A.Salary< B.SalaryA.Salary< B.Salary BB))

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Equi-JoinEqui-Join
R1R1 ⋈⋈ < join condition>< join condition> R2R2
Specialization of JoinSpecialization of Join
Join condition only hasJoin condition only has equalityequality comparisons onlycomparisons only

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Equi-Join ExampleEqui-Join Example
16. List the names of the managers of16. List the names of the managers of
eacheach
department.department.
EMPLOYEEEMPLOYEE [ Ename,[ Ename, SSNSSN, DOB, Address, Sex, Salary,, DOB, Address, Sex, Salary,
SuperSSN, Dno]SuperSSN, Dno]
DEPARTMENTDEPARTMENT [ DName,[ DName, DNumberDNumber, MgrSSN, MgrStart], MgrSSN, MgrStart]
DEPT_MGRDEPT_MGR ←← DEPARTMENTDEPARTMENT ⋈⋈ MGRSSN = SSNMGRSSN = SSN EMPLOYEEEMPLOYEE
RESULTRESULT ←← ΠΠ EnameEname (DEPT_MGR)(DEPT_MGR)

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Natural JoinNatural Join
R1R1 ** R2R2
• Similar to equi-join except that the attributes thatSimilar to equi-join except that the attributes that
are used for the join are those that have theare used for the join are those that have the
same name in each relationsame name in each relation
• Consequently, they are not explicitly specifiedConsequently, they are not explicitly specified
• The duplicate column is eliminatedThe duplicate column is eliminated

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Natural Join ExampleNatural Join Example
Subject Student Degree
CS114 Anna BIT
CS114 Anna BA
CS115 Fred BSc
CS180 Anna BIT
CS180 Anna BA
Student Degree
Anna BIT
Anna BA
Fred BSc
* =
Subject Student
CS114 Anna
CS115 Fred
CS180 Anna
CS214 Bobby

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17. What is the result schema of the following17. What is the result schema of the following
query? What attributes is the join performedquery? What attributes is the join performed
on?on?
DEPARTMENTDEPARTMENT [ DName,[ DName, DNumberDNumber, MgrSSN, MgrStart ], MgrSSN, MgrStart ]
DEPT_LOCSDEPT_LOCS [[ DNumberDNumber,, DlocationDlocation ]]
DEPARTMENT * DEPT_LOCSDEPARTMENT * DEPT_LOCS

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18. What is the difference between the results of18. What is the difference between the results of
the following queries? What attributes arethe following queries? What attributes are
thethe
joins performed on?joins performed on?
EMPLOYEEEMPLOYEE [ Ename,[ Ename, SSNSSN, DOB, Address, Sex, Salary, SuperSSN,, DOB, Address, Sex, Salary, SuperSSN,
Dno ]Dno ]
DEPARTMENTDEPARTMENT [ DName,[ DName, DNumberDNumber, MgrSSN, MgrStart ], MgrSSN, MgrStart ]
EMPLOYEE * DEPARTMENTEMPLOYEE * DEPARTMENT
andand
EMPEMP(MgrSSN ,Dnumber)(MgrSSN ,Dnumber) ←← ΠΠ SSN,DnoSSN,Dno (EMPLOYEE)(EMPLOYEE)
RESULTRESULT ←← EMP * DEPARTMENTEMP * DEPARTMENT

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DivisionDivision
R1R1 ÷÷ R2R2
• Result relation contains columns in R1, but notResult relation contains columns in R1, but not
in R2in R2
• Relations R1 and R2 must beRelations R1 and R2 must be divisiondivision
compatiblecompatible, i.e last n columns of R1 must be, i.e last n columns of R1 must be
identically named to columns in R2, where n isidentically named to columns in R2, where n is
the degree of R2the degree of R2
• The result relation contains tuples t, such that aThe result relation contains tuples t, such that a
value in t appears in R1, in combination withvalue in t appears in R1, in combination with
every tuple in R2every tuple in R2

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Division ExampleDivision Example
Student Degree Subject
Anna BIT CS114
Anna BIT CS115
Anna BIT CS180
Fred BSc CS114
Fred BSc CS180
Student Degree
Anna BIT÷ =
Subject
CS114
CS115

56
Division ExampleDivision Example
22. Retrieve the names of employees who work22. Retrieve the names of employees who work
onon
all projects that John Smith works on.all projects that John Smith works on.
Dno]Dno]
WORKS_ONWORKS_ON [[ ESSNESSN,, PnoPno, Hours ], Hours ]
SMITHSMITH ←← σσ Ename=‘John Smith’Ename=‘John Smith’ (EMPLOYEE)(EMPLOYEE)
SMITH_PNOSSMITH_PNOS ←← ΠΠ PnoPno (WORKS_ON(WORKS_ON ⋈⋈ ESSN=SSNESSN=SSN SMITH)SMITH)
SSN_PNOSSSN_PNOS ←← ΠΠ ESSN, PnoESSN, Pno (WORKS_ON)(WORKS_ON)
SSNS(SSN)SSNS(SSN) ←← SSN_PNOSSSN_PNOS ÷÷ SMITH_PNOSSMITH_PNOS
RESULTRESULT ←← ΠΠ EnameEname (SSNS * EMPLOYEE)(SSNS * EMPLOYEE)

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ContentsContents

58
Query Formulation in RAQuery Formulation in RA
• Understand what the English query meansUnderstand what the English query means
• Identify which relations, tuples (SELECT) and attributesIdentify which relations, tuples (SELECT) and attributes
(PROJECT) that will be required for the query(PROJECT) that will be required for the query
• Identify the relationships between required relations andIdentify the relationships between required relations and
accordingly which binary operators can be usedaccordingly which binary operators can be used
(JOIN, PRODUCT, UNION, DIVISION, …)(JOIN, PRODUCT, UNION, DIVISION, …)
• Formulate the query keeping in mind operator propertiesFormulate the query keeping in mind operator properties
(Commutative/Associative, Order precedence, De(Commutative/Associative, Order precedence, De
Morgan’s Laws)Morgan’s Laws)

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Which RA Operator to use ?Which RA Operator to use ?
• SELECTSELECT ←←
• PROJECTPROJECT ←←
• UNIONUNION ←←
• DIFFERENCEDIFFERENCE ←←
• CARTESIAN PRODUCTCARTESIAN PRODUCT ←←
• JOINJOIN
• Use unary operators SELECT /Use unary operators SELECT /
PROJECT when choosing tuples /PROJECT when choosing tuples /
attributes respectively from a singleattributes respectively from a single
relationrelation
• Use binary operators UNION,Use binary operators UNION,
PRODUCT, JOIN, … when definingPRODUCT, JOIN, … when defining
the relationship between 2 or morethe relationship between 2 or more
relationsrelations
{{ σσ ,,ΠΠ,, U,U, ——,, × }× } Complete Set ofComplete Set of

60
Complete Set of RA OperatorsComplete Set of RA Operators
• It has been proved thatIt has been proved that {{σσ,, ΠΠ,, ∪∪,, ——,, ×}×} is ais a
complete set of RA operatorscomplete set of RA operators
• Each remaining relational algebra operator canEach remaining relational algebra operator can
be expressed as a sequence of operations frombe expressed as a sequence of operations from
this setthis set
• These remaining operators have been definedThese remaining operators have been defined
primarily for convenience !primarily for convenience !

61
Expressing other operatorsExpressing other operators
• IntersectionIntersection
RR ∩ S ≡ ( R ∪ S ) – (( R — S ) ∪ ( S — R ))∩ S ≡ ( R ∪ S ) – (( R — S ) ∪ ( S — R ))
• (Thieta/Equi) Join(Thieta/Equi) Join
RR <condition><condition> SS ≡≡ σσ <condition><condition> ( R( R ×× S)S)
• Natural JoinNatural Join
S(B1,B2, B3, … Bm)S(B1,B2, B3, … Bm)
R1 (B1, A2, A3, . . . An)R1 (B1, A2, A3, . . . An) ←← ΠΠ (A1, A2, A3, . . . An)(A1, A2, A3, . . . An) RR
R * SR * S ≡≡ ΠΠ (B1, A2, A3, ... An, B2, ... Bm)(B1, A2, A3, ... An, B2, ... Bm) σσ <R.B1 = S.B1><R.B1 = S.B1> ( R1( R1××S)S)

62
Expressing other operatorsExpressing other operators
• DivisionDivision
T1T1 ←← ΠΠ YY ( R )( R )
T2T2 ←← ΠΠ YY ( ( S( ( S ×× T1 )T1 ) —— R )R )
RR ÷ S ≡÷ S ≡ T1T1 — T2— T2

63
ContentsContents

64
Relational Algebra ExercisesRelational Algebra Exercises
These exercises use the Company database as anThese exercises use the Company database as an
example to illustrate relational algebra queries thatexample to illustrate relational algebra queries that
require the use of multiple relational algebra operatorsrequire the use of multiple relational algebra operators
• EMPLOYEEEMPLOYEE [[SsnSsn, Fname, Mit, Lname, Dob, Address,, Fname, Mit, Lname, Dob, Address,
Sex,Sex,
Salary, Dno, SuperSSN]Salary, Dno, SuperSSN]
• DEPARTMENTDEPARTMENT [[DnumberDnumber, Dname, MGRSSN,MgrStart], Dname, MGRSSN,MgrStart]
• PROJECTPROJECT [[PnoPno, PName, Plocation, DNum], PName, Plocation, DNum]
• DEPENDENTDEPENDENT [[ESSN,DepNameESSN,DepName, Sex, DOB,, Sex, DOB,
• WORKS_ONWORKS_ON [[ESSNESSN,, PNoPNo, Hours], Hours]
• DEPT_LOCSDEPT_LOCS [[DNumberDNumber,, DLocationDLocation]]

65
RA ExerciseRA Exercise
23. Retrieve the name and address of all23. Retrieve the name and address of all
employeesemployees
who work for the Research Department.who work for the Research Department.
Dno ]Dno ]
DEPARTMENTDEPARTMENT [ Dname,[ Dname, DnumberDnumber, MgrSSN, MgrStart ], MgrSSN, MgrStart ]
RESEARCH_DEPTRESEARCH_DEPT ←← σσ Dname=‘Research’Dname=‘Research’ (DEPARTMENT)(DEPARTMENT)
RESEARCH_DEPT_EMPSRESEARCH_DEPT_EMPS ←← (RESEARCH_DEPT(RESEARCH_DEPT⋈⋈ Dnumber=DnoDnumber=Dno EMPLOYEE)EMPLOYEE)
RESULTRESULT ←← ΠΠ Ename,AddressEname,Address (RESEARCH_DEPT_EMPS)(RESEARCH_DEPT_EMPS)

66
24. For every project located in Houston, list the project24. For every project located in Houston, list the project
number, the controlling department number, and thenumber, the controlling department number, and the
department manager’s name, address & birth datedepartment manager’s name, address & birth date..
Dno]Dno]
DEPARTMENTDEPARTMENT [ Dname,[ Dname, DnumberDnumber, MgrSSN, MgrStart], MgrSSN, MgrStart]
PROJECTPROJECT [ PName,[ PName, PnumberPnumber, Plocation, Dnum], Plocation, Dnum]
HOUSTON_PROJSHOUSTON_PROJS ←← σσ Plocation=‘houston’Plocation=‘houston’ (PROJECT)(PROJECT)
CONTR_DEPTCONTR_DEPT ←← (HOUSTON_PROJS(HOUSTON_PROJS ⋈⋈ Dnum=DnumberDnum=Dnumber DEPARTMENT)DEPARTMENT)
PROJ_DEPT_MGRPROJ_DEPT_MGR ←← (CONTR_DEPT(CONTR_DEPT⋈⋈ MgrSSN=SSNMgrSSN=SSN EMPLOYEE)EMPLOYEE)
RESULTRESULT ←← ΠΠ Pnumber,Dnum,Ename,Address,BdatePnumber,Dnum,Ename,Address,Bdate (PROJ_DEPT_MGR)(PROJ_DEPT_MGR)

67
25. Find the names of employees who work on all25. Find the names of employees who work on all
projects controlled by department 5.projects controlled by department 5.
Dno]Dno]
WORKS_ONWORKS_ON [[ ESSNESSN,, PnoPno, Hours, Hours]]
DEPT5_PROJS(Pno)DEPT5_PROJS(Pno) ←← ΠΠ PnumberPnumber ((σσ Dnum=5Dnum=5 (PROJECT))(PROJECT))
EMP_PROJ(SSN,Pno)EMP_PROJ(SSN,Pno) ←← ΠΠ ESSN,PnoESSN,Pno (WORKS_ON)(WORKS_ON)
RESULT_EMP_SSNSRESULT_EMP_SSNS ← E← EMP_PROJMP_PROJ ÷÷ DEPT5_PROJSDEPT5_PROJS
RESULTRESULT ←← ΠΠ EnameEname (RESULT_EMP_SSNS * EMPLOYEE )(RESULT_EMP_SSNS * EMPLOYEE )

68
26. List project numbers for projects that involve an employee26. List project numbers for projects that involve an employee
whose name is Smith, either as a worker or as a managerwhose name is Smith, either as a worker or as a manager
ofof
the department that controls the project.the department that controls the project.
EMPLOYEEEMPLOYEE [ Ename,[ Ename, SSNSSN, DOB, Address, Sex, Salary, SuperSSN, Dno], DOB, Address, Sex, Salary, SuperSSN, Dno]
WORKS_ONWORKS_ON [[ ESSN, PnoESSN, Pno, Hours], Hours]
SMITHS(ESSN)SMITHS(ESSN) ←← ΠΠ SSNSSN ((σσ Ename=‘Smith’Ename=‘Smith’ (EMPLOYEE))(EMPLOYEE))
SMITH_WORKER_PROJSSMITH_WORKER_PROJS ←← ΠΠ PnoPno (WORKS_ON * SMITHS)(WORKS_ON * SMITHS)
MGRSMGRS ←← ΠΠ Ename,DnumberEname,Dnumber (EMPLOYEE(EMPLOYEE ⋈⋈ SSN=MgrSSNSSN=MgrSSN DEPARTMENT)DEPARTMENT)
SMITH_MGRSSMITH_MGRS ←← σσ Ename=‘Smith’Ename=‘Smith’ (MGRS)(MGRS)
SMITH_MANAGED_DEPTS(Dnum)SMITH_MANAGED_DEPTS(Dnum) ←← ΠΠ DnumberDnumber (SMITH_MGRS)(SMITH_MGRS)
SMITH_MGR_PROJS(Pno)SMITH_MGR_PROJS(Pno) ←← ΠΠ PnumberPnumber (SMITH_MANAGED_DEPTS * PROJECT)(SMITH_MANAGED_DEPTS * PROJECT)
RESULTRESULT ←← SMITH_WORKER_PROJSSMITH_WORKER_PROJS ∪∪ SMITH_MGR_PROJSSMITH_MGR_PROJS

69
27. Retrieve the names of employees who have no27. Retrieve the names of employees who have no
dependents.dependents.
Dno]Dno]
DEPENDENTDEPENDENT [[ ESSNESSN,, Dep_NameDep_Name, Sex, DOB, Relationship], Sex, DOB, Relationship]
ALL_EMPSALL_EMPS ←← ΠΠ SSNSSN (EMPLOYEE)(EMPLOYEE)
EMPS_WITH_DEPS(SSN)EMPS_WITH_DEPS(SSN) ←← ΠΠ ESSNESSN (DEPENDENT)(DEPENDENT)
EMPS_WITHOUT_DEPSEMPS_WITHOUT_DEPS ←← ( ALL_EMPS( ALL_EMPS ——
EMPS_WITH_DEPS)EMPS_WITH_DEPS)
RESULTRESULT ←← ΠΠ EnameEname (EMPS_WITHOUT_DEPS * EMPLOYEE(EMPS_WITHOUT_DEPS * EMPLOYEE ))

70
28. List the names of managers who have at least28. List the names of managers who have at least
one dependent.one dependent.
Dno]Dno]
DEPENDENTDEPENDENT [[ ESSNESSN,, Dep_NameDep_Name, Sex, DOB, Relationship], Sex, DOB, Relationship]
MGR(SSN)MGR(SSN) ←← ΠΠ MgrSSNMgrSSN (DEPARTMENT)(DEPARTMENT)
EMPS_WITH_DEPS(SSN)EMPS_WITH_DEPS(SSN) ←← ΠΠ ESSNESSN (DEPENDENT)(DEPENDENT)
MGRS_WITH_DEPSMGRS_WITH_DEPS ←← (MGRS(MGRS ∩∩ EMPS_WITH_DEPS)EMPS_WITH_DEPS)
RESULTRESULT ←← ΠΠ EnameEname (MGRS_WITH_DEPS * EMPLOYEE)(MGRS_WITH_DEPS * EMPLOYEE)

71
ReviewReview
• Relational algebra gives the theoretical foundations forRelational algebra gives the theoretical foundations for
Relational Query LanguagesRelational Query Languages
–– Relational algebra operations operate on entire relations,Relational algebra operations operate on entire relations,
and produce results which are also relationsand produce results which are also relations
–– Relational algebra expressions, consisting of a sequenceRelational algebra expressions, consisting of a sequence
of relational algebra operators, specify a high-levelof relational algebra operators, specify a high-level
procedure to achieve a query resultprocedure to achieve a query result
• However, relational algebraic query formulation isHowever, relational algebraic query formulation is
procedural, and therefore focuses on how a query resultprocedural, and therefore focuses on how a query result
can be achievedcan be achieved
• Declarative query languages, e.g., SQL, allow the user toDeclarative query languages, e.g., SQL, allow the user to
specifyspecify whatwhat info the user wants rather thaninfo the user wants rather than howhow the resultthe result
is to be obtainedis to be obtained

72
Recommended ReadingsRecommended Readings
Elmasri & NavatheElmasri & Navathe
Chapter 7Chapter 7

73
Next ...Next ...
Module 7Module 7
Structured QueryStructured Query
Language (SQL)Language (SQL)

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