UNIT II – B. RELATIONAL MODEL - Strawberrystrawberrydevelopers.weebly.com/uploads/5/2/3/5/52354675/...$200,000 from the rela9on PROJ using select opera9on. Define the rela9ons PROJ

UNITII–B.RELATIONALMODEL

Agenda•  StructureofRela9onalDatabases•  Rela9onalAlgebra•  ExtendedRela9onal-Algebra-Opera9ons•  Modifica9onoftheDatabase•  Views•  TupleRela9onalCalculus•  DomainRela9onalCalculus•  Forma9onofQueries

Rela9onalModel•  Rela9onalModelincludes:Rela%ons,Tuples,A/ributes,keysandforeign

keys.

— Rela/on:Atwodimensionaltablemakeupoftuples(Thisisasimpledefini9onthatwewilldefinemorerigorouslyinalaterchapter).

— Tuple:Arowofdatainarela9onmadeupofoneormoreaRributes.— A4ribute:Acharacteris9coftherela9oncontainedinatuple.

Rela9onalModel

ASampleRela9onalDatabase

BasicStructure•  Formally,givensetsD1,D2,….Dn,arela/onrisasubsetof

D1xD2x…xDnThusarela9onisasetofn-tuples(a1,a2,…,an)whereeachai∈Di

•  Example:if customer-name={Jones,Smith,Curry,Lindsay}

customer-street={Main,North,Park}customer-city={Harrison,Rye,PiRsfield}

Thenr={(Jones,Main,Harrison),(Smith,North,Rye),(Curry,North,Rye),(Lindsay,Park,PiRsfield)}isarela9onovercustomer-namexcustomer-streetxcustomer-city

ARributeTypes•  EachaRributeofarela9onhasaname•  Thesetofallowedvalues foreachaRribute is called thedomain of the

aRribute•  ARributevaluesare(normally)requiredtobeatomic,thatis,indivisible

— E.g.mul9valuedaRributevaluesarenotatomic— E.g.compositeaRributevaluesarenotatomic

•  Thespecialvaluenullisamemberofeverydomain•  Thenullvaluecausescomplica9onsinthedefini9onofmanyopera9ons

—  weshallignoretheeffectofnullvaluesinourmainpresenta9onandconsidertheireffectlater

Rela9onSchema•  A1,A2,…,Anarea/ributes•  R=(A1,A2,…,An)isarela%onschema

E.g.Customer-schema=(customer-name,customer-street,customer-city)

•  r(R)isarela%onontherela%onschemaR E.g. customer(Customer-schema)

Rela9onInstance•  Thecurrentvalues(rela%oninstance)ofarela9onarespecifiedbya

table•  Anelementtofrisatuple,representedbyarowinatable

Jones Smith Curry

Lindsay

customer-name Main North North Park

customer-street Harrison

Rye Rye

Pittsfield

customer-city

customer

attributes (or columns)

tuples (or rows)

Keys•  LetK⊆R•  KisasuperkeyofRifvaluesforKaresufficienttoiden9fyauniquetuple

ofeachpossiblerela9onr(R)— by“possibler”wemeanarela9onrthatcouldexistintheenterprise

wearemodeling.— Example:{customer-name,customer-street}and

{customer-name}are both superkeys of Customer, if no two customers can possiblyhavethesamename.

•  KisacandidatekeyifKisminimalExample: {customer-name} isacandidatekeyforCustomer,since it isasuperkey(assumingnotwocustomerscanpossiblyhavethesamename),andnosubsetofitisasuperkey.

DeterminingKeysfromE-RSets•  Strongen/tyset.Theprimarykeyoftheen9tysetbecomestheprimary

keyoftherela9on.•  Weaken/tyset. Theprimarykeyoftherela9onconsistsoftheunionof

theprimarykeyofthestrongen9tysetandthediscriminatoroftheweaken9tyset.

•  Rela/onshipset. Theunionoftheprimarykeysoftherelated en9tysetsbecomesasuperkeyoftherela9on.— For binary many-to-one rela9onship sets, the primary key of the

“many”en9tysetbecomestherela9on’sprimarykey.— For one-to-one rela9onship sets, the rela9on’s primary key can be

thatofeitheren9tyset.— For many-to-many rela9onship sets, the union of the primary keys

becomestherela9on’sprimarykey

E-RDiagramfortheBankingEnterprise

SchemaDiagramfortheBankingEnterprise

Example•  Design a rela9onal database corresponding to the E-R diagram given

below.

•  Sol.Therela9onaldatabaseschemaisgivenbelow.— person(driver-id,name,address)— car(license,year,model)— accident(report-number,loca9on,date)— owns(driver-id,license)— par9cipated(report-number,driver-id,license,damage-amount)

QueryLanguages•  Languageinwhichuserrequestsinforma9onfromthedatabase.•  Categoriesoflanguages

— procedural— non-procedural

•  “Pure”languages:— Rela9onalAlgebra— TupleRela9onalCalculus— DomainRela9onalCalculus

•  Purelanguagesformunderlyingbasisofquerylanguagesthatpeopleuse.

Rela9onalAlgebra•  Procedurallanguage•  Sixbasicoperators

— select— project— Union—  Intersec9on— setdifference— Cartesianproduct— rename

•  The operators take one or more rela9ons as inputs and give a newrela9onasaresult.

SelectOpera9on

•  Nota9on:σp(r)•  piscalledtheselec9onpredicate•  Definedas:

σp(r)={t|t∈randp(t)} Where p is a formula in proposi9onal calculus consis9ng of termsconnectedby:∧(and),∨(or),¬(not)Eachtermisoneof: op or

whereopisoneof:=,≠,>,≥.

SelectOpera9on–Example1

SelectOpera9on–Example2&3•  SelectElectricalEngineersfromEmployeeRela9on.•  Sol.

•  SelectElectricalormechanicalengineersfromEmployeeRela9on.•  Sol.

ENO ENAME TITLE

E1 J. Doe Elect. Eng E6 L. Chu Elect. Eng.

σ 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

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

SelectOpera9on–Example4•  Findtheprojectswithbudgetlessthanequalto$200,000&greaterthan

$200,000 from the rela9on PROJ using select opera9on. Define therela9onsPROJ1&PROJ2basedonBudget.

•  Sol: PROJ1=σBUDGET200000(PROJ)

PROJ1

PNO PNAME LOC

P1 Instrumentation 150000 Montreal

P2 Database Develop. 135000 New York

BUDGET PNO PNAME BUDGET LOC

P3 CAD/CAM 250000 New York

P4 Maintenance 310000 Paris

P5 CAD/CAM 500000 Boston

PROJ2

New York New York

PROJ

PNO PNAME BUDGET LOC

P1 Instrumentation 150000 Montreal

P3 CAD/CAM 250000 P2 Database Develop. 135000

P4 Maintenance 310000 Paris P5 CAD/CAM 500000 Boston

SelectOpera9on–Example5•  If a new tuplewith a BUDGET value of, $600,000 is to be inserted into

PROJ of previous example. Define the rela9ons PROJ1, PROJ2 & PROJ3basedonBudget.

•  Sol: PROJ1=σBUDGET

SelectOpera9on–Example6•  Consider the rela9on PROJ. Using select opera9on define the rela9ons

PROJ1,PROJ2&PROJ3basedonLoca9on.

•  Sol: PROJ1=σLOC=“Montreal”(PROJ) PROJ2=σLOC=“NewYork”(PROJ) PROJ3=σLOC=“Paris”(PROJ)

New York New York

PROJ

PNO PNAME BUDGET LOC

P1 Instrumentation 150000 Montreal

P3 CAD/CAM 250000 P2 Database Develop. 135000

P4 Maintenance 310000 Paris

SelectOpera9on–Example6contd.

PROJ1

PNO PNAME BUDGET LOC PNO PNAME BUDGET LOC

P1 Instrumentation 150000 Montreal P2 Database Develop. 135000 New York

PROJ2

CAD/CAM 250000 New York

PNO PNAME BUDGET LOC

Maintenance P4 310000 Paris

P3

PROJ3

CompanyDatabaseSchema

P.Query2

P.Query3

U.Query3

SelectOpera9on–Example7&8•  Select the EMPLOYEE tuples whose department number is four. Sol. σDNO = 4 (EMPLOYEE)

•  Select the EMPLOYEE tuples whose salary is greater than $30,000. Sol. σSALARY > 30,000 (EMPLOYEE)

SelectOpera9on–Example9•  Select the EMPLOYEE tuples whose department number is four and

whose salary is greater than $25,000 or those employees whose department number is five and whose salary is greater than $30,000.

•  Sol.

ProjectOpera9on

•  Nota9on:∏A1,A2,…,Ak(r)

whereA1,A2areaRributenamesandrisarela9onname.•  Theresultisdefinedastherela9onofkcolumnsobtainedbyerasingthe

columnsthatarenotlisted•  Duplicaterowsremovedfromresult,sincerela9onsaresets•  E.g.Toeliminatethebranch-nameaRributeofaccount

∏account-number,balance(account)

ProjectOpera9on–Example1

ProjectOpera9on–Example2•  List each employee’s first and last name and salary from Employee

Rela9on.

Sol. πLNAME,FNAME,SALARY(EMPLOYEE)Companydatabaseschema

ProjectOpera9on–Example3•  List each employee’s sex and salary fromEmployeeRela9on.

Sol. πSEX,SALARY(EMPLOYEE) Companydatabaseschema

ProjectExample4•  SelectPNO&BUDGETfromtherela9onPROJ.

•  Sol:π PNO,BUDGET(PROJ)

PNO BUDGET

P1 150000 P2 135000 P3 250000 P4 310000 P5 500000

PROJ

PNO BUDGET

P2 135000

P3 250000 P4 310000 P5 500000

PNAME

P1 150000 Instrumentation Database Develop.

CAD/CAM Maintenance CAD/CAM

Selec9onwithProjec9onExample•  List each employee’s first and last name and salary from Employee

Rela9onwhoseDNOis5fromtheemployeerela9on.

•  Sol.

UnionOpera9on

•  Nota9on:r∪s•  Definedas:

r∪s={t|t∈rort∈s}•  Forr∪stobevalid.

1.r,smusthavethesamearity(samenumberofaRributes)2.TheaRributedomainsmustbecompa%ble(e.g.,2ndcolumnofrdealswiththesametypeofvaluesasdoesthe2ndcolumnofs)

•  E.g.tofindallcustomerswitheitheranaccountoraloan∏customer-name(depositor)∪∏customer-name(borrower)

UnionOpera9on–Example1

UnionOpera9on:Example2•  FindSTUDENT∪ ΙΝSTRUCTOR.

•  Sol:

UnionOpera9on:Example3•  Toretrievethesocialsecuritynumbersofallemployeeswhoeitherwork

in department 5 or directly supervise an employee who works indepartment5,usetheunionopera9on.

Sol. DEP5_EMPS←σDNO=5(EMPLOYEE) RESULT1←πSSN(DEP5_EMPS) RESULT2(SSN)←πSUPERSSN(DEP5_EMPS) RESULT←RESULT1∪RESULT2

Intersec9onOpera9on

Intersec9onOpera9on:Example1•  FindSTUDENT∩ ΙΝSTRUCTOR.

•  Sol:

SetDifferenceOpera9on

•  Nota9onr–s•  Definedas:

r–s={t|t∈randt∉s}•  Setdifferencesmustbetakenbetweencompa%blerela9ons.

— randsmusthavethesamearity— aRributedomainsofrandsmustbecompa9ble

SetDifferenceOpera9on–Example1

SetDifferenceOpera9on–Example2•  Find(a)STUDENT ΙΝSTRUCTOR

(b)INSTRUCTOR–STUDENT

•  Sol:(a) (b)

Cartesian-ProductOpera9on•  Nota9onrxs•  Definedas:

rxs={tq|t∈randq∈s}•  AssumethataRributesofr(R)ands(S)aredisjoint.(Thatis,

R∩S=∅).•  IfaRributesofr(R)ands(S)arenotdisjoint,thenrenamingmustbeused.

Cartesian-ProductOpera9on-Example

Composi9onofOpera9ons•  Canbuildexpressionsusingmul9pleopera9ons•  Example:σA=C(rxs)

RenameOpera9on•  Allows us to name, and therefore to refer to, the results of rela9onal-

algebraexpressions.•  Allowsustorefertoarela9onbymorethanonename.Example:

ρx(E)returnstheexpressionEunderthenameXIfarela9onal-algebraexpressionEhasarityn,then

ρx(A1,A2,…,An)(E)returnstheresultofexpressionEunderthenameX,andwiththeaRributesrenamedtoA1,A2,….,An.

Example1:Queries•  Consider the rela9onal database givenbelowwhere theprimary keys are

underlined.Giveanexpressionintherela9onalalgebratoexpresseachofthefollowingqueries:

a.FindthenamesofallemployeeswhoworkforFirstBankCorpora9on.Sol.Πperson-name(σcompany-name= FirstBankCorpora9on (works))b.Findthenamesandci9esofresidenceofallemployeeswhoworkforFirst

BankCorpora9on.Sol.

employee(person-name,street,city)works(person-name,company-name,salary)company(company-name,city)manages(person-name,manager-name)

Example1:Queriescontd.c. Find the names, street address, and ci9es of residence of all employeeswho work for First Bank Corpora9on and earn more than $10,000 perannum.Sol.d.Findthenamesofallemployeesinthisdatabasewholiveinthesamecityasthecompanyforwhichtheywork.Sol.

Example2:BankingQueries

branch(branch-name,branch-city,assets)customer(customer-name,customer-street,customer-only)account(account-number,branch-name,balance)loan(loan-number,branch-name,amount)depositor(customer-name,account-number)borrower(customer-name,loan-number)

ExampleQueries•  Selectallloansofover$1200

•  Find the loan number for each loan of an amount greater than $1200

ExampleQueries•  Find thenamesof all customerswhohave a loan, an account, or both,

fromthebank

•  Find the names of all customers who have a loan and an account at bank.

ExampleQueries•  Find the names of all customers who have a loan at the Perryridge

branch.

•  Find the names of all customers who have a loan at the Perryridge branch but do not have an account at any branch of the bank.

ExampleQueries•  FindthenamesofallcustomerswhohavealoanatthePerryridgebranch.

- Query 2 ∏customer-name(σloan.loan-number = borrower.loan-number( (σbranch-name = “Perryridge”(loan)) x borrower))

 Query 1 ∏customer-name(σbranch-name = “Perryridge” ( σborrower.loan-number = loan.loan-number(borrower x loan)))

ExampleQueriesFindthelargestaccountbalance•  Renameaccountrela9onasd•  Thequeryis:

∏balance(account) - ∏account.balance (σaccount.balance < d.balance (account x ρd (account)))

Summary:Rela9onAlgebra•  Abasicexpression in therela9onalalgebraconsistsofeitheroneof the

following:— Arela9oninthedatabase— Aconstantrela9on

•  Let E1 and E2 be rela9onal-algebra expressions; the following are allrela9onal-algebraexpressions:— E1∪E2— E1-E2— E1xE2— σp(E1),PisapredicateonaRributesinE1— ∏s(E1),Sisalistconsis9ngofsomeoftheaRributesinE1— ρx(E1),xisthenewnamefortheresultofE1

Addi9onalopera9onsWedefineaddi9onalopera9onsthatdonotaddanypowertotherela9onalalgebra,butthatsimplifycommonqueries.•  Naturaljoin•  Division•  Assignment

■  Notation: r s

Natural-JoinOpera9on

•  Letrandsberela9onsonschemasRandSrespec9vely.Then,rsisarela9ononschemaR∪Sobtainedasfollows:— Considereachpairoftuplestrfromrandtsfroms.—  IftrandtshavethesamevalueoneachoftheaRributesinR∩S,add

atuplettotheresult,where–  thasthesamevalueastronr–  thasthesamevalueastsons

NaturalJoinOpera9on–Example1•  Example1:

R=(A,B,C,D)S=(E,B,D)— Resultschema=(A,B,C,D,E)— rsisdefinedas:

∏r.A,r.B,r.C,r.D,s.E(σr.B=s.B∧r.D=s.D(rxs))

NaturalJoinOpera9on–Example1contd..

NaturalJoinOpera9on–Example2

NaturalJoinOpera9on–Example3

NaturalJoinOpera9on–Example4

NaturalJoinOpera9on–Example5

•  Cartesianproductofr1&r2.

NaturalJoins:Canbeincomplete–Example6

NaturalJoins:CanbeNull–Example7

DivisionOpera9on

•  Suitedtoqueriesthatincludethephrase“forall”.•  Letrandsberela9onsonschemasRandSrespec9velywhere

— R=(A1,…,Am,B1,…,Bn)— S=(B1,…,Bn)Theresultofr÷sisarela9ononschemaR–S=(A1,…,Am) r÷s={t|t∈∏R-S(r)∧∀u∈s(tu∈r)}

r ÷ s

DivisionOpera9on

AssignmentOpera9on•  The assignment opera9on (←) provides a convenientway to express

complexqueries.—  Writequeryasasequen9alprogramconsis9ngof

–  aseriesofassignments–  followedbyanexpressionwhosevalue isdisplayedasaresultof the

query.—  Assignmentmustalwaysbemadetoatemporaryrela9onvariable.

—  Theresulttotherightofthe←isassignedtotherela9onvariableontheleyofthe←.

—  Mayusevariableinsubsequentexpressions.

ExampleQueries•  Findallcustomerswhohaveanaccountfromatleastthe“Downtown”

andtheUptown”branches.

where CN denotes customer-name and BN denotes branch-name.

Query 1 ∏CN(σBN=“Downtown”(depositor account)) ∩ ∏CN(σBN=“Uptown”(depositor account))

Query 2 ∏customer-name, branch-name (depositor account) ÷ ρtemp(branch-name) ({(“Downtown”),

(“Uptown”)})

•  Find all customers who have an account at all branches located inBrooklyncity.

ExampleQueries

∏customer-name, branch-name (depositor account) ÷ ∏branch-name (σbranch-city = “Brooklyn” (branch))

ExtendedRela9onal-Algebra-Opera9ons•  GeneralizedProjec9on•  OuterJoin•  AggregateFunc9ons

GeneralizedProjec9on•  Extendstheprojec9onopera9onbyallowingarithme9cfunc9onstobe

usedintheprojec9onlist.

∏F1,F2,…,Fn(E)•  Eisanyrela9onal-algebraexpression•  EachofF1,F2,…,Fn are arithme9c expressions involving constants and

aRributesintheschemaofE.•  Givenrela9oncredit-info(customer-name,limit,credit-balance),findhow

muchmoreeachpersoncanspend: ∏customer-name,limit–credit-balance(credit-info)

AggregateFunc9onsandOpera9ons•  Aggrega/on func/on takes a collec9on of values and returns a single

valueasaresult. avg:averagevalue

min:minimumvaluemax:maximumvaluesum:sumofvaluescount:numberofvalues

•  Aggregateopera/oninrela9onalalgebra

G1,G2,…,GngF1(A1),F2(A2),…,Fn(An)(E)— Eisanyrela9onal-algebraexpression— G1,G2…,GnisalistofaRributesonwhichtogroup(canbeempty)— EachFiisanaggregatefunc9on— EachAiisanaRributename

AggregateOpera9on–Example•  Rela9onr:

A B α α β β

α β β β

C 7 7 3 10

g sum(c) (r) sum-C

27

AggregateOpera9on–Example

•  Rela9onaccountgroupedbybranch-name:

branch-name g sum(balance) (account)

branch-name account-number balance Perryridge Perryridge Brighton Brighton Redwood

A-102 A-201 A-217 A-215 A-222

400 900 750 750 700

branch-name balance Perryridge Brighton Redwood

1300 1500 700

AggregateFunc9ons(Cont.)•  Resultofaggrega9ondoesnothaveaname

— Canuserenameopera9ontogiveitaname— Forconvenience,wepermitrenamingaspartofaggregateopera9on

branch-name g sum(balance) as sum-balance (account)

OuterJoin•  Anextensionofthejoinopera9onthatavoidslossofinforma9on.•  Computes the join and then adds tuples form one rela9on that do not

matchtuplesintheotherrela9ontotheresultofthejoin.•  Usesnullvalues:

— nullsignifiesthatthevalueisunknownordoesnotexist— All comparisons involving null are (roughly speaking) false by

defini9on.– Willstudyprecisemeaningofcomparisonswithnullslater

OuterJoin–Example•  Rela9onloan

■  Relation borrower

customer-name loan-number Jones Smith Hayes

L-170 L-230 L-155

3000 4000 1700

loan-number amount L-170 L-230 L-260

branch-name Downtown Redwood Perryridge

OuterJoin–Example•  InnerJoin

loanBorrower

loan-number amount L-170 L-230

3000 4000

customer-name Jones Smith

branch-name Downtown Redwood

Jones Smith null

loan-number amount L-170 L-230 L-260

3000 4000 1700

customer-name branch-name Downtown Redwood Perryridge

■  Left Outer Join loan Borrower

OuterJoin–Example•  RightOuterJoinloanborrower

loan borrower ■  Full Outer Join

loan-number amount L-170 L-230 L-155

3000 4000 null

customer-name Jones Smith Hayes

branch-name Downtown Redwood null

loan-number amount L-170 L-230 L-260 L-155

3000 4000 1700 null

customer-name Jones Smith null Hayes

branch-name Downtown Redwood Perryridge null

NullValues•  Itispossiblefortuplestohaveanullvalue,denotedbynull,forsomeof

theiraRributes•  nullsignifiesanunknownvalueorthatavaluedoesnotexist.•  Theresultofanyarithme9cexpressioninvolvingnullisnull.•  Aggregatefunc9onssimplyignorenullvalues

—  Isanarbitrarydecision.Couldhavereturnednullasresultinstead.— Wefollowtheseman9csofSQLinitshandlingofnullvalues

•  For duplicate elimina9on and grouping, null is treated like any othervalue,andtwonullsareassumedtobethesame— Alterna9ve:assumeeachnullisdifferentfromeachother— Botharearbitrarydecisions,sowesimplyfollowSQL

NullValues•  Comparisonswithnullvaluesreturnthespecialtruthvalueunknown

—  Iffalsewasusedinsteadofunknown,thennot(A<5)wouldnotbeequivalenttoA>=5

•  Three-valuedlogicusingthetruthvalueunknown:— OR:(unknownortrue)=true,

(unknownorfalse)=unknown(unknownorunknown)=unknown

— AND:(trueandunknown)=unknown,(falseandunknown)=false,(unknownandunknown)=unknown

— NOT:(notunknown)=unknown—  InSQL“P isunknown”evaluatestotrue ifpredicatePevaluates

tounknown•  Resultofselectpredicateistreatedasfalseifitevaluatestounknown

Modifica9onoftheDatabase•  The content of the database may be modified using the following

opera9ons:— Dele9on—  Inser9on— Upda9ng

•  Alltheseopera9onsareexpressedusingtheassignmentoperator.

Dele9on•  A delete request is expressed similarly to a query, except instead of

displaying tuples to theuser, the selected tuplesare removed from thedatabase.

•  Can delete only whole tuples; cannot delete values on only par9cularaRributes

•  Adele9onisexpressedinrela9onalalgebraby: r←r–Ewhererisarela9onandEisarela9onalalgebraquery.

Dele9onExamples•  DeleteallaccountrecordsinthePerryridgebranch.

■ Delete all accounts at branches located in Needham. r1 ← σ branch-city = “Needham” (account branch) r2 ← ∏branch-name, account-number, balance (r1) r3 ← ∏ customer-name, account-number (r2 depositor) account ← account – r2 depositor ← depositor – r3

■ Delete all loan records with amount in the range of 0 to 50

loan ← loan – σ amount ≥ 0 and amount ≤ 50 (loan)

account ← account – σ branch-name = “Perryridge” (account)

Inser9on•  Toinsertdataintoarela9on,weeither:

— specifyatupletobeinserted— writeaquerywhoseresultisasetoftuplestobeinserted

•  inrela9onalalgebra,aninser9onisexpressedby: r←r∪Ewhererisarela9onandEisarela9onalalgebraexpression.

•  The inser9on of a single tuple is expressed by le{ng E be a constantrela9oncontainingonetuple.

Inser9onExamples•  Insert informa9oninthedatabasespecifyingthatSmithhas$1200in

accountA-973atthePerryridgebranch.

■  Provide as a gift for all loan customers in the Perryridge branch, a $200 savings account. Let the loan number serve as the account number for the new savings account.

account ← account ∪ {(“Perryridge”, A-973, 1200)} depositor ← depositor ∪ {(“Smith”, A-973)}

r1 ← (σbranch-name = “Perryridge” (borrower loan)) account ← account ∪ ∏branch-name, account-number,200 (r1) depositor ← depositor ∪ ∏customer-name, loan-number(r1)

Upda9ng•  Amechanismtochangeavalue inatuplewithoutchargingallvalues in

thetuple•  Usethegeneralizedprojec9onoperatortodothistask

r←∏F1,F2,…,FI,(r)•  EachFiiseither

— theithaRributeofr,iftheithaRributeisnotupdated,or,—  if the aRribute is to be updated Fi is an expression, involving only

constantsand theaRributesof r,whichgives thenewvalue for theaRribute

UpdateExamples•  Makeinterestpaymentsbyincreasingallbalancesby5percent.

■  Pay all accounts with balances over $10,000 6 percent interest and pay all others 5 percent

account ← ∏ AN, BN, BAL * 1.06 (σ BAL > 10000 (account)) ∪ ∏AN, BN, BAL * 1.05 (σBAL ≤ 10000 (account))

account ← ∏ AN, BN, BAL * 1.05 (account)

where AN, BN and BAL stand for account-number, branch-name and balance, respectively.

Views•  Insomecases, it isnotdesirableforalluserstoseetheen9relogical

model(i.e.,alltheactualrela9onsstoredinthedatabase.)•  Considerapersonwhoneedstoknowacustomer’sloannumberbut

hasnoneedtoseetheloanamount.Thispersonshouldseearela9ondescribed,intherela9onalalgebra,by ∏customer-name,loan-number(borrowerloan)

•  Anyrela9onthatisnotoftheconceptualmodelbutismadevisibletoauserasa“virtualrela9on”iscalledaview.

ViewDefini9on•  Aviewisdefinedusingthecreateviewstatementwhichhastheform

createviewvas where is any legal rela9onal algebra queryexpression.Theviewnameisrepresentedbyv.

•  Onceaviewisdefined,theviewnamecanbeusedtorefertothevirtualrela9onthattheviewgenerates.

•  Viewdefini9on isnot thesameascrea9nganewrela9onbyevalua9ngthequeryexpression— Rather, a view defini9on causes the saving of an expression; the

expressionissubs9tutedintoqueriesusingtheview.

ViewExamples•  Consider the view (named all-customer) consis9ng of branches and

theircustomers.

■  We can find all customers of the Perryridge branch by writing:

create view all-customer as ∏branch-name, customer-name (depositor account) ∪ ∏branch-name, customer-name (borrower loan)

∏customer-name (σbranch-name = “Perryridge” (all-customer))

UpdatesThroughView•  Database modifica9ons expressed as views must be translated to

modifica9onsoftheactualrela9onsinthedatabase.•  Consider thepersonwhoneeds to seeall loandata in the loan rela9on

exceptamount.Theviewgiventotheperson,branch-loan,isdefinedas: createviewbranch-loanas ∏branch-name,loan-number(loan)

•  Since we allow a view name to appear wherever a rela9on name isallowed,thepersonmaywrite:

branch-loan←branch-loan∪{(“Perryridge”,L-37)}

UpdatesThroughViews(Cont.)•  The previous inser9on must be represented by an inser9on into the

actualrela9onloanfromwhichtheviewbranch-loanisconstructed.•  Aninser9oninto loanrequiresavalueforamount.Theinser9oncanbe

dealtwithbyeither.— rejec9ngtheinser9onandreturninganerrormessagetotheuser.—  inser9ngatuple(“L-37”,“Perryridge”,null)intotheloanrela9on

•  Someupdates through views are impossible to translate into databaserela9onupdates— createviewvasσbranch-name=“Perryridge”(account))v←v∪(L-99,Downtown,23)

•  Otherscannotbetranslateduniquely— all-customer←all-customer∪{(“Perryridge”,“John”)}

–  Havetochooseloanoraccount,andcreateanewloan/accountnumber!

ViewsDefinedUsingOtherViews•  Oneviewmaybeusedintheexpressiondefininganotherview•  Aviewrela9onv1 issaidtodependdirectlyonaviewrela9onv2 ifv2 is

usedintheexpressiondefiningv1•  Aviewrela9onv1issaidtodependonviewrela9onv2ifeitherv1depends

directlytov2orthereisapathofdependenciesfromv1tov2•  Aviewrela9onvissaidtoberecursiveifitdependsonitself.

ViewExpansion•  Awaytodefinethemeaningofviewsdefinedintermsofotherviews.•  Letviewv1bedefinedbyanexpressione1thatmayitselfcontainusesof

viewrela9ons.•  Viewexpansionofanexpressionrepeatsthefollowingreplacementstep:

repeat Findanyviewrela9onviine1 Replacetheviewrela9onvibytheexpressiondefiningviun/lnomoreviewrela9onsarepresentine1

•  Aslongastheviewdefini9onsarenotrecursive,thisloopwillterminate

TupleRela9onalCalculus•  Anonproceduralquerylanguage,whereeachqueryisoftheform

{t|P(t)}•  ItisthesetofalltuplestsuchthatpredicatePistruefort•  tisatuplevariable,t[A]denotesthevalueoftupletonaRributeA•  t∈rdenotesthattupletisinrela9onr•  Pisaformulasimilartothatofthepredicatecalculus

PredicateCalculusFormula1. SetofaRributesandconstants2. Setofcomparisonoperators:(e.g.,,≥)3. Setofconnec9ves:and(∧),or(v)‚not(¬)4. Implica9on(⇒):x⇒y,ifxistrue,thenyistrue

x⇒y≡ ¬xvy5. Setofquan9fiers:

  ∃ t∈ r(Q(t))≡ ”thereexists”atupleintinrela9onrsuchthatpredicateQ(t)istrue

  ∀t∈ r(Q(t))≡ Qistrue“forall”tuplestinrela9onr

BankingExample•  branch(branch-name,branch-city,assets)•  customer(customer-name,customer-street,customer-city)•  account(account-number,branch-name,balance)•  loan(loan-number,branch-name,amount)•  depositor(customer-name,account-number)•  borrower(customer-name,loan-number)

ExampleQueries•  Findtheloan-number,branch-name,and amountforloansofover

$1200

■ Find the loan number for each loan of an amount greater than $1200

Notice that a relation on schema [loan-number] is implicitly defined by the query

{t | ∃ s ∈ loan (t[loan-number] = s[loan-number] ∧ s [amount] > 1200)}

{t | t ∈ loan ∧ t [amount] > 1200}

ExampleQueries•  Findthenamesofallcustomershavingaloan,anaccount,orbothat

thebank

{t | ∃s ∈ borrower( t[customer-name] = s[customer-name]) ∧ ∃u ∈ depositor( t[customer-name] = u[customer-name])

■  Find the names of all customers who have a loan and an account at the bank

{t | ∃s ∈ borrower( t[customer-name] = s[customer-name]) ∨ ∃u ∈ depositor( t[customer-name] = u[customer-name])

ExampleQueries•  FindthenamesofallcustomershavingaloanatthePerryridgebranch

{t | ∃s ∈ borrower( t[customer-name] = s[customer-name] ∧ ∃u ∈ loan(u[branch-name] = “Perryridge” ∧ u[loan-number] = s[loan-number])) ∧ not ∃v ∈ depositor (v[customer-name] = t[customer-name]) }

■  Find the names of all customers who have a loan at the Perryridge branch, but no account at any branch of the bank

{t | ∃s ∈ borrower(t[customer-name] = s[customer-name] ∧ ∃u ∈ loan(u[branch-name] = “Perryridge” ∧ u[loan-number] = s[loan-number]))}

ExampleQueries•  Find the names of all customers having a loan from the Perryridge

branch,andtheci9estheylivein

{t | ∃s ∈ loan(s[branch-name] = “Perryridge” ∧ ∃u ∈ borrower (u[loan-number] = s[loan-number]

∧ t [customer-name] = u[customer-name]) ∧ ∃ v ∈ customer (u[customer-name] = v[customer-name]

∧ t[customer-city] = v[customer-city])))}

ExampleQueries•  Findthenamesofallcustomerswhohaveanaccountatallbranches

locatedinBrooklyn:

{t | ∃ c ∈ customer (t[customer.name] = c[customer-name]) ∧ ∀ s ∈ branch(s[branch-city] = “Brooklyn” ⇒

∃ u ∈ account ( s[branch-name] = u[branch-name] ∧ ∃ s ∈ depositor ( t[customer-name] = s[customer-name] ∧ s[account-number] = u[account-number] )) )}

SafetyofExpressions•  It is possible to write tuple calculus expressions that generate infinite

rela9ons.•  Forexample,{t|¬t ∈r}results inaninfiniterela9onifthedomainof

anyaRributeofrela9onrisinfinite•  Toguardagainsttheproblem,werestrictthesetofallowableexpressions

tosafeexpressions.•  An expression {t | P(t)} in the tuple rela9onal calculus is safe if every

componentoftappearsinoneoftherela9ons,tuples,orconstantsthatappearinP— NOTE:thisismorethanjustasyntaxcondi9on.

–  E.g.{t|t[A]=5∨true}isnotsafe---itdefinesaninfinitesetwithaRribute values that do not appear in any rela9on or tuples orconstantsinP.

DomainRela9onalCalculus•  A nonprocedural query language equivalent in power to the tuple

rela9onalcalculus•  Eachqueryisanexpressionoftheform:

{<x1,x2,…,xn>|P(x1,x2,…,xn)}— x1,x2,…,xnrepresentdomainvariables— Prepresentsaformulasimilartothatofthepredicatecalculus

ExampleQueries•  Find the loan-number, branch-name, and amount for loans of over

$1200

{< c, a > | ∃ l (< c, l > ∈ borrower ∧ ∃b(< l, b, a > ∈ loan ∧ b = “Perryridge”))} or {< c, a > | ∃ l (< c, l > ∈ borrower ∧ < l, “Perryridge”, a > ∈ loan)}

■  Find the names of all customers who have a loan from the Perryridge branch and the loan amount:

{< c > | ∃ l, b, a (< c, l > ∈ borrower ∧ < l, b, a > ∈ loan ∧ a > 1200)}

■  Find the names of all customers who have a loan of over $1200

{< l, b, a > | < l, b, a > ∈ loan ∧ a > 1200}

ExampleQueries•  Findthenamesofallcustomershavingaloan,anaccount,orbothat

thePerryridgebranch:

{< c > | ∃ s, n (< c, s, n > ∈ customer) ∧ ∀ x,y,z(< x, y, z > ∈ branch ∧ y = “Brooklyn”) ⇒ ∃ a,b(< x, y, z > ∈ account ∧ < c,a > ∈ depositor)}

■  Find the names of all customers who have an account at all branches located in Brooklyn:

{< c > | ∃ l ({< c, l > ∈ borrower ∧ ∃ b,a(< l, b, a > ∈ loan ∧ b = “Perryridge”)) ∨ ∃ a(< c, a > ∈ depositor ∧ ∃ b,n(< a, b, n > ∈ account ∧ b = “Perryridge”))}

SafetyofExpressions {<x1,x2,…,xn>|P(x1,x2,…,xn)}

issafeifallofthefollowinghold: 1. All values that appear in tuples of the expression are values from dom(P) (that is, the values appear either in P or in a tupleofarela9onmen9onedinP).

2. For every “there exists” subformula of the form ∃ x (P1(x)), the subformula is true if and only if there is a value of x in dom(P1)suchthatP1(x)istrue.

3. For every “for all” subformula of the form ∀x (P1 (x)), the subformula is true if and only ifP1(x) is true for all values x fromdom(P1).

EndofChapter3

Result of σ branch-name = “Perryridge” (loan)

Loan Number and the Amount of the Loan

Names of All Customers Who Have Either a Loan or an Account

Customers With An Account But No Loan

Result of borrower × loan

Result of σ branch-name = “Perryridge” (borrower × loan)

Result of Πcustomer-name

Result of the Subexpression

Largest Account Balance in the Bank

Customers Who Live on the Same Street and In the Same City as Smith

Customers With Both an Account and a Loan at the Bank

Result of Πcustomer-name, loan-number, amount (borrower loan)

Result of Πbranch-name(σcustomer-city = “Harrison”(customer account depositor))

Result of Πbranch-name(σbranch-city = “Brooklyn”(branch))

Result of Πcustomer-name, branch-name(depositor account)

The credit-info Relation

Result of Πcustomer-name, (limit – credit-balance) as credit-available(credit-info).

The pt-works Relation

The pt-works Relation After Grouping

Result of branch-name ς sum(salary) (pt-works)

Result of branch-name ς sum salary, max(salary) as max-salary (pt-works)

The employee and ft-works Relations

The Result of employee ft-works

The Result of employee ft-works

Result of employee ft-works

Result of employee ft-works

Tuples Inserted Into loan and borrower

Names of All Customers Who Have a Loan at the Perryridge Branch

E-R Diagram

The branch Relation

The loan Relation

The borrower Relation