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RelationalModel

AdvancedTopicsinFoundationsofDatabases,UniversityofEdinburgh,2019/20

RelationalModel

• Manyadhocmodelsbefore1970

− Hardtoworkwith

− Hardtoreasonabout

• 1970:RelationalModelbyEdgarFrankCodd

− Dataarestoredinrelations (ortables)

− Queriedusingadeclarativelanguage

− DBMSconvertsdeclarativequeriesintoproceduralqueries thatare

optimizedandexecuted

• KeyAdvantages

− Simpleandcleanmathematicalmodel(basedonlogic)

− Separationofdeclarativeandprocedural

RelationalDatabases

Flight origin destination airline

VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh

Constants

VIE,LHR,…

BA,U2,…

Vienna,London,…

RelationalDatabases


VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh

Relations

Constants

VIE,LHR,…

BA,U2,…

Vienna,London,…

RelationalDatabases


VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh

Constants

VIE,LHR,…

BA,U2,…

Vienna,London,…

Relations

Tuples

Relationalatoms

Flight(LHR,EDI,BA)

Airport(LGW,London)

Querying:RelationalAlgebra


VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh

Listalltheairlines



VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh

Listalltheairlines

πairline Flight

{BA,U2,OS}



VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh

ListthecodesoftheairportsinLondon



VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh


πcode (σcity=‘London’Airport)

{LHR,LGW}



VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh

ListtheairlinesthatflydirectlyfromLondontoGlasgow



VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh


πairline ((Flight⋈origin=code(σcity=‘London’ Airport))⋈destination=code(σcity=‘Glasgow’Airport))


Aux origin destination airline code city code city

LGW GLA U2 LGW London GLA Glasgow


πairline ((Flight⋈origin=code(σcity=‘London’ Airport))⋈destination=code(σcity=‘Glasgow’Airport))

{U2}

definestheauxiliaryrelationAux

• Selection:σ

• Projection:π

• Crossproduct:×

• Naturaljoin:⋈

• Rename:ρ

• Difference:∖

• Union:∪

• Intersection:∩

RelationalAlgebra

inboldaretheprimitiveoperators

Formaldefinitionscanbefoundinanydatabasetextbook

Querying:DomainRelationalCalculus


VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh

Listalltheairlines

{z|∃x∃yFlight(x,y,z)}



VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh

Listalltheairlines

{BA,U2,OS}



VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh


{x|∃yAirport(x,y) ∧ y=London}



VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh


{LHR,LGW}



VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh


{z|∃x∃y∃u∃v Airport(x,u) ∧ u=London∧ Airport(x,u) ∧ u=London}∧ Flight(x,y,z)}



VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh


{U2}

{x|∀yR(x,y)}{x|¬R(x)}{x,y |R(x)∨ R(y)}

DomainRelationalCalculus

{x1,…,xk |φ}

first-orderformulawith

freevariables{x1,…,xk}

But,wecanexpress“problematic”queries,i.e.,dependonthedomain


domain ={1,2,3}

D ={R(1,1),R(1,2)}Ans={}

{x1,…,xk |φ}




{x|∀yR(x,y)}{x|¬R(x)}{x,y |R(x)∨ R(y)}


domain ={1,2}

D ={R(1,1),R(1,2)}Ans={1}

{x1,…,xk |φ}




{x|∀yR(x,y)}{x|¬R(x)}{x,y |R(x)∨ R(y)}


…thus,weadopttheactivedomainsemantics - quantifiedvariablesrangeover

theactivedomain,i.e.,theconstantsoccurringintheinputdatabase

{x1,…,xk |φ}




{x|∀yR(x,y)}{x|¬R(x)}{x,y |R(x)∨ R(y)}

Algebra=Calculus

Afundamentaltheorem(assumingtheactivedomainsemantics):

Theorem: Thefollowingquerylangaugesareequallyexpressive

• RelationalAlgebra(RA)

• DomainRelationalCalculus(DRC)

• TupleRelationalCalculus(TRC)

Note:Tuplerelationalcalculus isthedeclarativelanguageintroducebyCodd.Domainrelational

calculushasbeenintroducedlaterasaformalismclosertofirst-orderlogic

{|∃x∃y∃z∃w∃vAirport(x,Vienna) ∧ Airport(y,Glasgow) ∧

Flight(x,z,w)∧ Flight(z,y,v)

Quiz!


VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh

IsGlasgowreachablefromVienna?

London

Vienna

Larnaca

Glasgow

Edinburgh

YES

Quiz!

London

Vienna

Larnaca

Glasgow

Edinburgh


VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh


NO





Quiz!

London

Vienna

Larnaca

Glasgow

Edinburgh


VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh


YES∧ Flight(z,z1,w1) ∧

∃z1∃w1

Quiz!


VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh


NO

London

Vienna

Larnaca

Glasgow

Edinburgh



∧ Flight(z,z1,w1) ∧

∃z1∃w1

Quiz!


VIE LHR BA

LHR EDI BA

LGW GLA U2

LCA VIE OS

Airport code city

VIE Vienna

LHR London

LGW London

LCA Larnaca

GLA Glasgow

EDI Edinburgh


London

Vienna

Larnaca

Glasgow

Edinburgh

Recursivequery- notexpressibleinRA/DRC/TRC

(unlessweboundthenumberofintermediatestops)

ComplexityofQueryLanguages

• Thegoalistounderstandthecomplexityofevaluatingaqueryoveradatabase

• Ourmaintechnicaltooliscomplexitytheory

• Whattomeasure?Queriesmayhavealargeoutput,anditwouldbeunfairto

counttheoutputas“complexity”

• Wethereforeconsiderthefollowingdecisionproblems:

− QueryOutputTuple(QOT)

− BooleanQueryEvaluation(BQE)

AFewWordsonComplexityTheory

detailscanbefoundinthestandardtextbooks

seealsonotesonthewebpageofthecourse

ComplexityClasses

TIME(f(n)) = {Π |ΠisdecidedbysomeDTMintimeO(f(n))}

NTIME(f(n)) = {Π |ΠisdecidedbysomeNTMintimeO(f(n))}

SPACE(f(n)) = {Π |ΠisdecidedbysomeDTMusingspaceO(f(n))}

NSPACE(f(n)) = {Π |ΠisdecidedbysomeNTMusingspaceO(f(n))}

PTIME = ⋃k>0 TIME(nk)

NP = ⋃k>0 NTIME(nk)

EXPTIME = ⋃k>0 TIME(2nk)

NEXPTIME = ⋃k>0 NTIME(2nk)

LOGSPACE = SPACE(logn)

NLOGSPACE = NSPACE(logn)

PSPACE = ⋃k>0 SPACE(nk)

EXPSPACE = ⋃k>0 SPACE(2nk)

ComplexityClasses

• Wecannowrecallthestandardtimeandspacecomplexityclasses:

thesedefinitionsarerelyingontwo-

tapeTuringmachineswitharead-

onlyandaread/writetape

• ForeverycomplexityclassCwecandefineitscomplementaryclasscoC

RelationshipAmongComplexityClasses

LOGSPACE⊆ NLOGSPACE⊆ PTIME⊆ NP,coNP ⊆

PSPACE⊆ EXPTIME⊆ NEXPTIME,coNEXPTIME ⊆⋯

Someusefulnotes:

• ForadeterministiccomplexityclassC,coC =C

• coNLOGSPACE =NLOGSPACE

• ItisgenerallybelievedthatPTIME≠ NP,butwedon’tknow

• PTIME⊂ EXPTIME⇒ atleastonecontainmentbetweenthemisstrict

• PSPACE=NPSPACE,EXPSPACE=NEXPSPACE,etc.

• But,wedon’tknowwhetherLOGSPACE=NLOGSPACE

CompleteProblems

• Thesearethehardestproblemsinacomplexityclass

• AproblemthatiscompleteforaclassC,itisunlikelytobelonginalowerclass

• AproblemΠiscomplete foracomplexityclassC,orsimplyC-complete,if:

1. Π ∈ C

2. ΠisC-hard,i.e.,everyproblemΠ’∈ Ccanbeefficientlyreduced toΠ

• ToshowthatΠisC-harditsufficestoreducesomeC-hardproblemΠ’toit

thereexistsalogspace algorithmthatcomputesafunctionfsuchthat

w ∈ Π’ iff f(w) ∈ Π - inthiscasewewriteΠ’≤L Π

SomeCompleteProblems

• NP-complete

‒ SAT(satisfiabilityofpropositionalformulas)

‒ Manygraph-theoreticproblems(e.g.,3-colorability)

‒ Travelingsalesman

‒ etc.

• PSPACE-complete

‒ QuantifiedSAT(orsimplyQSAT)

‒ Equivalenceoftworegularexpressions

‒ Manygames(e.g.,Geography)

‒ etc.

BacktoQueryLanguages


• Thegoalistounderstandthecomplexityofevaluatingaqueryoveradatabase

• Ourmaintechnicaltooliscomplexitytheory

• Whattomeasure?Queriesmayhavealargeoutput,anditwouldbeunfairto

counttheoutputas“complexity”

• Wethereforeconsiderthefollowingdecisionproblems:

− QueryOutputTuple(QOT)

− BooleanQueryEvaluation(BQE)


QOT(L)

Input:adatabaseD,aqueryQ/k ∈ L,atupleofconstantst∈ adom(D)k

Question: t ∈ Q(D)?

Someusefulnotation:

• GivenadatabaseD,andaqueryQ,Q(D)istheanswer toQ overD

• adom(D)istheactivedomain ofD - theconstantsoccurringinD

• WewriteQ/k forthefactthatthearity ofQ isk≥ 0

L issomequerylanguage;forexample,RA,DRC,etc. - wewillseeseveralquerylanguages


BQE(L)

Input:adatabaseD,aBooleanqueryQ ∈ L

Question: isQ(D)non-empty?

Someusefulnotation:

• GivenadatabaseD,andaqueryQ,Q(D)istheanswer toQ overD

• adom(D)istheactivedomain ofD - theconstantsoccurringinD

• WewriteQ/k forthefactthatthearity ofQ isk≥ 0

L issomequerylanguage;forexample,RA,DRC,etc. - wewillseeseveralquerylanguages

(≡Lmeanslogspace-equivalent)

Theorem: QOT(L)≡L BQE(L),whereL ∈ {RA,DRC,TRC}

BQE(L)

Input:adatabaseD,aBooleanqueryQ ∈ L


QOT(L)

Input:adatabaseD,aqueryQ/k ∈ L,atupleofconstantst∈ adom(D)k

Question: t ∈ Q(D)?


Theorem: QOT(DRC)≡L BQE(DRC)

Proof: (≤L)ConsideradatabaseD,ak-ary queryQ ={x1,…,xk |φ},andatuple(t1,…,tk)

LetQbool ={|∃x1⋯∃xk (φ ∧ x1=t1 ∧ x2=t2 ∧⋯ ∧ xk=tk)}

Clearly,(t1,…,tk) ∈ Q(D) iff Qbool (D)isnon-empty

(≥L)Trivial- aBooleandomainRCqueryisadomainRCquery


(letusshowthisfordomainrelationalcalculus)

BQE[D](L)

Input:aBooleanqueryQ ∈ L


ComplexityMeasures

• Combinedcomplexity - bothD andQ arepartoftheinput

• Querycomplexity - fixedD,inputQ

• Datacomplexity - inputD,fixedQ

BQE[Q](L)

Input:adatabaseD


Theorem: ForL ∈ {RA,DRC,TRC}thefollowinghold:

• BQE(L)isPSPACE-complete(combinedcomplexity)

• BQE[D](L)isPSPACE-complete,forafixeddatabaseD(querycomplexity)

• BQE[Q](L)isinLOGSPACE,forafixedqueryQ ∈ L (datacomplexity)

ComplexityofRA,DRC,TRC

Proof hints:

• Recursive algorithm that uses polynomial space in Q and logarithmic space in D

• Reduction from QSAT (a standardPSPACE-hardproblem)

Evaluating(Boolean)DRCQueries

Eval(D,φ) - forbrevitywewriteφ insteadof{|φ}

• Ifφ= R(t1,…,tk), then YESiff R(t1,…,tk) ∈ D

• Ifφ= ψ1∧ψ2,then YESiff Eval(D,ψ1) =YESandEval(D,ψ2) =YES

• Ifφ= ¬ψ,thenNOiff Eval(D,ψ) =YES

• Ifφ= ∃xψ(x),then YESiff forsomet∈ adom(D),Eval(D,ψ(t)) =YES

Lemma: It holds that

• Eval(D,φ) always terminates - this is trivial

• Eval(D,φ) = YES iff Q(D) is non-empty, whereQ = { | φ}

• Eval(D,φ) uses O(|φ| ⋅ log |φ| + |φ|2 ⋅ log |D|) space

Theorem: ForL ∈ {RA,DRC,TRC}thefollowinghold:

• BQE(L)isPSPACE-complete(combinedcomplexity)

• BQE[D](L)isPSPACE-complete,forafixeddatabaseD(querycomplexity)

• BQE[Q](L)isinLOGSPACE,forafixedqueryQ ∈ L (datacomplexity)

ComplexityofRA,DRC,TRC

Proof hints:

• Recursive algorithm that uses polynomial space in Q and logarithmic space in D

• Reduction from QSAT (a standardPSPACE-hardproblem)

EQUIV(L)

Input:twoqueriesQ1 ∈ Land Q2 ∈ L

Question:Q1≡Q2? orQ1(D)=Q2(D)forevery(finite)databaseD?

SAT(L)

Input:aqueryQ ∈ L

Question: istherea(finite)databaseD suchthatQ(D)isnon-empty?

OtherImportantAlgorithmicProblems

CONT(L)


Question:Q1⊆Q2? orQ1(D)⊆Q2(D)forevery(finite)databaseD?

EQUIV(L)



SAT(L)

Input:aqueryQ∈ L

Question: istherea(finite)databaseDsuchthatQ(D)isnon-empty?


CONT(L)



theseproblemsareimportant

foroptimizationpurposes

SAT(L)

Input:aqueryQ ∈ L

Question: istherea(finite)databaseD suchthatQ(D)isnon-empty?


• If the answer is no, then the input query Q makes no sense

• Query evaluation becomes trivial - the answer is always NO!

EQUIV(L)




• Replace a query Q1 with a query Q2 that is easier to evaluate

• But, we have to be sure that Q1(D) = Q2(D) for every database D

• Approximate a query Q1 with a query Q2 that is easier to evaluate

• But, we have to be sure that Q1(D)⊆Q2(D) for every database D

CONT(L)




Theorem: ForL ∈ {RA,DRC,TRC},SAT(L)isundecidable

Proof hint: By reduction from the halting problem.

Given a Turing machine M, we can construct a query QM ∈ L such that:

Mhaltsontheemptystringiff thereexistsadatabaseD suchthatQ(D)isnon-empty

SATisUndecidable

Note:Actually,thisresultgoesbacktothe1950when

BorisA.Trakhtenbrotprovedthattheproblemofdeciding

whetherafirst-ordersentencehasafinitemodelisundecidable

AneasyconsequenceofthefactthatSATisundecidableisthat:

Theorem: ForL ∈ {RA,DRC,TRC},EQUIV(L)andCONT(L)areundecidable

Proof: By reduction from the complement of SAT(L)

• Consider a query Q ∈ L - i.e., an instance of SAT(L)

• LetQ’ beaquerythatisunsatisfiable,i.e.,Q’(D)isemptyforeveryD

• Forexample,whenL =DRC,Q’ canbethequery{|∃xR(x)∧¬R(x)}

• Clearly,Qisunsatisfiable iff Q≡Q’(orevenQ⊆Q’)

EQUIVandCONTareUndecidable

Recap

• The main languages for querying relational databases are:

− Relational Algebra (RA)

− Domain Relational Calcuclus (DRC)

− Tuple Relational Calculus (TRC)

• Evaluation is decidable, and highly tractable indata complexity

− Foundations of the database industry

− The core of SQL is equally expressive to RA/DRC/TRC

RA =DRC =TRC

(undertheactivedomainsemantics)

• Satisfiability, equivalence and containment are undecidable

− Perfect query optimization is impossible

relational model - homepages.inf.ed.ac.ukhomepages.inf.ed.ac.uk/apieris/courses/atfd2020/relational...

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