FINITESTATEAUTOMATA

States• AnFSAhasafinitesetofstates

– Asystemhasalimitednumberofconfigurations• Examples

– {On,Off},– {1,2,3,4,…,k}– {TVchannels}– …

• Statescanbegraphicallyrepresentedasfollows:

On Off

Input

• AnFSAisdefinedoveranalphabet• Thesymbolsofthealphabetrepresenttheinputofthesystem

• Examples– {switch_on,switch_off}– {incoming==0,0<incoming<=10,incoming>10}

Transitionsamongstates

• Whenaninputisreceived,thesystemchangesitsstate

• Thepassagebetweenstatesisperformedthroughtransitions

• Atransitionisgraphicallyrepresentedbyarrows:

On Off

Simpleexamples(flip-flop)

On OffT

T

On OffR

S

SR

FSA

• FSAsarethesimplestmodelofcomputation• ManyusefuldevicescanbemodeledusingFSAs

…buttheyhavesomelimitations

Formally

• AnFSAisatriple<Q,A,δ>,where– Qisafinitesetofstates– Aistheinputalphabet– δ isatransitionfunction(thatcanbepartial),givenby δ:Q× A→ Q

• Remarkifthefunctionispartial,thennotallthetransitionsfromallthepossiblestatesforallthepossibleelementsofthealphabetaredefined

PartialvsTotalTransitionFunction

switch_offswitch_on

On Off

switch_off

switch_on

An FSA with a total transition function is called complete

Recognizinglanguages

• InordertobeabletouseFSAsforrecognizinglanguages,itisimportanttoidentify:– theinitialconditionsofthesystem– thefinaladmissiblestates

• Example:– Thelightshouldbeoffatthebeginningandattheend

Elements

• Theelementsofthemodelare– States– Transitions– Inputandalso– Initialstate(s)– Finalstate(s)

Graphicalrepresentation

• Initialstate

• Finalstate

Off

Off

Formally

• AnFSAisatuple<Q,A,δ ,q0,F>,where– Qisafinitesetofstates– Aistheinputalphabet– δ isa(partial)transitionfunction,givenby

δ:Q× A→ Q– q0∈Qiscalledinitialstate– F⊆Qisthesetoffinalstates

Movesequence

• Amovesequence startsfromaninitialstateandisaccepting ifitreachesoneofthefinalstates

On Off

switch_offswitch_on

switch_on

switch_off

Formally

• Movesequence:– δ∗:Q× A∗ → Q

• δ∗ isinductivelydefinedfrom δ– δ∗(q,ε)=q– δ∗(q,yi)=δ(δ∗(q,y),i)

• Initialstate:q0∈ Q• Final(oraccepting)states:F⊆ Q• ∀x(x∈ L↔ δ∗ (q0,x)∈ F)

Apracticalexample

• RecognizingPascalidentifiers

q0 q1

<letter><letter>

<digit>qE

<digit>

<digit> <letter>

FINITESTATETRANSDUCERS

Automataaslanguagetranslators

AfinitestatetransducerisanFSAthatworksontwotapes.→itisakindof``translatingmachine''.

Input tape

Output tape

Control device (finite state)

Theidea• y=τ(x)

– x:inputstring– y:output– τ:functionfromL1 toL2

• Examples:– τ1 theoccurrencesof“1”aredoubled(1-->11)– τ2 ‘a’isswappedwith‘b’(a<--->b):

• butalso– Compressionoffiles– Compilingfromhighlevellanguagesintoobjectlanguages– TranslationfromEnglishtoItalian

Informally

• Transitionswithoutput

• Example:τ halvesthenumberof“0”sanddoublesthenumberof“1”s

q q’i/w

q0 q1

0/ε1/11 1/11

0/0

Formally

• Afinitestatetransducer(FST)isatupleT=<Q,I,δ,q0,F,O,η>– <Q,I,δ,q0,F>:justlikeacceptors– O:outputalphabet– η :Q× I→ O*

• Remark:theconditionforacceptanceremainsthesameasinacceptors– Thetranslationisperformedonlyonacceptedstrings

Translatingastring

• Aswedidforδ,wedefineη*inductively– η*(q,ε)=ε– η∗(q,y.i)=η∗(q,y).η(δ∗ (q,y),i)

• Remarkη*:Q× I* → O*

∀x(τ(x)=η∗(q0,x)iff δ∗ (q0,x)∈ F)

PUMPINGLEMMA

CyclesThereisacycle:q1----aabab--->q1

a

a

b

a

a

b

b

b

b

b

a

q1 q2

q3

q0

q4q5

q6 q7

q9

q8

If one goes through the cycle once, then one can also go through it 2,3, …, n times

Moreformally

• Ifx∈ L and|x|≥ |Q|,thenthereexistsaq∈Qandaw∈ I+ suchthat:– x=ywz– δ∗ (q,w)=q

• Thereforethefollowingalsoholds:–∀n≥0 ywnz∈ L

ThisisthePumpingLemma (onecan“pump”w)

Consequencesofpumpinglemma

• L=∅? ∃x∈ L↔∃y∈ L|y|<|Q|:Just“removeallcycles”fromtheFSAacceptingx

• |L|=∞? Checkbyasimilarargumentwhether

∃x∈ L|Q|<=|x|<2|Q|• Notethatingeneral knowinghowtoanswerthequestion“x∈ L?”foragenericx,doesnot entailknowinghowtoanswertheotherquestions– ItworksforFSAs,but…

Impactinpractice

• Areweinterestedinaprogramminglanguageconsistingof…0correctprograms?

• Areweinterestedinaprogramminglanguageinwhichonecanonlywriteafinitenumberofprograms?

• ...

Anegativeconsequenceofpumpinglemma

• IsthelanguageL={anbn |n>0}recognizedbysomeFSA?• Letussupposeitis.Then:• Considerx=ambm,m>|Q|andletusapplyP.L.• Possiblecases:

– x=ywz,w=ak ,k>0====>am+r⋅kbm∈ L,∀r:NO– x=ywz,w=bk ,k>0====>same– x=ywz,w=ak bs ,k,s>0====>am-kakbsakbsbm-s

∈ L:NO

Intuitively• Inorderto“count”anarbitraryn weneedaninfinite

memory!• Rigorouslyspeaking,everycomputerisanFSA,but…itisthe

wrongabstraction:intractablenumberofstates!(samethingasstudyingeverysinglemoleculeintheflightofanairplane)

• Importanceofanabstractnotionofinfinity• Fromthetoyexample{anbn}tomoreconcretecases:

– Checkingwell-balancingofbrackets(typicallyusedinprogramminglanguages)cannotbedonewithfinitememory

• Wethereforeneedmorepowerfulmodels

OPERATIONS ONFSA

Closureinmath

• AsetSisclosed w.r.t.anoperationOPif,whenoperationOPisappliedtoelementsofS,theresultisstillanelementofS

• Examples:– Naturalnumbersareclosedw.r.t.sum(butnotsubtraction)

– Integersareclosedw.r.t.sum,subtraction,multiplication(butnotdivision)

– Rationals…– Reals…– …

Closureforlanguages

• L ={Li}:family oflanguages

• L isclosedw.r.t.operationOPifandonlyif,foreveryL1,L2∈ L ,L1OPL2∈L .

• R:regularlanguages(recognizedbyFSAs)

• R isclosedw.r.t.set-theoreticoperations,concatenation, “*”,…

Intersection

The“parallelrun”ofAandBcanbesimulatedby“couplingthem”

a abq1 q2q0

q9

a abp1 p2

p9

A

B p0

p0a abq0<A,B> p1q1 p2q2 p9q9

33

Example

q0 q1 p0

a

a

a

∩

q0 p0

a

a

q1 p0

A1: A2:

Formally

• Given– A1 =<Q1,I,δ1,q01,F1>– A2 =<Q2,I,δ2,q02,F2><A1,A2 >=<Q1×Q2,I,δ,<q01,q02>,F1×F2>– δ(<q1,q2>,i)=<δ1(q1,i),δ2(q2,i)>

• Onecanshow(bysimpleinduction)thatL(<A1,A2>)=L(A1)∩L(A2 )

• Canwedothesameforunion?

Union

• Theunionisbuiltanalogously• Given

– A1 =<Q1,I,δ1,q01,F1>– A2 =<Q2,I,δ2,q02,F2><A1,A2>=<Q1xQ2,I,δ,<q01,q02>,F1xQ2UQ1xF2>– δ(<q1,q2>,i)=<δ1(q1,i),δ2(q2,i)>

Complement(1)

• BasicideaFc =Q-F

…butingeneralthetransitionfunctionispartial!

q0 q1

10 0

1

Complement(2)

• BeforeswappingfinalandnonfinalstatesitisnecessarytocompletetheFSA

q0 q1

10 0

q0 q1

10 0

qE

0,1q0 q1

10 0

qE0,1

1

1

Unionagain

• Another possibility is tousecomplement andDeMorgan’s laws:A∪ B=¬(¬A∩¬B)

Philosophyofcomplement

• IfIscantheentireinputstring,thenitsufficesto“swapyesandno”(FwithQ-F)

• IfIcannotreachtheendofthestring,thenswappingFwithQ-Fdoesnotwork

• InthecaseofFSAsthereisaneasyworkaround(completingtheFSA)

• Ingeneralwecannotconsiderthenegativeanswertoaquestionasequivalenttothepositiveanswertotheoppositequestion!