A finite state version of the Kraft-McMillan theorem

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Frédérique Bassino, Marie-Pierre Béal, Dominique Perrin

To cite this version:

Frédérique Bassino, Marie-Pierre Béal, Dominique Perrin. A finite state version of the Kraft-McMillan theorem. SIAM Journal on Computing, Society for Industrial and Applied Mathematics, 2000, 30 (4), pp.1211-1230. �hal-00619334�

FR

ED

ERIQUE BASSINO

, MARIE-PIERRE B

EAL

, AND DOMINIQUE PERRIN

Abstrat. Themainresultisanite-stateversionoftheKraft-MMillantheoremharaterizing

thegeneratingsequeneofak-aryregulartree.Theproofusesanewonstrutionalledthemultiset

onstrutionwhihisaversionwithmultipliitiesofthewell-knownsubsetonstrutionofautomata

theory.

Keywords.generatingseries,regulartrees,nonnegativematries.

AMSsubjet lassiations.68Q45,68R10,94A45,37B10

1. Introdution. TheKraftinequality P

n0 s

n k

n

1haraterizesthegen-

erating sequenes (s

n )

n0

of leaves in a k-ary tree. It is used in onnexion with

Humanalgorithm tobuildprexodesorsearhtreesandusuallyrestritedto the

ase of nite trees. We are interested here in the ase of innite sequenes orre-

sponding to innite trees. These innite trees arise for example as searh trees in

innitesets. Theyalsoappearintheontextofniteautomata havingnestedloops

torepresentthesetofrstreturnstoagivenstate. Thetreethusobtainedisalleda

regulartree. Ithasonlyanitenumberofnon-isomorphisubtreessinetwosubtrees

orresponding to the samestate of theautomaton are isomorphi. The generating

sequenes ofsuh innitetreesare of interestin the appliations ofnite automata

totextompressionorhanneloding.

Our main result is a haraterization of the generating sequenes of leaves of

regular k-ary trees. Its essene is that the two onditions of being the generating

sequeneof

(i) ak-arytree

(ii) aregulartree

are independent in the sense that their onjuntion is enough to guarantee that a

sequeneisthegeneratingsequeneofaregulark-arytree.

The proof uses a new onstrution on graphs alled the multiset onstrution

whihisaounterpartforautomatawithmultipliitiesofthewell-knownsubseton-

strutionofautomatatheory.

Our resultshaveaonnexion with symbolidynamis. Atually, in both ases,

theemphasisisonthespaeofpathsinanite graph. Evenifwedonotuseresults

fromsymbolidynamis,someofthemethodsused,likestate-splittingorthePerron

theory are similar. Using an expression of Lind and Marus [15℄, our treatmentis

\dynamial in spirit". The relationship with symboli dynamis is disussed more

loselyin[7℄and[8℄.

The paper is organized as follows. Setion 2 ontains preliminary results and

denitions on graphs, trees, regularsequenes and the Perron-Frobeniustheory. In

Setion 3,wepresentthe multisetonstrution. Setion 4ontainsthe proofof our

mainresult(Theorem4.2). Thefollowingsetion(Setion5)treatsasimilarproblem,

withthesetofleavesreplaedbytheset ofallnodes.

The results ontained in this paper represent the terminal point of a series of

steps. Inapreviouspaper[7℄(withapreliminaryversionin[5℄),weprovedTheorem

InstitutGaspardMonge,UniversitedeMarne-la-Vallee,77454Marne-la-ValleeCedex2Frane.

fbassino,beal,perringu niv- mlv .fr.

4.2 in the partiular ase of a strit inequality. The proof uses the tehnique of

state-splitting from symboli dynamis. In the same paper, we also give a proof

of Theorem 5.3 whih is dierent from theproof given here,whih is based on the

multiset onstrution and is more simple. Part of the results of the present paper

waspresentedattheonfereneLATIN'98[6℄. Finally,thesurveypaper[8℄givesan

overviewoflengthdistributionsand regularsequenes.

2. Denitionsand bakground. Inthissetion,wexournotationonern-

ing graphs, trees and regular sequenes. We also reall some notions onerning

positivematries.

Awordontheterminologyusedhere. Weonstantlyusethetermregular where

ariherterminologyisoftenused. Inpartiular,whatweallherearegularsequene

is,in Eilenberg'sterminology,anN-rational sequene(see[11℄,[19℄or[10℄).

2.1. Graphs and trees. Inthispaper,weusediretedmultigraphsi.e.graphs

with possibly several edgeswith thesame originandthe sameend. We simplyall

them graphs in all what follows. We denote G = (Q;E) agraph with Q as set of

vertiesandE asset ofedges. WealsosaythatGisagraphontheset Q.

AtreeT onasetofnodesN witharoot r2N isafuntionT :N frg !N

whih assoiatesto eah node distint from theroot its fatherT(n), in suh away

that, for eah node n, there is a nonnegativeinteger h suh that T h

(n) = r. The

integerhistheheight ofthenoden.

A treeis k-ary ifeahnode hasat mostk hildren. A node withouthildrenis

alled aleaf. Anode whih isnotaleafis alledinternal. Anode nisadesendant

of a node m ifm = T h

(n) for someh 0. A k-ary tree is omplete ifall internal

nodeshaveexatlykhildrenandhaveatleastonedesendantwhihisaleaf.

For eah node n of a tree T, the subtree rooted at n, denoted T

n

is the tree

obtainedbyrestritingtheset ofnodestothedesendantsofn.

TwotreesS;Tareisomorphi,denotedS T,ifthereisamapwhihtransforms

S into T by permuting thehildren ofeah node. Equivalently, S T ifthere is a

bijetivemap f :N !M from theset ofnodesofS ontotheset ofnodesofT suh

thatf ÆS =T Æf. Suhamapf isalledanisomorphism.

If T is a tree with N as set of nodes, the quotient graph of T is the graph

G=(Q;E)where QandE aredenedasfollows. Theset QisthequotientofN by

theequivalenenmifT

n T

m

. Letm denotethelassofanodem. Thenumber

ofedgesfromm to nisthenumberofhildrenofmequivalentton.

Conversely, the set of paths in a graphwith given originis atree. Indeed, let

G = (Q;E) be a graph. Let r 2 Q be a partiular vertex and let N be the set

of paths in G starting at r. The tree T having N as set of nodes and suh that

T(p

0

;p

1

;:::;p

n )=(p

0

;p

1

;:::;p

n 1

)isalled theovering tree ofGstartingat r.

BothonstrutionsaremutuallyinverseinthesensethatanytreeT isisomorphi

totheoveringtreeof itsquotientgraphstartingat theimageoftheroot.

Proposition 2.1. Let T bea tree with root r. Let Gbeits quotient graph and

letibethevertexofGwhih isthe lassofthe rootofT. ForeahvertexqofGand

for eah n0,the numberof paths oflengthn fromi toqisequal tothe numberof

nodesof T atheight nin thelass of q.

A tree is said to beregularif it admits only anite number of non-isomorphi

subtrees,i.e. ifitsquotientgraphisnite.

For example, the innite tree represented on Figure 2.1 is a regular tree. Its

Fig.2.1. Aregulartree.

1 3

4 2

Fig.2.2.Anditsquotientgraph.

There isalsoaloseonnexion betweentreesand sets ofwordsonanalphabet.

Let X bea set ofwordson thealphabet f0;1;:::;k 1g. The set X is said to be

prex-losed ifanyprex ofanelementofX isalso inX. WhenX isprexlosed,

we anbuild atree T(X) asfollows. Theset of nodes is X, the root is the empty

wordandT(a

1 a

2 a

n )=a

1 a

2 a

n 1 .

LetforexampleX=f;0;1;10;11g.ThetreeT(X)isrepresentedonFigure2.3.

Fig.2.3.ThetreeT(X).

2.2. Regularsequenes. Weonsidersequenesofnaturalintegerss=(s

n )

n0 .

We shall not distinguish between suh a sequene and the formal series s(z) =

P

n0 s

n z

n

:

WeusuallydenoteavetorindexedbyelementsofasetQ,alsoalledaQ-vetor,

withboldfaesymbols. Forv=(v

q )

q2Q

wesaythatvisnonnegative,denotedv0,

(resp. positive, denoted v > 0) if v

q

0 (resp. v

q

> 0) for all q 2 Q. The same

onventions are used for matries. A nonnegative QQ-matrix M is said to be

irreduible if, forallindies p;q, there is anintegerm suh that (M m

) >0. The

matrixisprimitive ifthereisanintegermsuhthat M m

>0.

Theadjaeny matrix of agraphG=(Q;E) istheQQ-matrixM suh that

foreahp;q2Q,theintegerM

p;q

isthenumberofedgesfromptoq. Theadjaeny

matrixofagraphGisirreduibleithegraphisstronglyonneted. Itisprimitive

if,moreover,theg..d oflengthsofylesin Gis1.

LetGbeanitegraphandletI,T betwosetsofverties. Foreahn0,lets

n

bethenumberofdistintpathsoflengthnfromavertexofI toavertexofT. The

sequenes=(s

n )

n0

isalled thesequenereognized by(G;I;T)oralso by GifI

and T are alreadyspeied. When I =fig and T =ftg,wesimply denote(G;i;t)

insteadof (G;fig;ftg).

A sequene s = (s

n )

n0

of nonnegative integers is said to be regular if it is

reognizedbysuhatriple(G;I;T),whereGisnite. Wesaythatthetriple(G;I;T)

isarepresentationofthesequenes. ThevertiesofI arealledinitial andthoseof

T terminal. Tworepresentationsaresaidtobeequivalent iftheyreognizethesame

sequene.

Arepresentation(G;I;T)issaidto betrim ifeveryvertexofGisonsomepath

fromI toT. Itislearthatanyrepresentationisequivalenttoatrimone.

A wellknownresult in theory of nite automata allowsoneto use apartiular

representationofanyregularsequenessuhthats

0

=0. Oneanalwayshoosein

this ase arepresentation (G;i;t) of s with aunique initial vertex i, a uniquenal

vertext6=isuhthatnoedgeisenteringvertexiand noedge isgoingoutofvertex

t. Suh a representationis alled anormalized representation (see for example[17℄

page14).

Let (G;i;t) be atrim normalizedrepresentation. If we merge theinitial vertex

i and thenal vertext in a singlevertex still denoted byi, weobtaina newgraph

denotedbyG,whihisstronglyonneted. Thetriple(G ;i;i)isalledthelosureof

(G;i;t).

Let sbea regularsequene suh that s

0

=0. The star s

of thesequene sis

denedby

s

(z)= 1

1 s(z) :

Proposition 2.2. If (G;i;t) is a normalized representation of s, its losure

(G ;i;i)reognizesthe sequenes

.

Proof. Thesequene s isthe lengthdistribution ofthe pathsof rst returnsto

vertexiinG,thatisofnitepathsgoingfromitoiwithoutgoingthroughvertexi.

Thelengthdistributionof theset ofall returnsto iis thus 1+s(z)+s 2

(z)+:::=

1=(1 s(z)).

An equivalent denition of regular sequenes uses vetors instead of sets I;F.

Let i be a Q-rowvetorof nonnegativeintegers and let tbe aQ-olumn vetorof

nonnegativeintegers. Wesaythat(G;i;t)reognizesthesequenes=(s

n )

n0 iffor

eahintegern0

s

n

=iM n

t;

whereM istheadjaenymatrixofG. Theproofthatbothdenitionsareequivalent

followsfromthefatthatthefamilyofregularsequenesislosedunderaddition(see

[11℄). Atriple(G;i;t)reognizingasequenesisalsoalledarepresentationofsand

A sequenes=(s

n )

n0

ofnonnegativeintegersis rational ifitsatisesareur-

rene relationwithintegraloeÆients. Equivalently,s isrational ifthereexist two

polynomialsp(z);q(z)withintegraloeÆientsandwithq(0)=1suhthat

s(z)= p(z)

q(z) :

Anyregular sequeneis rational. Theonverseis howevernottrue(seeSetion

5). Forexample,thesequenesdenedbys(z)= z

1 z z 2

isthesequeneofFibonai

1 2

Fig.2.4.TheFibonaigraph.

numbersalsodenedbys

0

=0;s

1

=1ands

n+1

=s

n +s

n 1

. Itisreognizedbythe

graphofFigure 2.4withI =f1gandT =f2g.

2.3. Regular sequenes and trees. If T is atree, itsgenerating sequeneof

leaves is thesequene of numberss=(s

n )

n0

, where s

n

isthe numberof leavesat

heightn. Wealsosimplysaythat sisthegenerating sequene ofT.

Thefollowingresultisadiretonsequeneofthedenitions.

Theorem 2.3. The generatingsequeneof aregulartreeisaregularsequene.

Proof. LetT bearegulartreeandletGbeitsquotientgraph. SineT isregular,

G is nite. The leaves of T form an equivalene lass t. By Proposition 2.1, the

generating sequeneof T is reognizedby(G;i;t) where i isthelass of therootof

T.

Wesaythatasequenes=(s

n )

n1

satisestheKraft inequalityfortheinteger

kif

X

n0 s

n k

n

1;

i.e.usingtheformalseriess(z)= P

n0 s

n z

n

,if

s(1=k)1:

WesaythatssatisesthestritKraftinequalityforkifs(1=k)<1. Thefollowing

resultiswell-known(see[3℄page35forexample).

Theorem 2.4. A sequene s is the generating sequene of a k-ary tree i it

satisesthe Kraft inequalityfor the integerk.

Proof. LetrstTbeak-arytreeandletsbeitsgeneratingsequene. Itisenough

toprovethat,foreahn0,thesequene(s

0

;:::;s

n

)satisestheKraftinequality.

ItisthegeneratingsequeneofthenitetreeobtainedbyrestritingT tothenodes

atheightatmostn. WemaythussupposeT tobeanitetree. Wehave

s(z)=zt

1

(z)+:::+zt

k (z)

wheret

1

;:::;t

k

arethegeneratingsequenesofleavesofthe(possiblyempty)subtrees

rooted at the hildrenof the root of T. By indution on the number of nodes, we

havet (1=k)1whenethedesiredresult.

Conversely,weuseanindution onntoprovethatthereexistsak-arytreewith

generatingsequene(s

0

;:::;s

n

). Forn=0,wehaves

0

1andT iseitheremptyor

reduedtoonenode. SupposebyindutionhypothesistohavealreadybuiltatreeT

withgeneratingsequene(s

0

;s

1

;:::;s

n 1

). Wehave

n

X

i=0 s

i k

i

1;

then

n

X

i=0 s

i k

n i

k n

;

andthus

s

n k

n n 1

X

i=0 s

i k

n i

:

Thisallowsustoadds

n

leavesatheightnto thetreeT.

LetusonsidertheKraft'sequalityase. Ifs(1=k)=1,thenanytreeT havings

asgeneratingsequeneisomplete. Theonversepropertyisnottrueingeneral(see

[11℄p. 231). However,itis alassialresultthatwhenT is aomplete regulartree,

itsgeneratingsequenesatisess(1=k)=1(seeProposition2.8).

Forthesakeofaompletedesriptionoftheonstrutiondesribedaboveinthe

proof of Theorem 2.4, we have to speify the hoie made at eah step among the

leavesatheightn. Apossiblepoliyistohoosetogiveasmanyhildrenaspossible

tothenodeswhiharenotleavesandofmaximalheight.

IfwestartwithanitesequenessatisfyingKraft'sinequality,theabovemethod

builds a nite tree with generating sequene equal to s. It is not true that this

inremental method givesa regular tree when we start with a regular sequene, as

shownin thefollowingexample.

Lets(z)=z 2

=(1 2z 2

). Sines(1=2)=1=2,wemayapplytheKraftonstrution

to buildabinarytree withlengthdistribution s. Theresultisthe treeT(X)where

X istheset ofprexesoftheset

Y = [

n0 01

n

0f0;1g n

:

whihisnotregular.

Ifsisaregularsequenesuh thats

0

=0,thereexists aregulartreeT havings

asgeneratingsequene. Indeed,let(G;i;t)beanormalizedrepresentationofs. The

generatingsequeneoftheoveringtreeofGstartingatiiss. Ifssatisesmoreover

theKraftinequalityforanintegerk,itishowevernottruethat theregularovering

treeobtainedisk-ary,asshownin thefollowingexample.

Let sbethe regularsequene reognized bythe graphof Figure 2.5on theleft

withi=1andt=4. Wehaves(z)=3z 2

=(1 z 2

). Furthermores(1=2)=1andthus

ssatisesKraft'sequalityfork=2. Howevertherearefouredgesgoingoutofvertex

2and its regularoveringtreestarting at 1is 4-ary. A solution forthis exampleis

given by thegraph of Figure 2.5 onthe right. It reognizess and its overingtree

startingat1is theregularbinarytreeofFigure2.1.

TheaimofSetion4istobuildfromaregularsequenesthatsatisestheKraft

inequalityfor aninteger k a tree with generating sequene s whih is both regular

1 2 3

4

1 3

4 2

Fig.2.5.Graphsreognizings(z)=3z 2

=(1 z 2

).

2.4. Approximate eigenvetor. Let M be the adjaeny matrix of a graph

G. Bythe Perron-Frobenius theorem (see [12℄, for ageneral presentation and [15℄,

[14℄or[9℄forthelinkwithgraphsandregularsequenes),thenonnegativematrixM

hasanonnegativereal eigenvalueof maximalmodulus denotedby ,alsoalled the

spetralradiusofthematrix.

WhenGisstronglyonneted,thematrixisirreduibleandthePerron-Frobenius

theoremassertsthat thedimensionoftheeigenspaeofthematrixM orresponding

toisequalto one,andthat thereisapositiveeigenvetorassoiatedto .

Letkbeaninteger. Ak-approximateeigenvetorofanonnegativematrixM is,

bydenition,anintegralvetorv0suhthat

Mvkv :

Onehasthefollowingresult(see[15℄p.152).

Proposition 2.5. An irreduible nonnegative matrix M with spetral radius

admitsa positive k-approximate eigenvetorik.

Foraproof,see[15℄p.152. WhenMistheadjaenymatrixofagraphG,wealso

saythatvisak-approximateeigenvetorofG. Theomputationofanapproximate

eigenvetoranbeobtainedbytheuseofFranaszek'salgorithm(seeforexample[15℄).

Itanbeshownthatthereexistsak-approximateeigenvetorwithelementsbounded

abovebyk 2n

wheren isthedimensionof M [4℄. Thusthesize oftheoeÆientsof

ak-approximate eigenvetoris bounded abovebyanexponentialinn andanbein

theworstaseofthisorderofmagnitude.

Thefollowingresultiswell-known. Itlinkstheradiusofonvergeneofasequene

withthespetralradiusoftheassoiatedmatrix.

Proposition2.6. Letsbearegularsequenereognizedbyatrimrepresentation

(G;I;T). LetM bethe adjaeny matrixof G. Theradiusofonvergeneof sisthe

inverseof the maximaleigenvalueof M.

Proof. ThemaximaleigenvalueofM is=limsup

n0 n p

kM n

k,where kkis

anyof theequivalentmatrixnorms. Letbetheradiusofonvergeneofsand, for

eahp;q2Q, let

pq

betheradiusof onvergeneofthe sequeneu

pq

=(M n

pq )

n0 .

Then 1== min

pq

. Sine (G;I;T)is trim, we have

pq

for allp;q2 Q. On

the other hand, min

pq

sine s is a sum of some of the sequenes u

pq . Thus

s

=min

pq

whih onludestheproof.

Asaonsequeneofthisresult,theradiusofonvergeneofaregularsequene

sisapole. Indeed,withtheabovenotation,s(z)=i(1 Mz) 1

t. Thendet(I Mz)

is adenominator ofthe rationalfration s, thepolesof s areamong theinversesof

theeigenvaluesofM. Andsine1=istheradiusofonvergeneofs,ithasto bea