Publicações do PESC A Distributed Algorithm to Find <I>k</I>-Dominating Sets

Lucia D.Penso

Valmir C.Barbosa

UniversidadeFederaldoRiodeJaneiro

ProgramadeEngenhariadeSistemaseComputac~ao,COPPE

CaixaPostal68511

21945-970RiodeJaneiro-RJ,Brazil

draque@cos.ufrj.br, valmir@cos.ufrj.br

Abstract

WeconsideraconnectedundirectedgraphG(n;m)withnnodesandmedges. Ak-dominatingsetDinGis

asetofnodeshavingthepropertythateverynodeinGisatmostkedgesawayfromatleastonenodeinD.

Findingak-dominatingsetofminimumsize isNP-hard. Wegiveanewsynchronousdistributed algorithm

tondak-dominatingsetinGofsizenogreaterthan 

n=(k+1) 

. OuralgorithmrequiresO(klog 

n)time

andO(mlogk+nlogklog 

n)messagestorun. Ithasthesametimecomplexityasthebestcurrentlyknown

algorithm,but improvesonthat algorithm'smessagecomplexityandis,in addition,conceptuallysimpler.

Keywords: k-dominatingsets,distributed algorithms,graphalgorithms.

1. Introduction

Let G(n;m) be aconnectedundirected graphwith n nodes and m edges. Fork n 1,ak-dominating

set D in Gisaset ofnodeswith thepropertythat everynodein Gis atmostk edgesawayfromat least

oneofthenodesof D. Theproblem ofndingk-dominatingsets ofrelativelysmallsizes isimportantina

varietyofcontexts,includingmulticastsystems[11],theplacementofserversinacomputernetwork[2],the

cachingof replicasindatabaseandoperatingsystems[8],andmessageroutingwithsparsetables[9].

Findingak-dominatingsetinGwiththeleastpossiblenumberofnodesisthoughttobeanintractable

problem(itisNP-completewhenformulatedasadecisionproblem[4]),soone normallysettlesforasetof

small size that is notnecessarily optimal. In general, the small size to be sought is at most 

n=(k+1) 

,

sinceit canbeargued that a k-dominatingset withno morethan this numberof nodesalwaysexists [6],

and likewise that aconnectedgraphonnnodesnecessarilyexists forwhicheveryk-dominatingset hasat

least 

n=(k+1) 

nodes[10].

The argument for 

n=(k+1) 

as an upper bound is instructive in the present context, and goes as

follows[6]. LetT bearootedspanning tree ofGand D

1 ;:::;D

k +1

apartition ofits nodes such that,for

0`k,everynodeinD

`+1

isawayfromtherootanumberx

`

oftreeedgessuchthatx

`

mod(k+1)=`.

ThispartitioncanbeconstructedeasilybytraversingT breadth-rstfromtherootandassigningeverynew

layerofnodescircularly tothe setsD

1 ;:::;D

k +1

. Clearly, everyoneof these setsis ak-dominatingset in

G. Also, because theypartition the graph'snode set, and considering that n k+1, it must be that at

leastoneofthem hasnomorethan 

n=(k+1) 

nodes.

Our topic in this paper is nding a k-dominating set in G having no more than 

n=(k+1) 

nodes

in lockstepat theoccurrence ofclock pulses, and its edges arebidirectionalcommunicationchannels that

deliver messagesbetween their end nodes before the clock pulse that follows the sending of the message

occurs. Timeismeasuredbycountingclockpulses.

Thecurrentbest synchronousalgorithm tondak-dominating setin Gis from[6], andishenceforth

referredtoasAlgorithmKP.Itproceedsintwostages: therststagepartitionsGintothetreesofarooted

spanningforestF,eachhavingat leastk+1nodesandheightO(k);thesecondstage approacheseachtree

U 2F asdescribedearlierfor thespanning treeT and partitionsits nodesinto thesets D U 1 ;:::;D U k +1 . If

f isthenumberoftreesin F, thenthek-dominatingset outputbythealgorithmisD =D U1 `1 [


[D U f `f , whereU 1 ;:::;U f

arethetreesofF andD U

i

`

i

isthesmallestset ofD U i 1 ;:::;D U i k +1 for1if. Ifn U isthe

numberofnodesof U 2F,then

jDj=jD U 1 ` 1 j+


+jD Uf ` f j  X U2F  n U k+1    P U2F n U k+1  =  n k+1  ; sincen U k+1forallU 2F.

While thesecond stage of Algorithm KP canbeeasily implemented within the bounds of O(k) time

and O(n) messages, its rst stage is based on an arcane combination of previously developed algorithms

for related problems [3, 5, 7], resulting in a time complexity of O(klog 

n) and a message complexity of

O(mlogk+n 2

log 

n). The latter, incidentally, is our best estimate of what is really involved, in terms

of communication needs, in Algorithm KP|such needs are thoroughly ignored in [6], but this message

complexityseemsto followfromthemessagecomplexities ofthealgorithm's buildingblocks.

Inthispaper,weintroduceanewsynchronousdistributed algorithmforndingak-dominatingset of

no more than 

n=(k+1) 

nodes in G. Like Algorithm KP, our algorithm toocomprises twosubsequent

stages, each having the samegoalas itscounterpartin AlgorithmKP. The second stage,in particular, is

exactlythesameasAlgorithmKP's.

OurcontributionistheintroductionofanewalgorithmforthepartitionofGintothetreesofF. When

comparedtoAlgorithmKP,ouralgorithmhasthesamecomplexityofO(klog 

n)timewhileimprovingon

themessagecomplexity,whichinourcaseisofO(mlogk+nlogklog 

n). Wealsondouralgorithmtobe

conceptuallysimplerthanAlgorithmKP,whichcanprobablybeattributedtothefactthatitwasdesigned

from scratch with the partitioning problem in mind. While ouralgorithm simply generates asequence of

\meta-graphs,"thelastofwhichhasnodesthatdirectlygivetherootedtreesofF,AlgorithmKPreducesthe

partitionproblemtoother,relatedproblemsandthencombinesalgorithmsforthoseproblemsintobuilding

asolutiontothepartitionproblem. Henceforth,weletthealgorithmweintroducebecalled AlgorithmPB.

The following is how the remainder of the paper is organized. The rst stage of Algorithm PB is

introducedin Section2and analyzedforcorrectnessandcomplexityin Section3. Concluding remarksare

givenin Section4.

2. The algorithm

InthissectionweintroducetherststageofAlgorithmPB.ThisstagendsarootedspanningforestF inG,

PartitionG startsby letting thenode set of G be thenode set of adirected graph G 0 , and proceeds fromtherein  log (k+1)  phases. For0i  log(k+1) 

1,phaseirstbuildstheedgesetof ~

G

i

andthen

beginsthetransformationof ~

G

i

intoanotherdirectedgraph, ~

G

i+1

,byclusteringthenodesof ~

G

i

togetherto

formthenodesof ~ G i+1 . Each nodeof ~ G i

standsforarootedtreein G,andthisclustering isperformedin

suchawaythatnotonlyiseachresultingnodeof ~

G

i+1

alsoarootedtreeinG,butonethathasatleast2 i+1

nodesandO(2 i+1

)height. Aftercompletionofphase 

log(k+1) 

1,thenodesetof ~

G

dlog (k +1)e

represents

thedesiredrootedspanning forestF (earlierterminationis alsopossible,aswediscussshortly).

Ifx andy arenodesof ~

G

i

, thenwesaythat xand y arepotential neighbors in ~

G

i

ifanedgeexists in

Gjoining somenode intherootedtree representedbyx tosomenodein therootedtreerepresentedbyy.

Wesaythat theyareneighbors in ~

G

i

ifadirectededgeexistsbetweenthem. Anodethathasnoneighbors

isisolated. Ifx andy areneighbors in ~

G

i

, thenweusex!y to indicatethat theedgebetweenx andy is

directedfromxtoy. Inthiscase,wesaythatxisanupstreamneighborofy,whichinturnisandownstream

neighborofx.

PartitionGisbasedonmanipulationsofnodeidentiers,whichweassumetobeadistinctnonnegative

integerforeverynodeinG. Theidentierofnodexin ~

G

i

,denoted byid(x), istheidentieroftherootof

itstree. If anode's identieris lessthan thoseofallitsneighbors,thenwecallthenodealocal minimum.

Ifitisgreater,thenwecallitalocalmaximum. ThefollowingishowPartition Gworksduringphasei. We

uselog (t)

nto denotelog:::logn,wherelog isrepeatedttimes.

Step1. Find theedgesof ~

G

i :

1a. Leteachnodeof ~

G

i

beinactive,iftheheightofthecorrespondingrootedtreeisatleast2 i+1

,oractive,

otherwise.

1b. For everyactive node x of ~

G

i

, nd thepotentialneighbors of x in ~

G

i

. If no potential neighbors are

foundforanynode,thenhaltandexitPartition G.

1c. Foreachactivenodex of ~

G

i

,lety bethe activepotentialneighborofx with theleastidentier. If x

hasnoactivepotentialneighbors,thenlety bethe(inactive) potentialneighborof xhavingtheleast

identier. Addx!y totheedgesetof ~

G

i

,thusmakingxandy neighborsin ~

G

i .

Remark. If no neighbors are found for any node in Step 1b, then in reality ~

G

i

has one single node that

encompasses all the nodes of G and therefore corresponds to a rooted spanning tree of G. In this case,

PartitionGterminatesprematurely,that is,before completingall 

log(k+1) 

phases.

Remark. Attheendof Step1c,everyactivenodeof ~

G

i

hasexactlyonedownstreamneighbor,whileevery

inactivenodehasnone. Similarly,bothactivenodeswhosedownstreamneighborisactiveandinactivenodes

mayhavebetweenzeroandsomepositivenumberofupstreamneighbors. Anactivenodewhosedownstream

neighborisinactivehasnoupstreamneighbors.

Step2. Find thenodesof ~ G i+1 : 2a. Ifx!y isanedgeof ~ G i

such thatx isanactivenodeandy aninactivenode, thencombine x intoy

bycreatingasingle node whoseidentier remainsid(y). Let X be theset of activenodes of ~

G

i that

arenotisolated.

2b. For x2X,letZ(x)X betheset ofupstreamneighbors ofx. IfZ(x)6=;,thenletz bethemember

of Z(x) having the least identier. For y 2 Z(x) such that y 6= z, check whether Z(y) = ;. In the

aÆrmativecase,combiney into x. Otherwise(i.e., Z(y)6=;),eliminateedgey !x. LetX betheset

ofactivenodesof ~

G

i

byeliminatingedgesfrom ~

G

i

appropriately. Also,combine intothenewly-formednodeanynodein X

thatmayhavebecomeisolated. LetX bethesetofactivenodesof ~

G

i

thatarenotisolated.

2d. RepeatStep2cforlocalmaxima,thenletX betheset ofactivenodesof ~

G

i

thatarenotisolated. For

x2X,letl x =id(x). 2e. For x2X,let l x = 8 < : l y ; ify!xisanedgeof ~ G i ; l x 1; ify!xisnotanedgeof ~ G i andl z >l x ; l x +1; ify!xisnotanedgeof ~ G i andl z <l x ,

wherez isthedownstreamneighborof x,and

l + x = 8 < : l y ; ifx!y isanedgeof ~ G i ; l x +1; ifx!y isnotanedgeof ~ G i andl z <l x ; l x 1; ifx!y isnotanedgeof ~ G i andl z >l x ,

wherez istheupstreamneighborofx. Nowconsider thebinaryrepresentationsofl

x ,l x , andl + x ,and

let A(x) be the set of positive natural numbers psuch that l

x and l

x

have the same bit at the pth

positionwhile l

x andl

+

x

do not. Likewise, letB(x) betheset of numberspsuch that l

x and l

x have

dierentbitsat thepthposition whilel

x andl

+

x

havethesamebit. Letp 

(x)bethegreatestmember

ofA(x)[B(x). Ifx!y isanedgeof ~ G i suchthatp  (x)=p 

(y),thencombinexintoy andmakethe

resultingnodeisolatedbyeliminating edgesappropriately(ifanynodein X becomes isolatedbecause

of this, then combine that node into the newly-formednodeas well). Now letX bethe set of active

nodesof ~

G

i

thatarenotisolated,thenrepeatSteps2cand2dwithp 

'sinplaceofid's,andonceagain

letX betheset ofactivenodesof ~

G

i

thatarenotisolated. IfX 6=;,thenletl

x =p



(x)forallx2X

andrepeatStep2e. IfX =;,thenlettheset ofisolatednodesof ~

G

i

bethenodesetof ~

G

i+1 .

Remark. InStep2a, itispossibleformorethanonex toexist forthesamey. Inthiscase,everysuchx is

combinedintothesingleresultingnodeofidentierid(y). Notethatfornosuchxtheremayexistanodez

suchthat z!x ory !z isanedge of ~

G

i

. Thisisso,respectively,becausexhasan inactivedownstream

neighborandbyStep1chasnoactiveneighbors,andbecausey,beinginactive,hasnodownstreamneighbors.

As aconsequence,thenewly-formed node isisolated in ~

G

i

. Atthe end ofStep 2a, thesingledownstream

neighborofeverymemberofX isactive,andthereforealsoamemberofX.

Remark. InStep2b,theremayexist morethanoney2Z(x)suchthat y6=z andZ(y)=;. Everysuchy

getscombinedintonodex. AttheendofStep2b,everymemberofX hasexactlyonedownstreamneighbor

and at most oneupstream neighbor. That is, the membersof X are arranged into groupsof nodes, each

grouphavingatmostonenodewithnoupstreamneighborandexactlyonenodethathasthesameneighbor

for both upstream and downstream neighbor. Except for these two-node directed cycles, such groups of

nodesmayberegardedasdirectedchains.

Remark. AttheendofStep2d,themembersofX arearrangedintodirectedchainsofnodeswhoseidentiers

arestrictlyincreasingordecreasingalongthechains. Eachsuchchainhasatleasttwonodes,ofwhichexactly

onehasnoupstreamneighborandexactlyonehasnodownstreamneighbor.

Remark. Step 2e repeatedly manipulates the node labels l

x

so that the nding of minima and maxima,

respectively as in Steps 2c and 2d, can once again be used to combine nodes in X into isolated nodes.

integershaveanever-dwindlingrange: ifj 1identiesaniterationwithinStep2e,thentherangeoflabels

during iteration j is 0;:::;log (j)

n. Eventually, during a certain iteration j  log 

n, this rangebecomes

f0;1gandconsequentlythetaking ofminima andmaxima asin Steps 2cand 2disguaranteedto produce

anemptyX. Atthebeginningofeachiteration,themembersofX arearrangedintodirectedchainswhose

labelsarestrictlyincreasingordecreasingalongthechains. Eachsuchchainhasatleasttwonodes,ofwhich

exactlyone hasnoupstreamneighborand exactlyonehasnodownstreamneighbor.

Steps1and2specifytheithphaseofPartitionGasthemanipulationofdirectedgraphs,rsttondthe

edgesof ~

G

i

inStep1,theninStep2tondthenodesof ~

G

i+1

. Ofcourse,therealizationofsuchoperations

ongraphsrequires communicationamongthenodesof ~

G

i

,whichultimatelytranslatesinto communication

amongthenodesofG. However,theassumptionofasynchronousmodelofdistributedcomputationmakes

thecommunicationneedsofPartitionGratherstraightforwardtorealize[10].

Becauseeachnodein ~

G

i

standsforarootedtreeinG,Steps1and2canberegardedasbeingexecuted

bythetrees'roots, which in turncoordinatetheremainingnodesin theirtreesin carryingoutthevarious

tasksprescribedbythealgorithm. Forexample,Step1ais abroadcastwithfeedbackontreeedgesstarted

by the root, which sends out the upper bound of 2 i+1

1 on the tree height for the ith phase. This is

propagatedbytheothernodesinthetree,whichsendonwhattheyreceive,ifnonzero,afterdecrementingit

byone. Thefeedbackisstartedbytheleaves,whichclearlyneverhappensifatleastoneleafisnotreached

bythebroadcast,thus signalingtotherootthatthetreeisoversized.

In the same vein, by simply letting every node in G that belongs to the same node x in ~

G

i

havea

recordofid(x),ndingthepotentialneighborsofxinStep1bandthedirectededgesincidenttoitin ~

G

i in

Step 1carealso simpleprocedures thatfunction byprobing theconnectionsofx in G. Wheneveranedge

isdeployed betweentwonodes in ~

G

i

, acorrespondingedgein G,referredto asthepreferrededge between

thosetwonodes,canalsobeeasilyidentied andrecordedforlateruse.

All the remainingactionsin Partition Gcanberealized viacommunication betweentheroots of two

treeswhosenodesin ~

G

i

areconnectedbyanedge. Wheneveramessageneedstobesent,itcanberoutedon

treeedgesonly,except tomovefromonetreeto theother,at which timeitmust gothroughthepreferred

edge between the twotrees. This is, for example, the basis for realizing the combination of a node into

another: combininga node x into a node y that is connected to it by anedge in ~

G

i

involvesmaking the

preferrededgebetweenthemanedgeofthenewtreeandthenpropagatingthroughx'streetheinformation

thatanewroot existsandhasidentierid(y).

3. Correctness and complexity

MostofourcorrectnessandcomplexityargumentshingeonhowwellStep2esucceedsin breakingdirected

chainsofnodesin ~

G

i

asneeded. It istothepropertiesofStep2ethatweturnrst.

Lemma1. Letx!y beanedgeatthebeginningofaniterationofStep2eofPartition G. Thefollowing

holds:

(i) Ifp 

(x)2A(x)andp 

(y)2A(y),thenp  (x)6=p  (y); (ii) Ifp  (x)2B(x)andp 

(y)2B(y),thenp  (x)6=p  (y); (iii) Ifp  (x)2B(x)andp 

(y)2A(y),thenp 

(x)6=p 

(y).

Proof: BySteps2athrough2e,edgex!yisinachainofnodeswhoselabelsareeitherstrictlyincreasing

orstrictlydecreasingalongthechain. Supposetheformercaserst. Thenl

x <l x <l y <l + y .

Ifp (x)2A(x), then l

x andl

x

havethesamebitat positionp (x)while l

x and l

y

havedierentbits

at that sameposition. Ifp 

(y)2A(y), thenl

x andl

y

havethe samebitat position p  (y)while l y andl + y

havedierentbits atthatsameposition. Sop 

(x)andp 

(y)cannotbethesameposition,thusproving(i).

Ifp  (x)2B(x), then l x andl x

havedierent bitsat position p 

(x) whilel

x andl

y

havethesamebit

at thatsameposition. If p 

(y)2B(y), thenl

x andl

y

havedierentbits at position p  (y)while l y andl + y

havethesamebitatthatsameposition. Sop 

(x)andp 

(y)cannotbethesameposition,whichproves(ii).

Wenowprove(iii). If p 

(x) 2B(x), thenl

x and l

x

havedierent bitsat position p  (x) whilel x and l y

havethesamebit at that sameposition. Suppose thesebits are100, respectivelyforl

x ,l x , and l y . By denitionof p 

(x),at allotherpositions totheleftofp 

(x) inthebinaryrepresentationsofl

x ,l x ,l y (that

is, positions correspondingto higher powersof two) wemust havethesamebit for allthree labels orbits

thatdierfrom l

x to l

x

andalsofroml

x to l

y

. Inotherwords,theonlypossibilitiesare000,111,010,and

101. Butthesepossibilitieshaveallthesamebitforl

x andl

y

,whichcontradictsthefactthatl

x <l

y . Then

thebitsforl

x ,l x ,andl y at position p  (x)mustbe011. Ifp  (x)=p 

(y),thenthebitsofl

x andl

y

areboth1atposition p 

(y),whichisinagreementwiththe

denition ofA(y). Bythis samedenition,at position p 

(y) thebitof l +

y

must be0. Totheleft ofp 

(y),

thepossibilitiesforl

x ,l y ,l + y

are000,111,010,and101,againfollowingthedenitionofp 

(y). Thisimplies

the same bit for l

x and l

+

y

at all those positions, which like before contradicts the fact that l

x < l + y . So p  (x)6=p  (y).

Ifx!yisinachainofnodeswhoselabelsarestrictlydecreasingalongthechain,thenl

x >l x >l y >l + y .

Inthiscase,theargumentsthatprove(i)and(ii)remainunchanged,whilein theproofof(iii)itsuÆcesto

complementeverybitin thetripleswedisplayed(thisleadstocontradictionsofl

x >l y andl x >l + y ).

Lemma2. Letx!ybeanedgeatthebeginningofaniterationofStep2eofPartitionG. Ifp  (x)=p  (y), thenp  (x)2A(x)andp  (y)2B(y). Proof: If p  (x) 2B(x)with either p  (y)2B(y) or p 

(y)2 A(y), then byLemma 1, parts (ii) and (iii),

p  (x) 6=p  (y). If p  (x) 2A(x) and p 

(y)2A(y), then by Lemma 1,part(i), p 

(x)6=p 

(y). Thence the

lemma.

Lemma3. Letx!y !zbepartofachainatthebeginningofaniterationofStep2eofPartitionG. If

p  (x)=p  (y),thenp  (y)6=p  (z). Proof: ByLemma2,p  (x)2A(x) andp 

(y)2B(y). ByLemma1,parts(ii)and(iii),p 

(y)6=p 

(z).

NowletI be thelast phaseofPartitionGin which prematuretermination inStep 1bdoesnotoccur.

Then1I  

log(k+1) 

1andwehavethefollowing.

Lemma 4. For i = 0;:::;I, every node of ~

G

i+1

that is not an inactive node of ~

G

i

is formed by the

combinationofatleasttwonodesof ~

G

i .

Proof: ByStep2e,thenodesetof ~

G

i+1

isthesetofisolatednodesin ~

G

i

attheendofStep2. Thelemma

followseasilyfromthefactthateveryisolatednodeproducedduringStep2(thatis,isolatednodesthatare

notinactiveduring phasei)resultfromthecombinationofatleasttwonodesof ~

G

i .

WearenowinpositiontodemonstratethatPartitionGdoesindeedachieveitsgoals.

Theorem 5. Fori =0;:::;I+1,the nodesof ~

G

i

form arooted spanning forestof G. Each tree in this

foresthasatleast2 i

nodesandO(2 i

inductionhypothesisisthatthenodesof ~

G

i

formarootedspanningforestofG,eachofwhosetreeshaving

atleast2 i

nodesandO(2 i

)height.

Inordertoshowthatthenodesof ~

G

i+1

doindeedformarootedspanningforestofG,bytheinduction

hypothesis itsuÆces toarguethat theset X isemptyat theend ofphasei. Thereasonforthis isthat it

isthe setof isolatednodes atthe endof phaseithat wetaketo bethenodeset of ~

G

i+1

,and that X =;

indicatesthateverynodeinGispartofthetreerepresentedbysomeisolatednode. Butthisfollowsdirectly

fromthefactthattherangeoflabelsduringStep2edecreasessteadilyastheiterationsprogress. Eventually,

thisrangebecomessuch thateverylabeliseither 0or1,atwhichtime thendingofminimaandmaxima

makesX empty.

Havingshownthis,weconsiderthenumberofnodesand heightofeachofthetreesin thenodeset of

~ G i+1 . Anodeof ~ G i+1

iseither aninactivenode of ~

G

i

or results,by Lemma4,from thecombinationof at

least two nodesof ~

G

i

. In theformer case,by Step 1athe nodecorrespondsto arooted treein Gwith at

least2 i+1

nodes. Inthelattercase,bythe inductionhypothesis,it correspondsto arooted treein Gwith

atleastq2 i

nodesforq2,that is,atleast2 i+1

nodes.

Asfortheheight,wereasonsimilarly. Ifanodein ~ G i+1 isaninactivenodeof ~ G i

,thenitsheightis,by

theinductionhypothesis,O(2 i

),whichinturnisO(2 i+1

). Ifitisacombinationofatleasttwonodesof ~

G

i ,

theneitherthiscombinationtakesplaceinoneofSteps2athrough2dorinStep2e. Intheformercase,the

combination iseither performed overasingleedge (Steps2aand2b), orit isperformedoverachainof at

mostsixedges(rstin Step2c, theninStep2d). Ineithercase,theinductionhypothesisleadstoaheight

ofq2 i

forq aconstant,whichisO(2 i+1

). ThecaseofStep2e isentirelyanalogous,sinceby Lemma3itis

eitherperformedoverasingleedge,orelsebythetakingofminimaandmaxima,asinSteps2cand2d.

Corollary 6. Thenodes of ~

G

I+1

form arooted spanning forest of G, and in this forest each tree has at

leastk+1nodesandO(k)height.

Proof: IfPartitionGterminatesinStep1bofsomephase,thenI < 

log(k+1) 

1andthecorollaryholds,

because ~

Ghasinthiscaseonesinglenodeencompassingallthenk+1nodesofG,andfurthermorethe

heightofthetreethatcorrespondstothissinglenodeisbyTheorem5O(2 I+1

),whichisO(k). IfPartitionG

runs throughall the phases,then I = 

log (k+1) 

1and the corollaryfollowsdirectly from Theorem5

withi=I+1= 

log (k+1) 

.

WenownalizethesectionbydiscussingthetimeandnumberofmessagesneededbyPartitionGand

byAlgorithmPBas awhole.

Theorem7. PartitionGrequiresO(klog 

n)timeandO(mlogk+nlogklog 

n)messagestocomplete.

Proof: Duringtheith phase,i=0;:::; 

log (k+1) 

1,each ofSteps1a through2drequires aconstant

numberofcommunication\rounds,"eachinturnrequiringanumberoftimeunitsproportionaltotheheight

of arootedtree in that phase,which byTheorem 5is O(2 i

). Thesameholds for each of theiterationsof

Step2e,ofwhichthereareat mostlog 

n. ThenthetimerequiredforPartitionGtocompletegrowswith

dlog (k +1)e 1 X i=0 2 i log  n dlog (k +1)e 1 X i=0 2 log(k +1) 2 i log  n =(k+1)log  n dlog (k +1)e 1 X i=0 1 2 i <2(k+1)log  n;

soPartitionGrequires O(klog n)time.

The number of messages that PartitionG requires canbe estimated likewise for phase i, as follows.

Thenumberof messagessentduring Step1is dominatedbyStep 1bto determinethepotentialneighbors

in Gofanodein ~

G

i

,which requiresO(m)messages. Steps2a and2brequireO(n)messages,whichisthe

totalnumberoftree edges,becauseseveralnodesmaybecombinedintothesamenodeduringthosesteps.

However,eachoftheO(log 

n)communication\rounds"inSteps2cthrough2eismoreeconomical,because

the chain structure of ~

G

i

in those stepsallows communication to take place along single paths from the

trees'roots, andsothenumberofmessages owingin each rootedtreeisproportionaltoits height,which

during phasei isO(2 i

)byTheorem5. AlsobyTheorem5,each rooted treein ~ G i hasat least2 i nodes, so

there areat mostn=2 i

rooted treesin ~

G

i

. Itfollowsthat thenumberofmessagesrequiredbyPartitionG

forcompletiongrowswith

dlog (k +1)e 1 X i=0 m+ n 2 i 2 i log  n=m  log(k+1)  +nlog  n  log (k+1)  ;

soPartitionGrequires O(mlogk+nlogklog 

n)messages.

Corollary8. AlgorithmPBrequiresO(klog 

n)timeandO(mlogk+nlogklog 

n)messagestocomplete.

Proof: Immediate from Theorem7,considering that thealgorithm'ssecond stage requiresO(k) timeand

O(n)messages.

4. Concluding remarks

Wehaveconsideredthe problemofnding ak-dominatingsetwith nomorethan 

n=(k+1) 

nodesin G,

andgivenanewsynchronousdistributedalgorithmtosolveitinO(klog 

n)timewhilerequiringO(mlogk+

nlogklog 

n)messages. Our algorithm follows thesame overall strategy of [6], accordingto which rsta

rootedspanning forestisfoundin Gwithcertaincharacteristics,andthen thedesiredk-dominatingseton

thatforest.

Our algorithm introduces a new approach to nding the rooted spanning forest, which we think is

conceptually simplerthan theone of [6], and shares withthe algorithm of [6] theadditional computation

that is requiredto nd the k-dominating set. In both algorithms, theoverall complexity isdominated by

theforest-ndingstage. Bothhavethesametimecomplexity,butourshasbettermessagecomplexity.

Acknowledgments

The authors acknowledge partial support from CNPq, CAPES, the PRONEX initiative of Brazil's MCT

undercontract41.96.0857.00,andaFAPERJBBPgrant.

References

1. V.C.Barbosa,AnIntroductiontoDistributedAlgorithms,TheMITPress,Cambridge,MA,1996.

2. J.Bar-Ilan,G.Kortsarz,andD. Peleg,\Howto allocatenetworkcenters,"J. ofAlgorithms 15(1993),

385{415.

3. J. A. Garay, S. Kutten, and D. Peleg, \A sub-linear time distributed algorithm for minimum-weight

Completeness,W. H.Freeman&Co.,NewYork,NY, 1979.

5. A. V.Goldberg, S. A.Plotkin, andG.E. Shannon,\Parallelsymmetry-breakingin sparsegraphs," in

Proc. of the NineteenthAnnualACMSymposiumonTheoryof Computing,315{323,1987.

6. S.KuttenandD.Peleg,\Fastdistributedconstructionofsmallk-dominatingsetsandapplications,"J.

of Algorithms 28(1998),40{66.

7. A.PanconesiandA.Srinivasan,\Improveddistributedalgorithmsforcoloringandnetwork

decomposi-tionproblems,"inProc. of the Twenty-FourthAnnual ACMSymposiumon the Theory of Computing,

581{592,1992.

8. D.Peleg,\Distributeddatastructures:acomplexity-orientedview,"inProc.oftheFourthInt.Workshop

onDistributedAlgorithms,71{89,1990.

9. D. Pelegand E. Upfal,\A tradeobetweensize andeÆciency for routingtables,"J. of the ACM 36

(1989),510{530.

10. L. D.Penso,ADistributedAlgorithm toFind k-dominating Setsin Graphs,M.Sc. thesis,Federal

Uni-versityofRiodeJaneiro,Brazil,December1999.(In Portuguese.)

11. R.WittmannandM.Zitterbart,MulticastCommunication:Protocols andApplications, Morgan