On the status sequences of trees

(1)

Contents lists available atScienceDirect

Theoretical Computer Science

www.elsevier.com/locate/tcs

On the status sequences of trees

Aida Abiad

^a

^,

^b

^,

^c

^,∗ , Boris Brimkov

^d

, Alexander Grigoriev

^e

aDepartmentofMathematicsandComputerScience,EindhovenUniversityofTechnology,TheNetherlands bDepartmentofMathematics:Analysis,LogicandDiscreteMathematics,GhentUniversity,Belgium cDepartmentofMathematicsandDataScience,VrijeUniversiteitBrussel,Belgium

dDepartmentofMathematicsandStatistics,SlipperyRockUniversity,SlipperyRock,PA,USA eDepartmentofDataAnalyticsandDigitalization,MaastrichtUniversity,Maastricht,TheNetherlands

a rt i c l e i nf o a b s t ra c t

Articlehistory:

Received11September2020

Receivedinrevisedform11November2020 Accepted15December2020

Availableonline28December2020 CommunicatedbyT.Calamoneri

Keywords:

Tree

Statussequence Complexity Statusinjective Graphpartition

The statusofavertex vinaconnectedgraphisthe sumofthedistancesfromv toall other vertices. The status sequence ofa connected graph is the list of the statuses of all theverticesofthegraph. Inthispaper weinvestigatethe statussequencesoftrees.

Particularly,weshowthatitisNP-completetodecidewhetherthereexistsatreethathasa givensequenceofintegersasitsstatussequence.Wealsopresentsomenewresultsabout trees whosestatussequencesarecomprised ofafew distinctnumbersormanydistinct numbers.Inthisdirection,weshowthatanystatusinjectivetreeisuniqueamongtrees.

Finally, weinvestigatehow orbitpartitions and equitable partitions relate tothe status sequence.

©²⁰²⁰^TheÂuthor(s).^Published^byÊlsevier^B.V.^Thisîsânôpenâccessârticleûnder^the CCBYlicense(http://creativecommons.org/licenses/by/4.0/).

1. Introduction

Sequencesassociatedwithagraph,suchasthedegreesequence,spectrum,andstatussequence,containusefulinforma- tionaboutthegraph’sstructureandgiveacompactrepresentationofthegraphwithoutusingvertexadjacencies.Extracting andanalyzingtheinformationcontainedinsuchsequencesisacrucialissueinmanyproblems,suchasgraphisomorphism.

In thispaper, westudy thestatussequences oftrees,andanswer severalquestions aboutstatusrealizability, uniqueness, andtheirrelationtovariousgraphpartitions.

Thegraphsconsideredinthispaperareﬁnite,simple,looplessandconnected.LetG

= (

V

,

E

)

beagraph.Thestatusvalue (alsocalledtransmissionindex)ofavertexv inG,denoteds

(

v

)

,isthesumofthedistancesbetweenvandallothervertices, i.e., s

(

v

) =

u∈Vd

(

v

,

u

)

,whered

(

v

,

u

)

istheshortestpathdistancebetweenv andu inG.Thisconcept wasintroduced by Hararyin1959 [18].Thestatussequenceof G,denoted

σ (

G

)

,isthelistofthe statusvaluesofall verticesarrangedin nondecreasingorder.Statussequencesofgraphshaverecentlyattractedconsiderableattention,see,e.g.,[26,28,31,33].Status sequenceshavealsobeenusedininterestingpracticalapplications,suchaseﬃcientlydeterminingwhetheracompanysells aparticulartypeofcircuitboard;in[2],itwasshownthatalgorithmsforboardequivalencebasedongraphinvariantssuch asthestatussequencerequiremuchlesstimeandmemorycomparedtoalgorithmsthatusethecompletetopologyofthe circuit boards.Similarapproachescould beapplied toother graphisomorphismapplications,such asidentifying whether a chemicalcompoundisfound withina chemicaldatabase. Statussequenceshavealsobeenstudied andused inspectral graph theory,e.g., in thedeﬁnition of thedistance Laplacianmatrix andthe study of its eigenvalues (cf. [4,6,27]). More

*

Correspondingauthor.

E-mailaddresses:a.abiad.monge@tue.nl(A. Abiad),boris.brimkov@sru.edu(B. Brimkov),a.grigoriev@maastrichtuniversity.nl(A. Grigoriev).

https://doi.org/10.1016/j.tcs.2020.12.030

0304-3975/©²⁰²⁰^TheÂuthor(s).^Published^byÊlsevier^B.V.^Thisîsânôpenâccessârticleûnder^the^CC^BY^license (http://creativecommons.org/licenses/by/4.0/).

(2)

generally,status sequenceshavebeen implicitlyorexplicitlyusedin studyingmetricdimension [16,19,20,24],variants of theCops&Robbersgame[9,30],andrelatedproblems.

AconnectedgraphG isstatusinjectiveifthestatusvaluesofitsverticesarealldistinct.Wedenotebyk

(

G

)

thenumberof distinctstatusvaluesof

σ (

G

)

.Theparameterk

(

G

)

isalsoknownastheWienerdimensionofG,see[3].Agraphistransmission- regular ifk

(

G

) =

^1,î.e., îf^the^status^valuesôfâllîts ^verticesâreêqual.Transmission-regular graphshavebeenstudiedby severalauthors(see,e.g.,[1,5,21,22]).Asequence

σ

ofintegersisstatusrealizableifthereexistsagraphG with

σ (

G

) = σ

. LetF^beâ^familyôf^connected^graphsând^G^beâ^graphⁱⁿF^.Â^graph^Gîs^statusûniqueⁱⁿFîf^forâny^H

∈

F^,

σ (

H

) = σ (

G

)

impliesthatH

^G,î.e.,^Hând^G âreisomorphic.Forexample,pathsarestatusuniqueinthefamilyofallconnectedgraphs, sinceapathofordernistheonlygraphoforderncontainingavertexofstatusvaluen

(

n

−

¹

)/

2.

Ingeneral, comparedtotheadjacencylistortheadjacencymatrixofagraph,thereissomelossofinformationinthe statussequence.Forexample,onecannotobtainthedistancedegreesequenceofagraphfromitsstatussequence[28].Itis alsowellknownthatnon-isomorphicgraphsmayhavethesamestatussequence.Nevertheless,thestatussequencecontains important informationabout the graph,andthe study ofgraphs with specialstatus sequences has produced interesting results. For instance,it is known that spidersare statusunique intrees [31]. In ([7], p. 185) the authors proposed the problemofﬁndingstatusinjectivegraphs.ThisproblemwasaddressedbyPachter [28],whoproved thatforanygraph G thereexistsastatusinjectivegraphHthatcontains Gasaninducedsubgraph.

Buckley andHarary [8] posed the following problemin a paperthat discusses various problemsconcerning distance conceptsingraphs.

Problem 1.1 (Statussequence recognition).Characterize status sequences, i.e., ﬁnd a characterization that determines whetheragivensequenceofpositiveintegersisthestatussequenceofagraph.

Motivatedbytheabovefacts,inthispaperweconsiderthefollowingquestionraisedbyShangandLinin[31]:

Problem1.2.Arestatusinjectivetreesstatusuniqueinallconnectedgraphs?

Theaboveproblemwas recentlyansweredinthenegativein[29],wheretheauthorsprovideaconstructionofpairsof a treeandanontreegraphwiththe samestatussequence.Inthispaper,we providenewresultsonthe directionofthe aboveproblemforextremaltrees.Inparticular,weshowthatstatusinjectivetreesarestatusuniqueintrees.

Moreover,wealsoinvestigatethefollowingspecialcaseofProblem1.1:

Problem1.3(Treestatusrecognition).Givenasequence

σ

ofpositiveintegers,doesthereexistatreeT suchthat

σ (

T

) = σ

?

Recognitionproblemssimilarto1.1and1.3havebeenstudiedforsome othergraphsequences.Forexample,Erdösand Gallai[12] andHakimi[17] gaveconditionstodeterminewhetheragivensequenceisthedegreesequenceofsomegraph.

Sequencesrelatedtodistancesinagraphhavealsobeenstudied,see,e.g.,[8].Inthosepapers,theauthorsnotonlytackle therecognition problems,butalsoconstructfastalgorithmsforﬁndinggraphs thatrealize givensequences.Following the samedirection,weaddressthefollowingquestionregardingstatussequences:

Problem1.4(Statusrealizabilityintrees).Givenasequenceofintegers

σ

,eitherconstructatreeT such that

σ (

T

) = σ

orreportthatsuchatreedoesnotexist.

Wearealsointerestedinthefollowingproblemrelatedtothestatusuniquenessofagraph:

Problem1.5.Whatarenecessaryand/orsuﬃcientconditionsforasetofverticesofagraphtohavethesamestatus?

The main results of thispaper are as follows.We ﬁrst consider the problemof determining whether ornot a given sequenceisthestatussequenceofatree,andinvestigatethecomplexityofthisproblemintermsofthesymmetryofthe constructedtree.Inparticular,weshow thatforhighlysymmetrictreeswithk

(

G

) =

2

,

3,andforhighlyasymmetrictrees withk

(

G

) =

^n,^the^problem^ispolynomiallysolvable,whileforgeneralmid-symmetricclassestheproblemisNP-complete.

Furthermore,weshowthatinthehighlyasymmetriccaseswithk

(

G

) =

^n,îfâ^tree^realizingâ^given^status^sequenceêxists, thistreeis uniquelydefined. Inthissense, weexplore thecomplexity landscapeofthe statussequenceproblemintrees basedontreesymmetries.Finally,wealsoinvestigatewaystoapproachtheproblemusingpartitionsanddecompositions, whichisoftenanaturalplantoattackhardproblems.

Thepaperisorganizedasfollows.InSection2,weprovethatTreestatusrecognitionisNP-complete.InSection3,we presentseveralpolynomiallysolvablespecialcasesofStatusrealizabilityintrees,speciﬁcallyforsymmetricandasymmet- ric trees.InSection4,we explorehow variouswell-known graphpartitionsrelate tothe statussequence.Weﬁnishwith someconcludingremarksinSection5.

(3)

Fig. 1.Gadget of the reduction from3-Partition.

2. Complexityof TREESTATUSRECOGNITION

The eccentricityofa vertexv ina graphG,denoted

(

v

)

,isthe maximumdistancefromv to anyvertex,i.e.,

(

v

) =

max

{

^d

(

v

,

w

) :

^w

∈

^V

(

G

)}

^.^The^diameter^of^G,^denoted^diam

(

G

)

,isthemaximumeccentricityofavertexofG,i.e.,diam

(

G

) =

max

{ (

v

) :

^v

∈

^V

(

G

) }

^.^The^radius^of^G,^denoted^rad

(

G

)

,istheminimumeccentricityofavertexofG,i.e.,rad

(

G

) =

^min

{ (

v

) :

v

∈

^V

(

G

)}

^.Equivalently,if T

= (

V

,

E

)

isatree,theradius(sometimesalsocalledthedepthof T)isthesmallestnumberr suchthatthereexistsavertexv

∈

^V ^such^that^d

(

v

,

u

) ≤

^r^for^all^u

∈

^V^;^thisîs^the^sameâs^saying^that^the^diameterôf^the treeisatmost2r.ThecenterofGisthesetofverticeswhoseeccentricityisequaltorad

(

G

)

.

In this section, we show that determining whether or not a given sequence is the status sequence of a tree is NP- complete.Inparticular,westudy thecomplexityofTreestatusrecognitionwhenthetreesarerestrictedtohaveradius3.

WerefertothelatterproblemasSRT-R3. Theorem2.1.SRT-R3isNP-complete.

Proof. SRT-R3is clearly in NP, asthestatus sequence ofa tree ofradius 3 canbe computed inlinear time. We reduce the well-known stronglyNP-complete problem3-Partitionto SRT-R3. 3-Partition(cf. [13])reads:Given amultiset A

= {

^a1

,

a2

, . . . ,

an

}

ôf^positiveîntegers,^does^thereêxistâ^partitionôfAînto^triplets^such^thatâll^triplets^have^the^same^sum?

Without loss of generality, assume n is divisible by 3 and letm

=

ⁿ

/

3, A

=

n

i=1ai and B

=

^A

/

m. Since 3-Partition is stronglyNP-complete,wemayassumethattheinputoftheproblemisgiveninunaryencoding,i.e.,linearin A.Notethat 3-PartitionremainsstronglyNP-completeevenifB

/

4

<

ai

<

B

/

2 foralli

∈ {

¹

,

2

, . . . ,

n

}

^.

Givenan instance I of3-Partition,we constructan instanceof SRT-R3whoseinput sequence

σ

is comprisedof the following1

+

^m

+

ⁿ

+

^A ^integers:

•

^3A

+

^7m ^withmultiplicity1;

•

^4A

−

^2B

+

^11m

−

^{7 with}multiplicitym;

•

^5A

−

^2B

−

^2ai

+

^15m

−

^{8 with}multiplicity1,foreachi

∈ {

¹

,

2

, . . . ,

n

}

^;

•

^6A

−

^2B

−

^2ai

+

^19m

−

^{9 with}multiplicityai,foreachi

∈ {

¹

,

2

, . . . ,

n

}

^.

Clearly,thesize ofthe sequence

σ

issubquadraticin A,andthuspolynomial intheunaryencodedsizeof I.Wewill show that I is a yes-instanceof 3-Partitionif andonly if

σ

isa yes-instance ofSRT-R3. Suppose firstthat I is a yes- instance of3-Partition, i.e., thereisa partition ofA înto^m ^triplets^such ^that ^the^sumôf întegersⁱⁿ êach^triplet êquals B.Consider thefollowingtreeT.Takearootvertexwithexactlym child-nodeswhichrepresentmtriplets.Foreachchild- noderepresentingatriplet,createexactlythreedescendantsrepresentingtheelementsofthetriplet.Finally,foreachvertex representinganelementai ofatriplet,createexactlyai descendants,whicharetheleavesofthetree.Byconstructionthe treeisofradius3.SeeFig.1foranillustration.

Thestatusoftherootequals1

·

^m

+

²

·

^3m

+

³

·

iai

=

³^A

+

^7m.^The^status^of^any^vertexrepresentingatripletequals 1

·

⁴

+

²

· (

B

+

^m

−

¹

) +

³

·

³

(

m

−

¹

) +

⁴

· (

m

−

¹

)

B

=

^4A

−

^2B

+

^11m

−

^7.^Thereâre^m^such^vertices.^The^statusôfâ^vertex representinganelementa_iequals1

·(

^ai

+

¹

) +

²

·

³

+

³

·(

^B

−

^ai

+

^m

−

¹

) +

⁴

·

³

(

m

−

¹

) +

⁵

·(

^m

−

¹

)

B

=

^5A

−

^2B

−

^2ai

+

^15m

−

^8.

Finally,thestatusofaleafadjacenttoavertexrepresentingai equals1

·

¹

+

²

·

^ai

+

³

·

³

+

⁴

· (

B

−

^ai

+

^m

−

¹

) +

⁵

·

³

(

m

−

1

) +

⁶

· (

m

−

¹

)

B

=

^6A

−

^2B

−

^2ai

+

^19m

−

^9.^Thus,^tree^T ^of^radius³^realizes

σ

.

Before tackling the opposite direction ofthe proof, we ﬁrst recall some useful properties ofthe status values ofthe verticesinatree.ThemedianofaconnectedgraphG isthesetofverticesofG withthesmalleststatus.

Lemma2.2([11]).If v1 isavertexinthemedian ofa treeT , v1

,

v2

, . . . ,

vr isapathin T ,andv2 isnot amedianofT ,then s

(

v₁

) <

s

(

v₂

) < · · · <

s

(

v_r

)

.

Lemma2.3([31]).Supposethatv1andv2areadjacentverticesinatreeT

= (

V

,

E

)

.LetT1andT2betwocomponentsofT after deletionoftheedge

(

v₁

,

v₂

)

,andletv₁

∈

^V

(

T₁

)

andv₂

∈

^V

(

T₂

)

.Then,s

(

v₁

) −

^s

(

v₂

) = |

^V

(

T₂

)| − |

^V

(

T₁

)|

^.

Now,suppose,

σ

isayes-instanceofSRT-R3andatreeT isitsrealization.Withoutlossofgenerality, wemayassume A

>

3B

+

^19m

+

^9, ^forôtherwise^we âdd^to êveryêlement ôf^the^multisetân âdditive ^constant^3B

+

^19m

+

9 preserving

(4)

thehardnessofthe3-partitioncaseandrepeatthearguments.Forthesesuﬃcientlylargevalues A,thevertexwithstatus 3A

+

^7m îs^theônly ^medianôf ^T^.^Moreover, ^by^Lemma^2.2^forâllⁱ

∈ {

¹

,

2

, . . . ,

n

}

^,^the^vertices^of^status^6A

−

^2B

−

^2ai

+

19m

−

^{9 are}^the^leaves^of^T^.Furthermore,byLemma2.3foralli

∈ {

¹

,

2

, . . . ,

n

}

^,^any^vertex^of^status^6A

−

^2B

−

^2ai

+

^19m

−

⁹ isadjacenttoavertexofstatus5A

−

^2B

−

^2ai

+

^15m

−

^{8 as}^the^differenceⁱⁿ^status^betweenâ^leafândîtsâncestorêquals the size of the tree minus 2, which in T equals A

+

^4m

−

^1. ^Next, ^we ^notice ^that ^for ^any ⁱ

∈ {

¹

,

2

, . . . ,

n

}

^, ^the ^vertex of status 5A

−

^2B

−

^2ai

+

^15m

−

^{8 cannot} ^be ^directly ^adjacent ^to ^the ^median^with^status ^3A

+

^7m ^as ^the ^difference ⁱⁿ statuses istoolarge contradictingLemma2.3.Thisimpliesthat foranyi

∈ {

¹

,

2

, . . . ,

n

}

^,^theâncestorôfâ^vertexôf^status 5A

−

^2B

−

^2ai

+

^15m

−

^{8 can}ônly^beâ^vertexôf^status^4A

−

^2B

+

^11m

−

^{7 and}âll^verticesôf^the^latter^statusâreâdjacent tothemedian.Sinceallverticesadjacenttothemedianhavethesamestatus4A

−

^2B

+

^11m

−

^7,^each^of^these^vertices^has exactlythesametotalnumberofdescendants,namelyB

+

^3.

Consider anedge betweenavertexv ofstatus5A

−

^2B

−

^2ai

+

^15m

−

⁸

,

i

∈ {

¹

,

2

, . . . ,

n

}

^,ândîts âncestorû ôf^status 4A

−

^2B

+

^11m

−

^7.^By ^Lemma^2.3, ^the^number ôf^leaves^x âdjacent^to ^v êquals ^the^size ôf^the ^treeâfter^deletion ôf ^v together withallits xdescendantsminusthestatusdifferencebetween v anduandminusoneforcountingv itself.The sizeofthetreeafterdeletionofv withallitsdescendantsisA

+

^4m

+

¹

−

^x

−

¹

=

^A

+

^4m

−

^x.^The^differenceⁱⁿ^status^of^v anduis5A

−

^2B

−

^2ai

+

^15m

−

⁸

− (

4A

−

^2B

+

^11m

−

⁷

) =

^A

−

^2ai

+

^4m

−

^1.^Therefore,^x

=

^A

+

^4m

−

^x

− (

A

−

^2ai

+

^4m

−

¹

) −

¹

=

2ai

−

^x,^yielding ^x

=

^ai.

Summarizingtheﬁndings,thefollowingpropertiesnecessarilyholdforT: 1. ThemedianofT isthevertexofstatus3A

+

^7m;

2. Therearemverticesofstatus4A

−

^2B

+

^11m

−

^{7 adjacent}^to^the^median;

3. ThereareexactlyB

+

3 descendantsforeveryvertexofstatus4A

−

^2B

+

^11m

−

^7;

4. Onlyverticesofstatus5A

−

^2B

−

^2ai

+

^15m

−

⁸

,

i

∈ {

¹

,

2

, . . . ,

n

}

^,^can^be^immediatesuccessors/descendantsofavertex ofstatus4A

−

^2B

+

^11m

−

^7;

5. Foreveryvertexv_iofstatus5A

−

^2B

−

^2ai

+

^15m

−

⁸

,

i

∈ {

¹

,

2

, . . . ,

n

}

^,^thereâreêxactlyâileavesofstatus6A

−

^2B

−

2ai

+

^19m

−

^{9 adjacent}^to^vi.

Finally, considera vertex u ofstatus 4A

−

^2B

+

^11m

−

^7. Âs^stated âbove, ^thereâre êxactly ^B

+

3 descendants of u. By assumption (in 3-Partition) that B

/

4

<

ai

<

B

/

2 for all i

∈ {

¹

,

2

, . . . ,

n

}

^, ^vertex û ^has êxactly ^three îmmediate ^succes- sors/descendants, say vi

,

vj

,

vk where i

,

j

,

k

∈ {

¹

,

2

, . . . ,

n

}

^, ând ^these^vertices ^haveêxactly âi

,

aj

,

ak descendant leaves, respectively.Thus,a_i

+

^aj

+

^ak

=

^B,^which^completes^the^proof.

Theorem2.1straightforwardlyimpliesthefollowingcorollary.

Corollary2.4.TreestatusrecognitionisNP-complete.

Proof. InTheorem2.1,thereisnolossofgenerality:whenrealizingatreefromtheconstructedsequenceofintegers,we derivedproperties1–5implyingthattheonlytreespossiblyrealizingthesequencearetreesofradius3.

Notably,Theorem2.1hasalsostrongalgorithmicimplicationsforthefollowingoptimizationproblem.

Problem2.5(MinimumStatusCorrection).Givena sequence ofintegers

σ

,what is theminimumchange on

σ

(under anynormthatmeasuresdistancebetweensequences)thatmakes

σ

statusrealizableintrees?

Corollary2.6.UnlessP

=

^{N P ,}^there^is^no^polynomial^time^constantapproximationalgorithmforMinimumStatusCorrection. Proof. Any polynomial time constant approximation algorithm for Minimum Status Correction should recognize 0- correctioninstancesincluding“yes”-instancesofSRT-R3.ByTheorem2.1,SRT-R3isN P-complete.Thus,thealgorithmsolves an N P-completeproblemandP

=

^{N P.}

3. Polynomialspecialcasesof STATUSREALIZABILIT YINTREES

Inthissection,weinvestigatesomespecialcaseswherestatusrealizabilityintreescanbesolvedinpolynomialtime.

Inparticular,an efficientalgorithmforfinding suchatreeisprovidedinthecasethatallelements ofthegivensequence are distinct.Asasideresult, wefindthatthereisa uniquetreewiththegivenstatussequenceinthiscase,ifoneexists.

ThelatteraddressesanextremalcaseofProblem1.2,showingthatstatusinjectivetreesarestatusuniqueintrees.

Webeginwitharesultaboutasymmetrictrees.Thefollowingresultshowsthatgivenanintegersequencewithndistinct values,itcanbe determinedinpolynomialtimewhetherornotthissequenceisthestatussequenceofatreeT

= (

V

,

E

)

of ordern. Notethat such a tree, ifrealized,is highly asymmetric, asit cannot haveanynontrivial automorphisms; see Corollary4.2fordetails.Independentlyofus,thefollowingresulthasrecentlybeenprovenin[32].Wenotethattheproof ofTheorem3.1inthepresentpaperisconstructive,providinganexplicitlineartimealgorithmforthetreeconstruction.

(5)

Algorithm 1:Statusinjectivetree.

1 Input:Asequenceofintegersσ= {a1,a2,. . . ,an},witha1>a2> . . . >an.

2 Output:AtreeTwhosestatussequenceisσ,oranaﬃrmationthatsuchatreedoesnotexist.

3 fori=1,. . . ,ndo 4 Createavertexvi;

5 pi← ∅^; ^piistheparentofviinT

6 ci←^0; ^ciisthenumberofdescendantsofviinT

7 fori=1,. . . ,n−1do

8 Findj∈ {i+1,. . . ,n}suchthataj=ai−n+2(ci+1);

9 ifsuchindex jdoesnotexistthenreturn“σisnotthestatussequenceofatree”else 10 pi←^vj;

11 cj←^cj+^ci+^1;

12 returnT←({1,. . . ,n},{(vi,pi):i=1,. . . ,n−1})

Fig. 2.Illustration of Algorithm1for inputσ= {¹⁹,18,15,14,13,11,10}^.

Theorem3.1.Statusinjectivetreesarestatusuniqueintrees.

Proof. Theproofisconstructive, i.e.,wepresentan algorithmwhoseinputisasequence

σ

ofndistinctpositiveintegers andthe output iseither a unique tree T realizing

σ

ora conclusion that thesequence is not astatus sequenceof any tree.

Algorithm1attemptstoconstructatreewhoseverticesv1

, . . . ,

vncorrespondtothestatusvaluesintheinputsequence

σ

(whichbyassumptionarealldistinct);thetreeisspeciﬁedbyassigningaparent pi toeachvertexvi,exceptthevertex vn whichistreatedastheroot.Thealgorithmalsostoresandupdatesthenumberofdescendants ofvertex vi (excluding v_i itself)asthe variablec_i.Inthe mainloop ofthealgorithm, theverticesare considered accordingtodecreasing status values.

We will show by induction that after iteration i ofthe algorithm, the parents ofthe vertices v1

, . . . ,

vi are uniquely identiﬁedasp1

, . . . ,

pi,andthatthenumberofdescendantsofvi+¹ among

{

^v1

, . . . ,

vi

}

^is^ci+¹.

In theﬁrst iteration ofthe algorithm,by Lemma2.2,vertex v1 (correspondingto statusvalue a1) must be aleaf. By Lemma2.3,a₁

−

^ap1

= (

n

− (

c₁

+

¹

)) − (

c₁

+

¹

)

,soa_p₁

=

^a1

−

ⁿ

+

²

(

c₁

+

¹

)

.Ifthisvalueisnotin

σ

,then

σ

isnotthestatus sequenceofatree.Otherwise,thevertexv_j withstatusvaluea₁

−

ⁿ

+

²

(

c₁

+

¹

)

istheparentofv₁ andvertexv₁becomes a descendantofvertex p1.Thus, afterthe ﬁrstiteration,theparentof v1 isuniquelyidentiﬁedas p1,andthenumberof descendantsofv2 among

{

^v1

}

^is^c2 (becauseitiseither0,duethewaythevariableisinitialized(inline5),oritis1,i.e., v1,if p1

=

^v2).

Suppose the statement is truefor iterations1

, . . . ,

i

−

^1,ând ^considerîteration î.^By ^Lemma^2.3ând ^by ^the âssump- tion that the numberofdescendants of v_i among

{

^v1

, . . . ,

v_i₋₁

}

^is ^ci, itfollows that a_i

−

^api

= (

n

− (

c_i

+

¹

)) − (

c_i

+

¹

)

, so ap_i

=

^ai

−

ⁿ

+

²

(

ci

+

¹

)

. If this value is not in

σ

, then

σ

is not the status sequence of a tree (line 8). Other- wise, the vertex vj corresponding to statusvalue ai

−

ⁿ

+

²

(

ci

+

¹

)

isthe parent pi of vi (line 9) and thedescendants of vi, including vi, become descendants of pi (line 10). See Fig. 2 for an illustration. Moreover, since the input of the equation used to compute p_i (in line 7) is uniquely determined by assumption, and the equation is determinis- tic, it follows that p_i is uniquely identiﬁed. By induction, it follows that Algorithm 1 uniquely constructs a tree from a sequence of distinct integers, if such a tree can be constructed. Thus, status injective trees are status unique among trees.

NotethatAlgorithm1terminatesin O

(

nlogn

)

timeandcanthereforebeusedconstructivelyandveryeﬃciently.

While Algorithm 1 is designedto tacklethe asymmetric (status injective) cases of Statusrealizabilityintrees, the remainder ofthissection addressespolynomiallysolvablesymmetriccasesofSRT-R3,whenthenumberofdistinct status valuesisboundedbyaconstantandthedegreeofthecentervertexisboundedbyaconstant.Notethatby Theorem2.1, theproblemSRT-R3isNP-completeingeneral,butthereductionrequiresahighnumberofdistinctstatusvaluesandahigh

(6)

degreecentervertexofthetree.Recallthatk

(

G

)

denotesthenumberofdistinctstatusvaluesofagraphG.Thefollowing resultprovidesatightlowerboundonk

(

T

)

foratreeT,whichwillbeusedinthesequel.

Recall thatwedeﬁnethemedian ofaconnectedgraph G asthesetofverticesofG withthesmalleststatus.Itiswell knownthat atreeT haseitheronecenterortwoadjacentcenters;ifT hasone center,thendiam

(

T

) =

^2rad

(

T

)

,if T has twocenters,thendiam

(

T

) =

^2rad

(

T

) −

^1.

Lemma3.2.LetT beatreeonn vertices.Then,

k

(

T

) ≥

diam

(

T

) +

¹ 2

,

andthisboundistight.

Proof. T hasatmosttwomedians,andifT hastwomedians thenthey arenecessarilyadjacent,since otherwiseapath fromonemediantoanothernonadjacentmedianwillviolatethestrictinequalitiesofLemma2.2.Letv beamedianvertex of T. Suppose v is acenter of T,andthat T hasa single center.We willﬁrst show that thereare atleast twovertices that areatdistancerad

(

T

)

fromv.Supposeforcontradictionthat wistheonlyvertexatdistancerad

(

T

)

fromv,andlet wbethevertexfurthestfromv besides w.Letv bethevertexadjacenttov alongthepathfromv to w.LetT₁ bethe componentofT

−

^{v v}^containing^v^,^and^T2bethecomponentofT

−

^{v v}^containing^v^.^Then,^rad

(

T

) ≥

^d

(

v

,

u

) =

^d

(

v

,

u

) +

¹ for every u

∈

^V

(

T2

)

, andd

(

v

,

u

) =

^d

(

v

,

u

) +

¹

<

rad

(

T

) +

^{1 for} ^every ^u

∈

^V

(

T1

)

(where d

(

v

,

u

) <

rad

(

T

)

follows from the assumption that w is the only vertex at distance rad

(

T

)

from v, andthe fact that T1 does not contain w). Thus, d

(

v

,

u

) ≤

^rad

(

T

)

forevery u

∈

^V

(

T

)

,which means v isalso acenter of T, acontradiction. Thus, there are atleasttwo vertices, w and w, that are at distance rad

(

T

)

from v. Hence, without loss of generality, we can suppose the vertex adjacentto v along thepath fromv to w isnot amedian(otherwise we wouldconsiderthepath fromv to w).Then, d

(

v

,

w

) =

^rad

(

T

) =

^diam₂⁽^T⁾^,ând^by^Lemma^2.2,êachôf^the^verticesôn^the^path^from^v ^to^w,încluding^v ândîncluding^w, hasadifferentstatus.Thus,k

(

T

) ≥

^d

(

v

,

w

) +

¹

=

^diam₂⁽^T⁾

+

¹

≥

diam(T)+¹ 2

.

Now suppose v isacenterofT andthat T hastwocentervertices v andv.Let wbe avertexatmaximumdistance fromv.Sincethetwocentersofatreeareadjacent,thepathbetweenvandwpassesthroughv(otherwisethepathfrom vto w wouldbe longer).Then,d

(

v

,

w

) =

^rad

(

T

) =

^diam⁽₂^T⁾⁺¹^,ând^by ^Lemma^2.2,êach ôf^the^verticesôn^the^path^from^v to w,excluding v andincluding w,hasadifferentstatus.Thus,k

(

T

) ≥

^d

(

v

,

w

) =

^diam⁽₂^T⁾⁺¹^,^and^since^k

(

T

)

isaninteger,it followsthatk

(

T

) ≥

diam(T)+¹ 2

.Finally,suppose visnotacenterofT andlet wbeavertexatmaximumdistancefromv. Then,d

(

v

,

w

) ≥

^rad

(

T

) +

¹

≥

^diam₂⁽^T⁾

+

^1,ând^by^Lemma^2.2,êachôf^the^verticesôn^the^path^from^v ^to^w,êxcluding^v ând including w,hasadifferentstatus.Thus,k

(

T

) ≥

^d

(

v

,

w

) ≥

^diam₂⁽^T⁾

+

¹

≥

diam(T)+1 2

.Thebound istight,e.g., forstarsand paths.

Now,wearereadytoaddresspolynomialsolvabilityofSRT-R3withaﬁxednumberofdistinctstatusvalues;thesetypes oftreesarehighlysymmetric.

Theorem3.3.Let

σ

beamultisetofn integersandk bethenumberofdistinctvaluesin

σ

.Then,itcanbedeterminedwhetheror not

σ

isthestatussequenceofatreewhichhasradiusatmost3anddegreeofthecentervertexatmost

δ

in2^O⁽^k³^δ³⁾n^O^(δ)time.

Proof. Letr be a center vertex. Let usremind that the center ofa tree is the middle vertexor middle two vertices in everylongestpath.Notice,thecentervertexisnotnecessarilyamedianvertex.Letdeg

(

r

) ≤ δ

.Sincethecentervertexrhas boundeddegree

δ

, wecan enumerateoverall possibilities fortheset I of verticesatdistance1 fromr.There are _n

δ

of suchsets,implyingn^O^(δ) timeforthisenumeration.Fromnowon,weassumethesetI isgiven.

Ina

σ

-realizedtreeT,considertwo vertices,i

∈

Î ândânâdjacent ^toît^(non-root)^vertex^having ^status ^j,^see ^Fig.^3.

Forsimplicityofthenotation werefertothelattervertexas j,thoughitcanbeanyvertexofthatstatusandtherecould be many such vertices.We refer to a subtree of T induced by i

∈

Î, â ^vertexôf^status ^j ândâll ^leaves âdjacent^to ^that vertexas

(

i

,

j

)

-branch.Letai j be thenumberofstatus

verticesinan

(

i

,

j

)

-branch.ByLemma2.3,giventhestatuses of i and j,theparametersai j are uniquelydeﬁnedforall

=

¹

, . . . ,

k.Now,letanintegervariablexi j

≥

^{0 be}^the^number^of

(

i

,

j

)

-branchesina

σ

-realizedtree T withagivenset I.Thefollowingintegerlinearprogram(ILP)solves theproblemof ﬁndinga

σ

-realizedtreeT withagivensetI:

(7)

Fig. 3.Example of a tree with three(i,j)-branches.

i∈^I k

j=¹

a_{i j}x_{i j}

=

ⁿ

∀ =

¹

, . . . ,

k

;

k

=¹ k

j=¹

a_{i j}x_{i j}

=

ⁿ

+

^s

(

r

) −

^s

(

i

) −

¹

2

∀

ⁱ

∈

^I

;

x_{i j}

∈ Z

⁺

∀

ⁱ

∈

^I

,

j

=

¹

, . . . ,

k

,

where n is the multiplicity of element

in

σ

\ ({

^r

} ∪

^I

)

. Here, the ﬁrst equation preserves the given multiplicities of integers andstatuses in

σ

and T,respectively. Thesecond equation guarantees that forall i

∈

^I,^the ^necessity^condition of Lemma2.3forthe edge

(

r

,

i

)

is satisﬁed.Thus, iftheILP fails toﬁnd afeasible solution,the necessityconditions for realizabilityof

σ

inatreewithagivensetIalsofail.Ontheotherhand,givenasolutionxi j

,

i

∈

^I

,

j

=

¹

, . . . ,

k,totheILP, thestraightforwardassignment ofx_{i j} numberof

(

i

,

j

)

-branchesto i

∈

^I ^provides^arealizationof

σ

in T.Thisimpliesthe correctnessoftheILP.

TheILPhasatmostk

δ

variables.Thus,byLenstra [25],theILPcanbesolvedin2^O⁽^k³^δ³⁾time.

Westress thattheassumptions ofTheorem3.3donotrestrictthe hardnessresultfromSection 2.InTheorem2.1,the treealsohasradius3,butthenumberofdistinctstatusvalueskandthemaximumdegree

δ

ofthecentervertexarenot bounded.Thisgivesrisetothefollowingtwoopenquestions.First,isthereanintuitivecombinatorialalgorithm,notrelying onthe “black-box”ILPmachinery, thatsolves SRT-R3withk and

δ

ﬁxed?Second,ifonlyone oftheparametersk or

δ

is ﬁxed(notboth),isthereapolynomialtimealgorithmforsolvingSRT-R3?Below,wepartiallyaddressthissecondquestion byprovingthatfork

=

^{2 and}^for^k

=

^{3 the}^problem^Statusrealizabilityintreescanbesolvedinpolynomialtimewithno additionalrestrictions.Notethatsuchtreesarealsohighlysymmetric.

Deﬁnition3.4.Adoublestarisagraphthatcanbe obtainedbyappendinga

≥

^{1 pendent}^vertices^to^one^vertex^of ^K2 and b

≥

^{1 pendent}^vertices^to^the^other^vertex^of ^K2.Abalanceddoublestar isagraphthatcanbeobtainedbyappendinga

≥

¹ pendent verticesto eachvertexof K₂.Let T ^be ^the ^familyôf^trees ^which^can ^be ôbtained^by âppending ^b

≥

^{1 pendent} verticestoeachleafofastarortoeachleafofabalanceddoublestar.

GivenagraphG,theverticesinasetS

⊂

^V

(

G

)

arecalledsimilar ifforanyu

,

v

∈

^S,^there^exists^anautomorphismofG thatmapsu tov.

Nextwecharacterizetreeshavingsmallnumberk ofdistinct valuesinthestatussequence.Thegeneralcaseofgraphs withstatus sequencehaving a uniquevalue, that is,k

(

G

) =

^1, ^was ^studiedⁱⁿ ^[1]. În^that ^paper^the âuthors ôbtain^tight upperandlowerboundsfortheuniquestatusvalueofthestatussequenceintermsofthenumberofverticesofthegraph.

Fortrees T havingk

(

T

) =

^2,^acharacterizationappearedin[23].However,weaddaproofforcompleteness.

Proposition3.5.LetT beatreeonn vertices.Thenk

(

T

) =

2ifandonlyifT isastarorabalanceddoublestar.

Proof. IfT isastarorabalanceddoublestar,thenalltheleavesofT aresimilarandthereforehavethesamestatus,and allnon-leafverticesarealsosimilarandhavethesamestatus.Moreover,itcaneasilybeveriﬁedthatthestatusvaluesofa leafanditsnon-leafneighboraredifferent.Thus,k

(

T

) =

^2.

If T isa treewithk

(

T

) =

^2,^then^by ^Lemma^3.2,^diam

(

T

) ≤

^3.^If^diam

(

T

) =

^{0 or}^diam

(

T

) =

^1,^then ^T ^is^isomorphic ^to K₁ and K₂, respectively,andin bothcasesk

(

T

) =

^1. ^If^diam

(

T

) =

^2,^then ^T îsâ ^star.Îf^diam

(

T

) =

^3, ^then ^T ^is^a ^double star, i.e., T can be obtainedby starting froma path P4 withvertices v1

,

v2

,

v3

,

v4 (in pathorder) andappending a

≥

⁰ pendentverticesto v2 andb

≥

^{0 pendent}^vertices^to^v3.Then,s

(

v1

) =

¹

+

²

(

a

+

¹

) +

³

(

b

+

¹

)

,s

(

v2

) =

¹

(

a

+

²

) +

²

(

b

+

¹

)

, and1

(

b

+

²

) +

²

(

a

+

¹

) =

^s

(

v3

)

.Ifa

=

^b,^then^noneôf^these^numbersâreêqual,^so^k

(

T

) >

2.Ifa

=

^b,^then ^T ^is^a^balanced doublestar.

Thefollowingresultextendsthecharacterizationin[23] fortreeswithk

(

T

) =

^3.

(8)

Fig. 4.Structure of a tree with 3 distinct status values and diameter 4.

Theorem3.6.LetT beatreeonn vertices.Thenk

(

T

) =

³îfândônlyîf^T

∈

T^.

Proof. IfT

∈

T^,^thenâll^leavesôf^T âre^similar,âll^vertices^whichâre^neighborsôf^leavesâre^similar,ândâll^vertices^which are neither leaves nor neighborsof leavesare similar. Moreover, it can be verified that the status valuesof thesethree similarityclassesofverticesaredifferent.Thus,k

(

T

) =

^3.

LetT beatreewithk

(

T

) =

^3.^By^Lemma^3.2,^diam

(

T

) ≤

^5.^If^diam

(

T

) ≤

^3,^then^T îsêither^K1,orK₂,orastar,oradouble star.In thefirstthreecases, T hasatmosttwo similarityclasses.IfT isa balanceddoublestar,then by Proposition3.5, k

(

T

) =

^2.Îf^T îsâ^double^star^whichîs^not^balanced,^let ^v1 andv2 bethenon-leafverticesofT,let

1 bealeafadjacent tov₁,and

2 bealeafadjacentto v₂.Then,sincev₁ andv₂ haveadifferentnumberofleafneighbors,thestatusvaluesof v1,v2,

1,and

2 arealldistinct,contradictingtheassumptionthatk

(

T

) =

^3.^Thus,^diam

(

T

) =

^{4 or}^diam

(

T

) =

^5.

Supposeﬁrstthatdiam

(

T

) =

^4.^Let^P ^beâ^pathⁱⁿ^T ôf^length^4.^Let^the^verticesôf ^P ^beû1

,

u2

,

r

,

u4

,

u5 inpathorder.

Then, becauseofthe diameterrestriction,u₁ andu₅ are leavesof T,all neighborsofu₂ andu₄ besides r, u₁ andu₅ (if any)areleaves,andallneighborsofrbesidesu2 andu4(ifany)areeitherleavesorverticeswhoseonlyneighborsbesides rareleaves. Lett

≥

^{2 be}^the^numberôf^non-leaf^neighborsôf^r^(including û2 andu4)andletb

≥

^{0 be}^the^number^of^leaf neighborsofr.SeeFig.4foranillustration.

Supposeﬁrstthatthereexisttwonon-leafneighborsofr,sayv₁andv₂,whichhavedifferentnumbersofleafneighbors.

Let a1 be the number ofleaf neighbors of v1 anda2 be the number ofleaf neighbors of v2. Let

1 be one of the leaf neighbors of v1 and

2 be one of the leafneighbors of v2. Let A be the total numberof leaves whichare adjacentto verticesotherthanv1,v2,andr.Then,

s

(

r

) =

¹

(

t

+

^b

) +

²

(

a₁

+

^a2

+

^A

) =

^t

+

^b

+

^2a1

+

^2a2

+

^2A

s

(

v₁

) =

¹

(

a₁

+

¹

) +

²

(

t

+

^b

−

¹

) +

³

(

a₂

+

^A

) =

^2t

+

^2b

+

^a1

+

^3a2

+

^3A

−

¹ s

(

v₂

) =

¹

(

a₂

+

¹

) +

²

(

t

+

^b

−

¹

) +

³

(

a₁

+

^A

) =

^2t

+

^2b

+

^3a1

+

^a2

+

^3A

−

¹ s

(

1

) =

1

+

2

(

a1

) +

3

(

t

+

b

−

1

) +

4

(

a2

+

A

) =

3t

+

3b

+

2a1

+

4a2

+

4A

−

2 s

(

2

) =

1

+

2

(

a2

) +

3

(

t

+

b

−

1

) +

4

(

a1

+

A

) =

3t

+

3b

+

4a1

+

2a2

+

4A

−

2

.

Sincea₁

=

^a2,wehavethats

(

v₁

) =

^s

(

v₂

)

,s

(

1

) =

^s

(

2

)

,s

(

v₁

) =

^s

(

1

)

,s

(

v₂

) =

^s

(

2

)

,s

(

r

) =

^s

(

1

)

,ands

(

r

) =

^s

(

2

)

. Supposea1

=

^t

+

^b

+

^a2

+

^A

−

^1.^Then,^the^status^values^of^r,^v2,

1,and

2arealldistinct,acontradiction.

Nowsupposea1

=

^t

+

^b

+

^a2

+

^A

−

^1.^If^a2

=

^t

+

^b

+

^3a1

+

^A

−

^1,^then^the^status^values^of^v1,

1,

2,andrarealldistinct, acontradiction.Thus,a₂

=

^t

+

^b

+

^3a1

+

^A

−

^1.^Then,^the^status^values^of^v2,

1,

2,andrarealldistinct,acontradiction.

Thus,allnon-leafneighborsofrmusthavethesamenumberofleafneighbors,saya.Supposerhasb

≥

^{1 leaf}^neighbors, andlet wbealeafneighborofr.Letv beanynon-leafneighborofr,andlet

beanyleafneighborofv.Then,

s

() =

1

+

2a

+

3

(

t

+

b

−

1

) +

4

(

at

−

a

) =

3t

+

3b

−

2

+

4at

−

2a s

(

w

) =

1

+

2

(

t

+

b

−

1

) +

3

(

at

) =

2t

+

2b

−

1

+

3at

s

(

v

) =

¹

(

a

+

¹

) +

²

(

t

+

^b

−

¹

) +

³

(

at

−

^a

) =

^2t

+

^2b

−

¹

+

^3at

−

^2a s

(

r

) =

¹

(

t

+

^b

) +

²

(

at

) =

^t

+

^b

+

^2at

.

Sincet

≥

^{2 and}^a

≥

^1,^it^follows^that^s

() >

s

(

w

) >

s

(

v

) >

s

(

r

)

.Thiscontradictstheassumptionthatk

(

T

) =

^3.^If^r^has^no^leaf neighbors,thenT

∈

T ând^the^verticesôf^T ^have^three^similarity^classes:^r,^the^neighborsôf^r,ând^the^leavesôf^T^.Ît^can beeasilyverifiedthatthestatusvaluesofverticesfromthesethreeclassesaredistinct.Thus,ifk

(

T

) =

^{3 and}^diam

(

T

) =

^4, itfollowsthat T

∈

T.Bysimilararguments,itcanbeshownthatifk

(

T

) =

^{3 and}^diam

(

T

) =

^5,^then^T

∈

T.

4. Statussequenceandgraphpartitions

Inthissectionweexplorehowthewell-known conceptsofequitableandorbitpartitionsrelate tothestatussequence ofagraph.