On restrictions of balanced 2-interval graphs

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On restrictions of balanced 2-interval graphs

Philippe Gambette, Stéphane Vialette

To cite this version:

Philippe Gambette, Stéphane Vialette. On restrictions of balanced 2-interval graphs. 33rd

Interna-tional Workshop on Graph-Theoretic Concepts in Computer Science (WG’07), ?, Jun 2007, Dornburg,

Germany. pp.55-65, �10.1007/978-3-540-74839-7_6�. �hal-00141338v2�

hal-00141338, version 2 - 11 Jun 2007

Philipp e Gamb ette

LIAFA,UMR CNRS7089, Université Paris 7, Fran e

DépartementInformatique,ENS Ca han, Fran e

gambetteliafa.jussieu.fr

Stéphane Vialette

LRI,UMR CNRS8623, UniversitéParis-Sud 11, Fran e

vialettelri.fr

Abstra t

The lassof2-intervalgraphshasb eenintro du edformo dellings hedulingandallo ation

problems,andmorere entlyforsp e i bioinformati problems. Someofthoseappli ations

implyrestri tions on the 2-interval graphs, and justify the intro du tion of ahierar hy of

sub lasses of2-intervalgraphs thatgeneralizeline graphs: balan ed2-intervalgraphs,unit

2-intervalgraphs, and (x,x)-intervalgraphs. We provide instan es that show that all the

in lusionsarestri t. WeextendtheNP- ompletenesspro ofofre ognizing2-intervalgraphs

to the re ognition of balan ed 2-intervalgraphs. Finallywe give hintson the omplexity

of unit 2-interval graphs re ognition, by studying relationships with other graph lasses:

prop er ir ular-ar ,quasi-linegraphs,K

1;5

-free graphs,...

Keywords: 2-interval graphs, graph lasses, line graphs, quasi-line graphs, law-free

graphs, ir ularintervalgraphs,prop er ir ular-ar graphs,bioinformati s,s heduling.

1 2-interval graphs and restri tions

Theinterval numb er ofa graph,andthe lasses ofk -intervalgraphshave b eenintro du ed asa

generalizationofthe lassofintervalgraphsbyM Guigan[M G77℄inthe ontextofs heduling

and allo ation problems. Re ently, bioinformati s problems have renewed interest in the lass

of 2-interval graphs (ea h vertex is asso iated to a pair of disjoint intervals and edges denote

interse tion b etween two su h pairs). Indeed, a pair of intervals an mo del two asso iated

tasks in s heduling [BYHN

+

06℄, but alsotwo similar segmentsof DNA in the ontextof DNA

omparison [JMT92℄, or two omplementary segments of RNA for RNA se ondary stru ture

predi tion and omparison[Via04℄.

(a) (b) ( )

Figure 1: Heli es in a RNA se ondary stru ture (a) an b e mo deled as a set of balan ed

2-intervals among all 2-intervals orresp onding to omplementary and inverted pairs of letter

b onds. UnlikeDNA,RNAsareusuallysinglestranded,butmanyRNAmole uleshavese ondary

stru ture in whi h intramole ular lo ops areformed by omplementarybase pairing. RNA

se -ondarystru ture is generallydivided intoheli es ( ontiguous basepairs), and variouskinds of

lo ops (unpaired nu leotides surrounded by heli es). The stru tural stability and fun tion of

non- o ding RNA (n RNA) genes are largely determined by the formationof stable se ondary

stru tures through omplementary bases, and hen e n RNA genes a ross dierent sp e ies are

mostsimilarinthepatternof nu leotide omplementarityratherthanin thegenomi sequen e.

This motivates the use of 2-intervals for mo delling RNA se ondary stru tures: ea h helix of

the stru tureis mo deled by a 2-interval. Moreover, the fa tthat these 2-intervals are usually

required to b e disjoint in the stru ture naturally suggests the use of 2-interval graphs.

Fur-thermore,aimingatb ettermo delling RNAse ondarystru tures,itwassuggestedin [CHLV05℄

to fo us on balan ed 2-interval sets (ea h 2-intervalis omp osed of two equal length intervals)

and their asso iated interse tion graphsreferred as balan ed 2-interval graphs. Indeed, heli es

in RNA se ondary stru tures are mostof the time omp osed of equal length ontiguous base

pairsparts. Totheb est ofourknowledge,nothing is known onthe lassof balan ed 2-interval

graphs.

Sharp er restri tions have also b een intro du ed in s heduling, where it is p ossible to

on-sider tasks whi h all have the same duration,that is 2-intervalwhose intervals have the same

length [BYHN

+

06, Kar05℄. This motivates the study of the lasses of unit 2-interval graphs,

and(x;x)-intervalgraphs. Inthispap er,we onsiderthesesub lasses ofintervalgraphs,andin

parti ularwe addresstheproblemof re ognizingthem.

AgraphG=(V;E)oforder nisa 2-intervalgraphif itis theinterse tiongraphof aset of

nunions of two disjoint intervals onthe realline, thatis ea hvertex orresp ondstoaunion of

twodisjoint intervalsI

k =I k l [I k r

,k2J1;nK(lforleft andrforright),andthereisanedge

b etween I j andI k iI j \I k

6=;. Notethatforthesakeofsimpli ityweusethesameletterto

denote a vertexand its orresp onding 2-interval. A setof 2-intervals orresp onding toa graph

Gis alled a realization ofG. The setof all intervals,

S n k =1 fI k l ;I k r

g,is alled thegroundset of

G(or theground setof fI

1

;:::;I

n g).

The lass of 2-interval graphs is a generalization of interval graphs, and also ontains all

ir ular-ar graphs(interse tion graphsof ar s of a ir le), outerplanargraphs(have a planar

emb edding with all verti es around one of thefa es [KW99℄), ubi graphs (maximum degree

3 [GW80℄),andline graphs(interse tiongraphsofedges ofa graph).

Unfortunately, most lassi al graph ombinatorial problems turn out to b e NP- omplete

for 2-interval graphs: re ognition [WS84℄, maximum indep endent set [BNR96, Via01℄,

ol-oration [Via01℄, ...Surprisingly enough, the omplexity of the maximum lique problem for

2-interval graphs is still op en (although it has b een re ently proven to b e NP- omplete for

3-intervalgraphs[BHLR07℄).

Forpra ti alappli ation,restri ted 2-intervalgraphsareneeded. A 2-intervalgraphis said

to b e balan ed if it has a 2-interval realization in whi h ea h 2-interval is omp osed of two

intervalsofthesamelength[CHLV05℄,unit ifithasa2-intervalrealizationinwhi hallintervals

ofthegroundsethavelength1[BFV04℄,andis alleda(x;x)-intervalgraphifithasa2-interval

realization in whi hall intervals ofthegroundset areop en,haveintegerendp oints,and length

x [BYHN

+

06, Kar05℄. In the following se tions, we will study those restri tions of 2-interval

graphs,and theirp osition inthe hierar hyof graph lasses illustrated in Figure2.

Note that all (x;x)-interval graphsare unit 2-interval graphs, and that all unit 2-interval

graphsarebalan ed 2-intervalgraphs. We analsonoti e that(1;1)-intervalgraphsareexa tly

line graphs: ea h interval of length 1 of the ground set an b e onsidered as the vertex of a

anotherstri tly ontainsit,andthedashedlinesmeanthatthereisnoin lusionrelationshipb

e-tweenthetwo. Dark lasses orresp ondto lassesnotyetpresentintheISGCIDatabase[BLS

+

℄.

olorationproblemisalso NP- ompletefor(2;2)-intervalgraphsandwider lasses ofgraphs. It

isalso knownthatthe omplexityofthemaximumindep endent setproblemis NP- ompleteon

(2;2)-intervalgraphs[BNR96℄. Re ognitionof(1;2)-uniongraphs,arelated lass(restri tionof

multitra k intervalgraphs),wasalso re ently provenNP- omplete [HK06℄.

2 Useful gadgets for 2-interval graphs and restri tions

For proving hardness of re ognizing2-intervalgraphs, West and Shmoys onsidered in [WS84℄

the omplete bipartitegraph K

5;3

asa useful 2-interval gadget. Indeed, all realizationsof this

graphare ontiguous,thatis,foranyrealization,theunionofallintervalsinitsgroundsetisan

interval. Thus, by puttingedgesb etween someverti es of a K

5;3

and anothervertexv,we an

for eone intervalof the2-intervalv (orjustoneextremityofthis interval)tob eblo kedinside

therealization of K

5;3

. It is notdi ult tosee thatK

5;3

has a balan ed 2-intervalrealization,

forexample theone in Figure3.

(a) (b) ( )

Figure 3: The omplete bipartite graph K

5;3

(a,b) has a balan ed 2-interval realization ( ):

verti esofS

5

areasso iatedtobalan ed2-intervalsoflength7,andverti esofS

3

areasso iated

tobalan ed 2-intervals oflength 11. Anyrealization ofthis graphis ontiguous,i.e.,theunion

5;3 l r

toadegree 5vertexinterse t5 disjoint2-intervals,and hen eone ofI

l

orI

r

interse tatleast3

intervals,whi hisimp ossible forunitintervals. Therefore,weintro du ethenewgadgetK

4;4 e

whi his a(2;2)-intervalgraphwithonly ontiguousrealizations.

(a) (b) ( )

Figure 4: The graph K

4;4

e (a),a ni er representation(b),and a 2-interval realization with

op enintervalsof length 2( ).

Prop erty 1. Any2-interval realization of K

4;4

eis ontiguous.

Proof. WriteG=(V;E)thegraphK

4;4

e. Tostudyallp ossible realizationsofG,letus study

all p ossible realizationsofG[V I

8 ℄. As2-intervals I 1 , I 2 , I 3 and I 4

are disjoints, their ground set I

xed = f[l i ;r i ℄; 1 i  8; r i <l i+1

gis a setof eight disjoint intervals. The ground setI

mobile ofI 5 ,I 6 and I 7 isa set of

six disjoint intervals. Let x

i

b e thenumb er ofintervals ofI

mobile interse ting i8intervals of I xed . We havedire tly: x 0 +x 1 +x 2 +x 3 +x 4 +x 5 +x 6 +x 7 +x 8 =jI mobile j=6: (1)

Asthere are12edgesin G[Vnfv

8

g℄ whi hisbipartite, we alsohave:

x 1 +2x 2 +3x 3 +4x 4 +5x 5 +6x 6 +7x 7 +8x 8 12: (2)

Finally, tobuilda realizationofGfromarealization ofG[Vnfv

8

g℄,one mustpla eI

8

soas

to interse t three disjoint intervals of I

xed

. Thus one of the intervals of I

8

interse ts atleast

twointervals℄l

k ;r k [and℄l l ;r l [(k<l )ofI xed

. Sothere isaholeb etweenthose twointervals,

forexample [r

k

;l

k +1

℄, whi h is in luded in one of theintervals of I

8

. So we noti e that I

8 has

to ll one of the seven holes of I

xed

. Thus, the intervals of I

mobile

an not ll morethan six

holes, and the observation that an interval interse ting i onse utive intervals (for i 1) lls

i 1holes, weget: x 2 +2x 3 +3x 4 +4x 5 +5x 6 +6x 7 +7x 8 6: (3)

Equations1,2and3arene essaryforanyvalidrealizationofG[Vnfv

8

g℄whi hgivesavalid

realization ofG.

Let's supp ose by ontradi tion that the union of all intervals of the ground set of G is

not an interval. Then there is a hole, that is an interval in luded in the overing interval of

fI 1 ;:::;I 8 g, whi h interse t no I i

. We pro eed like for equation 3, with the onstraint that

anotherhole annotb e lledbytheintervalsof I

mobile ,sowe getinstead: x 2 +2x 3 +3x 4 +4x 5 +5x 6 +6x 7 +7x 8 5: (4)

Byadding1 and4,andsubtra ting2,wegetx

0

 1 : imp ossible! Sowe haveprovedthat

We show in this se tion that the lass of balan ed 2-interval graphsis stri tly in luded in the

lass of 2-interval graphs, and stri tly ontains ir ular-ar graphs. Moreover, we prove that

re ognizingbalan ed 2-intervalgraphsisashardasre ognizing(general)2-intervalgraphs.

Prop erty 2. The lass ofbalan ed2-intervalgraphs isstri tlyin ludedinthe lassof2-interval

graphs.

Proof. We build a 2-intervalgraph that has no balan ed 2-interval realization. Let's onsider

a hainof gadgetsK

5;3

(intro du edin previous se tion) towhi h we add three verti es I

1 , I 2 , and I 3 asillustratedin Figure5. (a) (b)

Figure 5: An example of unbalan ed 2-interval graph (a) : any realization groups intervals of

thesevenK

5;3

inablo k,andthe hainofsevenblo ks reatessixholes b etweenthem,whi h

makeit imp ossible tobalan e thelengthsofthethree 2-intervalsI

1 ,I 2 ,andI 3 .

Inanyrealization,thepresen e ofholes showedby rossesin theFiguregivesthe following

inequalities for any realization: l (I

l 2 ) < l (I l 1 ), l (I l 3 ) < l (I r 2 ), and l (I r 1 ) < l (I r 3 ) (or if the

realizationof the hainofK

5;3

app earsin thesymmetri alorder: l (I

l 1 )<l (I l 3 ), l (I r 3 )<l (I l 2 ), and l (I r 2 ) < l (I r 1

)). If this realization was balan ed, then we would have l (I

l 1 ) = l (I r 1 ) < l (I r 3 )=l (I l 3 ) <l (I r 2 ) =l (I l 2

) (orfor thesymmetri al ase: l (I

r 1 )=l (I l 1 ) <l (I l 3 )=l (I r 3 )< l (I l 2 ) = l (I r 2

)) : imp ossible! So this graph has no balan ed 2-interval realization although it

hasa2-intervalgeneralization.

Theorem 1. Re ognizingbalan ed 2-intervalgraphs is an NP- ompleteproblem.

Proof. Wejustadaptthepro ofofWestandShmoys[WS84,GW95℄. Theproblemofdetermining

if there is a Hamiltonian y le in a 3-regular triangle-free graph is proven NP- omplete, by

redu tionfrom themoregeneralproblem withouttheno triangle restri tion. So we redu ethe

problemofHamiltonian y leina3-regulartriangle-freegraphtobalan ed2-intervalre ognition.

Let G = (V;E) b e a 3-regular triangle-free graph. We build a graph G

0

whi h has a

2-intervalrealization (asp e ial one,very sp e i , alled H-representationand whi hweproveto

b ebalan ed)i Ghasa Hamiltonian y le.

The onstru tion of G

0

, illustrated in Figure 6(a) is almost identi al to the one by West

and Shmoys, so we just prove that G

0

has a balan ed realization, shown in Figure 6 (b), by

omputing lengthsforea hintervaltoensure it. All K

5;3

Figure6: Thereis abalan ed 2-intervalofG (whi hhasb eendilatedin thedrawingtoremain

readable) i there is an H-representation (that is a realization where the left intervals of all

2-intervals aregroup edtogether ina ontiguousblo k)foritsindu ed subgraphGi thereis a

Hamiltonian y lein G.

in se tion1oftotallength79, inparti ularH

3

. We anthusae tlength83totheintervalsof

v 0

. The intervals of theother v

i

an havelength 3,and their M(v

i

) length 79, sothroughthe

omputationillustrated in Figure6,intervals of z an have length 80+82+2(n 1)+3,that

is 163+2n. We dilate H

1

until a hole b etween two onse utive intervals of its S

3

an ontain

anintervalofz,thatisuntiltheholehaslength165+2n: soafterthis dilating,H

1

haslength

79(165+2n). Finally if G hasa Hamiltonian y le, thenwe have found a balan ed 2-interval

realization ofGof totallength 13;273+241n.

Itisknownthat ir ular-ar graphsare2-intervalgraphs,they arealsobalan ed 2-interval.

Prop erty 3. The lass of ir ular-ar graphs is stri tly in luded in the lass of balan ed

2-intervalgraphs.

Proof. The transformation is simple: if we have a ir ular-ar representation of a graph G =

(V;E),thenwe ho osesomep oint P ofthe ir le. WepartitionV inV

1

[V

2

,whereP interse ts

all the ar s orresp onding to verti es of V

1

and none of the ar s of the verti es of V

2

. Then

we ut the ir le atp oint P to mapit to a line segment: every ar of V

2

b e omes an interval,

and every ar of V

1

b e omes a 2-interval. To obtain a balan ed realization we just ut in half

the intervals of V

2

to obtain two intervals of equal length for ea h. And for ea h 2-interval

[g(I l );d(I l )℄[[g(I r );d(I r )℄ of V 1

,as b oth intervalsare lo ated on one ofthe extremities of the

realization,we anin reasethelength oftheshortestsothatitrea hesthelengthofthelongest

without hanginginterse tionswiththeother intervals. Thein lusionisstri tb e auseK

2;3

isa

balan ed2-intervalgraph(asasubgraphofK

5;3

forexample)butisnota ir ular-ar graph(we

an ndtwoC

4

in K

2;3

,and onlyone an b erealized witha ir ular-ar representation).

4 Unit 2-interval and (x,x)-interval graphs

Prop erty 4. Let x 2 N;x  2. The lass of (x;x)-intervalgraphs is stri tly in luded in the

k(and integerop enb ounds)has arepresentationwhereall intervalshave lengthk+1.

We use the following algorithm. Let S b e initialized asthe set of all intervals of length k ,

and let T b e initially theempty set. As longas S is not empty,let I = [a;b℄ b ethe left-most

intervalof S, remove from S ea h interval [;℄ su h that  < b (in luding I), add [;+1℄

to T, and translate by +1all the remainingintervals in S. When S is empty,the interse tion

graph of T, whereall intervals have length k+1 is the sameasthe interse tion graph forthe

original S.

We also build for ea h x  2 a (x+1;x+1)-interval graph whi h is not a (x;x)-interval

graph. We onsiderthebipartitegraphK

2x

andap erfe tmat hingf(v

i ;v 0 i );i2J1;xKg. We all K 0 x

thegraphobtainedfrom K

2x

withthefollowing transformations,illustrated in Figure7(a):

removeedges(v

i

;v

0 i

)ofthep erfe tmat hing,addfourgraphsK

4;4 e alledX 1 ,X 2 ,X 3 ,X 4 (for ea hX i ,we all v i l andv i r

theverti esofdegree 3),link v

2 r and v 3 l , link allv i tov 1 r and v 4 l ,link all v 0 i to v 2 l andv 3 r

,and nallyadd a vertexa (resp. b)linked toall v

i

,v

0 i

, andtotwoadja ent

verti es of X

1

(resp. X

4

) of degree 4. We illustrate in Figure 7(b) that K

0 x

has a realization

withintervalsof lengthx+1. We anprovebyindu tiononx thatK

0 x

hasno realizationwith

intervalsoflengthx: itisratherte hni al,sowejustgivetheidea. AnyrealizationofK

0 x

for es

the blo k X

2

to share an extremity with the blo k X

3

, so ea h 2-interval v

0 i

has one interval

interse ting the other extremity of X

2

, and the other interse ting the other extremity of X

3 .

Then onstraintson thep ositionof verti esv

i

for e theirintervalstoapp earastwostairways

asshowninFigure7(b). Sov

1 r

must ontainxextremitiesofintervalswhi hhavetob edierent,

soitmusthavelength x+1.

(a)

(b)

Figure7: The graphK

0 4

(a)is (5,5)-intervalbut not(4,4)-interval.

The omplexityofre ognizingunit2-intervalgraphsand(x;x)-intervalgraphsremainsop en,

howeverthefollowing showsarelationship b etween those omplexities.

Lemma 1. funit 2-interval graphsg=

S x2N



arealization withjVj=n2-intervals,thatis 2nintervals ofthegroundset. Sowe onsiderthe

intervalgraphofthegroundset,whi hisaunitintervalgraph. Thereisalineartimealgorithm

basedonbreadth-rstsear hto omputearealization ofsu h agraphwhereintervalendp oints

arerational, withdenominator2n [CKN

+

95℄. So bydilating bya fa tor2n su ha realization,

weobtain arealization ofGwhere intervals ofthegroundset havelength 2n.

Theorem2. Ifre ognizing(x;x)-intervalgraphsispolynomialforanyintegerxthenre ognizing

unit 2-interval graphs is polynomial.

5 Investigating the omplexity of unit 2-interval graphs

Inthis se tionwe showthatallprop er ir ular-ar graphs( ir ular-ar graphssu hthatnoar

is in luded in anotherin therepresentation)areunit 2-intervalgraphs,and we studya lassof

graphswhi hgeneralizesquasi-line graphsand ontainsunit 2-intervalgraphs.

Re all that, a ording to Prop erty 3, ir ular-ar graphs are balan ed 2-interval graphs.

However, ir ular-ar graphsarenot ne essarilyunit2-intervalgraphs.

Prop erty 5. The lass of proper ir ular-ar graphs is stri tly in luded in the lass of unit

2-interval graphs.

Proof. Asin thepro ofof Prop erty3,we ho oseap oint P on the ir leof therepresentationof

aprop er ir ular-ar graphG,andmapsthe ut ir leintoalinesegment. Weextendtheouter

extremitiesofintervalsthathaveb een utsothatnointerval ontainsanother. Thusweobtain

aset of2-intervalsforar s ontainingP, andaset I of intervals forar s not ontainingP. For

ea hintervalofI,weadd anewintervaldisjoint ofanyothertogeta 2-interval. Ifwe onsider

theinterse tion graph ofthe ground setof su ha representation,it isa prop erinterval graph.

So itisalsoa unitintervalgraph[Rob69℄,whi h providesaunit 2-intervalrepresentationof G.

To ompletethepro of,wenoti ethatthedomino(two y lesC

4

havinganedgein ommon)

isa unit2-intervalgraph butnot a ir ular-ar graph.

Quasi-linegraphsarethose graphswhoseverti es arebisimpli ial, i.e., the losed neighb

or-ho o dofea hvertexistheunionoftwo liques. Thisgraph lasshasb eenintro du edasa

gener-alization of linegraphsanda useful sub lassof law-freegraphs[Ben81, FFR97,CS05,KR07℄.

Following theexample of quasi-linegraphsthatgeneralize line graphs,weintro du eherea new

lass of graphs for generalizing unit 2-interval graphs. Let k 2 N



. A graph G = (V;E) is

all-k -simpli ial if the neighb orho o d of ea h vertex v 2 V an b e partitioned into at most k

liques. The lassofquasi-line graphsisthus exa tlythe lassofall-2-simpli ial graphs. Noti e

thatthisdenitionisequivalent tothefollowing: in the omplementgraphofG,forea hvertex

u,the verti esthatarenot intheneighb orho o d ofu arek - olorable.

Prop erty6. The lassofunit2-intervalgraphsisstri tlyin ludedinthe lassofall-4-simpli ial

graphs.

Proof. Thein lusion istrivial. What isleft istoshowthat thein lusion isstri t. Considerthe

following graph whi h is all-4-simpli ial but not unit 2-interval: start with the y le C

4 , all its verti es v i , i 2 J1;4K, add four K 4;4 e gadgets alled X i

, and for ea h i we onne t the

vertex v

i

to two onne ted verti es of degree 4 in X

i

. This graph is ertainly all-4-simpli ial.

But if wetryto build a 2-intervalrealization ofthis graph,thenea hof the 2-intervals v

k has

an interval trapp ed into the blo k X

k

. So ea h 2-interval v

k

has only one interval to realize

the interse tionswith the other v

i

: this is imp ossible as we have torealize a C

4

whi h has no

Proof. The Kneser Graph KG(7;2) is triangle-free, but not 4- olorable [Lov78℄. We onsider

the graph obtained by adding an isolated vertex v and then taking the omplement graph,

i.e., KG(7;2)℄fvg. It is law-free as KG(7;2) is triangle-free. And if it wasall-4-simpli ial,

then the neighb orho o d of v in KG(7;2)℄fvg, that is KG(7;2), would b e a union of at most

four liques, so KG(7;2) would b e 4- olorable: imp ossible so this graph is law-free but not

all-4-simpli ial.

Prop erty 8. The lass of all-k -simpli ial graphs is stri tly in luded in the lass of K

1;k +1 -free graphs.

Proof. If a graph G ontains K

1;k +1

, then it has a vertex with k+1 indep endent neighb ors,

and hen e G is notall-k -simpli ial. The wheel W

2k +1

is a simple example of a graph whi h is

K 1;k +1

-freebut in whi h the enter an nothaveits neighb orho o d(aC

2k +1

) partitionedintok

liques orless.

Unfortunately, all-k -simpli ial graphs do not have a ni e stru ture whi h ould help unit

2-intervalgraphre ognition.

Theorem 3. Re ognizingall-k -simpli ialgraphs isNP- ompletefor k3.

Proof. We redu e from the Graph k - olorability problem, whi h is known to b e

NP- ompletefork3 [Kar72℄. LetG=(V;E)b ea graph,and letG

0

b ethe omplement graphof

Gtowhi h weadda universalvertexv. We laimthatGisk - olorableiG

0

isall-k -simpli ial.

If G is k - olorable, then the non-neighb orho o d of any vertex in G is k - olorable, so the

neighb orho o d of anyvertexin G isa unionof atmostk liques. And theneighb orho o d of v is

also aunionof atmostk liques, soG

0

is all-k -simpli ial.

Conversely,if G

0

is all-k -simpli ial, then in parti ular the neighb orho o d of v is a union of

atmostk liques. Let'spartitionitintokvertex-disjoint liquesX

1

;:::;X

k

. Then, oloringG

su hthattwoverti eshavethesame olori theyarein thesameX

i

leadstoa validk - oloring

ofG.

6 Con lusion

Motivatedbypra ti alappli ationsin s hedulingand omputationalbiology,wefo usedin this

pap eronbalan ed2-intervalgraphsandunit2-intervalsgraphs. Also,weintro du edtwonatural

new lasses: (x;x)-intervalgraphsand all-k -simpli ial graphs.

Wementionheresomedire tionsforfutureworks. First,the omplexityofre ognizingunit

2-interval graphs and (x;x)-interval graphs remains op en. Se ond, the relationships b etween

quasi-line graphs and sub lasses of balan ed 2-intervals graphs still have to b e investigated.

Last,sin emostproblemsremainsNP-hardforbalan ed2-intervalgraphs,thereisthusanatural

interestininvestigatingthe omplexityandapproximationof lassi aloptimizationproblemson

unit2-intervalgraphsand (x;x)-intervalgraphs.

A knowledgments

We are grateful to Vin ent Limouzy in parti ular for bringing to our attention the lass of

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We give thedetailedpro ofsof Theorem1 andProp erty4.

Proof of Theorem1. Let G = (V;E) b e a 3-regulartriangle-free graph. We build a graph G

0

whi hhasa2-intervalrealization(asp e ialone,verysp e i ,andwhi hweprovetob ebalan ed)

iG hasaHamiltonian y le.

First we will detailhow we build G

0

startingfrom the graph G,and adding some verti es,

in parti ular K

5;3

gadgets. The idea is that the edges of G will partition into a Hamiltonian

y le anda p erfe t mat hingi all 2-intervals of therealization of G

0

an have their left

inter-val realizing the Hamiltonian y le, and their right interval realizing the p erfe t mat hing. A

realization withsu ha pla ementof theintervalsis alled an H-representation of G.

Wepro eedasillustratedinFigure6. We ho osesomevertexofGthatwe allv

0

(whi hwill

b e theorigin of the Hamiltonian y le), and the other are alled v

1

;:::;v

n

. For ea h vertex

v i

ofGwe link ittoa vertexof theS

5

of aK

5;3

alled M(v

i

) (whi h willblo k one ofthefour

extremities of the 2-intervalv

i

). We link all verti es toa new vertex z, whi h is linked to no

M(v) ex ept M(v

0

) (thus the intervalof ea h v

i

interse ting M(v

i

), for i6=0, won't interse t

z). We addthreeK 5;3 ,H 1 ,H 2 andH 3

: twoverti esoftheS

5

ofH

1

arelinkedtoz,athird one

islinked toone vertexoftheS

5

ofH

2

,one vertexoftheS

5

ofH

3

islinkedtoz,andall verti es

ofH

3

tov

0 .

To explain this onstru tion in detail, we study the realization of G

0

, if we supp ose it is a

(balan ed)2-intervalgraph,and weprovethatitleads us tonda Hamiltonian y lein G.

As the realization of H

1

and H

2

are two ontiguous blo ks of intervals then one of their

extremities must interse t. Asz is linked totwodisjoint verti es ofH

1

,b oth intervalsof z are

used to realize those interse tions. But one interval of z that we all z

r

, also has to interse t

one vertex of H

3

whi h is notlinked to H

1

, so z

r

interse ts the se ond extremityof theblo k

H 1

(the rstextremityb eing o upied bytheextremity of H

2

). And asz

r

interse tsonly one

intervalofH

3

,itmustb etheextremityofH

3

. Theother intervalof zis ontainedin theblo k

H 1

, thus an't interse tM(v

0

) neither all the verti es v

i

, so all those 2-intervals interse t z

r .

And as none of them interse t H

3

ex eptv

0

,all of them ex ept v

0

have an interval ontained

in z

r

, thatwe all v

i;g

. The other interval of ea hv

i

is linked to aK

5;3

soithas one extremity

o upiedbyK

5;3

,andtheother one isfree.

Conversely, if G has a Hamiltonian y le, then it is p ossible to nd a H-representation,

su h that all the onstraintsindu ed by the edges of G

0

are resp e ted, asillustrated with the

realization in Figure6. We havealready proved thatthis realization an b ebalan ed.

Proof of Property4. In thefollowing, aswe only onsideringthe interval ofv

i l orv i r lo atedat

one extremity of the blo k X

i

, and not the one inside, we will use v

i l and v i r to denote those

extremityintervals. Forea hvertexv

i

,we allv

i;l

itsleftintervalandv

i;r

itsrightinterval. We

do thesameforv

0 i

,and all l (I) theleftextremityofanyintervalI.

We prove byindu tion that the graph K

0 x

is (x+1;x+1)-intervalbut not (x;x)-interval,

and thatforanyunit 2-intervalrealization,there existsanorder  2S

x su h that:  either l (v  (x);l ) < ::: < l (v  (1);l ) < l (v 0  (x);l ) < ::: < l (v 0  (1);l ) and l (v 0  (x);r ) < ::: < l (v 0  (1);r )<l (v  (x);r )<:::<l (v  (1);r ),

 or the symmetri ase: l (v

 (1);l ) < ::: < l (v  (x);l ) < l (v 0  (1);l ) < ::: < l (v 0  (x);l ) and l (v 0  (1);r )<:::<l (v 0  (x);r )<l (v  (1);r )<:::<l (v  (x);r ).

Those two equalities orresp ondin fa ttothe two stairwaysstru ture whi h app ears in

Base ase : we study all p ossible unit 2-interval realizations of K 2

to prove that one of the

exp e ted inequalitiesis alwaystrue. We alsoprovethatK

0 2

hasno(2,2)-intervalrealization.

Firstre allthatrealizationsofX

i

subgraphs anonlyb eblo ksof ontiguousintervals. The

edge b etween v 2 r and v 3 l

for es the two blo ks of X

2

and X

3

to b e ontiguous, with intervals

v 2 l and v 3 r

attheir extremities. Ea h 2-interval v

0 i

must interse tb oth v

2 l and v 3 r , so one of its intervalsinterse tsv 2 l

andtheotherinterse tsv

3 r

. Thus,onesameintervalofv

0 i

annotinterse t

b othaandbwhi haredisjoint,soainterse tsoneintervalofv

0 i

(saytheoneinterse tingv

2 l

,the

other ase b eing treatedsymmetri ally) andb interse tstheother one (so,theone interse ting

v 3 r

). Ea h v

i

has tointerse t b oth a and b, so it has to interse t a with its rst interval and

b withthe se ond. But 2-interval v

i

must also interse tv

1 r

and v

4 l

whi h areb oth disjoint and

disjoint toaandb. Soone intervalofea hv

i

mustinterse tv

1 r

andtheotheronemustinterse t

v 4 l

.

Sowe haveshownthatanyunit2-intervalrealization ofK

0 2

hasthefollowingasp e t(orthe

symmetri ) : the extremity of the blo k X

1

interse ting all v

i

whi h interse t a (or b) whi h

interse tsall v

0 i

,whi h interse ttheextremityX

2

(orX

3

) whi h interse tstheextremityofX

3

(orX

2

),whi h interse ts allv

0 i

,whi h interse tb (ora),whi h interse tsall v

i

,whi h interse t

theextremityofX

4 .

Nowwe supp ose, by ontradi tion,that thereexistsa (2,2)-intervalrealization of K

0 2 . v 1 r is

anintervaloflength 2,but oneof itstwopartsof lengthone hastointerse tanelementof X

1 .

Theother hasto interse tb othv

1 and v 2 . Asneither v 1 norv 2

an interse totherintervals of

X 1

,thentherstintervalofv

1

and v

2

isthesameinterval. Bypro eeding thesamewayonX

4

andv

4 l

,weobtainthatthese ondintervalofv

1

andv

2

isthesameinterval, sov

1

and v

2

should

orresp ond to the same 2-interval: it ontradi tswith the fa t that verti es v

1

and v

2

have a

dierent neighb orho o d. SoK

0 2

hasno(2,2)-intervalrealization.

To obtain the exp e ted inequalities, we have to analyzethe p ossible p ositions of all v

i and v

0 i

. We only treattherst two inequalitiesasthese ond aseis symmetri .

Supp osethat l (v

2;l ) <l (v 1;l ). Asv 1 and v 0 1

arenonadja ent,thenintervalv

1;l isstri tly on theleftofv 0 1;l ,sov 2;l

isstri tlyontheleftofv

0 1;l

. Thusthosetwointervalsdonotinterse t. But

v 2 and v 0 1 are adja ent, so v 2 and v 0 1

musthave interse ting right intervals. But then we have

l (v 0 2;r ) < l (v 0 1;r ) < l (v 2;r ) < l (v 1;r

), and the right intervals of v

0 2

and v

1

an not interse t. We

dedu etheir left intervals interse t,sol (v

2;l )<l (v 1;l )<l (v 0 2;l )<l (v 0 1;l ).

Ifwesupp osethatl (v

1;l

) <l (v

2;l

),wegetaswellthatl (v

0 1;r )<l (v 0 2;r )<l (v 1;r )<l (v 2;r ) and l (v 1;l ) < l (v 2;l ) < l (v 0 1;l ) < l (v 0 2;l

). So for anyunit 2-interval realization of K

0 2

there exists an

order =12or=21su hthat:

 either l (v  (2);l ) < l (v  (1);l ) < l (v 0  (2);l ) < l (v 0  (1);l ) and l (v 0  (2);r ) <l (v 0  (1);r ) <l (v  (2);r ) < l (v  (1);r ),

 orthesymmetri inequalities.

Re ursion: supp osethatforsomex,K

0

x 1

isnot(x 1;x 1)-intervalbutis(x;x)-interval,

and thatany(x;x)-intervalrealizationveriesone of theexp e tedinequalities.

Graph K 0 x 1 is an indu e subgraph of K 0 x =(V;E): K 0 x 1 =K 0 x [V nfv x ;v 0 x g℄. So by the

indu tionhyp othesis,thereexistsanorder 2S

x 1

su hthatforanyunit2-intervalrealization

ofK 0 x :  eitherl (v  (x 1);l )<:::<l (v  (1);l )<l (v 0  (x 1);l )<:::<l (v 0  (1);l ) and l (v 0  (x 1);r )<:::< l (v 0  (1);r )<l (v  (x 1);r )<:::<l (v  (1);r ),

 or the symmetri ase: l (v  (1);l ) < ::: <l (v  (x 1);l ) < l (v  (1);l ) < :::< l (v  (x 1);l ) and l (v 0  (1);r )<:::<l (v 0  (x 1);r )<l (v  (1);r ) <:::<l (v  (x 1);r ). Thep ositionofv x andv 0 x

remains tob edetermined. We treatonlythe ase wheretherst

twoinequalitiesaretrue,asthese ond aseis symmetri .

Asv

x

and v

1 r

areadja ent,and v

0

 (x 1)

and v

1 r

arenot, thenl (v

1 r )<l (v x;l )<l (v 0  (x 1);l ). So

wedenej thefollowingway: v

 (j);l

istheleftmostintervalsu hthatl (v

x;l

)l (v

 (j);l

). if there

isnone, we sayj=0. Then we all

0

2S

x

thep ermutationdened by:

8 < :  0 (i)=(i)if i<j;  0 (j+1)=x;  0 (i)=(i 1)if i>j:

Thenwedire tlygetinequalities:

 l (v 1 r ) < l (v  0 (x);l ) < ::: < l (v  0 (j+1);l )  l (v x;l ) < l (v  0 (j 1);l ) < ::: < l (v  0 (1);l ) < l (v 0  0 (x);l )<:::<l (v 0  0 (j+1);l )<l (v 0  0 (j 1);l )<:::<l (v 0  0 (1);l )  l (v 0  0 (x);r ) < ::: < l (v 0  0 (j+1);r ) < l (v 0  0 (j 1);r ) < ::: < l (v 0  0 (1);r ) < l (v  0 (x);r ) < ::: < l (v  0 (j+1);r )<l (v  0 (j 1);r )<:::<l (v  0 (1);r )

We obtain the exp e ted inequalities by reasoning the same wayas in the end of the base

ase. So in parti ular we have l (v  (x);l ) < :::< l (v  (1);l ) and v 1 r

must interse t all those v

i for

i 2 J1;xK, but also an interval of X

1

whi h interse ts none of the v

i

. So it must have length

x+1,thusK

0 x

isnota(x;x)-intervalgraph

Con lusion: As the base ase and the re ursion has b een proved, exp e ted prop erties of

thegraphK

0 x