Even-hole-free graphs part II: Recognition algorithm

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Conforti, M, Cornuejols, G, Kapoor, A et al. (1 more author) (2002) Even-hole-free graphs

part II: Recognition algorithm. Journal of Graph Theory, 40 (4). 238 - 266 . ISSN

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https://doi.org/10.1002/jgt.10045

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Part II: Reognition Algorithm

Mihele Conforti

Gerard Cornuejols y

Ajai Kapoor

and KristinaVuskovi z

September 1997,revised April 2000, November 2001,February 2002

DipartimentodiMatematiaPuraedAppliata,UniversitadiPadova,ViaBelzoni7,35131Padova,Italy. y

CarnegieMellonUniversity,ShenleyPark,Pittsburgh,PA15213,USA. z

ShoolofComputerStudies,UniversityofLeeds,LeedsLS29JT,UnitedKingdom.

We presentan algorithm that determines in polytime whether a graph ontainsan evenhole. Thealgorithmisbasedonadeomposition theoremforeven-hole-freegraphs obtainedin PartIofthis paper. Wealsogiveapolytimealgorithmto ndanevenhole inagraphwhenoneexists.

1 Introdution

In a graph, a yle is even if it ontains an even number of nodes, and odd otherwise. A

hole is a hordless yle with at least four nodes. A graph that ontains no even hole is

alledeven-hole-free. (GraphGontainsgraphH meansthatH appearsinGasan indued

subgraph. Graph GisH-freemeansthatG doesnot ontaingraph H.)

In thispart, we present a polytimereognition algorithm foreven-hole-free graphs. The

algorithm buildson a strutural theorem proved in [4℄. The algorithm is notpratialsine

thedegreeofthepolynomialishigh: ourmainontributionisinshowingthatthisreognition

problemisintheomplexitylassP.Previously,itwasnotevenknownwhetherthisproblem

was in NP (it is trivially in o-NP, however). It was known (Bienstok [1℄) that it is

NP-omplete to reognize whether a graph ontains an even hole passing through a speied

node. On the positive side, Porto [10 ℄ solved the even hole reognition problem in linear

time for planar graphs and Markossian, Gasparian and Reed [9℄ solved it in polytime for

diamond-and-ap-freegraphs. Adiamondisayleoflengthfourwithasinglehord. A ap

isayleoflengthgreaterthanfourwithasinglehordthatformsatrianglewithtwoedges

of the yle. In [5 ℄ we extended this lastresult to ap-free graphs. Here we give a solution

forall graphs.

Finding an Even Hole

Notethatourreognitionalgorithmforeven-hole-freegraphsan beusedto ndaneven

hole in graph G, if one exists: Let v 1

;:::;v n

denote the nodes of G and let H = G. In

iteration i, test whether H nv i

ontains an even hole. If the answer is yes, set H = Hnv i and otherwisekeepH unhanged. Performniterations. At termination, thegraph H isthe

desired even hole.

With 2 alls to thereognition algorithm, we an also hek inpolytime whether, given

agraphGand anodev ofG,all theevenholesofGontain v. Byontrast, asstatedabove

[1 ℄, given a graph G and a node v of G, it isNP-omplete to hek whether there exists an

even hole thatontains v.

Cutsets

The deompositiontheoremof [4 ℄ whihwe useherehastwotypesof utsets. We dene

these now.

ForS V(G), wedenote byGnS thesubgraphobtainedfromthegraphGbyremoving

the nodes of S and all theedges with at leastone node inS. The node set S isa utset of

thegraphGifthegraphGnS ontainsmoreonnetedomponentsthanG. ForS V(G),

N(S) denotesthesetofnodesinV(G)nS withatleastone neighborinS andN[S℄denotes

N(S)[S. Node set S is a k-star if S is omprised of a lique C of size k and nodes with

a doublestarand to a3-star asatriple star. IfS isomprisedof aliqueC andall nodesof

Gwith at leastone neighborinC,itis alleda full k-star.

A graph G has a 2-join V 1

jV 2

, with speial sets (A 1

;A 2

;B 1

;B 2

), if its nodes an be

partitionedintosetsV 1

andV 2

insuhawaythat,fori=1;2,V i

ontainsdisjoint,nonempty

nodesets A i

andB i

,suh thatevery nodeof A 1

is adjaent to every nodeof A 2

,every node

ofB 1

isadjaent toeverynodeof B 2

,andthere arenootheradjaeniesbetweenV 1

and V 2

.

Furthermore jV i

j>2 fori=1;2, and ifA i

and B i

arebothof ardinality1,then thegraph

induedby V i

isnota hordlesspath.

Star utsets were introdued byChvatal [2 ℄ and 2-joinsby Cornuejolsand Cunningham

[8 ℄. In [6 ℄ and [3℄, 2-joins, star and double star utsets were used to onstrut reognition

algorithmsforbalaned0;1 matries andbalaned0;1matries. Reently,they wereused

to deomposeBerge graphs[7 ℄.

Base Classes

The deomposition theoremof [4℄ shows that every even-hole-free graph exeptthose in

two base lasses ontains a 2-join or a k-star utset. These two base lasses are the

ap-free graphs and basi graphs. Cap-free graphs have been dened already. In [5 ℄, polytime

algorithms are given for reognizing ap-free graphs and for reognizing even-hole-free

ap-freegraphs. Theseond baselassof graphsusedinthedeompositiontheorem of[4 ℄ isthe

lass of basi graphs. We do not dene basi graphs here. We just note that every basi

graph is obtained from the linegraph of a tree by adding two adjaent nodes x and y, and

asaonsequene we an hek inpolytime whether agraphis basi. Sine thereis aunique

hordlesspathbetweenanytwo nodesinthelinegraphof atree,it alsofollowsthat we an

hekinpolytimewhether a basigraphis even-hole-free.

Deomposition Theorem

The following theorem follows from the main result proved in [4 ℄. (In [4 ℄, the result is

proved forodd-signablegraphs, alass ofgraphsthat ontainseven-hole-free graphs.)

Theorem 1.1 A onneted even-hole-free graph is ap-free or basi or ontains a 2-join or

a k-star utset, k=1;2;3.

Idea of the Algorithm

The abovedeompositiontheoremisthebasisofourreognitionalgorithmfor

even-hole-free graphs. Whenever a 2-joinora k-star utset is present in a graph G, we deompose G

into two ormore smaller orsimpler graphs,alled bloks. When Gontainsa k-star utset,

thisisdoneasfollows.

Denition 1.2 LetSbeanodeutsetinagraphGandC 1

;:::;C n

theonnetedomponents

of GnS. We denethe bloksof the deomposition tobegraphs G 1

;:::;G n

,whereG i

is the

subgraph of G indued by V(C i

)[S.

1 2 1 2 1 2 2 2 are in dierent onneted omponents of G(V

2

), dene blok G 1

to be the subgraph of G

indued by nodeset V 1

[fa 2

;b 2

g, wherea 2

2A 2

and b 2

2B 2

. If G(V 2

) ontains a path from

A 2

to B 2

, let Q bea shortest suh path and dene blokG 1

to bethe subgraph of G indued

by node setV 1

plusa markerpathP 2

=a 2

;:::;b 2

that ishordless andsatises thefollowing

properties. Nodea 2

is adjaent to all the nodes in A 1

, node b 2

isadjaent to all the nodes in

B 1

and these are the only adjaeniesbetween P 2

and V 1

. Furthermore, the marker path P 2 has length 4 if Q has even length, and length 5 otherwise. Blok G

2

isdened similarly. See

Figure1.

P1

V1

V2

A1

B2

A1

V1

V2

B2

G

G1

G2

a2

a1

b2

b1

B1

A2

B1

A2

P2

Figure 1: 2-JoinDeomposition

Ifweweretofollowthestandardparadigmforreatinganalgorithmfromadeomposition

theorem,wewouldnowshow that

() whenthedeompositionisappliedreursivelyto thebloks,thetotal numberofbloks

reatedis polynomial.

Unfortunately, although (a) is true for the two utsets of Theorem 1.1, neither (b) nor

()holds.

The problemwith()isthat, ifwedonottake areofdominatednodesproperly,wean

getanexponentialnumberofbloksevendeomposingjustwithstarutsets. (We saythatu

isdominated byvifuisadjaenttov andN(u)N[v℄.) Anotherproblemisthatwedonot

know how toboundthenumberof bloksifwe mixk-starutsetand 2-join deompositions.

Our solutionto ()is to do k-starutsets rst,then2-joins, and to dealwith dominated

nodesspeially.

In Setion5,we disussthe2-joindeompositionof agraphG thathasnok-star utset,

k = 1;2;3. We show that G is even-hole-free if and only if the two bloks G 1

and G 2

of

thedeompositionareeven-hole-free. Furthermore,weshowthatthebloksG 1

andG 2

have

no k-star utsets, k = 1;2;3. Finally, if the 2-join deompositionis applied reursively, we

show that only a linear number of bloks is reated overall. By Theorem 1.1, G is

even-hole-free if and only if all these bloks belong to a base lass and are even-hole-free. This

yieldsa polytime algorithm forheking whether agraph withoutk-starutsets,k =1;2;3,

iseven-hole-free.

A major diÆultythat needs to be addressed when deomposingby a star, double star

ortriplestarutsetis thefatthat(b)above doesnothold. Consider,forexample, agraph

G onsistingof an even holeH and a node x with exatlytwo nonadjaent neighborsin H,

sayu;v,where bothpaths of H from u to v have an oddnumberof edges. Ifwe deompose

G by the star utset N[x℄ onsisting of x and its two neighbors u;v, the two bloks of the

deomposition are even-hole-free, whereas G ontains the even hole H. Thus star utset

deompositionisnoteven-hole-free preserving.

To address thisdiÆulty,we rst applya ertain leaning proedure to theinputgraph

G. This proedure transforms G into a polynomial family of indued subgraphs of G with

thepropertythat, ifGontainsan even hole,then at leastone graphinthefamilyontains

anevenholethatwilleithernotbebrokenbyk-starutsetdeompositionorwillbedeteted

whileperforming thedeomposition.

Clean Graphs

Denition 1.4 Let H be an even hole and u2V(G)nV(H). We say that u isgood w.r.t.

H if ithas atmostthreeneighborsin H andthe graph indued by N(u)\V(H) isonneted.

Otherwise, u isalled bad.

Denition 1.5 An evenhole H of G islean if there is no badnode w.r.t. H.

Denition 1.6 Let u be a good node w.r.t. an even hole H. We say that u is of Type gi

w.r.t. H ifjN(u)\V(H)j=i.

an edge uv suhthat nodeuisa Typeg1nodew.r.t. H, node visa Typeg2nodew.r.t.

H, the neighbor x of u in H is distint from the neighbors v 1

;v 2

of v in H and x;v 1 havea ommon neighbor y 6=v

2

in H (speialtent).

Denition 1.8 Let H bean evenhole and u a Typeg3 node w.r.t. H, withneighbors u 1

;u 2 andu

3

inH suhthatu 1

u 2

andu 2

u 3

areedges. LetH 0

betheholeinduedby(V(H)nfu 2

g)[

fug. We say that H 0

is obtained fromH through a Typeg3 node substitution.

Consider a speial tentuv w.r.t. an evenhole H. Let H 0

bethe hole indued by the node

set(V(H)[fu;vg)nfy;v 1

g. Wesaythatsuh ahole H 0

isobtainedfromH throughaspeial

tent substitution.

A tent substitution is either a Type g3 node substitution or a speial tent substitution.

Note that holes H andH 0

areof the same length.

y

u v

x v

1 v

2

Figure 2: SpeialTent

Denition 1.9 Let G be a graph ontaining an even hole H. We dene C G

(H) to be the

family of all holes of Gobtained from H through a sequeneof tent substitutions.

Denition 1.10 An even hole H

of G is spotless if all the holes in C G

(H

) are lean.

Denition 1.11 A graph G is lean if it is either even-hole-free or it ontains a spotless

smallest even hole H

.

Given agraphG, Setion4presentsaleaningproedurewiththefollowingproperty: it

onstrutsinpolytimealeangraphG 0

thatiseven-hole-freeifandonlyifGiseven-hole-free.

The graph G 0

onsists of a polynomial number of indued subgraphs of G, at least one of

whihislean. Thedeompositionoflean graphsbyk-starutsetsispresentedinSetion3.

The mainresult of that setion isthat a lean graph G an be deomposed reursivelyinto

a family of bloks that have no k-star utsets and satisfy the following property: (i) either

Gis identiedas ontaining an even hole duringthedeompositionproess or(ii)when the

deompositionproess isompleted, all bloks inthefamilyare even-hole-free graphsifand

TheotherdiÆultywithk-starutsetsisthat()doesnothold. Asmentionedearlier,our

approahto () isto remove dominatednodes. We prove inSetion3 thatthetotal number

of bloks generated by reursive deompositionwith k-star utsets ispolynomialifone rst

removes dominated nodes and uses full k-star utsets. For this reason, in our reognition

algorithm,wewillatuallyuse thefollowingrenementof Theorem1.1.

A gem is a graph on ve nodes, suh that four of the nodes indue a hordless path of

lengththree and thefthnodeis adjaent to all of thenodesofthispath.

Theorem 1.12 Let G be a onneted even-hole-free graph. If G ontains no gem or

domi-natednode,then Gisap-freeorbasior ontains a2-joinor afull k-starutset, k =1;2;3.

Proof: Followsfrom Theorem1.1 andthe nexttwolemmas. 2

Lemma 1.13 AssumeGontainsnogemandno4-hole. LetC bealiqueandu2V(G)nC.

If N[u℄N[C℄,then u isdominated by some node in C.

Proof: Suppose N[u℄ N[C℄, but no node of C dominates u. Let K C be a minimal

set suh that N[u℄ N[K℄, i.e. for eah v 2 K, N[u℄ 6N[K nfvg℄. Sine u 2 N[K℄, u is

adjaent to a node of K, say x. Sine u is not dominatedby x there exists v 2N(u) suh

that v is not adjaent to x. Sine v 2 N[K℄, v is adjaent to some node of K nfxg, say

y. Sine x;y;v;u is nota 4-hole, u is adjaent to y. Sine N[u℄ 6N[K nx℄, there exists a

node w adjaent to u and x butnot y. Now eitherw;x;y;v;u indues a gem orw;x;y;v is

a 4-hole. 2

Lemma 1.14 Assume Gontains nodominated nodes,no gemandno 4-hole. IfGontains

a k-star utset, k=1;2;3, then G ontains a full k-star utset.

Proof: Let C be the lique enter of a k-star utset S of G, where k = 1;2;3. Suppose

S 0

=C[N(C)isnotautsetofG. Thensome omponentofGnS,sayC 1

,mustbeentirely

ontained inS 0

nS. Then u 2 C 1

satises the onditions of Lemma 1.13 and therefore u is

dominatedby anode inC,ontraditingthe assumption. 2

Dominated nodes an be identied in polytime and we will show in Setion 3 that, in

lean graphs, their removal is even-hole-preserving. In Setion 3, we also show that, when

G has a gem, there is a rather simpledeomposition result. So Theorem 1.12 provides the

basis forourreognition algorithm of even-hole-free graphs. The outline of thealgorithm is

as follows: hek for 4-holes and a few other graphs that ontain even holes and that an

be identied in polytime (to simplify the analysis, later), then lean G, remove dominated

nodes,deomposebyfullk-starutsets,k =1;2;3, thenby2-joins, andnallyhekthatall

thebloks areeither basiorap-free,and ontain no even holes.

2 The Algorithm

A wheel (H;x) is a graph induedby a hole H and a node x 2= V(H) havingat least three

neighborsinH,sayx 1

;:::;x n

. AsubpathofH onnetingx i

andx j

l

isa setoroflengthat least2. A wheelisevenifitontainsan even numberofsetors. Itis

easy to seethat an even wheelalways ontainsan even hole.

A 3PC(x;y) is a graph indued by three hordless paths from node x to y, having no

ommon or adjaent intermediate nodes. Note that x and y are not adjaent. It is easy to

seethat a 3PC(x;y) always ontainsan even hole.

A3PC(x 1

x 2

x 3

;y 1

y 2

y 3

)isagraphinduedbythreehordlesspaths,P 1

=x 1

;:::;y 1

,P 2

=

x 2

;:::;y 2

and P 3

=x 3

;:::;y 3

,havingno ommonnodes and suh thattheonlyadjaenies

betweennodesof distintpathsare theedgesof thetwo liquesof sizethree induedbythe

disjoint node sets fx 1

;x 2

;x 3

g and fy 1

;y 2

;y 3

g. It is easy to see that a 3PC(x 1

x 2

x 3

;y 1

y 2

y 3

)

alwaysontainsan even hole.

A 3PC(x 1

x 2

x 3

;y) is a graph indued by three hordless paths P 1

= x 1

;:::;y, P 2

=

x 2

;:::;y and P 3

=x 3

;:::;y,havingno ommon nodesother than y and suhthat the only

adjaenies between nodes of P i

ny and P j

ny, for i;j 2 f1;2;3g distint,are the edges of

theliqueof sizethreeinduedbyfx 1

;x 2

;x 3

g.

WesaythatagraphGontainsa3PC(:;:)ifitontainsa3PC(x;y)forsomepairofnodes

x;y2V(G). We say thata graphG ontainsa 3PC(;)ifforsome x 1

;x 2

;x 3

;y 1

;y 2

;y 3

2

V(G)there existsa 3PC(x 1

x 2

x 3

;y 1

y 2

y 3

). Similarlywesay thatitontainsa 3PC(;:)ifit

ontainsa 3PC(x 1

x 2

x 3

;y) forsome x 1

;x 2

;x 3

;y2V(G).

As mentionedabove, an even-hole-free graph annotontains an even wheel,a 3PC(:;:)

nor a 3PC(;). Our reognition algorithm for even-hole-free graphs starts by heking

whetherthegraphontainsoneofthetwofollowingstrutures(thisanbedoneinpolynomial

time).

Denition 2.1 A wheel (H;x) is a short 4-wheel if it ontains four setors and one of the

following holds: thewheelhas threeshort setors, orit hastwo nonadjaent short setorsand

a setor of length three.

Denition 2.2 A 3PC(:;:) is short if one path has length 2 and one has length 3. A

3PC(;) is short if one path has length one and one has length two. A short 3PC is

either a short 3PC(:;:) or a short 3PC(;).

RECOGNITION ALGORITHM FOR EVEN-HOLE-FREE GRAPHS

Input: A graphG.

Output: YESif Giseven-hole-free,and NOotherwise.

Step 1: IfG ontainsa4-hole, a 6-hole, ashort4-wheelora short3PC,outputNO.

Step 2: ApplytheCleaningAlgorithm ofSetion4to Gandlet L 1

betheoutputfamilyof

graphs(so,ifGhasanevenhole,thensomegraphinL 1

hasanevenholeandislean).

Step 3: Start with L 2

= ;. For eah L 2 L 1

, perform the Node Cutset Deomposition

AlgorithmofSetion3. IfthealgorithmidentiesLasnotbeingeven-hole-free,output

NO. Otherwise, union the output with L 2

(so the graphs in L 2

have no full k-star

3 2 of Setion5 andunion theoutput withL

3

(sothegraphsinL 3

have no 2-join).

Step 5: Start with L 4

= L 5

= ;. For eah L 2 L 3

, hek whether L ontains a ap. If it

does,add L to L 4

. Otherwise,add Lto L 5

.

Step 6: For eah L 2 L 4

, hek whether L is a basi graph. If some L 2 L 4

is not basi,

output NO. Otherwise, for eah L 2 L 4

, hek whether L ontains an even hole. If

some L2L 4

ontainsan even hole,output NO.Otherwise,goto Step7.

Step 7: For eah L2L 5

,hek whether Lontains an even hole. Ifsome L2L 5

ontains

an even hole,output NO.Otherwise,outputYES.

The CleaningAlgorithm, theNode CutsetDeompositionAlgorithmand the2-Join

De-ompositionAlgorithm will be shown to be polynomial in the next three setions. Steps 6

and 7 hek ap-free and basigraphs. This an be performed in polytime, aspointedout

already. So,theabove reognitionalgorithman beimplementedto runinpolynomialtime.

In thenext three setions,wewillshowthatthefollowingstatements areequivalent.

(i) Gis even-hole-free,

(ii) allthegraphsinL 1

are even-hole-free,

(iii)all thegraphsinL 2

areeven-hole-free,

(iv) all thegraphsinL 3

are even-hole-free.

We will also show that the graphs in L 3

do not ontain a 4-hole, a dominated node, a

gem, afullk-starutset,k =1;2;3;nora2-join. So,byTheorem1.12, ifGiseven-hole-free,

all the graphsin L 3

must be either ap-free and even-hole-free, or basi and even-hole-free.

The algorithm heks this in Steps 6 and 7. This establishes the validity of the algorithm

(subjetto beingable to performSteps2, 3and 4 aslaimed).

3 k-Star Cutsets in Clean Graphs

Throughout this setion, unless otherwise stated, we assume that G is a lean graph with

spotless smallesteven hole H

. In addition, we assume that Gontains no 4-hole, no short

4-wheeland no short3PC.

Lemma 3.1 If node u isdominated by node v, then Gnfug ontains a hole in C G

(H

).

Proof: Assume that H

ontains u. Let u 1

and u 2

be theneighbors of u inH

. Sine u is

dominatedbyv,v is adjaent to u, u 1

and u 2

. Sine H

is lean,v is of Type g3 w.r.t. H

,

andhenetheholeinduedbythenode set(V(H

)nfug)[fvg isinC G

(H

)and inGnfug.

2

Beforeprovingthemainresultsof thissetion, letusprovethefollowingusefullemma.

Lemma 3.2 Suppose C is a lique and C S N[C℄ is a utset breaking all the holes of

C G

(H

). Then,for eah H 2C G

(H

Proof: Suppose H 2 C G

(H ) is hosen suh that the set P = V(H)\C has maximum

ardinality. AsH isbroken byS, thereexistsa node x2V(H)\S thathasno neighborin

P. Let wbea neighborof xinC. Now, ifP 6=;,thenw mustbea Typeg3 nodew.r.t. H.

AftersubstitutingwintoH,we wouldgetahole inC G

(H

)havingmorenodesfromC than

H,a ontradition. 2

This lemma,together withthedenitionofC G

(H

), impliesthefollowing.

Corollary 3.3 SupposeC isa liqueandC S N[C℄isa utset breakingallthe holes of

C G

(H

). Then,for eah H 2C G

(H

), the tents w.r.t. H are disjoint fromC.

Inthedeompositionalgorithm,we treatthedeompositionofgems inaspeialway. Let

usonsiderthisaserst.

Lemma 3.4 Let G be an even-hole-free graph and fx;y 0

;y;z;z 0

g a node set that indues a

gem, suh that y 0

;y;z;z 0

isa hordless path. ThenS =(N(x)[N(y)[N(z))nfy 0

;z 0

g is a

triple star utset breaking y 0

from z 0

.

Proof: Supposenot. Then,inGnS,letP beahordlesspathonnetingy 0

toz 0

. Thenodes

of P together with y and z induea hole H. Node x hasfour neighbors on H, so (H;x) is

an even wheel. 2

Remark 3.5 Ifatriple starutsetS fromLemma3.4issuh thatthe onneted omponents

ofGnS that ontain y 0

andz 0

respetivelyarebothof sizegreaterthan1, then N(x)[N(y)[

N(z) isa fulltriple star utset.

Lemma 3.6 Let fx;y 0

;y;z;z 0

g be a node set that indues a gem, suh that y 0

;y;z;z 0

is a

hordless path. Let S =N(x)[N(y)[N(z)nfy 0

;z 0

g and C 1

(resp. C 2

) be the onneted

omponent of GnS that ontains y 0

(resp. z 0

). If jC 1

j=1 (resp. jC 2

j=1), then Gnfy 0

g

(resp. Gnfz 0

g) ontains a hole in C G

(H

).

Proof: Suppose that jC 1

j = 1. If H

does not ontain y 0

then we are done, so suppose it

does. Let H

=y 0

;h 1

;:::;h n

;y 0

. Then sineN(y 0

)S,h 1

;h n

2S.

Case 1: h 1

orh n

is infx;yg.

W.l.o.g. assume that h 1

2 fx;yg. Assume h 1

= x. Sine H

is a hole, h n

does not

oinide with y and it annot be a neighborof x. Sine h n

2S, it must be a neighborof y

orz. Ifh n

isa neighborof ztheny 0

;x;z;h n

;y 0

isa4-hole. Heneh n

isaneighborof y. But

theny isofTypeg3 w.r.t. H

and sotheholeinduedbythenodeset(V(H

)nfy 0

g)[fyg

isinC G

(H

) and inGnfy 0

g.

When h 1

=y,thesame argument holdsbyinterhangingtherolesof x and y.

Case 2: h 1

;h n

2Snfx;y;zg

Assume rst that one ofthe nodesx ory,isadjaent to bothnodesh 1

and h n

. Assume

w.l.o.g. that x is adjaent to both h 1

and h n

. Then x isof Type g3 w.r.t. H

and thehole

induedby thenodeset (V(H

)nfy 0

g)[fxgisinC G

(H

) and inGnfy 0

g.

If x is adjaent to h 1

but not to h n

, and y is adjaent to h n

but not to h 1

, then sine

the node set V(H

V(H )nfy 0 ;h 1 ;h n

g. W.l.o.g. assume that x has a neighbor in V(H )nfy 0 ;h 1 ;h n g. Sine H

is lean, x is adjaent to h 2

. Sine the hole indued by (V(H

)nfh 1

g)[fxg is lean,

nodes h 3

;:::;h n 1

are not adjaent to y or x. But then the hole indued by the node set

(V(H

)nfy 0

;h 1

g)[fx;yg isinC G

(H

) andin Gnfy 0

g.

So we may assume that one of h 1

or h n

is adjaent to z. Assume w.l.o.g. that h n

is

adjaent to z. Then y isadjaent to h n

, sineotherwise y 0

;y;z;h n

;y 0

is a 4-hole. Alsox is

adjaent to h n

, sine otherwisey 0

;x;z;h n

;y 0

is a 4-hole. Node z annot be adjaent to h 1

,

sine H

is lean and of length greater than 4. Hene h 1

is adjaent to either x or y. But

thenone of x ory isadjaent to bothh 1

and h n

,whihisnotpossible. 2

The above result is all we need when G ontainsa gem. So, forthe next result, we will

assume thatGontainsno gem.

Denition 3.7 A 3PC(x;y), with paths P 1

, P 2

and P 3

, is deomposition detetable w.r.t.

the node utset S if one of the following holds:

(i) P 1

isof length 2 or 3, V(P 1

)S and the intermediate nodes of P 2

andP 3

are in two

dierent omponents of GnS.

(ii) P 1

isof length 3, V(P 1

) S and there are three distint omponents of GnS, C 1

, C 2 and C

3

, suh that for some z2 Snfx;yg, the intermediate nodes of P 2

are ontained

in V(C 1

)[V(C 2

)[fzg and the intermediate nodesof P 3

are ontained in V(C 3 ). A 3PC(x 1 x 2 x 3 ;y 1 y 2 y 3

), with the three paths P 1

, P 2

and P 3

, isdeompositiondetetable

w.r.t. the node utset S if fx 1 ;x 2 ;x 3 ;y 1 ;y 2 ;y 3

g S, P 1

is an edge and the intermediate

nodesof P 2

and P 3

are ontained in two dierent omponents of GnS.

A deomposition detetable 3PC is either a deomposition detetable 3PC(:;:) or a

de-omposition detetable 3PC(;).

Inorderto showthatweendupwithapolynomialnumberofpieeswhen wedeompose

a graph using our node utsets, we need to rene the bloks. Let S be a k-star utset,

k = 1;2;3, with lique enter C. Let C 1

;:::;C n

be the onneted omponents of GnS

and G 1

;:::;G n

thebloks of thedeomposition. We dene therened bloks G 0 1

;:::;G 0 n

as

follows: for i=1;:::;n, remove from G i

all nodes of SnC that do not have a neighborin

C i

.

Theorem 3.8 Suppose that G ontains no 4-hole, no short 3PC, no gem and that G is a

lean graph with spotless smallest even hole H

. When deomposing G with a full k-star

utset S =N[C℄, k=1;2;3; then either somehole in C G

(H

) isentirely ontained in one of

the rened bloksof the deomposition or there exists a deomposition detetable 3PC w.r.t.

S.

Proof: Consider thefollowingtwo ases.

Case 1: Alltheholesof C G

(H

) arebroken byS.

Then,byLemma3.2,foreahH 2C G

(H

),V(H)\C=;. Furthermore,byCorollary3.3,

no nodeof C isof Type g3. Let C=fv 1

;:::;v k

g,wherek =jCj. Denote byP 1

;:::;P m

Hislean,eahnodeofCisadjaenttoatmostonepathP 1

;:::;P m

. Hene2mk 3.

Case 1.1: m=k=3.

Then wemay assumethat V(P i

)=N(v i

)\V(H), i=1;2;3.

IfallthenodesofC areofTypeg2w.r.t. H,letu i

andw i

betheneighborsofv i

inH and

assume w.l.o.g. thatthe nodesu 1 ;w 1 ;u 2 ;w 2 ;u 3 ;w 3

appear inthisorder whentraversing H.

Let Q 1

be thew 1

u 2

-subpath of H that doesnotontain u 1 ;w 2 ;u 3 ;w 3

. LetQ 2

(respetively

Q 3

) be the w 2

u 3

-subpath (respetivelyw 3

u 1

-subpath) of H that does not ontain nodes of

Q 1

. Sine H is an even hole, at least one of the three paths Q i

is of odd length, say Q 1

.

But then the hole indued by V(Q 1

)[fv 1

;v 2

g is an even hole of length smaller than H,

ontraditing ourhoie of H.

If all the nodesof C areof Type g1 w.r.t. H,let u i

be the neighborof v i

in H. Let Q 1 be theu

1 u

2

-subpath of H that doesnot ontain u 3

. Dene Q 2

and Q 3

ina similarfashion.

SineH isbrokenbyS,someonnetedomponentofGnS ontainstheintermediatenodes

of one of these paths, say Q 1

, but not of the other two paths. So we get a deomposition

detetable3PC(u 1

;u 2

)satisfying(i) or(ii) ofDenition3.7.

IfC hasbothTypeg1 andTypeg2nodesw.r.t. H,assumew.l.o.g. thatv 1

isofTypeg1

and v 2

is of Type g2. Sine H is a smallest even hole, v 1

v 2

is a speial tent w.r.t. H.

Now a tent substitution would produe a smallest even hole in C G

(H

) that intersets C,

ontraditing Corollary3.3.

Case 1.2: m=2.

First,supposethatk =3. AssumethatN[v 1

℄\V(H)=V(P 1

)and N[fv 2

;v 3

g℄\V(H)=

V(P 2

),where jN[v 2

℄\V(H)jjN[v 3

℄\V(H)j. Ifv 2

andv 3

bothhave a neighborinH but

do nothave a ommonneighborinH,then Gontains a4-hole. Hene, sinev 2

and v 3

are

of Type g1 or g2 or v 3

does not have a neighbor in H, jV(P 2

)j 3. If jV(P 2

)j = 3, then

G(P 2 [fv 2 ;v 3

g) isa gem. Itfollows thatV(P 2

)=N[v 2

℄\V(H).

Now, if v 1

and v 2

are of the same type, we get a deomposition detetable 3PC(;)

or 3PC(:;:). If one is of Type g1 and theother of Type g2, v 1

v 2

is a speial tent. But this

ontraditsCorollary 3.3.

If k=2,thearguments fromthe previousparagraphhold.

Case 2: A blokG i

ontainsa hole ofC G

(H

).

SupposeH 2 C G

(H

) is a hole in G i

suh that V(H)\C has maximum ardinality. If

H 62 G 0 i

, it follows from the denition of rened blok that some node x 2

2 V(H)\N(C)

hasnoneighborinHnN[C℄. So,there existsahordlesspathP 0 =x 1 ;x 2 ;x 3

inH suhthat

x 1

;x 2

2N(C) and x 1

is adjaent to some w 1

2CnV(H). IfV(H)\C6=; orw 1

2N(x 3

),

then w 1

is of Type g3 and, after substitutingw 1

into H,we would obtaina hole of C G (H ) in G i

with larger intersetion with C than H, a ontradition. It follows that, for eah

H2C G

(H

),V(H)\C=;and w 1

x 3

isnotan edge.

By the hoie of x 2

, thisimpliesx 3

2N(C). In fat, by thesame argument, no node of

C is of Type g3 w.r.t. H. As Gis4-hole-free and gem-free,x 2

is adjaent toneither w 1

nor

w 3

. Sox 2

isadjaent to somenodew 2

2C. SineGis4-hole-free, w 2

isadjaent tobothx 1 and x

3

. Hene w 2

Input: A graphG thatdoesnotontain a4-hole, a short3PC nora short4-wheel.

Output: EitherGisidentiedasnot being even-hole-free,ora listL ofinduedsubgraphs

of Gwiththefollowingproperties:

The graphs in L do not ontain a gem, a full k-star utset, k = 1;2;3, nor any

dominatednodes.

If theinputgraph Gontainsan even hole andis lean, withspotless smallesteven

holeH

,then one ofthe graphsinthelistontainsa hole inC G

(H

).

Step 1: InitializeM=fGg,L=;.

Step 2: IfMisempty,returnL andstop. Otherwise,remove agraphF from M. IfF has

no hordless path of length 4, go to Step 2. Otherwise, remove all dominated nodes

from F and go to Step3.

Step 3: If F ontains a gem fx;y 0

;y;z;z 0

g, suh that y 0

;y;z;z 0

is a hordless path, go to

Step4. IfF ontainsa fullk-starutset S,k =1;2;3,go to Step5. Otherwise,addF

to L andgo to Step 2.

Step 4: IfS =(N(x)[N(y)[N(z))nfy 0

;z 0

gisnotautsetbreakingy 0

fromz 0

,gotoStep

6. IftheonnetedomponentofF nS thatontainsy 0

isofsize1,addgraphF nfy 0

g

to Mand goto Step2. Iftheonnetedomponentof F nS thatontainsz 0

isofsize

1,add graphF nfz 0

gto Mand goto Step2. Otherwise,let S=N(x)[N(y)[N(z)

and goto Step 5.

Step 5: Chek whether there exists a deomposition detetable 3PC(:;:) or 3PC(;)

w.r.t. S. If yes, go to Step 6. Otherwise, onstrut the rened bloks of

deompo-sitionbyS,add them to Mand go to Step2.

Step 6: Return thatGis noteven-hole-free and stop.

Lemma 3.9 TheNode CutsetDeomposition Algorithm produes the desired output.

Proof: Firstsupposethat the algorithm terminatesin Step6. Then by Lemma3.4 and the

fat that 3PC(:;:)'s and 3PC(;)'s ontain even holes, the algorithm orretly identies

G as notbeing even-hole-free. Now suppose that the algorithm outputs the list L, i.e. the

algorithm doesnotterminate inStep 6. Then learly, bySteps 2and 3, thegraphs inL do

not ontain any dominatednode, gem orfullk-star utset, k =1;2;3. Nowfurther assume

that theinputgraph G is lean and ontains a spotless smallesteven hole H

. We want to

show thatsome graphin listL ontainsahole inC G

(H

).

LetF beagraphtakenolistMinStep2. ItisenoughtoshowthatifF ontainsahole

inC G

(H

) then at leastone of thegraphs thatgets puton list M orL in Steps 3, 4 and 5

also ontainsahole inC G

(H

). Thisfollows fromLemma3.1, Lemma 3.6andTheorem 3.8.

omposition Algorithm isbounded by jV(G)j 5

.

Proof: LetF bea graph taken olist Min Step2. Supposethat F isdeomposed inStep

5 bya fullk-starutsetS,k =1;2;3. LetC 1

;:::;C n

be theonnetedomponentsofF nS

andlet F 1

;:::;F n

betherenedbloks ofdeompositionbyS. LetC be theliqueenterof

S.

Claim: No two of the graphs F 1

;:::;F n

ontain the same hordless path of length 4.

Proof of Claim: Let P be a hordless path of length 4 and suppose that P appears in F 1 and F

2

. Then V(P) V(F 1

)\S and V(P)V(F 2

)\S. Sine V(P) S,it ontains two

nonadjaent nodes a;b2 SnC,suh that there exists a hordlesspath P 0

from a to b that

uses only nodes in C asintermediate nodes. Sine a2 V(F 1

)\V(F 2

), by denitionof the

renedbloks,ahasneighborsinbothC 1

and C 2

. SimilarlybhasneighborsinbothC 1

and

C 2

. Note that by denitionof S,nodesof C do not have neighbors inC 1

and C 2

. Butnow

there is a 3PC(a;b) that uses P 0

and paths in C 1

and C 2

. This 3PC(:;:) is deomposition

detetablew.r.t. S andhene wouldhave beendetetedinStep5. Thisompletes theproof

of thelaim.

By Step 2, the algorithm only adds to L subgraphs of G that have a hordless path of

length4. So,itfollowsfrom thelaimthatthenumberof graphsinLis atmostjV(G)j 5

. 2

4 Cleaning

ThissetionisdevotedtotheonstrutionoftheCleaningAlgorithm. Weassumethroughout

thissetionthat Gontainsno 4-hole, no6-hole, no short4-wheeland noshort3PC (reall

Denitions2.1and2.2). TheCleaningAlgorithmwilltakeasinputthegraphGandprodue

apolynomialfamilyL ofinduedsubgraphsofGsuh that, ifGontainsan evenhole,then

at leastone of the graphsin L ontains an even hole and is lean. Given a hole H, a node

v 62 H is strongly adjaent to H if v has at least two neighbors in H. Reall that an even

hole H islean ifit hasno badstrongly adjaent nodes(Denitions1.4and 1.5).

Lemma 4.1 Let u be a bad node w.r.t. a smallest even hole H of G. Then either u has

exatly two neighbors in H and these nodesare nonadjaent, or (H;u) is an even wheel and

all the setors of the wheel areodd.

Proof: If u has two neighbors in H, then they are nonadjaent sine u is bad. So assume

that u has at least three neighbors in H. If u has an odd numberof neighbors in H, then

sine H is an even hole, one of the setors of the wheel (H;u) must be even. That setor

together withu indues aneven holeand sine thathole annot be smaller thanH,u must

be of Type g3, ontraditing theassumption that u is bad. By a similarargument, ifu has

an even numberofneighborsinH,thenall thesetors of (H;u)mustbeodd. 2

Denition 4.2 Let v be a bad node w.r.t. a smallest even hole H of G. For i = 1;2;3,

we say that v is of Type bi w.r.t. H if V(H)\N(v) indues a graph G 0

with exatly two

onneted omponents, jV(G 0

)j 4 and the largest onneted omponent of G 0

has exatly i

Suppose H is a smallest even hole in G and v 1

and v 2

are two nonadjaent bad nodes

w.r.t. H. Consider thefollowingthree typesof subpathsofH.

e-path We all a subpathQ i

of H an edge-path (or e-path) if one of its endnodesis adjaent

to v 1

, theother is adjaent to v 2

, at mostone endnode is adjaent to bothv 1

;v 2

, and

no intermediate nodeof Q i

is adjaent to v 1

orv 2

.

n-path We all a subpath P i

of H a node-path (or n-path) if it is a maximal path with the

followingproperty: theendnodesof P i

areadjaenttov 1

andnonodeof P i

isadjaent

tov 2

,ortheendnodesofP i

areadjaenttov 2

andnonodeofP i

isadjaenttov 1

. Note

thatan n-pathan have length0.

z-path We all asubpathP 0

ofH a zero-path (orz-path) if itis a maximalpath withall the

nodes adjaent to both v 1

and v 2

. As G is 4-hole-free, there is at most one z-path.

Furthermore,ifthe z-pathexists, it hasat mosttwo nodes.

We onstrutthe graphH 0

fromH denedasfollows:

Contrat eahe-path Q i

of H to a singleedge q i

.

Contrat eahn-path P i

of H to a singlenode p i

.

IfH hasa z-path P 0

,ontrat it to asingle nodep 0

alled thez-nodeof H 0

.

Sine H has at leastone node adjaent to v 1

butnotv 2

and another adjaent to v 2

but

not v 1

, the graph H 0

has at least two nodes distint from the z-node. Moreover, if H 0

has

no z-node, it hasat least fournodes. Tosee this, notethat, sine H has no z-path,it must

haveanevennumberofe-paths. IfHhasexatlytwoe-paths,thenV(H)[fv 1

;v 2

gontains

an even hole smaller thanH. So H has at leastfour e-pathsand hene H 0

hasat leastfour

nodes.

WeallanedgeoranodeofH 0

even(odd)iftheorrespondingpathofHhaseven (odd)

numberof edges. We allan edgeora node of H 0

real ifthe orresponding path ofH is an

Lemma 4.3 Let q i

and q i+1

betwo onseutiveedges ofH suhthat their ommon endnode

p i

isdistintfromp 0

. Thenq i

andq i+1

havethesameparity ifandonlyifp i

isodd. Moreover,

the edges of H 0

inident with p 0

are odd.

Proof: Indeed, otherwise either (H;v 1

) or (H;v 2

) would have an even setor, ontraditing

Lemma4.1. 2

Lemma 4.4 Supposethat H 0

has a z-node p 0

andthat q i

=p i

p i+1

is an even edge. Then q i has a real endnode that isadjaent to p

0

by a real edge. Moreover, p 0

isa real node and H 0

has at least four edges.

Proof: By Lemma 4.3, p 0

is not an endnode of q i

. If P 0

has a node u 0

that is adjaent to

neitherendnodeof Q i

,thenV(Q i

)[fu 0

;v 1

;v 2

gindues anevenhole. SineH isa smallest

even hole of G,V(H)nV(Q i

) ontainsthree nodes. Butnow, sinev 1

and v 2

arebad w.r.t.

H, they are of Type b1. This impliesthat Gontains a 4-hole, a ontradition. Hene,we

mayassumew.l.o.g. thatp i

is areal node and isadjaent to p 0

byareal edge. Asv 1

and v 2 are bad nodes w.r.t. H, it follows that p

i+1

is not adjaent to p 0

in H 0

. Hene, sineevery

node ofP 0

must beadjaent toan endnodeof Q i

,p 0

isa real node. Finally,sinev 1

and v 2 arebad, H

0

hasat leastfouredges. 2

Lemma 4.5 Let q i

and q j

betwo nononseutive edges of H 0

withthe same parity. Suppose

that p 0

is not an endnode of q i

nor q j

. Then q i

and q j

havereal endnodes that are adjaent

by a real edge.

Proof: Supposenot. SineV(Q i

)[V(Q j

)[fv 1

;v 2

gdoesnotindueasmallerevenholethan

H, it follows that H 0

has four edges, say i= 1 and j = 3, the paths Q 2

and Q 4

eah have

length 2, and v 1

, v 2

are bothof Type b1. Sine G has no short3PC(:;:), bothQ 1

and Q 3 have length greater 1. It follows that V(Q

2

)[V(Q 4

)[fv 1

;v 2

g is an 8-hole. Sine H is a

smallesthole, Q 1

and Q 3

bothhave length2. But now V(Q 1

)[V(Q 2

) forms a 6-holewith

v 1

orv 2

,ontraditing theassumption thatG ontainsno6-hole. 2

Lemma 4.6 If p i

is a node of H 0

that is not adjaent to p 0

, then either p i

is even or P i

is

an edge.

Proof: The result holdswhen i=0, sowe assume nowi6= 0. Suppose p i

is odd. Then, by

Lemma 4.3, the two edges of H 0

that have p i

asa ommon endnode, say q i

and q i+1

, must

have thesame parity. So,if P i

is notan edge, V(Q i

)[V(Q i+1

)[fv 1

;v 2

g induesa smaller

even hole thanH. 2

Theorem 4.7 Let v 1

and v 2

be nonadjaent bad nodes w.r.t. a smallest even hole H of G.

Then either v 1

and v 2

have a ommon neighbor in H, or exatly one of v 1

;v 2

is of Type b2

w.r.t. H.

Proof: LetH 0

bedened from H asabove. Assume v 1

and v 2

have no ommon neighborin

H. ThenH 0

hasno z-node. Letp 1

;:::;p m

be thenodesofH 0

appearinginthisorder when

traversing H 0

and assumew.l.o.g. that v 1

isadjaent to p 1

. Then p k

is adjaent to v i

1 m 1 1 whih impliesthat p

m

is adjaent to v 2

.

Case 1: m6.

It follows from Lemma 4.5 that H 0

annot have three onseutive even edges. Hene

H 0

has two odd edges, the endnodes of whih are not adjaent by a real edge. But this

ontraditsLemma 4.5.

Case 2: m=4.

Supposev 1

is nota Typeb2 node w.r.t. H. Then,byLemmas 4.1 and 4.6, bothp 1

and

p 3

mustbe even. Now, ifp 2

and p 4

arealso even,then byLemma 4.3, theedges ofH 0

must

be alternately odd and even. Thus H 0

hastwo odd edges whose endnodes are notadjaent

byareal edge, ontraditing Lemma4.5. Henev 2

isof Typeb2.

If bothv 1

and v 2

are of Type b2,then all the nodes of H 0

are odd and, by Lemma 4.3,

all the edges of H 0

must have the same parity. But then, any two nonadjaent edges of H 0

ontradit Lemma4.5. 2

Lemma 4.8 Let H be a Type b2 node free smallest even hole and let v 1

and v 2

be two

nonadjaent bad nodesw.r.t. H. ThenH =u 0

;u 1

;:::;u r

where v 1

and v 2

areboth adjaent

to u 0

. If v 1

and v 2

have exatly one ommon neighbor in H, then w.l.o.g. v 1

is adjaent to

u 1

andthe two setors of(H;v 1

) withommon endnode u 1

, ontain allthe neighbors ofv 2

in

H. Otherwise,v 1

and v 2

are both adjaent tou 1

andw.l.o.g. the two setors of (H;v 1

) with

ommon endnode u 1

, ontains all the neighbors of v 2

in H.

Proof: By Theorem 4.7, H has a z-path. Consider H 0

= p 0

;p 1

;:::;p m

obtained from H

as before, where p 0

is the z-node. Assume w.l.o.g. that q i

= p i

p i+1

where 0 i m and

m+1 0. Furthermore,assume w.l.o.g. that v 1

is adjaent to p 1

, i.e. theendnodesof P 1 are adjaent to v

1

. By Lemmas 4.3 and 4.4, all the edges of H 0

are odd, exept maybe q 1 and q

m 1 .

Case 1: H 0

hasaneven edge.

W.l.o.g. q 1

is even. By Lemma 4.4, q 0

is a real edge, bothp 0

and p 1

are real nodes and

m 3. If m = 3, we are done. Assume m =4. As p 0

and p 1

are real nodes, Lemma 4.1

impliesthatp 3

mustbeodd. Butthen,byLemma4.6,v 1

wouldbeofTypeb2. Henem5.

As bothq 2

and q 3

are odd by Lemma 4.3, it follows that p 3

is odd. Hene, by Lemma 4.5

appliedto q 2

and q 4

,q 4

is even. But thenq 1

and q 4

ontraditLemma 4.5.

Case 2: Alltheedges ofH 0

areodd.

By Lemma 4.3, p 2

is odd. If m 4, then the pair q 1

and q 3

ontradits Lemma 4.5. If

m=3, then, by Lemmas 4.1and 4.6, v 2

would beof Type b2 w.r.t. H. Hene m=2 and,

byLemma4.1applied to H and v 2

,P 0

hastwo nodesu 0

and u 1

. So we aredone. 2

This lemma impliesthenext result.

Theorem 4.9 Let H be a Type b2 node free smallest even hole. Let v 1

be a Type b3 node

w.r.t. H and N(v 1

)\V(H)=fu 1 ;u 2 ;u 3 ;u 4

g, whereu 2

is adjaent tou 1

and u 3

. If v 2

is a

badnode w.r.t. H,then N(v 2

)\fu 2

;u 4

;v 1

Input: AgraphGthatdoesnotontaina4-hole,a6-hole, ashort4-wheelnorashort3PC.

Output: AfamilyL of induedsubgraphsof Gthat satisesthefollowing: IfG ontainsa

smallesteven hole H, then, forsome G 0

2L ontaining H, thefamily C G

0

(H) has no

Type b2 nodes. Moreover, if there is a Type b1 or b3 node w.r.t. H but no Type b2

node w.r.t. a hole in C G

(H), then H is a spotless smallest even hole in some graph

G 00

2L.

Step 1: SetL=fGg.

Step 2: For every (P 1

;P 2

;u), where P 1 = x 0 ;x 1 ;x 2 ;x 3 and P 2 = y 0 ;y 1 ;y 2 ;y 3 are disjoint

hordlesspathsinGand u2N(x 1

)\N(y 1

),add to L thegraphsobtainedfrom Gby

removingthenode setN(fx 1 ;x 2 ;y 1 ;y 2

;ug)n(V(P 1

)[V(P 2

)).

Theorem 4.10 Proedure BAD produes the desired output.

Proof: Let u be a Type bi node w.r.t. a smallest even hole H, where i 3. Take

P 1 = x 0 ;x 1 ;x 2 ;x 3 and P 2 = y 0 ;y 1 ;y 2 ;y 3

to be disjoint subgraphs of H suh that N(u)\

fx 1 ;x 2 ;y 1 ;y 2

ghasmaximumardinality. Denote byG 0

thegraphGn(N(fx 1 ;x 2 ;y 1 ;y 2

;ug)n

(V(P 1

)[V(P 2

))). Then G 0

2L andH G 0

.

Claim: In G, node u isa Type bi node w.r.t. all the holesin C G

0 (H).

ProofofClaim: Indeed,inG 0

,thenodesx 1 ;x 2 ;y 1 andy 2

havedegree2. Sinetheybelong

to H,they also belong to all theholesin C G

0

(H). It follows thatP 1

and P 2

are subpathsin

all theholesof C G

0

(H). This ompletestheproofof thelaim.

ByTheorem4.7,ifi=2,theneveryholeinC G

0

(H)isTypeb2nodefree. ByTheorem4.9,

ifi=3andall holesinC G

(H) areTypeb2nodefree, thenH isa spotlesssmallestevenhole

inG 0

. Finally,byTheorem 4.7, ifi=1 and all holesinC G

(H) are Type b2 node free, then

H is aspotlesssmallesteven hole inG 0

. 2

Lemma 4.11 Let H be a Type b2 node free smallest even hole and v 1

, v 2

and v 3

be three

pairwise nonadjaent badnodesw.r.t. H. Thenthereexistsanode u2V(H) thatisadjaent

to v 1 , v 2 and v 3 .

Proof: ByTheorem4.7,there existsanodeu2V(H)thatisadjaentto v 1 and v 2 . Suppose v 3

isnotadjaentto u.

As v 1

and v 2

are two nonadjaent bad nodes w.r.t. H, by Lemma 4.8, we may let

H =u 0

;u 1

;:::;u m

where u = u 0

, node u 1

is adjaent to v 1

(and possibly v 2

) and the two

setors of (H;v 1

) with ommon endnode u 1

ontain all theneighbors of v 2

inH. Consider

thefollowingtwo ases.

Case 1: v 3

is adjaent to u 1

.

As v 3

is not adjaent to u 0

, and v 1

is adjaent to u 0

but not to u 2

, it follows from

Lemma 4.8 that the two setors of v 1

sharing u 0

ontain all the neighbors of v 3

in H. By

Theorem 4.7, nodes v 2

and v 3

have a ommon neighbor in H. The onlypossibilityis node

u 1

. So u 1

3 1 Supposethatv

1 ,v

2 and v

3

do nothave a ommonneighborinH. Letu i

be adjaent to

v 1

and v 3

, and let u j

be adjaent to v 2

and v 3

. Then i>j. Firstassume that u i

=u m

. It

follows from Lemma 4.8applied to v 1

and v 3

that N(v 1

)\V(H) =fu 0 ;u 1 ;u m 1 ;u m g. But

then(H;v 1

) isa short4-wheel, aontradition.

Itfollowsthati<m. Theni=j+1,otherwisethesetfu i ;u j ;v 1 ;v 2 ;v 3 ;u 0

gwouldindue

a6-hole. Ifv 3

isnotadjaent tou j 1

,thenbyLemma4.8, thetwosetors of(H;v 3

)sharing

u i

mustontainalltheneighborsofv 2

. Butthenv 3

isnotadjaenttou i+1

andthetwosetors

of (H;v 3

) sharing u j

must ontain all the neighbors of v 1

in H, a ontradition. Hene v 3 is adjaent to both u

j 1 and u

i+1

. Now, by Lemma 4.8, the setors of (H;v 3

) sharing u i+1 (u

j 1

)ontain all theneighbors ofv 1

(v 2

). So (H;v 3

) isa short4-wheel,a ontradition. 2

Theorem 4.12 Let H bea Typeb2node free smallesteven hole. If thereexist three

nonad-jaent badnodes w.r.t. H, then thereexists a node u in H suh that all the bad nodes w.r.t.

H are adjaent to node u or to oneof the neighbors of u in H.

Proof: Supposev 1

, v 2

and v 3

are three nonadjaent bad nodesw.r.t. H and u is a ommon

neighborinH (suh a node existsbyLemma 4.11). Let u 1

;u 2

denote theneighborsof u in

H. Supposev is a bad node w.r.t. H that is not adjaent to a node in fu;u 1

;u 2

g. Then,

v is adjaent to at most one of the nodes v 1

;v 2

;v 3

, else G ontains a 4-hole. Say v is not

adjaent to v 1

and v 2

. Now, by Lemma4.11, nodesv 1

;v 2

;v have a ommon neighborin H,

sayw. But thenw;v 1

;u;v 2

isa 4-hole, aontradition. 2

Foranode set S,denote by(S) theardinalityofa largest stableset inS.

Theorem 4.13 Let H be a Typeb2 node free smallest even hole and S bethe set of all bad

nodesw.r.t. H.

a. If (S) =1, then there are two nonadjaent nodes u 1

;u 2

in H suh that either S = N 0

whereN 0

=N(u 1

)\N(u 2

),orthereexistsa2SnN 0

withthepropertythat,ifN denotes

thesetof nodesofGn(N 0

[fag)adjaenttoallnodes inN 0

[fag,thenjV(H)\Nj3

and SN [N 0

[fag.

b. If (S) = 2, then there are two nonadjaent nodes u 1

;u 2

in H, and a third node w 1

in

H (not neessarily distint from u 1

or u 2

) suh that, if A = S nN(w 1

) and N 00

=

(N(u 1

)\N(u 2

))nN(w 1

), then either (AnN 00

)1, or thereexists a node a2AnN 00

and a node v 1

adjaent to u 1

, u 2

and w 1

with the property that, if N is the set of

nodes of Gn(N 00

[fa;v 1

g) that are adjaent to all the nodes in N 00

[fa;v 1

g, then

jV(H)\Nj3 and (An(N [N 00

[fag)) 1.

Proof: a. Letu 1

and u 2

betwo nodesof H suh that

(i) theshortestpath ofH onnetingu 1

and u 2

hasat leastthree edges,

(ii) N 0

=N(u 1

)\N(u 2

) hasmaximumardinality.

By (i), N 0

S. If N 0

= S, we are done. So, suppose a 2 S nN 0

. Denote by N the

nodesofGn(N 0

[fag)adjaentto allnodesinN 0

[fag. Then,sineS isaliqueontaining

N 0

[fag, S N [N 0

1 2 more ommonneighborsinS thanu

1 and u

2

,whih ontradits(ii).

b. Supposev 1

;v 2

2S are nonadjaent.

By Theorem4.7, nodes v 1

and v 2

have a ommon neighbor inH, say w 1

. LetA be the

setof badnodesthat arenotadjaent to w 1

. AsG is4-holefree, eahnode ofA isadjaent

to exatlyoneof v 1

;v 2

. Fori=1;2,denote byA i

thesetofnodesofAadjaentto v i . Then A 1 \A 2

= ; and A 1

[A 2

= A. As (S) = 2, it follows that both A 1

and A 2

are liques

(possiblyempty). Nowassume thatu 1

and u 2

aretwo nodesof H suhthat

(i) v 1

isadjaentto bothu 1

and u 2

,

(ii) theshortestpathinH onneting u 1

and u 2

hasat leastthree edges,

(iii)N 00

=(N(u 1

)\N(u 2

))nN(w 1

) hasmaximumardinality.

As v 1

is a bad node w.r.t. H, suh a pair of nodes u 1

;u 2

always exists. (ii) and (iii)

imply that N 00

A. As G is 4-hole free and N(v 1

)\A 2

= ;, it follows that N 00 A 1 . If A 1 =N 00

,then AnN 00

=A 2

,so(AnN 00

)1 and we aredone. So,supposea2A 1

nN 00

.

Denote byN thenodesof Gn(N 00

[fa;v 1

g)adjaentto allthenodesinN 00

[fa;v 1

g. Then,

sine A 1

is a lique ontaining N 00

[fag, it follows that A 1

N [N 00

[fag, and hene

(An(N[N 00

[fag))1.

If jV(H)\Nj4,thenN wouldontain twonodesx 1

and x 2

satisfying(i) and (ii)and

havingmore ommonneighborsinA 1

thanu 1

and u 2

,whihontradits(iii). 2

PROCEDURE b4

Input: AgraphGthatdoesnotontaina4-hole,a6-hole,ashort4-wheelnorashort3PC.

Output: AfamilyL of induedsubgraphsof Gthat satisesthefollowing: IfG ontainsa

smallesteven hole H suh that C G

(H) is Typebi node freefor i=1;2;3, then H is a

spotless smallesteven holeinsome G 0

2L.

Step 1: SetL=L 2

=fGg and L 1

=L 3

=;.

Step 2: Forevery hordless pathP =w 0 ;w 1 ;w 2 ;w 3 ;w 4

inG, add to L the graphobtained

from Gbyremovingthenodeset ( S

3 i=1

N(w i

))nV(P).

Step 3a: Forevery hordlesspathP =w 0

;w 1

;w 2

inGand v 1

6=w 0

;w 2

adjaent tow 1

,add

to L 1

thegraphobtained fromG byremovingthe node set N(w 1

)nfw 0 ;w 2 ;v 1 g.

For k=1 to 2, do

begin

Step 3b: ForeveryL2L k

andforeverynonadjaent u 1

;u 2

2V(L),add toL k+1

thegraph

obtainedfrom Lbyremovingthenode setN(u 1

)\N(u 2

).

Step3: ForeveryL2L k

andforeverynonadjaentu 1

;u 2

2V(L),letN 0

=N(u 1

)\N(u 2

).

Foreverya2V(L)nN 0

,letN denotethesetofnodesofLn(N 0

[fag)thatareadjaent

to allthenodesinN 0

[fag. Fori=0;1;2;3, letN i

denote thefamilyof allsubsets of

N withardinalityjNj i. ForeveryM 2N i

,add toL k+1

thegraphobtainedfromL

by removingthenodeset M[N 0

[fag.

3

Theorem 4.14 Proedure b4 produes the desired output.

Proof: Let H bea smallesteven holeinGthat isTypeb2 nodefree, and S be thesetof all

bad nodesw.r.t. H.

If (S)3,then, by Theorem4.12, Step2 produesa graph G 0

inL whereH islean.

If (S) = 1, then, by Theorem 4.13a, Steps 3b and 3 applied to G 2 L 2

when k = 2,

produes agraph G 0

2L 3

whereH is lean.

Finally, if (S) = 2, then Step 3a produes a graph L 2 L 1

where the nodes of G in

N(w 1

)nfw 0

;w 2

;v 1

gareremoved. Thebadnodesthatremainarev 1

and A=SnN(w 1

). By

Theorem4.13b,Steps3band3appliedtoLwhenk=1produeagraphinL 2

thatontains

H and suhthat theset A 2

of remainingbad nodesw.r.t. H satises(A 2

)1 (Note that

N 0

in Step 3 (k=1) of the algorithm is equal to N 00

[fv 1

g as dened in Theorem 4.13b

wheneverv 1

isadjaent to u 1

and u 2

.) Now, byTheorem4.13a, Steps3b and3 whenk =2

produesome graph G 0

2L 3

whereH is lean.

So, in all ases, the algorithm produes a graph G 0

inL whereH is lean. To omplete

the proof it remainsto showthat, ifC G

(H) is Type bi node free fori=1;2;3, thenH is a

spotless smallesteven holein G 0

. This followsfrom thenext two laims.

Claim 1: IfH

isaleansmallestevenholeand C G

(H

) isTypebinodefree,fori=1;2;3,

thenanyhole obtainedfrom H

throughone speialtent substitution isalso lean.

Proof of Claim 1: Let xy be a speial tent w.r.t. H

, with intermediate paths P 1

and P 2

,

whereP 1

isoflength2,andletH betheholeinduedbythenodesetV(P 2

)[fx;yg. W.l.o.g.

assume that x is of Type g2 w.r.t. H

, with neighbors x 1

and x 2

in H

, and node y has a

uniqueneighbory 1

inH

. Let p 1

be theintermediatenode of P 1

,and w.l.o.g. let x 2

and y 1 bethe endnodes of P

1

. We willshow thatthe strongly adjaent nodes to H areof Type g2

org3.

Suppose not and let u be a strongly adjaent node to H that is not of Type g2 or g3.

Then u must have at least one neighbor in P 2

. Let u 1

be the neighbor of u in P 2

that is

losesttox 1

,andletP 0

bethex 1

u 1

-subpathofP 2

. SineH

islean,uiseithernotstrongly

adjaent toH

orisofTypeg2org3w.r.t. H

. Alsoumustbeadjaentto anode infx;yg,

sowehave thefollowingthree ases to onsider.

Case 1: Node u isadjaent to bothx and y.

First assume that u is adjaent to y 1

. Then u must have at least two neighbors in P 2

,

sineotherwiseuis ofTypeg3 w.r.t. H. Ifu hastwo neighborsinP 2

thenu 1

isadjaent to

y 1

and (H;u) is a short4-wheel. If u has three neighbors inP 2

then it is of Type g3 w.r.t.

H

and the hole induedbythe node set V(P 0

)[fx;ug is even of lengthsmaller than H

,

ontraditingourhoieofH

. Heneuisnotadjaenttoy 1

. Byasimilarargumentuisnot

adjaent to x 1

either. Sine u musthave a neighborinP 2

and sineit iseither notstrongly

adjaent to H

or it is of Type g2 or g3 w.r.t. H

, this impliesthat u does not have any

neighborsinP 1

. Node u 1

is notadjaent to y 1

,sineotherwiseu;y;y 1

;u 1

;u is a4-hole. Let

H 0

be thehole indued by the node set V(P 0

)[V(P 1

)[fy;ug. But now (H 0

;x)is a short

4-wheel.

1 1 1 thenu isof Typeg3 w.r.t. H

,with allneighborsinP 2

. Butthen(H;u)isa short4-wheel.

Heneu isnotadjaent tox 1

nory 1

,whih impliesthatit annothave anyneighborsinP 1

.

Butnowthereis ashort3PC(x;y 1

),wheretwoof thepathsarex;P 1

;y 1

andx;y;y 1

andthe

thirdpathpasses throughu.

Case 3: Node u isadjaent to y but nottox.

Nodeuisnotadjaenttox 1

,sineotherwiseu;y;x;x 1

;u isa4-hole. Ifuisadjaenttoy 1 thenu isof Typeg3 w.r.t. H

,with allneighborsinP 2

. Butthen(H;u)isa short4-wheel.

Heneu isnotadjaent tox 1

nory 1

,whih impliesthatit annothave anyneighborsinP 1

.

If u is of Type g1 or g3 w.r.t. H

, then u is of Type b1 or b3 w.r.t. H, ontraditing the

assumptionthat C G

(H

) isType b1and b3 node free. SineH

islean,u must be ofType

g2 w.r.t. H

,ontraditing Lemma4.1 appliedto H and u.

Claim 2: IfH

isaleansmallestevenholeand C G

(H

) isTypebinodefree,fori=1;2;3,

thenanyhole obtainedfrom H

throughone Typeg3 node substitutionisalso lean.

Proof of Claim 2: Letx be a Typeg3 node w.r.t. H

,with neighbors x 1

, x 2

and x 3

in H

.

Assume that x 2

is themiddleneighborof x in H

and let H be the holeobtained from H

bysubstitutingx forx 2

. Wewillshowthatthestronglyadjaent nodestoH areof Typeg2

org3. Let ube astrongly adjaent node to H. We onsiderthefollowingtwoases.

Case 1: Node u isnotadjaentto x.

Then u annot be adjaent to both x 1

and x 3

, sine otherwisex;x 1

;u;x 3

;x is a 4-hole.

Sine u is strongly adjaent to H, it is also strongly adjaent to H

. Sine H

is lean, u

is of Type g2 org3 w.r.t. H

. But then, sineu is not adjaent to bothx 1

and x 3

,u is of

Typeg2 org3 w.r.t. H aswell.

Case 2: Node u isadjaent to x.

If u is not adjaent to x 1

nor x 3

then it is also not adjaent to x 2

, sine otherwise u

would bea bad stronglyadjaent node w.r.t. H

. ByLemma 4.1 appliedto H and u,node

u annot be of Type g2 w.r.t. H

, and hene it is of Type g1 or g3 w.r.t. H

. But then u

is of Type b1 or b3 w.r.t. H, ontraditing the assumption thatC G

(H

) is Type b1 and b3

node free. Therefore umust beadjaent to x 1

or x 3

.

First assumethat u is adjaent to both x 1

and x 3

. Then u must also be adjaent to x 2

,

sine otherwise u;x 1

;x 2

;x 3

;u is a 4-hole. Sine H

is lean, u is of Type g3 w.r.t. H

and

hene w.r.t. H aswell.

Now assume that u is adjaent to x 1

but not to x 3

. Note that sine H

is lean, u an

have atmostthree neighbors inV(H

)nfx 2

g. Ifu hastwoneighborsinV(H

)nfx 2

g,then

u is of Type g2 org3 w.r.t. H

and hene of Type g3 w.r.t. H. If u hasthree neighborsin

V(H

)nfx 2

g,then(H;u)isa short4-wheel. Thisompletes theproofofClaim2 andofthe

theorem. 2

CLEANING ALGORITHM

Input: AgraphGthatdoesnotontaina4-hole,a6-hole,ashort4-wheelnorashort3PC.

Output: AfamilyL of induedsubgraphs ofG suh that, ifGontains an even hole,then

some G 0

Step 2: ApplyProedure BAD to Gand let L 0

be theresultingoutput family.

Step 3: ApplyProedure BAD to eah graphinL 0

and unionthe outputwith L.

Step 4: ApplyProedure b4to eah of thegraphsinL 0

and uniontheoutput withL.

IfGontainsaneven holethen,afterStep2,L 0

ontainsagraphG 0

withasmallesteven

hole H suh that C G

0(H) isType b2 node free. Now, ifH has a Type b1 orb3 node inG 0

,

we get the desired output in L after Step 3 and otherwise we get it after Step 4. So the

CleaningAlgorithmproduesthedesiredoutput. Thesizeoftheoutputan beestimated to

beO(n 25

).

5 2-Join Deompositions

Inthissetion,weassume thatG doesnotontain a 4-hole,a dominatednode,agem nora

fullk-starutset, k=1;2;3. So,by Lemma1.14, Gontains nok-star utset.

LetV 1

jV 2

bea2-joinwithspeialsets(A 1 ;A 2 ;B 1 ;B 2

). Fori=1;2,letP i

bethefamilyof

hordlesspathsP =x 1

;:::;x n wherex 1 2A i ,x n 2B i andx j 2V i n(A i [B i

),2jn 1.

Lemma 5.1 Thesets P i

are nonempty and ontain no path of length 1, for i=1;2.

Proof: Let u2A 1

and v2B 1

.

First, suppose that there is no path in V 1

from A 1

to B 1

. Then, sine jV 1

j > 2, either

fug[A 2

orfvg[B 2

isa star utset. Hene P 1

6=;. Similarly,P 2

6=;.

Now, if uv is an edge, then no node of A 2

an be adjaent to a node of B 2

(sine G is

4-hole-free). As P 2

6=;,it followsthat V 2

n(A 2

[B 2

)6=;. Butthenfu;vg[A 2

[B 2

would

bea doublestarutset. 2

The bloks of a2-joindeompositionare graphsG 1

and G 2

denedas follows. BlokG 1 onsistsofthesubgraphofGinduedbynodesetV

1

plusamarker pathP 2

=a 2

;:::;b 2

that

ishordlessand satisesthefollowingproperties. Node a 2

isadjaent to allthenodesinA 1

,

nodeb 2

is adjaent toall thenodesinB 1

andthese aretheonlyadjaeniesbetweenP 2

and

the nodesof V 1

. Furthermore, let Q2 P 2

. The marker path P 2

haslength 4 if Qhas even

length, andlength5 otherwise. BlokG 2

is denedsimilarly.

Theorem 5.2 Let G 1

and G 2

be the bloks of a 2-join deomposition of G. Then, G is

even-hole-freeif and onlyif G 1

and G 2

areeven-hole-free.

Proof: FirstassumethatG 1

orG 2

hasanevenhole,sayG 1

does. ReplainginG 1

themarker

path P 2

by a pathQ2P 2

of thesame parityyields agraph G 0 1

thatontains an even hole.

Sine G 0 1

is asubgraphof G, thisholeis also aneven hole of G.

Conversely,supposethatGontainsan even hole. IfP 1

(resp. P 2

) haspathsof dierent

parities then, learly, G 2

(resp. G 1

) has an even hole. If all the paths of P 1

[P 2

have the

same parity, then both G 1

and G 2

have even holes. So, we may assume that all the paths

of P 1

are odd and all the paths of P 2

are even. But then eah even hole H of G must be

ontainedinV 1 [A 2 [B 2 or V 2 [A 1 [B 1

. Hene H belongseither toG 1

orG 2

of a 2-join deomposition of G.

Proof: LetG 1

and G 2

bethebloksof a 2-joindeompositionofG and supposethat one of

them, say G 1

,ontains a fullk-star utsetS, k =1;2;3. We willobtain a ontradition by

showingthatthisimpliesthatGalsoontainsa fullk-starutset. Weonsiderthefollowing

three ases.

Case 1: S=N[x℄

IfxisnotanodeofthemarkerpathP 2

,thenSisalsoautsetinG. Firstassumethatx

oinideswitha 2

orb 2

,sayx=a 2

. SineP 2

isnotanedge, thenodesofB 1

areallontained

inthesameomponentofG 1

nS. LetubeanodeofG 1

nS thatisnotinthesameomponent

as B 1

. But then N(a)[fag, where a 2 A 2

, is a full star utset in G breaking u from B 1

.

Nowassume thatxis anintermediatenode ofP 2

. Notethatthegraph induedbythenode

setV 1

[fa 2

;b 2

gisonnetedsineotherwiseGwouldhaveastar utset. Henexisadjaent

to a 2

or b 2

, say a 2

. Let u 2 A 1

and v 2 B 1

be the endnodes of a path in P 1

. Sine P 2

is

of lengthgreater than 2, the nodes of B 1

[fug are all ontainedin thesame omponent of

G 1

nS. Lety beanode ofG 1

nS thatisnotinthesameomponent asB 1

. Then N(u)[fug

isa fullstar utsetinGbreakingy from v.

Case 2: S=N(x)[N(y)

If P 2

ontainsneither x nory, thenS is also autset inG. IfP 2

ontains bothx and y,

thensineP 2

isoflengthgreaterthan3,eitherN(x)[fxgorN(y)[fygisa fullstarutset

in G 1

, and we are donebyCase 1. So assumew.l.o.g. that x =a 2

and y2 A 1

. Let u be a

node ofA 2

. Then N(u)[N(y) is afulldoublestarutset inG.

Case 3: S=N(x)[N(y)[N(z)

IfP 2

doesnotontainanodeinfx;y;zg,thenS isalsoautsetinG. Sow.l.o.g. assume

thatx=a 2

and y;z2A 1

. But thenN(x)[N(y) isa fulldoublestar utsetinG. 2

We now present analgorithm thatdeomposes a graphusing2-joins.

Remark 5.4 In [8℄, a setofforing rulesis giventhat deides inpolytimewhethera pair of

edgesa 1

a 2

andb 1

b 2

belongtoa 2-joinwithspeial sets (A 1

;A 2

;B 1

;B 2

) suh that for i=1;2

a i

2A i

and b i

2B i

. Thealgorithm either outputs suh a 2-joinor it onludes that no suh

2-join exists. We outline here this algorithm for the sake of ompleteness. As pointed out

to us by Jim Geelen and Paul Seymour, these foring rules an be formulated as a 2-SAT

problem,thusprovidinganalternate,andelegant,proofthata2-joinanbefoundinpolytime.

Let a 1

;a 2

;b 1

;b 2

;u be ve distint nodes suh that a 1

a 2

and b 1

b 2

are edges but neither

a 1

b 2

nora 2

b 1

isan edgeandu isadjaent to at mostone ofthenodesa 2

;b 2

(possiblynone).

The following rules yield a 2-join V 1

jV 2

with a 1

;b 1

;u 2 V 1

and a 2

;b 2

2 V 2

or show that no

suh 2-joinexists.

During thealgorithm,thenodesh inV 1

are partitionedintothree sets:

Node hbelongsto A 1

ifitis adjaent to a 2

butnot b 2

,

Node hbelongsto B 1

ifit isadjaent to b 2

butnota 2

,

Node hbelongsto S 1

ifitis adjaent to neither a 2

norb 2