perfectly contractile diamond-free graphs

Perfectly ContractileDiamond-Free Graphs

Irena RusuUNIVERSITE D’ORLEANS

L.I.F.O., B.P. 675945067 ORLEANS CEDEX 2, FRANCE

Received September 24, 1998; revised July 8, 1999

Abstract: Everett et al. [Everett et al., Discr Math, 1997] conjectured that agraph with no odd hole and no stretcher is perfectly contractile, i.e., it can bereduced to a clique by successively contracting even pairs. We show that thisconjecture is true for diamond-free graphs, and propose a polynomial algorithmto perform the successive contractions. c© 1999 John Wiley & Sons, Inc. J Graph Theory 32:

359–389, 1999

Keywords: perfect graph; even pair; perfectly contractile graph

1. INTRODUCTION

In a graphG = (V,E), let x be an arbitrary vertex, and denote byNG(x) the set ofvertices adjacent to x. This set (also denoted N(x) when no confusion is possible)is called the neighborhood of x, while its elements are the neighbors of x. Forall the definitions not given here (path, cycle, chromatic number, clique number,etc.), the reader is referred to [5]. The length of a path or a cycle is its numberof edges.

Following Meyniel [9], we say that two nonadjacent vertices u, v of G form aneven pair if no odd chordless path joining them exists in G. The contraction of theeven pair (u, v) is the operation that consists in replacing the two vertices u, v by aunique vertex t whose neighborhood is NG(u) ∪ NG(v) − {u, v}; the resultinggraph is denoted Guv. Meyniel [9] (see also Fonlupt, Uhry [4]) proved that con-

c© 1999 John Wiley & Sons, Inc. CCC 0364-9024/99/040359-31

360 JOURNAL OF GRAPH THEORY

tracting an even pair preserves chromatic number χ and clique number ω, thussuggesting that successive contractions of even pairs could possibly be used to re-duce a given graph G to a smaller, simpler graph with the same parameters χ andω (immediate applications to perfect graphs, or parameters computing may be con-sidered). In the case where the final graph is a clique, G is called even contractile;whenever this reduction can be performed not only for the graph G itself, but alsofor every one of its induced subgraphs, G is called perfectly contractile (see [1]).

A hole is a chordless cycle of length at least five, and an antihole is the com-plement graph of a hole. A hole is even or odd depending on its number of edges;the parity of an antihole is the one of the corresponding hole. No odd hole andno antihole contain an even pair, but they are not the only minimal such graphs.Call stretcher a graph whose edge set may be partitioned into two triangles andthree vertex-disjoint chordless paths, each with an endpoint in both triangles (seeFig. 1). In a graph with no odd hole, every induced stretcher has the three vertex-disjoint chordless paths of the same parity; therefore, a stretcher is called odd oreven depending on the parity of the paths (odd, respectively, even).

Everett et al. [2] proposed the two conjectures below (unless any specification,anytime we say that a graph G contains, or does not contain, another graph H weunderstand as an induced subgraph):Conjecture 1. A graph with no odd hole, no antihole, and no odd stretcher isperfectly contractile.Conjecture 2. A graph with no odd hole, no antihole, and no stretcher is perfectlycontractile.

It is easy to see that Conjecture 1 is stronger than Conjecture 2. Both theseconjectures are still open, but partial results (concerning particular classes of graphs)have been found (see [3, 6–8]).

We will be concerned, in this article, with the graphs containing no diamond, i.e.,no induced subgraph with four vertices and five edges (sometimes denotedK4−e).We show that diamond-free graphs with no odd hole and no stretcher are perfectlycontractile (the antiholes are automatically excluded), thereby proving Conjecture2 for diamond-free graphs.

2. MAIN RESULT

An arbitrary graphH will be assumed to have vertex set V (H) and edge setE(H),except for the particular graph G we consider all along in the article. Its vertex set(respectively, its edge set) will be always denoted V (respectively, E).

FIGURE 1. A stretcher.

PERFECTLY CONTRACTILE DIAMOND-FREE GRAPHS 361

A path with vertices v1, v2, . . . , vp and edges vivi+1 (i = 1, . . . , p − 1) isdenoted v1v2 · · · vp, while a cycle with vertices v1, v2, . . . , vp and edges vivi+1(i = 1, . . . , p, where vp+1 = v1) is denoted v1v2 · · · vpv1. Often, several consecu-tive vertices of a path (which induce a smaller path called, for instance, X) will bereplaced in the enumeration above by the nameX of the smaller path. An exceptionwill be accepted to these rules of notation: when a vertex of a path is identified bytwo symbols (for instance, v1 = w) that must be simultaneously indicated, the pathwill be denoted v1 = w, v2, . . . , vp instead of v1 = wv2 · · · vp (which is less clear).

For two vertices u, v on a path P , denote by Puv the subpath of P joining u andv. Say that a vertex of a graph is special, if its neighborhood in the graph induceseither a clique or a stable set. Our result is based on the following theorem.

Theorem 1. A diamond-free graph with no odd hole and no stretcher contains atleast one special vertex.

The proof of this result, which is both long and technical, is given in Sections3 and 4. Section 3 presents the outlines of the reasoning, while Section 4 containsthe entire, detailed proof.

In the rest of this section, we prove that Theorem 1 is sufficient to deduceConjecture 2 for diamond-free graphs. Before starting the proof, we need someadditional results. Let us begin with this easy, but important remark.

Claim 1. Let H be a graph with no odd hole and C : c1c2 · · · c2k+1c1 (k > 1)be an odd cycle of H. Then C contains at least one triangle, i.e., there exist threevertices ci, cj , cl of C (with 1 ≤ i < j < l ≤ 2k + 1) such that cicj , cjcl,cicl ∈ E(H).

Proof of Claim 1. Obviously, the claim is true for k = 2. By induction, sup-pose that it is true for every index k0 < k (where k ≥ 3 is a fixed integer), andlet cacb (1 < a + 1 < b ≤ 2k + 1) be a chord of C. Then one of the cyclesc1c2 · · · cacbcb+1 · · · c2k+1c1, caca+1 · · · cb−1cbca is odd. If it is a triangle, we aredone; otherwise, the induction hypothesis implies the existence in this smaller cycleof a triangle, which is also a triangle of the initial cycle C.

The next result may be found in [10] (an expansive cycle is the graph obtainedfrom a chordless cycle with at least four vertices by adding a vertex adjacent toexactly three consecutive vertices on the cycle).

Claim 2. Let H be a graph with no odd hole and w be a vertex of H. Assumethat w is contained in some chordless cycle (of length at least four), and let C besuch a cycle with minimum number of vertices. Then at least one of the statementsbelow is true:

1. The neighbors of w on C form an even pair;2. H contains a stretcher;3. H contains an expansive cycle.

Now, we inquire in which cases the contraction of an even pair can create adiamond in the new graph.


Claim 3. LetH be a diamond-free graph and (a, b) be an even pair ofH. ThenHab contains a diamond if and only if there exist three vertices q, s, t in H suchthat q, s, t, a, b induce the structure in Fig. 2.

Proof of Claim 3. Let q, r, s, t be the vertices of a diamond in Hab, whoseedges are qr, rs, st, tq, tr. Then, as H is diamond-free, the new vertex obtainedafter the contraction of a and b must be in this diamond:

—If it is q (and similarly for s), then in H none of a, b can be adjacent to both rand t, otherwise a diamond can be found in H , a contradiction. Then, say that a isadjacent to r and b to t. Now, the path artb is an odd chordless path joining a andb in H , a contradiction.

—If it is r (and similarly for t), then inH at least one vertex among a, bmust beadjacent to t (say b). If a is also adjacent to t, then neither a nor b can be adjacentto both q, s (or H contains a diamond), so we can suppose that a is adjacent toq and b to s; we obtain the graph in Fig. 2. Now, if a is nonadjacent to t, as bcannot be adjacent to both q and s, we can suppose that bq /∈ E(H); then we musthave aq ∈ E(H), therefore aqtb is an odd chordless path joining a and b in H , acontradiction.

Now, call quasi-stretcher a graph obtained from a stretcher (with trianglesx1y1z1,x2y2z2 and chordless paths X,Y, Z) by adding some edges (at least one), whichjoin vertices on different paths (among X,Y, Z). These edges are called chords.Then we have the following.

Claim 4. LetH be a diamond-free graph with no odd hole and no stretcher. If,in a quasi-stretcherS ofH with trianglesx1y1z1, x2y2z2, all the chords have a com-mon endpoint w, then w is adjacent to every vertex in {x1, y1, z1, x2,y2, z2} − {w}.

Proof of Claim 4. We preserve the notation in the definition of a quasi-stretcherand assume, without loss of generality, that w lies on X , and is not identical to x1.Let us prove that w is adjacent to x1, y1, z1.

Without loss of generality, we can assume that w has a neighbor on Y . Let cbe such a neighbor, which is closest to y1 (possibly c = y1). Then either the cycleXx1wYcy1x1 is of length three and we are done (as H has no diamond), or it is of

FIGURE 2. The contraction of (a, b) yields a diamond.


length at least four and it must be even (otherwise, it would contain an odd hole,by Claim 1). So the paths Xx1w, Yy1c have the same parity.

Moreover, w must have some neighbor d /= z2 on Z, otherwise the cycle givenby Xwx2Zz2z1Yy1cw would be odd and without chords, since it would be obtainedfrom the even (since chordless) cycle Xx1x2Zz2z1x1 by replacing the part z1Xx1w

by the part z1Yy1cw, which has different parity. Thus, let d be the neighbor of wclosest to z1. The cycle Yy1cwZdz1y1 either is a triangle (so we are done) or is even(since it has no chords), and in this last case the parities of the paths Yy1c, Zz1d aredifferent.

But then the parities of the paths Xx1w, Zz1d are different (since Xx1w, Yy1c

have the same parity), and the cycle Xx1wZdz1x1 is odd and has no chords, acontradiction.

So, the only possibility is that w is adjacent to x1, y1, z1. If w = x2, the claimis proved. Otherwise, a similar reasoning shows that w is adjacent to x2, y2, z2 andthe claim is proved.

Now, we can show the following.

Theorem 2. Diamond-free graphs with no odd hole and no stretcher are perfectlycontractile.

Proof of Theorem 2. Let G be such a graph and consider the following al-gorithm (a suitable even pair is an even pair whose contraction does not create adiamond in the resulting graph):

Algorithm CONTRACT

1. G′: = G;G′′: = G;

2. while G′ is not a clique do

3. let x be a special vertex of G′;4. if NG′(x) induces a clique then

5. G′: = G′ − x;

6. else

7. let (a, b) be a suitable even pair of G′;8. G′: = G′ab;G

′′: = G′′ab.

Denote by integers going from 1 to k the iterations of the while loop in step 2.We prove the following by induction on r (r = 1, 2, . . . , k).

Claim 5. At any iteration r of the while loop in step 2 the following statementshold:

1. the suitable even pair in step 7 exists;2. the graph G′ obtained at the end of step 8 is a diamond-free graph with no

odd hole and no stretcher.


Proof of Claim 5. By hypothesis (in the case r = 1) and by induction hypo-thesis (in the case r > 1) we have that, just before starting the r-th iteration, thegraph G′ has no diamond, no odd hole, and no stretcher. Obviously, the removaloperation in line 5 does not change the nature of the graph. But a contraction canchange it, as shown by Claim 3.

In order to find an even pair that does not change the nature of the graph aftercontraction, notice that, in line 7 of the algorithm, NG′(x) induces a stable set anddoes not induce a clique, thus NG′(x) contains at least two nonadjacent verticesu and v. They form an even pair, otherwise the odd chordless path joining themtogether with xwould give an odd cycle containing at least one odd hole (by Claim1, since all the chords must contain x, and x is in no triangle).

Case 1. If the contraction of u, v does not create a diamond, then we renameu, v with a, b.

Case 2. If a diamond appears, there exist vertices q, t, s such that u, v (insteadof a, b), q, t, s induce the structure in Fig. 2. Then x /= t and xt /∈ E (since x iscontained in no triangle). We claim that (x, t) is an even pair. To prove this, weonly have to apply Claim 2 as follows.

In our case, G′ has no odd hole, and the vertex v is contained in a chordlesscycle of length four. Then by Claim 2 we obtain that statement 1 is true, sinceG′ has no stretcher (by hypothesis) and no expansive cycle (such a cycle con-tains a diamond). Thus, x and t form an even pair. Moreover, its contraction doesnot yield a diamond, since x is in no triangle. Then, rename x, t, respectively,with a, b.

In both cases before, we have obtained an even pair (a, b) whose contractioncreates no diamond (and no odd hole, since a, b are joined by no odd chordlesspath). We only have to show that its contraction creates no stretcher either. Recallthat, both in Case 1 and Case 2, a, b have at least one neighbor in common, whichwe denote w.

By contradiction, assume a stretcher is obtained after contraction; so let x1y1z1,x2y2z2 be the two triangles of a stretcher S induced in G′ab, and denote X,Y, Zits three chordless paths joining, respectively, x1 to x2, y1 to y2, z1 to z2. AsG′ contains no stretcher, one of the vertices of S is the vertex c resulting aftercontraction.

—If c is an internal vertex of a path amongX,Y, Z, then each of a, b is adjacentinG′ to exactly one neighbor of c on S (otherwiseG′ contains a stretcher), so usingw to join a and b we have in G′ a quasi-stretcher whose chords have a commonendpoint w.

—If c is some endpoint of a path, we can suppose it is x1. At most one of thevertices y1, z1 is adjacent to both a and b in G′ (otherwise we have a diamond). Ifsuch a vertex exists, assume it is y1. Now, z1 must be adjacent to exactly one ofa, b, say a (thus, z1b /∈ E by the preceding assumption). Then y1 must be adjacentto a, otherwise it is adjacent only to b and az1y1b is an odd path joining a and b inG′, a contradiction. Now, b must be adjacent to the neighbor d of x1 on X , since a


cannot be adjacent to the three neighbors of x1 on S (a stretcher would then existin G′). In this case, if we join a, b using w, we have in G′ a quasi-stretcher withtriangles ay1z1, x2y2z2 and paths X ′ : awbXdx2 , respectively, Y, Z, where all thechords, except possibly for y1b, contain w. But, in fact, the chord y1b cannot exist,since then the cycle X ′bx2

Zz2z1y1b is odd and has no chord.So, independently of the position of the vertex c on S in G′ab, in G′ we have a

quasi-stretcher whose chords have a common vertex w. By Claim 4, w is adjacentto every vertex in every triangle (since it is none of these vertices). In Case 1 before(i.e., where the even pair (a, b) consists of two vertices in the neighborhood of thespecial vertex x), w is identical to x and is contained in no triangle, so we havea contradiction. In Case 2 before, vertices a, b are on a chordless cycle on fourvertices x = a, v, t = b, u, so both u and v can be considered instead of w. ByClaim 4, they are both adjacent to every vertex in every triangle, so G′ contains adiamond, a contradiction.

Claim 5 is proved.

Now, denote by (ar, br) (r = 1, 2, . . . , k) the suitable even pair found in step7 during the r-th iteration. We know that it is an even pair in the graph G′, but G′is obtained by contractions and removals. To prove that G is perfectly contractile,we need to show that these pairs are also even pairs in the graph G′′ obtainedexclusively by contractions.

Claim 6. For an arbitrary iteration r (r = 1, 2, . . . , k), the pair (ar, br) is aneven pair in G′′, too. Moreover, at the end of the algorithm, G′′ is a triangulatedgraph, i.e., it contains no chordless cycle on at least four vertices.

Proof of Claim 6. We use induction on r to prove that in step 7 every pair(ar, br) is an even pair of the current graph G′′, too.

If (a1, b1) is not an even pair of G′′ (which is equal to G at this moment of theexecution), then an odd chordless path P of G′′ joins a1, b1. As (a1, b1) is an evenpair of G′, P must contain some vertices that have been removed from G′. Then,the vertex z with this property that was removed earliest had, before its removal,at least two nonadjacent vertices in its neighborhood (its neighbors on P ). Thiscontradicts the choice of the vertices we remove in step 5.

Now, suppose that the statement is true for all the pairs (aj , bj) (1 ≤ j < r).In the graph G′′ obtained after contracting all these pairs, if (ar, br) is not an evenpair of G′′, then an odd chordless path P ′ of G′′ joins ar, br. As before, we find avertex that has been removed although its neighborhood was not a complete graph,a contradiction.

It remains to show that G′′ is a triangulated graph. By contradiction, supposethat it contains at least one chordless cycle C with at least four vertices. Then atmost one vertex of C is in G′ (which is now a clique), the others are in G′′ − G′,i.e., they have been removed in step 5. Take the one that has been removed earliest.Then at the moment of its removal, in its neighborhood in G′ at least two verticeswere nonadjacent: its neighbors on C. We have a contradiction with the conditionthat NG′(x) induces a clique.


Theorem 2 is now proved, since triangulated graphs are perfectly contractile, asshown in [7] (in fact, Hertz proves that Meyniel graphs, which contain triangulatedgraphs, are perfectly contractile).

3. THEOREM 1—OUTLINE OF THE PROOF

We give here the definitions and the main steps that must be performed to proveTheorem 1. The proofs of the intermediate lemmas are presented in detail in Section4. The reader is strongly encouraged to carefully read this section before startingthe next one.

All along in the rest of the article, G will be a diamond-free graph with no oddhole and no stretcher. Moreover, as the proof of Theorem 1 is done by contradiction,G will be supposed to have no vertex whose neighborhood induces a clique or astable set.

Notice that, since G is diamond-free, for every edge xy of G there exists aunique clique containing both x and y. We define a sound path to be a chordlesspath P : v1v2 · · · vp such that, for every l = 2, 3, . . . , p, each vertex x ∈ N(vl) −N(vl−1) has exactly one of the properties below:

(P1) vlx is in no triangle;(P2) vlx is in some triangle, but x is in no Tj (j < l − 2),

where Tj = N(vj) ∩ N(vj+1) (then Tj ∪ {vj , vj+1} is the unique clique con-taining both vj , vj+1). Obviously, every vertex v1 can start a sound path.

Remark 1. According to the definition, whenever a vertex u ∈ N(vl) −N(vl−1) does not extend the sound path Pv1vl , exactly one of the following state-ments holds:

i) u has a neighbor on Pv1vl−2 ;ii) u has no neighbor on Pv1vl−2 , but it has a neighbor s /∈ N(vl) such that us is

in some triangle and s is in some Tj (j < l − 1).

Remark 2. The preceding remark allows us to say that whenever a vertexu ∈ N(vl) − N(vl−1) does not extend the sound path Pv1vl , a chordless path(edge disjoint from P ) of length 1 or 2 exists joining u to some vertex vi onPv1vl−1 : this path is uvi if (i) holds and usvi if (ii) holds. Obviously, in this lastcase, s induces a triangle with two consecutive vertices on Pv1vl−1 (by (ii), s is insome Tj).

Remark 3. By the definition of a sound path, no vertex of G can be in morethan one set Tj , (1 ≤ j ≤ p− 1). Indeed, if some vertex w is in Ta ∩ Tb (1 ≤ a <b ≤ p − 1), then a < b − 1 (otherwise we have a diamond) and neither (P1) nor(P2) in the definition of a sound path is verified for vb and w ∈ N(vb)−N(vb−1).

The index k such that Tk−1 /= ∅ and k is maximum with this property is calledthe s-length of the sound path. All along in the rest of the article, P : v1v2 · · · vpwill be a sound path with maximum s-length, and as many edges as possible.


To prove Theorem 1, we will firstly show the following.

Lemma 1. For the sound path P (chosen as indicated), the s-length k is equalto p.

In other words, the last edge of the path is on a triangle. Then, we considerv′k ∈ Tk−1, and notice that the sound path Pv1vk−1 could be extended by v′k, butit cannot be extended beyond v′k (otherwise, the new path would have either ans-length larger than the one of P , or the same s-length, but larger number of edges,a contradiction).

Now, if v′k does not extend the sound path Pv1vk−1 , by Remark 2 there exists achordless path W : w0 = v′k, vi (of length 1) or W : w0 = v′k, w1, vi (of length 2)joining v′k to some vertex vi on Pv1vk−2 .

If, on the contrary, v′k extends the sound path Pv1vk−1 , then recall the hypothesisthat (by contradiction) v′k is not a special vertex, thus its neighborhood is not iden-tical to the clique Tk−1 ∪ {vk−1, vk} − {v′k}. Then, v′k has a neighbor w1 not inN(vk−1) (and thus not in N(vk), otherwise a diamond may be found), which doesnot extend the sound path Pv1vk−1v

′k. Consequently, by Remark 2, there exists a

chordless pathw1vi (of length 1) orw1w2vi (of length 2) joiningw1 to some vertexvi on Pv1vk−1 . If we add to this path (of length 1 or 2) the vertex v′k, we obtaineither a path W : w0 = v′k, w1, vi (of length 2) or a path W : w0 = v′k, w1, w2, vi(of length 3) joining v′k to some vertex vi on Pv1vk−1 .

Then we have the following.

Remark 4. If P is a sound path with maximum s-length and as many edges aspossible, then a path W : w0 = v′k, . . . , wh, vi(0 ≤ h ≤ 2), edge disjoint from P ,exists joining v′k to some vertex vi of Pv1vk−1 (i is assumed as large as possible).Similarly (but taking into account the fact that vk extends the sound path Pv1vk−1),we have that a chordless path U : u0 = vk, . . . , ur, vj (1 ≤ r ≤ 2) exists joiningvk to some vertex vj of Pv1vk−1 (j is assumed as large as possible). Moreover, byRemark 2, whenever w2 (respectively, u2) exists, it induces a triangle with twoconsecutive vertices on Pv1vk−1 .

The criterion of shortness. We can assume that no internal vertex of the pathW , respectively, U , has a neighbor on P by choosing the shortest path joining v′k,respectively, vk to some vertex on Pv1vk−1 (consequently W has no chord, exceptpossibly for viv′k, and this can occur only if i = k− 1; similarly for U ). Moreover,we can assume that among all the vertices in Tk−1, we chose v′k to be the vertexthat gives the shortest path W . This criterion for the choice of W and U will becalled the criterion of shortness.

Now, say that the sound path P : v1v2 · · · vk−1vk has a regular construction, ifthere exists a chordless path vks1s2 with s1, s2 /∈ V (P ) such that

(R1) s1 has no neighbor on P but vk;(R2) s2vk−1, s2vk−2 ∈ E;(R3) there exists a vertex s3 such that s1s2s3 is a triangle;(R4) the path v1 · · · vk−2s2s1 is a sound path.


The construction is called semi-regular if only (R1) and (R2) hold (notice that wemust have s2vk /∈ E, otherwise s2, vk−2, vk−1, vk induce a diamond). Obviously,when (R3) holds, the vertex s3 is not on P (otherwise (R1) is contradicted).

Then we have the following (see Fig. 3).

Lemma 2. If P,W,U are as defined, then either P has a regular constructionusing u1, u2, or v′k extends Pv1vk−1 to a sound path P ′ : Pv1vk−1v

′k, which has a

regular construction using w1, w2.

This lemma shows that we can always find a sound path with maximum s-lengthand as many edges as possible, which has a regular construction. Without lossof generality, we will suppose that this path is P itself. To find a contradictionin this case (thus proving Theorem 1), it is sufficient to show that there existsa chordless path joining v′k and u3 that induces with vk−1, vk, u1, u2 a stretcher.This will be done using successive regular constructions, in the following way.Denote u(0)

2 = vk−1, u(0)3 = vk, u

(0)1 = v′k, u

(1)1 = u1, u

(1)2 = u2, u

(1)3 = u3.

As long as possible, do the following two steps (the current index is i ≥ 1; seeFig. 4):

Step 1. Given the sound path Pv1vk−2u(i)2 u

(i)1 , we find u(i)

3 ∈ N(u(i)2 )∩N(u(i)

1 )such that Pv1vk−2u

(i)2 u

(i)3 is a sound path.

Step 2. Find a regular construction on Pv1vk−2u(i)2 u

(i)3 using two vertices

u(i+1)1 , u

(i+1)2 .

The execution of these steps can be finite or not (this last case could appear ifsome vertices are considered an infinite number of times, for different indices i).We wish to show that, in fact, the infinite execution is not possible. To this end,in the following we consider integers s, j, l, q for which the vertices u(s)

1 , u(s)2 , u

(s)3

(and similarly for the others) have been defined using Steps 1 and 2.Denote Ss = {u(s)

1 , u(s)2 , u

(s)3 } and, for j < l, let the multi-set Sjl be defined

as Sjl = ∪ls=jSs (so, in Sjl an element can appear several times, if it is containedin several sets Ss). The multi-set Sjl is said to induce an extended stretcher, if thetwo conditions below hold:

FIGURE 3. Regular constructions on P and P ′.


FIGURE 4. Successive regular constructions.

a) the vertices in Sjl are pairwise distinct, and distinct from the vertices onPv1vk−2 ;

b) the set of edges with both extremities inSjl is{u(a)1 u

(a)2 , u

(a)2 u

(a)3 , u

(a)3 u

(a)1 , j ≤

a ≤ l} ∪ {u(a)3 u

(a+1)1 , j ≤ a ≤ l − 1} ∪ {u(a)

2 u(b)2 , j ≤ a < b ≤ l} (see

Fig. 4).

Then we can show that:

Lemma 3. For a fixed q ≥ 1, the multi-set S0q induces an extended stretcher.

Since the graph has a finite number of vertices, there will exist some index fsuch that the path v1 · · · vk−2u

(f)2 u

(f)3 either is not a sound path, or it is a sound

path but has no more regular construction (see Fig. 4). In any case, Remark 4 forthe path Pv1vk−2u

(f)2 u

(f)1 (where u(f)

3 plays the role of v′k) guarantees the existence

of a chordless path U (f+1) joining u(f)3 to some vertex of Pv1vk−2u

(f)2 . As before,

we assume that U (f+1) is chosen following the criterion of shortness.Steps 1 and 2 have been described starting with a path P , which has maximum

s-length and as many edges as possible. We can remark now that, in fact, it can bechosen in a more particular way. Indeed, inG there exist sound pathsP (that we callextremal sound paths) with same s-length and same number of edges asP for whichin Lemma 2 only the regular construction on P exists (i.e., independently of thechoice of the vertex similar to v′k, the path P ′ admits no regular construction). The

path Pv1vk−2u(f)2 u

(f)1 is such a path. There exists no u(f)

3 such that Pv1vk−2u(f)2 u

(f)3

admits a regular construction, otherwise the successive constructions would nothave been stopped.

Now, if we start the execution of Steps 1, 2 with an extremal sound path (forwhich we preserve the notation P ), we can still perform the successive contractionsin order to arrive to another extremal path (for which we preserve the notation


Pv1vk−2u(f)2 u

(f)1 ). We can still find the path U (f+1) and, moreover, we can be sure

that the path W given by Remark 4 for the path P and v′k (using the criterion ofshortness) does not form a regular construction onP ′ : Pv1vk−1v

′k. These properties

will be used to show the following.

Lemma 4. There exist two pathsW, respectively,U , joining v′k (respectively,u(f)3 )

to some vertex va (respectively, vb) on Pv1vk−3 and such that the internal vertices

of W, respectively, of U , are nonadjacent to every vertex in {u(i)1 , u

(i)2 , u

(i)3 , 0 ≤

i ≤ f}.Then the trianglesvk−1vku

′k, u

(1)2 u

(1)1 u

(1)3 give a stretcher with the pathsvk−1u

(1)2 ,

vku(1)1 and a chordless path extracted fromWv′

kvaPvavbUvbu(f)

3u

(f)1 u

(f−1)3 u

(f−1)1 · · ·

u(2)1 u

(1)3 . Obviously, this contradicts the hypothesis thatG has no stretcher, so The-

orem 1 is proved.

4. DETAILS OF THE PROOF

The proofs of the four lemmas do not use very complicated arguments, but requirethe analysis of a large number of cases, which can appear at different stages of thereasoning. We start by a claim that will be frequently used in our proof.

Claim 7. Let S be a quasi-stretcher in G, with triangles x1y1z1, x2y2z2 andpathsX,Y, Z. IfS contains at most one more triangle (in addition tox1y1z1, x2y2z2),then this triangle really exists, and it has exactly one vertex on each path X,Y, Zand on each triangle x1y1z1, x2y2z2 (see Fig. 5).

Proof of Claim 7. We firstly consider the case where no triangle exists in ad-dition to x1y1z1, x2y2z2 and we show that, in fact, it cannot appear (a contradictionis obtained). Then, if such a triangle exists, we prove it has the required properties.

Case 1. No triangle exists but x1y1z1, x2y2z2.Assume, without loss of generality, that there exists a chord cd joining the vertex

c ofX to the vertex d ofY (so cd /= x1y1, cd /= x2y2). We take c as close as possibleto x1 and, after the choice of c, we take d as close as possible to y1. Then the cycleXx1cYdy1x1 (of length at least four, otherwise we have another triangle) must be

FIGURE 5. Quasi-stretcher with unique additional triangle.


even (an odd hole would be induced otherwise), so the parities of Xx1c, Yy1d areidentical.

Now, if d /= y2, the cycle Xcx2Zz2z1Yy1dc is odd (it is obtained from the even,since triangle-free, cycleXx1x2Zz2z1x1 by replacing the part z1Xx1c with z1Yy1dc,which has different parity). Moreover, the indicated cycle has no triangle, so wehave a contradiction. If d = y2, we necessarily have d /= y1, and a reasoning similarto the previous one gives a contradiction.

Case 2. Precisely one triangle cde exists, which is different fromx1y1z1, x2y2z2.AsX,Y, Z are chordless paths, all the vertices c, d, e cannot lie on the same path.

Necessarily, there will exist a path (sayX) that will contain exactly one vertex (sayc) of the triangle.

We first prove that d, e cannot be on the same path among Y and Z. By con-tradiction and without loss of generality, assume that they are both on Y , with dbetween y1 (included) and e. Both the cycles Xx1cYdy1x1 and Xcx2Yy2ec are even(by Claim 1, since no triangle is induced inside), so the paths Xx1c, Yy1d have thesame parity, and the same holds for the paths Xcx2 , Yey2 , thus the paths X and Yhave different parities. Moreover, Y and Z have the same parity, since no triangleis induced by the chords of the cycle Yy1y2Zz2z1y1, which must be even. Then, XandZ have different parities, which implies that the cycleXx1x2Zz2z1x1 is odd andits chords form no triangle. Then it contains an odd hole, a contradiction.

Now, the three vertices lie on the three paths: c on X, d on Y, e on Z. If none ofthem is in the set {x1, y1, z1}, then we are in the situation of Case 1 for the trianglesx1y1z1, cde and the paths Xx1c, Yy1d, Zz1e (which cannot induce a stretcher, byhypothesis), so we have a contradiction. Thus, cde has at least one vertex on thetriangle x1y1z1. A similar reasoning shows that cde has at least one vertex on thetriangle x2y2z2.

Claim 7 is proved.

We are going to prove the first lemma.Lemma 1. For the sound path P : v1v2 · · · vp, the s-length k is equal to p.

Proof of Lemma 1. If this is not the case, consider the vertex vk+1 and noticethat vkvk+1 is in no triangle (since k is the s-length of P ). As N(vk+1) cannotbe a stable set (we have supposed by contradiction that G has no special vertex),vk+1 must be contained in a triangle vk+1tu such that none of t, u extends thesound path v1v2 · · · vk+1 (if this was the case, the path v1v2 · · · vk+1t, respectively,v1v2 · · · vk+1u would have larger s-length than P , a contradiction). But this is notpossible, by the following claim.

Claim 8. Let vs be a vertex of P (1 ≤ s ≤ p) such that vs is in some tri-angle vscd, with c, d /∈ V (Pv1vs) ∪ Ts−1. Then either c or d extends the soundpath Pv1vs .

Proof of Claim 8. If this is not the case, then by Remark 2 for c (respectively,for d) there exists a pathC (respectively,D), of length 1 or 2, joining them to somevertex of Pv1vs−1 . Denote vi (respectively, vj) the vertex of Pv1vs−1 , which is the


other endpoint of C (respectively, of D) and assume vi, vj are chosen as close aspossible to vs. If the pathC (respectively,D) has length 2, denote by c′ (respectively,d′) the internal vertex. We can assume that (whenever c′ exists) c′vs /∈ E, otherwisethe vertices c′, c, d, vs (which cannot induce a diamond) imply that dc′ ∈ E, andwe can make our reasoning for c′, d instead of c, d, with the advantage that the pathjoining c′ to a vertex of P has no more internal vertex. Similarly, we can assumethat d′vs /∈ E (whenever d′ exists).

Then the cycles PvivsCcvi , respectively PvjvsDdvj must be even, since they haveno chords. This implies that the parities of the pathsCcviPvivj , cPvsvj , dPvsvj , Ddvjare the same (it is sufficient to consider them by pairs in this order and to count), sothe path CcviPvivjDvjd is even. Therefore, the cycle with these vertices (obtainedusing the edge cd) is odd, so it must contain at least one triangle. Without loss ofgenerality, we will treat only the case i < j.

Notice that (whenever the indicated vertices exist) d′ /= c′ and d′c, c′d /∈ E(otherwise, we have a diamond). By the choice of the paths C,D, the only vertexamong c, d, c′, d′ that can form chords with some vertex on Pvivj is the last ver-tex on D (d, if d′ does not exist, respectively, d′ in the contrary case). So everytriangle in the cycle CcviPvivjDvjdc must contain this vertex. We deduce that thislast vertex is not d (by the property (P2) for vs), so d′ exists and all the trian-gles in the odd cycle CcviPvivjDvjdc contain d′. Moreover, there are at most twosuch triangles: one containing two consecutive vertices on Pvivj (by Remark 3,d′ cannot be in more than one Ts, 1 ≤ s ≤ k) and (if c′ exists) d′vic′. Let d′xybe one of these triangles such that x is on Pvivj (as close as possible to vj) andy is either on Pvivj (in the first case) or equal to c′ (in the second case). The tri-angles d′xy, dvsc and the paths d′d, Pxvs , PyviCvic (in the first case), respectively,the paths d′d, Pxvs , c′c (in the second case) form a quasi-stretcher that satisfiesthe hypothesis of Claim 7. Then, by Claim 7 we deduce that the second trianglecontaining d′ really exists, and it has a vertex on the triangle dvsc. But this isimpossible.

Lemma 1 is proved.

The following claim will be intensely used in the proof of Lemma 2. The soundpath P below is not necessarily of maximum s-length, nor with maximum numberof edges. Still, we use the notation vi for its vertices in order to avoid too complicated(and too different) names for objects that are almost the same.

Claim 9. Let P = v1v2 · · · vl be a sound path such that vl−1vl is in sometriangle vl−1vlv

′l. And let X : x0 = v′l, . . . , xm, vi (0 ≤ m ≤ 3), Y : y0 =

vl, y1, . . . , yn, vj (1 ≤ n ≤ 3) be two paths (with i, j < l taken as large aspossible) such that the following conditions hold:

(C1) X,Y are edge-disjoint from P, and no internal vertex of X or Y is on P ;(C2) every chord of the path Pv1vl−1Xv′

lxm

(respectively, of the path Pv1vl−1

Yvlyn) is of type xmvq with 1 ≤ q ≤ i (respectively, of type ynvs, with 1 ≤s ≤ j);

(C3) x2vl /∈ E, y2v′l /∈ E (whenever x2, respectively, y2, exists).


Then we have:

(S1) (V (X)− {vi}) ∩ (V (Y )− {vj}) = ∅, and the only possible edges betweenthe two pathsX− vi, Y − vj are y1x1, y2x2, the last one being possible onlyif m = 2 and n = 2;

(S2) either m = 2 and x2vl−1 ∈ E (i.e., i = l − 1), or n = 2 and y2vl−1 in E(i.e., j = l − 1); furthermore, if m = n = 2 and y2x2 ∈ E, then we haveboth x2vl−1 ∈ E, y2vl−1 ∈ E;

(S3) if, moreover, when m = 2 (respectively, when n = 2) the vertex x2 (respec-tively, y2) induces a triangle with two consecutive vertices onP , then at leastone of P ′ : v1v2 · · · vl−1v

′l and P has a semi-regular construction (which

uses the vertices x1, x2 in the case of P ′ and the vertices y1, y2 in the caseof P). Furthermore, if n = m = 2 and y2x2 ∈ E, then both P ′ and P havesemi-regular constructions (using x1, x2, respectively, y1, y2).

Proof of Claim 9. We are going to show that the vertices inX− vi and Y − vjare distinct; by convention, whenever we treat a vertex xq or yq we understand ‘‘ifthis vertex exists.’’

By (C2), y1 /= v′l and y1vl−1, y1v′l /∈ E. Similar statements hold for x1. Con-

sequently, x1 /= y1 because of v′l (which is adjacent to x1 but not to y1). Wedo not need to prove that y1x1 /∈ E (since in (S1) this chord is accepted) sowe inquire about y1x2. Before that, we remark that x3vl, y3v

′l /∈ E (and sim-

ilarly x3vl−1, y3vl−1 /∈ E), because of the cycles x3x2x1v′lvlx3, respectively,

y3y2y1vlv′ly3, which would then be odd and chordless. Now, we successively con-

sider the vertices on the path Y and show they are distinct and nonadjacent (exceptfor the indicated cases) to the vertices on X .

(A) y1 /= x2 and y1x2 /∈ E.Indeed, y1 /= x2 because of vl (by (C3), x2vl /∈ E). If y1x2 ∈ E, then the cycle

x2x1v′lvly1 is odd and the only chord it can have is y1x1. But then the triangles

v′lvl−1vl, x1x2y1 and the paths v′lx1,Pvl−1viXvix2 , vly1 induce a quasi-stretcherwhose chords form at most one triangle (given by y1 and two consecutive verticesofPvivl−1 ; notice that y1x3 /∈ E, since then y1, x3, x2, x1 would induce a diamond).Claim 7 may be applied to obtain a contradiction.

(B) y1 /= x3 and y1x3 /∈ E.By (A), y1 /= x3 because of x2, which is adjacent to x3 but not to y1. If, by

contradiction, y1x3 ∈ E, then the cycle x3Pvivly1x3 is odd, since the parity ofPvivl is the same as the parity ofPvivl−1v

′l, and the same as the parity of vix3x2x1v

′l

(the cycle given by X and Pvivl−1 must be even as it has no chord). Recall alsothat x3vl /∈ E, so all the chords in the cycle x3Pvivly1x3 contain y1 (and this canhappen only if n = 1). Thus, y1 must form a triangle y1ztwhose two other verticesare on the path x3Pvivl−2 . Claim 7 for the quasi-stretcher given by the trianglesvlv′lvl−1, y1zt and the paths that can be found along Y,XP and, respectively, P

give a contradiction.


Now, it is easy to see that y2 /= v′l, so we inquire about y2x1.

(C) y2 /= x1 and y2x1 /∈ E.Since y2v

′l /∈ E andx1v

′l ∈ E, we already have that y2 /= x1. If, by contradiction,

y2x1 ∈ E, then the cycle x1v′lvly1y2x1 is odd and the only possible chord is

y1x1. Claim 7 for the quasi-stretcher given by the triangles x1y2y1, v′lvl−1vl and

the paths x1v′l, Yy2vjPvjvl−1 , y1vl gives a contradiction, since all the additional

triangles contain x1 and have the other two vertices on Pvjvl−1 (thus, once again,there exists at most one such triangle, since two setsTa, Tb, a /= b do not intersect byRemark 3).

Let us now treat x3 before x2:

(D) y2 /= x3 and y2x3 /∈ E.The inequality y2 /= x3 is guaranteed by y1. If y2x3 ∈ E, then the cycle

x3x2x1v′lvly1y2x3 is odd and the only possible chords are y1x1, y2x2 (once again,

x3vl /∈ E, by a previous remark). The first chord is useless (it is contained inno triangle), while the second one gives a triangle x2x3y2, which, together withv′lvl−1vl and the paths x2x1v

′l, x3Pvivl−1 , y2u1vl, gives a quasi-stretcher satisfying

the hypothesis of Claim 7 (all the triangles contain y2, if n = 2, and y2 cannot bein more than one set Ta). But Claim 7 yields a contradiction.

(E) y2 /= x2 and if y2x2 ∈ E, then n = m = 2, x2vl−1 ∈ E, y2vl−1 ∈ E.Statement (D) guarantees that y2 /= x2 because of x3. If y2x2 ∈ E, consider the

cycle x2x1v′lPvl−1vjYvjy2x2, which is odd since it is obtained from the even cycle

x2x1v′lvly1y2x2 by replacing the odd path v′lvly1y2 by the path v′lPvl−1vjYvjy2 ,

which is even, like the path vlPvl−1vjYvjy2 (which is completed in the even cyclePvlvjYvjvl by the even path vly1y2). In the indicated odd cycle, all the trianglescontain x2 (they are either of the form x2vsvs+1, or x2y2z, where z is y3 or vjdepending on the existence or not of y3; notice that y3x2x1 cannot be a triangle,since then y3, x2, x1, y2 would induce a diamond). By considering one of thesetriangles, the triangle v′lvl−1vl and the corresponding paths joining pairs of vertices,we can apply Claim 7 to obtain a contradiction, except if the triangle we chooseis not disjoint from v′lvl−1vl. This can happen only if x2vl−1 ∈ E (so m = 2 andi = l−1), and x2vl−1 forms a triangle either with y2 (so n = 2) or with vl−2. In thefirst case, (E) is proved. In the second one, x2vl−1vly1y2x2 is an odd cycle, whichimplies y2vl−1 ∈ E (so n = 2 and (E) is true).

We are going to investigate now the vertex y3. Notice that we cannot have x2y2 ∈E, since then, by (E), we should also have n = 2 and this is not the case.

(F) y3 /= x1 and y3x1 /∈ E.Because of (C), x1 is nonadjacent to y2, while y3 is adjacent to y2, so y3 /= x1.

Moreover, if x1y3 ∈ E, then the cycle y3Pvjvl−1v′lx1y3 is odd, since y3Pvjvl−1v

′l

has the same parity as y3Pvjvl and as vly1y2y3, i.e., odd. The only possibility for this


cycle to contain a triangle is that m = 1 and x1 induces some triangle either withy3, vj or with two consecutive vertices on Pvjvl−1 . Anyway, x1 will be containedin at most two triangles, and we can build a quasi-stretcher by considering one ofthem and the triangle v′lvl−1vl. Then the hypothesis of Claim 7 is verified and wehave a contradiction.

(G) y3 /= x2 and y3x2 /∈ E.The first assertion is guaranteed by the vertex y2, while the second one must

be necessarily true, otherwise the cycle x2x1v′lvly1y2y3x2 is odd and contains no

triangle.

(H) y3 /= x3 and y3x3 /∈ E.Indeed, y3 /= x3 since y3x2 ∈ E, while x3x2 /∈ E. If y3x3 ∈ E, then the cycle

y3Pvjvl−1v′lx1x2x3y3 is odd and, like in (F), we obtain a contradiction.

Statement (S1) is proved. To prove (S2), we can assume that y2x2 /∈ E, otherwiseby (E) we are done. Now, the parities of the pathsXv′

lviPvivj ,Pvjvl−1v

′l are the same

(the cycle formed by these two paths is even since it is chordless), and the same asthe parities of the paths Pvjvl , Yvjvl . Then the path Xv′

lviPvivjYvjvl is even, so the

cycle with the same vertices is odd, therefore, it must contain some triangle. Twocases can occur.

Case 1. The triangle v′lvl−1vl is not a triangle of this cycle (equivalently, Pvivjdoes not contain vl−1). Then, except for the possible chord y1x1, all the chordscontain either xm (if j < i) or yn (if i < j). Without loss of generality, we assumethat j < i, so that all the chords but y1x1 contain xm. As xm can form at most onetriangle with consecutive vertices onP , Claim 7 can be applied for this triangle, thetriangle v′lvl−1vl and the paths joining pairs of vertices along Y,P and, respectively,X . Notice that these triangles must be disjoint, since Pvivj does not contain vl−1,i.e., i < l − 1. We have a contradiction.

Case 2. The triangle v′lvl−1vl is a triangle of this cycle (equivalently, the largestof i and j is equal to l− 1). None of x3, x1, y3, y1 (whenever they exist) is adjacentto vl−1 (we would have either an odd chordless cycle on five vertices or a diamond),so, if i = l − 1, then m = 2, while if j = l − 1, then n = 2. Statement (S2) isproved.

It remains to prove (S3). By (S2), we have that either m = 2 (and i = l − 1) orn = 2 (and j = l − 1).

Case y2x2 ∈ E (and thus m = n = 2). By (S2), we have that i = j = l− 1, sox2y2vl−1 is a triangle containing both x2vl−1 and y2vl−1. If the triangle x2vsvs+1induced by x2 with two consecutive vertices on P is not identical to x2vl−2vl−1,then s < l − 3. Therefore, for x2 ∈ N(vl−1)−N(vl−2), the edge vl−1x2 satisfiesnone of the conditions (P1), (P2) in the definition of a sound path, since vl−1x2 isin a triangle and x2 ∈ Ts. We have a contradiction, thus x2vl−2 ∈ E must hold.Similarly, y2vl−2 ∈ E must hold.

Case y2x2 /∈ E. We treat only the case n = 2, 0 ≤ m ≤ 3 (the reasoning issimilar for m = 2, 1 ≤ n ≤ 3).


We have that y2vl−1 ∈ E and that y2 (which induces a triangle with two con-secutive vertices on P) necessarily has at least one other neighbor on Pv1vl−2 .Denote vf such a neighbor that is closest to vl−2. By (S1), no other edge but possiblyx1y1 exists between the two paths X and Y . Then the two paths y2vl−1Xv′

lxm

andy2y1vlXv′

lxm

(or y2y1Xx1xm , if x1y1 ∈ E) are chordless paths of different paritiesjoining y2 and xm.

To each of these paths we can try to add a chordless path extracted fromy2Pvfvixmin order to obtain two cycles of different parities.

• If vi /= vl−1, then this can be correctly done, since the vertices of the twopaths that are joined each time to form a cycle are distinct. As no vertex buty2 and xm has neighbors on Pv1vl−1 , and the subpath of y2Pvfvixm is chosento have no chords, these cycles will be chordless, except if vl−2 is on Pvfvi(in this case, the chord vl−2vl−1 will exist in the cycle formed with the pathy2vl−1Xv′

lxm

). Obviously, both of the cycles cannot be chordless, since theyhave different parities and are of length at least five (an odd hole would thenexist G). So, we must have the chord vl−2vl−1 in the cycle formed with theindicated path, i.e., either vf or vi is vl−2. If vf = vl−2, then P has a semi-regular construction using y1, y2 and we are done. Otherwise, i.e., if vi = vl−2and vf /= vl−2, then the odd cycle (among the two cycles of different parities)still contains no triangle, so it contains an odd hole, a contradiction.

• If vi = vl−1, then m = 2 (otherwise either a diamond or an odd chordlesscycle is induced), so, by hypothesis,x2 also forms a triangle with two adjacentvertices of P . We show that it is not possible to have both x2vl−2 /∈ E andy2vl−2 /∈ E. If this was the case and vf , vg (f, g < l − 2) are the neighborsof y2, respectively, x2, closest to vl−2, then by adding to each of the pathsx2vl−1y2, x2x1v

′lvly1y2 (of different parity) a chordless subpath of the path

x2Pvgvf y2, we obtain two chordless cycles of different parities, one of whichis an odd hole, a contradiction. Then at least one of x2 and v2 is adjacent tovl−2 and we are done.

Claim 9 is proved.

We now prove Lemma 2.Lemma 2. If P,W,U are as defined, then either P has a regular constructionusing u1, u2, or v′k extends v1v2 · · · vk−1 to a sound path P ′ : Pv1vk−1v

′k, which

has a regular construction using w1, w2 (see Fig. 3).

Proof of Lemma 2. Recall that, by Remark 4 for the paths W : w0 =v′k, . . . , wh, vi, U : u0 = vk, . . . , ur, vj , we have 0 ≤ h ≤ 2, 1 ≤ r ≤ 2 and,whenever it exists, each of u2, w2 induces a triangle with two consecutive verticeson P . Moreover, vi, vj are taken such that the indices i, j < k are as large aspossible.

Now, the conditions (C1), (C2) in Claim 9 are satisfied for P,W,U instead ofP, X, Y and, moreover, the supplementary condition in (S3) is satisfied, too. Toapply Claim 9, we need to verify the condition (C3).


Notice that the condition (C3) is similar for u2 and w2. Indeed, whenever w2exists, v′k extends the sound path v1 · · · vk−1 (otherwise (i) or (ii) in Remark 1 forv′k would imply at most the existence of w1, but in no case the existence of w2). SoP and P ′ play similar roles; therefore, u2 and w2 play similar roles. Therefore, weonly prove that u2v

′k /∈ E.

Suppose, by contradiction, that u2v′k ∈ E. As Pvjvk−1 is even (because of the

chordless even cycle Pvjvku1u2vj), the cycle Pvjvk−1v′ku2vj is odd, therefore it

must contain at least one triangle. It is easy to see that all the triangles contain v′kand, since Tk−1 cannot intersect some other Ta by Remark 3, the only possibilityis to have v′kvj ∈ E (so that v′ku2vj is a triangle). Now, let vs, vs+1 be the twoconsecutive vertices of P , which are adjacent to u2. By the definition of j, wehave s + 1 ≤ j and we even deduce that s + 1 < j, otherwise the subgraphinduced by u2, vj−1, vj , v

′k (which cannot be a diamond) implies v′kvj−1 ∈ E, so

v′k ∈ Tj−1 ∩ Tk−1, a contradiction. Since s + 1 < j, we have that u2 is a vertexof N(vj) − N(vj−1) such that vju2 is in some triangle, vju2v

′k, and such that

u2 ∈ Ts, s < j. Property (P2) is contradicted for vj .We can now apply Claim 9 for P,W,U . By (S3), either P or P ′ : v1 · · · vk−1v

′k

has a semi-regular construction. Without loss of generality, we will suppose that Phas a semi-regular construction (this includes the case where both P, P ′ have semi-regular constructions); the reasoning is similar when P ′ has such a construction (itis even simpler, since the case h = 0, r = 2 that we are going to encounter herebelow has no equivalent when h = 2, since r ≥ 1). So u1, u2 exist, u1 has noneighbor on P while u2 is adjacent to vk−2, vk−1, and possibly other vertices on P .

We first inquire about the condition (R4) in the definition of a regular construc-tion, which is more difficult to solve than (R3). We have two cases treated in Claims10 and 11.

Claim 10. In the case where P has a semi-regular construction, but P ′ does nothave a semi-regular construction, Pv1vk−2u2u1 is a sound path.

Proof of Claim 10. By hypothesis, only P has a semi-regular construction, so,by Claim 9 for P,W,U we have that (whenever w2 exists) w2u2 /∈ E (otherwisethe statement (S1) implies that h = r = 2 and statement (S3) implies that bothP, P ′ have semi-regular constructions). We show first that u2 extends the soundpath Pv1vk−2 , and second that u1 extends the sound path Pv1vk−2u2.

Part 1. u2 extends the sound path Pv1vk−2 .Otherwise, by Remark 2, there exists a chordless path of length 1 or 2, which

joins u2 to some vertex vj (j < k − 2) that we take closest to k − 2. If this pathis of length 2, the internal vertex in N(u2) − N(vk−2) is called u4; by the sameremark, u4 is in some Ta, a < k − 1.

Then we can consider the sound pathPv1vk−1 such that vk−2vk−1 is on the trianglevk−2vk−1u2 and we apply Claim 9. In Claim 9, we will have Pv1vk−1 instead ofP, u2 instead of v′k, and the path joining u2 to vj instead of X (then m = 0 orm = 1 depending on which of (i) or (ii) in Remark 1 holds for u2). Instead of Ywe must take a path according to the value of i:


• If i /= k − 1, i.e., the path vk−1Wv′kvi

is chordless, then we take this pathinstead of Y (it will have 1, 2, or 3 internal vertices).

• If i = k − 1, then we also have h = 2 (so w2 is the last internal vertex ofW ) and, moreover, we know that w2 must have at least another neighbor vqon Pv1vk−2 , since w2 induces a triangle with two consecutive vertices on P .Furthermore, q, which is taken closest to k − 2, has the property q /= k − 2,since P ′ has no semi-regular construction, by hypothesis. Thus, we can takethe path vk−1w2vq (with one internal vertex) instead of Y .

Thus, in Claim 9, n has values 1, 2, or 3. We verify that (C1)–(C3) hold. We havenothing to prove for (C1), (C2). For (C3), as we have no vertex corresponding tox2, only the relation corresponding to y2v

′k /∈ E must be verified, i.e., w1u2 /∈ E.

This is true indeed, since for the paths U,W we have already applied Claim 9, and(S1) in this case guarantees that w1u2 /∈ E.

Now, Claim 9 implies by (S2) that at least one of the two paths has only thevertices with indices 0, 1, and 2. This is not possible for the one corresponding toX , so we must have it for the path corresponding to Y , i.e., w2 does not exist, sothat w1 is the last vertex ofW (and the path we chose instead of Y is vk−1v

′kw1vi).

Moreover w1vk−2 ∈ E.The path Pv1vk−2w1v

′kvk cannot be a sound path, since it would have larger s-

length than P (which is supposed to have maximum s-length). Thus, w1 or v′k orvk (in this order) satisfy (i) or (ii) in Remark 1 for the corresponding subpath.

• If w1 does not extend the sound path Pv1vk−2 , we have a path of length 1or 2 joining w1 to some vertex vi (i < k − 2) that we take closest to vk−2.The path cannot have length 1, since in this case the two paths (with differentparities) w1v

′kvk−1u2 and w1v

′kvku1u2 (or w1u1u2 if w1u1 ∈ E) may be

added to a chordless subpath of the path w1Pvivju2 (or w1Pvivju4u2 if u4exists) to obtain two cycles with different parities. Moreover, these cycles arechordless, since Claim 9 that we applied twice guarantees the existence of nochord, thus G contains an odd hole, a contradiction.

Thus, there must exist some vertex w4 ∈ N(w1) − N(vk−2) on the pathjoining w1 to vi. If w4 has no neighbor among the vertices v′k, vk−1, u2, thenthe reasoning above may be performed to obtain a contradiction, using thepaths with different parities w1vk−2u2 and w1v

′kvk−1u2 (we can notice that

even if i = k − 3 or j = k − 3, we have only one chord in a cycle, nota triangle, as needed to avoid odd holes). But w4v

′k /∈ E, since w4 would

then play a role similar to the one of w1 and the same reasoning would givew4vk−2 ∈ E; this would contradict the choice ofw4 ∈ N(w1)−N(vk−2) by(ii) forw1 with respect to the pathPv1vk−2 . Moreover,w4vk−1 /∈ E, otherwiseClaim 9 for Pv1vk−1 (instead of P), the path u2vj (or u2u4vj) instead of X ,and the path vk−1w4vi instead of Y gives a contradiction, because none ofthe two paths has a vertex with index 2. Finally, w4u2 /∈ E, otherwise thecyclew4w1v

′kvk−1u2w4 is odd and chordless. We have a contradiction, sow1

extends the sound path Pv1vk−2 .


• In the case where w1 extends the sound path Pv1vk−2 , but v′k does not extendthe sound path Pv1vk−2w1, v

′k cannot satisfy (i) in Remark 1, since it has no

neighbors on Pvk−2vk . If it satisfies (ii), it has a neighborw′1 /∈ N(w1), whichforms a triangle with two consecutive vertices on P . Then w′1 /= vk−1, w

′1 /=

vk and w′1 /∈ Tk−1 (by the criterion of shortness, the path W , which joins v′kto a vertex of Pv1vk−1 is as short as possible among all the paths joining avertex in Tk−1 to a vertex in Pv1vk−1 , see Section 3). Then w′1 plays the samerole as w1, with the supplementary property that it has at least two neighborson Pv1vk−1 . Then we have, as forw1, thatw′1vk−2 ∈ E and, furthermore, thatw′1 does not extend the sound path Pv1vk−2 (since it has some neighbor onthis path). As for w1, we obtain a contradiction.

• In the case where v′k extends the sound path Pv1vk−2w1, but vk does notextend the sound path Pv1vk−2w1v

′k, vk cannot satisfy (i) so it must satisfy

(ii). Therefore, there must exist a vertex u′1 ∈ N(vk)−N(v′k), which inducesa triangle with two adjacent vertices on the path Pv1vk−2w1v

′k. The vertex u′1

plays, with respect to P , the same role as u1. As u1 was chosen, by thecriterion of shortness, such that the path U is the shortest joining vk to avertex on Pv1vk−1 , u

′1 can have no neighbor on Pv1vk−2 (a shortest path would

be found). But then u′1 can form no triangle with two consecutive vertices ofPv1vk−1w1v

′k, a contradiction.

We deduce that u2 extends the sound path Pv1vk−2 .Part 2. u1 extends the sound path Pv1vk−2u2.If, by contradiction, u1 does not extend the indicated path, then (since u1 has no

neighbor on P ), there exists a vertex u′2 ∈ N(u1) − N(u2), which is in some Ta(a < k−1). By the criterion of shortness, vku′2 /∈ E; we can also assumeu′2v′k /∈ E,since the proof is the same as for u2. Call vq the neighbor of u′2 on Pv1vk−2 withmaximum index, and apply Claim 9 for P,W, vku1u

′2vq (instead, respectively, of

P, X, Y ). The supplementary condition in (S3) is satisfied. Then we must have asemi-regular construction. It cannot be on P (since then u′2, vk−2, vk−1, u2 wouldinduce a diamond), neither on P ′ (if w2 is adjacent to both vk−2, vk−1, the verticesvk−2, vk−1, u2, w2 would induce a diamond, since we are in the case wherew2u2 /∈E). We have a contradiction.

Claim 10 is proved.

Claim 11. In the case where both P, P ′ have semi-regular constructions, Pv1vk−2

u2u1 or Pv1vk−2w2w1 is a sound path.

Proof of Claim 11. Since both u2, w2 are adjacent to both vk−2, vk−1, we musthave w2u2 ∈ E, otherwise w2, vk−2, vk−1, u2 induce a diamond.

To show that at least one of u2, w2 extends the sound path Pv1vk−2 , we onlyhave to apply Claim 8 with i = k − 2 and the triangle vk−2w2u2. Without lossof generality (since the reasoning is exactly the same), we assume that u2 extendsthe sound path Pv1vk−2 (this includes the case where both u2, w2 extend the soundpath Pv1vk−2).


Now, ifu1 extends the sound pathPv1vk−2u2, then the claim is proved. Otherwise,by Remark 2 and since u1 has no neighbor onP , there exists a vertex u′2 ∈ N(u1)−N(u2), which is in someTa (a < k−1). By the criterion of shortness, vku′2 /∈ E; wecan also assumeu′2v′k /∈ E, since the proof is the same as foru2. Call vq the neighborof u′2 on Pv1vk−2 with maximum index, and apply Claim 9 for P,W, vku1u

′2vq

(instead, respectively, of P, X, Y ) to deduce that u′2w1 /∈ E.Moreover, u′2w2 /∈ E (by (S3) u′2 would be, as u2, adjacent to vk−2, vk−1 andG

would contain a diamond, since u′2u2 /∈ E, by the choice of u′2). By Claim 10 forP with W and vku1u

′2vq, we obtain that w2, w1 extend the sound path Pv1vk−2 .

Claim 11 is proved.

In both cases where Pv1vk−2w2w1 or Pv1vk−2u2u1 is a sound path, the last edgeof the path is on a triangle, because the existence of both w2, u2 is guaranteed by(ii) in Remark 1 (so (R3) in the definition of a regular construction is verified).The third vertex in the triangle containing u2u1 (respectively, w2w1) is denoted u3(respectively, w3).

Lemma 2 is proved.

Now, as indicated in the previous section, we can assume that P is a sound pathwith a regular construction, and denote u(0)

1 = v′k, u(0)2 = vk−1, u

(0)3 = vk, u

(1)j =

uj (j = 1, 2, 3). By induction, if the sound path Pv1vk−2u(i)2 u

(i)3 is known and if it

has a regular construction using three vertices denoted u(i+1)1 , u

(i+1)2 , u

(i+1)3 , then

we find another sound path Pv1vk−2u(i+1)2 u

(i+1)1 , and so on (see Fig. 4).

Let us prove (recall the notation in Section 3) the following.

Lemma 3. For a fixed q ≥ 1, the multi-set S0q induces an extended stretcher.

Proof of Lemma 3. It is easy to see that the vertices in S0q must be dis-tinct from the vertices on Pv1vk−2 because of the sound paths Pv1vk−2u

(a)2 u

(a)1 (for

each a).Notice that, since G has no diamond and all the vertices u(j)

2 (j = 1, . . . , q) are

adjacent to vk−1, vk−2, the vertices vk−1, vk−2 together with all the vertices u(j)2

(j = 1, . . . , q) induce a clique.We use induction on the distance k = l − j between two indices l > j ≥ 0 to

prove the following claims.

Claim 12. Assume thatSjl induces an extended stretcher. Then a vertex z adjacentto u

(l)3 , but nonadjacent to u

(l)1 , u

(l)2 is distinct from all the vertices in Sjl and

nonadjacent to each of them, except for u(l)3 .

Proof of Claim 12. By induction, consider firstly the case where k = l−j = 1.Obviously, z /= u

(l−1)2 , z /= u

(l−1)3 , since z is nonadjacent tou(l)

2 , u(l)1 . By contradic-

tion, if zu(l−1)2 ∈ E, then the cycle zu(l−1)

2 u(l−1)3 u

(l)2 u

(l)3 z implies that zu(l−1)

3 ∈ E,

so the triangles u(l)3 u

(l)2 u

(l)1 , zu

(l−1)2 u

(l−1)3 and the paths u(l)

3 z, u(l)2 u

(l−1)2 , u

(l)1 u

(l−1)3


form an odd stretcher, a contradiction. We deduce that zu(l−1)2 /∈ E. Then zu(l−1)

3 /∈E, because of the cycle zu(l)

3 u(l)2 u

(l−1)2 u

(l−1)3 z. Now, z /= u

(l−1)1 (a stretcher would

be induced) and zu(l−1)1 /∈ E, otherwise the cycle zu(l)

3 u(l)1 u

(l−1)3 u

(l−1)1 z is odd and

chordless.Assume the claim is true for a distance k ≥ 1 and let us prove it for k + 1.

Then we have, by induction for l and j + 1, that z is distinct from all the ver-tices in S(j+1)l and nonadjacent to any of them, except u(l)

3 . We have to prove

that z is distinct from and nonadjacent to u(j)1 , u

(j)2 , u

(j)3 . Obviously, z is dis-

tinct from u(j)1 , u

(j)2 , u

(j)3 , otherwise u(l)

3 would have neighbors in u(j)1 , u

(j)2 , u

(j)3 ,

and this is impossible, by hypothesis. Moreover, zu(j)2 /∈ E, otherwise the cycle

zu(j)2 u

(j)3 u

(j+1)1 u

(j+1)3 · · ·u(l)

1 u(l)3 z is odd and the only possible chord is zu(j)

3 . But

then the triangles zu(j)2 u

(j)3 , u

(j+1)3 u

(j+1)2 u

(j+1)1 and the paths zu(l)

3 u(l)1 · · ·u(j+2)

1

u(j+1)3 , u

(j)2 u

(j+1)2 , u

(j)3 u

(j+1)1 form a stretcher, a contradiction.

Now, zu(j)3 /∈ E, otherwise the cycle zu(l)

3 u(l)2 u

(j)2 u

(j)3 is odd and chordless.

Similarly, zu(j)1 /∈ E.

Claim 13. Assume thatSjl induces an extended stretcher, and there exists a vertexz as in Claim 12. Then a vertex t adjacent to z and u(l)

2 , but nonadjacent to u(l)3 , is

distinct from all the vertices in Sjl and:—either it is nonadjacent to every vertex in Sjl except for u(l)

2 ;—or it is adjacent to every vertex u(s)

2 (s = 0, 1, . . . , l) and nonadjacent to anyother vertex in Sjl.

Proof of Claim 13. In the case k = l−j = 1, we have t /= u(l)1 , t /= u

(l−1)1 , t /=

u(l−1)2 , t /= u

(l−1)3 , since z has no neighbor among these vertices. Moreover, tu(l)

1 /∈E, otherwise a diamond can be found. Furthermore, tu(l−1)

3 /∈ E because of the

cycle tzu(l)3 u

(l)1 u

(l−1)3 t and tu(l−1)

1 /∈ E because of the cycle tu(l)2 u

(l)1 u

(l−1)3 u

(l−1)1 t.

Let us now suppose the claim true for some distance k ≥ 1, and let us prove it fork+1. Using induction for l and j+1, we only have to prove that t is nonadjacent tou

(j)1 , u

(j)3 , and that tu(j)

2 ∈ E if and only if tu(j+1)2 ∈ E (t is obviously distinct from

u(j)1 , u

(j)2 , u

(j)3 , since z has no neighbor among these vertices). Now, if tu(j)

3 ∈ E,

then the cycle tzu(l)3 u

(l)1 u

(l−1)3 u

(l−1)1 · · ·u(j+1)

1 u(j)3 t is odd and chordless, a contra-

diction. Moreover, if tu(j)1 ∈ E, then the cycle tu(l)

2 u(l)1 u

(l−1)3 · · ·u(j)

3 u(j)1 t is odd

and chordless, a contradiction. Furthermore, tu(j)2 ∈ E if and only if tu(j+1)

2 ∈ E,

otherwise the vertices t, u(l)2 , u

(j+1)2 , u

(j)2 induce a diamond (as k ≥ 2 we have

j + 1 /= l).

Claim 14. Assume that Sjl induces an extended stretcher and there exist verticesz, t as in Claims 12, 13. Then a vertexw adjacent to t, z is distinct from every vertexin Sjl and nonadjacent to any vertex in Sjl.


Proof of Claim 14. As usual, we start by the case k = l − j = 1 and easilynotice that w is distinct from u

(l)1 , u

(l)2 , u

(l)3 , u

(l−1)1 , u

(l−1)2 , u

(l−1)3 . Now, wu(l)

2 /∈E,wu

(l)3 /∈ E (otherwise a diamond can be found) and wu

(l)1 /∈ E (otherwise

a stretcher can be found). Moreover, the proof that wu(l−1)3 /∈ E,wu

(l−1)2 /∈

E,wu(l−1)1 /∈ E is the same as in the case of an arbitrary j (instead of l − 1),

so we give it only once, in the general case (here below).Assume the claim is true for k ≥ 1 and let us prove it for the distance k +

1. By induction hypothesis for l and j + 1, we only have to prove that w isdistinct from and nonadjacent to any vertex in {u(j)

1 , u(j)2 , u

(j)3 }. As before, w

is distinct from all these vertices because of z, which is nonadjacent to them.Now, wu(j)

3 /∈ E, otherwise the cycle wzu(l)3 u

(l)1 u

(l−1)3 u

(l−1)1 · · ·u(j+1)

1 u(j)3 w is

odd and chordless, a contradiction. Moreover, wu(j)2 /∈ E, otherwise the cycle

wu(j)2 u

(l)2 u

(l)3 zw is odd and chordless, a contradiction. Finally, if wu(j)

1 ∈ E,

then the cycle wtu(l)2 u

(j)2 u

(j)1 w implies tu(j)

2 ∈ E, so the triangles twz, u(j)2 u

(j)1

u(j)3 and the paths tu

(j)2 , wu

(j)1 , zu

(l)3 u

(l)1 · · ·u(j+1)

1 u(j)3 form a stretcher, a

contradiction.

We can now prove Lemma 3 by induction on q. For q = 1, it is true by Lemma2. If it is true for some fixed q, then it is also true for q + 1, by Claims 12, 13, 14for j = 0, l = q and, respectively, z = u

(q+1)1 , t = u

(q+1)2 , w = u

(q+1)3 . Lemma 3

is proved.

As indicated in Section 3, Lemma 3 allows us to deduce the existence of extremalsound paths with the same s-length and the same number of edges as P , so we canassume that the path P we worked with is an extremal sound path. Therefore, forthe pathW : w0 = v′k, w1, . . . , wh, vi (0 ≤ h ≤ 2) built as in Remark 4 (and usingthe criterion of shortness) we know that w1, w2 do not form a regular constructionon P ′ : Pv1vk−1v

′k. We also consider (see again Section 3) the chordless path

U (f+1) : d0 = u(f)3 , d1, . . . , dr′ , vj , with 0 ≤ r′ ≤ 2, which joins u(f)

3 to a vertex

vj (as close as possible to u(f)2 ) ofPv1vk−2u

(f)2 . If it exists, d2 induces a triangle with

two adjacent vertices on Pv1vk−2u(f)2 , since its existence is due to (ii) in Remark 1

for d1.We prove now the following.

Lemma 4. There exist two paths W, respectively, U , disjoint from P joining v′k(respectively, u(f)

3 ) to some vertex va (respectively, vb) on Pv1vk−3 and such thatthe internal vertices of W, respectively, of U , are nonadjacent to every vertex in{u(i)

1 , u(i)2 , u

(i)3 , 0 ≤ i ≤ f}.

Proof of Lemma 4. The hypothesis of Claim 9 (the supplementary condition in(S3) included) are verified for the sound path Pv1vk−2u

(f)2 u

(f)1 and the pathsU (f+1)

(instead ofX) and u(f)1 u

(f−1)3 u

(f−1)2 (instead of Y ), for the same reasons they were


verified for P,U , andW . Then, we deduce that all the vertices involved are distinctand no edge exists between the two paths but u(f−1)

3 d1 and, possibly, u(f−1)2 d2 (by

(S1), in this last case we also have r′ = 2, and by (S3) we have a semi-regularconstruction on Pv1vk−2u

(f)2 u

(f)3 ).

By Claim 12, we have that, whenever it exists, d1 is nonadjacent to any vertexin {u(i)

1 , u(i)2 , u

(i)3 , 0 ≤ i ≤ f − 1}.

Notice also that, if we perform Steps 1, 2 (see Section 3) starting with the pathPv1vk−2u

(f)2 u

(f)1 , the vertices u(f−1)

3 , u(f−1)2 , u

(f−1)1 form a regular construction,

and that, in general, for the path Pv1vk−2u(i)2 u

(i)1 , the vertices u(i−1)

3 , u(i−1)2 , u

(i−1)1

form a regular construction. That means that, in fact, P and Pv1vk−2u(f)2 u

(f)1 are

symmetrical with respect to the construction, since, independently of the fact that westart the construction with P or with Pv1vk−2u

(f)2 u

(f)1 , we obtain the same vertices,

but in reverse order. Thus, the roles played by u(f)3 and v′k are symmetrical, as

well as the roles played by U (f+1) and W . Thus, exactly in the same way asfor d1, we can deduce that (whenever it exists) w1 is adjacent to no vertex in{u(i)

1 , u(i)2 , u

(i)3 , 0 ≤ i ≤ f}.

Claim 15. If w2 exists, then it cannot be adjacent to both vk−2, vk−1.

Proof of Claim 15. By contradiction, assume that wvk−2, wvk−1 ∈ E. The

vertex w2 must be adjacent to every u(i)2 (otherwise a diamond may be found).

Sincew1, w2 do not form a regular construction on P ′, the path Pv1vk−2w2w1 is nota sound path, so eitherw2, orw1 satisfies (i) or (ii) with respect to the correspondingpath. Anyway, by Claim 13 for w2, we have that w2 has no neighbor among thevertices {u(i)

1 , u(i)3 } (0 ≤ i ≤ f).

Case 1. If w2 does not extend Pv1vk−2 , then there exists a path W ′ of length 1or 2 (the possible internal vertex is called w4), which joins w2 to a vertex vt onPv1vk−3 (t is taken as large as possible). Then d2 (if it exists) cannot be adjacent

to both vk−2 and u(f)2 , since in this case it would be adjacent to w2 and Claim 8

for vk−2 on P and w2, d2 would give a contradiction. Moreover, if d2u(f)2 ∈ E,

then Claim 9 for Pv1vk−2u(f)2 and the pathsW ′ (instead ofX), u(f)

2 d2vj (instead of

Y ) would give a contradiction ((S2) cannot be satisfied). We deduce that d2u(f)2 /∈

E, i.e., the path u(f)2 U (f+1) is chordless. Now, Claim 9 for Pv1vk−2u

(f)2 and the

paths W ′ (instead of X), u(f)2 U (f+1) (instead of Y ) implies that d2 does not exist

(but d1 exists) and d1vk−2 ∈ E (the hypothesis (C3) in Claim 9 is d1w2 /∈ E

and this must be true, otherwise the cycle w2w1v′kvku

(1)1 u

(1)3 · · ·u(f)

3 d1w2 impliesd1w1 ∈ E and the triangles w1w2d1, v

′kvk−1vk form a stretcher with the paths

w1v′k, w2vk−1, d1u

(f)3 u

(f)1 · · ·u(1)

1 vk).

Furthermore, the path Pv1vk−2d1u(f)3 u

(f)1 cannot be a sound path (it would have

larger s-length than P ), so we have three possibilities:


• d1 does not extend the sound path Pv1vk−2 .

Then d1 is joined by a path D of length 1 or 2 (with possible internal vertex d3)to some vertex vs of Pv1vk−2 chosen as close as possible to vk−2.

Let us prove that (whenever it exists) w4 is adjacent to none of the vertices{u(i)

1 , u(i)3 , 0 ≤ i ≤ f}. Indeed, if w4w1 ∈ E, then this is true by Claim 14

by denoting u(−1)3 = w1, u

(−1)2 = w2, u

(−1)1 = w4 and performing a change of

notation (in order to have indices going from 0 to f + 1 instead of indices goingfrom −1 to f ). On the other hand, if w4w1 /∈ E, then w4v

′k /∈ E (by the criterion

of shortness), w4vk /∈ E (w4w2w1v′kvkw4 would be an odd hole), w4vk−1 /∈ E

(w4, w2, vk−1, vk−2 would be a diamond) and then we can use induction. If we

suppose that w4 is nonadjacent to any vertex in {u(i)1 , u

(i)3 , 0 ≤ i ≤ i0} (for a fixed

i0 ≥ 0), then w4u(i0+1)1 /∈ E because of the cycle w4w2u

(i0)2 u

(i0)3 u

(i0+1)1 w4 and

w4u(i0+1)3 /∈ E because of the cycle w4w2w1v

′kvku

(1)1 · · ·u(i0+1)

3 w4.

Thus, we can show that the paths w2vk−2d1, w2u(f)2 u

(f)3 d1 (chordless and with

different parities) may be added to a chordless subpath of Du

(f)3 vs

PvsvtW′vtw2

in

order to form cycles with no other possible chord but vtvk−2 or vsvk−2. Indeed, thechords of these cycles should join either w4 or d3 to vertices of the two indicatedpathsw2vk−2d1, w2u

(f)2 u

(f)3 d1. But d3vk−2 /∈ E (by the definition of d3), d3u

(f)2 /∈

E (as for d2), d3u(f)3 /∈ E (else d3 it would be similar to d1, so it would be adjacent

to vk−2, a contradiction). In its turn, w4 is nonadjacent to vk−2, u(f)2 (we would

have a diamond) and u(f)3 (as previously shown).

Then the two paths w2vk−2d1 and w2u(f)2 u

(f)3 d1 are of different parities and

may be added to a chordless subpath of Du

(f)3 vs

PvsvtW′vtw2

to form two cycles of

different parities, one of which, at least, must contain a triangle (otherwise the oddone gives a contradiction). The only possible triangle is given by vk−3, vk−2 (sot = k − 3 or s = k − 3) and one of w2, d1 (which will be, in this case, the last onits path). This is not possible for w2 (a diamond w2, vk−3, vk−2, vk−1 would exist),so d1 is adjacent to vk−3 (i.e., s = k − 3).

Denote now u(f+1)1 = d1, u

(f+1)2 = vk−2, u

(f+1)3 = vk−3. Then the reason-

ing we used before to show that w1, w2, w4 are nonadjacent to {u(i)1 , u

(i)3 , 0 ≤

i ≤ f} is valid now for {u(i)1 , u

(i)3 , 0 ≤ i ≤ f + 1}, thus w1, w2, w4 are non-

adjacent to d1. Finally, if w4 does not exist or is nonadjacent to w1, the cyclePvk−3vtW

′vtw2

w1v′kvku

(1)1 u

(1)3 · · ·u(f)

3 d1vk−3 is odd and with no triangle (but possi-ble with chords between d1 andPvtvk−3 ; otherwise (i.e., ifw4 exists andw4w1 ∈ E)

the triangles w2w4w1, vk−2vk−3d1 and the paths w2vk−2, w4Pvtvk−3 , w1v′kvku

(1)1

u(1)3 · · ·u(f)

3 d1 give a stretcher, a contradiction.

• d1 extends the sound path Pv1vk−2 , but u(f)3 does not extend the sound path

Pv1vk−2d1.


Then u(f)3 either has a neighbor on Pv1vk−2 (and this is not the case, since then

U (f+1) would have only one vertex, d0 = u(f)3 , by the criterion of shortness), or it

has a neighbor d′1, which induces a triangle with two consecutive vertices on thepath. Then, as for d1, we deduce d′1vk−2 ∈ E and, as previously, we should havethat d′1 extends the sound path Pv1vk−2 ; but this is not true, since d′1 has at least twoconsecutive neighbors on Pv1vk−2 .

• u(f)3 extends the sound path Pv1vk−2d1, but u(f)

1 does not extend the sound

path Pv1vk−2d1u(f)3 .

Then, since (i) in Remark 1 cannot be satisfied by u(f)1 for the indicated path,

we have (ii): a vertex z adjacent to u(f)1 but not to u(f)

3 (so nonadjacent to u(f)2 )

exists such that z forms a triangle with d1, vk−2 (this is the only possibility to have

(ii) since the path Pv1vk−2u(f)2 u

(f)1 is a sound path). But in this case the triangles

zd1vk−2, u(f)1 u

(f)3 u

(f)2 and the edges joining pairs of vertices in different triangles

form a stretcher.Case 2. Ifw2 extends the sound path Pv1vk−2 , butw1 does not extend the sound

path Pv1vk−2w2, then (since w1 cannot satisfy (i)) there exists a path W ′′ of length2 (the internal vertex inN(w1)−N(w2) is denotedw5), which joinsw1 to a vertexvi on Pv1vk−2 (i is taken as large as possible). Then w5v

′k /∈ E (by the criterion of

shortness).The cycle Pvivk−2w2w1w5vi must be even (it has no triangles), so Pvivk−2 must

be even. Then Pvivk−1 is odd and the cycle Pvivk−1v′kw1w5vi is odd and the only

possible chord is w5vk−1. In order to have a triangle, we must have i = k − 2; butthen a diamond appears given by w5, w2, vk−2, vk−1, a contradiction.

Claim 15 is proved.

As the roles played by w2 and d2 are similar, as in Claim 15 we can prove that,if d2 exists, then it cannot be adjacent to both vk−2, u

(f)2 . We continue our proof

with the following result.

Claim 16. If w2 exists, then it cannot be adjacent to vk−1.

Proof of Claim 16. As shown before, the only possible case is that w2 isadjacent to vk−1 but not to vk−2. Notice that, since w2 is obtained because (ii)holds forw1 with respect to the sound path P ′ : Pv1vk−2v

′k, w2 must form a triangle

with two adjacent vertices on P ′, so it has some other neighbor on Pv1vk−2 ; wedenote vi the neighbor that is closest to vk−2.

If u(f)2 U (f+1) is not chordless, then d2 exists and d2u

(f)2 ∈ E. Now, Claim 9 can

be applied forPv1vk−2u(f)2 and the paths vk−1w2vi (instead ofX), u(f)

2 d2vj (insteadof Y ), if d2vk−2 /∈ E (condition (C2) in Claim 9). This condition is verified (byCase 1, for d2), so Claim 9 gives a contradiction ((S1) cannot be true).

Thenu(f)2 U (f+1) is chordless and we apply Claim 9 forPv1vk−2u

(f)2 and the paths

vk−1w2vi (instead ofX), u(f)2 U (f+1) (instead of Y ) to deduce that d2 does not exist


and d1vk−2 ∈ E. Once more, d1, u(f)3 , u

(f)1 cannot extend the path Pv1vk−2 , since

we would have a path with larger s-length than P and, as in Claim 15, we havethree cases:

• d1 does not extend Pv1vk−2 .

Then d1 satisfies (i) or (ii), so there exists a chordless path D (with possibleinternal vertex d3 ∈ N(d1) − N(vk−2)), which joins d1 to some vertex vs onPv1vk−2 taken as close as possible to vk−2.

We consider the two chordless paths w2vk−1vk−2d1 and w2vk−1u(f)2 u

(f)3 d1, of

different parities (w2u(f)2 /∈ E, otherwise w2vk−2 ∈ E or we have a diamond;

w2u(f)3 /∈ E, otherwise the cycle w2vk−1vk−2d1u

(f)3 w2 implies w2d1 ∈ E and

the triangles w2d1u(f)3 , vk−1vk−2u

(f)2 give a stretcher). Moreover d3 (whenever it

exists) is adjacent to none of the vertices vk−1 (then d3vk−1u(f)2 u

(f)3 d1d3 would im-

ply d3u(f)3 ∈ E, so Claim 12 with z = d3 would be contradicted, since d3vk−1 ∈

E), u(f)2 (otherwise Claim 9 for Pv1vk−1 and vk−1w2vi, instead of Y, u(f)

2 d3vs,

instead of X , would give a contradiction), u(f)3 (as before). Then the two indi-

cated chordless paths of different parities may be added to a chordless subpath ofw2Pvivsd3d1 (and similarly, if d3 does not exist) to obtain a chordless odd path,except if one of the cycles has a triangle. The only possibility is that s = k− 3 andd1vk−3 ∈ E (so d3 does not exist).

Moreover, w2 is adjacent to no vertex in {u(i)1 , u

(i)3 , 0 ≤ i ≤ f}, by Claim

13 using a change of notation. Also, w2d1 /∈ E because of w2vk−1u(f)2 u

(f)3 d1w2,

which would be odd and chordless.Then the triangles d1vk−2vk−3, u

(f)3 u

(f)2 u

(f)1 and the paths d1u

(f)3 , vk−2u

(f)2 ,

Pvk−3viw2w1v′kvku

(1)1 u

(1)3 · · ·u(f)

1 form a stretcher.

• d1 extends the sound path Pv1vk−2 , but u(f)3 does not extend the sound path

Pv1vk−2d1.

As in Claim 15, a neighbor d′1 of u(f)3 can be found, which induces a triangle with

two consecutive vertices on the path Pv1vk−2 . As d1, the vertex d′1 should extendthe path Pv1vk−2 , but this is not true.

• u(f)3 extends the sound path Pv1vk−2d1, but u(f)

1 does not extend the sound

path Pv1vk−2d1u(f)3 .

Once again, as in Claim 15, we find a vertex z adjacent to u(f)1 but not to u(f)

3

(so nonadjacent to u(f)2 ) such that z forms a triangle with d1, vk−2; a stretcher may

then be built, a contradiction.Claim 16 is proved.

Once again, since the roles played by w2 and d2 are similar, as in Claim 16 wecan prove that, if d2 exists, then it cannot be adjacent to u(f)

2 .


We finish now the proof of Lemma 4. Because of Claims 15, 16, neither w2 nord2 (whenever they exist) is adjacent to vk−2 or to vk−1 (otherwise an odd chordlesspath with five vertices would be found). We show that we can find a pathW and apath U that join v′k, respectively, u(f)

3 , to vertices of P such that no internal vertex

onW,U is adjacent to {u(i)1 , u

(i)2 , u

(i)3 , 0 ≤ i ≤ f}.

If w2 exists, then the vertex vi on P adjacent to w2 and closest to vk−1 hasthe property i < k − 2 and W may be taken equal to W , if we prove that w2 isnonadjacent to all the vertices {u(i)

1 , u(i)2 , u

(i)3 , 0 ≤ i ≤ f}. By Claim 9 for P,W

(instead of X), u(1)1 u

(1)2 vk−1 (instead of Y ), where the condition (C3) corresponds

tow2vk /∈ E and is true by the criterion of shortness for the choice of u(1)1 , we have

that w2 is nonadjacent to u(1)2 , u

(1)1 . Moreover, w2u

(1)3 /∈ E, otherwise the triangles

v′kvk−1vk, u(1)3 u

(1)2 u

(1)1 form a stretcher.

Furthermore, to prove that w2 is adjacent to none of the vertices {u(i)1 , u

(i)2 , u

(i)3 ,

0 ≤ i ≤ f}, we use induction: if this is true for the vertices {u(i)1 , u

(i)2 , u

(i)3 , 0 ≤ i ≤

i0} (for a fixed i0), then we prove the same for i0+1. Indeed,w2u(i0+1)2 /∈ E because

of w2w1v′kvk−1u

(i0+1)2 w2; w2u

(i0+1)1 /∈ E because of w2w1v

′kvku

(1)1 · · ·u(i0+1)

1 ;w2u

(i0+1)3 /∈ E because of the stretcher with triangles u(i0+1)

2 u(i0+1)1 u

(i0+1)3 ,

u(i0)2 u

(i0)3 u

(i0)1 and paths u(i0+1)

2 u(i0)2 , u

(i0+1)1 u

(i0)3 , u

(i0+1)3 w2w1v

′kvku

(1)1 · · ·u(i0)

1 .If w2 does not exist, then for w1 the neighbor vi could have the same property

i < k − 2 (and then we are done), or could have i = k − 2. In this last case, as thepath Pv1vk−2w1v

′kvk cannot be a sound path (it would have larger s-length than P ),

either w1 or v′k or vk satisfy (i) or (ii) for the corresponding subpath.

• Ifw1 does not extend Pv1vk−2 and has another neighbor vs on P , then we cantakeW to be the path v′kw1vs. Otherwise, there exists a vertexw4 ∈ N(w1)−N(vk−2), which has at least one neighbor va on Pv1vk−3 and we can takeW :v′kw1w4va, if we prove thatw4 has no neighbor in {u(i)

1 , u(i)2 , u

(i)3 , 0 ≤ i ≤ f}.

This is done as previously for w2.• If w1 extends the sound path Pv1vk−2 but v′k does not extend the sound pathPv1vk−2w1, then v′k cannot satisfy (i), so there exists a vertex w′1 ∈ N(v′k)−N(w1) such that w′1 induces a triangle with two consecutive vertices onPv1vk−2 . Thenw′1 has at least one neighbor va with a /= vk−2 and we can takeW equal to the path v′kw

′1va. By Claim 12,w′1 is nonadjacent to the indicated

vertices.• If v′k extends the sound path Pv1vk−2w1, but vk does not extend Pv1vk−2w1v

′k,

then the neighbor of vk, which is adjacent to some vertex onP , implies that thepath u(1)

1 u(1)2 vk−1 does not verify the criterion of shortness, a contradiction.

So we can findW as indicated. Similarly (for symmetric reasons), we can find Ujoining u(f)

3 to a vertex vb of Pv1vk−3 and such that no internal vertex has neighbors

in {u(i)1 , u

(i)2 , u

(i)3 , 0 ≤ i ≤ f}.

Lemma 4 is proved.


As indicated in Section 3, we can now build the stretcher with triangles vk−1vkv′k,

u(1)2 u

(1)1 u

(1)3 and paths vk−1u

(1)2 , vku

(1)1 , respectively, a chordless path extracted

fromWv′kvaPvavbUvbu(f)

3u

(f)1 u

(f−1)3 · · ·u(2)

1 u(1)3 . This contradicts the hypothesis on

G and Theorem 1 is proved. �

5. FURTHER CONSIDERATIONS

It is easy to see that the algorithm CONTRACT in Section 2 is polynomial. Indeed,finding a vertex whose neighborhood induces a clique or a stable set may be donein at most O(n3) operations (where n is the number of vertices in the graph);removing a vertex needs at most O(n) operations; by Claim 5, finding a suitableeven pair needsO(n3) operations (since we need to test the existence of the verticesq, t, s before choosing the pair as in Case 1 or as in Case 2); finally, the contractionof an even pair may be done in O(n). So every iteration of the while loop maybe performed in O(n3) time; therefore, the algorithm CONTRACT may be donein O(n4).

The graph G′′ resulting at the end of the execution is obtained from G after asequence of even pair contractions. It is triangulated, so it may be reduced to aclique using another sequence of contractions; if we use Hertz’s algorithm [7], thisis done in O(n3) time.

Consequently, reducing a diamond-free graph with no odd hole and no stretcherto a clique using successive contractions of even pairs needs O(n4) time.

We can notice here that a diamond-free graph with no odd hole can have nospecial vertex (stretchers are examples of such graphs). A weaker result exists inthis case (see [11]), claiming that a diamond-free graph with no odd hole and whosecliques are all of size of least three has a vertex contained in at most two cliques.The manner to find such a vertex (at the end of a path with particular properties)inspired us for the definition of a sound path we used all along in the article.

References

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[2] H. Everett, C. M. H. de Figueiredo, C. Linhares–Sales, F. Maffray, O. Porto,and B. Reed, Path parity and perfection, Discr Math 165/166 (1997), 233–252.

[3] C. M. H. de Figueiredo, F. Maffray, and O. Porto, On the structure of bull-freeperfect graphs, Graphs and Combin 13 (1997), 31–55.

[4] J. Fonlupt and J. P. Uhry, Transformations which preserve perfectness andh-perfectness of graphs, Ann Discrete Math 16 (1982), 83–85.

[5] M. C. Golumbic, Algorithmic graph theory and perfect graphs, AcademicPress, New York, 1980.


[6] R. Hayward, C. T. Hoang, and F. Maffray, Optimizing weakly triangulatedgraphs, Graphs and Combin 6 (1990), 33–35.

[7] A. Hertz, A fast algorithm for coloring Meyniel graphs, J Combin Theory SerB 50 (1990), 231–240.

[8] C. Linhares–Sales and F. Maffray, Even pairs in claw-free perfect graphs, JCombin Theory Ser B 74 (1998), 169–191.

[9] H. Meyniel, A new property of critical imperfect graphs and some conse-quences, Europ J Combin 8 (1987), 313–316.

[10] I. Rusu, Even pairs in Artemis graphs, Discrete Mathematics, to appear.[11] A. Tucker, Coloring perfect (K4 − e)-free graphs, J Combin Th Ser B 42

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