Berge Graphs withChordless Cycles ofBounded Length

Irena RusuUNIVERSITE D’ORLEANS, L.I.F.O., B.P. 6759

45067 ORLEANS CEDEX 2, FRANCE

Received November 19, 1997; revised March 5, 1999

Abstract: A graph is called weakly triangulatedif it contains no chordless cy-cle on five or more vertices (also called hole) and no complement of such a cycle(also called antihole). Equivalently, we can define weakly triangulated graphs asantihole-free graphs whose induced cycles are isomorphic either to C3 or to C4.The perfection of weakly triangulated graphs was proved by Hayward [Hayward, JCombin Theory B. 39 (1985), 200–208] and generated intense studies to efficientlysolve, for these graphs, the classical NP-complete problems that become polyno-mial on perfect graphs. If we replace, in the definition above, the C4 by an arbitraryCp (p even, at least equal to 6), we obtain new classes of graphs whose perfec-tion is shown in this article. In fact, we prove a more general result: for any eveninteger p ≥ 6, the graphs whose cycles are isomorphic either to C3 or to one ofCp, Cp+2, . . . , C2p−6 are perfect. c© 1999 John Wiley & Sons, Inc. J Graph Theory 32: 73–79, 1999

Keywords:perfectness; weakly triangulated graphs; dominating pair

1. INTRODUCTION

A perfect graphis a graphG = (V, E) whose chromatic numberχ and cliquenumberω are equal for the graph itself, as well as for all its induced subgraphs. Aholeis simply a chordless cycle of length at least five and anantiholea complementof such a graph. With these definitions, Berge’s strong perfect graph conjecture [1]

c© 1999 John Wiley & Sons, Inc. CCC 0364-9024/99/010073-07

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(abbreviated SPGC) claims that a graph is perfect if and only if it contains no oddhole and no odd antihole (as induced subgraphs).

The SPGC is still unsettled, but has been proved for particular classes of graphs(one of the most important of them being the class ofweakly triangulated graphs[3], in which no hole or antihole is induced), while it is still open for some otherparticular classes (for instance, theantihole-free graphs, in which no antihole isinduced). Speaking about the antihole-free graphs, which seem to be quite difficultto approach, we can notice that weakly triangulated graphs are exactly the antihole-free graphs whose only chordless cycles areC3 andC4 (whereCk is the classicalnotation for a chordless cycle of lengthk ≥ 3). Then, a natural question arises: whatabout the antihole-free graphs in which only the cyclesC3 andCp are allowed (for afixed even integerp ≥ 6)? Do they satisfy the SPGC? Do they have even pairs, i.e.,pairs(x, y) of vertices such that any chordless path joiningx, y is of even length?Do they have nice characterizations yielding efficient recognition algorithms? Forweakly triangulated graphs, the answers to all these questions are affirmative.

The aim of this article is to prove that each of the classesCp, obtained byconsidering the graphs whose cycles are isomorphic either toC3 or to one ofCp, Cp+2, . . . , C2p−6, is a class of perfect graphs (observe that the condition for thegraph to be antihole-free is, in fact, not necessary, since it is implied by the absenceof C4 in the graph).

2. MAIN RESULT

As usual, we use the notationN(x) to denote theneighborhoodof a vertexx, i.e.,the set of vertices (also calledneighborsof x) that are adjacent inG = (V, E) tox, and the notationN(S) (whereS ⊆ V ) to denote∪x∈SN(x) − S. Theclosedneighborhoodof a vertexx is defined asN [x] = N(x) ∪ {x}. We calldominatingpair in G a pair(x, y) of vertices such thatN [x] ⊇ N(y).

The announced result will be an easy consequence of the following lemma, whichestablishes that every graph in a classCp has either a special pair of vertices, or aspecial vertex.

Lemma 1. Let p ≥ 6 be a fixed even integer andG = (V, E) be a graphwhose only chordless cycles areC3, Cp, Cp+2, . . . , C2p−6. Then at least one of thestatements below is true:

(1) G contains an adjacent dominating pair;(2) G possesses a vertex contained in at most two maximal cliques.

Proof. Let c1c2 · · · ck be a longest chordless path ofG, chosen such that|N(ck) − N [ck−1]| is as large as possible (we can suppose thatk ≥ 3, otherwisethe graph is a union of cliques and we are done). In order to simplify further ex-planation, make a change of notation as follows: letai = ck−i (i ∈ {0, 1, . . . ,k − 1}). Thenak−1ak−2 · · · a1a0 is a longest chordless path ofG chosen such that|N(a0) − N [a1]| is as large as possible.

BOUNDED-LENGTH CHORDLESS CYCLES 75

Since the path cannot be extended, every vertexw ∈ N(a0) − {a1} must havea neighbor on the path. Two possibilities exist for a givenw:

(i) eitherwa1 ∈ E;(ii ) or k ≥ p − 1 and there existsj ∈ {p − 2, p, p + 2, . . . , 2p − 8} such that

wa1, wa2, . . . , waj−1 /∈ E, while waj ∈ E.We partition the neighborhood ofa0 into four setsN(a0) = {a1} ∪ S ∪ T ∪ Q,

whereS ∪ T = N(a0) ∩ N(a1), Q = N(a0) − N [a1], T = N(Q) ∩ N(a0), S =N(a0) ∩ N(a1) − T . Notice that, because of (ii ) here above, every vertex inQis adjacent to someaj , j ∈ {p − 2, p, p + 2, . . . , 2p − 8} (but nonadjacent toa2, a3, . . . , aj−1). If Q is empty, then statement (1) in the lemma is satisfied by thepair (a1, a0). In the contrary case, we can prove thatQ is a clique. To this end, weuse the following small result:

(3) If u, v ∈ N(a0) (nonadjacent) such that they both have neighbors on a(notnecessarily chordless) path of length at mostp − 5 included inV − N [a0], thenuv ∈ E.

If this is not the case, then letx, y be two neighbors ofu, respectively,v on theindicated path such that the distanced(x, y) (in number of edges) betweenx, y isminimum. The chordless cycle induced byu, a0, v and a chordless path betweenx, y is of lengthd(x, y) + 4 ≤ p − 1, a contradiction.

Now,Q is a clique, otherwise (3) is contradicted byq, q′ ∈ Q (such thatqq′ /∈ E)and the pathap−2ap−1 · · · a2p−8. We can easily deduce thatT is a clique, too(otherwise aC4 or aC5 may be built usinga1, two nonadjacent vertices ofT , andtwo neighbors inQ of these vertices). Moreover,

(4) Every vertex inS is adjacent toa2 (or statement(1) holds).If this is not the case, considers ∈ S such thatsa2 /∈ E. Thensa3, sa4, . . . ,

sap−2 /∈ E (a chordless cyclesa1a2 · · · ai of bad length would be found using thesmallesti such thatsai ∈ E). By (3), it is sufficient to considerq ∈ Q, s (thensq /∈ E), and the pathap−2ap−1 · · · a2p−7 to deduce thatsap−2, sap−1, . . . ,sa2p−7 /∈ E (whenever the indicated vertices exist). Also, the remaining verticesal (l > 2p− 7) on the path are nonadjacent tos, since in this case we would obtaina too long cyclesa1a2 · · · al.

Consequently, the pathak−1ak−2 · · · a1s is chordless and has the same length asthe initial pathak−1ak−2 · · · a0. Now, if N(s) ⊆ N [a1], then we have statement (1)in the lemma. Otherwise, letu ∈ N(s) − N [a1]. Notice thatua0 /∈ E, since in thecontrary caseu would be inQ, and no vertex ofQ sees any vertex ofS. Moreover,usatisfies (ii ) (withu in place ofw), otherwise the pathak−1ak−2 · · · a1su is chordlessand longer than the initial path. By (3), if we takeq ∈ Q, s and a suitable path(obviously of length at mostp−5) of the subgraph given byu, ap−2, ap−1, · · · a2p−8,we have a contradiction. So (4) is established.

(5) Every vertex inS is adjacent to every vertex inT .Let s ∈ S (sosa2 ∈ E) and suppose that there existst ∈ T such thatst /∈ E.

Thenta2 /∈ E (in this casea0sa2t would be aC4) andta3, ta4, . . . , tap−2 /∈ E(otherwise a too small chordless cycle may be found using{a1, a2, . . . , ap−2, t}).But t could have some neighbors in{ap−1, ap, . . . , a2p−7}. Let, if it exists,ap+l

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(−1 ≤ l ≤ p − 7) be the neighbor oft, which is closest toap−2, sol is odd (sincethe cyclea1a2 · · · ap+lt must be even).

On the other hand, fors (by (3), using someq ∈ Q and the pathap−2ap−1 · · ·a2p−7), we have thatsap−2, sap−1, . . . , sa2p−7 /∈ E. So letaj (2 ≤ j ≤ p − 3)be the neighbor ofs that is closest toap−2 (such a neighbor always exists, sincesa2 ∈ E). Thenj is even, since the cyclea0sajaj+1 · · · anq (whereq ∈ Q whoseneighbor closest toap−2 is an) must be even.

But then the chordless cyclea0sajaj+1 · · · ap+lt has length(p + l − j) + 4,which is odd, a contradiction. We deduce thatt has no neighborar with r ≤ 2p−7.Let us show that the only neighboraj (2 ≤ j ≤ p − 3) of s is a2. If this is not thecase, then, ast has no neighborar with r ≤ 2p − 7, the cyclea1sajaj+1 · · · anqt(with q andan as before) has odd length, but no chord. We have a contradiction.

So, ifs has some neighbors on the pathak−1ak−2 · · · a0, then it must be adjacentto a2p−6. We consider the two cases:

(i) sa2p−6 /∈ E (or a2p−6 simply does not exist).Thenak−1ak−2 · · · a2sa0 is a chordless path with the same length as the initial

path, so we can use fors, a0 the same reasoning we performed fora1, a0 to deducethatN(a0) = {s} ∪ S′ ∪ T ′ ∪ Q′, whereS′, T ′, Q′ have the same meanings andproperties asS, T, Q but with respect tos instead ofa1. Recall that the initial pathwas chosen such that|N(a0) − N [a1]| was as large as possible, and notice thatN(a0) − N [a1] ⊆ N(a0) − N [s]. We deduce that the equality holds, i.e.,Q = Q′.But then, by the definitions ofT, T ′, we have thatT = T ′, and sos is adjacent toevery vertex inT , since it is adjacent to every vertex inT ′.

(ii ) sa2p−6 ∈ E.Consider again the vertext ∈ T such thatst /∈ E and letq ∈ Q such that

tq ∈ E. The cyclea1tqan′an′+1 · · · a2p−6s (wherean′ is the neighbor ofq closesttoa2p−6, 2p−6 ≥ n′ ≥ p−2) is a cycle of length(2p−6−n′)+5, which is at mostequal top+1 (sincen′ ≥ p−2). If it is of lengthp, thenn′ = p−1, soqap−1 ∈ E;but then the cyclea0qap−1ap · · · a2p−6s has length(2p−6)− (p−1)+4 = p−1,a contradiction. If it is of lengthp + 1 (thenn′ = p − 2), this cycle must contain aunique chord, which will form aC3 and aCp (otherwise a chordless cycle smallerthanp could be found). Asan′ = ap−2 and we already know thattap−2 /∈ E (seebefore), the only possibility isqap−1 ∈ E. But this contradicts the choice ofan′ asthe neighbor ofq closest toa2p−6.

So (5) is proved. Then we denoteW = {a1} ∪ S and we haveN(a0) = W ∪T ∪ Q, where no edge is missing, neither betweenW andT , nor betweenW anda2. MoreoverW, T, Q are cliques (to see this forW , it is sufficient to consider anytwo nonadjacent vertices inW , anda0, a2 to obtain aC4). We can also supposethatT /= ∅, since in this case we already have statement (2) in the lemma, namelywith vertexa0.

Now, if any vertex inT is adjacent to every vertex inQ, then we are done: thepair(t, a0) is a dominating pair. Consider, therefore, a vertext ∈ T for which thereexist verticesq, q′ ∈ Q such thattq ∈ E, tq′ /∈ E.

(6) ak−1ak−2 · · · a1t is a chordless path.

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Once more, we can notice that none of the edgesta2, ta3, . . . , ta2p−7 existsin G. Indeed, by (3) (usingq′) u has no neighbors on the subpath with the ex-tremitiesap−2, a2p−7. Now, suppose thatan, am (n, m ≥ p − 2 and even) are,respectively, the neighbors ofq, q′ closest toap−3 and let (if, by contradiction, itexists)aj (2 ≤ j ≤ p−3) be the neighbor oft closest toap−2. The chordless cycletajaj+1 · · · anq must be even and imposes thatj is odd (sincen is even), whilethe cyclea0tajaj+1 · · · amq′ implies thatj is even (sincem is even). We have acontradiction. Soak−1ak−2 · · · a1t is a chordless path, since any neighborar of twith r > 2p − 7 would imply the existence of a too long chordless cycle.

(7) N(t) − N [a1] ⊂ Q.

Suppose that this is not true and letz ∈ N(t)−N [a1]−Q. Thenz /∈ N(a0) and(by (6) and (ii ) with z in place ofw) there exists somej ∈ {p − 2, p, . . . , 2p − 8}such thatza1, za2, . . . , zaj−1 /∈ E, while zaj ∈ E. Again by (3), usingt, q′and a suitable path (obviously of length at mostp − 5) of the subgraph given byz, ap−2, ap−1, . . . , a2p−8, we have a contradiction.

Because of (7), we know that either(a0, t) is a dominating pair (and statement(1) in the lemma holds) ort has a neighborr out of N(a0), which is necessarilyadjacent toa1. In this last case, we continue as follows:

(8) The pathak−1ak−2 · · · a1r is chordless.

Notice first that, because of (3),r has no neighbors in{ap−2, ap−1, . . . , a2p−8}(otherwise by consideringt, q′ and a suitable path of the graph induced byr and thepreceding vertices, we have, by (3), thattq′ ∈ E). Once more, suppose thatan, am

are, respectively, the neighbors ofq, q′ closest toap−3 and let (if, by contradiction,it exists)aj (2 ≤ j ≤ p − 3) be the neighbor ofr closest toap−2. Notice thatrq, rq′ /∈ E (otherwise we have aC4 with a0, a1). The cycletrajaj+1 · · · anqimposes thatj is even (sincen is even), while the cyclea0trajaj+1 · · · amq′ impliesthatj is odd (sincem is even). We have a contradiction. Moreover,ra2p−7 /∈ E,otherwise the chordless cycletra2p−7a2p−8 · · · ahq (whereah is the neighbor ofqclosest toa2p−7) has length at mostp − 1.

It remains to show thatra2p−6 /∈ E (then the indicated path must be chordless,otherwise a longer chordless cycle could be found). Suppose the contrary. Thechordless cyclea0a1ra2p−6a2p−7 · · · ahq has length(2p − 6 − h) + 5 ≤ p +1, sinceh ≥ p − 2. If the length is exactlyp, thenh = p − 1 and the cycletra2p−6a2p−7 · · · ahq has length(2p − 6 − h) + 4 = p − 1, a contradiction. If thelength is exactlyp + 1, we have an odd chordless cycle.

(9) The pair(a1, r) is a dominating pair.

If this is not the case, then letz ∈ N(r) − N [a1] and notice thatza0 /∈ E(otherwise,z must be inQ anda0a1rz is aC4). Now, ak−1ak−2 · · · a1rz cannotbe a chordless path, since it is longer than the initial path. Thenz must havesome neighboraj , j ∈ {p − 2, p, p + 2, . . . , 2p − 8}. Let al, ag (p − 2 ≤ l, g ≤2p − 8) be two neighbors ofz, respectively,q such that|l − g| is minimum (wecan supposel ≤ g). The cycletrzalal+1 · · · agq has lengthg − l + 5 (which is atmost(2p − 8) − (p − 2) + 5 = p − 1) and at most two chords,zt andzq. Then we

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must necessarily havel = g andzt, zq ∈ E (otherwise a too small chordless cycleexists); but thena0a1rzq is aC5.

Because of (9), statement (1) in the lemma is verified once again and the proofis finished.

It is easy now to establish the result related to perfectness.

Theorem 1. For any fixed even integerp ≥ 6, the graphs containing no chordlesscycle butC3, Cp, Cp+2, . . . , C2p−6 are perfect.

Proof. By contradiction, suppose that there exists a minimal imperfect graphG = (V, E) with this property. Then, by Lemma 1, eitherG has a dominating pairof vertices or a vertexv of G is contained in at most two maximal cliques. In thefirst case,G or G has a star-cutset, i.e., a disconnecting setS one of whose verticesis universal inS, and this contradicts the star-cutset lemma (see [2]); in the secondone, a result of Padberg [4], stating that in a minimal imperfect graphG any vertexis contained in exactlyω(G) cliques of sizeω(G), implies thatω(G) ≤ 2, thusGis isomorphic to someC2k+1 (k ≥ 2), another contradiction.

3. CONCLUDING REMARKS

It is worth noticing that the property established in Lemma 1 is not sufficient tocharacterize any of the classesCp, nor their union. Then the question of recog-nition complexity for these classes remains unsolved, as well as the question ofwhether graphs inCp (other than cliques) always have even pairs. Statement (2) inLemma 1 does not necessarily imply the existence of an even pair, as shown by thegraphC6.

Moreover, the initial similarity we found (with respect to chordless cycles)between the classes containing onlyC3, Cp (p ≥ 6 and even) and the class ofweakly triangulated graphs is not confirmed by the property in Lemma 1: thereexist weakly triangulated graphs satisfying neither (1) nor (2) (see the autocomple-mentary weakly triangulated graph defined by Hayward [3]).

ACKNOWLEDGMENTS

The author is indebted to Stefan Hougardy, Chinh Hoang, and Frederic Maffrayfor bringing this problem to her attention and for useful discussions concerning thesubject.

References

[1] C. Berge, Farbung von Graphen, deren samtliche bzw. deren ungerade Kreisestarr sind, Wiss. Z. Martin Luther Univ., Halle Wittenberg (1961), 114–115.

BOUNDED-LENGTH CHORDLESS CYCLES 79

[2] V. Chvatal, Star-cutsets and perfect graphs, J Combin Theory B 39 (1985),189–199.

[3] R. B. Hayward, Weakly triangulated graphs, J Combin Theory B 39 (1985),200–208.

[4] M. W. Padberg, Perfect zero-one matrices, Math Prog 6 (1974), 180–196.