on planar perfectly contractile graphs

Graphs and Combinatorics (1997) 13: 167-187

On Planar Perfectly Contractile Graphs

Graphs andCombinatorics© Springer-Verlag 1997

Claudia Linhares Sales!", Frederic Maffray/ and Bruce Reed?

I LSD2-IMAG, Grenoble, FranceCNRS, Laboratoire Leibniz, IMAG, BP 53, 38041 Grenoble Cedex 9, FranceCNRS, ER Combinatoire, Universite Paris 6, place Jussieu, 75005 Paris, France

Abstract. An even pair in a graph is a pair of verticessuch that every chordless path betweenthem has even length. A graph is called perfectly contractile when every induced subgraphcan be transformed into a clique through a sequence of even-pair contractions. In this paperwe characterite the planar graphs that are perfectly contractile by determining all theminimal forbidden subgraphs. We give a polynomial algorithm for the recognition ofperfectly contractile planar graphs.

Key words: Perfect graphs, even pairs, planar graphs.

I. Introduction

A graph G is called perfect if for every induced subgraph H of G, the chromaticnumber of H (X(H» is equal to the size of its largest clique (w(H». Giving a goodcharacterization of perfect graphs is a famous open problem. So far it is onlyknown that perfect graph recognition is in co-NP. Another question is: given aperfect graph, how fast can one find an optimal coloring of its vertices? It followsfrom work of Grotschel, Lovasz and Schrijver that the coloration problem ispolynomial for all perfect graphs, using the ellipsoid method. However, we wouldlike to have simpler algorithms based on the combinatorial nature of perfectgraphs. The following ideas may be a step in that direction.

Let G = (V, E) be a graph with vertex-set V and edge-set E. Two non-adjacentvertices of G are termed an even pair (resp. odd pair) if every chordless path betweenthem has an even (resp. odd) number of edges. Given a pair of non-adjacentvertices x, y in a graph G, one can build a new graph, denoted by Gjxy, obtainedfrom G by removing x and y and adding a new vertex xy with edges from thisvertex to all vertices adjacent to x or yin G. This is called identifying or contractingthe pair x, y. Bertschi [1] proposed to call perfectly contractile any graph G thatsatisfies the following property: for every induced subgraph H of G, there exists a

• On leave from UFF, Niteroi-Brazil and partially supported by CNPq-Brazil grantnumber 201420/92

168 C.L. Sales et al.

Fig. 1. An odd stretcher and an even stretcher

sequence of graphs H = Ho, HI' . .. , Hk such that Hi+l results from the contractionof one even pair in Hi' and H; is a complete graph. We are interested in thecharacterization of perfectly contractile graphs. Recall that a chordless cycle withat least five vertices is called a hole, and its complement an antihole. A hole on kvertices is denoted by Ck ; its complement by Ck •

Definition 1. A stretcher is any graph that is obtained from a C6 by subdividing theedges that do not lie in a triangle in such a way that the three subdivided paths havethe same length parity. A stretcher is called even (resp. odd) if the three paths haveeven length (resp. odd length) .

Conjecture 1 [9]. A graph is perfectly contractile if and only if it contains no oddhole, no antihole, and no odd stretcher.

We will prove this conjecture for planar graphs. Notice that planar graphs containno antihole of length at least seven, and that the antihole C6 is the smallest oddstretcher.

Theorem 1. A planar graph is perfectly contractile if and only if it contains no oddhole and no odd stretcher.

Our motivation for studying this conjecture is to understand better the structure of perfect graphs, in view of Berge's Strong Perfect Graph Conjecture: A graphis perfect if and only if it contains no odd hole and no odd antihole. The StrongPerfect Graph Conjecture is true for planar graphs, by results of Tucker [1OJ orHsu [5]. In other words a planar graph is perfect if and only if it contains no oddhole.

The following theorem is not difficult:

Theorem 2 [3, 7]. If x and y form an even pair in G, then X(G/xy) = X(G) andw(G/xy) = w(G) . Furthermore, there is a linear algorithm to compute an optimalcolouring for G and find a largest clique in G given the corresponding object in Gjx»,

o

Corollary 1. If there is a sequence of graphs H = Ho, .. . , Hk such that Hk is a cliqueand Hi+l arises from Hi via the contraction of an even pair, then X(H) = w(H). 0

Corollary 2. Perfectly contractile graphs are perfect. o

On Planar Perfectly Contractile Graphs 169

The proof of Theorem 1 is given in section 2. In section 3, we show how torecognize the perfectly contractile graphs among all planar graphs.

For an introduction to planar graphs see [8].

2. Proof of the Main Theorem

To justify the "only if" part of the theorem we need to show that every possiblesequence of even-pair contractions in an odd stretcher leads to a C6 . This turns outto be somewhat less easy to prove than we thought, and we defer this proof to theannex. The rest of this section is now devoted to proving the 'if ' part of thetheorem, which is the meat of this paper. Let us start with some remarks anddefinitions.

In the proof, when discussing any planar graph, we will always assume that thegraph is drawn in the plane. A diamond is a clique of size four with one edgeremoved; the two extremities of the missing edge are called the points of thediamond. Let C be a cutset of a graph G and B l , ••• , Bk be the components ofG - C. We call piece of C each subgraph induced by B, U C (i = 1, ... , k). Also, wewill say that a vertex x sees a vertex Y if xy is an edge.

We will frequently use the following easy fact without further reference:

Remark. Let G be a planar graph that contains no odd holes, and C = Pi + P2 be anycycle in G, such that every chord of C is inside C and links a vertex of Pi to a vertexof P2 • Then C is an even cycle if and only if the subgraph induced by the vertices ofC contains an even number of triangles , for otherwise an odd hole is found.

A pseudo-stretcher is a graph M consisting of two disjoint triangles with threevertex-disjoint chordless paths X , Y, Z joining the vertices of the first triangle to thevertices of the second one in a one-to-one fashion, so that these paths have thesame length parity and there is no other triangle in M (but there may be edges fromone of the three paths to another). We will always write X = Xl ' " Xp, Y = Yl ... Yq,Z = Zi · · · z" so that {x l' Yl',Zd and {x p,Yq,z,} are the two triangles in M . Forshort, to refer to a segment of the paths X, Y or Z, we will use the subscripts of itsextremities.

Lemma 1. Every pseudo-stretcher either is a stretcher or contains an odd hole.

Proof. Let M be a pseudo-stretcher as above. If there is no edge from a vertex onone of the paths X, Y, Z to a vertex on another path, other than the triangle edges,then M is a stretcher and we are done. So let us suppose that there is such an edge,sayan edge xjYj' and choose it so that i + j is minimized. We have either i > 1 orj > 1, and symmetry allows us to assumej > 1. IfXCI, i] and Y[I,j] have differentparities, then XCI, i] + XjYj + Y[j, 1] + Y1Xl is an odd hole. If they have the sameparity, then Zl X l + XCI, i] + xiYj and ZlYl + Y[I,j] are chordless paths of different parities and so the cycle C = ZlX l + X[I,i] + xiYj + Y[j,q] + Yqz, + Z is anodd cycle. However, C contains no triangle since x p and Yl are not in C, and thenC necessarily contains an odd hole. 0

170 c.L. Sales et al.

Lemma 2. Let G be a planar perfect graph containing no odd stretcher. If G containsan even stretcher M and one of the holes F of M is also a face of G, then F containstwo vertices which form a 2-cutset and an even pair of G.

Proof. Let us label M as below. We may assume that F = X + XlZl + Z + zrxp,and choose a planar representation of G in which F is the infinite face. We proposeto show that:

{xl ,xp} is a 2-cutset of G that separates X - {xl ,xp} from YU Z. (1)

Suppose that this is not true and let P be the shortest path between X - {xl,xp}and Y UZ. By planarity, P is from X - {Xl' X p } to Y. Write P = V l ... Vr so thatVl = Xi with 2 ~ i ~ P - 1, and Vr = Yj with 1 ~ j ~ q and t ~ 2. We also choose Pso that i and j are as small as possible. Hence Vr - l does not see any vertex inY[I,j - IJ, and V2 does not see any vertex in X [2, i-I]. Also notice that:

Either Vr- l misses every vertex of Y[j + 2, qJ or vr- l sees an even numberof edges of Y, (2)

for otherwise, letting I be the largest subscript such that Vr- l sees Y" one of thecycles Vr-lY, + Y[/,qJ + YqZr + Z + ZlYl + Y[I,YjJ + YjVr- l or vr-lYj + Y[j,/J +Y'Vr - l would contain an odd hole.

Case 1.j # 1.

Let P~ be the shortest path from Xl to Vr contained in X[I, iJ + P, and letP, = ZlXl + P~ . Let P2 = ZlYl + Y[I,j]' So P, + P2 is a cycle containing exactlyone triangle, so it is an odd cycle, and P, has a different parity from P2 • Therefore,C = P, + Y[j, qJ + YqZr + Z[r, IJ is an odd cycle. The only possible chords of Care from Vr - l to Y[j, qJ, and there must be an odd number of triangles in C. By (2),this means that Vr- l sees Yj+l and no other vertex of Y[j + 1,q]. Ifj is even, thenwe have an odd stretcher formed by the two triangles {Xl'Yl,zd, {vr-l,Yj,Yj+dand the three chains Pl [Xl' vr-lJ, Y[I,n , Z + ZrYq+ Y[q,j + IJ , a contradiction.Ifj is odd, let P3 be the shortest chain from x p to Vr - l contained in X[p, iJ + P. Thechain P3 is odd since P3 + vr-lYj+l + Y[j + l,qJ + Yqxp is an even hole. Now wehave an odd stretcher formed by the triangles {vr-l'Yj'Yj+d, {Xp, Z" Yq}, and thethree chains P3 , Y[j, IJ + YlZl + Z, Y[j + l,qJ, a contradiction.

Case2.j= 1.

Let P4 be the shortest path from x p to Vr contained in X[p, iJ + P. If Vr - l seesan even number of edges along Y, then the vertices of Y UP4 form an even cycle,but then P4 + YlZl + Z + zrxp is an odd hole , a contradiction. Therefore, and by(2), Vr - l sees Y2 and no vertex of Y[3, q]. Now we have an odd stretcher, formed bythe triangles {Vr- l 'Yl 'Y2}, {xp,z"yq} and the three chains P4 [ Vr- l , Xp J , YlZl + Z,Y[2, qJ, a contradiction.

So (1) is established. Note that X is an even chordless path between Xl and xpIt follows that {Xl'X p } is an even pair, for otherwise there would necessarily existtwo distinct pieces of G - {xl,xp} containing two Xl - xp chordless paths ofdifferent parities, hence G would contain an odd hole. 0


Now let G be any planar graph containing no odd hole and no odd stretcher. Weprove that G is perfectly contractile by induction on the number of its vertices. Thefact is clear for graphs with at most five vertices. Let us first examine some specialcutsets of G.

Lemma3. If G has a clique-cutset C, then G admits a sequence of even-pair contractions that leads to a clique.

Proof. Let B1 , ... , Bk be the connected components of G - C. Clearly, each pieceGi contains no odd hole and no odd stretcher. So, by the induction hypothesis,each piece G,admits a sequence of even-pair contractions that leads to a clique Qi'It is easy to see that any even pair in Gi is also an even pair in G.So, we can performthe sequence of contractions on G1 , then on G2 , etc. The graph that results from allthese contractions has a clique-cutset C whose pieces are the cliques Ql' .. . , Qk'This graph is triangulated, so it is perfectly contractile [2]. 0

Lemma4. If G contains a 2-cutset {a, b} which is also an even pair, then G is perfectlycontractile.

Proof. Clearly Glab is planar. If G/ab contains no odd stretcher then by theinduction hypothesis it is perfectly contractile, and then so is G. So it suffices toprove that Glob contains no odd stretcher. Call c the contracted vertex in Glob.Notice that c is a cut vertex in Glab. Suppose that Glab contains an induced oddstretcher H, then H must contain c; since H is 2-connected, it lies entirely in a pieceG1 /ab of Glab with respect to {c}. Considering another piece G2 /ab, there exists achordless a-b path P of G whose interior vertices are all in G2 ; moreover P is evensince {a,b} is an even pair of G. Call NH(c) the set of neighbours of c in H. Notethat in G each of a, b misses every vertex of H - (NH(c) U {c}), and that everyinterior vertex of P misses every vertex of H - c.

Suppose that c is a vertex of degree two in H, with NH(c) = {x, y}. If one of a, bsees both of x, y, then replacing c in H with that vertex yields an odd inducedstretcher in G. If each of a, b sees exactly one of x, y, then replacing c in H with Pyields an odd induced stretcher in G.

Now suppose that c is a vertex of degree three in H, with NH(c) = {x ,y,z} andxy is an edge. Ifone of a, b, say a, sees all of x, y, z in G, then replacing c in H witha yields an odd induced stretcher in G. If one of a, b, say a, sees x, z and misses y(or sees y, z and misses x) then replacing c with a gives an induced subgraph of Gisomorphic to H - yc (or H - xc), which is easily seen to contain an odd hole. Theonly remaining case is when one of a, b sees x, y and misses z and the other sees zand misses x, y. Then replacing c with P yields an induced odd stretcher in G. Inall cases a contradiction occurs. 0

Lemma 5. If abc is a P3 -cutset of G and {a, c} is an even pair, then G is perfectlycontractile.

Proof. Clearly G/ac is planar. As before, if Glac contains no odd stretcher then, bythe induction hypothesis, it is perfectly contractile and so is G. Suppose indirectly

172 C.L. Sales et at.

that Gjac contains an odd stretcher M. Observe that vertices band ac form aclique-cutset C of Glac. Since a stretcher cannot be separated by a clique-cutset, Mis entirely contained in a piece Gl/ac of C. Also, M contains ac. Let P be an induceda-c path in G - b contained in a piece G2 of {a,b,c}. Replacing ac with P in M, weobtain an induced odd stretcher in G, a contradiction. 0

Remark that if abc is a P3 -cutset of G, then all a-c chordless paths in G - b havethe same parity, or else an odd hole can he found in G. This implies the followinglemma:

Lemma 6. If abc is a P3 -cutset of G, and if there exists an even chordless a-c-pathin G - b, then {a, c} is an even pair.

From now on we will assume that:

(hl) G has no clique-cutset;(h2) G has no 2-cutset forming an even pair;(h3) G has no P3 -cutset whose extremities form an even pair.

Let F be any face of G. Note that F is a cycle, or else there would be a cut-vertexin G. Moreover, if F has a chord xy, then it is easily seen that {x,y} is a cutset ofG. Therefore, every face of G is an even hole or a triangle.

We will call facial any pair of vertices that lie on the same non-triangular face.Observe that if we contract a facial even pair, we obtain a planar graph (note thatthe even pairs of Lemmas 4 and 5 are facial). We want to find facial even pairswhose contraction leaves the graph odd stretcher-free.

Lemma 7. If G admits a facial even pair of vertices a, b that have a commonneighbour, then Gjab is planar and contains no odd hole and no odd stretcher.

Proof. Let us suppose that a, b is a facial even pair, with a common neighbour x.We know that Gtab is planar and contains no odd hole. Let us suppose that Glabcontains an odd stretcher M, with the same notation as above for M. We mayassume that ab = Xk, and that in G, among the vertices of GnM, vertex a sees Xk- l(or Yl and Zl if k = 1) and may also see Xk+l' and b sees Xk+l (or Yq and z, if k = p)and may also see Xk- l . In fact if a sees Xk+l (or if b sees xk-d then replacing in Mthe vertex ab with a (or b) yields an odd stretcher in G. So we may assume that amisses Xk+l and b misses X k- l•

Since a and b are on the same face, either there is no edge from x to Y or thereis no edge from x to Z; by symmetry we may assume there is no edge from x to Z.In G we have a path Xl'" Xk- l aXbXk+l .. . Xp- It contains a (minimal} chordlessxl-xp-path U. This path U must contain x, because there are no edges fromXl'" Xk- l a to bXk+l .. . xp' The path U must be odd, for otherwise U + X1Z l + Z +zrxp would be an odd chordless cycle and not a triangle, a contradiction. Since Uis odd, x does not see both Xl and x p; by symmetry we may assume that x does notsee Xl'

Note that C = U + xpYq + Y + Y1Xl is an even cycle, and the only possiblechords of C are from X and to Y + xp- Hence X must see an even (possibly null)


number of edges along Y + x p , or else an odd hole would be found. If x sees noedge on Y + x p , then the vertices of U U Y UZ form a pseudo-stretcher. Finally,assume that x sees at least two edges along Y + x p , hence at least one along Y. LetI be the smallest subscript such that y,X is an edge, with 1 < I ::;; q - 1. The cycleY[y"yq] + yqxp + U[xp,x] + xy, is even since it contains an even number oftriangles, namely all the triangles from x to Y + xp- Therefore, we have chordlesspaths y,X + U[x,x p] + xpzr and Y[y"yq] + YqZr of different parities. Since Y +YqZr + Z + ZlYl is an even hole, it follows that y,X + U[x,xp] + xpzr + Z +ZlYl + Y[Yl'Y,] is an odd hole, a contradiction. 0

The following even pairs satisfy the hypotheses of the preceding lemma.

Lemma 8. If the neighbourhood of some vertex x of G is a stable set, then any twoof its neighbours form an even pair and some two of these neighbours are a facialeven pair .

Proof. Clearly some two neighbours of x are a facial pair. If there exists an inducedodd path between any two neighbours of x then adding x we find an odd cycle withno triangle, which contains an odd hole. 0

Lemma 9. If G contains a diamond whose points form a facial pair, they also forman even pair.

Proof. Assume the diamond has vertices a, b, c, d, with points a, b, and supposethat there is an odd chordless path P from a to b. Both P U{c} and P U{d} are oddcycles. The planarity and the fact that a, b are on a face implies that one of thesetwo cycles is chordless, a contradiction. 0

Lemma 10. If G contains facial even pairs as described in Lemmas 8 and 9, then G isperfectly contractile.

Proof. By Lemma 7, the contraction of these even pairs does not introduce oddstretchers. Since the obtained graph is also planar and smaller than G, it is perfectlycontractile and so is G. 0

Now let us assume there is no such pair, by adding the following hypothesesabout G:

(h4) no vertex of G has a stable neighbourhood;(h5) G does not contain a diamond whose points are facial.

The hypothesis (h2) allow us to make another hypothesis about G which is atrivial consequence of Lemma 2:

(h6) G contains no even stretcher of which one hole is a face.

The next lemma tells us where to find a suitable pair in a non-triangular face(ifany) of G.

Lemma 11. Every non-triangular face of G contains a facial even pair.

174 CiL. Sales et al.

Proof. Let F be any non-triangular face. Recall that F is an even hole . Consider anyvertex z on F, and let x, y be the neighbours of z on F. We choose a planarrepresentation of G in which F is the infinite face. For any two vertices u, v ofF - {z}, we will let F[u , v] denote the segment of the path in F - {z} from U to v.

We claim that x, y form an even pair. By contradiction, assume there is an oddchordless path Q from x to y, and choose Q so that it has as many common verticeswith F as possible. Consider the last vertex U along Q with the property thatQ[x, u] = F[x, u], and let v be the first vertex of Q nF after u. We have pathsF' = F[u, v] = U 1 " ' upand Q' = Q[u,v] = w1 • • • wq, so that u = U 1 = W1, V = up=wq , p, q ~ 2, and the interior vertices of Q' are not in F. Note that if z sees v then,since F is chordless, it must be that v = y. The paths Q' and F' have differentparities or else, replacing Q' with F' in Q, we would get another odd chordless x -ypath with more vertices in common with F than Q. By the choice of Q, we actuallyhave Q = F[x, u] + Q' + F[v, y]. Note that C = Q' + F' is an odd cycle and not atriangle and by the remark, C must contain an odd number of triangles. Observethat Q + yz + zx is an odd cycle and not a triangle, so z sees an odd number ofedges of Q. It follows immediately that:

Let us show that:

z sees an odd number of edges of Q'. (3)

Each WE Q' sees at most two vertices of F', and, if it sees two,they must be consecutive. (4)

To prove this , suppose on the contrary that W sees three vertices U;, Uj , Uk of F'. Twoofi,j, k have the same parity, say i andj. So, UjWUj is a P3-cutset and by Lemma 6,{Uj, Uj} is an even pair. Therefore, UjWUj is a P3-cutset as in Lemma 5, a contradiction to (h3). Likewise, if W sees only two vertices Uj and Uj' then i and j must havedifferent parities, and we must have Ij - il = 1 or else {w} U F'[u j , uj ] would be anodd hole. So (4) is established.

Now we will prove that:

Each f E F' sees at most two vertices of Q', and, if it sees two,they must be consecutive. (5)

We suppose indirectly that f sees two non-consecutive vertices of Q'. Let i (j) bethe smallest (greatest) integer such that fWj (fw) is an edge. Now, let us define thetwo cycles C1 = F[u1,f] + fWj + Q[wj , w1 ] and C2 = F[up,f] + fWj + Q[wj , wq ] .

We claim that there is no vertex W of Q[wi, wj ] seen by both f and z. Indeed, inthat case {f,w, z} forms a P3 -cutset, If the chordless path zx + F[x,f] is even thenthis P3 -cutset would contradict (h3) because of Lemmas 5 and 6. If the path is oddthen adding W yields an odd cycle of length at least five in which the only possiblechords are from wand to F'. So, by 4, W sees the neighbour f' of f along F[u,f].But now {f',w,z} form a P3-cutset satisfying Lemmas 5 and 6 and violating (h3),a contradiction.

The path Q[w j , wj ] will be partitioned into f-segments and z-segments definedas follows . Starting from Wi' in the increasing order of subscripts along Q', an


f-segment (resp. z-segment) begins with a vertex seen by f (resp. z) and ends withthe vertex before the next one seen by z (resp. f). We claim that:

On each f-segment f sees an even number of edges. (6)

Suppose that this is false and consider an f-segment U that contradicts it. Let uscall k (1) the smallest (greatest) integer such that Wk and w, belong to U and fWk' fw,are edges. Let t be the nearest vertex from Wk in Q[x, wk ] that is seen by z. Let r bethe largest subscript such w, belongs to U (hence z sees W'+l)' Clearly C3 = zt +Q[t, W,+l] + ZW'+1 must be an even hole as it is not a triangle and it contains notriangle. If1oF k + 1, then Q[wk,w,] + fWk + fw. is an odd cycle and not a triangle,since f sees an odd number of edges on Q[wk,W,]. So (C3 - Q[wk,w,]) + »J + fw,is an odd hole, a contradiction . If1= k + 1, then (C3 - w,w,+d + wJ + fW'+1 is anodd cycle with exactly one triangle, which contains f Then there are two inducedpaths between z and f along this cycle; one of them forms an odd cycle withF[z, u1 ] + F[u1,f] and another odd cycle with F[z, up] + F[up,f]. By planarityonly one of these odd cycles can have chords, hence one of them is an odd hole, acontradiction.

Therefore, f sees an even number of edges along Q'. At least one of C1 , C2 , sayC1 has an even number of induced triangles . Indeed, if both had an odd number oftriangles then, since f sees an even number of edges along Q', the odd cycle Q' + F'would have an even number of triangles, a contradiction. Now (C1 - fwi) + fWj +Q[wj , w;] is an even cycle, since it contains an even number of triangles. Thisimplies that the paths PI = Q[wl , Wj] and P2 = F[ul,f] + fWj have the same parity. So, replacing PI with P2 in Q, we obtain another odd induced path between xand y with a larger intersection with F than Q, a contradiction. All these contradictions prove 5.

For now we will focus on the possible existence of the triangles of C differentfrom {v,up-l,wq-d and {U,V2'W2}, We claim that:

C contains only one triangle and it must be either {v, Up-I' W q - l} or{U,V2'W2}, (7)

Suppose indirectly that such a triangle T exists and let us call Wi and Uj the verticesof Q' and F' respectively such that WiUj is an edge of T and i, j are minimal. Thenexactly one of UjWi+1 or Uj+IWi is an edge i.e., either T= {Wi,Wi+I,Uj} or T={Wi' Uj , Uj +1 }.

Case 1. T = {Wi' Wi+l' uj } .

By the supposition and 5 we have j oF p - 1.Consider the two following cyclesC1 = Q[wl , w;] + wiuj + F[u1,uj] and C2 = Q[Wi+I' wq] + Wi+IUj + F[uj, up]. Thecycles C1 and C2 have the same parity. Also, C1 + T must be an odd cycle, forotherwise the paths PI = Q[wl, Wi+l] and P2 = F[u l, uj] + UjWi+1 would have thesame parity and there would be another odd induced path Q - PI + P2 from x toy with a larger intersection with F than Q, contradicting the choice of Q. So C2 + Tis also an odd cycle and C1, C2 are even cycles. Since, by (3), z sees an odd numberof edges along Q', we can assume without lost of generality that z sees an odd


number of edges along Q[wl, Wi] and an even number along Q[Wi+ l, wq]. Moreover the paths P3 = Q[wl , wa + wiuj and P4 = F[u l , uj] have the same parity sinceC l = P3 + P4 is an even hole. So F - P4 + P3 is also an even cycle, but it containsan odd number of triangles (formed from the edges seen by Z along P3 ), therefore itcontains an odd hole, a contradiction.

Case 2. T = {Wi' Uj, uj+d.

By the supposition and 4 we have i *2. Consider the two cycles C l =Q[wl , wa + wiuj + F[ul,uJ and C2 = Q[wj,wq] + WjUj+l + F[uj+l , up]. Again, Cl

and C2 have the same parity and must be odd cycles, for otherwise the paths P, =Q[wl,wa and P2 = F[ul,uJ + UjWj would have the same parity (Cl = P, + P2)and there would be another odd induced path Q - Pl + P2 from x to y with alarger intersection with F than Q, contradicting the choice of Q. SO C l + T andC2 + T are even cycles. As before, since Z sees an odd number of edges along Q' by(3), we can assume without lost of generality that z sees an odd number of edgesalong Q[wl , Wi] and an even number along Q[wj, wq]. Notice that the paths P3 =Q[wl, wa + WiUj+l and P4 = F[u l, uj+l] have the same parity, since C l + T is aneven cycle. So F - P4 + P3 is also an even cycle, but it contains an odd number oftriangles (formed from the edges seen by z along P3 ), therefore it contains an oddhole, a contradiction.

To finish this claim, we observe that C is an odd cycle and therefore it must contain an odd number of triangles. Hence exactly one of {Wj, Wj+1 ' uJ or {wj, Uj' uj+1 }

exists and (7) is proved.Now (7) allows us to assume, without lost of generality, that the only triangle

of Cis {v,up-l, wq-d.Now, we claim that:

The only vertices that z sees along Q' are Wl and W2• (8)

We suppose indirectly that z sees other vertices than Wl and W 2. Let j be thegreatest integer such that zWj is an edge. By (3), z sees an odd number of edges alongQ[Wl' wq-2]. So the paths PI = zWj + zx + F[x, u] and P2 = Q[u, wj] have different parities, since PI + P2 is a cycle with an odd number of triangles. Recall thatC l = P2 + Q[Wj'wq-l] + wq-l up-l + F[u, Up-I] is an even cycle by (7), so CI

P2 + P, is an odd cycle with an odd number of triangles, hence contains an oddhole, a contradiction.

Now, by (7), the only triangle of C is {wq-l,Up-l'V} and the only trianglecontaining Z along Q' is {z,x = U, w2 } . These two triangles together with the pathsQ[W2' wq - I ] , F[u, up-d and F[v, y] + yz form a pseudo-stretcher. By the originalhypothesis on G and Lemma 1, th is must be an even stretcher, which contradicts(h6). This completes the proof of Lemma 11. 0

Lemma 12. If G contains a non-triangular face, then G is perfectly contractile.

Proof. By Lemma 7, we can contract the vertices x, y found in Lemma 11, and weobtain a planar graph with no odd stretcher and smaller than G. By applying theinduction hypothesis, this graph is perfectly contractile and so is G.

The following lemma can be applied when every face of G is a triangle.


Lemma 13. When every face of G is a triangle then G is a comparability graph.

Clearly, this final lemma implies the main theorem, since all comparability graphsare perfectly contractile [2, 41.

Proof of Lemma 13.

This proof is rather similar to the proof of [6, Theorem 6.2] but will be easier sincewe are under a different hypothesis. We assume that G is neither K 4 nor theoctahedron, i.e., the complement of 3K2 , for which the desired conclusion holdseasily. Since every face is a triangle, and there is no triangle cutset, every neighbourhood must be an even hole. Let D be the set of all vertices of degree four.

Claim 1. Every triangle intersects D.

Proof Suppose that a triangle {x,y,z} does not intersect D, i.e., x, y and z havedegree greater than or equal to six. Let us denote by X = Xl X2 ... X2k the chordlesspath formed by N(x) - {y,z} and so that z sees Xl and y sees X2k ' Likewise, lety = Y1'" Y21 be the chordless path formed by N(y) - {x,z}, with Y1 = X2k' and sothat Y21 sees z; also let Z = Zl ... Z2m be the chordless path formed by N(z) - {x,y},with Y21 = Zl and Z2m = Xl' Note that X and Y have no common neighbour otherthan Yl' for otherwise this vertex with X and Y would form a triangle cutset. Now,X + Y + Z is an odd cycle, and not a triangle, so it must contain an odd numberof triangles. However, the cycle XiX + XX 2k + Y + Z is even, so Y + Z contains aneven number of triangles. By symmetry, X + Z, as well as X + Y, contains an evennumber of triangles. Therefore X + Y + Z must contain at least one triangle withone vertex in each one of X, Y and Z (and, by planarity, there is exactly one sucha triangle). Let x ', y' and z' be the vertices of this triangle with X' EX, y' E Yand z' E Z . Since N(x) is an even hole, the paths Pi = zX1 + X[x 1 ,x'] and P2 =X[X',X 2k] + X2kY have different parities. If Pi (resp . P2) is an even induced path,then Pi + x'y' + y'y + yz (resp. P2 + yz + zz' + z'x') is an odd hole, in both casesa contradiction. 0

Claim 2. Every triangle contains at most two vertices from D.

Proof Suppose on the contrary that some triangle consists of three vertices ofdegree four . Then it is a routine matter to check that the triangle and its neighboursinduce the octahedron. Since the octahedron is maximal planar, and G has noclique-cutset, G must be equal to the octahedron, which we have ruled out. 0

Claim 3. The components of G[D] are even cycles or paths of even length.

Proof Claim 2 implies that G[D] is a bipartite graph. Moreover, since every vertexof G[D] has a C4 for neighbourhood in G, there is no induced K 1, 3 in G[D].Therefore every component of G[D] is a chordless even cycle or path. Suppose thatsome component is an odd path P = Vi ' " v2l. Let us label the neighbours of VI asV2' X, Z, U, with edges XV2' UV2' ZX, and ZU. Since all vertices of P have degree four,they must all be adjacent to both X and u. Let us call y the fourth vertex of N(V21)'and notice that y is also adjacent to both X and u. We call Z = Zl ... Z2k the

178 CiL, Sales et al.

chordless path formed by N(z) - {x,vd, with Zl = u. Likewise, let X =Xl"'X2m

be the chordless path formed by N(x) - (PU {z,y}), with Xl = Z2k' Also, let Y =Yl... Y2q be the chordless path formed by N(y) - {x,V2l}' with Yl = X2m andY2q = u.

Notice that X nZ = {x.}, for otherwise, if there exists a vertex t in X nZ {x.}, then {t,x,z} would be a clique cutset. Similarly, X nY = {X 21}. Note thatz =F y, for otherwise P U{z} would induce an odd hole. Also, Z n Y is empty, forotherwise if there exists a vertex t E Z nY then P U {x,y, t} is either an odd hole ora triangle cutset. If z is adjacent to y, then P U{y, z} induces an even hole with X

and u adjacent to all vertices. But then either z and y have degree four or {z, y, x}or {z,y, u} is a clique cutset, in each case a contradiction. So z does not see y.

Now X + Y + Z is an odd cycle and so it has an odd number of triangles. Notethat XU Z U {y} induces an even cycle, and so X + Z must contain an evennumber of triangles. Likewise X + Y contains an even number of triangles. Thesame holds for Y + Z since Z U Y U {x} induces an even cycle. So, as in Claim 1,there is x' E X, y' E Y and z' E Z, such that {x',y',z '} is a triangle of X + Y + Z.Since X is an odd path, X[x',xlJ and X[X',X 2kJ have different parities.If X[x',xlJ (X[X',X 2kJ) is odd, then X[x',xlJ + xlz + zVl + P + v2ly +yy' + y'x ' (X[X',X2kJ + X2kY + YV21 + P + Vlz + zz' + z'x') is an odd hole, acontradiction. 0

Let S be a maximum stable set of G[DJ. By Claims 1 and 3, the set S containsevery second vertex on each cycle of D and it contains the two endpoints of everypath that is a component of G[D]. We observe that S meets every triangle T of G.Indeed, if T has two vertices of degree four then necessarily exactly one of them isin S; if T contains only one vertex of degree four, then this vertex cannot have twoneighbours in D (since its neighbourhood is a C4 and D is triangle-free), so it mustbe the endpoint of a path that is a component of G[DJ, and so it is in S.

Now, G - S has no triangle and so it is a bipartite graph (because any odd cycleof G - S would contain an odd hole of G), and every edge of G has at least oneextremity in G - S. To obtain a transitive orientation of G, make every vertex onthe left side of G - S a source and every vertex on the right side of G - S a sink.Observe that every edge of G is oriented because it has at least one extremity inG - S. To check that G is transitively oriented, assume that we have arcs ab andbe. Then necessarily b is in S, and so a, e are in G - S, a and b are adjacent(because N(b) is a C4 ), a is a source and e is a sink, hence the arc ae exists.So Lemma 13 holds. 0

Finally, the proof of Theorem 1 is complete. Note that this proof yields apolynomial algorithm which finds a sequence of contractions of a planar perfectlycontractile graph that leads to a clique.

3. Recognition of Planar Perfectly Contractile Graphs

We will use the algorithm of Hsu [6J for the decomposition of perfect planargraphs. Given a planar perfect graph G, Hsu associates with G a decomposition


tree as follows. The root of the tree is G. If H is a node of the tree, and H admits acutset Q which is of one of the four types defined below, then the children of Haresome specific graphs Hi built from the components of H - Q, as explained below.If H has no such cutset, then H is a leaf of the tree and it must belong to threespecial classes 9', 2, fC. (In fact such a tree can be defined for any planar graph, andHsu 's result is that the root graph is perfect if and only if the leaves are all in9' U 2 U fC.) Let us define precisely the types of cutsets. Let Q be a cutset of H, andB! , ... , B, the components of H - Q.

• Type I: Q is a clique-cutset.In this case the children are the graphs H[Bi U QJ for each i = 1, ... , k.

• Type II: Q = {a, b} with a, b not adjacent.In this case, let p be the parity of a chordless (a, b)-path in H (all such paths havethe same parity or else H contains an odd hole). Then the i-th child is defined bytaking H [B; U QJ and adding between a and b a chordless path of length two if pis even and three if it is odd. This path has one or two interior vertices which wecall artificial. (In fact we want to build these children only if none of them isisomorphic to H itself, i.e., only when either k > 2 or when k = 2 and each ofH[B! U QJ and H[B2 U QJ is different from a path P3 or P4 .)

• Type I II: Q = {a, b, c}, where ab is an edge and ac is not, and there is a chordlesseven (a, c)-path in some H [B j U {a, c}], and a chordless odd (b,c)-path in someH[B;U {b, c}].(In fact all (a,c)-paths in H - b must have the same parity or else we find an oddhole; likewise all (a, b)-paths in H - c have the same parity. Also k = 2 otherwiseH would contain a subdivision of K3,3 ')Then the i-th child is obtaind by taking H[B; U QJ and adding an artificial vertexa' with edges a'a and a'c and, if be is not an edge, a second artificial vertex b' withedges b'b and a'b'. (Again we do this decomposition only if none of the childrenis isomorphic to H itself.)

• Type IV: Q = {a, b, c, d } induces a C4 in H with edges ab, be, cd, da, there is achordless even (a,c)-path in some H[BjU {a,c}], and there is a chordless odd(b,d)-path in some H[B;U {b,d}]. Then the i-th child is obtaind by takingH[B; UQJ and adding two artificial vertices a', b' with edges a'a, a'b, a'c, a'b', b'a,b'c and b'd. (Again this decomposition step is applied only if none of the resultingchildren is isomorphic to H.)

One will apply the decomposition along a cutset of a given type only if there is nopossible decomposition along cutsets of lower types. Now let us define the classes9', 2, ~.

• Class ~ consists of (some) planar comparability graphs.

• Class 2 consists of planar line-graphs of bipartite graphs, such that the degreeof each vertex is two, three or four. The neighbourhood of such a vertex must beeither two non-adjacent vertices, or three vertices with exactly one edge betweenthem, or four vertices with exactly two non-incident edges.

180 Cd. , Sales et al.

Fig. 2. Graphs s. ,S2 and S3 of the Class !/

• Class!/' consists of the three graphs St , S2, S3 in Fig. 2, plus any of the fivegraphs obtained from Sl by replacing one or more of the edges e, f , 9 with achordless path of length three.

Hsu's decomposition tree can be built in time 0(n3 ) for every planar perfect graphwith n vertices.

We will show how to use Hsu's decomposition tree for the recognition ofperfectly contractile planar graphs.

The Algorithm

As proved in Section 2, Theorem 1, a planar perfect graph G is perfectly contractileif and only if it contains no odd stretcher. Our plan is to show that this can betested on the leaves of the tree.

Recall that for a stretcher M we will write X = Xl' " Xp , Y = YI . . .Yq , Z =ZI " 'Zr the three paths of M, so that {XI,YI ,zd and {xp,yq,zr} are the two triangles in M. For short, to refer to a segment of the paths X, Y or Z, we will use thesubscripts of its extremities.

Consider a non-leaf node H in the decomposition tree of G, and the corresponding cutset Q of H.

Since a stretcher has no clique cutset, the following lemma is immediate.

Lemma 14. If Q is of Type I, then H is perfectly contractile if and only if each childof H is perfectly contractile.

Now assume Q = {a,b} is a cutset of Type II with the above notation. If M isan odd stretcher in H, the two triangles of M must be in B,UQ for some i, forotherwise the three paths X, Y, Z of M should go across the 2-cutset, which isimpossible. Actually M is entirely in B;UQ, except if a, b are on one of these threepaths of M, say on X. In this case an odd stretcher can be found in Hi by replacingthe portion of X between a and b with the new artificial vertices of Hi' Conversely,if a child Hi has an odd stretcher M', then an odd stretcher M of H can be foundfrom M' by replacing the artificial path (if it does occur in M ') with any (a,b)-pathin the suitable component. Therefore we have:

Lemma 15. If Q is of Type II, then H is perfectly contractile if and only if each childof H is perfectly contractile.

Now assume Q = {a, b,c} is a cutset of Type III with the above notation. If Mis an odd stretcher of H, it could be that the two triangles of M are separated by Q,but only if (up to isomorphism) Q= {YI ,ZI'U} of M, where U is any vertex on


x - {Xl} ' In this case, using the path X, we see that in the component of H - Qcontaining the vertex X l there is a chordless (u'Yl)-path avoiding ZI and a chordless (u, Zd-path avoiding y l ' of the same parity. By virtue of the definition of TypeIII, one of these two paths could be extended to an odd hole of H, a contradiction.Therefore, the two triangles of M are again in the subgraph H[Bi UQ] for some i.Now, as above, M can be separated by Q only if a, c are on a path of M, say X. Infact we can conclude exactly as above by replacing the part of X between a and cwith the artificial path. So we have:

Lemma 16. If Q is of Type III, then H is perfectly contractile if and only if eachchild of H is perfectly contractile.

Now assume Q = {a, b,c,d} is a cutset of Type IV with the above notation, andassume M is an odd stretcher in H. Exactly the same argument as before thepreceding lemma shows that the two triangles of M must again be in H[Bi UQ] forsome i, and that vertices of M can be separated by Q only if either a, c or b, dareon one path X, of M. Exactly as above we can then replace the part of this Pibetween these two vertices with the corresponding artificial path, and obtain anodd stretcher in one child of H. Conversely, if a child of H has an odd stretcher M',then we can get an odd stretcher in H by replacing the artificial path (if it doesoccur in M') with any (a, c)- or (b,d)-path in the suitable component. So we have:

Lemma 17. If Q is of Type IV, then H is perfectly contractile if and only if each childof H is perfectly contractile.

The conclusion of these four lemmas is that a planar perfect graph G is perfectlycontractile if and only if every leaf of its tree is. Recall that each element of C(j isperfectly contractile [2, 4]. For the elements of [1', it is clear that 82 and 83 areperfectly contractile, while none of 81 and its subdivisions is perfectly contractilesince they contain an odd stretcher. So, there remains to test which elements of 2are perfectly contractile. The following will be useful.

Lemma 18 (Usu [6]). Let H be an element of 2 . Then two non-adjacent verticesform an even pair of H if and only if they are the neighbours of a vertex of degreetwo. 0

Let us call 2 0 the subclass of 2 consisting of the graphs C4 and C6 plus the twographs that we call M2 and M;. The graph M2 is the even stretcher whose threepaths have length two. The graph M; is what results from contracting an even pairin M2 and removing the pendant vertex. It is clear that these four graphs areperfectly contractile.

Lemma 19. A graph of 2 is perfectly contractile if and only if it is in 2 0 ,

Proof. Let H be a element of 2. IfH has a vertex of degree two then its neighboursform an even pair. We contract these two neighbours, and then remove the vertexz which has just become pendant. Repeating this procedure until there is no morevertex of degree two yields a unique graph H'. In other words, H' is obtained from

182 CiL. Sales et al.

H by shortening each path whose interior vertices have degree two to a path oflength one or zero according to the length parity of the path. (In fact such paths ofH cannot have length more than three, or else a cutset of Type II would exist.) ByLemma 18, it is clear that H is perfectly contractile if and only if H' is. If H has novertex of degree two then set H' = H.

Now we examine H'. IfH' is a clique we are done. IfH' is not a complete graph,we propose to show that H' has no even pair (from which it will follow that H', andhence H, contains an odd stretcher). First we remark that:

H' is an element of 2. (9)

Clearly H' is planar and has no odd hole. If H' = H there is nothing to prove. If His not C4 or C6 , let x, y be the endpoints of any path P of H whose interior verticeshave degree two and such that x, y have degree at least three. Note that suchvertices are the only ones whose neighbourhood may be different in H' and in H ,so the other vertices still have degree three or four and are not the center of a claw.The sets Qx = NH(x) - P and Qy = NH(y) - P are cliques of size two. If P has oddlength then in H' it is replaced by the edge xy and it is clear that both x, y still havedegree three and are not the center of a claw in H'. If P is even then x, yarecontracted, resulting in a contracted vertex xy whose neighbourhood is QxU Qy;so xy is not the center of a claw and has degree three (if QxnQy ¥- 0) or four(if QxnQy = 0). In consequence, (9) could fail only if H' contains a diamond.Suppose that H' does contain a diamond on vertices t, u, v, w with edges tu, tv, UV,

uw, vw. This is possible only if at least one of u, v (say u) is a vertex resulting fromthe contraction of the extremities of an even path P; whose interior vertices havedegree two in H (and maybe v too is a vertex resulting from the contraction of theextremities of an even path P; whose interior vertices have degree two in H). Thenwe see that {t,w} is a 2-cutset of H, with one component consisting of P; U Pv• SinceH has no cutset of Type II, it must be that this cutset has just two components andthat it would produce a child isomorphic to H, i.e., the other component B ofH - {t,w} is such that B U {t, w} is a chordless (t, w)-path oflength two or three. Inboth cases, one interior point of this path would still be a vertex of degree two inH', a contradiction. So (9) is established. Now Lemma 18 implies that H' has noeven pair.

To finish the lemma, observe that when H' is a clique of size two then H mustbe an even hole, and in fact H is a C4 or a C6 , because longer even holes havecutsets of Type II . IfH' is a clique of size three, then it is a routine matter to checkthat H can only.be M2 or M; . 0

In conclusion we have:

Theorem 3. A planar graph is perfectly contractile if and only if every leaf on itsdecomposition tree is an element of G&' U 2 0 U {82 , 83 } .

Using Hsu's algorithm and this theorem, we obtain an algorithm for testingwhether a given planar graph with n vertices is perfectly contractile in time O(n3

) .

If the graph is not perfectly contractile, the algorithm will return an odd stretcheror an induced odd hole.

On Planar Perfectly Contractile Graphs

Annex: Contracting an Odd Stretcher

183

In order to prove that no odd stretcher is a perfectly contractile graph, we need toknow the general type of graph that appears after one performs an arbitrarysequence of even-pair contractions starting from an odd stretcher. This leads us tothe following definition.

Definition 2. An odd b-stretcher is any graph G that can be obtained with thefollowing recipe.

(a) Take three pairwise vertex -disjoint non-trivial connected bipartite graphs B1,B2, B3. Mark their vertices pink or green so that each color is a stable set in each Bi.Pick one pink vertex Pi and one green vertex gi in each Bi, such that neither Pi nor gjis a cutpoint of Bi· Add the edges P1P2 , P2P3 , P1P3 and glg2, g2g3 , glg3 to make twotriangles.

(b) In addition, any of the following extensions may be added on:

• For each vertex u in {Pl ,P2,P3} U {gl ,g2,g3}' we may add a new non-trivialconnected bipartite graph B(u) and identify one vertex of B(u) with u. The verticesof B(u) are colored pink or green so that each color class is stable and u receives thesame color as' in (a).(So u becomes a cutpoint that separates B(u) - u from the rest of the graph, whichis connected.)

• For each pair u, v of vertices both in the triangle {Pl, P2, P3} or in the triangle{g l ' g2, g3 }' we may add a new connected bipartite graph B(u , v) having at least fourvertices, pick an edge of B(u , v) whose two endpoints are not cutpoints of B(u , v), andidentify the endpoints of this edge with u, v. (So {u, v} becomes a cutset that separatesB(u, v) - {u, v} from the rest of the graph .)

(In total there may be as many as twel ve such added-on bipartite extensions. Whenwe do not add on an extension we will say that it is null.)

It is clear that every odd stretcher is an odd b-stretcher, where the B;'s are oddchordless paths and the bipartite extensions are null.

Even b-stretchers can be defined similarly, the only difference being that wepick vertices Pi and gj of the same color in the three B;'s.

For eventual further reference, we would like to emphasize the fact that thedefinitions and claims in this section never involve planarity.

Claim 4. Let G be an odd b-stretcher with the above notation. Then two non-adjacentvertices x, y form an even pair of G if and only if they are of the following types:

(i) The vertices x, yare both in the bipartite subgraph B, U B(p;) U B(gi) and have thesame color , for any i == 1,2,3.

(ii) The vertices x, yare both in B(u, v) U B(u) U B(v) and are at even distance in thisbipartite subgraph (for some u, v E {Pl, P2' P3} or u, v E {gl' g2,g3})'

Proof First we show the if part.

When (i) holds, assume without loss of generality that x, yare both pinkvertices of B 1 U B(pd U B(g d = B~. Consider an arbitrary chordless path P be-

184 CiL, Sales et al.

tween them. If P is in B~ then it obviously has even length. In the alternate case, Phas vertices from G - B~. Since {Pl,gd is a cutset that separates B~ - {Pl,gdfrom the rest of the graph, we can write P = P; + Q+ Py , where P; (resp. Py ) is achordless path from x to PI (resp. from y to gd in B~, and Q is a chordless pathfrom PI to gl in G - (B~ - {Pl,gd). Since x, yare pink P, has even length and P,has odd length. After PI' the path Q does not go into B(Pl ,Pj) - Pj for j = 2, 3,because it would have to go out via Pj and then PlPj would be a chord. So we mayassume that P2 is next to PI along Q. Then Q does not go into B(P2) since P2 is acutpoint, and Q does not continue with P3 for then PlP3 would be a chord. HenceQ must go into B2 - P2 , and necessarily Q goes out via g2' So we can writeQ = PlP2 + Q' + g2gl' The path Q' has odd length since it is completely inside B2,and it follows that P has even length.

When (ii) holds, recall that {u, v} is a cutset that separates B(u, v)UB(u) UB(v) - {u, v} from the rest of the graph, and that u, v are adjacent. Therefore everychordless path between x and y stays entirely in B(u, v)UB(u) UB(v), and by theassumption any such path is even.

Now we prove the only if part of the claim. So we must show that for every pairof non-adjacent vertices x, y that are not of type (i) or (ii) there exists an oddchordless path between them (wecall such a path a killer). It is easy to see that suchpairs are of the following types:

(iii) The vertices x, yare both in the bipartite subgraph B, UB(Pi) UB(gi) andhave different colors (for some i = 1, 2, 3).

(iv) The vertices x, yare both in B(u, v)UB(u) UB(v) and are at odd distance inthis bipartite subgraph (for some u, v E {PI' P2,P3} or u, v E {gl,g2 ' g3})'

(v) x E B, and y E B2 •

(vi) Z E B(Pl) and y E B(P2,P3)'(vii) x E B(pd and y E B2.(viii) x E B(Pl) and y E B(g2)'(ix) x E B(pd and y E B(gl,g2)'(x) x E B(Pl) and y E B(g2,g3)'(xi) x E B(Pl, P2) and y E B(Pl ' P3)'(xii) x E B(Pl,P2) and y E Bl .

(xiii) x E B(Pl, P2) and y E B3.(xiv) x E B(Pl ,P2) and y E Biq«, g2)'(xv) x E B(Pl,P2) and y E B(gl,g3)'

(We omit all analogous types that could be obtained from the above by swappingthe "p's" and the "g's" or by permuting the subscripts 1, 2, 3.)

As a preliminary, for each i = 1,2,3 let R, be a chordless path between Pi andgi in Bi • Any such path is odd.

In cases (iii) or (iv) the existence of a killer is obvious since the bipartitesubgraphs in question are connected.

Assume case (v) occurs with x, y of the same color, say pink. Since gl (resp. g2)is not a cutpoint of B, (resp. B2 ) there exists a chordless path Px (resp. Py ) from x toPI (resp. from y to P2) in B, - gl (resp. B2 - 92)' Since x, yare pink Px , P, are even.So P; + PlP2 + Py is a killer.


Assume case (v) occurs with x pink and y green. Since gl is not a cutpoint of BI

there exists an even chordless path P; from x to PI in BI - gI' Likewise there existsan even chordless path P, from y to gz in Bz - Fz- So P; + PIP3 + R3 + g3gz + P,is a killer.

Now we consider cases (vi) to (x). Let P, be a chordless path from x to PI inB(pd·

In case (vi), we may assume y ¢ {PZ,P3} or else this would be case (ii) or (iv).Since pz (resp. P3) is not a cutpoint of B(PZ,P3) there exists a chordless path P,(resp. P;) from y to P3 (resp. to pz) in B(PZ,P3) - Pz (resp. in B(PZ,P3) - P3)' SinceB(pz , P3) is bipartite Pyand P; have different parities. So, one of the chordless pathsP; + PIPZ + P, or P; + PIP3 + P; is a killer.

In case (vii),we may assume y # pz, or else this would be case (ii) or (iv). Let Py

(resp. P;) be a chordless path from y to Fz (resp. to gz) in Bz - gz (resp. in Bz - pz).These two paths have different parities. Then one of the chordless paths P; +PIPZ + P, or P, + PIP3 + R 3 + g3gZ + P; is a killer.

In case (viii), let P, be a chordless path from y to gz in B(gz). The precedingparagraph entails the existence of two chordless paths of different parities from xto gz in G - (B(gz) - gz). Combining one of them with P; gives a killer.

In case (ix), we may assume y ¢ {gl ' gz} or else this would be case (i) or (vii).There exists a chordless path P, (resp. P;) from y to gl (resp. to gz) in B(gl ,gz) - gz(resp. in B(gl,gz) - gd. These two paths have different parities. Hence one of thechordless paths P; + PIP3 + R3 + g3g1 + P; or P; + PIP3 + R3 + g3gZ + P; is akiller.

In case (x), we may assume y ¢ {gZ,g3} or else this would be a previous case.Let Py (resp. P;) be a chordless path from y to gz (resp. to g3) in B(gZ,g3) - g3(resp. in B(gz,g3) - gz). These two paths have different parities. Then one of thechordless paths P; + PIPZ + R z + P; or P; + PIP3 + R3 + P; is a killer.

Now we consider cases (xi) to (xv). We may assume x ¢ {PI'PZ} or else thiswould be one of the cases (vi) to (ix) or an analogue. Since pz is not a cutpoint ofB(PI'Pz) there exists a chordless path P; from x to PI in B(PI'Pz) - Pz- Likewisethere exists a chordless path P~ from x to pz in B(PI' pz) - PI' The paths P; and P~

have different parities.In case (xi), we may assume y ¢ {PI,P3} or else this would be a previous case.

As usual there exists a chordless path P, from y to PI in B(PI' P3) - P3 and thereexists a chordless path P; from y to P3 in B(PI,P3) - Pl ' These two paths havedifferent parities. So one of the chordless paths P, + Py or P~ + PZP3 + P; is akiller.

In case (xii), we may assume y # PI' There exists a chordless path P; from y togl in BI - PI' Then one of the chordless paths Py + glgz + Rz + P~ .or P, +glg3 + R3 + P3PZ + P~ is a killer.

In case (xiii), let Py be a chordless path from y to P3 in B3. Then one of thechordless paths P, + P3PI + P; or P, + P3PZ + P~ is a killer.

In case (xiv), we may assume y # gz. Let P, be a chordless path from y to glin B(gl,gz) - gz. Then one of P, + R I + P, or P, + glg3 + R3 + P3PZ + P; is akiller.

In case (xv),there is a chordless path Py from y to gl in B(gl , g3) - g3 and there


is a chordless path P; from y to g3 in B(gt, g3) - gt . These two paths have differentparities, so one of Py + R, + P; or P; + R3 + P3P2 + P; is a killer. This completesthe proof. 0

Claim 5. Let G be an odd b-stretcher and {x, y} be any even pair of G. Then Glxy isan odd b-stretcher.

Proof. Assume x, yare as in case (i) of Claim 4, i.e., x, y E B~ = B, U B(pd U B(gt).Write G* = Gjxy. Clearly after the contraction B~ becomes a new bipartite graphBr (If one of x, y is Pt or gt we still call the contracted vertex Pt or gt .) The graphG* is "almost" a b-stretcher. The only changes that are relevant to the descriptionof G* are either that we may have contracted vertices from different connectedcomponents of B~ - {Pt ,gd (hence these two components become one, and maybePt or gt is no longer a cutpoint) or that we may have created new components(and maybe Pt or gt becomes a cutpoint). However, G is still an odd b-stretcherwhere Bt is modified as follows: the "new" B(pt) is Pt together with all theconnected components of Bt - Pt that are bipartite. The analogue holds for thenew B(gd. The new B; consists of {Pt,gd together with all bipartite componentsof Bt - {Pt,gd that are not already in the new B(pt) or the new B(gt). It followsthat neither Pt nor 9 t is a cutpoint of this new Bt .

Now assume x, yare as in case (ii) of Claim 4, more precisely x, y E C =

B(Pt ,P2)UB(pdUB(P2)' Write C* = Clxy. As in the preceding paragraph, G* =Glxv is an odd b-stretcher with the following modifications : the "new" B(pt) is Pttogether with all the bipartite connected components of C* - Pt . The analogueholds for the new B(P2)'The new B(Pt,P2) is {Pt,P2} together with all the bipartitecomponents of C* - {Pt,P2} that are not already in the new B(pd and the newB(P2)' Hence neither Pt nor P2 is a cutpoint of this new B(pt, P2)' 0

Claim 6. Let G be an odd b-stretcher having no even pair. Then G is a C6 .

Proof. Under the hypothesis, B(pd is null or else any neighbour of Pt in B(pt)together with P2 would form an even pair of G by (ii)of Claim 4. Likewise B(pt, P2)is null or else it would have two vertices that form an even pair. Hence all extensions are null. Likewise, each B, must be a clique (i.e., two adjacent vertices) or elseitem (i) of Claim 4 would again yield an even pair in G. These remarks show thatG is nothing but a C6 • 0

Lemma 20. Let G be any odd b-stretcher. Then any sequence of even-pair contractions that cannot be continued results in a C6 •

Proof. A direct consequence of the three claims that precede.

It is easier to prove the following (but we will not use it):

o

Lemma 21. Every even b-stretcher is perfectly contractile (and in particular can beeven-contracted into a triangle).


Acknowledgements. We are grateful to Claude Benzaken for many comments and suggestions during the writing of this paper.

References

1. Bertschi, M.: La colorabilite unique dans les graphes parfaits. PhD thesis , Institut deMathematiques, Universite de Lausanne, Switzerland, 1988

2. Bertschi, M.E .: Perfectly contractile graphs. J. Comb. Theory Ser. B SO, 222-230 (1990)3. Fonlupt, J., Uhry, J.P .: Transformations which preserve perfectness and h-perfectness of

graphs. Ann. of Disc. Maths. 16,83-85 (1982)4. Hertz, A., de Werra, D.: Perfectly orderable graphs are quasi-parity graphs: a short

proof. Disc. Math. 68, 111-113 (1988)5. Hsu , W.-L.: Coloring planar perfect graphs by decomposition. Combinatorica, 6,381

385 (1986)6. Hsu, W.L.: Recognizing planar perfect graphs. J. Assoc. Compo Mach. 34, 255-288

(1987)7. Meyniel , H.: A new property of critical imperfect graphs and some consequences.

European Journal of Combinatorics 8,313-316 (1987)8. Nishizeki, T., Chiba , N.: Planar Graphs: Theory and algorithms, volume 32 of Ann.

Disc. Math. North Holland, 19889. Everett, H., Reed, B.A.: Problem session on parity problems. Perfect Graphs Workshop,

Princeton University, New Jersey, June 199310. Tucker, A.: The validity of the perfect graph conjecture for K 4 -free graphs. In: C. Berge

and V. Chvatal: Topics on perfect graphs, pages pp . 149-157. Amsterdam: NorthHolland 1984

Received: March 8, 1995Revised: September 19, 1996

on planar perfectly contractile graphs

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