An algorithm for coloring some perfect graphs.
ABSTRACT We propose a sequential method with 3chromatic interchange for coloring perfect graphs.

Chapter: On the coloration of perfect graphs
05/2006: pages 6584;  SourceAvailable from: Claudia Linhares Sales[Show abstract] [Hide abstract]
ABSTRACT: In this note, the authors generalize the ideas presented by A. Tucker in his proof of the SPGC for K 4 Gamma efree graphs in order to find a vertex v (called here a Tucker vertex) in G whose special neighborhood allows to extend a previous coloring of G Gamma v. The search of such a vertex led us to define a property on the intersection of large cliques and a family of classes of graphs that satisfy this property. We prove that every graph G in this family has a Tucker vertex and we use this fact to give a polynomialtime algorithm to compute !(G). We give a proof of the SPGC for a new class: graphs where every edge of a maximal clique of size at least 4 belongs to precisely that clique. The proof directly yields a combinatorial polynomialtime algorithm for coloring perfect graphs in this new class with the size of a maximum clique colors. Key words: Perfect graphs, vertexcoloring algorithms, Tucker vertex. 1 Introduction A graph G is perfect if, for each induced subgraph H of G...04/1999;
Page 1
DISCRETE
MATHEMATICS
ELSEVIER Discrete Mathematics 183 (1998) 116
An algorithm for coloring some perfect graphs
Hac~ne Ait Haddadrne a, Sylvain Gravier b, Frederic Maffray c,*
a Dept. de Recherche Opdrationnelle, Institut de MathEmatiques, USTHB, BP 32, El Alia 16111,
Algiers, Aloeria
b Laboratoire Leibniz, IMAG, BP 53, 38041, Grenoble Cedex 9, France
c CNRS, Laboratoire Leibniz, IMAG, BP 53, 38041, Grenoble Cedex 9, France
Received 7 March 1996; received in revised form 18 March 1997; accepted 31 March 1997
Abstract
We propose a sequential method with 3chromatic interchange for coloring perfect graphs.
O. Introduction
In a graph G=(V,E)
e(u)¢e(v) for every edge uv. Each color class is a stable set, hence a kcoloring
can be seen as a partition into stable sets $1 ..... Sk. The chromatic number z(G) is
the smallest k such that G admits a kcoloring. A graph G is then called perfect [1]
if z(H)= co(H) for every induced subgraph H of G, where co(H) is the size of a
largest clique in H. Determining the value of x(G) is an NPhard problem in general.
It becomes polynomially solvable for perfect graphs, as proved by Gr6tschel et al. [4];
their algorithm however uses the ellipsoid method and is not very efficient in practice.
Tucker [10] found a combinatorial algorithm for coloring every 3colorable perfect
graph G with x(G) colors, in time O(]VI3). Note that the 3colorable perfect graphs
are exactly the K4free perfect graphs.
The following sequential method for coloring perfect graphs was proposed in [5,6].
Let v be a vertex of G. Assume that G  v has been colored with a coloring c using
co(G)/> 3 colors $1,..., S~(G). Suppose that the following property T(v, c) happens: there
exist three distinct colors i,j,k such that the subgraph induced by Si USj USk tO {v}
contains no K4. We can then apply Tucker's algorithm to this subgraph and color it
with three colors. Hence, keeping the other co(G) 3 colors, we get an co(G)coloring
of G. So it is interesting to find such vertices v that T(v,c) holds for every co(G)
coloring of G  v. We call such vertices Tucker vertices.
a kcolor&o is a mapping c:V*{1,2 ..... k} such that
* Corresponding author. Email: frederic.maffray@imag.fr.
0012365X/98/$19.00 Copyright (~) 1998 Elsevier Science B.V. All rights reserved
PH S0012365X(97)000824
Page 2
2 H. Ai't Haddaddne et al./Discrete Mathematics 183 (1998) 116
It would be interesting to characterize the class of graphs where every induced
subgraph has a Tucker vertex, but this seems to be too hard. In this paper we investigate
the following Property (P) of a vertex v: the union of any four K4's containing v
contains a Ks. Equivalently, the neighbourhood of v has the following property (P4):
the union of any four triangles contains a K4. The main result is:
Theorem 1. If the union of any four K4's containing v contains a K5 then v is a
Tucker vertex.
The proof of this theorem is based on the description of graphs with property (P4)
given in the next section.
We can define the class ~ of graphs such that every induced subgraph H has a
vertex with Property (P). Given such a graph we can order its vertices as Vl ..... v,
so that vl has Property (P) in G, v2 has Property (P) in G vl, etc. The existence
of such an ordering is a characteristic of graphs in qq; indeed, if H is an induced
subgraph of G and i is the smallest subscript such that vi E V(H) then it is clear
that vi has Property (P) in H. Finding such an ordering is algorithmically easy since
testing Property (P) is a polynomial task, as shown in the next section. Hence (~ is a
polynomially recognizable class.
Now recall Berge's Strong Perfect Graph Conjecture: a graph is perfect if and only
if it contains no odd chordless cycle (odd hole) and no complement C2k+1 of an
odd chordless cycle (odd antihole). Tucker proved that every K4free graph satisfies
Berge's conjecture, namely that every graph containing no K4, no odd hole and no C7
is perfect.
m
Corollary 1. Let G be a graph in f# and containing no odd hole and no C7. Then G
is perfect.
Proof. We prove this corollary by induction on [V[. By Theorem 1, the graph G has
a Tucker vertex v. By the induction hypothesis, G  v admits an og(G  v)coloring. If
N(v) contains an o~(G v)clique then og(G)= co(Gv)+ 1 and we assign to v a new
color. In the opposite case, we have co(Gv)= ~o(G). Then, since v is a Tucker vertex,
there exist three colors i,j,k such that the subgraph G ~ induced by Si USjUSkU {v}
is K4free. This G t contains no odd hole and no C7 and by Tucker's theorem it is
perfect, and so we can apply Tucker's algorithm to it. This yields an ~o(G)coloring
of G. []
Recall that a graph is chordal if it does not contain as an induced subgraph any hole
of length at least four. It is well known [2] that every chordal graph either is a clique
or has two nonadjacent simplicial vertices (a vertex is simplicial if its neighbourhood
is a clique). Hence computing co(G) is easy for a chordal graph (see [3]).
Clearly every 3colorable perfect graph is in f~, and every chordal graph too. Also
every linegraph of bipartite graph is in fg; this is because in a such a linegraph,
Page 3
H. A~'t HaddadOne et al./ Discrete Mathematics 183 (1998) 116 3
the neighbourhood of every vertex is one or two disjoint and nonadjacent cliques,
and so every vertex has Property (P). Every diamondfree perfect graph is also in c~
(the diamond is the graph K4 e); this is because, as shown by Tucker [9], every
diamondfree perfect has a vertex whose neighbourhood contains at most two disjoint
cliques of size at least three, and this vertex clearly has Property (P).
1. A characterization of property (P4)
Given a graph G, let T(G) be the graph whose vertices are the triangles (cliques of
size three) of G and whose edges are the pairs of triangles of G that have a common
edge. Then G is called tconnected if T(G) is connected. For a given component T/ of
T(G) we may consider the subgraph Gi of G whose edges are those edges that appear
in a triangle that is a vertex of T/. Clearly T(Gi)= ~. The subgraphs Gt ..... Gd such
that T(G1 ), T(G2) ..... T(Ga) are the connected components of T(G) will be called the
tcomponents of G. Two Gi's may have common vertices but they have no common
edge.
An edge is called flat if it does not lie in a triangle.
We consider the property (P4) of a graph: the union of any four triangles in this
graph contains a K4. Hence a graph has a vertex v with the Property (P) mentioned
in the introduction if and only if the neighbourhood of this vertex has property (P4).
We call forbidden configuration any graph that contains four triangles but does not
contain a K4. So G has property (P4) if and only if it does not contain a forbidden
configuration. In fact it is sufficient to consider the forbidden configurations that are
minimal (with respect to vertexset inclusion). Such configurations have at most twelve
vertices since in the worst case the four triangles are vertexdisjoint, and so there
is a finite number of them. This implies that the graphs with property (P4) can be
recognized in time o(IvI12). The characterization that we give below will lead to a
much faster recognition procedure.
Clearly, the fiat edges of a graph have no influence on property (P4) and on the
definition of T(G) and of the tcomponents; hence we may assume that a graph with
property (P4) has no fiat edges. Fig. 1 shows the minimal configurations containing
no fiat edges and whose tcomponents are vertexdisjoint. The minimal configurations
in which different tcomponents intersect are not drawn.
Lemma 1. Let G be a tconnected graph with property (P4) and having no flat edges.
Then G has no induced C4.
Proof. Suppose that G contains a C4 with vertices VO, Vl,/)2,v 3 and edges VOVl,Vll)2,
v2v3, V3Vo. Write C{vl, v2, v3, vo}. The subscripts on the vi's will always be under
stood modulo 4. By the hypothesis, every edge oivi+ 1 lies in a triangle. For each i,
choose a vertex ui such that ~)iOi+lUi is a triangle and such that U= {ul,u2,u3,uo} is
as small as possible (some ui's may be equal).
Page 4
H. Ai't Haddadkne et al./Discrete Mathematics 183 (1998) 116
+ F3
] FlO
o•11
Fig. 1. The minimal forbidden configurations.
Note that no vertex u can see all of vl, v2,/)3, l)0 for otherwise these five vertices
would form a forbidden configuration (a 4wheel). Hence ui ¢ ui+2 (i = 0, 1 ) and so
lull>2.
If IUI =2, then necessarily for some i we have ui = ui+l (and ui does not see vi+2)
and ui+2 =ui+3 (and ui+2 does not see vi). Here the pair uiui+ 2 may be an edge or
not, but in either case U to C is a forbidden configuration.
If IUI =4 then each ui misses /)i+2 and vi+3. If U does not induce a K4 then it
follows that U tO C does not induce a K4, and so U U C is a forbidden configuration. If
U does induce a K4 then ul, uo, Vl, v2, u2 induce a forbidden configuration (a 4wheel).
Now we may assume that I UI = 3, hence and by symmetry we may assume ul = uz
and u3 Cuo. It must be that u3 misses vl, for otherwise {Ul,U3} would contradict the
minimality of U. Likewise, Uo misses v3. Also, v2 must miss at least one of Uo, u3 or
else {uo, u3} would contradict the minimality of U. Assume that v2 misses u3. If v2 also
misses uo then U tA C cannot contain a K4, and so U tA C is a forbidden configuration.
So we assume that v2 sees uo. It follows that uo sees Ul or else CtO{uo, ul} would
induce a forbidden configuration. Note that uou3 is not an edge or else CU {uo, u3}
would be a forbidden configuration. Likewise ul u3 is not an edge. The same arguments
show that, letting U3 be the set of all vertices that see vo and v3, every vertex in U3
misses vl,v2, ul and uo. Moreover U3 is a clique or else CU {ul,u~3,u~3t } would induce
a forbidden configuration for any two nonadjacent vertices u 3~, u 3" of U3.
By the tconnectivity of G, there must exist a shortest path in T(G) from the triangles
induced by U3 tA {Vo, v3 } to the triangles induced by C tO {ul, uo}. Considering along
this path the first triangle ~ that is not included in {vo, v3} tO U3, by symmetry we find
!
two cases: either ~ = {z, u~, Vo} for some u 3 c U3 and z ~ U3; or z = {z, u~, u~ ~} for some
Page 5
H. A?t Haddadkne et al./Discrete Mathematics 183 (1998) 1 16 5
u3,u 3t tt C U3 and z ¢~ U3. Since z ff U3 we may assume that z misses v2 (necessarily in
the first case, by symmetry in the second case). Clearly z ~ C t3 U U U3. In the first
case, ifz sees both Vl,V2 then CU {u~,z} is a forbidden configuration, while ifz misses
one of Vl,V2 then CU{Ul,U~,Z} is a forbidden configuration. In the second case, if
! /!
z sees both ul,v2 then Ul,V2,V3,u3,u3,z induce a forbidden configuration; if z misses
one of Ul,V2 then Vl,V2,v3,Ul,U3,U3,Z induce a forbidden configuration. In all cases
the contradictions complete the proof. []
I I!
Lemma 2. Let G be a tconnected graph with property (P4) and having no flat edges.
Then G has no induced Ck (>~4).
Proof. Suppose that G contains a hole, and let k be the smallest integer such that G
contains a Ck, on vertices C={vo, Vl ..... V~l} and edges ViVi+ 1 (i =0 ..... k1 mod k).
By the previous lemma k >~ 5.
For each i there exists a vertex ui such that uil)ivi+ 1 is a triangle of G. It is easy to
check that every vertex of G  C can see either just one or two or three consecutive
vertices of C, for otherwise C tO {ui} would contain a shorter hole or a forbidden
configuration (a kwheel).
First suppose that, for all i, we have N(ui)NC={vi, vi+l}. Then {Uo, U~,U2,U3}
must form a K4 for otherwise Vo, Vl,V2,v3,va, uo, ul,u2,u 3 would induce a forbidden
configuration; but then Vo, Vl, v2, v3, uo, ul, u2 induce a forbidden configuration.
Now we may assume that uo sees vo, v~ and v2. If u3 sees v2 then Vo, vl,v2,v3,v4,uo, u 3
would induce a forbidden configuration (even when uo sees u3, and even when k = 5).
Hence u3 misses v2. Likewise u2 misses V4 (or else we could take u2 = u3). If u2
misses Vl then Vo, Vl, v2, v3, v4, uo, u2, u3 form a forbidden configuration (regardless of
the adjacency between uo, u2,u3). So u2 must see vj. Then u2 must see uo or else
vo, vl,v2,v3,uo, u2 induce a forbidden configuration. Now u4 must miss v3 or else
vl, v2, v3, v4, vs, u2, u4 induce a forbidden configuration (even when k = 5). Likewise u3
misses vs. But now vl, v2, v3, v4, vs, u2, u3, u4 induce a forbidden configuration (regardless
of the adjacency between u2, u3, u4). These contradictions complete the proof. []
Under the same hypothesis it is possible to show the stronger fact that G cannot
contain a chordless path on six vertices, but we will not use this.
Lemma 3. Let G=(V,E) be a tconnected graph with property (P4) and having
no fiat edges. Then we can partition V into cliques Q, $1 ..... Sk with the following
properties, where Bi = N ( Si ) N Q:
(sl) There are no edges between two Sf s;
(s2) For all i=1 .... ,k, [Bil>~2 and QBier;
(s3) For each i= 1 ..... k, there exists bi EBi such that SitO{bi} is a clique.
Any such partition will be called an acceptable partition. Note that Q is a maximal
clique of G by (s2). A graph may have several acceptable partitions. If SiUBi is a