The smallest hard-to-color graph

The smallest hard-to-color graph

Discrete Mathematics 96 (1991) IW 100-212 North-Holland he smallest hardPierre Hansen* Julio Kuplinsky Mathematics Department. Received 3 Decem...

1MB Sizes 3 Downloads 62 Views

Discrete Mathematics

96 (1991)

IW

100-212

North-Holland

he smallest hardPierre Hansen* Julio Kuplinsky Mathematics

Department.

Received 3 December

, NJ MWRZ, iJSA

Uhersity . New Rrmwick

Rutgen

1087

Revised I3 March 1990

Abstract Hansen P. and J. Kuplinsky.

The smallest hard-to-color

graph. Discrete Mathematics

96 (1991)

w-212. For a given approximate some implementation graph

vertex-coloring

of the algorithm

is said to be hard

non-optimal main

algorithm.

a graph is said to be slightly hard to color if

uses more colors than the minimum needed. Similarly.

to color

if every

implementation

of the algorithm

coloring. We study smallest such graphs for the sequential coloring algorithm.

result is that, for the largest-first

algorithm.

there

is a unique smallest

a

resuhs in a Our

hard-to-color

graph.

1. Introduction As graph coloring numerous

heuristic

is NP-hard, algorithms

perform well. Indeed, Furthermore,

have been proposed.

no heuristic

algorithm

with the best known bound,

These

optimally

colors is NP-hard

[3].

is known for which the ratio of the number

number is bounded

i.e. O(n(log

by a constant. The algorithm

log n)‘/(logrt)‘)

appears to be that of

(91. For most of the other heuristics the bound on this ratio is O(n) negative

and

In most cases they do not

finding a coloring in less than 2y(G)

of colors used to the chromatic Wigderson

only small graphs are colored

results suggest studying

when

and why heuristics

(4.

for graph

coloring perform

badly. In time this might lead on the one hand to the design of

better heuristics

and on the other

colorable optimally direction

in polynomial

to the recognition

of new classes of graphs

time. In this paper we make a first step in that

by studying smallest graphs for which sequential

not perform well, with special reference to the largest-first

coloring algorithms heuristic.

* Research supported in part by Grant 0271 of the Air Force Office of Scientific Research. cn)12-365x/91/$03.50

@J 1991 -

Elsevicr Science Publishers H.V.

All rtght~ reserved

do

P. Hansen, J. Kuplinsky

200

A related question is to study when heuristics perform extremely well. This was recently done by Chvatal et al. [2], who characterize graphs for which the largest-first and smallest-last heuristics provide optimal colorings for all induced subgraphs.

2. Notation and terminology All the graphs we consider are finite, with no multiple edges or loops. If G is a graph, we denote by V(G) its vertex-set, by E(G) its edge-set and by y(G) its chromatic number. The argument G will be omitted in those cases in which only one graph is under consideration. We will consider ‘partially specified’ algorithms, ori to be more precise, classes of algorithms. One such example is the largest-first algorithm. To apply this algorithm to a graph, one orders its vertices according to non-increasing degrees. Then, assuming inductively that a (possibly empty) subset of vertices has been colored, the first uncolored vertex is assigned the least possible color. An implementation of the algorithm must specify how ties in the ordering should be broken. By an instance of this algorithm we understand any coloring resulting from the application of this rule to a specific graph, with the only restriction that the vertices should be ordered as stated above. It is in this sense that we speak of a partially specified algorithm. Definition. A graph G is said to be slightly-hard-to-color (WC) with respect to an algorithm d if for some instance of it the number d(G) of colors used satisfies d(G) > y(G). We similarly define a hard-to-color (HC) graph as one for which any instance of the algorithm is a non-optimal coloring. We will determine smallest graphs for which a given algorithm produces non-optimal colorings. More precisely, in the case of SHC graphs, we are looking for a graph G which realizes min()E(G)i:

for some instance of .P$ d(G) > y(G), and IV(G)1 = no},

where no= min{ IV(G)l: for some instance of ti, A similar definition applies to HC graphs. We now discuss some algorithms.

3. SequentiaSalgorithm To apply this algorithm to a graph G one chooses a total ordering (u, , . . . , v,J of V. The co Ioring c is defined inductively for 1 6 i d n by C(Vi)= min(k E N: C(Vj)f k for every i < i such that (Vi, 1~~)E E).

The smallest hard-to-color graph

201

Definition. A total ordering of V such that the kiduced coloring of G is optimal will be called an optimal ordering. Example 3.1. Let G denote the following graph. 0 1

3

4

e 2

Fig. 1.

The indicated non-optimal.

ordering

results

in the following

e 1

2

3

coloring,

which is clearly

e 1

Fig. 2.

Therefore G is SHC for the sequential algorithm. For bipartite graphs, it is easy to prescribe optimal orderings . This is the next result, the proof of which is obvious.

in

Proposition 3.2. Let G be a bipartite and connected graph, and assume that v,) is a total ordering of V such that for any index i, 1~ i 6 n, the ( 2/l,-.=9 subgraph induced by (v,, . . . , Vi> is connected. Then this ordering is optimal. If G is not connected, one obtains an optimal coloring by considering such an ordering for each one of the components of G. Proposition 3.3. Let (vl, . . . , v,) be a total ordering of the vertex-set of the graph G, and let c be the coloring obtained by applying the sequential algorithm to G with this ordering. Suppose that at least 7 colors are used (i.e., a-@ j 3 pj. ihen, (i) for any i E (1, . . . , n> and any 5, 1 s s CCc(‘Ui), there exists an index j < i such that c(Vj) = s and Vj ti adjacent to Vi, (ii) IEI 2 0, (iii) F d lTlaX,,i,, (min{i, deg(vi) t l}}, (iv) if t > y, then there exists an index i, jj + 1 s i 6 n, such that deg(vi) 3 j+ 1.

my-

Proof. (i) This is obvious by the very definition of the sequential algorithm. (ii) For each color t, 1
- 1).

P. Himen, J. Kuplinsky

202

(iii) It is clear by induction that c(uJ c i for all i, 1s i s n. Moreover, I{C(Uj):{Uj,

Vi} E E)l -@(uj),

so {1,2,. . . , de&Q + 11 Q (C(uj): { Uj, vi) E E), and hence C(Ui)6 deg(ui) t 1. (vi) Choose an index i such that c(Ui) = jk Then by (i) there exist indices . . - 1) such that Uis is adjacent to Ui and c(Ug) =S for 11, . . . ) 1jL1 E{l,...,i 1s s =zjj - 1. The indices must be distinct, so y G i. cannot be a clique, since in that case {Vi, Vi,, . . . , Vi?_,) The set {Uil, . . . , u+} would be a clique too, contradicting the fact that t > y. There exist therefore two - 1 such that {Uis,vi,} $ E. indices l<:\<s would imply by (i) the existence of a vertex Uil, where 1s r s jj - 1, adjacent to Vi, and such that C(Q) = s. But c(uj,) = r, so r = s and {Vi,, Us,}E E, which is a contradiction. We therefore conclude that i 3 v + 1. 0 Remark 3.4. Result (iii) has been proved by Welsh and Powell additional assumption that the vertex ordering satisfies deg(ui+,) 16 i s n - 1. This is not necessary, but, as noted by Syslo [7], obtain a better bound by choosing an ordering other than that vertex degrees.

[8] under the 3 deg(ui) for one does not of decreasing

Proposition 3.5. A!! SHC graph with respect to the sequential algorithm contains a path of length 3 with non-adjacent endvertices. Proof. Let G be an SHC graph for the sequential algorithm, c a coloring using t > y colors. There exists a vertex u such that c(u) = j? By definition of the algorithm, there exist vertices ul, . . . , u7.-, adjacent to u and such that C(Ui)= i for 1
Corollary 3.6. The path of length three (denoted P4) ks the smallest SHC graph with respect to the sequential algorithm.

Proposition 3.7. Let ( Vk)rzl be a partition of V into stable sets, (u, , . . . , u,) a total ordering of V. For 1 G i < n, let k(i) be defined by Ui E VktiI. If k is a nondecreasing function, then the sequential coloring induced by the ordering uses at most m colors. Proof. We will prove inductively that c(uJ G k(i) for 1 G i d n. The result being clear for i = 1, assume that i > 1. Suppose that vi is adjacent to Uj, where j < i.

The smallest hard-to-color graph

203

Ihen c(uj) G k(j) 6 k(i), but in fact k(j) ( k(i), because Vi and Vi are adjacent. We conclude that c(Vi) d k(i). El The following result has been obtained before (see for instance Matula et al. [51). Coroky

3.8. There are no HC graphs for the sequential algorithm.

Proof. It is enough to apply Proposition sets. Cl

3.7 to a partition of V into y stable

4. Largest-first algorithm This is a special case of the sequential (v,, - - 9 v,) of V(G) is required to satisfy

algorithm,

in which the ordering

l

d(Vi+,)

ad(Vi)

for 1 si Qz - 1.

Any such ordering will be called a largest-first ordering. Examples 4.1. (ij Consider the following graph. a

0

5

1

3

4

2

6

Fig. 3.

It is SHC, for the indicated ordering induces the following coloring. e

w

2

1

l

2

3

1

Fig. 4.

(ii) Consider now the graph in Fig. 5.

U

t

Fig. 5.

2

P. Hansen, J. Kuplinsky

204

ekinition. This will be called the envelope graph. This graph has a unique optimal coloring, the one with color ciasses (s, u}, {t, v> and {x, y, w}. But any largest-first ordering will give rise to a coloring c with c(s) = c(v). We conclude that this graph is HC. We propose to prove that these are smallest SHC and HC graphs, respectively. Note first the following easy consequence of Proposition 3.3(iv). Corolary 4.2. Let G be an SHC graph for the largest-first algorithm, (d, , . . . , d,,) its degree-sequence. Then y -I-2 d n and y s dy+z. Proposition 4.3. Let G be an SHC graph for the largest-first algorithm. Denote by n the number of vertices of G. (i) !f y = 2, then G contains a path of length five. (ii) rf y > 2 and some instance of the algorithm uses at least v > y colors, then j%n-2. Proof; (i) Consider a largest-first ordering d of V inducing a non-optimal coloring c. There exists a vertex v such that c(v) = 3. By Proposition 3.3(i), there are vertices vl, v2 adjacent to v, and wI adjacent to v2 such that VI

-4

v2

=cv,

w,

-c v2

and c(v,) = I,

c(u2)

= 2,

c(w,)

=

1.

Notice that w, # vI because G contains no triangles. It follows from this that deg(v,) 3 2 and deg&) 2 2. Since G contains no triangles, there exist vertices X, y $ {v, vi, v2, wl} such that x is adjacent to v, and y is adjacent to wI. Notice that x # y because G does not contain odd cycles. Then (x, vl , v, vz, wI , y) is the required path. (ii) Let (vi, . . . , v,,) be a largest-first ordering of V such that the induced coloring c uses at least t colors. As in the proof of Proposition 3.3(iv), we argue that there is a vertex vi such that c(vI) = 7, and indices h, k E { 1, . . . , i - 1} such that vh and vk are adjacent to vi but not to each other, and moreover, C(Q) < C(Q)* But then, there exists a vertex w adjacent to vk and such that c(w) = c(vlt). Clearly w, vk, IJ~are distinct, and {v,, , w } , {v,, , vk > $ E, so deg(2Jj) 6 n - 3. On the other hand, 7 - 1 s deg( vi) 6 deg( v,,), so v 6 n - 2. Cl Proposition 4.4. A smallest SHC graph for the largest-first algorithm is connected. The proof is straightforward,

so we omit it.

3. Pr,is a smallest SHC graph for the largest-first algorithm.

The smallest hard-to-color

graph

205

Proof. Let G be a smallest SHC graph for the sequential algorithm; let n be the number of vertices of G. Suppose that y ~2. Then by I-roposition 4.3(h), y d n - 3, so n 2 6, also true if y = 2 by part (i) of the same proposition (obviously y > 1). Since Pb was shown to be an SHC graph, we have n = 6. By Proposition 4.4, IEI 2 5, so IEI = 5 and Pfiis a smallest SHC graph. E We now turn to the consideration

of HC graphs.

Proposition 4.6. An HC graph for the largest-first algorithm is not regular. Proof. This is a trivial consequence

of Corollary 3.8.

Cl

In order to reduce the number of cases to consider in determining HC graph, we determine next the smallest bipartite HC graph.

a smallest

Theorem 4.7. Any bipartite HC graph for the largest-first algorithm contains the partial subgraph depicted in Fig. 6. (The path joining the vertices of degree 3 in the subgraph has an odd number of edges-but three at least.) Proof. Let G be a bipartite HC graph for the largest-first algorithm. By Proposition 4.4 we can assume that G is connected. For every largest-first ordering of V there must be some vertex y such that, if x1 is the first vertex in the ordering, then x1 and y belong to different sets of the bipartition and c(y) = 1, where c is the coloring associated with the ordering. For brevity, we are going to denote the first such vertex (with respect to the ordering being considered) the ‘skew’ vertex of the ordering. Choose now the largest-first ordering d so that its skew vertex occurs in the last possible position. To be more precise, we assume that < satisfies the following condition: for any largest-first ordering <‘, if y’ is the ske*;; vertex of the ordering, then l(z

E

v: z d’y’}l

d

I{z

E

v: 2 =sy)l.

(4-l)

If all neighbors of y had degree 1, then, since G is connected, it would be a star. But clearly a star is not an HC graph, so we can assume that for ome neighbor x of y, deg(x) 2 2. We claim that y < X. In fact, since x is adjacent to y, it is in the same set of the bipartition as xl; if x < y, then-since y is skew-we would have c(x) = 1. This would contradict the fact that c(y) = 1.

Fig. 6.

P. Hamen. J. Kuplinsky

206

Assume now that deg(y) d 2. Since y < X, we have deg(x) = deg( y ) = 2. We can now interchange A: and y to obtain another largest-first ordering 6’ and associated coloring c’. We clearly have c’(z) - c(z) for z <‘.I-, i.e. for z < y. But this implies that the skew vertex of =? occurs after x, thus contradicting (4. I). The preceding argument shows that deg(y) 2 3 and hence that deg(x,) 3 3. Consider now a path (x,, 21, .

l

l

, Zhr

l

. . . &r

l

.

, z,~,. y)

from x1 to y. Notice that m > 0 since c(x,) = c(y) = 1. Moreover. we choose the notation in the following way: l zh is a neighbor of x, but zi is not a neighbor of x, for i > it, l zk is a neighbor of y but zi is not a neighbor of y for h < i < k (notice that this is possible because xl and y, being on different sets of the bipartition, do not have common neighbors). Now the path (x, , z{,, . . . , zk, y) contains exactly one neighbor of xl and exactly Cl one neighbor of y , as required. Corollary 4.8. The smallest bipartite HC graph for the largest-first algorithm is as in Fig. 7. Proposition 4.9. Let G be a smallest HC graph for the largest-first algorithm. Denote by n the number of vertices qf G, by 6 its minimum degree and by A its maximum degree. Then : (i) A d rt - 2, (ii) G is connected, (iii) IEI an - ? and 6 2 1. Proof. (i) Suppose G had a vertex v, of degree IZ- I. Let H be the graph obtained by removing vI from G. Since vl is adjacent to all vertices of H, we have y(H) = y(G) - 1. Let (II*, . . . , v,) be an optimal largest-first ordering of V(H). Consider the ordering (v, , . . . , v,) of V(G). Clearly, this is a largest-first ordering. Let cc; be the induced coloring. It is clear by induction that c,;(v,) = c,,(vj) + I for i > I. Therefore ,(;a(G) = &( li’ i +- i = y(H) + 1 = y(G), which is a contradiction. (ii) The proof is similar to thnt of Proposition 4z43 so we omit it. (iii) This is a direct consequencr of (ii). We want to point out however, that for graphs given by degree sequences (this is what we will consider later), condition (iii) retains all the information contained in condition (ii). More

Fig. 7.

precisely.

given a graphic

connected

graph having this as its degree sequence iff d,, 2 I and l 2: d, z 11- l

(see Berge [ 1, Chapter

sequence

6, Theorem

(n,,

. . . , d,,),

Yl).

whcrc

112 2. there

exists ;I

Cl

Proof. We have proved before that this graph is HC. Let 6 be a smallest HC . . . , d,,) its degree sequence. By Proposition

graph. (& 6

Also.

5

ii

7

-S

and

by Corollary

iElc11

4.8,

ifrt=7.

y 3 3.

cy=, d: L 24, and so we would would be regular, contradicting

4.5 and Example 4. I(ii).

If

(4.2)

y 2 4,

then

by Theorem

have H = 6. But then. Proposition

4.2.

(it,3 4,

so

4.0(i)

c;

by Proposition

4.6. We conclude that y = 3.

Let d be a largest-first ordering of V, c the coloring induced by it. There exists a vertex u such that C(V) = 4, and vertices w, w’, w” < u and adjacent that (c(w),

c(w’),

c(w”)}

= { 1, 2,3}.

argue that (w, w’, w”} cannot and c(w) < c(w’).

3.3(iv).

we

so we may assume that {w, w’} $ E

It follows from this that there is a vertex x < w’ adjacent to w’

and such that c(x) = c(w), w, w’,

As in the proof of Proposition

be a clique,

to CI such

so in particular

WI’, x are distinct,

{w, A-} $ E. Notice that the vertices U,

and also that deg( u) 2 3. We

contains the partial subgraph in Fig. 8. (A dotted joined by it are not adjacent

have shown

that (I‘

line indicates that the vertices

in (3.)

Case I : d, < 3. The preceding largest-first vertices

discussion shows that in this case the coloring

ordering

assigns color 4 to the fifth vertex.

we have denoted

21, w,

w’,

induced

by any

It is also clear thar the

w”, x arc the first five vertices

in any

largest-first ordering. Suppose that deg(u) = deg(w). have a largest-first ordering vertex

Then

we could interchange

tr and w and s;ill

inducing a coloring c’. Among the elements in the set

at most to 21 and w”, so r’(w) s 3, a {v, w’, w”, x}, contradiction, because w is the fifth vertex in the new ordering. We have then deg(w) > deg(z!).

w is adjacent

P. Hansen, J. Kuplinsky

208

Fig. 9.

We remark that this, together with the fact that n < 7, implies that w and w” must be adja.cent. If v and x were adjacent, we would have de&:+‘) 3 4, so deg(w) 2 5. But w is a contradiction. Therefore u and x are not not adjacent to w’ or x, so n ~8, + adjacent. Table 1 Code

Degree sequence

01

443333 444433 3333332 4333331 4333333 4433332 4443331 5333332 5433331

D2 D3

04 Ds Q5 D7 Q3 &I

Suppose now that deg(x) = deg(v). In that casz we would interchange x and V. But among the vertices in {v, w, w’, w”}, x is adjacent at most to w’ and w”, so if c’ denotes the new coloring c’(x) < 3. We reach again a contradiction, so deg(x) > deg(v). Using again the fact that rt < 7 we conclude that x and w’ must be adjacent. ’ Notice finally that, since c(w) < c( w ‘), we have c(w) = c(x) = 1 or 2, and therefore c(w’) = 3 or c(w”) = 3. Then the one of w’, w” which has color 3 must be adjacent to vertices colored 1 and 2 in {w, w ‘, w’, x}. In any case, w ’ and w’ must be adjacent. Therefore G must contain and hence be equal to the graph in Fig. 9. Case 2: d6 a 3. (4.3) By Proposition 4.9 d,H

and

d,en-2,

(4.4)

and by Proposition 4.6 G is not regular.

(4.5)

The graphic sequences satisfying conditions (4.2)-(4.5) are given in Table 1.

The smallest hard-to-color graph

209

Case 2a: The degree sequence of G is OS, Da, or D8. Let ~7 be the vertex of least degree. By removing v7 from G we obtain a graph H which is not HC. In the cases D3 and Ds, let (vi, . . , v6) be a largest-first ordering of V(H) inducing an optimal coloring c of H. Since y(H) 6 y(G) = 3, c uses at most 3 CO~O~S.We have degH(vi) 3 degH(vi+i) for i = 1, . . . ,5, and in view of the fact that IdegG - deg,&v)l # 1 for aI1 V, w E V(H), v # w, we also have deg&vi) 2 deg&vi+i) for i = 1, . . . ,6, i.e., (v,, . . . , ~7) is a largest-first ordering of G. If c’ is the corresponding coloring, we clearly have C(Vi) = c’(Vi) for i = 1, . . . ,6, and since deg(v7) = 2, c’ uses at most 3 colors, i.e., it is optimal, a contradiction. In the case D4, the same argument applies if v7 is adjacent to a vertex of degree 3. Assume therefore that v7 is adjacent to X, deg&) = 4, so that H has degree sequence (3,3,3,3,3,3). Choose a partition (V,‘,)~i~) of V(H) into y(H) stable sets such that x E VI. By Proposition 3.7, there is an ordering (v, , . . . , v6) of V(H) such that v1 -X and the corresponding coloring is optimal. Again 9 v7) is an optimal largest-first ordering of V(G). ( VI, Case 2b: The degree sequence of G is DS. This is similar to the case of D4, but now we apply Proposition 3.7 to a partition (Vk)i= y(G) 1 of V(G) such that the vertex of degree 4 belongs to VI. Case 2c: The degree sequence of G is D6. Let v7 be the vertex of degree 2. Arguing as in Case 2a, we see that v7 must be adjacent to the two vertices of degree 4 or to one of these and another of degree 3. Suppose first that v7 is adjacent to the vertices of degree 4. By removing v7 from G we obtain a graph H having degree sequence (3,3,3,3,3,3). There are two such graphs: KS,3 and the complement of C6. If H = K3,3, since G is not bipartite, v7 would be adjacent to vertices x and y on different sides of the bipartition. But then, by Proposition 3.2, there is an optimal ordering of H such that the first two vertices are x and y. As in Case 2a, we rcsch a contradiction. We conclude that H is as in Fig. 10. Qne can now check that the orderings (~1, ~2, ~3, ~4, ~5, Q), v3, v5, v,) give rise to optimal colorings. (211,v3, vs, v2, v4, vg), and (vl, ~4, 3~2, l

l

l

l

P. Hansen, J. Kuplinsky

210

So no matter whether u7 is adjacent in G to v1 and v2, v1 and v3 or v1 and v,, we always have optimal largest-first orderings of G. Suppose now that v7 is adjacent to a vertex of degree 3. By remtiving the corresponding edge we obtain a graph H with degree sequence (4,4,3,3,3,2,1) which is not HC. Let (vl, . . . , v7) be a optimal largest-first ordering of V(H). Again y(H) s y(G), and hence (vl, . . , v7) is an optimal largest-first ordering of G. Case 2d: The degree sequence of G is D7 or D9. We remove v7, the vertex of degree 1, from G. Let IY be the resulting graph. By arguing as in Case 2a, we conclude that we need only consider, for D9, that N has degree sequence (4,4,3,3,3,3) or (5,3,3,3,3,3). But the only graph having the latter as degree sequence is the five-wheel, which has chromatic number 4, so we conclude that H must have degree sequence (4,4,3,3,3,3), which is also the only possibility we have to discuss for the case of D7. There are 3 graphs with degree sequence (4,4,3,3,3,3); their complements are: (i) K3 U P3 (this is the disjoint union), (ii) C4Ll P2, (iii) P6. We now consider on H the action of its vertex-automorphism group. By considering the complementary graphs, it is clear that, in all three cases, the vertices of degree 4 are conjugate (i.e. there is a vertex-automorphism taking one vertex into the other). Therefore, in the case of Lb,, any optimal largest-first ordering of V(H) gives rise to an optimal largest-first ordering of G. The same argument applies to the case in which the degree sequence of G is D7 and I;i = C4 l_.iP2. The only remaining cases are those in which H is as in Fig. 11 and v7 is adjacent in G to v2 or v5, or else H is as in Fig. 12 and IV’~ is adjacent to ~~~or vSz In the first case, the orderings (v,, v3, v2, v4, us, v6) and (vI, v3, us, u2, u4, v~) result in optimal largest-first orderings of G. In the second case, the same result is obtained f!-om (US, 216,VI, 212,213,v4) and (v,, v6, v3, vl , v4, v2). Notice for later l

v6

v4

Vt

Fig. 11.

The smallest hard-tc)-color graph

211

Fig. 12.

reference that in fact we have proved a stronger result, namely that K3 U P3 and p6 are not HC. Case 2e: The degree sequence of G is D1. The three graphs with this degree sequence have been listed as (i), (ii), (iii) in the discussion of Case 2d. But we proved there that the first and the last are not HC. Therefore we need only consider C4 U P2, see Fig. 13. In order to prove that this graph is not HC, it is enough to consider the ordering

(w,

~2, ~3, ~4, ~5, ~6).

Case 2f: The degree sequence of G is D2. The graphs having this degree sequence are the complement of PJ U P3 and of P2Ll P4. The vertices of degree 1 in P3 U P3 form a stab!e set of size 4, so y (P3 u P3) a 4.

We can therefore assume that G = P2 U P4. But this is the graph in Fig. 13 with the edge (v3, v,} added. Again the ordering (v,, v2, v3, v4, vs, v6) defines an optimal coloring of G. Cl

v6

v5

Fig. 13.

References [l]

C. Berge, Graphs (North-Holland,

(21 V. Chvtital, C.T.

Ho@,

graphs, J. Graph Theory

N.V.R.

Amsterdam, Mahadev

11 (1987) 481495.

1985).

and D. de Werra,

Four classes of perfectly orderable

212

P. Hansen, J. Kuplinsky

[3] M.R. Garey and D.S. Johnson, The complexity of near-optimal graph coloring, J. Assoc.

Comput. Mach. 23 (1975) 43-49. [4] D.S. Johnson, Worst-case behavior of graph-coloring algorithms, in: Proc. 5th Southeast Conf. on Combinatorics, Graph Theory, and Computing (1974) 513-527. [5] D.W. Matula, G. Marble and J. D. Isaacson, Graph coloring algorithms, in: R.C. Read, ed., Graph Theory and Computing (Academic Press, New York, 1972). (61 J. Mitchem, On various algorithms for estimating the chromatic number of a graph, Comput. J. 19 (1976) 182-183. [7] M.M. Syslo, Sequential colorings ver us Welsh-Powell bound. Discrete Math. 74 (1989) 241-243. [8] D.J.A. Welsh and M.B. Powell, An upper bound for the chromatic number oT a graph and its application to timetabling problems, Comput. J. 10 (1967) 85-86. [9] A. Wigderson, Improving the performance guarantee for approximate graph coloring, J. Assoc. Comput. Mach. 30 (1983) 729-735.