On triangle-free graphs of order 10 with prescribed 1-defective chromatic number

0.15 MB·English

by  Nirmala Achuthan

On triangle-free graphs of order 10 with prescribed
5
1 1-defective chromatic number
0
2

n Nirmala Achuthan N.R. Achuthan
a
J G. Keady.

9 Department of Mathematics and Statistics, Curtin University
1 Bentley WA 6102, Australia

]
O January 20, 2015
C
.
h
t Abstract
a
m
A graph is (m,k)-colourable if its vertices can be coloured with m
[ colourssuchthatthemaximumdegreeofanysubgraphinducedonver-

ticesreceivingthesamecolourisatmostk. Thek-defectivechromatic
1
number for a graphis the leastpositive integer m for which the graph
v
4 is (m,k)-colourable. All triangle-free graphs on 8 or fewer vertices are
8 (2,1)-colourable. Thereareexactlyfourtriangle-freegraphsoforder9
3 which have 1-defective chromatic number 3. We show that these four
4
graphs appear as subgraphs in almost all triangle-free graphs of order
0
10 with 1-defective chromatic number equal to 3. In fact there is a
.
1
unique triangle-free (3,1)-critical graph on 10 vertices and we exhibit
0
this graph.
5
1
: Keywords. k-defectivechromaticnumberk-independencetriangle-freegraph
v
(3,1)−critical graph
i
X Math Review Codes. MSC 05C15 MSC 05C35
r
a
1 Introduction
Weconsiderinthispaperundirectedgraphswithnoloopsormultipleedges.
For all undefined concepts and terminology we refer to [4].
Given a graph G, d (u), N (u) and N [u] denote respectively the de-
G G G
gree, the neighbourhood, and the closed neighbourhood of a vertex u in G.
The union of graphs G and G is denoted by G ∪G . For convenience we
1 2 1 2
write 2G in place of G∪G.
1
Let k be a nonnegative integer. A subset U of the vertex set V(G) is
k-independent if ∆(G[U]) ≤ k. A 0-independentset is an independentset in
the usual sense. A graph G is (m,k)-colourable if it is possible to assign m
colours, say 1,2,...,m to the vertices of G, one colour to each vertex, such
that the set of all vertices receiving the same colour is k-independent. The
smallest integer m for which G is (m,k)-colourable is called the k-defective
chromatic number of G and is denoted by χ (G). A graph G is said to be
k
(m,k)-critical if χ (G) = m and χ (G−u) < m for every u in V(G). A
k k
graph G is said to be (m,k)-edge-critical if χ (G) = m and χ (G−e) < m
k k
for every e in E(G).
It is easy to see that the following statements are equivalent.
(i) G is (m,k)-colourable.
(ii) There exists a partition of V(G) into m sets each of which is k-
independent.
(iii) χ (G) ≤ m.
k
Note that χ (G) is the usual chromatic number. It is easy to see that
0
|V(G)|
χ (G) ≤ ⌈ ⌉. The concept of k-defective chromatic number has been
k k+1
extensively studied in the literature (see [2, 6, 7, 8, 10, 13, 14]). Given a
positive integer m, it is well known that there exists a triangle-free graph
G with χ (G) = m. A natural question that arises is: what is the smallest
k
order of a triangle-free graph G with χ (G) = m? We denote this smallest
k
orderbyf(m,k). Theparameterf(m,0)hasbeenstudiedbyseveralauthors
(see[3,5,11,9])andf(m,0)isdeterminedform ≤ 5. Ithasalsobeenshown
that f(3,1) = 9 and f(3,2) = 13. Furthermore the corresponding extremal
graphs have been characterized (see [13, 2]).
Inthispaperwecharacterizetriangle-freegraphsoforder10withχ (G) =
1
3. In a subsequentpaper [1] we build from the results of this paper to deter-
mine the smallest order of a triangle-free planar graph which has 1-defective
chromatic number 3.
Inall thefiguresinthispaperadoublelinebetween sets X andY means
that every vertex of X is adjacent to every vertex of Y.
2 Preliminary results
We need the following results, proofs of the theorems being in the papers
cited.
2
Theorem 2.1 ([10, 12]) Let G be a graph with maximum degree ∆. Then
∆+1 ∆
χ (G) ≤ ⌈ ⌉ = 1+⌊ ⌋.
k
k+1 k+1
Theorem 2.2 ([13])Thesmallestorderofatriangle-freegraphwithχ (G) =
1
3 is 9, that is, f(3,1) = 9. Moreover, G is a triangle-free graph of order
9 with χ (G) = 3 if and only if it is isomorphic to one of the graphs G ,
1 i
1 ≤ i ≤ 4 given in Figure 1.
u u z z u u z z
1 2 1 2 1 2 1 2
u u
u u z z u u z z
3 4 3 3 4 3
G G
1 2
u u z z u u z z
1 2 1 2 1 2 1 2
u u
z
u u z z u u u
3 4 3 3 4 5
G G
3 4
Figure 1: The critical graphs of order 9 with χ (G) = 3: G to G of [13].
1 1 4
3 Main results
Consider a graph G of order n. The following notation is used repeatedly
in the paper:
u is a vertex of degree ∆(G), A = N (u), B = V(G)−N [u], (1)
G G
H = G[B] and z ∈ B with d (z) = ∆(H). (2)
H
We henceforth denote the vertex set V(G) by V and the edge set E(G)
by E.
3
Lemma 3.1 Let G be a triangle-free graph. In the notation described above,
suppose that ∆(H) = |B|−1 and |A∩N (z)| ≤ 2k, where k is a nonnegative
G
integer. Then χ (G) ≤ 2.
k
Proof. Considerapartition ofA∩N (z)intotwosetsA andA suchthat
G 11 12
|A | ≤ k for i = 1and 2. SinceG is triangle-free, the sets N (z)∪{u}∪A
1i H 11
and (A−A )∪{z} are both k-independent. Hence χ (G) ≤ 2. (cid:3)
11 k
Lemma 3.2 Let G be a triangle-free graph of order 10 with χ (G) ≥ 3.
1
Then (i) ∆(H) ≥ 2 and (ii) 4 ≤∆(G) ≤ 6.
Proof. The lower bound for ∆(G) follows from Theorem 2.1. Let u ∈ V
with d (u) = ∆(G). If ∆(H) ≤ 1, then {u}∪B is 1-independent. Since A
G
is also 1-independent, this implies χ (G) ≤ 2. Thus ∆(H) ≥ 2 and hence
1
|B| ≥ 3 implying that ∆(G) = |A| ≤ 6. (cid:3)
Lemma 3.3 Let G be a triangle-free graph of order 10 with ∆(G) = 6. If
χ (G) = 3 then there exists a vertex u∗ in G such that G−u∗ ∼= G .
1 4
Proof. Assume that χ (G) = 3. Using the notation described before we
1
have |B|= 3. From (i) of Lemma 3.2 we have ∆(H) ≥ 2. Thus ∆(H) = 2.
Let z ∈ B with d (z) = 2. Using Lemma 3.1, we conclude that |A∩
H
N (z)| ≥ 3. Also, as d (z) ≤ 6, |A∩N (z)| ≤ 4.
G G G
Let A = A∩N (z), A = A−A and N (z) = {z ,z }. Since G is
1 G 2 1 H 1 2
K -free, the set A ∪{z ,z } is 0-independent. If z is adjacent to at most
3 1 1 2 1
one vertex of A , then
2
A∪{z }is 1-independent. So is V −(A∪{z }) = {u,z,z }.
1 1 2
It follows that χ (G) ≤ 2, a contradiction. Hence z (similarly z ) has at
1 1 2
least two neighbours in A . Since |A | ≤ 3, z and z have at least one
2 2 1 2
common neighbour in A .
2
Suppose that there is exactly one common neighbour, say x, of z and
1
z in the set A . This implies that |A | = 3 and X = (A−{x})∪{z ,z }
2 2 2 1 2
is 1-independent. Since V − X = {u,x,z} is also 1-independent we have
χ (G) ≤ 2, a contradiction. Thus A has at least two common neighbours,
1 2
say x and y, of z and z .
1 2
Now select a vertex u∗ from A as follows. If |A | = 4 then u∗ is any
1
vertex of A . Otherwise, that is, if |A | = 3 then u∗ is a vertex in A
1 1 2
(note that |A | = 3) different from x and y. Now it is easy to verify that
2
G−u∗ ∼= G . Hence the result. (cid:3)
4
4
Lemma 3.4 Let G be a triangle-free graph of order 10 with ∆(G) = 5. If
χ (G) = 3 then either there exists a vertex u∗ with G−u∗ ∼= G for 1 ≤ i ≤ 4
1 i
or G∼= G illustrated in Figure 2.
5
u
5
z
2
u
4 z
1
u
u
3
v
u
2
z
u
1
Figure 2: G
5
Proof. Suppose that χ (G) = 3. Using the notation described before, it
1
follows that |B| = 4. Now using Lemma 3.1 and Lemma 3.2(i), we have
∆(H) = 2. Let v ∈ B such that (z,v) 6∈ E, N (z) = {z ,z } and A =
H 1 2 1
A∩N (z). Note that |A |≤ 3.
G 1
Case i. |A | =3.
1
Let A−A = {x ,x }. Suppose that (z ,x ) 6∈ E.
1 1 2 1 1
Claim 3.4.1. (v,z )∈ E.
2
Since χ (G) = 3 and
1
A∪{z } is 1-independent,V−(A∪{z }) = {u,v,z,z } is not 1-independent.
1 1 2
This proves Claim 3.4.1.
Claim 3.4.2. (v,x )∈ E.
2
Since χ (G) = 3 and (A−{x })∪{z ,z } is 1-independent, it follows that
1 2 1 2
{u,z,v,x } is not 1-independent. This in turn implies that (v,x ) ∈E.
2 2
CombiningClaims3.4.1and3.4.2withtheassumptionthatGistriangle-
free, we have (z ,x ) 6∈ E. Now, note that the sets
2 2
X = A∪{z ,z } and V −X = {u,z,v} are both 1-independent
1 1 2 1
implying that χ (G) ≤ 2, a contradiction. Thus (z ,x )∈ E. Using similar
1 1 1
arguments we conclude that (z ,x ) ∈ E and (z ,x ) ∈E for i= 1, 2. Now,
1 2 2 i
clearly, G−v ∼= G . This completes Case i.
4
Case ii. |A | ≤ 2.
1
Since ∆(H)= 2 and |B|= 4, clearly H is either P ∪K or P or C .
3 1 4 4
5
Let us first consider the case that H ∼= P ∪K or P .
3 1 4
If |A | ≤ 1 then the sets X = A∪{z} and V −X partition the vertex set V
1
of G into two 1-independent sets implying that χ (G) ≤ 2, a contradiction.
1
Hence |A | = 2. Let A = {u ,u }. If (v,u ) 6∈ E then the sets X =
1 1 1 2 1 1
{u,u }∪(B −{z}) and V −X partition V into 1-independent sets. This
1 1
implies that χ (G) ≤ 2, a contradiction. Thus (v,u ) ∈ E. Similarly
1 1
(v,u )∈ E.
2
Now let us assume that H ∼= P and (v,z ) ∈ E(H). The arguments
4 2
used to conclude that v and z are both adjacent to u and u can now be
1 2
repeated with reference to the vertices z and z since d (z ) = 2. Thus
1 2 H 2
we conclude, without loss of generality, that z and z are both adjacent to
1 2
say u and u from A−{u ,u }. Let {u } = A−{u ,u ,u ,u }. Note that
3 4 1 2 5 1 2 3 4
G−u ∼= G .
5 2
Now let H ∼= P ∪K . If z has at most one neighbour in A−{u ,u }
3 1 1 1 2
then χ (G) ≤ 2 since
1
X = A∪{z } and V −X are both 1-independent.
1
Thusz andsimilarlyz haveatleasttwoneighboursinA−{u ,u }. Nowlet
1 2 1 2
{u ,u ,u } = A−A . Supposethat z and z have two common neighbours
3 4 5 1 1 2
in {u ,u ,u }, say u and u . Then clearly G−u ∼= G .
3 4 5 3 4 5 1
Now assumethat z andz have exactly one common neighbour. Specif-
1 2
ically, assume that z is adjacent to u and u ; z is adjacent to u and u .
1 3 4 2 3 5
Now
X = (A−{u })∪{z ,z } is 1-independent so that V −X is not
1 3 1 2 1
as χ (G) = 3. This implies that (v,u ) ∈ E. Similarly, by considering the
1 3
sets
X = {u ,u ,u ,u ,z } and X = {u ,u ,u ,u ,z }
2 1 2 3 4 2 3 1 2 3 5 1
we conclude that (v,u ) and (v,u ) are in E. Then G ∼= G given in Figure
5 4 5
2.
From now onwards we will assume that H ∼= C . Thus every vertex of
4
H has degree ∆(H) = 2 in H. Moreover we assume that z has the largest
number of neighbours in A. Recall that (v,z) ∈/ E(H). Since |A | ≤ 2, we
1
have |N (z)∩N (v)∩A|≤ 2.
G G
Firstly if |N (z)∩N (v)∩A| = 1 then the sets
G G
X = (A−(N (z)∩N (v)))∪{z,v} and V −X
1 G G 1
providea(2,1)-colouringofG,acontradictiontotheassumptionthatχ (G) =
1
3.
6
Next let |N (z) ∩ N (v) ∩ A| = 0. If |A | ≤ 1 then by the choice z,
G G 1
|N (v)∩A| ≤ 1. But then the sets Y = A∪{v,z} and V −Y = {u,z ,z }
G 1 1 1 2
provide a (2,1)-colouring of G, a contradiction. Hence |A | = 2 and let
1
A = {u ,u }. If v has atmost one neighbour in A then the sets
1 1 2
X = {v,z,u ,u ,u ,u } and V −X = {u,u ,z ,z }
2 2 3 4 5 2 1 1 2
form a (2,1)-colouring of G, a contradiction. If v has two neighbours in A,
say u and u , then the sets
3 4
X = {z ,z ,u ,u ,u ,u } and V −X = {u,u ,z,v}
3 1 2 1 2 3 4 3 5
provide a (2,1)-colouring of G, a contradiction.
Hence |N (z) ∩ N (v) ∩ A| = 2. Without any loss of generality we
G G
assume that N (z)∩N (v)∩A = {u ,u }. Similarly we can easily show
G G 1 2
that |N (z )∩N (z )∩A|= 2. Without any loss of generality, let N (z )∩
G 1 G 2 G 1
N (z )∩A= {u ,u }. Now let {u }= A−{u ,u ,u ,u }. It is easy to see
G 2 3 4 5 1 2 3 4
that G−u ∼= G .
5 3
This completes the proof of the lemma. (cid:3)
Lemma 3.5 Let G be a triangle-free graph of order 10 with ∆(G) = 4 and
3 ≤ ∆(H) ≤ 4. If χ (G) = 3 then there exists a vertex u∗ in G such that
1
G−u∗ ∼= G or G .
1 2
Proof. We will assume χ (G) = 3. Let A = {u ,u ,u ,u }. If ∆(H) = 4
1 1 2 3 4
then G is a subgraph of K and χ (G) ≤ χ (G) = 2, a contradiction.
5,5 1 0
Hence we assume ∆(H)= 3.
Let N (z) ={z ,z ,z }and v ∈ B suchthat (z,v) 6∈ E(H). We provide
H 1 2 3
a proof of this lemma by making and proving, a sequence of claims.
Claim 3.5.1. |N (v)| ≥ 2
H
Suppose that |N (v)| ≤ 1; then we can partition V into two 1-independent
H
sets, X = A ∪ {z} and V − X. Hence χ (G) ≤ 2, a contradiction. This
1
establishes Claim 3.5.1.
Without any loss of generality, assume that (v,z ) and (v,z ) are in
1 2
E(H). Note that |N (z)∩A| ≤ 1 and |N (v)∩A|≤ 2.
G G
Claim 3.5.2. If |N (z)∩A| = 1 then G−u ∼= G .
G 1 2
Supposethat|N (z)∩A| = 1andlet(z,u ) ∈E. If,inaddition, (v,u ) ∈ E
G 1 1
then the sets
X = {u ,u ,u ,z,v} and V −X
2 3 4
7
partition V into 1-independent sets implying χ (G) ≤ 2, a contradiction.
1
Hence (v,u ) 6∈ E. If |N (v)∩A| ≤ 1 then again χ (G) ≤ 2 since
1 G 1
X = A∪{v,z} and V −X are both 1-independent,
1 1
Hence |N (v)∩A| = 2. Let us assume that N (v)∩A= {u ,u }. The set
G G 2 3
X = {u ,u ,u ,z,v} is 1 independent, so V −X is not 1-independent
2 1 3 4 2
as χ (G) = 3. This implies that (u ,z ) ∈ E. Similarly we conclude that
1 2 3
(u ,z ) ∈ E.
3 3
Since the sets
Y = {u,z,v,z } and Y = {u ,u ,u ,z ,z } are 1-independent,
1 3 2 1 2 3 1 2
V −Y = A∪{z ,z } and V −Y = {z,z ,u,u ,v} are not 1-independent
1 1 2 2 3 4
as χ (G) = 3. Hence (u ,z ), (u ,z ) and (u ,z ) are all in E. Now G−u
1 4 1 4 2 4 3 1
is isomorphic to G given in Figure 3.
2
z z z u
1 2 4
v
u2 u3 u z3
Figure 3: G−u ∼= G
1 2
This establishes Claim 3.5.2. Henceforth we will assume that |N (z)∩A| =
G
0.
Claim 3.5.3. |N (v)∩A|= 2 and (v,z ) 6∈ E(H).
G 3
Otherwise, that is, if |N (v) ∩ A| ≤ 1, then X = A ∪ {z,v} and V − X
G
provideapartitionofV into1-independentsets,implyingχ (G) ≤ 2. Hence
1
|N (v)∩A| = 2. Since d (v) ≤ 4 we now have (v,z ) 6∈ E. This establishes
G G 3
Claim 3.5.3.
Withoutanylossofgenerality,wenowassumethatN (v)∩A = {u ,u }.
G 1 2
Clearly there are no edges between {z ,z } and {u ,u }.
1 2 1 2
Claim 3.5.4. For i = 1 and 2, (u ,z ) ∈E.
i 3
NownotethatthesetX = {u ,u ,u ,z,v}is1-independentwhileV−X =
1 2 3 4 1
{u,u ,z ,z ,z } is not as χ (G) = 3. This implies (u ,z ) ∈ E. Similarly
1 1 2 3 1 1 3
(u ,z ) ∈ E. This establishes Claim 3.5.4.
2 3
8
Since z is adjacent to u , u and z and d (z ) ≤ 4we can assume, with-
3 1 2 G 3
out any loss of generality, that (z ,u ) 6∈E. The set X = {u,u ,v,z,z } is
3 3 1 3 3
1-independent while V −X = {u ,u ,u ,z ,z } cannot be as χ (G) = 3.
1 1 2 4 1 2 1
This implies that (u ,z ) and (u ,z ) are both in E. Now if (z ,u ) 6∈ E,
4 1 4 2 3 4
we can similarly conclude that (u ,z ) ∈ E for i = 1 and 2. In this case we
3 i
can easily verify that G−z ∼= G (see Figure 4(a)). On the other hand,
1
that is if (z ,u ) ∈E, we can check that G−u ∼= G (see Figure 4(b)).
3 4 3 2
u u v z u u u z
1 2 3 1 2 3
u v
u3 u4 z1 z2 z1 z2 z u4
(a) G−z ∼= G (b) G−u ∼= G
1 3 2
Figure 4: Graph G−u∗
This proves the lemma. (cid:3)
Suppose that G is a triangle-free graph of order 10 with ∆(G) = 4 and
χ (G) = 3. As a consequence of Lemmas 3.2(i) and 3.5 we can assume that
1
∆(H) = 2. It is easy to see that H is isomorphic to one of the graphs (i)
P ∪2K (ii) P ∪K (iii) P ∪K (iv) P (v) C and (vi) C ∪K .
3 1 3 2 4 1 5 5 4 1
Lemma 3.6 Let G be a triangle-free graph of order 10 with ∆(G) = 4 and
∆(H) = 2. Furthermore, let H be isomorphic to P ∪2K or P ∪K . If
3 1 3 2
χ (G) = 3 then there exists a vertex u∗ in G such that G−u∗ ∼= G or G
1 1 2
or G .
3
Proof. Assume that χ (G) = 3. Let z ∈ B with d (z) = 2 and N (z) =
1 H H
{z ,z }. Forxin{z,z ,z }wehave|N (x)∩A| ≥ 2,otherwiseX = A∪{x}
1 2 1 2 G 1
andV −X providea(2,1)-colouring ofG,acontradiction. Sinced (z) = 2,
1 H
we have |N (z)∩A| = 2. Since G is K -free, this implies |N (z )∩A| = 2
G 3 G i
for i = 1 and 2. Without any loss of generality we can write N (z)∩A =
G
{u ,u } and N (z )∩A= {u ,u } for i = 1 and 2.
1 2 G i 3 4
Let {z ,z } = V(H)−{z,z ,z }. If (z ,u ) and (z ,u ) are in E then
3 4 1 2 3 1 3 2
G−z ∼= G or G or G according as the number of edges between {z }
4 1 2 3 3
and {u ,u } is 0 or 1 or 2. Hence we will assume, without loss of generality,
3 4
that (z ,u ) 6∈ E. Suppose (z ,u ) 6∈ E then
3 2 4 2
X = {u,u ,z ,z ,z ,z } and V −X
1 2 1 2 3 4 1
9
forma(2,1)-colouringofG,acontradiction. Hence(z ,u ) ∈ E. If(z ,u ) ∈
4 2 4 1
E then G−z ∼= G or G or G . Hence we assume that (z ,u ) 6∈ E. Now
3 1 2 3 4 1
since d (u ) ≤ 4, we can assume that (u ,z ) 6∈ E, from which it follows
G 3 3 3
that the sets
X = {u ,u ,u ,z,z } and V −X
1 2 3 4 3 1
form a (2,1)-colouring of G, a contradiction.
This proves the lemma. (cid:3)
Lemma 3.7 Let G be a triangle-free graph of order 10 with ∆(G) = 4 and
∆(H) = 2. Furthermore, let H be isomorphic to P ∪K . If χ (G) = 3 then
4 1 1
there exists a vertex u∗ in G such that G−u∗ ∼= G , for some i, 1 ≤ i≤ 3.
i
Proof. Let us suppose that χ (G) = 3. Let z and z be vertices in B with
1 1
d (z) = d (z ) = 2. Note that (z,z ) ∈ E(H). Let z (z ) be the other
H H 1 1 2 3
neighbour of z (z ) . Finally, let {z } =V(H)−{z,z ,z ,z }.
1 4 1 2 3
Claim 3.7.1. For x = z and z , |N (x)∩A| = 2.
1 G
This claim can be proved using arguments similar to the ones used in
Lemma 3.6.
Now, without any loss of generality, let N (z) ∩ A = {u ,u } and
G 1 2
N (z )∩A= {u ,u }. Sinceχ (G) = 3andV−A−{z ,z }is1-independent
G 1 3 4 1 2 3
it follows that A∪{z ,z } is not 1-independent. Note that z and z do not
2 3 2 3
have a common neighbour in A. Thus we conclude that either (z ,u ) ∈ E
2 i
for i = 3 and 4 or (z ,u ) ∈ E for i = 1 and 2. Suppose, without loss of
3 i
generality, (z ,u ) ∈ E for i = 3 and 4.
2 i
If z is adjacent to both u and u , then it is easy to verify that G−z ∼=
3 1 2 4
G .
2
Hence (z ,u )6∈ E for i= 1 or 2. Without any loss of generality assume
3 i
that (z ,u ) 6∈ E. Now
3 1
X = {u ,u ,u ,z} and X = {u ,u ,u ,z,z } are 1-independent.
1 2 3 4 2 1 3 4 3
Since χ (G) = 3, the sets V − X = {u,u ,z ,z ,z ,z } and V − X =
1 1 1 1 2 3 4 2
{u,u ,z ,z ,z } are not 1-independent. This in turn implies that (u ,z ) ∈
2 1 2 4 i 4
E for i = 1 and 2. Now it is easy to verify that G−z ∼= G or G or G .
3 1 2 3
Henceitfollowsthatthereexistsau∗suchthatG−u∗ ∼= G or G or G .
1 2 3
This proves the lemma. (cid:3)
Lemma 3.8 Let G be a triangle-free graph of order 10 with ∆(G) = 4 and
∆(H) = 2. Furthermore, let H be isomorphic to P . If χ (G) = 3 then
5 1
there exists a vertex u∗ such that G−u∗ ∼= G , for some i, 1≤ i ≤ 3.
i
10