New upper bound for a class of vertex Folkman numbers

0.12 MB·English

by  N. Kolev

New upper bound for a class of vertex Folkman
numbers
6
0
N.Kolev
0
2
Department of Algebra

n Faculty of Mathematics and Informatics
a
”St. Kl. Ohridski” University of Sofia
J
5 J. Bourchier blvd, 1164 Sofia
1
BULGARIA
1

] N.Nenov
O
C Department of Algebra
. Faculty of Mathematics and Informatics
h
t ”St. Kl. Ohridski” University of Sofia
a
5 J. Bourchier blvd, 1164 Sofia
m
BULGARIA
[
[email protected]

1
v
Submitted: Jan 1, 2005; Accepted: Jan 2, 2005; Published: Jan 3, 2005
9
4 Mathematics Subject Classifications: 05C55
2
1
0
6 Abstract
0
h/ Leta1,...,ar bepositiveintegers,m = Pri=1(ai−1)+1andp = max{a1,...,ar}.
t For a graph G the symbol G→ {a ,...,a } denotes that in every r-coloring of the
a 1 r
m vertices of G there exists a monochromatic a -clique of color i for some i = 1,...,r.
i
: The vertex Folkman numbers F(a ,...,a ;m−1) = min{|V(G)| : G → (a ...a )
v 1 r 1 r
i and Km−1 6⊆ G} are considered. We prove that F(a1,...,ar;m − 1) ≤ m + 3p,
X
p ≥ 3. This inequality improves the bound for these numbers obtained in [6].
r
a
1 Introduction
We consider only finite, non-oriented graphs without loops and multiple edges. We call a
p-clique of the graph G a set of p vertices, each two of which are adjacent. The largest
positive integer p, such that the graph G contains a p-clique is denoted by cl(G) . In this
paper we shall also use the following notations:
V(G)− vertex set of the graph G;
E(G)− edge set of the graph G;
G¯− the complement of G;
1
G[V], V ⊆ V(G)− the subgraph of G induced by V;
G−V− the subgraph induced by the set V(G)\V;
N (v),v ∈ V(G)− the set of all vertices of G adjacent to v;
G
K − the complete graph on n vertices;
n
C − simple cycle on n vertices;
n
P − path on n vertices;
n
χ(G)− the chromatic number of G;
⌈x⌉− the least positive integer greater or equal to x.
Let G and G be two graphs without common vertices. We denote by G +G the
1 2 1 2
′
graph G for which V(G) = V(G ) ∪ V(G ) and E(G) = E(G ) ∪ E(G ) ∪ E , where
1 2 1 2
′
E = {[x,y] | x ∈ V(G ),y ∈ V(G )}.
1 2
Definition Let a ,...,a be positive integers. We say that the r-coloring
1 r
V(G) = V ∪...∪V , V ∩V = ∅ , i 6= j
1 r i j
of the vertices of the graph G is (a ,...,a )-free, if V does not contain an a -clique for
1 r i i
each i ∈ {1,...,r}. The symbol G → (a ,...,a ) means that there is not (a ,...,a ) -
1 r 1 r
free coloring of the vertices of G.
We consider for arbitrary natural numbers a ,...,a and q
1 r
H(a ,...a ;q) = {G : G → (a ,...,a ) and cl(G) < q}
1 r 1 r
The vertex Folkman numbers are defined by the equalities
F(a ,...,a ;q) = min{|V(G)| : G ∈ H(a ,...,a ;q)}.
1 r 1 r
It is clear that G → (a ,...,a ) implies cl(G) ≥ max{a ,...,a }. Folkman [3] proved
1 r 1 r
that there exists a graph G such that G → (a ,...,a ) and cl(G) = max{a ,...,a }.
1 r 1 r
Therefore
F(a ,...,a ;q) exists if and only if q > max{a ,...,a }. (1)
1 r 1 r
These numbers are called vertex Folkman numbers. In [5] Luczak and Urbansky defined
for arbitrary positive integers a ,...,a the numbers
1 r
r
m = m(a1,...,ar) = X(ai −1)+1 and p = p(a1,...,ar) = max{a1,...,ar}. (2)
i=1
Obviously Km → (a1,...,ar) and Km−1 9 (a1,...,ar). Therefore if q ≥ m + 1 then
F(a ,...,a ;q) = m.
1 r
From (1) it follows that the number F(a ,...,a ;q) exists if and only if m ≥ p + 1.
1 r
Luczak and Urbansky [5] proved that F(a ,...,a ;q) = m + p. Later ,in [6], Luczak,
1 r
Rucinsky and Urbansky proved that Km−p−1+C¯2p+1 is the only graph in H(a1,...,ar;m)
with m+p vertices.
2
From (1) it follows that the number F(a ,...,a ;m−1) exists if and only if m ≥ p+2.
1 r
An overview of the results about the numbers F(a ,...,a ;m−1) was given in [1]. Here
1 r
we shall note only the general bounds for the numbers F(a ,...,a ;m − 1). In [8] the
1 r
following lower bound was proved
F(a ,...,a ;m−1) ≥ m+p+2 , p ≥ 2.
1 r
In the above equality an equality occurs in the case when max{a ,...,a } = 2 and m ≥ 5
1 r
(see[4,6,7]). ForthesereasonsweshallfurtherconsideronlythenumbersF(a ,...,a ;m−
1 r
1) when max{a ,...,a } ≥ 3.
1 r
In [6] Luczak, Rucinsky and Urbansky proved the following upper bound for the
numbers F(a ,...,a ;m−1) :
1 r
F(a ,...,a ;m−1) ≤ m+p2 , for m ≥ 2p+2
1 r
In [6] they also announced without proof the following inequality :
F(a ,...,a ;m−1) ≤ 3p2 +p−mp+2m−3 , for p+3 ≤ m ≤ 2p+1.
1 r
In this paper we shall improve these bounds proving the following
Main theorem Let a ,...,a be positive integers and m and p be defined by (2). Let
1 r
m ≥ p and p ≥ 3. Then
F(a ,...,a ;m−1) ≤ m+3p.
1 r
Remark This bound is exact for the numbers F(2,2,3;4) and F(3,3;4) because
F(2,2,3;4) = 14 (see [2])
and
F(3,3;4) = 14 (see [9]).
2 Main construction
We consider the cycle C . We assume that
2p+1
V(C ) = {v ,...,v }
2p+1 1 2p+1
and
E(C ) = {[v ,v ],i = 1,...,2p}∪{v ,v }.
2p+1 i i+1 1 2p+1
Let σ denote the cyclic automorphism of C , i.e. σ(v ) = v for i = 1,...,2p,
2p+1 i i+1
σ(v2p+1) = v1.Using this automorphism and the set M1 = V(C2p+1)\{v1,v2p−1,v2p−2} we
3
define M = σi−1(M ) for i = 1,...,2p + 1. Let Γ denote the extension of the graph
i 1 p
C¯ obtained by adding the new pairwise independent vertices u ,...,u such that
2p+1 1 2p+1
N (u ) = M for i = 1,...,2p+1 . (3)
Γp i i
¯
We easily see that cl(C ) = p .
2p+1
Now we extend σ to an automorphism of Γ via the equalities σ(u ) = u , for
p i i+1
i = 1,...,2p, and σ(u ) = u Now it is clear that
2p+1 1
σ is an isomorphism of Γ . (4)
p
The graph Γ was defined for the first time in [ 8]. In [8] it is also proved that Γ → (3,p)
p p
for p ≥ 3. For the proof of the main theorem we shall also use the following generalisation
of this fact.
Theorem 1. Let p ≥ 3 be a positive integer and m = p + 2. Then for arbitrary
positive integers a ,...,a (r is not fixed) such that
1 r
r
m = 1+X(ai −1)
i
and max{a ,...,a } ≤ p we have
1 r
Γ → (a ,...a ).
p 1 r
3 Auxiliiary results
The next proposition is well known and easy to prove.
Proposition 1 Let a ,...,a be positive integers and n = a +...+a . Then
1 r 1 r
a a n
1 r
⌈ ⌉+...+⌈ ⌉ ≥ ⌈ ⌉.
2 2 2
If n is even than this inequality is strict unless all the numbers a ,...,a are even. If n
1 r
is odd then this inequality is strict unless exactly one of the numbers a ,...,a is odd.
1 r
Let P be the simple path on k vertices. Let us assume that
k
V(P ) = {v ,...,v }
k 1 k
and
E(P ) = {[v ,v ],i = 1,...,k −1}.
k i i+1
4
¯
We shall need the following obvious facts for the complementary graphP of the graph
k
P :
k
k
cl(P¯ ) = ⌈ ⌉ (5)
k
2
cl(P¯ −v) = cl(P¯ ), for each v ∈ V(P¯ ) (6)
2k 2k 2k
cl(P¯2k −{v2k−2,v2k−1}) = cl(P¯2k+1) for k ≥ 2 (7)
cl(P¯ −v ) = cl(P¯ ) , i = 1,...,k , k ≥ 1. (8)
2k+1 2i 2k+1
The proof of Theorem 1 is based upon three lemmas.
Lemma 1 Let V ⊂ V(C ) and |V| = n < 2p+1. Let G = C¯ [V] and let G ,...,G
2p+1 2p+1 1 s
be the connected components of the graph G¯ = C [V].
2p+1
Then
n
cl(G) ≥ ⌈ ⌉. (9)
2
If n is even, then (9) is strict unless all |V(G )| for i = 1,...,s are even. If n is odd,
i
then (9) is strict unless exactly one of the numbers |V(G )| is odd.
i
Proof Let us observe that
G = G¯ +...+G¯ . (10)
1 s
Since V 6= V(C ) each of the graphs G is a path.From (10) and (5) it follows that
2p+1 i
s
n
i
cl(G) = X⌈ ⌉,
2
i=1
where n = |V(G )|, i = 1,...,s. From this inequality and Proposition 1 we obtain the
i i
inequality (9) . From Proposition 1 it also follows that if n is even then there is equality
in (9) if and only if the numbers n ,...,n are even , and if n is odd then we have equality
1 s
in (9) if and only if exactly one of the numbers n ,...,n is odd.
1 s
Corollary 1 It is true that cl(Γ ) = p.
p
Proof Itisobviousthatcl(C¯ ) = pandhencecl(Γ ) ≥ p.Letusdenoteanarbitrary
2p+1 p
maximal clique of Γ by Q. Let us assume that |Q| > p. Then Q must contain a vertex u
p i
for some i = 1,...,2p+1. As the vertices u are pairwise independent Q must contain at
i
most one of them. Since σ is an automorphism of Γ (see (4)) and u = σi−1(u ), we may
p i 1
assume that Q contains u . Let us assign the subgraph of Γ induced by N = M by
1 p Γp(u1) 1
H. The connected components of H are {v2,v3,...,v2p−3} and {v2p,v2p+1} and the both
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of them contain an even number of vertices. Using Lemma 1 we have cl(H) = p − 1.
Hence |Q| = p and this contradicts the assumption .
The next two lemmas follow directly from (10) , (6) , (7) , and (8) and need no proof.
Lemma 2 Let V ( V(C ) and G = C¯ [V]. Let P = {v ,v ,...,v } be a connected
2p+1 2p+1 k 1 2 k
component of the graph G¯ = C [V]. Then
2p+1
(a) if k = 2s then
cl(G−v ) = cl(G),i = 1,...,2s
i
and
cl(G−{v2s−2,v2s−1}) = cl(G).
(b) if k = 2s+1 then
cl(G−v ) = cl(G),i = 1,...,s
2i
Lemma 3 LetV ⊆ V(C ) andC¯ = G.LetP = {v ,...,v }and P = {w ,...,w }
2p+1 2p+1 2k 1 2k s 1 s
be two connected components of the graph G¯ = C [V]. Then
2p+1
(a) if s = 2t then
cl(G−{v ,w }) = cl(G),
i j
for i = 1,...,2k, j = 1,...,s and
cl(G−{v2k−2,v2k−1,wj}) = cl(G),
for j = 1,...,s.
(b) If s = 2t+1 then
cl(G−{v2k−2,v2k−1,w2i}) = cl(G) , for i = 1,...,t.
4 Proof of theorem 1
r
We shall prove Theorem 1 by induction on r. As m = (a − 1) + 1 = p + 2 and
Pi=1 i
max{a ,...,a } ≤ p we have r ≥ 2. Therefore the base of the induction is r = 2. We
1 r
warn the reader that the proof of the inductive base is much more involved then the proof
of the inductive step. Let r = 2 and (a −1)+(a −1)+1 = p+2 and max{a ,a } ≤ p.
1 2 1 2
Then we have
a +a = p+3. (11)
1 2
Since p ≥ 3 and max{a ,a } ≤ p we have that
1 2
a ≥ 3 , i = 1,2. (12)
i
We must prove that Γ → (a ,a ). Assume the opposite and let V(Γ ) = V ∪ V be a
p 1 2 p 1 2
(a ,a )-free coloring of V(Γ ). Define the sets
1 2 p
V′ = V ∩V(C¯ ) , i = 1,2
i i 2p+1
6
and the graphs
G = C¯ [V′] , i = 1,2.
i 2p+1 i
′
ByassumptionV doesnot containana -cliqueandhenceV doesnotcontainana -clique,
i i i i
′
too. Therefore from Lemma 1 we have V ≤ 2a −2, i=1,2. From these inequalities and
i i
the equality
′ ′
|V |+|V | = 2p+1 = 2a +2a −5
1 2 1 2
(as p = a +a −3− see (11) ) we have two possibilities:
1 2
′ ′
|V | = 2a −2 , |V | = 2a −3
1 1 2 2
or
′ ′
|V | = 2a −3 ,|V | = 2a −2.
1 1 2 2
Without loss of generality we assume that
′ ′
|V | = 2a −2 , |V | = 2a −3. (13)
1 1 2 2
From(13)andLemma 1 we obtaincl(G ) ≥ a −1 andbytheassumption that thecoloring
i i
V ∪V is (a ,a )-free we have
1 2 1 2
cl(G ) = a −1 for i = 1,2. (14)
i i
From (13) , (14) and Lemma 1 we conclude that
The number of the vertices of each connected
(15)
component of G¯ is an even number;
1
and
the number of the vertices of exactly one of the
(16)
connected components of G¯ is an odd number.
2
According to (15) there are two possible cases.
Case 1. Some connected component of G¯ has more then two vertices. Now from (15)
1
it follows that this component has at least four vertices. Taking into consideration (15)
and (4) we may assume that {v ,...,v }, s ≥ 2 is a connected component of G¯ . Since
1 2s 1
′
V does not contain an a -clique we have by Lemma 1 that s < a . Therefore 2s+2 ≤ 2p
1 1 1
and we can consider the vertex u .
2s+2
Subcase 1.a. Assume that u ∈ V . Let v ∈ V′. We have from (3) that
2s+2 1 2s+2 2
′
NΓp(u2s+2) ⊇ V1 −{v2s−2,v2s−1}. (17)
′
From (14) and Lemma 2 (a) we have that V1 −{v2s−2,v2s−1} contains an (a1 −1)-clique
Q . From (17) it follows that Q∪{u } is an a -clique in V which is a contradiction.
2s+2 1 1
7
′
Now let v ∈ V . From (3) we have
2s+2 1
′
NΓP(u2s+2) ⊇ V1 −{v2s−2,v2s−1,v2s+2}. (18)
According to (15) we can apply Lemma 3 (a) for the connected component {v ,...,v } of
1 2s
G¯ and the connected component of G¯ that contains v . We see from (14) and Lemma
1 1 2s+2
′
3(a) that V1 − {v2s−2,v2s−1,v2s+2} contains an (a1 − 1)-clique Q of the graph G1. Now
from (18) it follows that Q∪{u } is an a -clique in V , which is a contradiction.
2s+2 1 1
Subcase 1.b. Assume that u ∈ V . If v ∈/ V′ then from (3) it follows
2s+2 2 2s+2 2
′
N (u ) ⊇ V . (19)
Γp 2s+2 2
′
As V contains an (a −1)-clique Q (see (14) ). From (19) it follows that Q∪{u } is
2 2 2s+2
an a -clique in V , which is a contradiction.
2 2
′
Let now v ∈ V . In this situation we have from (3)
2s+2 2
′
N (u ) ⊇ V −{v }. (20)
Γp 2s+2 2 2s+2
We shall prove that
V −{v } contains an (a −1)-clique of Γ . (21)
2 2s+2 2 p
′
As v is the last vertex in the connected component of G , we have v ∈ V . Let L
2s 1 2s+1 2
be the connected component of G¯ containing v . Now we have L = {v ,v ,...}.
2 2s+2 2s+1 2s+2
Now (21) follows from Lemma 2(b) applied to the component L. From (20) and(21) it
follows that V contains aa a -clique, which is a contradiction.
2 2
Case 2. Let some connected component of G¯ have exactly two vertices.
1
From (12) and (13) it follows that G¯ has at least two connected components. It is
1
clear that G¯ also has at least two components. From (16) we have that the number of
2
the vertices of at least one of the components of G is even. From these considerations
2
and (4) it follows that it is enough to consider the situation when {v ,v } is a connected
1 2
component ofG¯ and{v ,...,v }isacomponent ofG¯ ,and{v ,v }isacomponent
1 3 2s 2 2s+1 2s+2
of G¯ . We shall consider two subcases.
1
Subcase 2.a. If u ∈ V .
2s+2 1
Let s = 2. We apply Lemma 3(a) to the components {v ,v } and {v ,v }. From (14)
1 2 5 6
we conclude that
′
V −{v ,v } contains an (a −1)-clique. (22)
1 2 6 1
From (3) we have
′
N (u ) ⊇ V −{v ,v }. (23)
Γp 6 1 2 6
Now (22) and (23) give that V contains an a -clique.
1 1
8
Let s ≥ 3. From (3) we have
′
N (u ) ⊇ V −{v }. (24)
Γp 2s+2 1 2s+2
′
According to Lemma 2(a)V −{v } contains an(a −1)-clique. Nowusing (24)we have
1 2s+2 1
that this (a −1)-clique together with the vertex u gives an a -clique in V . Subcase
1 2s+2 1 1
2.a. is proved.
Subcase 2.b. Let u ∈ V .
2s+2 2
′
Let s = 2. From (3) we have N (u ) ⊇ V − {v }. According to Lemma 2(a) and
Γp 6 2 3
′
(14) V −{v } contains an (a −1)-clique. This clique together with u ∈ V gives an
2 3 2 2s+2 2
a -clique in V , which is a contradiction.
2 2
′
Let s ≥ 3. Here from (3) we have NΓp(u2s+2) ⊇ V2 − {v2s−2,v2s−1}. According to
′
Lemma 2 (a) and (14) we have that V2 −{v2s−2,v2s−1} contains an (a2 −1)-clique. This
clique together with u ∈ V gives an a -clique in V , which is a contradiction. This
2s+2 2 2 2
completes the proof of case 2 and of the inductive base r = 2.
Now we more easily handle the case r ≥ 3. It is clear that
G → (a ,...,a ) ⇔ G → (a ,...,a )
1 r ϕ(1) ϕ(r)
for any permutation ϕ ∈ S . That is why we may assume that
r
a ≤ ... ≤ a ≤ p. (25)
1 r
We shall prove that a +a −1 ≤ p. If a ≤ 2 this is trivial : a +a −1 ≤ 3 ≤ p. Let
1 2 2 1 2
a ≥ 3. From (25) we have a ≥ 3, i = 2,...,r. From these inequalities and the statement
2 i
of the theorem
r
X(ai −1)+1 = p+2
i=1
we have
p+2 ≥ 1+(a −1)+(a −1)+2(r−2).
2 1
From this inequality and r ≥ 3 it follows that a +a −1 ≤ p. Thus we can now use the
1 2
inductive assumption and obtain
Γ → (a +a −1,a ,...,a ). (26)
p 1 2 3 r
Consider an arbitrary r-coloring V ∪...∪V of V(Γ ). Let us assume that V does not
1 r p i
contain an a -clique for each i = 3,...,r. Then from (26) we have V ∪ V contains
i 1 2
(a +a −1)-clique. Now from the pigeonhole principle it follows that either V contains
1 2 1
an a -clique or V contains an a -clique. This completes the proof of Theorem 1.
1 2 2
5 Proof of the main theorem
Let m and p be positive integers p ≥ 3 and m ≥ p + 2. We shall first prove that for
arbitrary positive integers a ,...,a such that
1 r
r
m = 1+X(ai −1)
i=1
9
and max{a ,...,a } ≤ p we have
1 r
Km−p−2 +Γp → (a1,...,ar). (27)
We shall prove (27) by induction on t = m− p−2. As m ≥ p+ 2 the base is t = 0
and it follows from Theorem 1. Assume now t ≥ 1. Then obviously
Km−p−2 +Γp = K1 +(Km−p−3 +Γp).
Let V(K1) = {w}. Consider an arbitrary r-coloring V1 ∪...∪Vr of V(Km−p−2 +Γp).
Let w ∈ V and V , j 6= i does not contain a a -clique.
i j j
In order to prove (27) we need to prove that V contains an a -clique. If a = 1 this is
i i i
clear as w ∈ V . Let a ≥ 2. According to the inductive hypothesis we have
i i
Km−p−3 +Γp → (a1,...,ai−1,ai −1,ai+1,...,ar). (28)
We consider the coloring
V1 ∪...∪Vi−1 ∪{Vi −w}∪...∪Vr
of V(Km−p−3+Γp). As Vj, j 6= i do not contain aj-cliques, from(28) we have that Vi−{w}
contains an (a −1)-clique. This (a −1)-clique together with w form an a -clique in V .
i i i i
Thus (27) is proved.
From Corollary 1 obviously follows that cl(Km−p−2+Γp) = m−2. From this and (27)
we have Km−p−2 + Γp ∈ H(a1,...,ar;m − 1). The number of the vertices of the graph
Km−p−2 +Γp is m+3p therefore F(a1,...,ar;m−1) ≤ m+3p.
The main theorem is proved.
References
[1] J. Coles, Algorithms for bounding Folkman numbers. Master thesis (see
http//www.jpcoles.com/uni/rit/thesis/jpc-thesis.folkman.pdf ).
[2] J. Coles, S. Radziszovski, Computing the Folkman number F (2,2,3;4),
v
http//www.cs.rit.edu/spr/ publ/paper 50.pdf, submitted.
[3] J. Folkman, Graphs with monochromatic complete subgraphs in every edge coloring,
SIAM. J. Appl. Math. 18, 1970, 19-24.
[4] P. Guta. On the structure of k-chromatic critical graphs of order k +p, Stud. Cern.
Math., 50 , 1988, N 3-4, 169-173.
[5] T.Luczak, S. Urbansky, A note on restricted vertex Ramsey numbers, Period. Math.
Hung., 33, 1996, 101- 103.
[6] T. Luczak, A. Rucinsky, S. Urbansky, On minimal vertex Folkman graphs, Discrete
Math., 236, 2001, 245-262.
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