Stability and Ramsey numbers for cycles and wheels

0.16 MB·English

by  Nicolás Sanhueza

Stability and Ramsey numbers for cycles and wheels
Nicolás Sanhueza
[email protected]
Departamento de Ingeniería Matemática, Universidad de Chile
Abstract
5 Westudythestructureofred-blueedgecoloringsofcompletegraphs,withnocopies
1
of the n-cycle C in red, and no copies of the n-wheel W = C ∗K in blue, for an
0 n n n 1
2 odd integer n. Our first main result is that in any such coloring, deleting at most two
n vertices we obtain a vertex-partition of G into three sets such that the edges inside the
a partition classes are red, and edges between partition classes are blue. As a second
J
0 result, we obtain bounds for the Ramsey numbers of r(C2k+1,W2j) for k < j integers,
which asymptotically confirm the values of 4j +1, as it were conjectured by Zhang et
3
al.

]
O
C 1 Introduction
.
h
t
a We study the structure of red-blue edge-colorings in complete graphs, which avoid certain
m
monochromatic subgraphs. More concretely, we consider the case of odd positive integer n,
[
and the forbidden monochromatic graphs given by the red n-cycle Cn and the blue n-wheel
1 W := C ∗K . Our main result is the following:
v n n 1
6 Theorem 1. Let k ≥ 6 and N ≥ 5k + 3. Suppose G := K has a red-blue coloring of its
N
8
edges in a way such that C is not a red subgraph of G and W is not a blue subgraph of
7 2k+1 2k+1
7 G. Then, there is a partition of V(G) given by {U ,U ,U ,U } such that |U | ≤ 2, |U | ≤ 2k
0 1 2 3 0 i
0
for 1 ≤ i ≤ 3; and every edge in G−U inside the partition classes {U ,U ,U } is red, and
. 0 1 2 3
1
blue otherwise.
0
5
A similar result was obtained by Nikiforov and Schelp [NS08], considering the case where
1
: the forbidden monochromatic subgraphs are odd cycles. More precisely, they proved that
v
Xi given k ≥ 2 and N ≥ 3k +2, if G := KN has a red-blue coloring of its edges in a way such
that C is neither a red nor a blue subgraph of G, then there is a partition of V(G) given
r 2k+1
a by {U ,U ,U } such that |U | ≤ 1 and the edges inside the partition classes U and U have
0 1 2 0 1 2
one color; and are colored with the remaining color otherwise.
Our proof of Theorem 1 depends on certain bounds on asymmetric Ramsey numbers.
We focus on the case where G is the n-cycle C , and H is the m-wheel W , for n,m
n m
integers. Some Ramsey numbers of C and W areknown ([CCMN09], [CCNZ12], [ZZC14]),
n m
depending on the parity of n and m and their relative size. In particular, it is known that
2n−1 for even m, with m ≥ 4, n ≥ 3m/2−1,

3n−2 for odd m, with n ≥ m ≥ 3, (n,m) 6= (3,3),
r(Cn,Wm) =  2m+1 for odd n, with m ≥ 3(n−1)/2, (n,m) 6= (3,3),(n,m) 6= (3,4),
 3n−2 for odd n and m; with n < m ≤ 3(n−1)/2.
Notice that r(C ,W ) is not known for odd n and even m with n < m < 3(n−1)/2. Zhang
n m
et al. [ZZC14] raised a conjecture concerning these values.
1
Conjecture 2 (Zhang et al. [ZZC14]). Let n < m integers, with n odd and m even. Then
r(C ,W ) = 2m+1.
n m
From now on, suppose n < m are integers, with n = 2k+1 and m = 2j. We confirm the
previous conjecture asymptotically in terms of j.
Theorem 3. Let 2 < k < j be integers. We have the following bounds for the Ramsey
number of the (2k +1)-cycle versus the 2j-wheel.
(a) If k ≥ 3, then r(C ,W ) ≤ 9j +1.
2k+1 2j 2
(b) If k ≥ 3, then r(C ,W ) ≤ 4j +334.
2k+1 2j
In particular, we will make use of the upper bounds of r(C ,W ) for our proof of
2k+1 2k+2
Theorem 1. Both bounds of Theorem 3 follow from a more general type of bound that we
state and prove in Section 4.
2 Preliminaries
We fix a little bit of notation. For every graph G, we write |G| and kGk for its number of
vertices and edges respectively. The length of a path P is kPk, its number of edges. For
disjoint sets of vertices A and B, an (A,B)-path is a path with one endpoint in A, the other
in B and no other vertices in A∪B.
Given a red-blue coloring of the edges of a graph G, let GR be the graph on V(G) only
containing the red-colored edges, similarly define GB as the graph on V(G) only containing
theblue-colorededges. LetER(G)andEB(G)bethesetofedgesofGR andGB, respectively.
Definition 1. Let G be a graph. A hedgehog is a tuple (W,X) where X ⊆ W ⊆ V(G),
X induces a complete subgraph and the edges in E(W \X,X) induce a complete bipartite
subgraph.
The notion of hedgehogs will be useful in the proof of Theorem 1. The main property of
hedgehogs is that every pair of vertices can be joined by paths of various lengths, and that
allows us to find cycles of various sizes.
Lemma 4. Let (W,X) a hedgehog. Then every pair of distinct vertices in W can be joined
by red paths of every length between 2 and |X|−1.
Proof. As X induces a complete subgraph, every pair of distinct vertices in X can be joined
by paths of every length between 1 and |X| − 1. For distinct pair of vertices in W, not
necessarily contained in X, we can use the edges in E(W \X,X) to extend the mentioned
paths or to find a pair of length 2 connecting these vertices, and conclude the result.
Corollary 5. For every i ∈ {1,2}, let (R ,S ) be hedgehogs in a graph G, such that R ∩R =
i i 1 2
∅. Suppose that min{|S |,|S |} ≥ 3 andthat there existtwo disjointedgesin E(R ,R ). Then
1 2 1 2
G contains cycles of every length between 6 and |S |+|S |.
1 2
Proof. Using Lemma 4 we can join every two vertices in an hedgehog with paths of various
lengths. Choosing these vertices to be the endpoints of two disjoint edges in E(R ,R ) we
1 2
find cycles of the desired lengths.
2
We shall make use of the values of Ramsey numbers for cycles, which are completely
known.
Theorem 6 (Faudree-Schelp, [FS74]). We have
6 (n,m) ∈ {(3,3),(4,4)},

r(Cn,Cm) =  2nn+−m1−1 34 ≤≤ mm <≤ nn,boodthd nm,,m(ne,vmen),6=(n(,3m,3)),6= (4,4),
2
 maxnn+ m2 −1,2m−1o 3 ≤ m ≤ n, even m and odd n.
Theorem 7 (Surahmat et al. [SBT06]). We have that r(C ,W ) = 6k + 1, for all
2k+1 2k+1
integers k ≥ 1.
Next, we need some results on the stability of cycle-forbidding red-blue colorings, as
shown by Nikiforov and Schelp [NS08].
Theorem 8 (Nikiforov-Schelp, [NS08]). Let G be a hamiltonian graph of order 2n such
that C * G and C * G. Then there exists a partition of V(G), {U ,U } such that
2n−1 2n−1 1 2
|U | = |U | = n and U ,U are independent. Moreover, there exists a vertex u ∈ V(G) such
1 2 1 2
that G−u ∼= K .
n,n−1
We shall make use of an intermediate result of the same authors, in the same vein.
Lemma 9 (Nikiforov-Schelp, [NS08]). Let n = 2k + 1 ≥ 5 and N ≥ 3k + 2, and a graph
G := K with an associated red-blue coloring of its edges c : E(G) → {R,B} such that
N
C * GR and C * GB. Then there exists a color C ∈ {R,B} and a partition {Y ,Y }
2k+1 2k+1 1 2
of V(G) such that E(Y ,Y ) ⊆ EC(G) for all i ∈ {1,2}, and there are no disjoint C-colored
i i
edges in E(Y ,Y ).
1 2
A graph G is pancyclic if it contains cycles of every length between 3 and |V(G)|. The
girth of a graph g(G) is the length of its shortest cycle, the circumference of a graph c(G)
is the length of its longest cycle. A graph G is weakly pancyclic if it contains cycles of every
length between g(G) and c(G). We shall make use of various theorems that assure that,
under certain conditions, a graph is pancyclic or weakly pancyclic. The following lemma has
a trivial proof.
Lemma 10. Let k ≥ 1 and G be a graph on 2k + 1 vertices. Let {V ,V } be a partition
1 2
of V(G) such that |V | = k + 1 and |V | = k and E(V ,V ) form the edges of a complete
1 2 1 2
bipartite subgraph. If there is an edge e = {x,y} ⊆ V , then G is pancyclic.
1
Theorem 11 (Bondy, [Bon71]). Let G be a graph on n ≥ 3 vertices with minimum degree
at least n/2. Then G is pancyclic, or n = 2k and G ∼= K .
k,k
Corollary 12 (Dirac, [Dir52]). Let G be a graph on n ≥ 3 vertices with minimum degree
greater than n/2. Then G is pancyclic.
Theorem 13 (Brandt, [Bra97]). Let G be a non bipartite graph on n vertices with more
than (n−1)2/4+1 edges. Then G is weakly pancyclic and contains a triangle.
3
More theorems and results about weakly pancyclic graphs will be stated in Section 4.
The next simple lemma ensures a bound on the girth of a graph given a lower bound on the
minimum degree.
Lemma 14. Let G be a graph on n ≥ 9 vertices with δ(G) ≥ n. Then g(G) ≤ 5.
4
Proof. Let C be a shortest cycle in G. Suppose |C| ≥ 6. By the choice of C, each vertex
in V(C) has exactly two neighbors in V(C). So, |N(x) \ V(C)| ≥ δ(G) − 2 for each x ∈
V(C). Furthermore, for each pair of distinct vertices x,y ∈ V(C), the sets N(x)\V(C) and
N(y)\V(C) are disjoint, for the same reason. Thus,
n ≥ |C|+ |N(x)\V(C)| ≥ |C|+|C|(δ(G)−2) = |C|(δ(G)−1),
X
x∈V(C)
implying that n/6+1 ≥ ⌈n/4⌉, a contradiction for n ≥ 9.
Theorem 15 (Dirac, [Dir52]). Let G be a 2-connected graph in n ≥ 3 vertices. Then
c(G) ≥ min{2δ(G),n}.
3 Proof of Theorem 1
InthissectionweproveTheorem 1. WeassumeTheorem 3,postponingitsprooftoSection4.
Let k ≥ 6, so that n = 2k + 1 ≥ 13. Let N ≥ 5k + 3 and G := K . Suppose there
2k+1
exists a red-blue edge-coloring of K in a way such that C is not a red subgraph of G
N 2k+1
and W is not a blue subgraph of G. By Theorem 7, we may asumme that N ≤ 6k.
2k+1
Theorem 3 implies that r(C ,W ) ≤ 5k+3 when k ≥ 5. Hence, G contains a blue
2k+1 2k+2
copy of W as a subgraph. Choose such a copy, and let C be the “rim” of the wheel (the
2k+2
(2k +2)-cycle) and let w be the “hub” of the wheel (the vertex not in C).
Consider the graph G[C] with the induced edge-coloring of G. This graph does not
contain a red copy of C , as it would be present in G as well. It also does not contain
2k+1
a blue copy of C , as otherwise, adding w, we would create a blue copy of W in the
2k+1 2k+1
graph G. So, the graph GR[C] (the graph induced by vertices of C, but only considering
the red-colored edges) satisfies the hypotheses of Theorem 8. Hence, there exists a partition
{U′,U′} of the vertices of C such that |U′| = |U′| = k+1, both U′ and U′ induce complete
1 2 1 2 1 2
red subgraphs, and there exists a vertex v ∈ V(C) such that E(U′ − v,U′ −v) ⊆ EB(G).
1 2
Without loss of generality, we suppose that v ∈ U′. We define U = U′ − {v} for each
2 i i
i ∈ {1,2} and U = {w}, as defined in the previous paragraph. So |U | = k+1 and |U | = k.
3 1 2
Recall that every edge contained in U is red, for i ∈ {1,2,3} and every edge between
i
different pairs in {U ,U ,U } is blue. We choose a triple (X ,X ,X ) of pairwise disjoint
1 2 3 1 2 3
sets such that U ⊆ X and every edge in E(X ,X ) is blue; for each distinct i,j ∈ {1,2,3}.
i i i j
Assume (X ,X ,X ) maximizes the sum |X | + |X | + |X | among all possible 3-tuples
1 2 3 1 2 3
satisfying the previous conditions. With this choice of (X ,X ,X ), we get the following
1 2 3
lemma.
Lemma 16. Every edge contained in one of the sets X , X or X is red.
1 2 3
4
Proof. Let i ∈ {1,2} and suppose that X contains a blue edge e. As |U | ≥ k and U is a
i i i
complete red subgraph, this means that |X | ≥ k +1. Also, |X | ≥ k. So, by Lemma 10,
i 3−i
there is a blue monochromatic (2k +1)-cycle in X ∪X . As |X | ≥ 1 and every vertex in
1 2 3
X is joined by a blue edge to each vertex in X ∪X , we get a blue copy of W in G, a
3 1 2 2k+1
contradiction.
Now, suppose that X contains a blue edge e = x x . As C ⊆ K , we have a
3 1 2 2k+1 k,k,1
blue copy of C contained in X ∪X ∪{x }. But x is joined with a blue edge to every
2k+1 1 2 2 1
vertex in X ∪ X ∪ {x }, so we find a blue copy of W as a subgraph in G, which is a
1 2 2 2k+1
contradiction. This proves the lemma.
Lemma 16 together with our assumption that G contains no red copy of C imply
2k+1
that |X | ≤ 2k for each i ∈ {1,2,3}. If 3 X = V(G), then we are done. So, from now on,
i i=1 i
S
we suppose that V(G)\ 3 X 6= ∅.
i=1 i
S
Let v ∈/ 3 X . We show that there exists an i ∈ {1,2,3} such that E(v,X ) ⊆ ER(G).
i=1 i i
S
If itwere not thecase, thereexist vertices x ∈ X such that vx isblue, forevery i ∈ {1,2,3}.
i i i
As E(X ,X ) ⊆ EB(G) and |X |,|X | ≥ k, we can find a blue (x ,x )-path in X ∪X of
1 2 1 2 1 2 1 2
length 2k−1. Along with the edges vx and vx , we get a blue copy of C as a subgraph
1 2 2k+1
of G. But x is joined with a blue edge to every vertex in X ∪X ∪{v}, so we find a blue
3 1 2
copy of W as a subgraph in G, which is a contradiction.
2k+1
This allows us to define
W := X ∪{v ∈ V(G) : E(v,X ) ⊆ ER(G)},
1 1 1
W := X ∪{v ∈ V(G) : E(v,X ) ⊆ ER(G)}\W , and
2 2 2 1
W := X ∪{v ∈ V(G) : E(v,X ) ⊆ ER(G)}\(W ∪W ).
3 3 3 1 2
By the previous observations, {W ,W ,W } is a partition of V(G).
1 2 3
Lemma 17. Let i ∈ {1,2,3} and v ∈ W \X . Then v has at least one red neighbor in X
i i j
for some j ∈ {1,2,3}\{i}.
Proof. Immediate from the maximality of |X |+|X |+|X |.
1 2 3
We say that a hedgehog in G is red if it is present in GR, and blue otherwise. Then each
of the tuples (W ,X ) for i ∈ {1,2,3} are red hedgehogs. In particular, as |X | ≥ k +1 and
i i 1
|X | ≥ k, we have that both (W ,X ) and (W ,X ) are disjoint red hedgehogs that satisfy
2 1 1 2 2
the hypothesis of Corollary 5, and so, if there are disjoint red edges in E(W ,W ), there
1 2
would be a red (2k +1)-cycle. This result can be strengthened, according to the following
lemma.
Lemma 18. There are no two vertex-disjoint red (W ,W )-paths.
1 2
Proof. Suppose otherwise. Let P ,P be two vertex-disjoint (W ,W )-paths of minimum
1 2 1 2
joint length, and such that kP k ≤ kP k. Let p1 ∈ W and p1 ∈ W be the endpoints of P ;
1 2 1 1 2 2 1
and p2 ∈ W and p2 ∈ W be the endpoints of P .
1 1 2 2 2
As (W ,X ) is a red hedgehog and |X | ≥ k + 1, by Lemma 4 we can join the vertices
1 1 1
p1 and p2 in W with red paths of every length between 2 and k. Similarly, we can join p1
1 1 1 2
and p2 in W with red paths of every length between 2 and k−1. Joining the paths P and
2 2 1
5
P with the (p1,p2)-paths and (p1,p2)-paths previously mentioned, we obtain red cycles of
2 1 1 2 2
every length between kP k+kP k+4 and kP k+kP k+2k−1. As C is not contained
1 2 1 2 2k+1
as a red subgraph in G, necessarily the bound
2k −2 ≤ kP k+kP k (1)
1 2
holds. For each i,j ∈ {1,2}, we define qj as the vertex in P adjacent to pj in the path
i j i
P . Note that if kP k ≥ 3, then the vertices qj and qj are distinct. As (W ,X ) is a red
j j 1 2 3 3
hedgehog, there exists x ∈ X such that xv is a red edge for every other vertex v ∈ W .
3 3
Suppose the path P contains x in its vertices. As every vertex in V(P ) ⊆ W is a
j j 3
red neighbor of x, using x we can find a strictly shorter path among the vertices of P ,
j
contradicting the minimality of kP k+kP k. We deduce that
1 2
if a path P contains x in its vertices, then kP k ≤ 4. (2)
j j
Now, suppose that P does not contain the vertex x and kP k > 4. Then, using (2)
j 3−j
we deduce that x does not belong in V(P ), and by hypothesis, x ∈/ V(P ). Then
3−j j
p3−jq3−jxq3−jp3−j is a path of length 4 that is vertex-disjoint with P , which again con-
1 1 2 2 1
tradicts the minimality of kP k+kP k. We have proved that
1 2
if a path P does not contain x in its vertices, then kP k ≤ 4. (3)
j 3−j
Using (2) and (3) together, the shorter of P and P has length at most 4, and so it
1 2
follows that kP k ≤ 4. Recalling the equation (1), we get that
1
2k −6 ≤ kP k. (4)
2
As 2k +1 ≥ 13, we have that kP k > 4. Therefore, using (3) we deduce that x ∈ P .
2 1
The edge xy is red for every y in W , and x is contained in P , while the path P is long;
3 1 2
the idea is to use x to construct shorter cycles using the vertices of P . Concretely, for every
2
vertex y ∈ V(P )∩W , we get that p1P xyP p2 is a red (p1,p2)-path with no edge contained
2 3 1 1 2 1 1 1
in W . We can choose y ∈ V(P ) ∩ W in kP k − 1 ways (every vertex in P , except its
1 2 3 2 2
endpoints), hence, xyP p2 can be chosen of every length between 2 and kP k. Using (4), this
2 1 2
implies the existence of red (x,p2)-paths of every length between 2 and 2k −6.
1
Using these paths and the (p2,p1)-paths in W , we deduce the existence of red cycles in
1 1 1
G of every length between 6 and 3k −5. We conclude that 3k −5 < 2k +1, which is false
for k ≥ 6. This contradiction proves the lemma.
Now we show that both W and W have no more than 2k vertices each.
1 2
Lemma 19. |W | ≤ 2k for all i ∈ {1,2}.
i
Proof. Recall that (W ,X ) is a red hedgehog. As |X | ≥ k+1, we can choose R ,R ⊆ X
1 1 1 1 2 1
disjoint in a way such that |R | = 1 and |R | = k. Let R = W \ (R ∪ R ). Then these
1 2 3 1 1 2
three subsets are disjoint and the edges between E(R ,R ) are all red if i 6= j in {1,2,3}. If
i j
|W | > 2k, then |R | ≥ k so we can easily find a red copy of C , a contradiction.
1 3 2k+1
If |X | ≥ k +1 the conclusion follows from the same argument just presented, replacing
2
(W ,X ) with (W ,X ). So it suffices to study the case where |X | = k but |W | ≥ 2k +1.
1 1 2 2 2 2
6
By Lemma 10, every edge contained in W \X must be blue, and we get a complete blue
2 2
subgraph of size at least k +1.
We claim that
each vertex outside W has at most one red neighbor in W \X . (5)
2 2 2
Indeed, if this were not the case, we could find a red (2k + 1)-cycle, using the complete
bipartite red subgraph in E(X ,W \ X ) and an arbitrary red edge in X . Let x be any
2 2 2 2 3
vertex in X , and let Y be the set of blue neighbors of x in W . Because of Lemma 18, there
3 2 3 2
are no two disjoint red edges between Y and X . By construction there are no red edges
2 1
between X and X , and by (5) we conclude that no vertex in X has two red neighbors
1 2 1
in Y . Then the red edges in E(Y ,X ) form a (possibly empty) star with its center in Y .
2 2 1 2
Deleting this (possible) center of the star, we obtain a set Y′ ⊆ Y such that both E(Y′,x )
2 3
and E(Y′,X ) only contain blue edges. Furthermore, E(X ,X ) and E(x ,X ∪ X ) also
1 1 2 3 1 2
contain only blue edges. We have that |Y′| ≥ 2. Using at most two vertices of Y′ we find a
blue cycle of length 2k +1 in X ∪X ∪Y′, and joining x we obtain a blue copy of W ,
1 2 3 2k+1
a contradiction.
We now prove a lemma that will allow us to conclude Theorem 1 afterwards.
Lemma 20. Let W,W′ be disjoint sets of vertices in V(G) such that
A. there exists X ⊆ W of size at least k −1 such that (W,X) is a red hedgehog,
B. W′ has size at least 3k +2,
C. there exists X′ ⊆ W′ of size at least k such that X′ induces a red clique,
D. at least one of the following two hypothesis holds:
D1. W and W′ cover all vertices of V(G) except at most one, and every edge between
W and W′ is blue, or
D2. W and W′ cover all vertices of V(G), and there exists a vertex in W that only
sends blue edges outside W.
Then there exists a partition {V ,V ,V ,V } of V(G) such that |V | ≤ 2, every edge inside the
0 1 2 3 0
partition classes {V ,V ,V } is red, and every edge between the partition classes {V ,V ,V }
1 2 3 1 2 3
is blue.
Proof. Let H be the graph formed by the vertices in W′ with the induced edge coloring from
G. By hypothesis B, |V(H)| ≥ 3k +2. By Theorem 6, r(C ,C ) = 3k +2, so there is
2k+2 2k+2
a monochromatic copy of C in H.
2k+2
Note that both hypothesis D1 and D2 imply the existence of a vertex in W that only
sends blue edges to H. Now, H does not contain a monochromatic C , neither in red nor
2k+1
in blue (any vertex v ∈ W that only sends blue edges to H together with a blue C in
2k+1
H would form a blue W ). So H satisfies the hypotheses of Lemma 9, and there exists a
2k+1
partition {Y ,Y } of V(H) and a color C ∈ {R,B} such that
1 2
1. Y and Y both induce complete C-colored subgraphs, and
1 2
2. E(Y ,Y ) does not contain two disjoint C-colored edges.
1 2
Hence the edges inE(H) that arenot C-coloredforma bipartitegraph. By hypothesis C,
there exists a red complete subgraph in H of size at least k ≥ 3, so the red edges in H cannot
7
form a bipartite subgraph. So, C = R. Using the fact that |Y |,|Y | ≤ 2k and hypothesis
1 2
B, we deduce that min{|Y |,|Y |} ≥ k +2. Define Y := W. We have that (Y ,Y ), (Y ,Y )
1 2 3 1 1 2 2
and (Y ,X) are red hedgehogs satisfying the hypothesis of Corollary 5. So, there are no two
3
disjoint red edges between each pair in {Y ,Y ,Y }, and so, the red edges induce a (possibly
1 2 3
empty) red star between each of the pairs.
Case A: Hypothesis D1 holds: Then there can only be red edges between Y and
1
Y . So, choosing V as the vertices not covered by W ∪W′ (at most one) together with the
2 0
(possible) center of the red star between Y and Y , we get that the edges between each pair
1 2
in {Y \V ,Y \V ,Y \V } are all blue.
1 0 2 0 3 0
Case B: Hypothesis D2 holds: Deleting the centers of the three red stars between
the pairs in {Y ,Y ,Y } eliminates every red edge between these sets. We want to select at
1 2 3
most two vertices in V , so it suffices to study the case where there is a red edge between
0
each pair in {Y ,Y ,Y }.
1 2 3
Supposefirstthatwecanfindthreerededges,onebetweeneachdistinctpairof{Y ,Y ,Y },
1 2 3
such that the graph induced by these three edges is disconnected. Let e be the selected
ij
edges between Y and Y , for each distinct i,j ∈ {1,2,3}. The red hedgehogs Y ,Y and Y
i j 1 2 3
contain complete subgraphs of size at least k+2, k+2 and k−1, respectively. So, each edge
e is adjacent to two hedgehogs with complete subgraphs of size at least k −1 and k +2.
ij
If the edges {e }3 induce a disconnected subgraph, we can choose an edge disjoint to the
ij i6=j
other two, and using Corollary 5 we can join the endpoints of the disjoint edges in {e }3
ij i6=j
with paths of every length between 2 and k−2 or k+1, respectively. Using these paths we
can find red cycles of every length between 9 and 2k+2, in particular, a red copy of C ,
2k+1
a contradiction.
So, for every pair of three red edges e ∈ E(Y ,Y ) with i < j ∈ {1,2,3}; the graph
ij i j
induced by these three edges is connected. Every red edge between {Y ,Y ,Y } is part of one
1 2 3
of the three red stars, and so, contains at least one of the centers of these stars. If no edge
contains the three centers, then we easily find three red edges between pairs in {Y ,Y ,Y }
1 2 3
inducing adisconnected subgraph, acontradiction. Ifoneoftheseedges containthreecenters
of the stars, then choosing V as the vertices of this edge, we get that every edge between
0
each pair in {Y \V ,Y \V ,Y \V } is blue, as required.
1 0 2 0 3 0
In every case: define V := Y \V . As Y and Y are red complete graphs, it only remains
i i 0 2 3
to show that Y is a complete red subgraph. If this were not the case, there is a blue edge
3
e = xy in Y . As x only sends blue edges to V ∪V , and each of V and V has size at least
3 1 2 1 2
k + 1, we find a blue C in x ∪ Y ∪ Y . Adding y, we obtain a blue copy of W , a
2k+1 1 2 2k+1
contradiction.
Recall that we have a partition of V(G) in {W ,W ,W } and there exists sets X for
1 2 3 i
each i ∈ {1,2,3} such that (W ,X ) are red hedgehogs for i ∈ {1,2,3}; we also have that
i i
|X | ≥ k + 1, |X | ≥ k and |X | ≥ 1. By Lemma 19, the sets W = W , X = X ,
1 2 3 1 1
W′ = V(G) \ W and X′ = X satisfy hypothesis A, B and C of Lemma 20. The same
1 2
holds if we replace the role of W with W and X with X . So we can conclude Theorem 1
1 2 1 2
immediately if hypothesis D2 of Lemma 20 is satisfied by W or W . So, we may assume
1 2
that
every vertex in W has a red neighbor outside W , for each i ∈ {1,2}. (6)
i i
8
By Lemma 18, there are no two disjoint red (W ,W )-paths. By Menger’s Theorem
1 2
applied to the graph GR, we obtain that the size of a minimum (W ,W )-separator in GR is
1 2
at most one. Let S ⊆ V(G) be such a separator. We separate the rest of the proof in two
cases: there exists a red edge in E(W ,W ) or not.
1 2
Case A: There exists a red edge in E(W ,W ). Let e = e e be such an edge,
1 2 1 2
with e ∈ W for i ∈ {1,2}. As there are no two disjoint (W ,W )-paths by Lemma 18, the
i i 1 2
red edges in E(W ,W ) form a non-empty star. Let e be the center of this star. Every
1 2 i
vertex in W , other than e , must send a red edge outside W because of (6), so it must be
i i i
sent to W . The red edges in W induce a connected graph, so if there were a red edge in
3 3
E(W ,W ) disjoint from e , we could find two disjoint red (W ,W )-paths, which is not
3 3−i 3−i 1 2
possible. Using (6) again, we see that every vertex in W must send a red edge to e ∈ W .
3−i i i
So the sets W = W , X = X , W′ = V(G)−W −e and X′ = X −e satisfy hypotheses
3−i 3−i i i i
A, B, C and D1 of Lemma 20, and we are done.
Case B: Every edge in E(W ,W ) is blue. Remember that S is a (W ,W )-separator
1 2 1 2
in GR of size at most one. It is not empty, because by (6) every vertex in W and W has a
1 2
red neighbor in W andthe red edges in W forma connected subgraph, and this implies that
3 3
there exists, at least, one red (W ,W )-path. Furthermore, S cannot be contained outside
1 2
W , because deleting any vertex in W ∪W does not eliminate all the red (W ,W )-paths.
3 1 2 1 2
Therefore, we have S = {s} ⊆ W .
3
Suppose that GR[W ]\{s} is connected. Then there exists i ∈ {1,2} such that E(W \
3 3
{s},W ) only contains blue edges, as otherwise there would still be red (W ,W )-paths in
i 1 2
GR − S. Then the sets W = W , X = X , W′ = (W ∪ W ) \ S and X′ = X satisfy
i i 3 2 3−i
hypotheses A, B, C and D1 of Lemma 20, therefore concluding Theorem 1. So, we may
suppose that GR[W ]\{s} is disconnected. Hence, S = X .
3 3
By Lemma 17, every vertex in W \S has a red neighbor in W or W . It cannot have
3 1 2
red neighbors in both W and W , as that would form a red (W ,W )-path in G−S. So,
1 2 1 2
Y := {v ∈ W \S : v has a red neighbor in W } and
1 3 1
Y := {v ∈ W \S : v has a red neighbor in W },
2 3 2
together partition W \ S. Furthermore, we have that every edge in E(Y ,W ), E(Y ,W )
3 1 2 2 1
and E(Y ,Y ) is blue. So, we have a partition {W ∪Y ,W ∪Y ,S} of V(G) such that every
1 2 1 1 2 2
edge between W ∪Y and W ∪Y is blue. Suppose |W ∪Y | ≤ |W ∪Y | (otherwise, the
1 1 2 2 1 1 2 2
proof is similar).
If |W ∪ Y | ≤ 2k, then |W ∪ Y | ≥ N − 1 − 2k ≥ 3k + 2. Applying Lemma 20, with
1 1 2 2
W = W ∪Y , X = X , W′ = W ∪Y and X′ = X ; we conclude the theorem.
1 1 1 2 2 2
Suppose then that |W ∪Y | > 2k. Since W ∪Y cannot contain a red C , it contains
1 1 1 1 2k+1
a blue edge e = v v . Consider H := G[W ∪ Y ]. If there exists a vertex w ∈ V(H) with
1 2 2 2
at least k blue neighbors in H, then by Lemma 10, we could find a blue (2k + 1)-cycle
which together with w form a blue (2k + 1)-wheel, which is impossible. Then, we have
∆B(H) ≤ k −1, and δR(H) ≥ |H|−k.
We have that |H| ≥ |W ∪ Y | > 2k. So, δR(H) > 1|H| and by Corollary 12, HR is
1 1 2
pancyclic, and thus contains a red copy of C , a contradiction.
2k+1
9
4 Proof of Theorem 3
Both of the bounds of Theorem 3 will follow from a more general type of bound.
Definition 2. Given two reals α ∈ [1,1) and β > 0, we say that (α,β) is an admissible pair
4
if every 2-connected non-bipartite graph G = (V,E) with |V| = n and
δ(G) ≥ αn+β
contains every cycle C , for every t such that 6 ≤ t ≤ c(G).
t
Note that the non-bipartiteness of the graph is useful in the definition, because otherwise
K is a graph with δ(G) ≥ n/2 that is not weakly pancyclic. Given Lemma 14 and
n/2,n/2
α ≥ 1/4, the condition of containing every cycle with length between 6 and c(G) is slightly
weaker thanbeing weakly pancyclic, for graphs with at least 9 vertices, and it can be checked
by inspection that the cycle condition is also satisfied by non-bipartite graphs with 8 vertices
or less.
Brandt et al. [BFG98] proved some theorems concerning the values of (α,β) that assure
weak pancyclicity of the graph, with or without the requirement of 2-connectedness.
Theorem 21 (Brandt et al. [BFG98]). Every non-bipartite graph of order n with minimum
degree δ(G) ≥ n+2 is weakly pancyclic with girth at most 4.
3
This implies that (1, 2) is an admissible pair.
3 3
Theorem 22 (Brandt et al. [BFG98]). Every non-bipartite 2-connected graph of order n
with minimum degree δ(G) ≥ n+1000 is weakly pancyclic unless G has odd girth 7, in which
4
case it has every cycle from 4 up to its circumference except the 5-cycle.
This implies that (1,250) is an admissible pair. Brandt et al. also give an example to
4
show that no admissible pair (α,β) has α < 1/4. Take two copies of K intersecting in
m,m
one vertex and join one vertex on the opposite side of the intersection vertex in one K to
m,m
such a vertex in the other K . Then this graph has n := 4m−1 vertices, minimum degree
m,m
(n+1)/4, is 2-connected, Hamiltonian and has a triangle, but it is not weakly pancyclic as
it does not contain any even cycle of length more than (n+1)/2.
We prove a bound on the Ramsey number r(C ,W ) for k < j, that depends on the
2k+1 2j
existence of an admissible (α,β) pair.
Theorem 23. Let (α,β) be an admissible pair, and 2 < k < j integers. Then
3j +β
r(C ,W ) ≤ .
2k+1 2j
1−α
Then, using j = k+1withtheadmissible pairsimplied fromTheorem 21andTheorem22
respectively we obtain easily the bounds of Theorem 3.
Now let us prove Theorem 23. Let 2 < k < j be integers and (α,β) an admissible pair
with α ≥ 1/4. Let G be a graph with |G| ≥ (3j + β)/(1 − α) and c : E(G) → {R,B} a
red-blue coloring of its edges. Suppose, for the sake of contradiction, that C * GR or
2k+1
W * GB.
2j
The bound α ≥ 1/4 implies that |G| ≥ 4j + 4β > 4j, and thus, |G| ≥ 4j +1. We start
3
with a series of lemmas.
10