Anti-Ramsey numbers of paths and cycles in hypergraphs
Ran Gu1, Jiaao Li2 and Yongtang Shi3
1College of Science, Hohai University,
Nanjing, Jiangsu Province 210098,
P.R. China
2School of Mathematical Sciences and LPMC
Nankai University, Tianjin 300071, China
3Center for Combinatorics and LPMC
Nankai University, Tianjin 300071, China
Emails: [email protected]; [email protected]; [email protected]
Abstract
The anti-Ramsey problem was introduced by Erdős, Simonovits and Sós in 1970s. The anti-Ramsey number of a hypergraph H, ar(n,s,H), is the smallest integer c such that in any coloring of the
edges of the s-uniform complete hypergraph on n vertices with exactly c colors, there is a copy of H whose edges have distinct colors. In this paper, we determine the anti-Ramsey numbers of linear paths and loose paths in hypergraphs for sufficiently large n, and give bounds for the anti-Ramsey numbers of Berge paths. Similar exact anti-Ramsey numbers are obtained for linear/loose cycles, and bounds are obtained for Berge cycles. Our main tools are path extension technique and stability results on hypergraph Turán problems of paths and cycles.
Keywords: anti-Ramsey; rainbow; hypergraphs; paths; cycles; Turán number
AMS Subject Classification (2010): 05D05, 05C35, 05C65
1 Introduction
The
anti-Ramsey number of a graph G, denoted by ar(n,G), is the minimum number of colors needed to color the edges of the complete graph Kn so
that, in any coloring, there exists a copy of G whose edges have distinct colors. The Turán number of a graph G, denoted by ex(n,G), is the maximum
number of edges in a graph on n vertices that does not contain G as a subgraph. It is easy to observe that
[TABLE]
for any graph H.
In 1973, Erdős, Simonovits and Sós [11] showed a remarkable result that ar(n,Kp)=ex(n,Kp−1)+2 for sufficiently large n.
Montellano-Ballesteros and Neumann-Lara [35] extended this result to all values of n and p with n>p≥3. In [11], it was shown that ar(n,H)−ex(n,{H−e,e∈E(H)})=o(n2) when n→∞. Furthermore, Jiang [29] proved that if H is a graph such that each edge is incident to a
vertex of degree two, then ar(n,H)−ex(n,{H−e,e∈E(H)})=O(n). A history
of results and open problems on this topic was given by Fujita, Magnant, and Ozeki [18].
A hypergraph H consists of a set V(H) of vertices and a family E(H) of nonempty subsets of V(H) called edges of H. If
each edge of H has exactly s vertices, then H is s-uniform and H is called an s-graph. A complete s-uniform hypergraph is a hypergraph whose edge
set consists of all s-subsets of the vertex set. In an edge-coloring of a (hyper)graph H, a sub(hyper)graph F⊆H is rainbow if all edges of F have distinct colors.
The anti-Ramsey number and Turán number are naturally extended from graphs to hypergraphs. The anti-Ramsey number of an s-uniform hypergraph H,
denoted by ar(n,s,H), is the minimum number of colors needed to color the edges of a complete s-uniform hypergraph on n
vertices so that there exists a rainbow H in any coloring. The Turán number of H, denoted by ex(n,s,H),
is the maximum number of edges in an s-uniform hypergraph on n vertices that contains no H. Özkahya and Young [39] investigated the anti-Ramsey number of matchings in hypergraphs, where a matching is a set of edges in a (hyper)graph in which no two edges have
a common vertex. A k-matching, denoted by Mk, is a matching with k edges. Özkahya and Young [39] gave the lower and upper bounds for ar(n,s,Mk) in terms of ex(n,s,Mk−1). They proved that
[TABLE]
where the lower bound holds for every n, and the upper bound holds for n≥sk+(s−1)(k−1). For s=2, Schiermeyer [40] proved that ar(n,2,Mk)=ex(n,2,Mk−1)+2 for k≥2 and n≥3k+3, and this condition was further released to all n≥2k+1 by Chen, Li and Tu [5] and by Fujita, Kaneko, Schiermeyer and Suzuki [17], independently.
In fact, for k-matchings, the Turán number ex(n,s,Mk) is still not known for k≥3 and s≥3. Let [n] denote the set {1,2,…,n}, (k[n]) denote the set consisting of all the k-sets of [n].
Erdős put forward a conjecture in 1965 that ex(n,s,Mk)=max{∣As∣,∣Bs(n)∣}, where
As=(s[sk−1]) and Bs(n)={F∈(s[n])∣F∩[k−1]=∅}.
This conjecture is true for s=2, which was
shown by Erdős and Gallai [10]. In [9], Erdős proved that there exists a constant n0(s,k) such that for n>n0(s,k), the conjecture holds. Then Bollobás, Daykin and Erdős [1] improved the bound for n0(k,s) such that n0(k,s)≤2s3(k−1). It was improved to n0(k,s)≤3s2(k−1) by Huang, Loh and Sudakov [28] later.
For the anti-Ramsey number of k-matching, Özkahya and Young [39] conjectured
that when k≥3, ar(n,s,Mk)=ex(n,s,Mk−1)+2 if n>sk, and
[TABLE]
if n=sk, where cs is a constant dependent on s. They proved that the conjecture is true when k=2,3 for sufficiently large n. Later, Frankl and Kupavskii [16] proved that ar(n,s,Mk)=ex(n,s,Mk−1)+2 for n≥sk+(s−1)(k−1)
and k≥3. For more results on matchings, we refer to [13, 15].
For paths, Simonovits and Sós [41] proved that ar(n,P2t+3+ϵ)=tn−(2t−1)+1+ϵ for large n, where ϵ=0,1 and Pk is a
path on k vertices. Comparing with the Turán number of paths
[TABLE]
given by Erdős and Gallai [10], it follows that ar(n,Pk)=ex(n,Pk−1)+O(1) when k is odd, and ar(n,Pk)=ex(n,Pk−2)+O(1) when k is even. For a cycle Ck of order k, Erdős, Simonovits and Sós [11] conjectured that
ar(n,Ck)=n(2k−2+k−11)+O(1). This conjecture was confirmed by Montellano-Ballesteros and Neumann-Lara [36], and they gave the exact value of ar(n,Ck) for all n≥k≥3.
It would be interesting to investigate the relation between the anti-Ramsey number and the Turán number for paths and cycles in hypergraphs. The Turán numbers of paths and cycles are extensively studied, see [12, 19, 20, 33] or Section 2 below for details. Motivated by this, we will study the anti-Ramsey numbers of paths and cycles and compare it with the Turán numbers of paths and cycles in hypergraphs.
There are several possible ways to define paths and cycles in hypergraphs as generalization of paths and cycles in graphs from different aspects.
Definition 1.1
Let H be an s-uniform hypergraph.
(i) A Berge path of length k in H is a family of k
distinct edges e1,…,ek and k+1
distinct vertices v1,…,vk+1 such that for each
1≤i≤k, ei contains vi and vi+1. Let
Bk denote the family of Berge
paths of length k. A Berge cycle of length k in H is a cyclic list of k distinct edges e1,…,ek and k distinct vertices v1,…,vk such that ei contains vi and vi+1 for each 1≤i≤k, where vk+1=v1. Denote the family of all Berge cycles of
length k by BCk.
(ii) A loose path of length k in H
is a collection of distinct edges
{e1,e2,…,ek} such
that consecutive edges intersect in at least one element and
nonconsecutive edges are disjoint. Denote the family of loose paths of
length k by Pk. A loose cycle is defined similarly in a cyclic order, and denote the family of all loose cycles of
length k by Ck.
(iii) A linear path of length k in H
is a collection of distinct edges
{e1,e2,…,ek} such
that consecutive edges intersect in exactly one element and
nonconsecutive edges are disjoint. Let Pk denote the linear
path of length k. A
linear cycle is defined similarly in a cyclic order, and let Ck denote the collection of linear
path of length k.
We first give the exact anti-Ramsey numbers of short paths Pi, Bi, Pi for i=2,3.
Theorem 1.1
(i) For s≥3 and n≥3s−4, ar(n,s,P2)=2.
(ii) For s≥4 and sufficiently large n, ar(n,s,P3)=(s−2n−2)+2.
(iii) For n≥3s−4, ar(n,s,B2)=ar(n,s,P2)=2.
(iv) For n≥4s−3, ar(n,s,B3)=ar(n,s,P3)=3.
For linear paths and loose paths, we obtain the exact anti-Ramsey numbers for sufficiently large n.
Theorem 1.2
For any integer k, if k=2t≥4 and s≥3, then for sufficiently large n,
[TABLE]
if k=2t+1>5 and s≥4, then for sufficiently large n,
[TABLE]
Theorem 1.3
For any integer k, if k=2t≥4 and s≥3, then for sufficiently large n,
[TABLE]
if k=2t+1≥5 and s≥3, then for sufficiently large n,
[TABLE]
We remark that, due to some technique obstruction, our proof of Theorem 1.2 does not work directly for the case k=5 or the case k is odd and s=3. However, those special cases are handled in Theorem 1.3 for loose path with a refined analysis.
The methods developed in proving Theorems 1.2 and 1.3, with additional effort and some new ideas, allow us to determine the anti-Ramsey numbers of linear cycles and loose cycles as well, if k and s are not too small. We obtain the following exact results for linear cycles and loose cycles.
Theorem 1.4
For any integer k, if k=2t≥8 and s≥4, then for sufficiently large n,
[TABLE]
if k=2t+1≥11 and s≥k+3, then for sufficiently large n,
[TABLE]
Theorem 1.5
For any integer k, if k=2t≥8 and s≥4, then for sufficiently large n,
[TABLE]
if k=2t+1≥11 and s≥k+3, then for sufficiently large n,
[TABLE]
For a Berge path Bk, Györi, Katona and Lemons in [25] proved that ex(n,s,Bk)≤kn(sk) when k>s+1>3, and ex(n,s,Bk)≤s+1n(k−1) when 2<k≤s, which are sharp for infinitely many n. Then Davoodi, Györi, Methuku and Tompkins [8] proved that ex(n,s,Bs+1)≤n. We apply those results to obtain the bounds for the anti-Ramsey number ar(n,s,Bk) as follows.
Theorem 1.6
If k>2s+1, then for sufficiently large n,
[TABLE]
If s+2≤k≤2s+1, then for sufficiently large n,
[TABLE]
If k≤s+1, then for sufficiently large n,
[TABLE]
Theorem 1.6 indicates that the anti-Ramsey number ar(n,s,Bk) varies for different s and k. This may suggest that determining the exact value of ar(n,s,Bk) would be very difficult. Note that the Turán number of Berge path is still not clear at this moment.
However, it seems that the anti-Ramsey numbers of Berge cycles have different behavior with Berge paths, and we obtain the following bounds, similar to the Özkahya-Young result [39] on matchings.
Proposition 1.7
For any fixed integers s≥4, k≥3,
[TABLE]
The next section will be focused on introducing results on Turán numbers of paths and cycles in hypergraphs, which are needed tools to derive our main results. A useful lemma obtained from the stability results on Turán problems will be given in the next section as well, which will be frequently used to find certain desired paths in later proofs. The proof of the main results will be presented in later sections.
2 Preliminaries
Note that the s-uniform Berge path B2 and loose path P2 are the same definitions, and the determination of ex(n,s,P2) is trivial.
In [20], Füredi, Jiang and Seiver determined ex(n,s,Pk) for s≥3.
Theorem 2.1
[20]**
Let s, t be positive integers with s≥3. For sufficiently
large n, we have
[TABLE]
and
[TABLE]
For P2t+1, the unique extremal family consists of all the s-subsets of [n] which meet some fixed set S of size t. For P2t+2, the unique extremal family consists of all the above edges plus one additional s-set disjoint from S.
The determination of ex(n,s,Pk) is nontrivial even for k=2.
Frankl [12] gave the value of ex(n,s,P2)
for s≥4 and sufficiently large n. Then Keevash, Mubayi and Wilson [32] determined the value of ex(n,4,P2) for all n. Note
that when s=3, ex(n,3,P2)≤n, which can be achieved when n is divisible
by 4 by taking n/4 vertex disjoint copies of K4(3) (i.e. the complete 3-graph on
4 vertices). For k≥3, Füredi, Jiang and Seiver [20] provided the exact Turán number of Pk for sufficiently large n, where s≥4, k≥3. Kostochka, Mubayi and Verstraëte [33] considered ex(n,s,Pk) for s≥3, k≥4 and sufficiently large n. Later, Jackowska, Polcyn and Ruciński [31] determined ex(n,3,P3) for all n. We summarize those results (only for the sufficiently large n) as follows.
Theorem 2.2
[12, 20, 31, 33, 32]**
For sufficiently
large n, we have
-
ex(n,s,P2)=(s−2n−2)* for s≥4, and ex(n,3,P2)≤n.*
2. 2.
ex(n,s,P2t+1)=(sn)−(sn−t)* for s≥3 and t≥1.*
3. 3.
ex(n,s,P2t+2)=(sn)−(sn−t)+(s−2n−t−2)* for s≥3 and t≥1.*
The unique extremal family for P2 consists of all the s-subsets of [n] containing some two fixed vertices for s≥4.
For P2t+1, the unique extremal family consists of all the s-subsets of [n] which meet some fixed set S of size t. For P2t+2, the unique extremal family consists of all the above edges plus all the s-sets in [n]\S containing some two fixed vertices not in S.
For linear cycles, Frankl and Füredi [14] showed that
the unique extremal n-vertex s-graph (s≥3) containing no C3 consists of all edges containing
some fixed vertex x, for large enough n. For s=3, Csákány and Kahn [7] obtained the same result
for all n≥6. Füredi and Jiang [19], and Kostochka, Mubayi and Verstraëte [33] determined the Turán number of Ck for all k≥3, s≥3 and sufficiently large n as follows.
Theorem 2.3
[19, 33]**
Let s,t be positive integers with s≥3. For sufficiently large n, we have
[TABLE]
and for (s,t)=(3,1),
[TABLE]
For C2t+1, the only extremal family
consists of all the s-sets in [n] that meet some fixed t-set
L. For C2t+2, the only extremal
family consists of all the s-sets in [n] that intersect some
fixed t-set L plus all the s-sets in [n]∖L that
contain some two fixed elements.
For the exceptional case of C4 in 3-uniform hypergraphs,
Kostochka, Mubayi and Verstraëte [33] showed that
[TABLE]
and they also characterized the extremal graphs.
The Turán number of a loose cycle was initially studied by Chvátal [6]. Then Mubayi and Verstraëte [38] proved that ex(n,s,C3)=(sn)−(s−1n−1) for all s≥3 and n≥3s/2. Füredi and Jiang [19]
determined ex(n,s,Ck) for k≥3, s≥4 and sufficiently large n. This confirms (in a
stronger form) a conjecture proposed by Mubayi and Verstraëte [37] for
k≥3, s≥4. Kostochka, Mubayi and Verstraëte [33] extended the results above and determined ex(n,s,Ck) for all
s≥3 and large n. We summarize their results as follows.
Theorem 2.4
[19, 33]**
Let t≥2, s≥3 be fixed integers. For sufficiently large n, we have
[TABLE]
[TABLE]
and
[TABLE]
For C2t+1 (t≥2), the only extremal
family consists of all the s-sets in [n] that meet some fixed
t-set L. For C2t+2 (t≥2), the only
extremal family consists of all the s-sets in [n] that intersect
some fixed t-set L plus one additional s-set outside L. For C4, the only
extremal family consists of all the s-sets in [n] that intersect
some fixed t-set L plus ⌊sn−1⌋ disjoint edges outside L.
For Turán number of a Berge cycle, Győri and Lemons proved that ex(n,3,BC2k+1)≤O(k4)⋅n1+1/k in [27]
and ex(n,3,BC2k+1)≤O(k2)⋅ex(n,C2k) in [26]. Füredi and Özkahya [21] obtained better constant factors (depending on k). Further improvements were obtained for even k by Gerbner, Methuku and Vizer [24], also by
Gerbner, Methuku and Palmer [23], and for odd k by Gerbner [22].
For s≥4, Győri and Lemons [26] showed that ex(n,s,BC2k+1)≤Os(ks−2)⋅ex(n,3,BC2k+1) and ex(n,s,BC2k)≤Os(ks−1)⋅ex(n,C2k), i.e., ex(n,s,BCk)=O(n1+1/⌊k/2⌋). The constant factors were improved
by Jiang and Ma [30], also for even k by Gerbner, Methuku and Vizer [24].
There are many other results on the Turán numbers of paths and cycles, in graphs [2, 3, 34, 42] or hypergraphs [4, 21, 25]. The readers are referred to these references for details.
The following stability result on linear paths and linear cycles will be needed in our proofs. Let ∂H denote
the (s−1)-graph consisting of sets contained in some edge of H.
Theorem 2.5
[33]**
For fixed s≥3 and k≥4, let ℓ=⌊2k−1⌋ and H be an n-vertex s-graph with ∣H∣∼ℓ(s−1n)
containing no Pk or containing no Ck. Then there exists G∗⊂∂H with ∣G∗∣∼(s−1n) and a set L of ℓ vertices of H such that L∩V(G∗)=∅ and e∪{v}∈H for any (s−1)-edge e∈G∗ and any v∈L. In particular,
∣H−L∣=o(ns−1).
Notice that the stability result above considers the case k≥4. For k=3, s≥4, Füredi et al. [20] provided another version of stability result on linear paths, and as the authors in [31] pointed out, the similar stability result holds for k=3 and s=3 as well. We rewrite their results for k=3 with the similar notations in Theorem 2.5 (in a slightly weaker form). Note that when k=3, ℓ=⌊2k−1⌋=1.
Theorem 2.6
For fixed s≥3, let H be an n-vertex s-graph with ∣H∣∼(s−1n)
containing no P3. Then there exists G∗⊂∂H with ∣G∗∣>21(s−1n) and a vertex v of H such that v∈/V(G∗) and e∪{v}∈H for any (s−1)-edge e∈G∗. In particular,
∣H−v∣=o(ns−1).
Considering the structure of the (s−1)-graph G∗, we present
the following lemma, which is frequently used in our proofs.
Lemma 2.7
For fixed s≥3 and k≥3, let t=⌊2k−1⌋ and H be an n-vertex s-graph with ∣H∣∼t(s−1n), which
contains no Pk, or contains no Ck when k≥4. Let G∗⊂∂H be the (s−1)-graph as defined in Theorem 2.5 or 2.6 above.
Given a vertex set W of d vertices in H, where d is a fixed constant.
Then for sufficiently large n, there are max{t−1,1} pairs of (s−1)-edges in G∗,
say {ai,bi}, i=1,…,t−1,
such that each of the following holds.
(i) For every i, ai and bi have exactly one common vertex (i.e. ∣ai∩bi∣=1),
(ii) for any i=j,
ai∪bi and aj∪bj are vertex disjoint, and moreover,
(iii) all these (s−1)-edges are disjoint from W.
Proof.
The number of (s−1)-edges incident with some vertices in W is at
most ∣W∣⋅(s−2n−1), so in G∗ the number of (s−1)-edges disjoint from W is
at least
[TABLE]
By Theorem 2.2 and Eq.(2), we get that ∣G∗∣−d(s−2n−1)>ex(n,s−1,P2) for sufficiently large n. So we can find a pair {a1,b1} of
(s−1)-edges with exactly one common vertex.
Since
[TABLE]
we can repeat the argument above to find {a2,b2},
…,{at−1,bt−1} satisfying the properties described in Lemma
2.7.
□
Given a path P, if a vertex v belongs to more than one edge in P, we call v a cross vertex of P, or say v is a cross(P) vertex. If v belongs to exactly one edge in P, we call v a free vertex of P, or say v is a free(P) vertex.
3 Short Path–Proof of Theorem 1.1
(i). Let H be a complete s-uniform hypergraph on n vertices. It is clear that ar(n,s,P2)≥2. Suppose that there exists a 2-coloring of H without a rainbow P2. Then there must be two edges e1 and e2 satisfying that the colors of e1 and e2 are different and ∣e1∩e2∣>1. Let u∈e1∖e2 and v∈e2∖e1. Consider the edge e3 consisting of u, v and s−2 vertices in V(H)∖V(e1∪e2). Since there is no rainbow P2, e3 can not be colored with either of the two colors, a contradiction. So any 2-coloring of H admits a rainbow P2.
(ii). Let H be a complete s-uniform hypergraph on n vertices. Consider the following coloring of H with (s−2n−2)+1 colors. Take two vertices u and v in H, then the number of edges containing both u and v is (s−2n−2). Coloring each of these edges with different colors and the remaining edges of H with an additional color. We can see that this coloring of H yields no rainbow P3. Thus, ar(n,s,P3)≥(s−2n−2)+2.
To prove that ar(n,s,P3)≤(s−2n−2)+2, we argue by contradiction. Suppose that there exists a coloring of H without a rainbow P3, which uses (s−2n−2)+2 colors. Let G be a spanning subgraph of H with (s−2n−2)+2 edges such that each color appears on exactly one edge of G. Since ∣G∣=(s−2n−2)+2>ex(n,s,P2) for sufficiently large n, there is a linear path P of length two with edges e1 colored by α1 and e2 colored by α2 in G. Let v be the common vertex of e1 and e2. Since H contains no rainbow P3, any edge which contains only one vertex from (e1∪e2)∖{v}, must be colored with α1 or α2 in H.
Denote by F the subgraph obtained by deleting e1 and e2 from G. If there is a linear path P′ of length two with edges f1 and f2 in F, let us say the colors of f1 and f2 are β1 and β2, respectively. If f1 or f2 contains a free(P) vertex w of e1∪e2 and w is not a cross(P′) vertex in f1∪f2, then the edge consisting of w and some s−1 vertices in V(H)∖V(e1∪e2∪f1∪f2), along with f1 and f2 form a rainbow P3. Suppose f1∪f2 contains exactly one free(P) vertex w of e1∪e2 and w is the cross(P′) vertex. Take an edge e3 consisting of a free(P) vertex x=w of e1, a free(P′) vertex y of f1 and s−2 vertices of V(H)∖V(e1∪e2∪f1∪f2), then the color of e3 is either α1 or α2. Hence the path with edges e3, f1 and f2 is a rainbow P3. If f1∪f2 contains no free(P) vertex of e1∪e2, then the edge e4 formed by a free(P) vertex x of e1, a free(P′) vertex y of f1 and s−2 vertices of V(H)∖V(e1∪e2∪f1∪f2), must be colored with α1 or α2. So the path with edges e4, f1 and f2 is a rainbow P3, a contradiction. Therefore, we can assume that there is no P2 in F. By Theorem 2.2, F consists of all the (s−2n−2) edges containing two fixed vertices x and y. Note that {x,y}⊈e1, and {x,y}⊈e2 since e1,e2∈/F.
We divide our discussion into the following cases depending on the relationship between vertices x, y and edges e1, e2.
Case 1. x, y∈/e1∪e2.
Consider the edge e consisting of x, y, a free(P) vertex of e1∪e2, and s−3 vertices in V(H)∖V(e1∪e2∪{x,y}). Then, by the structure of F, we have e∈F, and thus e has a different color with α1 and α2. Therefore, we find a rainbow P3 with edges e,e1 and e2 in H.
Case 2. x∈e1∪e2, y∈/e1∪e2, and x is not the cross(P) vertex in e1∪e2.
The edge e, which consists of x, y and s−2 vertices in V(H)∖V(e1∪e2∪{x,y}), has a different color with α1 and α2. Hence, e, e1, e2 form a rainbow P3 in H. Note that if y∈e1∪e2, x∈/e1∪e2, and y is not the cross(P) vertex in e1∪e2, we can also find a rainbow P3 in H similarly.
Case 3. x is a cross(P) vertex in e1∪e2, y∈/e1∪e2.
Let e be an edge
with a free(P) vertex in e1 and s−1 vertices in V(H)∖V(e1∪e2∪{x,y}). If e has color α2, then we have a rainbow P2 with edges e and e1. Similar to Case 2, we can find a rainbow P3 in H. So the color of e is α1.
Pick an edge e′ consisting of x, y, a vertex z in e∖e1 and s−3 vertices in V(H)∖V(e1∪e2∪{x,y,z}), then e, e′, e2 form a rainbow P3 in H. And by symmetry, if y is a cross(P) vertex in e1∪e2 and x∈/e1∪e2, we can find a rainbow P3 in H as well.
Case 4. x∈e1∖e2, y∈e2∖e1.
Take an edge e consisting of a free(P) vertex w=x in e1, and s−1 vertices in V(H)∖V(e1∪e2), the color of e is α1
or α2. If the color of e is α1, then the edge e′ consisting of w, x, y and s−3 vertices of V(H)∖V(e1∪e2∪e∪{x,y}), along with e and e2 form a rainbow P3. If the color of e is α2, then the edge e′′ consisting of x, y and s−2 vertices of V(H)∖V(e1∪e2∪e∪{x,y}), along with e1 and e form a rainbow P3.
We have examined all the cases in the above discussion. In conclusion, any coloring of H with (s−2n−2)+2 colors admits a rainbow P3. Hence, we have that ar(n,s,P3)=(s−2n−2)+2.
(iii). Since ar(n,s,B2)≤ar(n,s,P2)≤ar(n,s,P2), we can obtain that ar(n,s,B2)=ar(n,s,P2)=2 for n≥3s−4.
(iv).
Let H be a complete s-uniform hypergraph on n vertices. Consider a 3-coloring of H such that there is no rainbow P3 in H. Since ar(n,s,P2)=2<3 by (iii), there is a rainbow loose path P of length 2 with edges e1 and e2, colored by, say, α1 and α2. Suppose that the number of free(P) vertices in e1 is a, so the number of free(P) vertices in e2 is equal to a. Let
[TABLE]
Assume that there is an edge f with color α3∈/{α1,α2} such that f∩(e1∪e2)=∅. Consider an edge e consisting of all the free(P) vertices in e1,
p vertices in f, and s−a−p vertices in V(H)∖V(e1∪e2∪f). Note that the color of e is either α1 or α2.
If e is colored with α2, then e1,e,f is a rainbow P3. So e can only be colored with α1. Similarly, let
[TABLE]
Consider the edge e′ consisting of all the free(P) vertices in e2,
q vertices in V(f)∖V(e), and s−a−q vertices in V(H)∖V(e1∪e2∪f∪e),
then e′ is colored with α2. Thus, e,f,e′ is a rainbow P3.
So each of the edges colored with α3 contains vertices in e1∪e2. Take an edge h with color α3. Note that h∩e1=∅ and h∩e2=∅.
Case 1. Either e1 or e2 contains a free(P) vertex not belonging to h.
Without loss of generality, suppose that there are b free(P) vertices in e1∖h, where b≥1. Take an edge e with b free(P) vertices in e1∖h and s−b vertices in V(H)∖V(e1∪e2∪h). Then e is colored with α1 or α2. If e is colored with α2, then e, e1, h form a rainbow P3. So e is colored with α1. Then we have a rainbow P2 with edges h and e2, which are colored with α3 and α2, respectively. And we have an edge e colored with α1 and e∩(h∪e2)=∅. It is the same situation as we analysed before, we can also find a rainbow P3 in H.
Case 2. All the free(P) vertices in e1∪e2 belong to h.
Recall that there are a free(P) vertices in e1. Take an edge e consisting of a free(P) vertices in e1 and s−a vertices in V(H)∖V(e1∪e2∪h). If e is colored with α2, then we have a loose path P′ of length two with edges e and e1, which are colored with α2 and α1, respectively, and e contains at least one free(P′) vertex not belonging to h. It is just similar to Case 1, in which we can find a rainbow P3 in H. Thus e is colored with α1. Similarly, take an edge e′ consisting of a free(P) vertices in e2 and s−a vertices V(H)∖V(e1∪e2∪h∪e), we can obtain that e′ is colored with α2. Now, there is a rainbow P3 consisting of edges e, h and e′.
Therefore, ar(n,s,P3)≤3, for n≥4s−3. Since ar(n,s,P3)≥3 trivially holds, we have that ar(n,s,P3)=3 for n≥4s−3.
Since ar(n,s,B3)≤ar(n,s,P3), we obtain that ar(n,s,B3)=ar(n,s,P3)=3 for n≥4s−3.
4 Linear Path–Proof of Theorem 1.2
Let H be a complete s-uniform hypergraph on n vertices.
For the lower bounds, we construct a coloring of H by using the extreme s-graphs in Theorem 2.2.
Proposition 4.1
(a) For k=2t, we have
[TABLE]
(b) For k=2t+1, we have
[TABLE]
Proof.
(a) If k=2t, we pick a vertex set S with t−1 vertices. Take all the edges that meet S and color each of these edges with different colors. Then color the remaining edges of H with one additional color. This gives a coloring of H with (sn)−(sn−t+1)+1 colors. Since each vertex is contained in at most two edges of a rainbow linear path and a rainbow linear cycle, it is easy to see that any rainbow linear path or rainbow linear cycle in H has length at most 2(t−1)+1<2t. So we have ar(n,s,P2t)≥(sn)−(sn−t+1)+2 and ar(n,s,C2t)≥(sn)−(sn−t+1)+2.
(b) If k=2t+1, we pick a copy of the extreme P2t-free graph obtained in Theorem 2.2. Then color each edge of this extreme P2t-free graph with a distinct color, and color the remaining edges of H with one additional color to obtain a coloring of H with (sn)−(sn−t+1)+(s−2n−t−1)+1 colors. It is routine to check that there is no rainbow P2t+1 and no rainbow C2t+1 in the above coloring, and thus ar(n,s,P2t+1)≥(sn)−(sn−t+1)+(s−2n−t−1)+2 and ar(n,s,C2t+1)≥(sn)−(sn−t+1)+(s−2n−t−1)+2.
□
For the upper bounds, let
[TABLE]
We argue by contradiction and suppose that there is a coloring of H using D colors yielding no rainbow Pk.
Let G be a spanning subgraph of H with D edges such that each color appears on exactly one edge
of G. By Theorem 2.2, we obtain that there is a linear path P of length k−1 in G. Denote by e1,e2,…,ek−1 the edges of P, and α1,α2,…,αk−1 the colors of e1,e2,…,ek−1, respectively.
Since H contains no rainbow Pk, we obtain the following fact.
Observation 4.1
Let v be a free(P) vertex in e1∪ek−1. Then for any edge g satisfying g∩P={v}, the edge g must be colored with a color of {α1,α2,…,αk−1}.
Denote by F the subgraph obtained by deleting e1,e2,…,ek−1 from G.
We divide the remaining proof into two cases according to the parity of k.
4.1 Completing the proof when k=2t is even
In this subsection, we assume that k=2t≥4 is even.
Claim 4.1
When k=2t≥4, there is no Pk−1 in F.
Proof.
By contradiction,
suppose there is a linear path P′ of length k−1 in F. Denote the edges of P′ by f1,f2,…,fk−1 with colors β1,β2,…,βk−1, respectively. Since there is no rainbow Pk in H, every edge g with g∩V(P′)={u}, where u is a free(P′) vertex in f1∪fk−1, must be colored with a color of {β1,β2,…,βk−1}.
We obtain an s-graph Fe by deleting f1,f2,…,fk−1 and all the edges containing at least two vertices of P∪P′ from F. Let c denote the number of vertices of P∪P′. Then c≤2[(k−1)s−(k−2)], and so we have
[TABLE]
for sufficiently large n. Thus, we have a linear path P′′ of length k−2 in Fe. Denote by h1,h2,…,hk−2 the edges of P′′. Moreover, every edge in P′′ contains at most one vertex from P∪P′. So it follows from Observation 4.1 that P′′ contains no free(P) vertex of e1, ek−1, and no free(P′) vertex of f1, fk−1. Take an edge e consisting of a free(P) vertex x of e1, a free(P′′) vertex of h1∖V(P∪P′) (since s≥3, such vertex does exist), and s−2 vertices in V(H)∖V(P∪P′∪P′′), then e is colored with one color in {α1,α2,…,αk−1} by Observation 4.1. Take another edge e′ consisting of a free(P′) vertex of f1∖{x}, a free(P′′) vertex of hk−2∖V(P∪P′) and s−2 vertices in V(H)∖(P∪P′∪P′′∪e), then Observation 4.1 indicates e′ is colored with one color in {β1,β2,…,βk−1}. Hence the path with edges e,h1,h2,…,hk−2,e′ is a rainbow Pk, a contradiction. This proves the claim.
□
Note that ∣F∣∼(t−1)(s−1n). By Claim 4.2, Theorems 2.5 and 2.6 are applied to F. So we can find an (s−1)-graph G∗⊂∂F with ∣G∗∣∼(s−1n) for k≥6 and with ∣G∗∣≥21(s−1n) for k=4, and there is a vertex set L={v1,v2,…,vt−1} such that L∩V(G∗)=∅ and e∪{v}∈F for any (s−1)-edge e∈G∗ and any v∈L. Moreover,
∣F−L∣=o(ns−1). We point out that all the vertices of L are not free(P) vertices in e1∪ek−1.
Otherwise, let W be the vertex set of P. By Lemma 2.7, we can find an (s−1)-edge disjoint with W in G∗, and it gives an s-edge in F containing only a free(P) vertex of P. This edge together with P form a rainbow Pk, a contradiction.
Claim 4.2
When k=2t≥4, there is no edge in F−L.
Proof.
Suppose to the contrary there exists an edge h∈F−L with color λ say. By Lemma 2.7, we can find two (s−1)-edges a0, b0 in G∗, such that a0 and b0 have exactly one common vertex u and are disjoint from P and h. Let W be the vertex set of P∪h∪a0∪b0. By Lemma 2.7, we can find (s−1)-edges {ai,bi} disjoint from W for i=1,…,t−1, such that for every i, ai and bi have exactly one common vertex, and for any j=i,
{ai,bi} and {aj,bj} are vertex disjoint. Then,
[TABLE]
form a Pk−2 in F, denoted by P′. In the rest of the paper, this kind of path would be abbreviated as
[TABLE]
Let β1,β2,…,βk−2 be the colors of f1,f2,…,fk−2 respectively. Note that b0∪{v1} and bt−1∪{vt−1} are edges of F, so both of them have colors distinct from any other edges in F. The edges, which consist of one free(P′) vertex in f1, one free(P) vertex in e1 and s−2 vertices disjoint with P and P′, must be colored with colors from {α1,α2,…,αk−1} by Observation 4.1. Let f be an edge consisting of the free(P′) vertex u in f1, a vertex in h and s−2 vertices disjoint with P, P′, b0 and h. Then the color of f is in {λ,β1,β2,…,βk−2}, because otherwise h∪f∪P′ is a rainbow Pk. If the color of f is λ, then we can extend f∪P′ to a rainbow Pk with an additional edge, which containing one free(P′) vertex in fk−2 and a free(P) vertex in ek−1.
Assume the color
of f is βj for some j. Let W be the vertex set of f∪P′∪h∪b0. By Lemma 2.7, we can find (s−1)-edges {ai′,bi′} in G∗, which are disjoint from W for i=1,…,t−1. Furthermore,
[TABLE]
is a rainbow Pk, a contradiction. This shows that F−L contains no edge, i.e., all the edges in F contain vertices in L.
□
Notice that
[TABLE]
and there are (sn)−(sn−t+1) edges in H which intersect L.
Therefore Claim 4.2 implies that F contains no isolated vertices, and
[TABLE]
We will derive the final contradiction from the following claim.
Claim 4.3
When k=2t≥4, there exist at most one edge in P which is disjoint with L.
Proof.
Suppose that there are two edges ei and ej (j>i) in P, which are disjoint with L. If j>i+1, we find an edge f in F containing a vertex in ei and disjoint with ej, and an edge g in F containing a vertex in ej and disjoint with ei and f. Let f∩L=vp, g∩L=vq. Without loss of generality, we suppose that vp=v1, vq=vt−1. Let W consist of the vertices in ei∪ej∪f∪g. By Lemma 2.7, we can find (s−1)-edges {ai′,bi′} disjoint from W for i=1,…,t−1. Then,
[TABLE]
is a rainbow Pk in H, a contradiction.
If j=i+1, we find an edge h in F such that h contains exactly one vertex in ej∖ei, and is disjoint with ei. Without loss of generality, we suppose that h∩L=v1. Let W consist of the vertices in ei∪ej∪h. By Lemma 2.7, we can find (s−1)-edges {ai′′,bi′′} disjoint from W for i=1,…,t−1. Then,
[TABLE]
is a rainbow Pk in H, a contradiction.
□
Since P has k−1 edges, Claim 4.3 shows there are at least k−2 edges containing vertices of L in P.
As F=G−E(P), we conclude that there are at least k−2 edges containing vertices in L which are not belonging to F, contradicting (3). This completes the proof for even k.
4.2 Completing the proof when k=2t+1 is odd
In this subsection, we assume that k=2t+1 is odd.
Recall that F denotes the subgraph obtained from G by deleting e1,e2,…,ek−1. If there is no Pk−1 in F, then we can use Theorem 2.5 to characterize the structure of F. However, this may not be the case when k is odd. Fortunately, we can prove that after deleting a few edges, the remaining subgraph of F contains no Pk−1.
Claim 4.4
If there is a linear path P1 of length k−1 in F, then F−E(P1) contains no Pk−1.
Proof.
Suppose to the contrary that there is a linear path P2 of length k−1 in F−E(P1). Notice that the colors used in P1 and P2 are pairwise distinct by the selection of F. Let f1,f2,…,fk−1 be the edges of P1 with colors β1,β2,…,βk−1, respectively, Denote by g1,g2,…,gk−1 the edges of P2 with colors γ1,γ2,…,γk−1, respectively.
Let c denote the number of vertices of P∪P1∪P2. Then we have c≤3[(k−1)s−(k−2)]. Note that the number of edges which contain at least two vertices in P∪P1∪P2 is at most ∑i=2s(ic)(s−in−c). Since
[TABLE]
for sufficiently large n, there exists a linear path P3 of length k−3, such that every edge in P3 has at most one vertex of P∪P1∪P2.
Hence, all the free(P) vertices in e1∪ek−1, free(P1) vertices in f1∪fk−1 and free(P2) vertices in g1∪gk−1 are not in P3 by Observation 4.1. Denote by h1,h2,…,hk−3 the edges of P3. Consider an edge e, which consists of a free(P) vertex x in e1, a free(P3) vertex in h1∖(P1∪P2) and s−2 vertices disjoint with P∪P1∪P2∪P3, it follows from Observation 4.1 that the color of e is in {α1,α2,…,αk−1}.
And consider an edge e′, which consists of a free(P1) vertex y=x in f1∪fk−1 (we can find such a vertex y since s>3), a free(P3) vertex in hk−3∖(P1∪P2) and s−2 vertices disjoint with P1∪P2∪P3∪P∪e, then the color of e′ is from {β1,β2,…,βk−1} by Observation 4.1. Moreover, e∪P3∪e′ is a rainbow Pk−1. Now consider another edge e′′, which consists of a free(P2) vertex z=x,y in g1∪gk−1, a vertex in e′∖(P1∪P3) and s−2 vertices disjoint with P1∪P2∪P3∪P∪e∪e′, then e′′ has a color appeared in e∪P3∪e′ by Observation 4.1. However, to prevent extending P2 to a rainbow Pk, the color of e′′ should be one of {γ1,γ2,…,γk−3}, a contradiction.
□
So if F has a Pk−1, denote by F0 the subgraph obtained by deleting all the k−1 edges of that Pk−1 in F. Then we have F0 is Pk−1-free by Claim 4.4 and
[TABLE]
Since ∣F0∣∼(t−1)(s−1n) and by Theorem 2.5, we can find an (s−1)-graph G∗⊂∂F0 with ∣G∗∣∼(s−1n) and a set L of t−1 vertices of F0 such that L∩V(G∗)=∅ and e∪{v}∈F0 for any (s−1)-edge e∈G∗ and any v∈L. Moreover,
∣F0−L∣=o(ns−1).
If F dose not contain a Pk−1, then by Theorem 2.5 again, we can find an (s−1)-graph G∗⊂∂F with ∣G∗∣∼(s−1n) and a set L of t−1 vertices of F such that L∩V(G∗)=∅ and e∪{v}∈F for any (s−1)-edge e∈G∗ and any v∈L. Additionally,
∣F−L∣=o(ns−1). Since the number of edges meeting L is at most (sn)−(sn−t+1), we have ∣F−L∣≥∣F∣−[(sn)−(sn−t+1)]=(s−2n−t−1)+2−(k−1)>k−1. Now we delete any k−1 edges of F−L from F and still denote the remaining subgraph F0.
Therefore, in either case, we can find an (s−1)-graph G∗⊂∂F0 with ∣G∗∣∼(s−1n) and a set L of t−1 vertices of F0 such that L∩V(G∗)=∅ and e∪{v}∈F0 for any (s−1)-edge e∈G∗ and any v∈L. Moreover,
∣F0−L∣=o(ns−1). We select a G∗ with the maximum number of (s−1)-edges.
Let the vertices in L be v1,v2,…,vt−1. We point out that all the vi for i=1,2,…,t−1 are not free(P) vertices in e1∪ek−1. Otherwise, let W be the vertex set of P. Then by Lemma 2.7 we can find an (s−1)-edge disjoint with W in G∗, and this together with vi will form an s-edge which extends P to a rainbow Pk.
Since the number of edges containing vertices of L is at most
(sn)−(sn−t+1) in F0, we have ∣F0−L∣>(s−2n−t−1)+2−2(k−1).
We further claim the following.
Claim 4.5
∣F0−L∣≤(s−2n−t−1)+(s2s−2)+(s−12s−1)n.
Proof.
By contradiction, assume that ∣F0−L∣>(s−2n−t−1)+(s2s−2)+(s−12s−1)n. Note that the number of vertices of F0−L is n−t+1, by Theorem 2.2, we can find a P2 in F0−L, denoted by P1. Let h1, h2 be the edges of P1 with colors γ1, γ2, respectively. The number of edges containing at least s−1 vertices in P1 is less than (s2s−1)+(s−12s−1)n. Since ∣F0−L∣−[(s2s−2)+(s−12s−1)n]>ex(n−t+1,s,P2) for sufficiently large n, there is another linear path P2 of length two in F0−L such that each edge of which has at least two vertices not in P1. Let h3, h4 be the edges of P2 with colors γ3, γ4. So h4 contains a free(P2) vertex x∈/P1. Furthermore, one of h1,h2 contains a free(P1) vertex not belonging to P2. Let us say h2 has a free(P1) vertex y∈/P2. Take two (s−1)-edges a0, b0 in G∗ that are disjoint from P1, P2 and P, such that a0∩b0=u. Let W be the vertex set of P∪P1∪P2∪a0∪b0. By Lemma 2.7, we can find (s−1)-edges {ai,bi} disjoint from W for i=1,…,t−1, and so
[TABLE]
is a P2t−2=Pk−3 in F0. Denote by f1,f2,…,fk−3 the edges of P′ with colors β1,β2,…,βk−3 the colors of each edges in P′, respectively. Note that f1=a0∪{v1}.
Consider the edge g consisting of x, y, u and s−3 vertices disjoint with P, P1, P2 and b0. Then the color of g is in {γ1,…γ4,β1,…,βk−3}, otherwise we can easily extend P′ to a rainbow Pk by adding g, h1 and h2. If the color of g is in {γ1,γ2}, then h3∪h4∪g∪P′ is a rainbow Pk. If the color of g is in {γ3,γ4}, then h1∪h2∪g∪P′ is a rainbow Pk. So the color of g must be in {β1,…,βk−3}. Let W be the vertex set of P∪P1∪P2∪P′∪b0. By Lemma 2.7, we can find (s−1)-edges {ai′,bi′} disjoint from W for i=1,…,t−1,
and
[TABLE]
is a rainbow Pk, a contradiction.
□
Now Claim 4.5 provides some further structural properties of F0.
Claim 4.6
*(a) There is no isolated vertex in F0.
(b) There are at most
(s2s−2)+(s−12s−1)n+2(k−1)−2 edges meeting L but not in F0.
(c) Every vertex in V(F0)∖L belongs to G∗, and is not an isolated vertex in G∗.*
Proof.
(a) In fact, if F0 contains an isolated vertex, then the number of edges meeting L in F0 is at most (sn−1)−(sn−1−(t−1)). Thus, we have
[TABLE]
a contradiction to Claim 4.5. This indicates that F0 contains no isolated vertices.
(b) By Claim 4.5, there are
[TABLE]
edges in F0 containing vertices in L. Since there are (sn)−(sn−t+1) edges containing vertices of L in H, we have that there are at most
(s2s−2)+(s−12s−1)n+2(k−1)−2 edges meeting L but not in F0.
(c) If there exists a vertex v∈V(F0)∖L but v∈/G∗ or v is an isolated vertex in G∗, then we have at least (s−2n−(t−1)−1)>(s2s−2)+(s−12s−1)n+2(k−1)−2 edges meeting L but not belonging to F0, which is a contradiction to (b). Hence, every vertex in V(F0)∖L is contained in some edges of G∗ .
□
Now we focus on the edges which are disjoint with L, namely, the edges in (F∪P)−L=G−L and more generally, the edges in H−L. Considering the relationship between edges in H−L,
we make the following claim.
Claim 4.7
*Assume that there exist three edges f,g,h in H−L with distinct colors such that one of the following holds:
(i) f,g,h form a P3;
(ii) f,g form a P2, and h is disjoint with f∪g;
(iii) f,g,h are disjoint with each.
Then we can find a rainbow Pk in H.*
Proof.
Notice that for each e∈{f,g,h}, there is a unique edge e′ in G having the same color with e. So we denote f′,g′,h′ to be the edges in G with the same color with f,g,h, respectively. If the edge e is in G for some e∈{f,g,h}, we have e′=e.
(i) Assume that there are three edges f,g,h in H−L with distinct colors such that
f,g,h form a P3.
Realize that in G∗, there exists an (s−1)-edge a0 containing a vertex x in h∖g and disjoint with (f′∪g′∪h′∪f∪g∪h)∖{x}. Let W consist of the vertices in f′∪g′∪h′∪f∪g∪h∪a0. By Lemma 2.7, we can find (s−1)-edges {ai,bi} disjoint from W for i=1,…,t−1. Then,
[TABLE]
is a rainbow Pk in H.
(ii) Assume that there are three edges f,g,h in H−L with distinct colors such that f,g form a P2, and h is disjoint with f∪g.
In G∗, there exists an (s−1)-edge a0 containing a vertex x in g∖f and disjoint with (f′∪g′∪h′∪g∪h∪f)∖{x}. And there exists an (s−1)-edge b0 containing a vertex y in h and disjoint with (f′∪g′∪h′∪f∪g∪h∪a0)∖{y}. Let W consist of the vertices in f′∪g′∪h′∪f∪g∪h∪a0∪b0. By Lemma 2.7, we can find (s−1)-edges {ai,bi} disjoint from W for i=1,…,t−1. Then,
[TABLE]
is a rainbow Pk in H.
(iii) Assume that there are three edges f,g,h in H−L with distinct colors such that they are disjoint with each other.
In G∗, there exists an (s−1)-edge a0 containing a vertex x in f and disjoint with (f′∪g′∪h′∪f∪g∪h)∖{x}. And we can find (s−1)-edges a0′, b0′ in G∗, such that the (s−1)-edge a0′ contains a vertex y1 in g and disjoint with (f′∪g′∪h′∪f∪g∪h∪a0)∖{y1}; and the (s−1)-edge b0′ contains a vertex y2=y1 in g and disjoint with (f′∪g′∪h′∪f∪g∪h∪a0∪a0′)∖{y2}. Moreover, there exists an (s−1)-edge b0 containing a vertex z in h and disjoint with (f′∪g′∪h′∪f∪g∪h∪a0∪a0′∪b0′)∖{z}. Let W consist of the vertices in f′∪g′∪h′∪f∪g∪h∪a0∪b0∪a0′∪b0′. By Lemma 2.7, we can find (s−1)-edges {ai,bi} disjoint from W for i=1,…,t−1. Then,
[TABLE]
is a rainbow Pk in H. Note that in this case, we require that k=2t+1≥7.
□
As H is a counterexample which contains no rainbow Pk, we know that each of the conditions (i)(ii)(iii) in Claim 4.7 cannot exist. Hence there are at most two edges in P−L by Claim 4.7. In fact, if there are more than two edges in P−L, then one of conditions (i), (ii), (iii) in Claim 4.7 must occur. On the other hand, since ∣L∣=t−1, there are at most 2(t−1)=k−3 edges in P containing vertices in L. Therefore, there are exactly two edges in P−L, denoted by ei and ej. Since the number of edges meeting L in F is at most (sn)−(sn−t+1)−(k−3), we have
[TABLE]
We shall derive the final contradiction depending on F−L contains a P2 or not.
Case A. F−L contains a P2.
We take such a P2 in F−L and denote its edges by h1 and h2 with colors γ1 and γ2, respectively. Select an edge e in H−L such that e is disjoint with {ei,ej,h1,h2}. If the color of e is αi or αj, then h1,h2 form a P2 and e is disjoint with them, which satisfies condition (ii) of Claim 4.7, and so we can find a rainbow Pk in H. Assume instead that the color of e is not in {αi,αj}, then e,ei,ej have distinct colors. Furthermore, either e,ei,ej are pairwise disjoint, or ei,ej form a P2 and e is disjoint with them. This satisfies one of the conditions (ii) and (iii) of Claim 4.7, which we can find a rainbow Pk in H.
Case B. F−L does not contain a P2.
By (4) and Theorem 2.2, F−L is the extreme P2-free hypergraph on n−t+1 vertices. Namely, F−L consists of all the (s−2n−t−1) edges containing two fixed vertices x and y. Note that {x,y}⊈ei, and {x,y}⊈ej since ei,ej∈/F. If ei,ej are not consecutive in P, then we select an edge h in F−L such that h intersects ei∪ej as small as possible. Since F−L consists of all the (s−2n−t−1) edges containing x and y, we have that h intersects ei∪ej in at most two vertices, namely, some of x and y. Then ei,ej,h must satisfy one of the conditions (i), (ii) and (iii) in Claim 4.7, which we can find a rainbow Pk in H. Assume instead that ei,ej are consecutive in P in the following.
If either x∈ei∖ej, y∈ej∖ei or ei∩ej∈{x,y}, then we can select an edge h in F−L such that h∩({ei,ej}∖{x,y})=∅.
Take an edge e in H such that e is disjoint with {ei,ej,h} and L. If the color of e is αi or αj, then ej,h form a rainbow P2 and e is disjoint with them, or ei,h form a rainbow P2 and e is disjoint with them. Thus e,ej,h or e,ei,h satisfy condition (ii) of Claim 4.7, which we can obtain a rainbow Pk in H. If the color of e is neither αi nor αj, then e,ei,ej have distinct colors. Moreover, ei,ej form a P2 and e is disjoint with them. This shows e,ei,ej satisfy condition (ii) of Claim 4.7, which we can find a rainbow Pk in H.
Finally,
assume instead that
either {x,y}∩{ei,ej}=∅, or x∈ei∖ej, y∈/ej. Then we select an edge h in F−L such that h intersects ei∪ej as small as possible. Thus h intersects ei∪ej in at most one vertex, namely x, which is a free vertex in ei∪ej. This indicates ei,ej,h satisfy one of the conditions (i) and (ii) of Claim 4.7, and hence we can find a rainbow Pk in H.
Therefore, we have established the upper bound.
5 Loose Path–Proof of Theorem 1.3
Let H be a complete s-uniform hypergraph on n vertices.
The lower bound in Theorem 1.3 follows from a similar construction in Theorem 1.2 by applying the extreme s-graphs obtained from Theorem 2.1.
For the upper bound, if k=2t, since ar(n,s,Pk)≤ar(n,s,Pk)=(sn)−(sn−t+1)+2, we have done.
If k=2t+1,
we consider, by contradiction, a coloring of H using (sn)−(sn−t+1)+3 colors yielding no rainbow Pk.
Let G be a spanning subgraph of H with (sn)−(sn−t+1)+3 edges such that each color appears on exactly one edge
of G. By Theorem 2.2, we obtain that there is a loose path P of length k−1 in G. Denote by e1,e2,…,ek−1 the edges of P, and α1,α2,…,αk−1 the colors of e1,e2,…,ek−1, respectively.
Denote by F the subgraph obtained by deleting e1,e2,…,ek−1 from G. Similar to the proof of Theorem 1.2, we show that after deleting few edges from F, the remaining subgraph contains no Pk−1.
Actually, we prove something stronger. Call a loose path P′ bad, if the number of free(P′) vertices in the two end edges of P′ is at least three. Since s≥3, it is easy to get that a linear path is also a bad loose path.
Claim 5.1
There are no edge-disjoint bad loose paths of length k−1 in F.
Proof.
By contradiction, suppose that there are two edge-disjoint bad loose paths P1 and P2 of length k−1 in F. Denote by f1,f2,…,fk−1 the edges of P1 with colors β1,β2,…,βk−1, respectively. And denote by g1,g2,…,gk−1 the edges of P2 with colors γ1,γ2,…,γk−1, respectively.
Let c denote the number of vertices of P∪P1∪P2. Then c≤3[(k−1)s−(k−2)]. Note that the number of edges which contain at least two vertices in P∪P1∪P2 is at most ∑i=2s(ic)(s−in−c). Since
[TABLE]
for sufficiently large n, there exists a linear path P3 of length k−3, such that every edge in P3 has at most one vertex of P∪P1∪P2.
Hence, for the same reason as in Observation 4.1, all the free(P) vertices in e1∪ek−1, free(P1) vertices in f1∪fk−1 and free(P2) vertices in g1∪gk−1 are not in P3. Denote by h1,h2,…,hk−3 the edges of P3. Consider the edge e, which consists of a free(P) vertex x in e1, a free(P3) vertex in h1∖(P1∪P2) and s−2 vertices disjoint with P∪P1∪P2∪P3. Then the color of e must be from {α1,α2,…,αk−1}.
And consider the edge e′, which consists of a free(P1) vertex y=x in f1∪fk−1 (we can find such a vertex y since P1 is bad), a free(P3) vertex in hk−3∖(P1∪P2) and s−2 vertices disjoint with P1∪P2∪P3∪P∪e, then the color of e′ is from {β1,β2,…,βk−1}. Moreover, e∪P3∪e′ is a rainbow Pk−1. Now consider another edge e′′, which consists of a free(P2) vertex z=x,y in g1∪gk−1, a vertex in e′∖(P1∪P3) and s−2 vertices disjoint with P1∪P2∪P3∪P∪e∪e′, then e′′ must have a color appeared in the rainbow loose path e∪P3∪e′. However, to avoid extending P2 to a rainbow Pk, the color of e′′ should be one of {γ1,γ2,…,γk−3}, a contradiction. Therefore, F contains no edge-disjoint bad loose paths of length k−1.
□
So if F has a bad Pk−1, denote by F0 the subgraph obtained by deleting all the k−1 edges of that Pk−1 in F; if F dose not contain a bad Pk−1, then we delete any k−1 edges of it and denote the subgraph remained by F0. Then in either case,
[TABLE]
Therefore, F0 is Pk−1-free and ∣F0∣=(sn)−(sn−t+1)+3−2(k−1).
Note that ∣F0∣∼(t−1)(s−1n), by Theorem 2.5, we can find an (s−1)-graph G∗⊂∂F0 with ∣G∗∣∼(s−1n) and a set L of t−1 vertices of F0 such that L∩V(G∗)=∅ and e∪{v}∈F0 for any (s−1)-edge e∈G∗ and any v∈L. Moreover,
∣F0−L∣=o(ns−1). Select a G∗ with the maximum number of (s−1)-edges. Let the vertices in L be v1,v2,…,vt−1. Note that all the vi for i=1,2,…,t−1 are not free(P) vertices in e1∪ek−1. Otherwise, let W be the vertex set of P, by Lemma 2.7, we can find an (s−1)-edge disjoint with W in G∗, and then this together with vi will form an s-edge which extends P to a rainbow Pk.
We divide the edges of F0−L into two types. Let Q denote the set of free(P) vertices in e1∪ek−1. For an edge e∈F0−L, we call it of Type I if Q⊆e, and of Type II otherwise. Now we estimate the number of edges of each type.
Claim 5.2
There is no P2 in F0−L whose edges are all of Type II. Therefore, the number of edges of Type II is at most ⌊n/s⌋.
Proof.
Suppose that there is a P2 with two edges of Type II in F0−L, whose edges are denoted by h1 and h2 with colors γ1 and γ2, respectively.
We can take (s−1)-edges a0, b0 in G∗, such that a0 and b0 have exactly one common vertex u and are disjoint from P, h1 and h2. Let W be the vertex set of P∪h1∪h2∪a0∪b0. By Lemma 2.7, we can find (s−1)-edges {ai,bi} disjoint from W for i=1,…,t−1, such that for every i, ai and bi have exactly one common vertex, and for any j=i,
{ai,bi} and {aj,bj} are vertex disjoint. Then,
[TABLE]
is a P2t−2=Pk−3. Let the edges of P′ be f1,f2,…,fk−3, and the colors of edges be β1,β2,…,βk−3 respectively.
Consider an edge g, which consists of a vertex in h2∖h1, the common vertex u of a0 and b0, and s−2 vertices disjoint from P, P′, h1, h2, b0 and bt−1. So to prevent extending P′ to a rainbow Pk, the color of g is in {γ1,γ2,β1,β2,…,βk−3}.
If the color of g is γ1, consider the edge e, which consists of a vertex in Q∖h2, a vertex in at−1 and s−2 vertices disjoint from P, P′, h1, h2, g, b0 and bt−1, then the color of e is from {α1,α2,…,αk−1}, and hence h2∪g∪P′∪e is a rainbow Pk.
If the color of g is γ2, we pick a vertex w in h1, such that if ∣Q∣<s, let w∈h1∖Q, and if ∣Q∣≥s, let w be an arbitrary vertex in h1. Consider the edge e′ consisting of w, the common vertex of at−1 and bt−1, and s−2 vertices disjoint from P, P′, h1, h2, g, b0 and bt−1. Then the color of e′ is from {γ1,γ2,β1,β2,…,βk−3}, because otherwise, g∪P′∪e′∪h1 is a rainbow Pk. If the color of e′ is γ2, then h1∪e′∪P′ is a rainbow Pk−1. We obtain a rainbow Pk by adding an edge e′′, which consists of a vertex in Q∖h1, a free(P′) vertex in f1∖{u} and s−2 vertices disjoint from P, P′, h1, h2, g, e′, b0 and bt−1. If the color of e′ is γ1, consider the edge e′′′ consisting of a vertex in Q∖(g∪e′), a vertex in g∖(h2∪{u}) and s−2 vertices disjoint from P, P′, h1, h2, g, e′, b0 and bt−1. Then the color of e′′′ is from {α1,α2,…,αk−1}, and hence e′′′∪g∪P′∪e′ is a rainbow Pk. If the color of e′ is βj for some j. Let W be the vertex set of h1∪h2∪e′∪a0∪b0∪at−1∪bt−1∪g, by Lemma 2.7, we can find (s−1)-edges {ai′,bi′} in G∗, which are disjoint from W for i=1,…,t−1, and
[TABLE]
is a rainbow Pk.
So assume instead the color of g is one of {β1,β2,…,βk−3}. Let W be the vertex set of h1∪h2∪b0∪bt−1∪g. By Lemma 2.7, we can find (s−1)-edges {ai′′,bi′′} disjoint from W for i=1,…,t−1, and
[TABLE]
is a rainbow Pk. Hence, there is no P2 with two edges of Type II, and we have proved Claim 5.2.
□
Now, we move to Type I edges. Recall that Q denotes the set of free(P) vertices in e1∪ek−1. So if s≤∣Q∣≤2(s−1), the number of Type I edges is at most 1; if 2≤∣Q∣≤s−1, then a rough counting shows the number of Type I edges is at most (s−∣Q∣n−(t−1)−∣Q∣)≤(s−2n−t−1).
We further prove that there is no isolated vertex in F0. Indeed, if F0 has an isolated vertex, then combining with Claim 5.2,
[TABLE]
which is less than (sn)−(sn−t+1)+3−2(k−1) for sufficiently large n, a contradiction.
For the edges of Type I, we have the following claim.
Claim 5.3
The number of edges of Type I is at most 1.
Proof.
If s≤∣Q∣≤2(s−1), then Claim 5.3 follows, and so we may assume 2≤∣Q∣≤s−1. Suppose to the contrary there are at least two edges of Type I. Then we can find a P2 with two edges of Type I, denoted by h1 and h2. Pick vertices x∈h2∖h1, and y∈h1∖h2. The number of edges, which containing exactly one of {x,y}, one vertex in L, and disjoint with (h1∪h2)∖{x,y}, is at least
[TABLE]
which is at least (s−∣Q∣n−(t−1)−∣Q∣)+⌊sn⌋+2(k−1)−3, and so some of them must belong to F0. Suppose e∈F0 is such an edge and vj∈e∩L. Let W be the vertex set of h1∪h2∪e. By Lemma 2.7, we can find (s−1)-edges {ai,bi} disjoint from W for i=1,…,t−1, and
[TABLE]
is a bad Pk−1 in F0, a contradiction to (5).
□
By Claims 5.2 and 5.3, we get that, in F0, there are at most ⌊n/s⌋+1 edges disjoint with L. So there are at least ∣F0∣−(⌊n/s⌋+1)=(sn)−(sn−t+1)+3−2(k−1)−⌊n/s⌋−1=(sn)−(sn−t+1)−2(k−1)−⌊n/s⌋+2 edges in F0 containing vertices in L. Since there are (sn)−(sn−t+1) edges containing vertices of L in H, we have that there are at most
⌊n/s⌋+2(k−1)−2 edges meeting L but not in F0.
If there exists a vertex v∈V(F0)∖L but v is an isolated vertex in G∗, then we have at least (s−2n−(t−1)−1)>⌊n/s⌋+2(k−1)−2 edges meeting L but not belonging to F0, which is a contradiction. Hence, every vertex in V(F0)∖L is contained in some edges of G∗.
In fact, to find a rainbow Pk in H, we can make use of the suitable edges in F0−L and P−L to extend a Pk−3. We shall prove the following claim, which is analogous to Claim 4.7.
Claim 5.4
(i) For k≥7, if there are three edges f,g,h in H−L with distinct colors such that f,g,h form a P3, or f,g form a P2 and h is disjoint with f∪g, or f,g,h are disjoint with each other, then we can find a rainbow Pk in H.
(ii) For k=5, if there exists either a rainbow P3 or a P2 plus a disjoint edge with all three edges having distinct colors in H−L, then we can find a rainbow Pk in H.
Proof.
Similar to the proof of Claim 4.7, we denote f′,g′,h′ to be the edges in G with the same color of f,g,h, respectively. If the edge e is in G for some e∈{f,g,h}, we have e′=e.
(i) Let k≥7. Suppose that there are three edges f,g,h in H−L with distinct colors such that f,g,h form a P3. In G∗, there exists an (s−1)-edge a0 containing a vertex x in h∖g and disjoint with (f′∪g′∪h′∪f∪g∪h)∖{x}. Let W consist of the vertices in f′∪g′∪h′∪f∪g∪h∪a0. By Lemma 2.7, we can find (s−1)-edges {ai,bi} disjoint from W for i=1,…,t−1. Then,
[TABLE]
is a rainbow Pk in H.
If f,g form a P2, h is disjoint with f∪g.
In G∗, there exists an (s−1)-edge a0 containing a vertex x in g∖f and disjoint with (f′∪g′∪h′∪g∪h∪f)∖{x}. And there exists an (s−1)-edge b0 containing a vertex y in h and disjoint with (f′∪g′∪h′∪f∪g∪h∪a0)∖{y}. Let W consist of the vertices in f′∪g′∪h′∪f∪g∪h∪a0∪b0. By Lemma 2.7, we can find (s−1)-edges {ai,bi} disjoint from W for i=1,…,t−1. Then,
[TABLE]
is a rainbow Pk in H.
Assume that the three edges f,g,h are disjoint with each other.
In G∗, there exists an (s−1)-edge a0 containing a vertex x in f and disjoint with (f′∪g′∪h′∪f∪g∪h)∖{x}. And we can find (s−1)-edges a0′, b0′ in G∗, such that a0′ contains a vertex y1 in g and disjoint with (f′∪g′∪h′∪f∪g∪h∪a0)∖{y1}, b0′ contains a vertex y2=y1 in g and disjoint with (f′∪g′∪h′∪f∪g∪h∪a0∪a0′)∖{y2}. Moreover, there exists an (s−1)-edge b0 containing a vertex z in h and disjoint with (f′∪g′∪h′∪f∪g∪h∪a0∪a0′∪b0′)∖{z}. Let W consist of the vertices in f′∪g′∪h′∪f∪g∪h∪a0∪b0∪a0′∪b0′. By Lemma 2.7, we can find (s−1)-edges {ai,bi} disjoint from W for i=1,…,t−1. Then,
[TABLE]
is a rainbow Pk in H.
(ii) For k=5, the proof is identical to (i), and thus omitted.
□
Actually, in P, there are at most 2(t−1) edges containing vertices of L, so we can find at least 2 edges in P−L. However, there are not three edges of P satisfying the condition described in Claim 5.4, and so we derive that there are exactly two edges in P−L.
Case A. k≥7.
If ∣F0−L∣>0, then the two edges in P−L must be consecutive by Claim 5.4. Let ei and ei+1 be such two edges. We take an edge h∈F0−L, then select an edge g∈H−L, such that g is disjoint with P and h. If the color of g is αj for some j, then either the color of g is different with ei or different with ei+1. Suppose the colors of g and ei are different, then we have three edges ei,g,h satisfying the condition of Claim 5.4, and so we can find a rainbow Pk in H.
If the color of g is different with both ei and ei+1, then the three edges ei, ei+1 and g satisfy condition of Claim 5.4, which we can find a rainbow Pk in H similarly.
Assume instead that ∣F0−L∣=0. Then all the (sn)−(sn−t+1)+3−2(k−1) edges in F0 contain vertices in L, and P has k−3 edges containing vertices in L. Since the number of edges containing vertices of L is at most (sn)−(sn−t+1), in F∖F0 there are at most k−2 edges containing vertices in L. Since ∣F∖F0∣=k−1, there is an edge f in F∖F0 such that f∩L=∅. Furthermore, the color of f is different with any other edges in F.
Applying the same proof to the case that ∣F0−L∣>0, by replacing the edge h∈F0−L with f, we can find a rainbow Pk in H as well.
Case B. k=5.
As noticed above, there are exactly two edges, say ei and ej, in P−L.
If ∣F0−L∣=0, then all the (sn)−(sn−t+1)+3−2(k−1) edges in F0 contain vertices in L. Also P has k−3 edges containing vertices in L. Since the number of edges containing vertices of L is at most (sn)−(sn−t+1), in F∖F0 there are at most k−2 edges containing vertices in L. Since ∣F∖F0∣=k−1, there is an edge h in F∖F0 such that h∩L=∅. Then the color of h is different with any other edges in F, and different with the colors appeared in P. If ∣F0−L∣>0, then there exists an edge f1∈F0−L. We set f to be an edge such that f=h if ∣F0−L∣=0 and f=f1∈F0−L if ∣F0−L∣>0.
Then by Claim 5.4, f satisfies that
[TABLE]
For the former case of (6), pick an edge g∈H−L, such that g∩ei=∅, g∩f=∅ and g is disjoint with ej. Consider the color of g. If the color of g is αi, then the three edges ej, g, f are applied for Claim 5.4; if the color of g is αj, then the three edges ei, g, f are applied for Claim 5.4; if the color of g is different with both αi and αj, then the three edges ei, ej, g are applied for Claim 5.4. Therefore, we can always find a rainbow Pk in this case.
For the latter case of (6) that f∩ei=∅ and f∩ej=∅, we must have ei and ej are consecutive in P by Claim 5.4. Let g be an edge in H−L such that g is disjoint with ei, ej and f. If the color of g is αi or αj, then the three edges f, ej, g, or the three edges f, ei, g are applied for Claim 5.4. If the color of g is neither αi nor αj, then the three edges ei, ej, g are applied for Claim 5.4, and so we can still find a rainbow Pk in H.
This completes the proof of Theorem 1.3.
6 Linear Cycle–Proof of Theorem 1.4
Let H be a complete s-uniform hypergraph on n vertices. Denote by V the vertex set of H.
Let
[TABLE]
To prove Theorem 1.4, we show that g(n,s,k) is both the lower and upper bound for ar(n,s,Ck).
The lower bound follows from Proposition 4.1 by constructing a coloring of H using the extreme s-graphs without a Pk−1 in Theorem 2.2.
For the upper bound,
we argue by contradiction and suppose that there is a coloring of H using g(n,s,k) colors yielding no rainbow Ck. Since g(n,s,k)=ar(n,s,Pk) and by Theorem 1.2, there is a rainbow linear path P of length k in H. Let G be a spanning subgraph of H with P⊂G, such that ∣G∣=g(n,s,k) and each color appears on exactly one edge
of G. Denote by e1,e2,…,ek the edges of P, and let F=G−i=1⋃k−1ei.
Clearly, F is Ck-free. The following claim tells us more information about F when k=2t+1.
Claim 6.1
When k=2t+1, if there is a linear path P1 of length k−1 in F, then F−E(P1) is Pk−1-free.
Proof.
Assume that there is a linear path P1 of length k−1 in F. Suppose, by contradiction,
that there is a linear path P2 of length k−1 in F−E(P1). Denote the edges of P1 by f1,f2,…,fk−1, and the edges of P2 by g1,g2,…,gk−1, respectively.
We obtain an s-graph F′ by deleting edge set E(P1)∪E(P2) and all the edges containing at least two vertices of i=i⋃k−1ei∪E(P1)∪E(P2) from F. Let c denote the number of vertices of V(P)∪V(P1)∪V(P2). Then c≤ks−(k−1)+2[(k−1)s−(k−2)], and so we have
[TABLE]
for sufficiently large n. Thus, we have a linear path P3 of length k−3 in F′. Denote by h1,h2,…,hk−3 the edges of P3. Note that there are at most k−3 vertices in V(P3)∩(V(P)∪V(P1)∪V(P2)). Since s−1≥k−3+5 and every path has two disjoint end edges, we can always choose distinct vertices v1,v2,…,v6 such that the following holds: v1∈e1 and v2∈ek−1 are free(P) vertices, v3∈f1 and v4∈fk−1 are free(P1) vertices, v5∈g1 and v6∈gk−1 are free(P2) vertices, and vi∈/P3 for each 1≤i≤6.
Select u1∈h1∖(V(P)∪V(P1)∪V(P2)) and u2∈hk−3∖(V(P)∪V(P1)∪V(P2)). Consider the edge e′ consisting of v1,v2,u1 and s−3 vertices disjoint with P∪P1∪P2∪P3. Then e′ has a color appeared in ⋃i=1k−1ei; otherwise ⋃i=1k−1ei∪e′ is a rainbow Ck, a contradiction. Similarly, the edge e′′, which consists of v3,v4, a vertex x in e′∖{v1,v2,u1} and s−3 vertices disjoint with P∪P1∪P2∪P3∪e′, is colored with a color appeared on P1. In addition, consider the edge e′′′, which consists of v5,v6,u2, a vertex in e′′∖{v3,v4,x} and s−4 vertices disjoint with P∪P1∪P2∪P3∪e′∪e′′. We get that e′′′ is colored with a color appeared on P2. Now it follows that P3∪e′∪e′′∪e′′′ forms a rainbow Ck, a contradiction. This proves the claim.
□
So when k=2t+1, if F has a Pk−1, then we denote by F0 the subgraph obtained by deleting all the k−1 edges of that Pk−1 from F, and so F0 is Pk−1-free by Claim 6.1. If there is no Pk−1 in F, we delete any k−1 edges of F, and denote the subgraph remained by F0. When k=2t, we obtain F0 by deleting any k−1 edges from F. In any case, we obtain a subgraph F0 with ∣F0∣=∣F∣−[(k−1)s−(k−2)]∼(t−1)(s−1n). Moreover, F0 is Ck-free for k=2t, and F0 is Pk−1-free for k=2t+1. Thus we can apply Theorem 2.5 to F0 whenever k is even or odd. By Theorem 2.5, we can find an (s−1)-graph G∗⊂∂F0 with the maximum number of edges, such that ∣G∗∣∼(s−1n) and there is a set L of t−1 vertices of F0 such that L∩V(G∗)=∅ and e∪{v}∈F0 for any (s−1)-edge e∈G∗ and any v∈L. Moreover,
∣F0−L∣=o(ns−1). Denote L={v1,v2,…,vt−1}.
An s-edge e is called a missing-edge if e contains vertices of L and e∈/F0. Let M be the set of all the missing-edges, and let m=∣M∣ denote the number of missing-edges.
Since ∣F0∣−∣F0−L∣+m=(sn)−(sn−t+1), we have
[TABLE]
So it follows from ∣F0−L∣=o(ns−1) that
[TABLE]
We divide the remaining proof into two parts depending on the value of m. We will derive contradictions whenever m≤(s−2n−8s−t+1)−1 or
m>(s−2n−8s−t+1)−1.
6.1 The case when m is small: m≤(s−2n−8s−t+1)−1
In this subsection, we assume that m≤(s−2n−8s−t+1)−1. The proof applies similar ideas as the proof of Theorem 1.2, where we manage to find certain rainbow path of large length obtained from Lemma 2.7 and then extend to a rainbow Ck by selecting some specific edges in G−L. However, the differences with Theorem 1.2 is big enough in many details, leading us to rewrite a complete proof of this case.
We start to prove claims below similar to Claims 4.3, 4.6 and 4.7.
Claim 6.2
Every vertex v∈V∖L belongs to G∗. Moreover, for any vertex subset S of V with ∣S∣≤8s and v∈/S, there is an (s−1)-edge g∈G∗ such that v∈g and g is disjoint with S.
Proof.
If there is a vertex v such that v∈V∖L but v∈/G∗, then we have at least (s−2n−(t−1)−1) edges meeting L, but not belonging to F0. This implies the number of missing-edges is at least (s−2n−(t−1)−1)>(s−2n−8s−t+1)−1, a contradiction to our assumption that m≤(s−2n−8s−t+1)−1.
For the ‘moreover’ part, if in G∗ every (s−1)-set containing v meets S, then there are at least (s−2n−∣S∣−t+1)>m missing-edges, a contradiction. Therefore, there must exist an (s−1)-edge g in G∗ containing v and disjoint with S.
□
Claim 6.3
(a) If k=2t≥8, then there are no two edges e, f in G−L, such that ∣e∩f∣=1 or e∩f=∅.
*(b) If k=2t+1≥11, then there are no three edges e,f,h in G−L satisfying one of the following conditions:
(i) e,f,h form a P3;
(ii) e,f form a P2, and h is disjoint with e∪f;
(iii) e,f,h are pairwise disjoint.*
Proof.
(a) Let k=2t.
Suppose to the contrary that there exist two edges e, f in G−L such that ∣e∩f∣=1. Let u∈e∖f, v∈f∖e. By Claim 6.2, u,v∈V(G∗) and we can find an (s−1)-edge a0 in G∗ such that u∈a0 and a0 is disjoint with e∪f∖{u}. Applying Claim 6.2 again, there is an (s−1)-edge b0 in G∗ such that v∈b0 and b0 is disjoint with (e∪f∪a0)∖{v}. Let W be the vertex set of e∪f∪a0∪b0. By Lemma 2.7, we can find (s−1)-edge pairs {ai,bi} disjoint from W for i=1,…,t−1, such that for every i, ai and bi have exactly one common vertex, and for any j=i,
{ai,bi} and {aj,bj} are vertex disjoint. Then
[TABLE]
is a Pk−2 in F0.
Adding e,f to P′, we obtain a rainbow Ck, which is a contradiction.
Assume, by contradiction, that there are two edges e, f in G−L such that e∩f=∅. Select four distinct vertices x,y,z,w such that x,y∈e and z,w∈f. By Claim 6.2, we can find an (s−1)-edge a in G∗ such that x∈a and a is disjoint with (e∪f)∖{x}. Applying Claim 6.2 repeatedly, we can find (s−1)-edges b, a′, b′ one by one in G∗ such that y∈b and b is disjoint with (e∖{y})∪f∪a; z∈a′ and a′ is disjoint with e∪(f∖{z})∪a∪b; w∈b′ and b′ is disjoint with e∪(f∖{w})∪a∪b∪a′. Let W be the vertex set of e∪f∪a∪b∪a′∪b′. By Lemma 2.7, we can find (s−1)-edge pairs {ai,bi} disjoint from W for i=1,…,t−1. Then
[TABLE]
is a P4 in G, and
[TABLE]
is a Pk−4 in F0. Furthermore, P′∪P′′ forms a rainbow Ck, a contradiction. This proves (a).
(b) Let k=2t+1. We shall derive a contradiction by assuming one of conditions (i)(ii)(iii) holds.
(i) Assume that there is a linear path P1 with three consecutive edges e,f,h in G−L. Take two free(P1) vertices u,v such that u∈e and v∈h. By Claim 6.2, u,v∈V(G∗) and we can find an (s−1)-edge a0 in G∗ such that u∈a0 and a0 is disjoint with (e∖{u})∪f∪h. Also, there is an (s−1)-edge b0 in G∗ such that v∈b0 and b0 is disjoint with e∪f∪(h∖{v})∪a0. Let W be the vertex set of e∪f∪h∪a0∪b0. By Lemma 2.7, we can find (s−1)-edge pairs {ai,bi} disjoint from W for i=1,…,t−1. Then
[TABLE]
is a Pk−3 in F0.
Adding e,f,h to that Pk−3, it results a rainbow Ck, a contradiction.
(ii) Suppose there are three edges e,f,h in G−L, satisfying that e,f form a P2 and h is disjoint with e∪f. Take four distinct vertices x,y,z,w such that x∈e∖f, y∈f∖e, and z,w∈h. By Claim 6.2, we can find an (s−1)-edge a in G∗ such that x∈a and a is disjoint with (e∖{x})∪f∪h. Applying Claim 6.2 repeatedly, we can find (s−1)-edges b, a′, b′ in G∗ such that y∈b and b is disjoint with e∪(f∖{y})∪a∪h; z∈a′ and a′ is disjoint with e∪f∪(h∖{z})∪a∪b; w∈b′ and b′ is disjoint with e∪f∪(h∖{w})∪a∪b∪a′. Let W be the vertex set of e∪f∪h∪a∪b∪a′∪b′. By Lemma 2.7, we can find (s−1)-edge pairs {ai,bi} disjoint from W for i=1,…,t−1. Then
[TABLE]
is a P5 in G, and
[TABLE]
is a Pk−5 in F0. Thus P′∪P′′ is a rainbow Ck, a contradiction.
(iii) Suppose that there are three pairwise disjoint edges e,f,h in G−L. Take distinct vertices x,y,z,w,u,v such that x,y∈e, z,w∈f, and u,v∈h . By applying Claim 6.2 repeatedly, we can find (s−1)-edges a, b, a′, b′, a′′, b′′ in G∗ such that x∈a and a is disjoint with (e∖{x})∪f∪h; y∈b and b is disjoint with (e∖{y})∪f∪h∪a; z∈a′ and a′ is disjoint with e∪(f∖{z})∪h∪a∪b; w∈b′ and b′ is disjoint with e∪(f∖{w})∪h∪a∪b∪a′; u∈a′′ and a′′ is disjoint with e∪f∪(h∖{u})∪a∪b∪a′∪b′; v∈b′′ and b′′ is disjoint with e∪f∪(h∖{v})∪a∪b∪a′∪b′∪a′′. Note that the size of vertex set S in applying Claim 6.2 is at most 8s as required. Let W be the vertex set of e∪f∪h∪a∪b∪a′∪b′∪a′′∪b′′. By Lemma 2.7, we can find (s−1)-edges {ai,bi} disjoint from W for i=1,…,t−1. Then
[TABLE]
is a P7 in G, and
[TABLE]
is a Pk−7 in F0. Therefore, P′∪P′′ is a rainbow Ck, a contradiction. This completes the proof of Claim 6.3.
□
With the aid of Theorem 1.2, it is ready to finish the proof of this case now.
Recall that P is a linear path of length k in H. Since there are at most 2(t−1) edges meeting L in P, we have that there are at least 2 edges in P−L⊂G−L when k=2t, and at least 3 edges in P−L when k=2t+1. Hence we obtain edges satisfying the conditions of Claim 6.3, which is a contradiction.
6.2 The case when m is relatively large: m>(s−2n−8s−t+1)−1
In this subsection, we assume that m>(s−2n−8s−t+1)−1 to complete the proof.
In this case, the edges meeting L in F0 may not be enough to make similar arguments as Claim 6.2, and it seems that we can not use Lemma 2.7 directly to find a rainbow Ck as before. Our new strategy is to search a dense structure playing similar role as Lemma 2.7, which is motivated by some ideas in [33]. The key ingredient is to find some (s−2)-sets of V such that each of them can form rainbow edges with every vertex in L and a large number of other vertices. We shall eventually use these substructures to establish certain desired paths or cycles.
For any vertex set Z of a hypergraph H, the degree of Z in H, denoted by dH(Z), is the number of edges containing the entire set Z in H.
In the following, we denote K=2s+k−3 for convenience.
Claim 6.4
There are K pairwise disjoint (s−2)-sets Ti (i=1,…,K) in V∖L, such that for every i∈[K] and j∈[t−1] we have
[TABLE]
Proof.
For any i∈[K] and j∈[t−1], let dF0(Ti∪{vj}) denote the number of s-sets e such that Ti∪{vj}⊂e, but e∈/F0. That is the number of missing-edges containing Ti∪{vj}. Thus we have dF0(Ti∪{vj})+dF0(Ti∪{vj})=n−s+1.
Consider an (s−2)-set R of V∖L selected uniformly randomly from all (s−2)-sets of V∖L. Let Xi=dF0(R∪{vi}). For every s-set e∈/F0, let
[TABLE]
Then the expectation of Xi is
[TABLE]
By Markov’s inequality, we have
[TABLE]
Hence
[TABLE]
which implies that there are at least 21(s−2n−t+1) such (s−2)-sets R’s that satisfying dF0(R∪{vi})=n−s+1−dF0(R∪{vi})≥n−s+1−k(s−2n−t+1)(s−1)m for all i∈[t−1].
Among those 21(s−2n−t+1) R’s, we pick pairwise disjoint (s−2)-sets greedily as many as possible. Let ℓ be the largest number that we can pick pairwise disjoint R1,R2,…,Rℓ. We show that ℓ≥K. In fact, if ℓ<K, then the number of (s−2)-sets meeting
j=1⋃ℓRj is at most
[TABLE]
So we can select Rℓ+1 from the remained R’s such that Rℓ+1 is disjoint with j=1⋃ℓRj, a contradiction. Hence ℓ≥K and we can find K (s−2)-sets described in Claim 6.4.
□
Let T=j=1⋃KTj, and let U denote the vertex subset of V∖(L∪T) such that for every u∈U,
[TABLE]
for all i∈[K], j∈[t−1].
Claim 6.5
We have
∣U∣≥n−K(s−2)−(t−1)−(s−2n−t+1)K(t−1)(s−1)km.
Proof.
By Claim 6.4, for every vj and Ti, the number of vertex x such that the edge {{x}∪{vj}∪Ti}∈/F0 is bounded by n−s+1−dF0(Ti∪{vj})≤k(s−2n−t+1)(s−1)m. So we have ∣U∣≥n−K(s−2)−(t−1)−K(t−1)(s−2n−t+1)(s−1)km as required.
□
The following Claim 6.6 plays the role as Claim 6.3 in the previous subsection. The difference is that, in Claim 6.3 we assemble s-edges with (s−1)-sets in G∗ and vertices in L, but now we use Ti (i∈[K]) and some vertices in L and U to form desired rainbow s-edges. Since F0⊆G−(⋃i=1k−1ei), the colors appeared in F0 are distinct with colors appeared in ⋃i=1k−1ei. Thus, we can also use edges which have colors appeared in ⋃i=1k−1ei, along with edges in F0, to build rainbow paths and cycles.
Let J denote the set of edges in E(H−L)∖E(P) which are received colors appeared in ⋃i=1k−1ei.
Claim 6.6
*(1) If k=2t, then there are no two edges e,f∈(F0−L)∪J such that e,f form a rainbow P2 with ∣(f∖e)∩U∣≥1 and ∣(e∖f)∩U∣≥1.
(2) If k=2t+1, then
there are no two edges e,f∈F0−L such that e,f form a P2, with ∣(f∖e)∩U∣≥1 or ∣(e∖f)∩U∣≥1.*
Proof.
(1) For k=2t, if there are two edges e,f∈(F0−L)∪J such that e,f form a rainbow P2 with ∣(f∖e)∩U∣≥1 and ∣(e∖f)∩U∣≥1. Let x∈(f∖e)∩U and y∈(e∖f)∩U. Since T is disjoint with U, there are at most (2s−3) such Ti’s that contain vertices of e or f. Hence, there are at least K−(2s−3)=k>k−2 such Ti’s that are disjoint with e and f. Suppose, without loss of generality, that Ti is disjoint with e and f for each i∈[k−2]. Then by the definitions of T and U, there is a rainbow Pk−2 in F0 with edges h1,h2,…,hk−2, such that
[TABLE]
where u1,…,ut−2 are distinct vertices selected in U∖(e∪f).
Adding edges e,f to that Pk−2, we obtain a rainbow Ck, a contradiction.
(2) For k=2t+1, suppose to the contrary that there are edges e,f∈F0−L such that ∣e∩f∣=1 and x∈(f∖e)∩U. As there are at most (2s−2) Ti’s containing vertices of e or f, we obtain that there are at least K−(2s−2)=k−1>k−3 such Ti’s that are disjoint with e and f. Without loss of generality, assume that for i∈[k−3], Ti is disjoint with e and f. Then there is a Pk−3 with the first edge containing x, and all the k−3 edges are of the form Ti∪{vj,uℓ} similar as above, where i∈[k−3],j∈[t−1] and uℓ∈U.
Adding edges e,f to that Pk−3, we obtain a Pk−1 in F0, a contradiction to Claim 6.1 that F0 is Pk−1-free for k=2t+1.
□
Now we treat the edges of F0−L in details. For 0≤i≤s, let
[TABLE]
and let B(2+)=B(2)∪B(3)∪…∪B(s).
By Eq.(8), we have m<ϵ(n−t−s)s−1<ϵns−1 for every fixed positive constant ϵ when n is sufficiently large. Let δ=8(t+1)[2Kk(t−1)⋅(s−1)!]sϵs−2. Here we set the constant δ to satisfy 0.5<δ<1 by selecting an appropriate small constant ϵ>0.
As a consequence of Claim 6.5, we show the following inequality holds:
[TABLE]
In fact, it follows from Claim 6.5 that
[TABLE]
where the third line of the inequality holds since the constant
[TABLE]
for sufficiently large n.
Therefore, Eq.(9) holds, and we will use it to bound the size of B(0),B(1),B(2+) below.
Claim 6.7
*(a) ∣B(0)∣<4δm<4m. In particular, if m=O(ns−2), then ∣B(0)∣≤O(1).
(b) ∣B(1)∣<2δm<2m. In particular, if m=O(ns−2), then ∣B(1)∣≤O(n).*
Proof.
(a) Since the edges in B(0) can not form a Ck, by Eq.(9), we have
[TABLE]
If m=O(ns−2), then by Claim 6.5, there exists a positive real number C such that
∣U∣≥n−C. Thus we have
[TABLE]
(b) Let Q be the collection of (s−1)-sets h∈V∖(U∪L) such that there exists e∈B(1) and h⊂e.
We divide Q into two sets Q1 and Q2, such that for every h∈Q1 there is only one vertex u∈U satisfying h∪{u}∈B(1), and Q2=Q∖Q1. Hence we have
[TABLE]
Clearly, ∣Q1∣≤(s−1n−∣U∣)<4δm<4m by Eq.(9).
To bound ∣Q2∣, notice that there are no h1,h2∈Q2 such that ∣h1∩h2∣=1. Otherwise, we obtain two s-edges e,f∈F0−L containing h1,h2, respectively, where e,f form a rainbow P2 with ∣(f∖e)∩U∣=1 and ∣(e∖f)∩U∣=1. This contradicts to Claim 6.6.
Thus we have
[TABLE]
Therefore, by Eq.(9),
[TABLE]
In particularly, if m=O(ns−2), then by Claim 6.5, we have n≥∣U∣≥n−C for a positive constant number C, and so
[TABLE]
□
Note that
[TABLE]
Next, we explore properties of ∣B(2+)∣ and m.
Claim 6.8
*(a) We have ∣B(2+)∣≤ex(n−t+1,s,P2)=(s−2n−t−1).
(b) We must have k=2t and m=O(ns−2). In fact, we have m<(s−2)!4ns−2.*
Proof.
(a) If ∣B(2+)∣>ex(n−t+1,s,P2), then there is a P2 with two edges e,f∈B(2+). Moreover, by definition of B(2+), we have ∣(f∖e)∩U∣≥1 and ∣(e∖f)∩U∣≥1, a contradiction to Claim 6.6.
(b) If k=2t+1, then it follows from Eq.(7) (10) and Claim 6.7 that
[TABLE]
a contradiction to Claim 6.8 (a).
So we must have k=2t. Similar as inequality above, by Eq.(7) (10) and Claims 6.7 and 6.8 (a) for k=2t, we have
[TABLE]
which shows that m<(s−2)!4ns−2 as required.
□
Note that, applying Eq.(7) (10) again, Claims 6.7 and 6.8 provide a further estimation of ∣B(2+)∣ as follows:
[TABLE]
Now we are preparing to find certain edges aiming to lead a contradiction to Claim 6.6.
Claim 6.9
*(i) For any j∈{1,k−1}, there are at least two free(P) vertices in ej, which are not belonging to L.
(ii) There exist a free(P) vertex x∈e1 and a free(P) vertex y∈ek−1, such that x,y∈/L and not all edges in B(2+) containing both x and y.*
Proof.
(i) In fact, for any free(P) vertex v∈ej∩L, and any free(P) vertex u∈ej′, where j,j′∈{1,k−1} and j=j′, the edge g consisting of u,v and s−2 vertices in V∖V(P), must be colored with a color appeared in ⋃i=1k−1ei; otherwise, we have a rainbow Ck. This indicates that g is a missing-edge. Hence, if Claim 6.9 (i) dose not hold, then we count the number of missing-edges as
[TABLE]
violating Eq.(11). Hence Claim 6.9 (i) holds.
(ii) By Claim 6.9 (i), assume that x,z are free(P) vertices in e1∖L, and y is a free(P) vertex in ek−1∖L. By contradiction, suppose that all the edges of B(2+) contain the vertex pair {x,y}, and all the edges of B(2+) contain the vertex pair {z,y} as well. Then all the edges in B(2+) contain {x,y,z}, and so ∣B(2+)∣≤(s−3n−t+1−3), a contradiction to Eq.(11). This proves Claim 6.9 (ii).
□
Finally, we are ready to complete the proof.
By Claim 6.9 (ii), there exist a free(P) vertex x∈e1 and a free(P) vertex y∈ek−1, such that x,y∈/L and not all edges in B(2+) containing both x and y. Let e∗∈B(2+) be such an edge that not containing both x and y.
Assume that x,y∈/e∗. Since e∗∈B(2+), we select a vertex u∈e∗∩U, and so u∈/{x,y}. Consider an s-edge f consisting of x,y,u and s−3 vertices disjoint with P, L and e∗, such that ∣f∩U∣≥2. Then we have f∈J, otherwise f∪P forms a rainbow Ck. Thus f and e∗ form a rainbow P2 with ∣(f∖e∗)∩U∣≥1 and ∣(e∗∖f)∩U∣≥1, which contradicts to Claim 6.6 (1).
Assume instead that one of x,y belongs to e∗. Without loss of generality, suppose that {x}=e∗∩{x,y}. Consider the edge g consisting of x,y and s−2 vertices disjoint with P, L and e∗, such that ∣(g∖{x})∩U∣≥1. Then we have g∈J with the same reason as above. Moreover, g and e∗ form a rainbow P2 with ∣(e∗∖f)∩U∣≥1, again contradicting to Claim 6.6 (1).
Therefore, we establish the upper bound and complete the proof of Theorem 1.4.
7 Loose cycle–Proof of Theorem 1.5
Since the proof of Theorem 1.5 is similar to Theorem 1.4, in this section, we omit some details and pay more attention to the difference between the proofs of Theorem 1.4 and Theorem 1.5.
Let H be a complete s-uniform hypergraph on n vertices. Denote by V the vertex set of H.
The lower bound in Theorem 1.5 follows from a similar construction as Theorem 1.4 by applying the extreme s-graphs without Pk−1 obtained from Theorem 2.1.
For the upper bound, when k=2t, since a loose cycle is also a linear cycle, we have ar(n,s,Ck)≤ar(n,s,Ck)=(sn)−(sn−t+1)+2, and we are done.
For k=2t+1, we shall show below that how to modify the proof of the upper bound for anti-Ramsey number of linear cycles to obtain the upper bound for anti-Ramsey number of loose cycles.
For loose cycles, we again consider, by contradiction, a coloring of H using (sn)−(sn−t+1)+3 colors yielding no rainbow Ck.
Since ar(n,s,Pk)=(sn)−(sn−t+1)+3 by Theorem 1.3, there is a rainbow loose path P of length k in H. As before, let G be a subgraph of H with ∣G∣=(sn)−(sn−t+1)+3, such that P⊂G and each color appears on exactly one edge
of G. Denote by e1,e2,…,ek the edges of P, and let F=G−i=i⋃k−1ei.
With similar argument as the proof of Claim 6.1, we have the following claim.
Claim 7.1
If F contains a linear path P1 of length k−1, then F−E(P1) contains no Pk−1.
If there is a linear path P1 of length k−1 in F, then we let F0=F−E(P1); if there is no linear path of length k−1 in F, we delete any k−1 edges of F, and denote the subgraph remained by F0. So we have
[TABLE]
and F0 is Pk−1-free.
Note that ∣F0∣∼(t−1)(s−1n). By Theorem 2.5, we can find an (s−1)-graph G∗⊂∂F0 with ∣G∗∣∼(s−1n) and a set L of t−1 vertices of F0 such that L∩V(G∗)=∅ and e∪{v}∈F0 for any (s−1)-edge e∈G∗ and any v∈L. Moreover,
∣F0−L∣=o(ns−1). Select a G∗ with the maximum number of (s−1)-edges. Denote L={v1,v2,…,vt−1} as before.
We still call an s-edge e a missing-edge if e contains vertices of L and e∈/F0. Let M be the set of all the missing-edges, and let m=∣M∣. We have ∣F0∣−∣F0−L∣+m=(sn)−(sn−t+1), and so
[TABLE]
If m≤(s−2n−8s−t+1)−1, then Claim 6.2 still holds. Instead of Claim 6.3, we have the following similar claim.
Claim 7.2
*When k=2t+1≥11, there are no three edges e,f,h in G−L satisfying that one of the following conditions:
(i) e,f,h form a P3;
(ii) e,f form a P2, and h is disjoint with e∪f;
(iii) e,f,h are pairwise disjoint.*
The only difference between Claim 6.3 and Claim 7.2 is to construct a Ck rather than Ck to obtain a contradiction, with essentially the same argument (see also Claim 5.4 for details).
Then, as P has length k, we can derive that P−L must contain edges satisfying one of the conditions in Claim 7.2, giving the finial contradiction in the case m≤(s−2n−8s−t+1)−1.
If
m>(s−2n−8s−t+1)−1,
with the arguments that are identical to the linear cycles, Claim 6.4 and Claim 6.5 hold. By replacing P2 with P2, we obtain the following result similar to Claim 6.6 (2).
Claim 7.3
For k=2t+1,
there are no two edges e,f∈F0−L such that e,f form a P2, ∣(f∖e)∩U∣≥1 or ∣(e∖f)∩U∣≥1.
We still let B(i)={e∈F0−L: ∣e∩U∣=i} and B(2+)=B(2)∪B(3)∪…∪B(s). Then the counting arguments in Claim 6.7 still holds that ∣B(0)∣<4m and ∣B(1)∣<2m.
Hence we have
[TABLE]
By Claim 7.3, there are no two edges f1,f2∈B(2+) such that ∣f1∩U∣=i and ∣f2∩U∣=j with i=j. Moreover, for fixed r, if ∣f1∩U∣=∣f2∩U∣=r for two edges f1,f2∈B(2+), then f1∩U=f2∩U, i.e., all the edges in B(2+) contain exactly the same r vertices of U. Then it follows that
[TABLE]
By Claim 6.5 and a similar inequality as (9), we have
[TABLE]
which contradicts to Eq.(7).
Hence, we obtain the final contradiction, which proves
Theorem 1.5.
8 Berge Path and Berge Cycle
We shall present the proofs of Theorem 1.6 and Proposition 1.7 on Berge paths and Berge cycles in this section.
8.1 Berge Path–Proof of Theorem 1.6.
For the lower bounds, we will prove that
ar(n,s,Bk)≥k2n(s⌊k/2⌋) if k>2s+1, and
ar(n,s,Bk)≥s+1n⌊2k−2⌋ if 3<k≤2s+1.
For k>2s+1, we partition the n vertices into sets of size ⌊k/2⌋ (possibly one of those sets has size smaller than ⌊k/2⌋). Denote by S1,S2,…,Sℓ those obtained sets of size ⌊k/2⌋. Then for each k-set Si, color each edge contained in Si with a distinct color. The rest edges are colored with one additional color. It is routine to check that there is no rainbow Bk in the above coloring. So we have ar(n,s,Bk)≥k2n(s⌊k/2⌋).
For 3<k≤2s+1, we partition the n vertices into sets of size s+1. Then we select ⌊k/2⌋−1 edges in each (s+1)-set and color each of those edges with a different color. The rest edges are colored with one additional color. Similarly, this provides a s+1n⌊2k−2⌋-coloring without a rainbow Bk. Hence ar(n,s,Bk)≥s+1n⌊2k−2⌋.
For the upper bounds, we will show that
if k≥s+2, then for sufficiently large n, ar(n,s,Bk)≤k−1n(sk−1)+1; and if k≤s+1, then
ar(n,s,Bk)≤s+1(k−2)n for sufficiently large n.
(I) For k≥s+2, let H be a complete s-uniform hypergraph on n vertices. Consider a coloring of H using k−1n(sk−1)+1 colors and yielding no rainbow Bk. Let G be a subgraph of H with k−1n(sk−1)+1 edges such that each color appears on exactly one edge
of G. So the number of edges of G is ∣G∣=k−1n(sk−1)+1>ex(n,s,Bk−1). Hence there is a rainbow Berge path P of length k−1 in G. Denote by e1, e2, …, ek−1 the edges of P with colors α1, α2, …, αk−1, respectively. And there are k vertices w1, w2, …, wk in P such that wi,wi+1∈ei for i=1,…,k−1. Let F be the hypergraph obtained by removing all the edges of P from G. We have that ∣F∣=k−1n(sk−1)+1−(k−1)=k−1n(sk−1)−k+2.
If there is a Berge path P∗ of length k−1 in F. Denote by g1, g2, …, gk−1 the edges of P∗. And there are k vertices z1, z2, …, zk in P∗ such that zi,zi+1∈gi for i=1,…,k−1. Then either w1=z1 or w1=zk. Without loss of generality, suppose that w1=z1. Consider the edge e consisting of w1,z1 and s−2 vertices in V(F)∖(V(P)∪V(P∗)). If e is colored with a color not in {α1,α2,…,αk−1}, then e∪P is a rainbow Bk. So e is colored with a color belonging to {α1,α2,…,αk−1}, then e∪P∗ is a rainbow Bk. Therefore, we have showed that
[TABLE]
We further claim that the minimum degree δ(F) of F satisfying
[TABLE]
Indeed, if there is a vertex v having degree dF(v)<k−11(sk−1)−k+1 in F, then the number of edges in F−v is more than ∣F∣−(k−11(sk−1)−k+1)=k−1n−1(sk−1)+1≥ex(n−1,s,Bk−1)+1 for sufficiently large n. So there is a Bk−1 in F−v, which contradicts (13). This proves (14).
Since ∣F∣>ex(n,s,Bk−2) for sufficiently large n, there is a Berge path P′ of length k−2 in F. Denote by f1, f2, …, fk−2 the edges of P′ with colors β1, β2, …, βk−2, respectively.
And there are k−1 vertices u1, u2, …, uk−1 in P′ such that ui,ui+1∈fi for i=1,…,k−2. Since F contains no Bk−1 by (13), the neighbors of u1 and uk−1 must belong to {u1,u2,…,uk−1}. In fact, we shall further show in the following claim that
the neighbors of each vertex in {u2,…,uk−2} also belong to {u1,u2,…,uk−1}.
Before that, we need the definition of Berge cycles to state the following claim. An
s-uniform Berge cycle of length ℓ is a cyclic list of
distinct s-sets a1,…,aℓ and ℓ distinct vertices
v1,…,vℓ such that for each i=1,2,…,ℓ, ai
contains vi and vi+1 (where vℓ+1=v1).
[TABLE]
Suppose that there is a Berge cycle C containing the vertices u1,u2,…,uk−1. If an edge f in the C contains some vertex x other than u1,u2,…,uk−1, then deleting f from C, we have a Bk−2, which can be extended to a Bk−1 with edge f, contradicting to (13). Thus every edge in the cycle must be contained within
the vertices u1,u2,…,uk−1. Moreover, for each
vertex ui in C, the neighbors of ui must belong to {u1,u2,…,uk−1}. Suppose to the contrary that ui has a neighbor y
other than u1,u2,…,uk−1. Then the edge containing both ui and y is not an
edge of C, as shown in the argument above. Thus, removing an appropriate edge of C
so that we get a path of length k−2 with ui as an endpoint, and hence we can extend this
to a Bk−1 with y as an endpoint, a contradiction to (13). This proves (15).
Now we show that one can always find a Berge cycle of length k−1 containing the vertices u1,u2,…,uk−1. If there is an edge in F containing both
u1 and uk−1, then we can obtain a Berge cycle of length k−1. If not, recall that by (14) we have δ(F)≥k−11(sk−1)−k+1>(s−12k−1−2). That implies there exist edges f′ and f′′ in F, such that for some i, u1,ui+1∈f′ and ui,uk−1∈f′′. Thus, we have a Berge cycle of length k−1 on the vertices
[TABLE]
Hence, we can find a Berge cycle of length k−1 containing the vertices u1,u2,…,uk−1 in F. By (15), u1,u2,…,uk−1 constitute a component of F.
Let R denote the hypergraph obtained by deleting vertices u1,u2,…,uk−1 from F. Then ∣R∣≥k−1n(sk−1)−k+2−(sk−1)>ex(n−(k−1),s,Bk−2) for sufficiently large n. Hence there is a Berge path P′′ of length k−2 in R. Denote by h1, h2, …, hk−2 the edges of P′′ with colors γ1, γ2, …, γk−2, respectively.
And there are k−1 vertices v1, v2, …, vk−1 in P′′ such that vi,vi+1∈hi for i=1,…,k−2. Note that {u1,u2,…,uk−1}∩{v1,v2,…,vk−1}=∅. Since we have either w1∈/{u1,v1} or w1∈/{uk−1,vk−1}, suppose, without loss of generality, that w1∈/{u1,v1} holds. Consider the edge e′ with w1,u1,v1 and s−3 vertices in V(H)∖(V(P)∪V(P′)∪V(P′′)).
If s>3, e′ can only be colored with a color in {α1,α2,…,αk−1}, then h1∪e′∪P′ is a rainbow Bk. If s=3, then e′={w1,u1,v1}. If the color of e′ is not belonging to {β1,β2,…,βk−2}∪{γ1,γ2,…,γk−2}, then h1∪e′∪P′ is a rainbow Bk. If the color of e′ is in {β1,β2,…,βk−2}, let P~=e′∪P′′, then P~ is a rainbow Bk−1 in H. Consider an edge e′′={w1,u1,x}, where x∈/V(P)∪V(P′)∪V(P~). To prevent extending P, the color of e′′ must be in {α1,α2,…,αk−1}. However, to prevent extending P~, e′′ must be colored with a color from {β1,β2,…,βk−2}∪{γ1,γ2,…,γk−2}, a contradiction. By symmetry, if the color of e′ is from {γ1,γ2,…,γk−2}, we can deduce a similar contradiction as well. In conclusion, any coloring of H using k−1n(sk−1)+1 colors yields a rainbow Bk.
(II) For k≤s+1, let H be a complete s-uniform hypergraph on n vertices. Consider a coloring of H using s+1n(k−2)+1 colors and yielding no rainbow Bk. Let G be a subgraph of H with s+1n(k−2)+1 edges such that each color appears on exactly one edge
of G. So the number of edges of G is ∣G∣=s+1n(k−2)+1. Denote by C1, C2, …, Ct the components of G, and n1, n2, …, nt the number of vertices of each components, respectively. Then there is a component Ci, such that ∣Ci∣>s+1ni(k−2)≥ex(ni,s,Bk−1). Hence there is a rainbow Berge path P of length k−1 in Ci. Denote by e1, e2, …, ek−1 the edges of P with colors α1, α2, …, αk−1, respectively. And there are k vertices w1, w2, …, wk in P such that wi,wi+1∈ei for i=1,…,k−1. Let F be the hypergraph obtained by removing all the edges of P from G. We have that ∣F∣=s+1(k−2)n+1−(k−1)=s+1n(k−2)−k+2>s+1(k−3)n.
We will make use of the following result given in [25].
Proposition 8.1
[25]**
Fix ℓ and s such that s≥ℓ>2. Let H be a connected s-uniform hypergraph with
[TABLE]
edges, where n is the number of vertices in H. Then for each edge e∈H, there is a Berge path of length ℓ in H
starting with e.
Let the components of F be C1∗, C2∗, …, Cμ∗, and n1∗, n2∗, …, nμ∗ the number of vertices of each components, respectively. Then there is a component Cj∗ satisfying that ∣Cj∗∣>s+1k−3nj∗≥ex(nj∗,s,Bk−2). Now we focus on finding a Berge path of length k−2 containing some new vertices in Cj∗.
If there exists such a Cj∗ satisfying that ∣Cj∗∣>s+1k−3nj∗ and Cj∗∩Ci=∅, then we can find a Berge path of length k−2 in Cj∗, and its vertices are disjoint with P.
If every such Cj∗ with ∣Cj∗∣>s+1k−3nj∗ satisfying that Cj∗⊆Ci,
then we have that nj∗≥s+1≥k since the number of vertices of a Bk−2 is at least s−1+2=s+1. Furthermore, we claim that
[TABLE]
In fact, if nj∗<2k, then ∣Cj∗∣≤(snj∗)<(s2k). Delete the component Cj∗ from F, we have n−nj∗ vertices and more than s+1k−2n−k+2−(s2k)>s+1k−3(n−k)≥s+1k−3(n−nj∗) edges. So there is a component Ct∗, such that ∣Ct∗∣>s+1k−3nt∗≥ex(nt∗,s,Bk−2) edges and Ct∗∩Ci=∅, a contradiction. So we have nj∗≥2k, which proves (16). Hence, there is a vertex u1∈/{w1,w2,…,wk} in Cj∗. We take an edge e in Cj∗ containing u1, by Proposition 8.1, there is a Bk−2 starting with e.
In both cases above, we denote by P′ the Bk−2 we obtained, and denote by f1, f2, …, fk−2 the edges of P′ with colors β1, β2, …, βk−2, respectively. There are k−1 vertices u1, u2, …, uk−1 in P′ such that ui,ui+1∈fi for i=1,…,k−2. Note that u1∈/{w1,w2,…,wk}.
Let R denote the hypergraph obtained by deleting f1,f2,…,fk−2 from F. Then ∣R∣=s+1k−2n−k+2−(k−2)>s+1k−3n. Let the components of R be C1∗∗, C2∗∗, …, Cτ∗∗, and n1∗∗, n2∗∗, …, nτ∗∗ the number of vertices of each component, respectively. Then there is a component Cℓ∗∗, such that ∣Cℓ∗∗∣>s+1k−3nℓ∗∗≥ex(nℓ∗∗,s,Bk−2).
If there exists such a Cℓ∗∗ satisfying that Cℓ∗∗∩Cj∗∩Ci=∅, then we can find an edge e′ containing a vertex v1∈/{u1,u2,…,uk−1,w1,w2,…,uk}, and the color of e′ is different with α1,α2,…,αk−1,β1,β2,…,βk−2. Otherwise, Cℓ∗∗⊆Cj∗ or Cℓ∗∗⊆Ci. We have nℓ∗∗≥s+1≥k. With the argument similar to the proof of (16), we can obtain that
[TABLE]
Thus, there is a vertex not belonging to {u1,u2,…,uk−1,w1,w2,…,uk}. We still denote it by v1. Take an edge e′ in Cℓ∗∗ containing v1. Denote this edge by e′. So the color of e′ is different with α1,α2,…,αk−1,β1,β2,…,βk−2. Consider an edge e′′ containing w1,u1,v1, it must be colored with a color from {α1,α2,…,αk−1} since otherwise P can be extended. But then e′∪e′′∪P′ is a rainbow Bk, a contradiction. Therefore, we have proved the upper bound.
8.2 Berge Cycle–Proof of Proposition 1.7
Note that the lower bound in Proposition 1.7 is obvious, which follows from a similar observation as in Eq.(1) for hypergraphs. Now we prove the upper bound in Proposition 1.7.
Let H be a complete s-uniform hypergraph on n vertices. By contradiction, consider a coloring of H using ex(n,s,Bk−1)+k colors yielding no rainbow BCk. Let G be a subgraph of H with ex(n,s,Bk−1)+k edges such that each color appears on exactly one edge
of G. So the number of edges of G is ∣G∣=ex(n,s,Bk−1)+k>ex(n,s,Bk−1). Hence there is a rainbow Berge path P of length k−1 in G. Denote by e1, e2, …, ek−1 the edges of P with colors α1, α2, …, αk−1, respectively. And there are k vertices w1, w2, …, wk in P such that wi,wi+1∈ei for i=1,…,k−1.
Let F be the hypergraph obtained from G by removing all the edges of P. Then we have that ∣F∣=ex(n,s,Bk−1)+1.
Therefore, there is a Berge path P∗ of length k−1 in F. Denote by g1, g2, …, gk−1 the edges of P∗, where
there are k vertices z1, z2, …, zk in P∗ such that zi,zi+1∈gi for i=1,…,k−1. Consider an s-edge e containing w1,wk,z1,zk.
If e is colored with a color not in {α1,α2,…,αk−1}, then e∪P is a rainbow BCk. So e is colored with a color belonging to {α1,α2,…,αk−1}, but now e∪P∗ is a rainbow BCk, a contradiction. Hence, ar(n,s,BCk−1)≤ex(n,s,Bk−1)+k for any possible n. This proves Proposition 1.7.
Acknowledgement. Ran Gu was partially supported by
Natural Science Foundation of Jiangsu Province (No. BK20170860), National Natural Science Foundation of China (No. 11701143), and the Fundamental Research Funds for the Central Universities. Jiaao Li was partially supported by National Natural Science Foundation of China (No. 11901318), Natural Science Foundation of Tianjin (No. 19JCQNJC14100) and the Fundamental Research Funds for the Central Universities, Nankai University (No. 63191425). Yongtang Shi was partially supported by National Natural Science Foundation of China,
Natural Science Foundation of Tianjin (No. 17JCQNJC00300), the China-Slovenia bilateral project “Some topics in modern graph theory” (No. 12-6),
Open Project Foundation of Intelligent Information Processing Key Laboratory of Shanxi Province (No. CICIP2018005),
and the Fundamental Research Funds for the Central Universities.
