On the anti-Ramsey number of forests

TL;DR

This paper determines the exact and approximate values of the anti-Ramsey number for various forests in complete graphs, advancing understanding of rainbow subgraph avoidance in edge-colored graphs.

Contribution

It provides exact values for star forests and double stars, and approximate values for linear forests, expanding the knowledge of anti-Ramsey numbers for forests.

Findings

Exact anti-Ramsey number for star forests.

Approximate anti-Ramsey number for linear forests.

Exact value of ar(K_n, 2P_4) for n ≥ 8.

Abstract

We call a subgraph of an edge-colored graph rainbow subgraph, if all of its edges have different colors. The anti-Ramsey number of a graph $G$ in a complete graph $K_{n}$, denoted by $ar(K_{n}, G)$, is the maximum number of colors in an edge-coloring of $K_{n}$ with no rainbow subgraph copy of $G$. In this paper, we determine the exact value of the anti-Ramsey number for star forests and the approximate value of the anti-Ramsey number for linear forests. Furthermore, we compute the exact value of $ar(K_{n}, 2P_{4})$ for $n\ge 8$ and $ar(K_{n}, S_{p,q})$ for large $n$, where $S_{p,q}$ is the double star with $p+q$ leaves.

Equations

$$\lfloor \frac{(p-1)n}{2}\rfloor +\lfloor \frac{n}{n-p+1}\rfloor \le ar(K_n,K_{1,p+1})\le \lfloor \frac{(p-1)n}{2}+\frac{1}{2}\lfloor \frac{2n}{n-p+1}\rfloor \rfloor .$$

$$ar(K_n,P_{k+1})=\left(\lfloor \frac{k}{2}\rfloor -1\right)n-\left(\genfrac{}{}{0ex}{}{\lfloor \frac{k}{2}\rfloor }{2}\right)+1+r.$$

$$ar(K_n,t{K}_2)=ex(n,(t-1)K_2)+1=(t-2)n-\left(\genfrac{}{}{0ex}{}{t-1}{2}\right)+1.$$

$$ex(n,F)=\max_{1\leq i\leq t}\bigg{\{}(i-1)n-\binom{i}{2}+\bigg{\lfloor}\frac{p_{i}-1}{2}(n-i+1)\bigg{\rfloor}\bigg{\}}.$$

$$ex(n,F)=\left(\sum _{i=1}^k\lfloor \frac{p_i}{2}\rfloor -1\right)n-\left(\genfrac{}{}{0ex}{}{{\sum }_{i=1}^k\lfloor \frac{p_i}{2}\rfloor }{2}\right)+c,$$

$$ar(K_{n},F)=\max\bigg{\{}\max_{1\leq i\leq s}\bigg{\{}(i-1)n-\binom{i}{2}+\bigg{\lfloor}\frac{p_{i}-2}{2}(n-i+1)\bigg{\rfloor}+1\bigg{\}},(t-2)n-\binom{t-1}{2}+r\bigg{\}},$$

$$ar(K_n,F)=\left(\sum _{i=1}^k\lfloor \frac{p_i}{2}\rfloor -ϵ\right)n+O(1),$$

$$ar(K_n,tG)=(t-1)n-\left(\genfrac{}{}{0ex}{}{t}{2}\right)+ar(K_{n-t+1},G)$$

$$\bigg{\lfloor}\frac{k-2}{2}\bigg{\rfloor}\frac{n}{2}+O(1)\leq ar(K_{n},T_{k})\leq(k-1)n.$$

$$ex(n,T_k)\le \frac{k-1}{2}n.$$

$$ar(n,T_k)\le \frac{k-2}{2}n+O(1).$$

$$ar(K_n,S_{p,q})=\{\begin{array}{l}\lfloor \frac{(p-1)n}{2}\rfloor +1,1\le q\le p-1;\\ \lfloor \frac{p(n-1)}{2}\rfloor +1,q=p.\end{array}$$

$$ar(K_{n},F)=\max\bigg{\{}\max_{1\leq i\leq s}\bigg{\{}(i-1)n-\binom{i}{2}+\bigg{\lfloor}\frac{p_{i}-2}{2}(n-i+1)\bigg{\rfloor}+1\bigg{\}},(t-2)n-\binom{t-1}{2}+r\bigg{\}},$$

$$ar(K_n,F)\ge ar(K_n,t{K}_2)=ex(n,(t-1)K_2)+1=(t-2)n-\left(\genfrac{}{}{0ex}{}{t-1}{2}\right)+1.$$

$$ar(K_n,F)\ge ar(K_n,(t-1)K_{1,2}\cup K_2)=(t-2)n-\left(\genfrac{}{}{0ex}{}{t-1}{2}\right)+2.$$

$$f(n,F)=\max\bigg{\{}\max_{1\leq i\leq s}\bigg{\{}(i-1)n-\binom{i}{2}+\bigg{\lfloor}\frac{p_{i}-2}{2}(n-i+1)\bigg{\rfloor}+1\bigg{\}},(t-2)n-\binom{t-1}{2}+r\bigg{\}}.$$

$$ar(K_n,k{K}_{1,2}\cup l{K}_{1,1})=\{\begin{array}{l}(k+l-2)n-\left(\genfrac{}{}{0ex}{}{k+l-1}{2}\right)+1,k\ge 0,l\ge 2;\\ (k-1)n-\left(\genfrac{}{}{0ex}{}{k}{2}\right)+2,k\ge 1,l=1;\\ (k-1)n-\left(\genfrac{}{}{0ex}{}{k}{2}\right)+1,k\ge 1,l=0.\end{array}$$

$$f(n,F)=\max\bigg{\{}\bigg{\lfloor}\frac{p_{1}-2}{2}n\bigg{\rfloor}+1,(s-1)n-\binom{s}{2}+1,(t-2)n-\binom{t-1}{2}+r\bigg{\}}.$$

$$\begin{split}|c^{\prime}|&\geq f(n,F)+1-(n-1)\\ &\geq\max\bigg{\{}(s-1)n-\binom{s}{2}+1,(t-2)n-\binom{t-1}{2}+r\bigg{\}}+1-(n-1)\\ &=\max\bigg{\{}(s-2)(n-1)-\binom{s-1}{2}+1,(t-3)(n-1)-\binom{t-2}{2}+r\bigg{\}}+1\\ &\geq ar(K_{n-1},F-K_{1,p_{1}})+1,\end{split}$$

$$|c^{\prime }|\ge f(n,F)+1-(n-1)\ge f(n-1,F-K_{1,p_1})+1=ar(K_{n-1},F-K_{1,p_1})+1.$$

$$\begin{split}|E(G)|&\leq|E_{G}(V(F-K_{1,p_{t}}),V)|+|E(G[W])|\\ &\leq\big{(}\sum_{i=1}^{t-1}p_{i}+t-1\big{)}\big{(}\sum_{i=1}^{t}p_{i}+t-1\big{)}+\frac{(p_{t}-1)[n-(\sum_{i=1}^{t-1}p_{i}+t-1)]}{2}\\ &\leq\big{(}\sum_{i=1}^{t}p_{i}+t-1\big{)}^{2}+\frac{p_{t}-1}{2}n.\end{split}$$

$$(\sum_{i=1}^{t}p_{i}+t-1)^{2}\geq|E(G)|=f(n,F)+1\geq\bigg{\lfloor}\frac{p_{1}-2}{2}n\bigg{\rfloor}+2\geq\frac{n}{2}+1,$$

$$\begin{split}&\big{(}\sum_{i=1}^{t}p_{i}+t-1\big{)}^{2}+\frac{p_{t}-1}{2}n\geq|E(G)|=f(n,F)+1\\ &\geq(t-1)n-\binom{t}{2}+\bigg{\lfloor}\frac{p_{t}-2}{2}(n-t+1)\bigg{\rfloor}+2\\ &\geq\frac{(2t+p_{t}-4)n}{2}-\frac{(t-1)(p_{t}+t-2)}{2}.\end{split}$$

$$ar(K_{n},F)\geq\max\biggm{\{}\binom{\sum_{i=1}^{k}p_{i}-2}{2}+1,sn-\binom{s+1}{2}+r\biggm{\}},$$

$$sn-\left(\genfrac{}{}{0ex}{}{s+1}{2}\right)+1\le ar(K_n,F)\le sn-\left(\genfrac{}{}{0ex}{}{s+1}{2}\right)+2.$$

$$ar(K_n,F)=(s+1)n-\left(\genfrac{}{}{0ex}{}{s+2}{2}\right)+1.$$

$$ar(K_n,F)\le ar(K_n,P_{{\sum }_{i=1}^k{p}_i})=sn-\left(\genfrac{}{}{0ex}{}{s+1}{2}\right)+2.$$

$$ar(K_n,F)\le ex(n,F)=(s+1)n-\left(\genfrac{}{}{0ex}{}{s+2}{2}\right)+1.$$

$$ar(K_n,F)=\left(\sum _{i=1}^k\lfloor \frac{p_i}{2}\rfloor -2\right)n+O(1).$$

$$\text{(a)}ex(n,P_r)\le \frac{r-2}{2}n;$$

Full text

Keywords: Anti-Ramsey number, star forest, linear forest, double star.

1. Introduction

Let $G$ be a simple undirected graph. For $x\in V(G)$, we denote the neighborhood (the set of neighbors of $x$) and the degree of $x$ in $G$ by $N_{G}(x)$ and $d_{G}(x)$, respectively. The maximum degree and the minimum degree of $G$ are denoted by $\Delta(G)$ and $\delta(G)$, respectively. For $\emptyset\not=X\subset V(G)$, $G[X]$ is the subgraph of $G$ induced by $X$ and $G-X$ is the subgraph of $G$ induced by $V(G)\setminus X$. If $X=\{x\}$, then $G-X$ will be denoted by $G-x$ for short. Given a graph $G=(V,E)$, for any (not necessarily disjoint) vertex sets $A,B\subset V$, let $E_{G}(A,B):=\{uv\in E(G)|u\neq v,u\in A,v\in B\}$. A star forest is a forest whose components are stars and a linear forest is a forest whose components are paths. We use $tG$ and $\overline{G}$ to denote $t$ vertex-disjoint copies and the complement of $G$, respectively. Given two vertex disjoint graphs $G_{1}$ and $G_{2}$, we denote by $G_{1}+G_{2}$ the join of graphs $G_{1}$ and $G_{2}$, that is the graph obtained from $G_{1}\cup G_{2}$ by joining each vertex of $G_{1}$ with each vertex of $G_{2}$.

We call a subgraph of an edge-colored graph rainbow, if all of its edges have different colors. Let $G$ be a graph. The anti-Ramsey number $ar(K_{n},G)$ is the maximum number of colors in an edge-coloring of $K_{n}$ which has no rainbow copy of $G$. The Turán number $ex(n,G)$ is the maximum number of edges of a simple graph on $n$ vertices without a copy of $G$. The anti-Ramsey number was first studied by Erdős, Simonovits and Sós [6]. They showed that the anti-Ramsey number is closely related to Turán number. Since then, there are plentiful results in this field, including cycles [1, 17], cliques [16, 19], trees [12, 13] and so on. See Fujita, Magnant and Ozeki [7, 8] for an abundant survey. Among these results, almost the considered graphs are connected graphs, and a few unconnected graphs are considered including matchings [4, 11], vertex-disjoint cliques [21]. In this paper, we will consider the cases that $G$ is a star forest or a linear forest.

We mention some of the results, which are relevant to our work.

Jiang [12] and Montellano-Ballesteros [18] independently found the anti-Ramsey number for stars.

Theorem 1. ([12],[18]) For $n\geq p+2\geq 3$,

[TABLE]

By Theorem 1, $ar(K_{n},K_{1,p})=\lfloor\frac{(p-2)n}{2}\rfloor+1$ for $n\geq 3p+4$ and $p\geq 2$.

Simonovits and Sós [20] considered the anti-Ramsey number for paths and obtained the following result.

Theorem 2. ([20]) Let $P_{k+1}$ be a path of length $k\geq 2$ and $k\equiv r$ $(\mod 2),0\leq r\leq 1$. For large enough $n$ $(n\geq\frac{5}{4}k+C$ for some universal constant $C)$,

[TABLE]

Let $\Omega_{k}$ denote the family of graphs that contain $k$ vertex-disjoint cycles. Jin and Li [14] computed the anti-Ramsey number for $\Omega_{2}.$

Theorem 3. ([14]) For any $n\geq 6$, $ar(K_{n},\Omega_{2})=\max\{2n-2,11\}.$

The Turán number of $tK_{2}$ was determined by Erdős and Gallai [5] as $ex(n,tK_{2})=\max\{\binom{2t-1}{2},\\ \binom{t-1}{2}+(t-1)(n-t+1)\}$ for $n\geq 2t\geq 2$. The anti-Ramsey number of matchings was first considered by Schiermeyer [19].

Theorem 4.([19]) For $n\geq 3t+3\geq 8$, we have

[TABLE]

Later, Chen, Li and Tu [4] and independently Fujita, Kaneko, Schiermeyer and Suzuki [7] showed that $ar(K_{n},tK_{2})=ex(n,(t-1)K_{2})+1$ for $n\geq 2t+1\geq 5$. The values $ar(K_{2t},tK_{2})=ex(2t,(t-1)K_{2})+1$ for $2\leq t\leq 6$ and $ar(K_{2t},tK_{2})=ex(2t,(t-1)K_{2})+2$ for $t\geq 7$ were determined in [4] and by Haas and Young [11], independently.

Gilboa and Roditty [9] considered the graphs with small connected components and proved inductive results of the form “if $ar(K_{n},G\cup t_{0}P_{s})\leq f(n,t_{0},G)$ for sufficiently large $n$, then $ar(K_{n},G\cup tP_{s})\leq f(n,t,G)$ for sufficiently large $n$ and $t\geq t_{0}$, where $s=2$ or $3$.” These results imply the following theorem.

Theorem 5. ([9]) For sufficiently large $n$,

(1) $ar(K_{n},P_{3}\cup tP_{2})=(t-1)(n-\frac{t}{2})+1$ for $t\geq 2;$

(2) $ar(K_{n},P_{4}\cup tP_{2})=t(n-\frac{t+1}{2})+1$ for $t\geq 1;$

(3) $ar(K_{n},C_{3}\cup tP_{2})=t(n-\frac{t+1}{2})+1$ for $t\geq 1;$

(4) $ar(K_{n},tP_{3})=(t-1)(n-\frac{t}{2})+1$ for $t\geq 1;$

(5) $ar(K_{n},P_{k+1}\cup tP_{3})=(t+\lfloor\frac{k}{2}\rfloor-1)(n-\frac{t+\lfloor k/2\rfloor}{2})+1+(k\mod 2)$ for $k\geq 3$ and $t\geq 0;$

(6) $ar(K_{n},P_{2}\cup tP_{3})=(t-1)(n-\frac{t}{2})+2$ for $t\geq 1;$

*(7) $ar(K_{n},kP_{2}\cup tP_{3})=(t+k-2)(n-\frac{t+k-1}{2})+1$ for $k\geq 2$ and $t\geq 2.$ *

The Turán number of star forests and linear forests are considered by Lidický, Liu and Palmer [15].

Theorem 6.([15]) Let $F=\bigcup^{t}_{i=1}K_{1,p_{i}}$ be a star forest and $p_{1}\geq p_{2}\geq\cdots\geq p_{t}\geq 1$. For $n$ sufficiently large,

[TABLE]

Theorem 7.([15]) Let $F=\bigcup^{k}_{i=1}P_{p_{i}}$ be a linear forest, where $k\geq 2$ and $p_{i}\geq 2$ for $1\leq i\leq k$. If at least one $p_{i}$ is not $3$, then for $n$ sufficiently large,

[TABLE]

where $c=1$ if all $p_{i}$ are odd and $c=0$ otherwise.

In Sections 2 and 3, we generalize Theorem 5 by considering the anti-Ramsey number of star forests and linear forests, respectively.

Theorem 8. Let $F=\bigcup^{t}_{i=1}K_{1,p_{i}}$ be a star forest, where $p_{1}\geq 3$, $p_{1}\geq p_{2}\geq\cdots\geq p_{t}\geq 1$. Let $s=\max\{i:p_{i}\geq 2,1\leq i\leq t\}$. For $n\geq 3t^{2}(p_{1}+1)^{2}$, we have

[TABLE]

where $r=1$ if $p_{t-1}=1$ and $r=2$ otherwise.

Theorem 9. Let $F=\bigcup^{k}_{i=1}P_{p_{i}}$ be a linear forest, where $k\geq 2$ and $p_{i}\geq 2$ for $1\leq i\leq k$. We have

[TABLE]

where $\epsilon=1$ if all $p_{i}$ are odd and $\epsilon=2$ otherwise.

We get the approximate value of the anti-Ramsey number for linear forests by Theorem 9 and it would be interesting to determine the exact value. Bialostocki, Gilboa and Roditty [2] and independently Gorgol and Görlich [10] showed that $ar(K_{n},2P_{3})=\max\{n,7\}$ for $n\geq 6$. Gorgol and Görlich showed that $ar(K_{n},3P_{3})=2n-2$ for $n\geq 13.$ In Section 4, we will use Theorem 3 to compute the exact value of $ar(K_{n},2P_{4})$ for $n\geq 8$.

Theorem 10. For any $n\geq 8$, $ar(K_{n},2P_{4})=\max\{2n-2,16\}.$

Another motivation of this paper is the following conjecture of Gorgol and Görlich [10]:

Let $G$ be a connected graph on $n_{0}\geq 3$ vertices and $t\geq 1$, then for large $n$,

[TABLE]

if and only if $G$ is a tree.

Statement (4) in Theorem 5(4) and Theorem 8 show that this conjecture is true for $P_{3}$ and $K_{1,p}$ ($p\geq 3$), respectively. However, from Theorem 9, some simple calculation shows that this conjecture fails for $P_{l},l\geq 4$.

Actually, for an arbitrary tree $T_{k}$ with $k$ edges, it is difficult to determine the (approximate) value of $ar(K_{n},T_{k})$. Jiang and West [13] showed that for $n\geq 2k$,

[TABLE]

The upper bound comes from the well-known bound of $ex(n,T_{k})\leq(k-1)n.$ Erdős and Sós gave the following conjecture.

Conjecture 1.

[TABLE]

If Conjecture 1 is true (Ajtai, Komlós, Simonovits, Szemerédi announced it for large $k$), then the upper bound of $ar(K_{n},T_{k})$ can also be reduced to $\frac{k-1}{2}n.$ Also, Jiang and West [13] conjectured that:

Conjecture 2.

[TABLE]

Notice that if $T_{k}$ is a star or a path of even length, then $ar(n,T_{k})=\frac{k-2}{2}n+O(1).$

The double star $S_{p,q}$, where $p\geq q\geq 1$, is the graph consisting of the union of two stars $K_{1,p}$ and $K_{1,q}$ together with an edge joining their centers. In Section 5, we compute the anti-Ramsey number of double stars.

Theorem 11. For $p\geq 2,1\leq q\leq p$ and $n\geq 6(p^{2}+2p),$ we have

[TABLE]

Notice that if we take $T_{k}=S_{p,p-1}$, then we have $ar(K_{n},T_{k})=\lfloor\frac{k-2}{2}\rfloor\frac{n}{2}+O(1).$

This paper is organized as follows. In Section 2, we give the proof of Theorem 8. The proof of Theorem 9 will be given in Section 3. The proof of Theorem 10 will be given in Section 4. The proof of Theorem 11 will be given in Section 5. Finally we will give a conjecture in Section 6.

Notation: Given an edge-coloring $c$ of $G$, we denote the color of an edge $uv$ by $c(uv)$. We denote the number of colors by $|c|$. For any $v\in V(G)$, let $C(v):=\{c(vw)|\,w\in N_{G}(v)\}$ and $d_{c}(v):=|C(v)|.$ Let $H$ be a subgraph of $G$. Denote $C(H)=\{c(uv)|\,uv\in E(H)\}$. A color $a$ is stared (at $x$) if all the edges with color $a$ induce a star $K_{1,r}$ (centered at the vertex $x$). We let $d^{c}(v)=|\{a\in C(v)|\,a$ is stared at $v\}|$. A representing subgraph in an edge-coloring of $K_{n}$ is a spanning subgraph containing exactly one edge of each color. In the rest of this paper, we will use $V$ to denote the vertex set of $K_{n}$ for short.

2. Star forests

In this Section, we use the idea of [9] to prove Theorem 8.

Theorem 8. Let $F=\bigcup^{t}_{i=1}K_{1,p_{i}}$ be a star forest, where $p_{1}\geq 3,p_{1}\geq p_{2}\geq\cdots\geq p_{t}\geq 1$. Let $s=\max\{i:p_{i}\geq 2,1\leq i\leq t\}$. For $n\geq 3t^{2}(p_{1}+1)^{2}$,

[TABLE]

where $r=1$ if $p_{t-1}=1$ and $r=2$ otherwise.

For another lower bound, it is enough to consider the case $t\geq 3$. If $p_{t-1}=1,$ then $F\supset tK_{2}$. By Theorem 4, we have

[TABLE]

If $p_{t-1}\geq 2$, we have $F\supset(t-1)K_{1,2}\cup P_{2}$. By Theorem 5(6), we have

[TABLE]

Now we consider the upper bound. If $t=1$, then the result holds obviously by Theorem 1. Suppose $t\geq 2$. Let

[TABLE]

By Theorems 4 and 5 (statements (1), (4), (6), (7)), we have

[TABLE]

Let $c$ be any edge-coloring of $K_{n}$ using $f(n,F)+1$ colors. We will find a rainbow $F$ by considering the following two cases.

Case 1. There is some vertex $v_{0}\in V$ such that $d_{c}(v_{0})\geq\sum_{i=1}^{t}p_{i}+t$.

Choose some color $c_{0}$. Consider an edge-coloring $c^{\prime}$ of $K_{n}-v_{0}$: $c^{\prime}(e)=c(e)$ if $c(e)\notin C(v_{0})$; else $c^{\prime}(e)=c_{0}$, where $e\in E(K_{n}-v_{0})$.

Claim 1. There is a rainbow copy of $F-K_{1,p_{1}}$ in $K_{n}-v_{0}$ with respect to $c^{\prime}$.

Proof of Claim 1. We consider the following two subcases.

Subcase 1.1 $p_{2}\leq 2.$

In this subcase, we have $F-K_{1,p_{1}}=(s-1)K_{1,2}\cup(t-s)K_{1,1}$ and

[TABLE]

If $F-K_{1,p_{1}}=K_{1,1}$, then the result holds obviously. Suppose $F-K_{1,p_{1}}\neq K_{1,1}$. Then

[TABLE]

by (*). Thus there is a rainbow $F-K_{1,p_{1}}$ in $K_{n}-v_{0}$ with respect to $c^{\prime}$.

Subcase 1.2 $p_{2}\geq 3.$

In this subcase, we have

[TABLE]

By induction hypothesis, there is a rainbow $F-K_{1,p_{1}}$ in $K_{n}-v_{0}$ with respect to $c^{\prime}$.

By Claim 1, there is a rainbow $F-K_{1,p_{1}}$ in $K_{n}-v_{0}$ with respect to $c^{\prime}$ (also respect to $c$). Since $d_{c}(v_{0})\geq\sum_{i=1}^{t}p_{i}+t$, we are surely left with at least $p_{1}$ edges, say $v_{0}w_{1},v_{0}w_{2},\ldots,v_{0}w_{p_{1}}$, such that $w_{1},w_{2},\ldots,w_{p_{1}}\notin V(F-K_{1,p_{1}})$ and $c(v_{0}w_{1}),c(v_{0}w_{2}),\ldots,c(v_{0}w_{p_{1}})\notin\{c(e)|e\in E(F-K_{1,p_{1}})\}$. By adding such $p_{1}$ edges to $F-K_{1,p_{1}}$, we get a rainbow $F$.

Case 2. $d_{c}(v)\leq\sum_{i=1}^{t}p_{i}+t-1$ for all $v\in V$.

By induction hypothesis, $K_{n}$ clearly contains a rainbow $F-K_{1,p_{t}}$. Assume, by contradiction, that $K_{n}$ does not contain a rainbow $F$. Let $G$ be a representing subgraph of $K_{n}$ such that $E(F-K_{1,p_{t}})\subseteq E(G)$. Then we have $|E(G)|=f(n,F)+1$ and $d_{G}(v)\leq\sum_{i=1}^{t}p_{i}+t-1$ for all $v\in V.$

Let $W=V-V(F-K_{1,p_{t}})$. Since $d_{G}(v)\leq\sum_{i=1}^{t}p_{i}+t-1$ for all $v\in V(F-K_{1,p_{t}})$ and $G[W]$ does not contain a copy of $K_{1,p_{t}}$, we have

[TABLE]

We will finish the proof by considering the following two subcases.

Subcase 2.1 $p_{t}=1.$

Since $p_{1}\geq 3$ and $p_{t}=1$, we have

[TABLE]

a contradiction with $n\geq 3t^{2}(p_{1}+1)^{2}.$

Subcase 2.2 $p_{t}\geq 2.$

In this subcase, we have $s=t$ and

[TABLE]

Thus we have $n\leq\frac{2}{2t-3}[(\sum^{t}_{i=1}p_{i}+t-1)^{2}+(t-1)(p_{t}+t-2)],$ a contradiction with $n\geq 3t^{2}(p_{1}+1)^{2}.$ ∎

3. Linear forests

First, we have the lower bound of the anti-Ramsey number for linear forests.

Proposition 1. Let $F$ be a linear forest with components of order $p_{1},p_{2},\ldots,p_{k}$, where $k\geq 2$ and $p_{i}\geq 2$ for $1\leq i\leq k$. Let $n\geq\sum_{i=1}^{k}p_{i}$. Then we have

[TABLE]

where $s=\sum_{i=1}^{k}\lfloor\frac{p_{i}}{2}\rfloor-\epsilon$; $\epsilon=1$ if all $p_{i}$ are odd and $\epsilon=2$ otherwise; $r=2$ if exactly one $p_{i}$ is even and $r=1$ otherwise.

For the second lower bound, we color $K_{s}+\overline{K}_{n-s}$ rainbow and color the edges of $K_{n-s}$ with $r$ new colors. Every copy of $F$ in $K_{n}$ have at least $(r+1)$ edges in $K_{n-s}$. In this way we do not obtain a rainbow $F$ and use exactly $s(n-s)+\binom{s}{2}+r=sn-\binom{s+1}{2}+r$ colors. ∎

If all the components of the linear forest are even paths or odd paths, we can get the following corollary from Theorems 2 and 7.

Corollary 1. Let $F$ be a linear forest with components of order $p_{1},p_{2},\ldots,p_{k}$, where $k\geq 2$ and $p_{i}\geq 2$ for $1\leq i\leq k$ and $n$ sufficiently large. Let $s=\sum_{i=1}^{k}\lfloor\frac{p_{i}}{2}\rfloor-2$. If all $p_{i}$ are even, we have

[TABLE]

If all $p_{i}$ are odd, we have

[TABLE]

For the upper bound, when all $p_{i}$ are even, by Theorem 2,

[TABLE]

When all $p_{i}$ are odd, if $p_{1}=p_{2}=\cdots=p_{k}=3$, by Theorem 5(4), $ar(K_{n},F)=ar(K_{n},kP_{3})=(k-1)n-\binom{k}{2}+1$. If at least one $p_{i}$ is not $3$, by Theorem 7,

[TABLE]

∎

It is enough to consider the linear forests with at least one even path.

Theorem 12. Let $F$ be a linear forest with components of order $p_{1},p_{2},\ldots,p_{k}$, where $k\geq 1$, $p_{i}\geq 2$ for $1\leq i\leq k$ and at least one $p_{i}$ is even. Then

[TABLE]

[TABLE]

[TABLE]

Since we can regard the union of even paths as the subgraph of one long even path (see Corollary 1), we just need to prove the upper bound is correct for linear forest with exact one even path and some odd paths. So we assume that $F=P_{2s}\cup P_{2t_{1}+1}\cup P_{2t_{2}+1}\cup\cdots\cup P_{2t_{k}+1}$, where $s\geq 1,k\geq 1$ and $t_{k}\geq t_{k-1}\geq\cdots\geq t_{1}\geq 1$. Let $s+t_{1}+t_{2}+\cdots+t_{k}=m$. In this proof, we just consider the case $s\geq 6$. The case $s<6$ can be proved by the similar arguments but need to distinguish more cases as in [20].

Consider an edge-coloring of $K_{n}$ with $ar(K_{n},F)$ colors such that there is no rainbow $F$. First we take a rainbow path $P_{l}=u_{1}u_{2}\ldots u_{l}$ with maximum length. If $l\geq 2m+k$, we can get a rainbow $F$, a contradiction. By Proposition 1 and Theorem 2, $ar(K_{n},F)\geq ar(K_{n},P_{2m-1})+1$ for large $n$, which implies there is a rainbow $P_{2m-1}$. Hence we assume that $2m-1\leq l\leq 2m+k-1.$ Take a representing subgraph $G$ of $K_{n}$ such that $P_{l}\subset G$. Then $|E(G)|=ar(K_{n},F)$. We would partition $V\setminus V(P_{l})$ into three sets $U_{1},U_{2}$ and $U_{3}$ as follows:

U_{1}$$U_{2}$$U_{3}$$P_{l}$$u_{1}$$u_{2}$$u_{l-1}$$u_{l}

$U_{1}$ is the subset of vertices of $V\setminus V(P_{l})$ which are not jointed to $P_{l}$ at all: neither by edges nor by paths;

$U_{2}$ is the set of isolated vertices of $V\setminus V(P_{l})$ which are jointed to $P_{l}$ by edges;

$U_{3}=V\setminus(V(P_{l})\cup U_{1}\cup U_{2}).$

Claim 1. $|E(G[U_{1}])|\leq(m-2)|U_{1}|$.

Proof of Claim 1 We first prove that there is an $P_{2s}\cup P_{2t_{1}+1}\cup\ldots\cup P_{2t_{k}+1}$ in $P_{2s+2t_{1}+\ldots+2t_{k}-1}\cup P_{2s+2t_{1}+\ldots+2t_{k}-2}$ by induction on $k$. The base case $k=1$ is correct since $P_{2s}\cup P_{2t_{1}+1}\subset P_{2s+2t-1}\cup P_{2s+2t_{1}-2}$. Suppose the statement holds for $k-1$. We divide $P_{2s+2t_{1}+\ldots+2t_{k}-1}$ and $P_{2s+2t_{1}+\ldots+2t_{k}-2}$ respectively into two parts $P_{2s+2t_{1}+\ldots+2t_{k-1}-1},P_{2t_{k}}$ and $P_{2s+2t_{1}+\ldots+2t_{k-1}-2},P_{2t_{k}}$. We can find an $P_{2s}\cup P_{2t_{1}+1}\cup\ldots\cup P_{2t_{k-1}+1}$ in $P_{2s+2t_{1}+\ldots+2t_{k-1}-1}\cup P_{2s+2t_{1}+\ldots+2t_{k-1}-2}$ by induction hypothesis. Since $(2s+2t_{1}+\ldots+2t_{k-1}-1)+(2s+2t_{1}+\ldots+2t_{k-1}-2)>2s+(2t_{1}+1)+\ldots+(2t_{k-1}+1)$, there is at least one vertex of either $P_{2s+2t_{1}+\ldots+2t_{k-1}-1}$ or $P_{2s+2t_{1}+\ldots+2t_{k-1}-2}$ which is not used in $P_{2s}\cup P_{2t_{1}+1}\cup\ldots\cup P_{2t_{k-1}+1}$. Hence, we can find an $P_{2s}\cup P_{2t_{1}+1}\cup\ldots\cup P_{2t_{k}+1}$ in the original two long paths.

Since $G$ contains no $F$, $G[U_{1}]$ contains no $P_{2m-2}$ by the statement above. Thus $|E(G[U_{1}])|\leq(m-2)|U_{1}|$ by (a).

Claim 2. $|\{v\in U_{2}:d_{G}(v)\geq m-1\}|\leq(m+k)\binom{l-2}{m-1}$.

Proof of Claim 2 It is obvious that $N_{G}(v)\subset V(P_{l})\setminus\{u_{1},u_{l}\}$ for all $v\in U_{2}$. Suppose $|\{v\in U_{2}:d_{G}(v)\geq m-1\}|\geq(m+k)\binom{l-2}{m-1}+1$. Note that there are $\binom{l-2}{m-1}$ subsets of order $m-1$ in $V(P_{l})\setminus\{u_{1},u_{l}\}$. By Pigeonhole Principle, there are $A\subseteq V(P_{l})\setminus\{u_{1},u_{l}\}$ with $|A|=m-1$ and $B\subseteq U_{2}$ with $|B|=m+k+1$ such that $A\subseteq N_{G}(v)$ for any $v\in B$. So there is a rainbow $K_{m-1,m+k+1}$. By adding one edge whose endpoints are in the large part of $K_{m-1,m+k+1}$ to $K_{m-1,m+k+1}$, we can find a rainbow $F$, a contradiction.

The following claim 3 is Lemma 1 in [20]. We include the proof for the sake of completeness.

Claim 3. $|E(G[U_{3}])|+|E_{G}(P_{l},U_{3})|\leq(m-2)|U_{3}|$.

Proof of Claim 3 Take a component $H$ of $G[U_{3}]$ and denote by $r$ the length of the longest cycle of $H$. If $H$ is a tree, let $r=2$. By (b), we have $|E(H)|\leq\frac{r(|V(H)|-1)}{2}$. For any $v\in V(H)$, we can find an $P_{r}$ in $H$ starting from it. Hence, $u_{1},\ldots,u_{r}$ and $u_{l-r+1},\ldots,u_{l}$ cannot be joined to $v$. Otherwise, there is a rainbow $P_{l+1}$, a contradiction. For any three consecutive vertices $\{u_{i},u_{i+1},u_{i+2}\}$, there is no two independent edges in $E_{G}(\{u_{i},u_{i+1},u_{i+2}\},V(H))$ by the maximality of $P_{l}$. Hence, we have

[TABLE]

Adding all the components of $G[U_{3}]$ up, we get $|E(G[U_{3}])|+|E_{G}(P_{l},U_{3})|\leq\frac{l+1}{3}|U_{3}|$. Note that $l\leq 2m+k-1$. Since $s\geq 6$, $m\geq s+k\geq 6+k$ which implies $|E(G[U_{3}])|+|E_{G}(P_{l},U_{3})|\leq(m-2)|U_{3}|$

By Claims 1, 2 and 3, we have

[TABLE]

∎

By Corollary 1 and Theorem 12, we can get Theorem 9.

4. The exact value of $ar(K_{n},2P_{4})$

In this Section, we will prove Theorem 10. We denote the complete graph on $n$ vertices minus one edge by $K_{n}^{-}$. The following fact is trivial.

Fact 1. Let $n\geq 8$. If there is an edge coloring of $K_{n}$ using $17$ colors such that there is a rainbow $K_{6}$ or $K_{6}^{-}$, then there is a rainbow $2P_{4}$.

We first prove the following lemma.

Lemma 1. $ar(K_{8},2P_{4})=16.$

Case 1. $l=5$.

We claim that $c(xy)\in C(T_{1}\cup T_{2})$ for all $x\in V(T_{1})$ and $y\in V(T_{2})$. Otherwise, say $c(x_{1}y_{1})\notin C(T_{1}\cup T_{2})$, then $x_{2}x_{3}x_{1}y_{1}\cup y_{2}y_{3}y_{4}y_{5}$ is a rainbow $2P_{4}$, a contradiction. Then the total number of colors is at most $3+\binom{5}{2}=13<17$, a contradiction.

Case 2. $l=4$.

Let the remaining vertex be $z$. We have $c(x_{1}z)=c(x_{2}x_{3})$; otherwise there must be a rainbow $2P_{4}$ in $T_{1}\cup T_{2}\cup\{x_{1}z\}.$ Similarly, we have $c(x_{2}z)=c(x_{1}x_{3})$ and $c(x_{3}z)=c(x_{1}x_{2})$. Now we have four rainbow $C_{3}$’s: $C_{3}^{1}=x_{1}x_{2}x_{3}x_{1}$, $C_{3}^{2}=zx_{2}x_{3}z$, $C_{3}^{3}=zx_{1}x_{2}z$ and $C_{3}^{4}=zx_{1}x_{3}x_{1}z$. If there are $u\in\{z,x_{1},x_{2},x_{3}\}$, say $u=z$, and $1\leq j\leq 4$, say $j=1$, such that $c(zy_{1})\not=c(zy_{2})$ and $c(zy_{1}),c(zy_{2})\notin C(C_{3}^{1}\cup T_{2})$ then we have a rainbow $C_{3}^{1}$ and a rainbow $C_{5}=y_{1}zy_{2}y_{3}y_{4}y_{1}$ and the situation is the same as Case 1. Hence we have $|E_{G}(u,T_{2})|\leq 2$ for any $u\in\{z,x_{1},x_{2},x_{3}\}$. Then $17=|E(G)|\leq 3+4\times 2+\binom{4}{2}=17$, which means $G[V(T_{2})]=K_{4}$ and $|E_{G}(u,T_{2})|=2$ for any $u\in\{z,x_{1},x_{2},x_{3}\}$. Thus we can find a rainbow $2P_{4}$, a contradiction.

Case 3. $l=3$.

Let the remaining vertices be $z_{1},z_{2}$. By Case 2 and $G$ containing no rainbow $2P_{4}$, we have $|E_{G}(\{z_{1},z_{2}\},T_{1}\cup T_{2})|\leq 2$ and then $|E(G[V(T_{1}\cup T_{2})])|\geq 17-2-1=14.$ Thus $G[V(T_{1}\cup T_{2})]\cong K_{6}$ or $K_{6}^{-}$ and we can get a rainbow $2P_{4}$ by Fact 1, a contradiction.∎

Now we will complete the proof of Theorem 10.

Theorem 10. For any $n\geq 8$, $ar(K_{n},2P_{4})=\max\{2n-2,16\}.$

Consider an $(2n-1)$-edge-coloring of $K_{n}$ for $n\geq 9$. Suppose there is no rainbow $2P_{4}$. By Theorem 3, there is a rainbow $C_{k}\cup C_{l}$. Assume that $k\leq l$. Then $k=3$. Let $T_{1}=C_{3}=x_{1}x_{2}x_{3}x_{1}$ and $T_{2}=C_{l}=y_{1}\ldots y_{l}y_{1}$. We finish the proof by considering the following four cases.

Case 1. $4\leq l\leq n-5.$

In this case, there are at least two vertices $z_{1},z_{2}\notin V(T_{1}\cup T_{2})$. Since there is no rainbow $2P_{4}$, we have $c(x_{1}z_{i})=c(x_{2}x_{3})$, $c(x_{2}z_{i})=c(x_{1}x_{3})$ and $c(x_{3}z_{i})=c(x_{1}x_{2})$ for $i=1,2$. But we can have a rainbow $2P_{4}$ whatever the color of $z_{1}z_{2}$ is, a contradiction.

Case 2. $l=n-4$.

Let $V(K_{n})\setminus V(T_{1}\cup T_{2})=\{z\}.$ Since there is no rainbow $2P_{4}$, we have that $c(x_{1}z)=c(x_{2}x_{3})$, $c(x_{2}z)=c(x_{1}x_{3})$, $c(x_{3}z)=c(x_{1}x_{2})$, $c(xy)\in C(T_{1}\cup T_{2})$ for all $x\in V(T_{1})\cup\{z\}$ and $y\in V(T_{2})$. By Case 1, we can assume that $c(y_{i}y_{j})\in C(T_{1}\cup T_{2})$ for all $y_{i},y_{j}\in V(T_{2})$. Hence the total number of colors is at most $3+n-4=n-1<2n-1$, a contradiction.

Case 3. $l=n-3$.

Since there is no rainbow $2P_{4}$, we have that $c(xy)\in C(T_{1}\cup T_{2})$ for all $x\in V(T_{1})$ and $y\in V(T_{2})$. By Case 1 and Case 2, we can assume that $c(y_{i}y_{j})\in C(T_{1}\cup T_{2})$ for all $y_{i},y_{j}\in V(T_{2})$. Hence the total number of colors is at most $3+n-3=n<2n-1$, a contradiction.

Case 4. $l=3.$

In this case we will consider the following two subcases. We choose a representing subgraph $G$ such that $G\supset T_{1}\cup T_{2}$.

Subcase 4.1 $n=9.$

If there is a vertex $v_{0}$ such that $d^{c}(v_{0})=0$, then $K_{9}-v_{0}$ uses exactly $17$ colors and there is a rainbow $2P_{4}$ in $K_{9}-v_{0}$ by Lemma 1. Hence we have $d^{c}(v)\geq 1$, for all $v\in V(K_{9})$. Thus $\delta(G)\geq 1$.

Let $Z=V(K_{9})\setminus V(T_{1}\cup T_{2})=\{z_{1},z_{2},z_{3}\}.$ We claim that $|E_{G}(z_{j},T_{s})|\leq 1$ for any $1\leq j\leq 3$ and $1\leq s\leq 2$. Otherwise $G$ contains an $C_{3}\cup C_{4}$ and the situation is the same as Case 1. If there are $z_{i}\in Z$ and $s\in\{1,2\}$ such that $|E_{G}(z_{i},T_{s})|=1$, then $|E_{G}(z,T_{t})|=0$ for any $z\in Z\setminus\{z_{i}\}$ and $t\in\{1,2\}\setminus\{s\}$ by $G$ having no $2P_{4}$. If $E_{G}(Z,T_{1}\cup T_{2})=\emptyset$, then $|E(G[V(T_{1}\cup T_{2})])|=17-|E(G[Z])|\geq 17-3=14$ which implies $G[V(T_{1}\cup T_{2})]\cong K_{6}$ or $K_{6}^{-}$ and there is a rainbow $2P_{4}$ by Fact 1, a contradiction. Hence we have $E_{G}(Z,T_{1}\cup T_{2})\neq\emptyset$. Assume $z_{1}x_{1}\in E(G)$ and we will consider the following two subcases.

Subcase 4.1.1 $E_{G}(Z,T_{2})=\emptyset$.

Recall $z_{1}x_{1}\in E(G)$ and $|E_{G}(z_{j},T_{1})|\leq 1$ for any $1\leq j\leq 3$. If there is $z\in\{z_{2},z_{3}\}$ such that $z_{1}z\in E(G)$, then $E_{G}(\{x_{2},x_{3}\},T_{2})$ $=\emptyset$ which implies $|E(G)|\leq 15$, a contradiction. Hence we have $|E(G[Z])|\leq 1$ and then $|E(G[V(T_{1}\cup T_{2})])|\geq 17-3=14.$ Thus $G[V(T_{1}\cup T_{2})]\cong K_{6}$ or $K_{6}^{-}$ and there is a rainbow $2P_{4}$ by Fact 1, a contradiction.

Subcase 4.1.2 $E_{G}(Z,T_{2})\neq\emptyset$.

In this case, we have $|E_{G}(Z,T_{1}\cup T_{2})|=|E_{G}(z_{1},T_{1}\cup T_{2})|=2$ and assume $z_{1}y_{1}\in E(G)$. If $E_{G}(z_{1},\{z_{2},z_{3}\})\neq\emptyset,$ say $z_{1}z_{2}\in E(G)$, then $E_{G}(T_{1},T_{2})\subseteq\{x_{1}y_{1}\}$. Hence $|E(G)|\leq 12$, a contradiction. So we have $E_{G}(z_{1},\{z_{2},z_{3}\})=\emptyset,$ which implies $z_{2}z_{3}\in E(G)$ by $\delta(G)\geq 1.$ Thus $|E(G[V(T_{1}\cup T_{2})])|=17-3=14$ and $G[V(T_{1}\cup T_{2})]\cong K_{6}^{-}$. We can get a rainbow $2P_{4}$ by Fact 1, a contradiction.

Subcase 4.2 $n\geq 10.$

If there is a vertex $v_{0}$ such that $d^{c}(v_{0})\leq 2$, then $K_{n}-v_{0}$ uses $2n-1-d^{c}(v_{0})\geq 2n-1-2=2(n-1)-1$ colors and we get that $K_{n}-v_{0}$ contains a rainbow $2P_{4}$ by induction hypothesis, a contradiction. So we assume that $d^{c}(v)\geq 3$ for all $v\in V(K_{n})$. Thus $\delta(G)\geq 3$.

Let $Z=V(K_{n})\setminus V(T_{1}\cup T_{2})=\{z_{1},z_{2},\ldots,z_{n-6}\}.$ Since $G$ contains no rainbow $2P_{4}$, we have that if there are $z\in Z$ and $s\in\{1,2\}$ such that $E_{G}(z,T_{s})\neq\emptyset$, then $E_{G}(z^{\prime},T_{t})=\emptyset$ for any $z^{\prime}\in Z\setminus\{z\}$ and $t\in\{1,2\}\setminus\{s\}$.

Subcase 4.2.1 $E_{G}(Z,T_{1})\neq\emptyset$ and $E_{G}(Z,T_{2})\neq\emptyset.$

In this case, there is exactly one vertex in $Z$, say $z_{1}$, such that $E_{G}(z_{1},T_{1}),E_{G}(z_{1},T_{2})\neq\emptyset$ and $E_{G}(Z\setminus\{z_{1}\},T_{1}\cup T_{2})=\emptyset$. Since $d_{G}(v)\geq 3$, $G[T_{1}\cup T_{2}\cup\{z_{1}\}]$ contains a cycle with length at least $4$ and $G[Z\setminus\{z_{1}\}]$ contains a cycle of length at least 3. Then we can have a contradiction by Cases 1 to 3.

Subcase 4.2.2 $E_{G}(Z,T_{1})=\emptyset$ or $E_{G}(Z,T_{2})=\emptyset.$

Assume that $E_{G}(Z,T_{1})=\emptyset.$ By Case 1, we can assume that $|E_{G}(z,T_{2})|\leq 1$ for any $z\in Z$. Then $\delta(G[Z])\geq 2$ and there is a cycle in $G[Z]$. Since $d_{G}(x_{i})\geq 3$, we have $|E_{G}(x_{i},T_{2})|\geq 1$ for any $1\leq i\leq 3$ and there is a cycle with length at least $4$ in $G[T_{1}\cup T_{2}]$. Then we can have a contradiction by Cases 1 to 3. ∎

5. Double stars

The following Lemma 2 is an extension of Theorem 1. The idea of the proof is the same as the idea used in [12]. We include the proof for the sake of completeness.

Lemma 2. Let $V(K_{n})=\{v_{1},v_{2},\ldots,v_{n}\}$. For $1\leq p_{1}\leq p_{2}\leq\cdots\leq p_{n}\leq\frac{n}{3}$, the maximum number of colors of an edge-coloring of $K_{n}$ such that $d_{c}(v_{i})\leq p_{i}$ for any $1\leq i\leq n$ is at most $\frac{\sum^{n}_{i=1}p_{i}-n}{2}+1$.

[TABLE]

and we are done. Hence we may assume that $S\not=\emptyset$. For $v\in V$, let $C_{G}(v)$ be the set of colors used on the edges incident to $v$ in $G.$ Clearly, we have $C_{G}(v)\subset C(v)$ for all $v\in V$ and $C_{G}(v)=C(v)$ for $v\in S$. Particularly, $\{c(uv)\}=C(u)\cap C(v)=C_{G}(u)\cap C_{G}(v)$ for $u,v\in S$.

Claim 1. $G[S]$ is a clique.

Proof of Claim 1. Let $u,v\in S$. Since $c(uv)\in C(u)\cap C(v)=C_{G}(u)\cap C_{G}(v)$, $uv\in E(G)$. Hence $G[S]$ is a clique.

Claim 2. Let $u,v\in S$ and $w\in V-S$. If $c(uw)=c(vw)$, then $c(uw)=c(vw)=c(uv)$.

Proof of Claim 2. Since $c(uw)=c(vw)\in C(u)\cap C(v)=C_{G}(u)\cap C_{G}(v)=\{c(uv)\}$, we have $c(uw)=c(vw)=c(uv)$.

Claim 3. Let $u\in S,v\notin S$ and $uv\notin E(G)$. Then $c(uv)\notin C_{G}(v)$.

Proof of Claim 3. Suppose $c(uv)\in C_{G}(v)$. Since $u\in S$, we have $c(uv)\in C_{G}(u)$. Hence $c(uv)\in C_{G}(u)\cap C_{G}(v)$ which implies $uv\in E(G)$, a contradiction.

Claim 4. For all $v_{i}\notin S$, $d_{G}(v_{i})\leq p_{i}-\frac{|S\setminus N_{G}(v_{i})|}{2}$.

Proof of Claim 4. Let $S\setminus N_{G}(v_{i})=\{u_{1},u_{2},\ldots,u_{k}\}.$ By Claim 3, $c(u_{j}v_{i})\notin C_{G}(v_{i})$ for all $1\leq j\leq k$. Furthermore, by Claim 2, if $c(u_{j}v_{i})=c(u_{l}v_{i})$, then $c(u_{j}v_{i})=c(u_{l}v_{i})=c(u_{j}u_{l})$. This implies that no three edges in the set $\{u_{1}v_{i},u_{2}v_{i},\ldots,u_{k}v_{i}\}$ have the same color; otherwise, by Claim 1, $G[S]$ would contain a monochromatic triangle, a contradiction. Hence at least $\frac{k}{2}$ distinct colors are used on the edges $v_{i}u_{1},v_{i}u_{2},\cdots,v_{i}u_{k}$, and those colors are not in $C_{G}(v_{i})$. So $|C(v_{i})|\geq|C_{G}(v_{i})|+\frac{k}{2}=d_{G}(v_{i})+\frac{k}{2}$. Since $|C(v_{i})|=d_{c}(v_{i})\leq p_{i}$, we have $d_{G}(v_{i})+\frac{k}{2}\leq p_{i}$, which yields $d_{G}(v_{i})\leq p_{i}-\frac{k}{2}=p_{i}-\frac{|S\setminus N_{G}(v_{i})|}{2}.$

By Claim 4, we have

[TABLE]

Notice that $\sum_{v_{i}\notin S}|S\setminus N_{G}(v_{i})|$ counts exactly the number of non-edges in $G$ between $S$ and $V\setminus S$. We have $\sum_{v_{i}\notin S}|S\setminus N_{G}(v_{i})|=\sum_{v_{i}\in S}(n-1-p_{i}).$ Hence,

[TABLE]

On the other hand, for $v_{i}\notin S,d_{G}(v_{i})\leq p_{i}-1$. Hence, we have

[TABLE]

We have

[TABLE]

Since $n\geq 3p_{n}$, we have $\min\{|S|,n-\frac{\sum_{v_{i}\in S}(n-1-p_{i})}{2}\}\leq 2$. Therefore,

[TABLE]

∎

Now we will prove Theorem 11.

Theorem 11. For $p\geq 2,1\leq q\leq p$ and $n\geq 6(p^{2}+2p),$

[TABLE]

[TABLE]

When $q=p,$ we color the edges of $K_{n}$ as follows. We color $K_{1,n-1}$ rainbow and color $K_{n-1}$ with new $ar(K_{n-1},K_{1,p})$ colors without producing a rainbow $K_{1,p}$. In such way, we use exactly

[TABLE]

colors and do not obtain a rainbow $S_{p,p}$.

Now we consider the upper bound. Let

[TABLE]

Let $c$ be any edge-coloring of $K_{n}$ using $g(n,p,q)+1$ colors. We will find a rainbow $S_{p,q}$ by considering the following two cases.

Case 1. There is some vertex $v_{0}$ such that $d_{c}(v_{0})\geq p+2q+1.$

Let $U=\{v\in V\setminus\{v_{0}\}:$ the color $c(v_{0}v)$ is stared at $v_{0}\}.$ Consider the induced edge-coloring $c^{\prime}$ of $c$ on $K_{n}-v_{0}$. We have

[TABLE]

Claim 1. $K_{n}-v_{0}$ contains a rainbow $K_{1,q}$ with center in $U$ or a rainbow $K_{1,q+1}$ with center in $V(K_{n}-v_{0})\setminus U$ with respect to $c^{\prime}$ (also $c$).

Proof of Claim 1. Suppose $d_{c^{\prime}}(u)\leq q-1$ for all $u\in U$ and $d_{c^{\prime}}(v)\leq q$ for all $v\in V(K_{n}-v_{0})\setminus U$, by Lemma 2, we have

[TABLE]

While $|c^{\prime}|\geq g(n,p,q)+1-|U|,$ we can get $|U|\geq n$, a contradiction.

By Claim 1, we get a rainbow $K_{1,q}$ in $K_{n}-v_{0}$ with respect to $c$ whose center is $u_{0}$ such that the color $c(v_{0}u_{0})$ does not present in $c(K_{1,q})$. The vertex $v_{0}$ is the endpoint of at least $p+2q+1$ edges with distinct colors (with respect to $c$). Since $d_{c}(v_{0})\geq p+2q+1,$ there are at least $p$ edges, say $v_{0}v_{1},v_{0}v_{2},\ldots,v_{0}v_{p}$, such that $v_{i}\notin V(K_{1,q})$ and $c(v_{0}v_{i})\notin C(K_{1,q})\cup\{c(v_{0}u_{0})\}$ for all $i=1,\ldots,p$. So we can get a rainbow $S_{p,q}$.

Case 2. $d_{c}(v)\leq p+2q$ for all $v\in V$.

Since $g(n,p,q)+1\geq ar(K_{n},K_{1,p+1})+1$, we can find a rainbow $K_{1,p+1}$. Choose some color $c_{0}$ which is not in $C(K_{1,p+1})$. Consider an edge-coloring $c^{\prime\prime}$ of $K_{n}-V(K_{1,p+1})$: for $e\in E(K_{n}-V(K_{1,p+1}))$, if $c(e)\notin C(K_{1,p+1})$, then $c^{\prime\prime}(e)=c(e)$; else $c^{\prime\prime}(e)=c_{0}$. Then

[TABLE]

So there is a rainbow $K_{1,q+1}$ in $K_{n}-K_{1,p+1}$ with respect to $c^{\prime\prime}$ (also $c$). $K_{1,q+1}$ has at most one edge $e$ with color $c_{0}$. By joining the centers $K_{1,p+1}$ and $K_{1,q+1}$ and deleting one edge of $K_{1,p+1}$ and $K_{1,q+1}$ respectively, we can find a rainbow $S_{p,q}$. ∎

6. Open problems

A spider is a tree with at most one vertex of degree more than $2$, called the center of the spider (if no vertex of degree more than two, then any vertex can be the center). A leg of a spider is a path from the center to a vertex of degree 1. Thus, a star with $p$ edges is a spider of $p$ legs, each of length $1$, and a path is a spider of $1$ or $2$ legs.

The number of edges in a maximum matching of a graph $G$ is called the matching number of $G$ and denoted by $\nu(G)$.

Observation 1. Let $p\geq 2$ and $T$ be a spider of $p$ legs, each of length at least $2$. We have

[TABLE]

the equality holds if and only if $T$ has exactly one leg with length even.

Let $p\geq 2$ and $T$ be a spider of $p$ legs, each of length at least $2$. Let $\beta(T)=\min\{\nu(T-e):e\in E(T)\}$.

Proposition 2. Let $p\geq 2$ and $T$ be a spider of $p$ legs, each of length at least $2$. We have

[TABLE]

where $r=2$ if there is exactly one leg of $T$ with length even and $r=1$ otherwise.

If we regard $P_{k+1}$ as a spider of $2$ legs, the lower bound of Proposition 2 is sharp for $p=2$ and large $n$ by Theorem 2. We conjecture that the lower bound is sharp for $p\geq 3$ and large $n$.

Conjecture 3. Let $p\geq 3$ and $T$ be a spider of $p$ legs, each of length at least $2$. For large $n$,

[TABLE]

where $r=2$ if there is exactly one leg of $T$ with length even and $r=1$ otherwise.