A note on rainbow saturation number of paths

TL;DR

This paper improves bounds on the rainbow saturation number of paths in edge-colored graphs, providing tighter upper bounds for certain path lengths and color counts.

Contribution

It presents improved upper bounds for the rainbow saturation number of paths, refining previous results for specific path lengths and color parameters.

Findings

Established new upper bounds for $sat_t(n,rak{R}(P_ ext{ell}))$

Showed bounds are tighter for $ ext{ell} extgreater= 5$ and $t extgreater= 2 ext{ell} - 5$

Enhanced understanding of rainbow saturation in colored graphs.

Abstract

For a fixed graph $F$ and an integer $t$, the \dfn{rainbow saturation number} of $F$, denoted by $sat_t(n,\mathfrak{R}(F))$, is defined as the minimum number of edges in a $t$-edge-colored graph on $n$ vertices which does not contain a \dfn{rainbow copy} of $F$, i.e., a copy of $F$ all of whose edges receive a different color, but the addition of any missing edge in any color from $[t]$ creates such a rainbow copy. Barrus, Ferrara, Vardenbussche and Wenger prove that $sat_t(n,\mathfrak{R}(P_\ell))\ge n-1$ for $\ell\ge 4$ and $sat_t(n,\mathfrak{R}(P_\ell))\le \lceil \frac{n}{\ell-1} \rceil \cdot \binom{\ell-1}{2}$ for $t\ge \binom{\ell-1}{2}$, where $P_\ell$ is a path with $\ell$ edges. In this short note, we improve the upper bounds and show that $sat_t(n,\mathfrak{R}(P_\ell))\le \lceil \frac{n}{\ell} \rceil \cdot \left({{\ell-2}\choose {2}}+4\right)$ for $\ell\ge 5$ and $t\ge 2\ell-5$.

Equations

$$sa{t}_t(n,\mathfrak{R}(F))=\min \{|E(G)|:|V(G)|=n,G\text{is (R(F),t)-saturated}\}.$$

Full text

Keywords: rainbow saturation number, edge-coloring, path

1 Introduction

Throughout this note, all graphs are simple, undirected, and finite. Throughout we use the terminology and notation of [8]. For a positive integer $t$, let $[t]$ denote the set $\{1,\ldots,t\}$. A $t$-edge-coloring of a graph $G$ is a function $f:E(G)\rightarrow[t]$, and a graph equipped with such a coloring is a ($t$-edge-colored graph).

A graph $G$ is called $F$-saturated if it is a maximal $F$-free graph. The classical saturation problem, first studied by Zykov [9] and Erdös, Hajnal and Moon [2], asks for the minimum number of edges in an $F$-saturated graph. For more results on saturation numbers, the reader should consult the excellent survey of Faudree, Faudree, and Schmitt [3]. A rainbow analog of this problem was recently introduced by Barrus, Ferrara, Vardenbussche and Wenger [1], where a $t$-edge-colored graph is defined to be rainbow $F$-saturated if it contains no rainbow copy of $F$, i.e., a copy of $F$ all of whose edges receive a different color, but the addition of any missing edge in any color creates such a rainbow copy, denoted by $(\mathfrak{R}(F),t)$-saturated. This minimum size of a $t$-edge-colored rainbow $F$-saturated graph, denoted by $sat_{t}(n,\mathfrak{R}(F))$, is the rainbow saturation number of $F$, i.e.,

[TABLE]

In [1], the authors proved some results on $sat_{t}(n,\mathfrak{R}(F))$ of various families of graphs including complete graphs, trees and cycles. Korándi [7] prove a conjecture of Barrus et al. on the rainbow saturation number of complete graphs. For more results on this topic we refer to [4, 5, 6]. Especially, Barrus et al. [1] proved the following results on paths.

Theorem 1.1** ([1]).**

(i). $sat_{t}(n,\mathfrak{R}(P_{\ell}))\geq n-1,\ \ell\geq 4$. (ii). $sat_{t}(n,\mathfrak{R}(P_{4}))=n-1,\ t\geq 8$. (iii). $sat_{t}(n,\mathfrak{R}(P_{\ell}))\leq\lceil\frac{n}{\ell-1}\rceil\cdot\binom{\ell-1}{2},\ t\geq\binom{\ell-1}{2}$.

In the proof of Theorem 1.1(iii), Barrus et al. use rainbow $K_{\ell-1}$ as construction components. Here we improve the upper bounds by changing the construction components, which yields Theorem 1.2.

Theorem 1.2**.**

For $\ell\geq 5$ and $t\geq 2\ell-5$, $sat_{t}(n,\mathfrak{R}(P_{\ell}))\leq\lceil\frac{n}{\ell}\rceil\cdot\left(\binom{\ell-2}{2}+4\right)$.

In Section 2, for all positive integers $\ell\geq 5$, we provide infinitely many $t$-edge-colored graphs with $sat_{t}(n,\mathfrak{R}(P_{\ell}))$ attaining the upper bounds. In Section 3, the proof of the main result is presented.

2 Construction

In this section, we construct several rainbow $P_{\ell}$-saturated graphs achieving the upper bounds in Theorem 1.2, for all positive integers $\ell\geq 5$.

Definition 2.1**.**

For any positive integer $\ell\geq 5$, let $V(K_{\ell-2})=\{v_{0},v_{1},\ldots,v_{\ell-3}\}$ and $c(v_{i}v_{j})=i+j$ as the edge coloring rule of $K_{\ell-2}$. Let $H$ be the graph obtained from $K_{\ell-2}$ by adding two vertices $\{v_{\ell-2},x\}$ and four edges $v_{\ell-2}v_{\ell-3},v_{\ell-2}v_{\ell-4},xv_{\ell-3},xv_{\ell-4}$. Set $c(v_{\ell-2}v_{\ell-3})=c(xv_{\ell-4})=2\ell-5$ amd $c(v_{\ell-2}v_{\ell-4})=c(xv_{\ell-3})=2\ell-6$. Let $G^{*}=kH$ for all positive integers $k\geq 2$.

In the following we will study some properties of $H$ and $G^{*}$.

Lemma 2.2**.**

The graph $H$ is $t$-proper-edge-colored, and $H$ does not contain any rainbow copy of $P_{\ell}$, where $t=2\ell-5$ and $\ell\geq 5$.

Proof.

By the edge coloring rule of $H$, for any $v\in H$, all edges incident to $v$ have different colors. Thus $H$ is $t$-proper-edge-coloring, and $t=\ell-2+\ell-3=2\ell-5$. Suppose $H$ contains a rainbow copy of $P_{\ell}$, denoted by $P_{\ell}^{\prime}$. For $|H|=\ell$, we have $x,v_{\ell-2}\in P_{\ell}^{\prime}$. Then there are at least two same colors in $c(P_{\ell}^{\prime})$. $\square$

For a graph $G$ and $v\in G$, let $P_{v}$ be a rainbow Hamilton path from $v$ of $G$ and $c(P_{v})$ be the color set of $E(P_{v})$. Let $G_{v}$ be the set of all such $P_{v}$ and $c(G_{v})=\bigcap c(P_{v})$.

Lemma 2.3**.**

For any positive integer $\ell\geq 5$, let $H^{*}=H\setminus x$ and $e\in E(\overline{G^{*}})$. Then $H^{*}+e$ contains a rainbow $P_{\ell}$ such that $c(e)$ is any color from $[t]$ and $e$ is incident to $v_{i}\in H^{*}$, where $i\notin\{\ell-4,\ell-3\}$.

Proof.

We only need to prove $c(H^{*}_{v})=\emptyset$ for any $v\in H^{*}\setminus\{v_{\ell-4},v_{\ell-3}\}$.

Suppose $\ell$ is odd. Assume $a\in N^{+}$, then $V(H^{*})=\{v_{0},v_{1},\ldots,v_{\ell-4},v_{\ell-3},v_{\ell-2}\}$. By Definition 2.1, $t=2\ell-5$ and $v_{\ell-2}$ is only adjacent to $v_{\ell-3},\ v_{\ell-4}$.

If $v=v_{0}$, for $H^{*}$, then let $P_{v_{0}}^{1}={v_{0},v_{1},v_{3},\ldots,v_{2a+1},\ldots,v_{\ell-4},v_{\ell-2},v_{\ell-3},\ldots,v_{2a},\ldots,v_{4},v_{2}}$ and $P_{v_{0}}^{2}=\{v_{0},v_{1},v_{2},v_{3},\ldots,v_{2a},v_{2a+1},\ldots,v_{\ell-4},v_{\ell-3},v_{\ell-2}\}$. Then all elements in $c(P_{v_{0}}^{1})$ are even except for $\{1,2\ell-5\}$ and all elements in $c(P_{v_{0}}^{2})$ are odd. So we have $c(H^{*}_{v_{0}})\subseteq\{1,2\ell-5\}$. Change the positions of a few vertices in $P_{v_{0}}^{2}$, we can get $P_{v_{0}}^{21}=\{v_{0},v_{2},v_{1},v_{3},\ldots,v_{\ell-5},v_{\ell-3},$ $v_{\ell-4},v_{\ell-2}\}$. There is no $\{1,2\ell-5\}$ in $c(P_{v_{0}}^{21})$. Thus $c(H^{*}_{v_{0}})=\emptyset$.

If $v=v_{2a}$, for $H^{*}$, then let $P_{v_{2a}}^{1}=\{v_{2a},v_{2a-2},\ldots,v_{4},v_{2},v_{0},v_{1},v_{3},\ldots,v_{\ell-4},v_{\ell-2},v_{\ell-3},\ldots,v_{2a+2}\}$ and $P_{v_{2a}}^{2}=\{v_{[}2a],v_{0},v_{1},v_{2},v_{3},\ldots,v_{2a-1},v_{2a+1},\ldots,v_{\ell-4},v_{\ell-3},v_{\ell-2}\}$. Then all elements in $c(P_{v_{2a}}^{1})$ are even except for $\{1,2\ell-5\}$ and all elements in $c(P_{v_{2a}}^{2})$ are odd except for $\{2a,4a\}$. So we have $c(H^{*}_{v_{2a}})\subseteq\{1,2\ell-5,2a,4a\}$. Change the positions of a few vertices in $P_{v_{2a}}^{2}$, we can get $P_{v_{2a}}^{3}=\{v_{2a},v_{2a-1},\ldots,v_{3},v_{1},v_{2},v_{0},v_{2a+2},v_{2a+1},v_{2a+3},v_{2a+4},\ldots,v_{\ell-5},v_{\ell-3},v_{\ell-4},v_{\ell-2}\}$. There is no $\{1,2a,4a,2\ell-5\}$ in $c(P_{v_{2a}}^{3})$. Thus $c(H^{*}_{v_{2a}})=\emptyset$.

If $v=v_{2a+1}$, for $H^{*}$, then let $P_{v_{2a+1}}^{1}=\{v_{2a+1},v_{2a-1},\ldots,v_{3},v_{1},v_{0},v_{2},v_{4},\ldots,v_{\ell-3},v_{\ell-2},v_{\ell-4},\ldots,$ $v_{2a+3}\}$ and $P_{v_{2a+1}}^{2}=\{v_{2a+1},v_{1},v_{0},v_{2},v_{3},\ldots,v_{2a},v_{2a+2},\ldots,v_{\ell-4},v_{\ell-3},v_{\ell-2}\}$. Then all elements in $c(P_{v_{2a+1}}^{1})$ are even except for $\{1,2\ell-5\}$ and all elements in $c(P_{v_{2a+1}}^{2})$ are odd except for $\{2,2a+2,4a+2\}$. So we have $c(H^{*}_{v_{2a+1}})\subseteq\{1,2\ell-5,2,2a+2,4a+2\}$. Change the positions of a few vertices in $P_{v_{2a+1}}^{2}$, we can get $P_{v_{2a+1}}^{3}=\{v_{2a+1},v_{2a},\ldots,v_{3},v_{1},v_{2},v_{0},v_{2a+2},v_{2a+3},\ldots,v_{\ell-3},$ $v_{\ell-4},v_{\ell-2}\}$ and $P_{v_{2a+1}}^{21}=\{v_{2a+1},v_{3},v_{2},v_{1},v_{0},v_{4},\ldots,v_{\ell-2}\}$. There is no $\{1,2,4a+2,2\ell-5\}$ in $c(P_{v_{2a+1}}^{3})$ and no $2a+2$ in $c(P_{v_{2a+1}}^{21})$. Thus $c(H^{*}_{v_{2a+1}})=\emptyset$.

If $v=v_{\ell-5}$, for $H^{*}$, then let $P_{v_{\ell-5}}^{1}=\{v_{\ell-5},v_{\ell-7},\ldots,v_{4},v_{2},v_{0},v_{1},v_{3},\ldots,v_{\ell-4},v_{\ell-2},v_{\ell-3}\}$ and $P_{v_{\ell-5}}^{2}=\{v_{\ell-5},v_{0},v_{1},v_{2},v_{3},\ldots,v_{\ell-6},v_{\ell-4},v_{\ell-3},v_{\ell-2}\}$. Then all elements in $c(P_{v_{\ell-5}}^{1})$ are even except for $\{1,2\ell-5\}$ and all elements in $c(P_{v_{\ell-5}}^{2})$ are odd except for $\{\ell-5,2\ell-10\}$. So we have $c(H^{*}_{v_{\ell-5}})\subseteq\{1,2\ell-5,\ell-5,2\ell-10\}$. Change the positions of a few vertices in $P_{v_{\ell-5}}^{2}$, we can get $P_{v_{\ell-5}}^{3}=\{v_{\ell-5},v_{\ell-6},\ldots,v_{3},v_{1},v_{2},v_{0},v_{\ell-3},v_{\ell-4},v_{\ell-2}\}$. There is no $\{1,2\ell-5,\ell-5,2\ell-10\}$ in $c(P_{v_{\ell-5}}^{3})$. Thus $c(H^{*}_{v_{\ell-5}})=\emptyset$.

If $v=v_{\ell-4},\ \ell\geq 7$, for $H^{*}$, then let $P_{v_{\ell-4}}^{1}=\{v_{\ell-4},v_{\ell-6},\ldots,v_{3},v_{1},v_{0},v_{2},v_{4},\ldots,v_{\ell-3},v_{\ell-2}\}$ and $P_{v_{\ell-4}}^{2}=\{v_{\ell-4},v_{1},v_{0},v_{2},v_{3},\ldots,v_{\ell-5},v_{\ell-3},v_{\ell-2}\}$. Then all elements in $c(P_{v_{\ell-4}}^{1})$ are even except for $\{1,2\ell-5\}$ and all elements in $c(P_{v_{\ell-4}}^{2})$ are odd except for $\{2,\ell-3,2\ell-8\}$. So we have $c(H^{*}_{v_{\ell-4}})\subseteq\{1,2\ell-5,2,\ell-3,2\ell-8\}$. Change the positions of a few vertices in $P_{v_{\ell-4}}^{2}$, we can get $P_{v_{\ell-4}}^{3}=\{v_{\ell-4},v_{\ell-2},v_{\ell-3},v_{0},v_{2},v_{1},v_{3},\ldots,v_{\ell-5}\}$ and $P_{v_{\ell-4}}^{31}=\{v_{\ell-4},v_{\ell-2},v_{\ell-3},v_{2},v_{0},v_{1},v_{3},\ldots,$ $v_{\ell-5}\}$. There is no $\{1,2,2\ell-8\}$ in $c(P_{v_{\ell-4}}^{3})$ and no $\ell-3$ in $c(P_{v_{\ell-4}}^{31})$. Thus $c(H^{*}_{v_{\ell-4}})=\{2\ell-5\}$.

If $v=v_{\ell-3},\ \ell\geq 7$, for $H^{*}$, then let $P_{v_{\ell-3}}^{1}=\{v_{\ell-3},v_{\ell-5},\ldots,v_{4},v_{2},v_{0},v_{1},v_{3},\ldots,v_{\ell-4},v_{\ell-2}\}$ and $P_{v_{\ell-3}}^{2}=\{v_{\ell-3},v_{0},v_{1},v_{2},\ldots,v_{\ell-5},v_{\ell-4},v_{\ell-2}\}$. Then all elements in $c(P_{v_{\ell-3}}^{1})$ are even except for $1$ and all elements in $c(P_{v_{\ell-3}}^{2})$ are odd except for $\{\ell-3,2\ell-6\}$. So we have $c(H^{*}_{v_{\ell-3}})\subseteq\{1,\ell-3,2\ell-6\}$. Change the positions of a few vertices in $P_{v_{\ell-3}}^{2}$, we can get $P_{v_{\ell-3}}^{21}=\{v_{\ell-3},v_{0},v_{2},v_{1},v_{3},\ldots,v_{\ell-2}\}$ and $P_{v_{\ell-3}}^{22}=\{v_{\ell-3},v_{2},v_{0},v_{1},v_{3},v_{4},\ldots,v_{\ell-2}\}$. There is no $1$ in $c(P_{v_{\ell-3}}^{21})$ and no $\ell-3$ in $c(P_{v_{\ell-3}}^{22})$. Thus $c(H^{*}_{v_{\ell-3}})=\{2\ell-6\}$.

If $v=v_{\ell-2}$, for $H^{*}$, then let $P_{v_{\ell-2}}^{1}=\{v_{\ell-2},v_{\ell-3},v_{\ell-5},\ldots,v_{4},v_{2},v_{0},v_{1},v_{3},\ldots,v_{\ell-4}\}$ and $P_{v_{\ell-2}}^{2}=\{v_{\ell-2},v_{\ell-3},v_{0},v_{1},v_{2},\ldots,v_{\ell-5},v_{\ell-4}\}$. Then all elements in $c(P_{v_{\ell-2}}^{1})$ are even except for $\{1,2\ell-5\}$ and all elements in $c(P_{v_{\ell-2}}^{2})$ are odd except for $\ell-3$. So we have $c(H^{*}_{v_{\ell-2}})\subseteq\{1,2\ell-5,\ell-3\}$. Change the positions of a few vertices in $P_{v_{\ell-2}}^{2}$, we can get $P_{v_{\ell-2}}^{21}=\{v_{\ell-2},v_{\ell-3},v_{0},v_{2},v_{1},v_{3},\ldots,$ $v_{\ell-4}\}$ and $P_{v_{\ell-2}}^{22}=\{v_{\ell-2},v_{\ell-4},v_{\ell-3},v_{2},v_{0},v_{1},v_{3},\ldots,v_{\ell-5}\}$. There is no $1$ in $c(P_{v_{\ell-2}}^{21})$ and no $\{\ell-3,2\ell-5\}$ in $c(P_{v_{\ell-2}}^{22})$. Thus $c(H^{*}_{v_{\ell-2}})=\emptyset$.

The proof is similar when $\ell$ is even. $\square$

Corollary 2.4**.**

The graph $H$ is an $(\mathfrak{R}(P_{\ell}),t)$-saturated graph for $\ell\geq 7$.

Proof.

By Lemma 2.3, for any $e\in E(\bar{H})$, then $H+e$ contains a rainbow $P_{\ell}$ such that $c(e)$ is any color from $[t]$ and $e$ is incident to $v_{i}\in H^{*}$, where $i\notin\{\ell-4,\ell-3\}$. By Definition 2.1, $c(v_{\ell-2}v_{\ell-3})=2\ell-5=c(xv_{\ell-4})$ and $c(v_{\ell-2}v_{\ell-4})=2\ell-6=c(xv_{\ell-3})$. Then we have $c(H_{x})=c(H_{v_{\ell-2}})=\emptyset$.

If $\ell\geq 7$, then $H+e$ contains a rainbow $P_{\ell}$ also holds for $e$ which is incident to $v_{\ell-4},v_{\ell-3}$ by Definition 2.1.

Together with Lemma 2.2, $H$ is $(\mathfrak{R}(P_{\ell}),t)$-saturated for $\ell\geq 7$. $\square$

Proposition 2.5**.**

For $\ell=5$, $G^{*}$ in Definition 2.1 is $(\mathfrak{R}(P_{5}),5)$-saturated and $|E(G^{*})|=\frac{7n}{5}$.

Proof.

Similar as the proof in Lemma 2.3, $c(H_{v_{0}})=\emptyset,\ c(H_{v_{3}})=c(H_{x})=\emptyset$.

If $v=v_{1}$, then there are only $P_{v_{1}}^{1}=\{v_{1},v_{0},v_{2},v_{3}\}$ and $P_{v_{1}}^{2}=\{v_{1},v_{3},v_{2},v_{0}\}$ in $H^{*}$. Then $c(H^{*}_{v_{1}})=\{2,5\}$.

If $v=v_{2}$, then there are only $P_{v_{2}}^{1}=\{v_{2},v_{0},v_{1},v_{3}\}$ and $P_{v_{2}}^{2}=\{v_{2},v_{3},v_{1},v_{0}\}$ in $H^{*}$. Then $c(H^{*}_{v_{2}})=\{1,4\}$.

For $c(v_{2}x)=4,\ c(v_{1}x)=5$, we have $c(H_{v_{1}})=\{2\}$ and $c(H_{v_{2}})=\{1\}$.

For $H^{1},H^{2}\subset G^{*}$, let $e_{1}=v_{1}^{1}v_{1}^{2}\in E(\overline{G^{*}})$ and $c(e_{1})=2$. We can get a rainbow $P_{5}=\{v_{2}^{1},v_{3}^{1},v_{1}^{1},v_{1}^{2},v_{2}^{2}\}$ in $G^{*}+e_{1}$.

Let $e_{2}=v_{2}^{1}v_{2}^{2}\in E(\overline{G^{*}})$ and $c(e_{2})=1$, we can get a rainbow $P_{5}=\{v_{1}^{1},v_{3}^{1},v_{2}^{1},v_{2}^{2},v_{1}^{2}\}$ in $G^{*}+e_{2}$.

Let $e_{3}=v_{1}^{1}v_{2}^{2}\in E(\overline{G^{*}})$ and $c(e_{3})\in\{1,2\}$, we can get a rainbow $P_{5}=\{v_{2}^{1},v_{3}^{1},v_{1}^{1},v_{2}^{2},v_{1}^{2}\}$ in $G^{*}+e_{3}$.

Therefore, $G^{*}$ is $(\mathfrak{R}(P_{5}),5)$-saturated and $|E(G^{*})|=\frac{7n}{5}$. $\square$

Proposition 2.6**.**

For $\ell=6$, $G^{*}$ in Definition 2.1 is $(\mathfrak{R}(P_{6}),7)$-saturated and $|E(G^{*})|=\frac{5n}{3}$.

Proof.

Similar as in the proof of Lemma 2.3, $c(H_{v_{0}})=\emptyset,\ c(H_{v_{1}})=\emptyset,\ c(H_{v_{4}})=c(H_{x})=\emptyset$.

If $v=v_{2}$, then there are only $P_{v_{2}}^{1}=\{v_{2},v_{0},v_{1},v_{3},v_{4}\}$, $P_{v_{2}}^{2}=\{v_{2},v_{4},v_{3},v_{0},v_{1}\}$ and $P_{v_{2}}^{3}=\{v_{2},v_{4},v_{3},v_{1},v_{0}\}$ in $H^{*}$. Then $c(H^{*}_{v_{2}})=\{1,7\}$.

If $v=v_{3}$, then there are only $P_{v_{3}}^{1}=\{v_{3},v_{1},v_{0},v_{2},v_{4}\}$, $P_{v_{3}}^{2}=\{v_{3},v_{4},v_{2},v_{1},v_{0}\}$ and $P_{v_{3}}^{3}=\{v_{3},v_{4},v_{2},v_{0},v_{1}\}$ in $H^{*}$. Then $c(H^{*}_{v_{3}})=\{1,6\}$.

For $c(v_{2}x)=7,\ c(v_{3}x)=6$, we have $c(H_{v_{2}})=\{1\}=c(H_{v_{3}})$.

For $H^{1},H^{2}\subset G^{*}$, let $e_{1}=v_{2}^{1}v_{2}^{1}\in E(\bar{G^{*}})$ and $c(e_{1})=1$. We can get a rainbow $P_{6}=\{v_{0}^{1},v_{3}^{1},v_{4}^{1},v_{2}^{1},v_{2}^{2},v_{0}^{2}\}$ in $G^{*}+e_{1}$.

Let $e_{2}=v_{3}^{1}v_{3}^{2}\in E(\bar{G^{*}})$ and $c(e_{2})=1$, we can get a rainbow $P_{6}=\{v_{1}^{1},v_{2}^{1},v_{4}^{1},v_{3}^{1},v_{3}^{2},v_{1}^{2}\}$ in $G^{*}+e_{2}$.

Let $e_{3}=v_{2}^{1}v_{3}^{2}\in E(\bar{G^{*}})$ and $c(e_{3})=1$, we can get a rainbow $P_{6}=\{v_{0}^{1},v_{3}^{1},v_{4}^{1},v_{2}^{1},v_{3}^{2},v_{1}^{2}\}$ in $G^{*}+e_{3}$.

Therefore, $G^{*}$ is $(\mathfrak{R}(P_{6}),7)$-saturated and $|E(G^{*})|=\frac{5n}{3}$. $\square$

From Corollary 2.4, Propositions 2.5 and 2.6, we get the following theorem.

Theorem 2.7**.**

The graph $G^{*}$ is an $(\mathfrak{R}(P_{\ell}),t)$-saturated graph and $|E(G^{*})|=\lceil\frac{n}{\ell}\rceil\cdot(\binom{\ell-2}{2}+4)$, where $t=2\ell-5$ and $\ell\geq 5$.

3 Proof of Theorem 1.2

Let $G$ be an $(\mathfrak{R}(P_{\ell}),t)$-saturated graph with $n$ vertices. We present the proof by considering the following cases.

Case 1. $\ell=5$.

If $t=5$ and $n\equiv 0\ (mod\ 5)$, then all components of $G$ are $H$ in Definition 2.1, where $|H|=5$, i.e., $G=G^{*}$, which yields $|E(G)|=\frac{7n}{5}$.

If $t=6$ and $n\equiv 1\ (mod\ 5)$, then all components of $G$ are $H$ except for two rainbow triangles, which yields $|E(G)|=\frac{7n}{5}-\frac{12}{5}$.

If $t=5$ and $n\equiv 2\ (mod\ 5)$, then all components of $G$ are $H$ except for an edge $e$ that $c(e)=3$, which yields $|E(G)|=\frac{7n}{5}-\frac{9}{5}$.

If $t=5$ and $n\equiv 3\ (mod\ 5)$, then all components of $G$ are $H$ except for a rainbow triangle, which yields $|E(G)|=\frac{7n}{5}-\frac{6}{5}$.

If $t=5$ and $n\equiv 4\ (mod\ 5)$, then all components of $G$ are $H$ except for a rainbow $K_{4}$, which yields $|E(G)|=\frac{7n}{5}+\frac{2}{5}$.

Thus $sat_{t}(n,\mathfrak{R}(P_{5}))\leq\lceil\frac{7n}{5}\rceil$.

Case 2. $\ell=6$.

If $t=7$ and $n\equiv 0\ (mod\ 6)$, then all components of $G$ are $H$ in Definition 2.1, where $|H|=6$, i.e., $G=G^{*}$, which yields $|E(G)|=\frac{5n}{3}$.

If $t=7$ and $n\equiv 1\ (mod\ 6)$, then all components of $G$ are $H$ except for a rainbow triangle and a rainbow $K_{4}$, which yields $|E(G)|=\frac{5n}{3}-\frac{8}{3}$.

If $t=7$ and $n\equiv 2\ (mod\ 6)$, then all components of $G$ are $H$ except for an edge $e$ that $c(e)=4$, then $|E(G)|=\frac{5n}{3}-\frac{7}{3}$.

If $t=7$ and $n\equiv 3\ (mod\ 6)$, then all components of $G$ are $H$ except for a rainbow triangle, which yields $|E(G)|=\frac{5n}{3}-2$.

If $t=7$ and $n\equiv 4\ (mod\ 6)$, then all components of $G$ are $H$ except for a rainbow $K_{4}$, which yields $|E(G)|=\frac{5n}{3}-\frac{2}{3}$.

If $t=7$ and $n\equiv 5\ (mod\ 6)$, then all components of $G$ are $H$ except for two rainbow $K_{4}$ and a rainbow triangle, which yields $|E(G)|=\frac{5n}{3}-\frac{10}{3}$.

Thus $sat_{t}(n,\mathfrak{R}(P_{6}))\leq\lceil\frac{5n}{3}\rceil$.

Case 3. $\ell\geq 7$.

If $t=2\ell-5$ and $n\equiv 0\ (mod\ \ell)$, then all components of $G$ are $H$ in Definition 2.1, where $|H|=\ell$, i.e., $G=G^{*}$, which yields $|E(G)|=\frac{n}{\ell}\cdot(\binom{\ell-2}{2}+4)$.

If $t=2\ell-5$ and $n\equiv 1\ (mod\ \ell)$, then all components of $G$ are $H$ except for a vertex $v$, which yields $|E(G)|=\frac{n-1}{\ell}\cdot(\binom{\ell-2}{2}+4)$.

If $t=2\ell-5$ and $n\equiv 2\ (mod\ \ell)$, then all components of $G$ are $H$ except for an edge $e$, where $c(e)\in[t]$, which yields $|E(G)|=\frac{n-2}{\ell}\cdot(\binom{\ell-2}{2}+4)+1$.

If $t=2\ell-5$ and $n\equiv 3\ (mod\ \ell)$, then all components of $G$ are $H$ except for a rainbow triangle, which yields $|E(G)|=\frac{n-3}{\ell}\cdot(\binom{\ell-2}{2}+4)+3$.

If $t=2\ell-5$ and $n\equiv 4\ (mod\ \ell)$, then all components of $G$ are $H$ except for a rainbow $K_{4}$, which yields $|E(G)|=\frac{n-4}{\ell}\cdot(\binom{\ell-2}{2}+4)+6$.

$\ldots$

If $t=2\ell-5$ and $n\equiv a\ (mod\ \ell)$, then all components of $G$ are $H$ except for a rainbow $K_{a}$, which yields $|E(G)|=\frac{n-a}{\ell}\cdot(\binom{\ell-2}{2}+4)+\binom{a}{2}$. It is easy to check $e_{a}\leq\lceil\frac{n}{\ell}\rceil\cdot(\binom{\ell-2}{2}+4)$, since $1\leq a\leq\ell-1$ and then $1\leq a\leq\sqrt{(\ell-3)^{2}+8}$.

From all the above cases, we have for $\ell\geq 5$ and $t\geq 2\ell-5$, $sat_{t}(n,\mathfrak{R}(P_{\ell}))\leq\lceil\frac{n}{\ell}\rceil\cdot\left(\binom{\ell-2}{2}+4\right)$. The proof is thus complete. $\square$

Acknowledgements. Shujuan Cao is partially supported by the National Natural Science Foundation of China (No. 11801412 ).