Gallai-Ramsey numbers for monochromatic $K_{4}^{+}$ or $K_{3}$

All graphs considered in this paper are finite, simple and undirected. For a graph $G$, we use $|G|$ to denote the number of vertices of $G$, say the order of $G$. The complete graph of order $n$ is denoted by $K_{n}$. For a subset $S\subseteq V(G)$, let $G[S]$ be the subgraph of $G$ induced by $S$. For two disjoint subsets $A$ and $B$ of $V(G)$, $E_{G}(A,B)=\{ab\in E(G)\leavevmode\nobreak\ |\leavevmode\nobreak\ a\in A,b\in B\}$. Let $G_{1}=(V_{1},E_{1})$ and $G_{2}=(V_{2},E_{2})$ be two graphs. The union of $G_{1}$ and $G_{2}$, denoted by $G_{1}+G_{2}$, is the graph with the vertex set $V_{1}\bigcup V_{2}$ and the edge set $E_{1}\bigcup E_{2}$. The join of $G_{1}$ and $G_{2}$, denoted by $G_{1}\vee G_{2}$, is the graph obtained from $G_{1}+G_{2}$ by adding all edges joining each vertex of $G_{1}$ and each vertex of $G_{2}$. For any positive integer $k$, we write $[k]$ for the set $\{1,2,\cdots,k\}$. An edge coloring of a graph is called monochromatic if all edges are colored by the same color. An edge-colored graph is called rainbow if no two edges are colored by the same color.

Given graphs $H_{1}$ and $H_{2}$, the classical Ramsey number $R(H_{1},H_{2})$ is the smallest integer $n$ such that for any 2-edge coloring of $K_{n}$ with red and blue, there exists a red copy of $H_{1}$ or a blue copy of $H_{2}$. A sharpness example of the Ramsey number $R(H_{1},H_{2})$, denoted by $C_{(H_{1},H_{2})}$, is a 2-edge colored $K_{R(H_{1},H_{2})-1}$ with red and blue such that there are neither red copies of $H_{1}$ nor blue copies of $H_{2}$. Given graphs $H_{1},H_{2},\cdots,H_{k}$, the multicolor Ramsey number $R(H_{1},H_{2},\cdots,H_{k})$ is the smallest positive integer $n$ such that for every $k$-edge colored $K_{n}$ with the color set $[k]$, there exists some $i\in[k]$ such that $K_{n}$ contains a monochromatic copy of $H_{i}$ colored by $i$. The multicolor Ramsey number is an obvious generalization of the classical Ramsey number. When $H=H_{1}=\cdots=H_{k}$, we simply denote $R(H_{1},\cdots,H_{k})$ by $R_{k}(H)$. The problem about computing Ramsey numbers is notoriously difficult. For more information on classical Ramsey number, we refer the readers to  [5, 6, 10]. In this paper, we study Ramsey number in Gallai-coloring. A Gallai-coloring is an edge coloring of a complete graph with no rainbow triangle. Gallai-coloring naturally arises in several areas including: information theory [14]; the study of partially ordered sets, as in Gallai’s original paper [9] (his result was restated in [12] in the terminology of graphs); and the study of perfect graphs [2]. More information on this topic can be found in [7, 8]. A Gallai $k$-coloring is a Gallai-coloring that uses at most $k$ colors. Given a positive integer $k$ and graphs $H_{1},H_{2},\cdots,H_{k}$, the Gallai-Ramsey number $gr_{k}(K_{3}:H_{1},H_{2},\cdots,H_{k})$ is the smallest integer $n$ such that every Gallai $k$-colored $K_{n}$ contains a monochromatic copy of $H_{i}$ in color $i$ for some $i\in[k]$. Clearly, $gr_{k}(K_{3}:H_{1},H_{2},\cdots,H_{k})\leq R(H_{1},H_{2},\cdots,H_{k})$ for any $k$ and $gr_{2}(K_{3}:H_{1},H_{2})=R(H_{1},H_{2})$. When $H=H_{1}=\cdots=H_{k}$, we simply denote $gr_{k}(K_{3}:H_{1},H_{2},\cdots,H_{k})$ by $gr_{k}(K_{3}:H)$. When $H=H_{1}=\cdots=H_{s}(0\leq s\leq k)$ and $G=H_{s+1}=\cdots=H_{k}$, we use the following shorthand notation

The authors in [15] and [17] determined the Gallai-Ramsey number $gr_{k}(K_{3}:s\cdot K_{4},(k-s)\cdot K_{3})$ and $gr_{k}(K_{3}:s\cdot K_{3},(k-s)\cdot C_{4})$, respectively. In this paper, we investigate the Gallai-Ramsey number $gr_{k}(K_{3}:s\cdot K_{4}^{+},(k-s)\cdot K_{3}),$ where $K_{4}^{+}=K_{1}\vee(K_{3}+K_{1})$. We will prove the following result in Section 2.

When $s=k$ and $s=0$, we can get the following Theorem 2 and Theorem 3 from Theorem 1, respectively. So Theorem 2 and Theorem 3 can be seen as two corollaries obtained from Theorem 1.

To prove Theorem 1, the following theorem is useful.

Given a Gallai-partition $(V_{1},V_{2},\cdots,V_{t})$ of a Gallai-colored complete graph $G$, let $H_{i}=G[V_{i}]$, $h_{i}\in V_{i}$ for each $i\in[t]$ and $R=G\left[\{h_{1},h_{2},\cdots,h_{t}\}\right]$. Then $R$ is said to be the reduced graph of $G$ corresponding to the given Gallai-partition. By Theorem 4, all edges in the reduced graph $R$ are colored at most two colors.

First, recall some known classical Ramsey numbers of $K_{4}^{+}$ and $K_{3}$.

For the sake of notation, now we define functions $f_{i}$ for $i\in[5]$ as follows.

Let

It is easy to check that

and the following inequations hold.

Now we prove Theorem 1.

Case a. If $i\leq s-2$, then construct $G_{i+2}$ by a blow-up of $Q_{2}$ colored by the two colors $i+1$ and $i+2$ on $G_{i}$. Then $G_{i+2}$ has no monochromatic copy $K_{4}^{+}$ in the first $i+2$ colors.

Case b. If $i=s-1$ and $k=s$, then construct $G_{i+1}$ by a blow-up of $K_{4}$ colored by the color $i+1$ on $G_{i}$. Then $G_{i+1}$ has no monochromatic copy $K_{4}^{+}$ in the first $i+1$ colors.

Case c. If $i=s-1$ and $k>s$, then construct $G_{i+2}$ by a blow-up of $Q_{3}$ colored by the two colors $i+1$ and $i+2$ on $G_{i}$. Then $G_{i+2}$ contains neither monochromatic copy of $K_{4}^{+}$ colored by one of the first $i+1$ colors nor monochromatic copy of $K_{3}$ colored by the color $i+2$.

Case d. If $i\geq s$ and $i=k-1$, then construct $G_{i+1}$ by a blow-up of $K_{2}$ colored by the color $i+1$ on $G_{i}$. Then $G_{i+1}$ contains neither monochromatic copy of $K_{4}^{+}$ colored by one of the first $s$ colors nor monochromatic copy of $K_{3}$ colored by one of the remaining $k-s$ colors.

Case e. If $i\geq s$ and $i\leq k-2$, then construct $G_{i+2}$ by a blow-up of $Q_{1}$ colored by the two colors $i+1$ and $i+2$ on $G_{i}$. Then $G_{i+2}$ contains neither monochromatic copy of $K_{4}^{+}$ colored by one of the first $s$ colors nor monochromatic copy of $K_{3}$ colored by one of the remaining $i+2-s$ colors.

By the above construction, it is clear that $G_{k}$ is Gallai $k$-colored and contains neither monochromatic copy of $K_{4}^{+}$ in any of the first $s$ colors nor monochromatic copy of $K_{3}$ in any of the remaining $k-s$ colors. If $s$ and $k-s$ are both even, then by Case a, we can first construct $G_{s}$ of order $17^{\frac{s}{2}}$. Next by Case e, we continue to construct $G_{s+2},G_{s+4},\cdots,G_{k}$. So we can get $|G_{k}|=f_{1}(k,s)$. If $s$ is even and $k-s$ is odd, then by Case a, we can first construct $G_{s}$ of order $17^{\frac{s}{2}}$. Next by Case e, we continue to construct $G_{s+2},G_{s+4},\cdots,G_{k-1}$ and we can get $|G_{k-1}|=17^{\frac{s}{2}}\cdot 5^{\frac{k-s-1}{2}}$. Finally by Case d, we can construct $G_{k}$ of order $f_{2}(k,s)$. Similarly, we can get that $|G_{k}|=f(k,s)$ in the remaining three cases. Therefore,

Now we prove that $gr_{k}(K_{3}:s\cdot K_{4}^{+},(k-s)\cdot K_{3})\leq f(k,s)+1$ by induction on $k+s$. Let $n=f(k,s)+1$ and $G$ be a Gallai $k$-colored complete graph of order $n$. The statement is trivial in the case that $k=1$. The statement holds in the case that $k=2$ by Lemma 2.1. The statement holds in the case that $s=0$ by Theorem 3. So we can assume that $s\geq 1$ and $k\geq 3$, and the statement holds for any $s^{{}^{\prime}}$ and $k^{{}^{\prime}}$ such that $s^{{}^{\prime}}+k^{{}^{\prime}}<s+k$. Then $f(k,s)\geq 16$ and $n\geq 17$. Suppose, to the contrary, that $G$ contains neither monochromatic copy of $K_{4}^{+}$ in any of the first $s$ colors nor monochromatic copy of $K_{3}$ in any of the remaining $k-s$ colors. By Theorem 4, there exists a Gallai-partition of $V(G)$. Choose a Gallai-partition with the smallest number of parts, say $(V_{1},V_{2},\cdots,V_{t})$ and let $H_{i}=G[V_{i}]$ for $i\in[t]$. Then $t\geq 2$. Choose one vertex $h_{i}\in V_{i}$ and set $R=G[\{h_{1},h_{2},\cdots,h_{t}\}]$. Then $R$ is a reduced graph of $G$ colored by at most two colors.

We first consider the case that $t=2$. W.L.O.G, suppose that the color on the edges between two parts is red. If red is in the last $k-s$ colors (so $s<k$), then $H_{1}$ and $H_{2}$ both have no red edges, otherwise, there exists a red $K_{3}$, a contradiction. Hence $|H_{i}|\leq gr_{k-1}(K_{3}:s\cdot K_{4}^{+},(k-s-1)\cdot K_{3})-1$ for each $i\in[2]$. By the induction hypothesis and (1),

a contradiction. If red is in the first $s$ colors (W.L.O.G, suppose that red is the $s$th color), then $H_{1}$ has no red edges or $H_{2}$ has no red edges, otherwise, there exists a red $K_{4}^{+}$ in $G$, a contradiction. W.L.O.G, suppose that $H_{1}$ has no red edges. Then $|H_{1}|\leq gr_{k-1}(K_{3}:(s-1)\cdot K_{4}^{+},(k-s)\cdot K_{3})-1$. If $H_{2}$ has a red $K_{3}$, then we can find a red $K_{4}^{+}$ in $G$, a contradiction. Hence $H_{2}$ contains no monochromatic copy of $K_{4}^{+}$ in any of the first $s-1$ colors and no monochromatic copy of $K_{3}$ in any of the remaining $(k-s+1)$ colors. Hence $|H_{2}|\leq gr_{k}(K_{3}:(s-1)\cdot K_{4}^{+},(k-s+1)\cdot K_{3})-1$. By the induction hypothesis and (5), we get that

a contradiction.

Then we can assume that $t\geq 3$ and since $t$ is smallest, $R$ is colored by exactly two colors. Suppose that the two colors appeared in the Gallai-partition $(V_{1},V_{2},\cdots,V_{t})$ are red and blue, that is, the reduced graph $R$ is colored by red and blue. If the edge $h_{i}h_{j}$ is red in $R$, then $h_{j}$ is said to be a red neighbor of $h_{i}$. Let $N_{r}(h_{i})=\{$all red neighbors of $h_{i}\}$, named the red neighborhood of $h_{i}$, and $N_{b}(h_{i})$ be the blue neighborhood of $h_{i}$ symmetrically. For each vertex $h_{i}\in V(R)$ $(i\in[t])$, let $d_{r}(h_{i})=|N_{r}(h_{i})|$, named the red degree of $h_{i}$ in $R$, and $d_{b}(h_{i})$ be the blue degree of $h_{i}$ in $R$. Then $d_{r}(h_{i})+d_{b}(h_{i})=t-1$. If there exists one part (say $V_{1}$) such that all edges joining $V_{1}$ to the other parts are colored by the same color, then we can find a new Gallai-partition with two parts $(V_{1},V_{2}\bigcup\cdots\bigcup V_{t})$, which contradicts with that $t$ is smallest. It follows that $t\neq 3$ and the following fact holds.

Now we can assume that $t\geq 4$. We consider the following three cases.

This means that there is neither red $K_{3}$ nor blue $K_{3}$ in $G$. Since $R(K_{3},K_{3})=6$, we have $t\leq 5$. By Fact 1, $H_{i}$ contains neither red edges nor blue edges for each $i$. Then $|H_{i}|\leq gr_{k-2}(K_{3}:s\cdot K_{4}^{+},(k-s-2)\cdot K_{3})-1$ for $1\leq i\leq t$. By the induction hypothesis and (2), we get that

a contradiction.

Now we only consider the case that red or blue are in the first $s$ colors in the following. Then we have the following Facts.

Otherwise, suppose that there is a red $K_{4}$ in $G[V_{j_{1}}\cup\cdots\cup V_{j_{p}}]$. If there exists one red edge in $E_{G}(V_{j_{1}}\cup\cdots\cup V_{j_{p}},V(G)\setminus(V_{j_{1}}\cup\cdots\cup V_{j_{p}}))$, then $G$ contains a red $K_{4}^{+}$, a contradiction. It follows that all edges in $E_{G}(V_{j_{1}}\cup\cdots\cup V_{j_{p}},V(G)\setminus(V_{j_{1}}\cup\cdots\cup V_{j_{p}}))$ are blue. Then $(V_{j_{1}}\cup\cdots\cup V_{j_{p}},V(G)\setminus(V_{j_{1}}\cup\cdots\cup V_{j_{p}}))$ is a new Gallai-partition which contradicts with that $t\geq 4$ and $t$ is smallest.

By Fact 1 and Fact 2, we have the following facts.

Let $I_{r}=\{h_{i}\in V(R)$ : $H_{i}$ contains at least one red edge$\}$ and $I_{b}=\{h_{i}\in V(R)$ : $H_{i}$ contains at least one blue edge$\}$. Clearly by Fact 5, if red is in the first $s$ colors, then the induced subgraph of $R$ by $I_{r}$ is a blue complete graph and if blue is in the first $s$ colors, then the induced subgraph of $R$ by $I_{b}$ is a red complete graph.

Suppose, to the contrary, that $h_{i}$, $h_{j}$$\in I_{b}\cap I_{r}$. We know that $h_{i}h_{j}$ is either red or blue. Then we can find either a red $K_{4}$ or a blue $K_{4}$ in $G[V_{i}\cup V_{j}]$, which contradicts with Fact 2.

Suppose that $|H_{i}|\geq 2$ for each $i\in[l]$ and $|H_{j}|=1$ for each $j$ with $l+1\leq j\leq t$. Then $l\geq 1$. Otherwise, $R=G$. Then $G$ is 2-edge colored, which contradicts with that $k\geq 3$. Let $p_{0}$ be the number of $H_{i}$($i\in[l]$) which contains neither red nor blue edges, $p_{1}$ the number of $H_{i}$ which contains either red or blue edges and $p_{2}$ the number of $H_{i}$ which contains both red and blue edges. Then $l=p_{0}+p_{1}+p_{2}$, $p_{2}=|I_{b}\cap I_{r}|$ and $p_{1}+p_{2}=|I_{b}\cup I_{r}|$.

W.L.O.G., suppose that red appears in the first $s$ colors, blue appears in the last $k-s$ colors and red is the $s$th color. This means that there is neither red $K_{4}^{+}$ nor blue $K_{3}$ in $G$. Then by Fact 1, we can get that every $H_{i}$ contains no blue edge, which implies that $p_{2}=0$. Since $R(K_{4}^{+},K_{3})=9$, we have that $t\leq 8$.

Otherwise, suppose that $p_{1}\geq 3$. Then W.L.O.G., suppose that $H_{1}$, $H_{2}$ and $H_{3}$ are three corresponding subgraphs which contain red edges. By Fact 5, $R[\{h_{1},h_{2},h_{3}\}]$ is a blue $K_{3}$. So $G$ has a blue $K_{3}$, a contradiction.

We now consider subcases based on the values of $l$ and $t$.

By Claim 1, $p_{1}\leq 2$. Since $p_{0}+p_{1}=l\leq t$ and by Claim 2, we have that

a contradiction.

Then by Claim 1 and Claim 2, we can get that

a contradiction.

Thus, if $t=6$, then $p_{1}\leq 1$. By Claim 2, we have that

a contradiction.

If $t\geq 7$, then by Claim 3 and Claim 2, we have that $p_{1}=0$ and

a contradiction. The proof of Case 2 is completed.

This means that there exists neither red $K_{4}^{+}$ nor blue $K_{4}^{+}$ in $G$. W.L.O.G., suppose that red and blue are the $(s-1)$th and $s$th color, respectively. Since $R(K_{4}^{+},K_{4}^{+})=18$, we know that $t\leq 17$. Since $s\geq 2$ and $k\geq 3$, $f(k,s)\geq 35$. First we prove some claims.

W.L.O.G., suppose, to the contrary, that $d_{r}(h_{i})\geq 9$. Since $R(K_{3},K_{4}^{+})=9$, then the subgraph of $R$ induced by $N_{r}(h_{i})$ contains either a red $K_{3}$ or a blue $K_{4}^{+}$. So $R$ contains a red $K_{4}^{+}$ or blue $K_{4}^{+}$. Thus $G$ contains also, a contradiction.

Suppose that $d_{r}(h_{i})\geq 4$. To the contrary, suppose that $H_{i}$ contains a red edge. If the induced subgraph $R[N_{r}(h_{i})]$ contains a red edge, say $h_{p}h_{q}$, then we find a red $K_{3}$ which is $h_{i}h_{p}h_{q}$, which contradicts with Fact 4. Otherwise, $R[N_{r}(h_{i})]$ contains a blue $K_{4}$. So $G[\bigcup_{h_{p}\in N_{r}(h_{i})}V_{p}]$ contains a blue $K_{4}$, which contradicts with Fact 2. So we have that $h_{i}\notin I_{r}$. The proof for blue is as same as the above one for red symmetrically.