How Ramsey theory can be used to solve Harary's problem for $K_{2,k}$

TL;DR

This paper explores how Ramsey theory can be applied to determine upper bounds for Ramsey numbers involving the bipartite graph $K_{2,k}$, extending previous results and providing sharp bounds for specific graph classes.

Contribution

It introduces new bounds for $r(K_{2,k},G)$ and generalizes Harary's problem for these graphs using Ramsey theory techniques.

Findings

Established that $r(C_4,G) \,\leq\, kq+1$ for certain graphs G.

Proved that equality holds when $G \cong qK_2$ or $K_3$.

Derived bounds for $r(C_4,G)$ involving the number of vertices and edges.

Abstract

Harary's conjecture $r(C_3,G)\leq 2q+1$ for every isolated-free graph G with $q$ edges was proved independently by Sidorenko and Goddard and Klietman. In this paper instead of $C_3$ we consider $K_{2,k}$ and seek a sharp upper bound for $r(K_{2,k},G)$ over all graphs $G$ with $q$ edges. More specifically if $q\geq 2$, we will show that $r(C_4,G)\leq kq+1$ and that equality holds if $G \cong qK_2$ or $K_3$. Using this we will generalize this result for $r(K_{2,k},G)$ when $k>2$. We will also show that for every graph $G$ with $q \geq 2$ edges and with no isolated vertices, $r(C_4, G) \leq 2p+ q - 2$ where $p=|V(G)|$ and that equality holds if $G \cong K_3$.

Full text

Introduction

At a meeting held at Kent State University in 1980, Harary posed the general problem of determining the relationship between $r(H,G)$ and the sizes (number of edges) of the given graphs. He conjectured that $r(K_{3},G)\leqslant 2q+1$ for every isolate-free graph $G$ with $q$ edges. This bound is sharp since $r(K_{3},T)=2q+1$ for any tree $T$ with $q$ edges; also $r(K_{3},qK_{2})=2q+1$. Harary’s conjecture was subsequently proved independently by Sidorenko [11, 12] and by Goddard and Kleitman [6]. More generally, we can take $H$ to be any fixed graph and seek a sharp upper bound for $r(H,G)$ over all graphs $G$ with $q$ edges. In this paper we deal with the case $H=K_{2,k}$ (with $k\geqslant 2$) and show that these bounds are sharp. We determine all graphs $G$ where the bound is achieved. Prior to this through private communication it is known that the main Theorem had been proved by Bollabás and Szemerédi, using results in extremal graph theory.

1 An upper bound to Ramsey number $r(C_{4},G)$

Theorem 1**.**

For every graph $G$ with $|E(G)|=q\geqslant 2$ edges and with no isolated vertices, $r(C_{4},G)\leqslant 2q+1$. Equality holds if $G\cong qK_{2}$ or $K_{3}$.

Proof.

As we know that $r(C_{4},P_{3})=4$, $r(C_{4},K_{3})=7$, $r(C_{4},C_{4})=6$, $r(C_{4},3K_{2})=7$, $r(C_{4},K_{1,3})=6$, $r(C_{4},K_{2}\cup P_{3})=6$, $r(C_{4},2P_{3})=7$, $r(C_{4},2K_{2})=5$, $r(C_{4},K_{1,4})=7$, $r(C_{4},K_{2}\cup C_{3})=7$, $r(C_{4},K_{2}\cup K_{1,3})=7$, $r(C_{4},2K_{2}\cup P_{3})=8$, $r(C_{4},K_{1,3}+e)=7$, $r(C_{4},T_{3})=6$ (here $T$ represents the tree on 5 vertices containing exactly one vertex of degree 3), $r(C_{4},4K_{2})=9$ (see [2, 3, 10]). Thus the result holds for $q\in\{2,3,4\}$, with equality for $q\in\{2,4\}$ corresponding to $G\cong qK_{2}$; with equality for $q=3$ corresponding to $G\cong K_{3}$ and $G\cong 3K_{2}$. Let $G$ be a graph with $q\geqslant 4$ edges and no isolated vertices. If $\Delta(G)=1$ then $G\cong qK_{2}$. We have $r(C_{4},qK_{2})\leqslant 2(q-2)+r(C_{4},2K_{2})=2q+1$, and $r(C_{4},qK_{2})\geqslant 2q+1$ as a consequence of the two-coloring of $E(K_{2q})$ in which $R\cong K_{1,2q-1}$. Thus $r(C_{4},qK_{2})=2q+1$ for $q>1$. Without loss of generality, we can assume $G$ is connected since $r(C_{4},G_{1}\cup G_{2})\leqslant r(C_{4},G_{1})+r(C_{4},G_{2})-1$.

Case 1 If $\Delta(G)=2$

Then $G$ a path or a cycle. Using known results, $r(C_{4},P_{n})\leqslant r(C_{4},C_{n})\leqslant n+2$ (see [5, 6, 8, 9]) it follows that $r(C_{4},G)\leqslant 2q$ for all such graphs $G$ with $q\geqslant 2$ edges and maximum degree $\Delta(G)=2$.

Case 2 If $\Delta(G)\geqslant 3$ and $\delta(G)>1$

Given a vertex $v$ in $G$ of degree $\Delta(G)$, let $H=G\setminus v$ and let $N_{v}$, denote the neighborhood of $v$ in $G$. By induction, then for any isolated vertex free graph $H^{\prime}$ obtained from $G$ by removing $q^{\prime}$ edges, we get that $r(C_{4},H^{\prime})\leqslant 2(q-q^{\prime})+1$.

In the first scenario, suppose that $(R,B)$ is a two-coloring of $E(K_{2q})$ in which there is no red $C_{4}$ and no blue copy of $G$. We claim that $\delta(R)\geqslant 2\Delta(G)-1$.

w$$X=N_{G}(v)Blue copy of graph H \deg_{R}(w)\leqslant 2\Delta(G)-2$$|X|=\Delta(G)-2

*Figure 1: If $w$ is a vertex of with degree $\leqslant 2\Delta(G)-2$ in $R$ *

If $w$ is a vertex of with degree $\leqslant 2\Delta(G)-2$ in $R$, we may delete this vertex and its neighborhood in $R$ and still have at least $2q-(2\Delta(G)-1)=2(q-\Delta(G))+1$ vertices. Thus, there is a blue copy of $H$ in the two colored complete graph that remains after $w$ and its neighborhood in $R$ are deleted. In this copy let $X$ denote the vertex set that plays the role of $N_{v}$. Since $w$ is adjacent to each vertex of $X$ in $B$ there is a blue copy of $G$, so the claim that $\delta(R)\geqslant 2\Delta(G)-1$ is justified.

In the next scenario, since $\Delta(G)\geqslant 3$, we have $2\Delta(G)-1\geqslant\Delta(G)+1$, so $\delta(R)\geqslant\Delta(G)+1$. Delete an arbitrary vertex $w$ and exactly $\Delta(G)+1$ of its neighbors in $R$. Let $Y$ denote the set of $\Delta(G)+1$ neighbors chosen for deletion. Since $\Delta(G)\geqslant 3$, the complete graph that remains has at least $2q-(\Delta(G)+2)\geqslant 2(q-\Delta(G))+1$ vertices, so it must contain a blue copy of $H$. As before, let $X$ be the set that plays the role of $N_{v}$. Consider the edges between $Y$ and $X$. Since there is no blue copy of $G$, each vertex in $Y$ is adjacent in $R$ to at least one vertex of $X$. Since $|X|=\Delta(G)$ and $|Y|=\Delta(G)+1$ , there must be a vertex $x\in X$ adjacent in $R$ to two or more vertices of $Y$.

w$$X=N_{G}(v)Blue copy of graph H \deg_{R}(w)\geqslant 2\Delta(G)-1$$|X|=\Delta(G)$$|Y|=\Delta(G)+1$$\geqslant\Delta(G)+1

*Figure 2: If $w$ is a vertex of with degree $\geqslant 2\Delta(G)-1$ in $R$ *

Then $x$, $w$ and the two appropriate vertices of $Y$ yield a red $C_{4}$, a contradiction.

Case 3 If $\Delta(G)\geqslant 3$ and $\delta(G)=1$

It should be also noted that in the case $\Delta(G)\geqslant 3$ and $\delta(G)=1$ works though unless $G\setminus v$ has isolated vertices. However, connectivity of $G$ together with the inequality $r(C_{4},T_{q+1})\leqslant\max\{4;q+2;r(C_{4},K_{1,q}))\leqslant 2q+1$ (see [1, 4]) gives that $G$ is a connected graph which is not a tree.

Therefore, we are left with the case when $G$ has a vertex $v$ of degree $\Delta(G)$ such that $v$ is adjacent to $s$ pendent vertices of $G$, where $1\leqslant s\leqslant q-3$. Let $H$ be the graph obtained by removing the $s$ pendent vertices from $G$.

Suppose that $(R,B)$ is a two-coloring of $E(K_{2q})$ in which there is no red $C_{4}$ and no blue copy of $G$. We claim that that $\Delta(R)\leqslant s$. If $w$ is a vertex of degree $\geqslant s+1$, let $Y$ consist of any $s+1$ red neighbors of $w$. Define $X=Y\cup\{w\}$. By induction hypothesis, there is a blue copy of $H$ in the two colored complete graph that remains after the vertices of $Y$ are deleted. In this copy let $x$ denote the vertex set that plays the role of $N_{v}$. As before then $H$ can be extended to a blue $G$ as in order to avoid a red $C_{4}$, $s$ vertices of $Y$ will be forced to be adjacent to $x$ in blue. Therefore, $\Delta(R)\leqslant s$. Let $w$ be a vertex with degree $\Delta(R)\leqslant s$. Let $Y$ be a set containing $N_{R}(w)$ along with $2s-1-\Delta(R)$ other vertices distinct from $w$. Define $X=Y\cup\{w\}$.

w$$\deg_{R}(w)=\Delta(R)\leqslant s$$\deg_{B}(w)=2s-1-\Delta(R)\geqslant s-1In the blue copy of $H,x$play the role of v\in G.$$x

*Figure 3: If $w$ is a vertex of red degree at least two *

By induction hypothesis, there is a blue copy of $H$ in the two colored complete graph that remains after the vertices of $Y$ are deleted. If $s>1$ by the above argument, $x$ can be adjacent in blue to at most $s-2$ vertices of $Y$ and can be adjacent in red to at most $s$ vertices in $Y$(as $\Delta(R)\leqslant s$). A contradiction as $|Y|=2s-1$. If $s=1$ as $(w,x)$ is blue, we get a copy of $G$, a contradiction.

Thus $r(C_{4},G)\leqslant 2q$ for every graph $G$ with $q$ edges and with no isolated vertices other than $2K_{2}$, and the proof is complete. ∎

2 An upper bound to Ramsey number $r(C_{4},G)$ if $G$ is connected.

Theorem 2**.**

For every isloated vertex free graph $G$ with $q\geqslant 2$ edges and $p\geqslant 3$ vertices, $r(C_{4},G)\leqslant 2p+q-2$. Equality holds if $G\cong K_{3}$.

Proof.

It is easy to verify the theorem for $q\leqslant 4$ by the results of the previous section. If $\Delta(G)\leqslant 2$ then $G$ is a path or a cycle, and using known results, again it is easy to verify that the theorem is true in this case. So it suffices to show $r(C_{4},G)\leqslant q+2p-3$, if $q\geqslant 5$. Now assume $\Delta(G)\geqslant 3$. Given a vertex $v$ in $G$ of degree $\Delta(G)$, let $H=G\setminus v$ and let $N_{v}$, denote the neighborhood of $v$ in $G$. First assume that $H$ has no isolated vertices. By induction, then for any isolated vertex free graph $H^{\prime}$ obtained from $G$ by removing $q^{\prime}$ edges (and the $p^{\prime}$ corresponding vertices), we get that $r(C_{4},H^{\prime})\leqslant(q-q^{\prime})+2(p-p^{\prime})-3$. In particular, $r(C_{4},H)\leqslant(q-\Delta(G))+2(p-1)-3$. Suppose that $(R,B)$ is a two-coloring of $E(K_{q+2p-3})$ in which there is no red $C_{4}$ and no blue copy of $G$. then using a similar argument as in the last proof we would get a contradition for all possible cases expect the third case when $\Delta(G)\geqslant 3$ and $\delta(G)=1$.

In this case let $v\in G$ represent vertex of degree 1 in $G$, and $N_{v}$ consist of $v_{1}$. Clearly, $|\delta(R)|\geqslant 1$. First assume that, $|\Delta(R)|=1$. Let $u\in K_{2q}$ be a vertex with degree 1 and suppose it it adjacent to $w$ in red. Then the graph obtained by removing $u$ and $w$ from $R$ (say $K\cong K_{2(q-1)}$) by induction hypothesis will have a blue copy of $G\setminus v$ in it. As before let $X=\{x\}$ be the set that play the role of $N_{v}$. As $u$ is adjacent to all vertices of $K$ we will get a blue copy of $G$ in $K_{2q}$. Therefore, we may assume that $|\Delta(R)|>1$. Let $u\in K_{2q}$ be a vertex with at least two neighbors in red say $w_{1}$ and $w_{2}$. Let $Y=\{w_{1},w_{2}\}$. Then as before the complete graph obtained by removing $Y\cup\{u\}$ will contain a blue copy of $G\setminus v$ as illustrated in the following figure(since $H$ has $p-1$ vertices and $q-1$ edges).

Since $|N_{v}|=1$, $|Y|=2$ and there is no blue copy of $G$, the vertex $x$ must be adjacent in $R$ to the vertices $w_{1}$ and $w_{2}$. But then $uw_{1}xw_{2}x$ will yield a red $C_{4}$, a contradiction.

w$$X=\{x\}Blue copy of graph G \v\deg_{R}(w)\geqslant 2$$|Y|=\{w_{1},w_{2}\}$$w_{1}$$w_{2}$$x

*Figure 4: If $w$ is a vertex of red degree at least two *

Thus, $r(C_{4},G)\leqslant q+2p-3$ for every graph $G$ with $q$ edges and with no isolated vertices other than $K_{3}$, and the proof is complete. ∎

3 An Upper Bound for the Ramsey Number $r(K_{2,k},G)$ where $k\geqslant 2$

Lemma 3**.**

$r(K_{2,k},G)\leqslant kq+1$* if $G$ is a path, star or triangle.*

Proof.

We showed before that $r(K_{2,2},K_{1,q})=r(C_{4},K_{1,q})\leqslant 2q+1$ if $q\geqslant 2$. Also [7] gives $r(K_{2,k},K_{1,2})=r(K_{2,k},P_{3})\leqslant 2k+1$. And thus we can conclude the result is true for stars as $r(K_{2,k},K_{1,q})\leqslant r(B_{k},K_{1,q})\leqslant\max\{2q+1,\lfloor\frac{3}{2}(k+q-1)\rfloor+1\}\leqslant kq+1$ for $k\geqslant 3$ and $q\geqslant 3$. For paths and triangles the result follows directly from $r(K_{2,k},P_{q+1})\leqslant k+q+1$ [7] and $r(B_{k},B_{1})\leqslant 2k+3$ [13] respectively. ∎

Lemma 4**.**

For every isolated vertex free graph $G$ with $q\geqslant 2$ edges, $r(K_{2,k},G)\leqslant kq+1$ if $k\geqslant 2$.

Proof.

We will use induction on $q$. The result is true for $k=2$ or $q\in\{2,3\}$. Also without loss of generality $G$ is connected and the result is true if $G$ is a path, star or triangle (these follow from previous lemmas). Thus we can restrict our attention to $G$ such that $G$ is not a path, star or triangle satisfying $\Delta(G)>1$. Using the argument in Theorem 1:case 3, we may assume that $\delta>1$(by considering the two cases, $\Delta(R)\leqslant s+k-2$ and $\Delta(R)>s+k-2$). Thus, given a vertex in $G$ of degree $\Delta(G)$, let $H=G\setminus v$ and let $N_{v}$ denote the neighborhood of $v$ in $G$. Then $H$ has $q-\Delta$ edges. Suppose that $(R,B)$ is a two-coloring of $E(K_{kq+1}$) in which there is no red $K_{2,k}$ and no blue copy of $G$. We claim that that $\delta(R)\geqslant(k-1)\Delta+1$. If $w$ is a vertex of with degree $\leqslant(k-1)\Delta$ in $R$, we may delete this vertex and its neighborhood in $R$ and still have at least $kq+1-((k-1)\Delta+1)\geqslant\max\{k(q-\Delta)+1,p-1\}$ vertices(as $G$ is not a star). Thus there is a blue copy of $H$ in the two-colored complete graph that remains after $w$ and its neighborhood in $R$ are deleted. In this copy let $X$ denote the vertex set that plays the role of $N_{v}$. Since $w$ is adjacent to each vertex of $X$ in $B$ there is a blue copy of $G$, so the claim that $\delta(R)\geqslant(k-1)\Delta+1$ is justified. Delete an arbitrary vertex $w$ and exactly $(k-1)\Delta+1$ of its neighbors in $R$. Let $Y$ denote the set of $(k-1)\Delta+1$ neighbors chosen for deletion. As before the complete graph that remains has at least $\max\{k(q-\Delta)+1,p-1\}$ vertices, so it must contain a blue copy of $H$. As before, let $X$ be the set that plays the role of $N_{v}$. Consider the edges between $Y$ and $X$. Since there is no blue copy of $G$, each vertex in $Y$ is adjacent in $R$ to at least one vertex of $X$. Since $|X|=\Delta$ and there must be a vertex $x\in X$ adjacent in $R$ to $k$ or more vertices of $Y$. Then $x$ and the $k$ appropriate vertices of $Y$ yield a red $K_{2,k}$, a contradiction. ∎

Theorem 5**.**

For every graph $G$ with no isolated vertices, $r(K_{2,k},G)\leqslant kq+2$ if $k\geqslant 3$ and equality holds if $G\cong K_{2}$.

Proof.

This result follows from the previous lemma together $r(K_{2,k},K_{2})=k+2$. ∎