text:
\section{Introduction}
Given a graph $H$, let $\hat{R}_r(H)$ be the minimum $m$ such that there exists a graph $G$ with $m$ edges such that in every $r$-coloring of $G$, there is a monochromatic copy of $H$.  When $r=2$, we drop the subscript and just write $\hat{R}(H)$.  We refer to $\hat{R}(H)$ as the \emph{size-Ramsey} number of $H$.

Let $P_n$ be the path with $n$ vertices. Erd\H{o}s \cite{Er} famously asked if $\hat{R}(P_n) / n \to \infty$ and $\hat{R}(P_n) / n^2 \to 0$. Beck \cite{B2} proved that in fact, $\hat{R}(P_n) \le 900n$ (for $n$ sufficiently large). The bound $900n$ was subsequently improved in \cite{B}, \cite{Bol}, \cite{DP1}, \cite{Let} and currently rests at $74n$ as proved by Dudek and Pra{\l}at in \cite{DP2}.

As for the lower bound, it is clear that $\hat{R}(P_n)> 2n-4$ since $P_n$ has $n-1$ edges.  Beck \cite{B2} proved $\hat{R}(P_n)\geq (9/4-o(1))n$, Bielak \cite{Bie} proved $\hat{R}(P_n)\geq 9n/4-3$,  Bollob\'as \cite{B} proved $\hat{R}(P_n)\geq (1+\sqrt{2}-o(1))n$, and finally Dudek and Pra{\l}at \cite{DP2} proved $\hat{R}(P_n)\geq 5n/2-15/2$.

The closest thing there is to a conjecture about the precise value of $\hat{R}(P_n)$ is Bollob\'as' \cite{B} comment, ``it would not be surprising if $\hat{R}(P_n)$ turned out to be about $8n$.'' It is not known what insight led to this comment, but together with the recent flurry of activity on the upper bound, it inspired us to make a determined effort to improve the lower bound.  We prove the following.

\begin{theorem}\label{thm:main-2-col}
For all $\epsilon>0$, there exists $n_0$ such that if $n\geq n_0$ and $G$ is a graph with at most $(3.75-\epsilon) n$ edges, there exists a 2-coloring of the edges of $G$ such that every monochromatic path has order less than $n$.  Thus $\hat{R}(P_n) \ge (3.75 - o(1))n$.
\end{theorem}

For the general, $r$-color version of the problem, the best upper bound is due to Krivelevich \cite{K} who proved $\hat{R}_r(P_n)=O(r^2\log (r) n)$ (Dudek and Pra{\l}at \cite{DP3} later gave a different proof).  In fact, both \cite{K} and \cite{DP3} prove the stronger ``density version'' of the theorem: there exists a graph $G$ (a binomial random graph) with $|E(G)| = O(r^2\log (r)n)$ such that every subgraph of $G$ with at least $e(G)/r$ many edges contains a monochromatic path of order $n$ (A recent paper of Balogh, Dudek, and Li \cite{BDL} shows that the factor $r^2\log r$ cannot be improved for this stronger density version in the setting of random graphs).

As for the lower bound, Dudek and Pra{\l}at \cite{DP2}  proved that for any $r\ge 2$, $\hat{R}_r(P_n) \ge \frac{(r+3)r}{4}n - O(r^2)$ and then Krivelevich \cite{K} proved that for any $r\ge 3$ such that $r-2$ is a prime power, $\hat{R}_r(P_n) \ge (r-2)^2n - o(n)$. We improve on each of these results by proving the following.

\begin{theorem}\label{thm:main-r-col}
Let $r \ge 2$ and let $q$ be the largest prime power such that $q\leq r-1$.  Then $$\hat{R}_r(P_n)\ge \max\left\{ \of{\frac{(r-1)r}{2} + 2.75-o(1)}n, (q^2-o(1))n\right\}.$$
\end{theorem}

Note that the prime number theorem guarantees that for any $\varepsilon>0$ and $r$ sufficiently large, there is a prime between $(1-\varepsilon)r$ and $r$, so for sufficiently large $r$, the second term in the maximum will dominate and we have $\hat{R}_r(P_n) \ge (r-1 - o_r(1))^2 n$. Determining whether $\hat{R}_r(P_n)=\Theta(r^2n)$ or not is perhaps the most interesting open problem regarding the size-Ramsey number of a path.

\subsection{Outline, Notation}

Our improvement in the lower bound stems from two main ideas.

1) If we can partition the graph $G$ into sets of order at most $n-1$ such that the number of edges crossing the partition is at most $n-2$, then we can color the edges inside the sets red and the edges between the sets blue so there are no monochromatic $P_n$'s.  This has some similarity to the problem of determining the bisection width of a graph -- in which case a result of Alon \cite[Proposition 3.1]{A} gives good bounds on the number of crossing edges in a balanced bipartition of graphs with bounded maximum degree and at most $2n-2$ vertices.  However, in our case, $G$ may not have bounded maximum degree, $G$ may have more than $2n-2$ vertices, and we don't necessarily want the partition to be balanced.  Nevertheless, with some extra work, we are able to use similar methods from the bisection width problem (e.g.\ \cite{A}, \cite{KM}) in our setting.

2) From the ordinary path Ramsey problem it is known that if $G$ has at most $\frac{3n}{2}-2$ vertices, then there exists a 2-coloring of $G$ such that every monochromatic path has order less than $n$.  We show that if $G$ has between roughly $3n/2$ and $5n/3$ vertices and few enough edges, then there exists a 2-coloring of $G$ such that every monochromatic path has order less than $n$.  This allows us to only consider graphs with at least $5n/3$ vertices.

In Section \ref{sec:lems} we prove a number of lemmas which we will use throughout the proof. We also show how some of these lemmas imply the previously known lower bounds on the size-Ramsey number of paths.
In Section \ref{sec:2-col} we prove Theorem \ref{thm:main-2-col}.
In Section \ref{sec:r-col} we prove Theorem \ref{thm:main-r-col}. In Section \ref{sec:concl}, we list a few observations and approaches that may helpful in trying to improve the lower bounds we have provided.

If $S$ is a subset of vertices of a graph $G=(V,E)$, then $G-S$ refers to $G[V \setminus S]$. When $G$ is a graph, we write $|G|$ for $|V(G)|$.  For any other notation we defer to \cite{Die}.
All logarithms are natural (base $e$) unless otherwise stated.  Throughout the paper, if we refer to an \emph{$r$-coloring} of $G$, we mean an $r$-coloring of the edges of $G$.

\section{Lemmas}\label{sec:lems}

When proving a lower bound on the $r$-color size-Ramsey number of $P_n$, we are given a graph $G=(V,E)$ and we must exhibit an $r$-coloring of the edges of $G$ so that $G$ has no monochromatic paths of order $n$.  It is often useful to break this into cases depending the number of vertices of $G$.  In Section \ref{sec:Nlower} we use the examples from the ordinary path Ramsey problem to determine a lower bound on $|V|$.  In Section \ref{sec:Nupper} we prove a general result which allows us, when proving a lower bound on $\hat{R}_r(P_n)$, to restrict our attention to graphs with minimum degree at least $r+1$, which in turn gives us an upper bound on $|V|$.  In Section \ref{sec:prune}, we prove a lemma which we use in the proof of Theorem \ref{thm:main-r-col}. In Section \ref{sec:mainlem}, we prove the main lemma of the paper needed for the proof of Theorem \ref{thm:main-2-col}.  Finally, in Section \ref{extend} we show how to deal with the case when $G$ has between roughly $3n/2$ and $5n/3$ vertices.

\subsection{Examples from the ordinary path Ramsey problem}\label{sec:Nlower}

\begin{proposition}[Gerencs\'er, \ Gy\'arf\'as \cite{GG}]\label{3n/2}
If $G$ has at most $\frac{3n}{2}-2$ vertices, then there exists a 2-coloring of $G$ such that every monochromatic path has order less than $n$.
\end{proposition}

\begin{proof}
Partition $V(G)$ into two sets $X_1, X_2$ with $|X_1|\leq \frac{n}{2}-1$ and $|X_2|\leq n-1$.  Color all edges incident with $X_1$ red and all edges inside $X_2$ blue.  Any pair of consecutive vertices on a red path must contain at least one vertex of $X_1$. Thus the longest red path is of order at most $2|X_1|+1\leq n-1$.
\end{proof}

\begin{proposition}[Yongqi, Yuansheng, Feng, Bingxi \cite{YYFB}]\label{3n/2_r}
Let $r\geq 3$. If $G$ has at most $2(r-1)(\frac{n}{2}-1)=(r-1)(n-2)$ vertices, then there exists an $r$-coloring of $G$ such that every monochromatic path has order less than $n$.
\end{proposition}

\begin{proof}
Partition $V(G)$ into $2r-2$ sets $X_1, X_2, \ldots, X_{2r-2}$ each of order at most $\frac{n}{2}-1$. In the following, addition is modulo $2r-2$. For $i = 1, \ldots, r-1$, color with color $i$, the edges between $X_i$ and $X_{i+1},\ldots, X_{i+r-2}$ and the edges between $X_{i+r-1}$ and $X_{i+r},\ldots X_{i+2r-3}$.
Use color $r$ for the edges between $X_i$ and $X_{i+r-1}$ for $i=1,\ldots r-1$. Color arbitrarily within the $X_i$'s. This coloring has no monochromatic  $P_n$ in color $i$ for $i=1,\ldots r-1$ for the same reason as in Proposition \ref{3n/2}. There is none in color $r$ since each component of color $r$ is of order less than $n$.
\end{proof}

\subsection{A reduction lemma}\label{sec:Nupper}

\begin{fact}\label{Nupper}
If $G=(V,E)$ is a graph with minimum degree at least $r+1$, then $|V|\leq \frac{2|E|}{r+1}$.
\end{fact}

The following lemma shows that in order to get a lower bound on the $r$-color size-Ramsey number of $P_{n}$, we can restrict our attention to graphs $G$ with minimum degree at least $r+1$, and consequently at most $\frac{2|E|}{r+1}$ vertices.  This generalizes an observation which is implicit in the proof of Beck's lower bound \cite{B2}.

\begin{lemma}\label{mindegree}
Let $r$ and $n$ be positive integers with $n\geq r+4$. If every connected graph with at most $m$ edges and minimum degree at least $r+1$ (and consequently at most $2m/(r+1)$ vertices) has an $r$-coloring such that every monochromatic path has order less than $n-2$, then every graph with at most $m$ edges has an $r$-coloring such that every monochromatic path has order less than $n$.
\end{lemma}

A \emph{star} is a tree having at most one vertex of degree at least 2.  A \emph{star forest} is a vertex disjoint collection of stars.  The \emph{star arboricity} of a graph $G$, denoted $sa(G)$, is the minimum number of star forests needed to partition the edge set of $G$. In order to keep this aspect of the proof self contained, we give a short proof of the fact that the star arboricity of graph is at most $\Delta(G)$.  We note that stronger statements are known (see \cite[Theorem 1]{BFHP} for instance), but not needed for our purposes.

\begin{fact}\label{stararbor}
$sa(G)\leq \Delta(G)$
\end{fact}

\begin{proof}
Clearly this is true for $\Delta=1$.  Suppose $\Delta\geq 2$ and the statement holds for all graphs $G$ with $\Delta(G)\leq \Delta-1$.  Let $G$ be a graph with $\Delta(G)=\Delta$.

\begin{claim}\label{starforest}
Every graph $G$ without isolated vertices contains a star-forest $S$ such that every vertex is incident with an edge of $S$.
\end{claim}

\begin{proof}
Let $F$ be a spanning forest consisting of a spanning tree of each component of $G$.  Let $z$ be a leaf of $F$ and let $y$ be the neighbor of $z$ in $F$.  Let $C$ be the star consisting of $y$ and all of the neighbors of $y$ which are leaves in $F$.  Note that in $F-C$, there are no isolated vertices and thus we may repeat this process until $F-C$ is empty at which point we have the desired star forest.
\end{proof}

Apply Claim \ref{starforest} to $G$ to get a star forest $S$ such that every non-isolated vertex in $G$ is incident with an edge of $S$.  After deleting the edges of $S$ we have a graph $G'$ with maximum degree at most $\Delta-1$ and we may apply induction to finish the proof.
\end{proof}

\begin{proof}[Proof of Lemma \ref{mindegree}]
Suppose that every connected graph with at most $m$ edges and minimum degree at least $r+1$ has an $r$-coloring such that every monochromatic path has order less than $n-2$.  Let $G$ be a graph with at most $m$ edges.  Let $S=\{v\in V(G): d(v)\leq r\}$.  We begin by describing how to color the edges of $G-S$ so that $G-S$ contains no monochromatic paths of order $n-2$.

If $G-S$ has fewer than $n-2$ vertices, then coloring the edges of $G-S$ arbitrarily we have an $r$-coloring of $G-S$ with no monochromatic paths of order $n-2$.  So suppose $G-S$ has at least $n-2\geq r+2$ vertices.  Let $v$ be a vertex in $G-S$ and suppose that $v$ has exactly $r+1-t$ neighbors in $G-S$ for some positive $t$.  This means $v$ had at least $t$ neighbors in $S$, so by making $v$ adjacent to $t$ vertices in $G-S$ (each of which was previously a non-neighbor of $v$) we make $v$ have degree at least $r+1$ and the total number of edges is still at most $m$.  We repeat this process for each vertex in $G-S$ which has degree less than $r+1$, updating on each step.  We end up with a graph $H$ such that $G-S\subseteq H$, $H$ has at most $m$ edges, and $\delta(G)\geq r+1$.  For each connected component of $H$, color the edges according to the hypothesis so that there are no monochromatic paths of order $n-2$.  This implies that $G-S$ has no monochromatic paths of order $n-2$.  Note that we are now done with graph $H$.

Since $\Delta(G[S])\leq r$, by Fact \ref{stararbor} we can color the edges of $G[S]$ with $r$ colors so that every color class is a star forest.  The only edges we have yet to color are the edges between $S$ and $G-S$.  Let $v$ be a vertex in $S$ and suppose that $v$ sends $t$ edges to $G-S$.  This means that $v$ has at most $r-t$ neighbors in $S$ and is incident with edges of at most $r-t$ different colors in $S$, which means there are at least $t$ colors which are not used on $v$.  Use these $t$ colors to color each of the edges from $v$ to $G-S$ so that each such edge receives a different color.  After doing this for each vertex in $S$, we have colored all of the edges of $G$.  Note that any monochromatic path which only uses edges from $G-S$ has order less than $n-2$, any monochromatic path which only uses edges from $G[S]$ and $[S, V(G)-S]$ has order at most 3.  If a monochromatic, say color 1, path uses an edge from $[S, V(G)-S]$, then since its endpoint in $S$ is not incident with any other edges of color 1, this edge must be a pendant edge of the path (of which there are only two) and thus the longest monochromatic path in $G$ has order less than $(n-2)+2=n$.
\end{proof}

\begin{remark}
Proposition \ref{3n/2} and Lemma \ref{mindegree} imply that $\hat{R}(P_n)> \frac{9}{4}(n-2)-3$.
\end{remark}

\begin{proof}
By Lemma \ref{mindegree} we may assume that $|V|\leq \frac{2|E|}{3}\leq \frac{3}{2}(n-2)-2$ and thus we are done by Proposition \ref{3n/2}.
\end{proof}

\begin{remark}\label{r^2-1}
Proposition \ref{3n/2_r} and Lemma \ref{mindegree} imply that for $r\geq 3$, $\hat{R}_r(P_n)> \frac{r^2-1}{2}(n-4)$.
\end{remark}

\begin{proof}
By Lemma \ref{mindegree} we may assume that $|V|\leq \frac{2|E|}{r+1}\leq (r-1)(n-4)$ and thus we are done by Proposition \ref{3n/2_r}.
\end{proof}

\subsection{Pruning a tree so that no long paths remain}\label{sec:prune}

The following is a slight generalization of the lemma used in \cite{B} and \cite{DP2} to give a lower bound on the size-Ramsey number of a path.

\begin{lemma}\label{snip}
For every tree $T$ with $|V(T)|\geq \floor{n/2}$, there exists a set $E'$ of at most $\floor{\frac{|V|}{\floor{n/2}}}-1$ edges such that $T-E'$ has no paths of order $n$.
\end{lemma}

\begin{proof}
If $T$ has no path of order $n$ we are done, so choose a path of order $n$ and delete the middle edge (or one of the two middle edges if $n$ is odd).  This separates $T$ into two subtrees, each with at least $\floor{n/2}$ vertices.  Now repeat on each subtree and call the set of deleted edges, $E'$.  When the process stops, every component of $T-E'$ has at least  $\floor{n/2}$ vertices and no paths of order $n$.
 Thus $T-E'$ has at most $\floor{\frac{|V|}{\floor{n/2}}}$ components, which means  $|E'| \le \floor{\frac{|V|}{\floor{n/2}}}-1$.
\end{proof}

\begin{remark}
Proposition \ref{3n/2} and Lemma \ref{snip} imply that $\hat{R}(P_n)> \frac{5}{2}n-7$.
\end{remark}

\begin{proof}
Let $G=(V,E)$ be a connected graph with at most $\frac{5n}{2}-7$ edges.  We may assume $G$ is connected and by Proposition \ref{3n/2}, we have $\frac{3n}{2}-1\leq |V|\leq \frac{5n}{2}-6$.  We let $T$ be a spanning tree of $G$ which contains at least $\frac{3n}{2}-2$ edges.  Applying Lemma \ref{snip} to $T$, we are left with a forest $F$ with at least $\frac{3n}{2}-5$ edges and no paths of order $n$, so we may color all of the edges of $F$ red.  There are at most $n-2$ edges remaining in $E(G)\setminus E(F)$, all of which we may color blue.
\end{proof}

\subsection{Main lemma}\label{sec:mainlem}

We will only use the following lemma in the case where $k=1$ or $k=2$, but we state it in general here.  Note that for instance when $k=1$, this says that if $G$ is a graph on $n-3<N\leq 2(n-3)$ vertices, then there is a bipartition of $V(G)$ into sets of order $n-3$ and $N-(n-3)$ such that the number of edges crossing the partition is approximately what we would get by taking a random such partition of a graph with $|E(G)|-N$ edges.

\begin{lemma}\label{partition}
Let $n\geq 4$, let $G=(V,E)$ be a graph on $N\geq n-2$ vertices, and let $k$ be a positive integer uniquely defined by $k(n-3)<N\leq (k+1)(n-3)$ where $k\leq n^{1/64}$.  Let $\alpha_1=\frac{n-3}{N}$ and $\alpha_2=\frac{N-k(n-3)}{N}$.  If every component of $G$ has at least $n-2$ vertices, $\Delta(G)\leq N^{1/16}$ and $|E|\leq 100N\leq 100(k+1)n$, then there exists a partition of $V$ into $k+1$ parts $V_1, \dots, V_{k+1}$ such that $|V_1|,\dots,|V_k|, |V_{k+1}|\leq n-3$ and $|V_{k+1}|\leq N-k(n-3)+N^{15/16}$ and the number of edges crossing the partition is at most $(1-k\alpha_1^2-\alpha_2^2)(|E|-N)+N^{15/16}$.
\end{lemma}

The first tool needed to prove Lemma \ref{partition} is the following fact mentioned by Alon \cite{A}, stated in general and made explicit here.

\begin{lemma}\label{treepartition}
Let $G$ be a connected graph on $p$ vertices with maximum degree $\Delta$. For any $1\le \ell < p$, we can find a collection of connected subgraphs $S_1, \ldots, S_t$ of $G$ such that
\begin{enumerate}[label=(T\arabic*)]
\item\label{t1} $V(S_1), \ldots, V(S_t)$ form a partition of $V(G)$ with $\ell < |S_i| \le 1+\Delta \ell $ for all $i\in [t-1]$ and $|S_t| \le 1 + \Delta \ell$
\item\label{t2} $\sum_{i=1}^t |E(S_i)| \ge p - t$
\item\label{t3} if $\ell  = \floor{\sqrt{p}},$ then $\frac{1}{\Delta+1} \sqrt{p} \le t \le \sqrt{p}+1$
\end{enumerate}
\end{lemma}

\begin{proof}
Let $T_0$ be a rooted spanning tree of $G$ with (arbitrary) root $r$.  For a rooted tree $T$ and vertex $v$, let $s(T,v)$ denote the subtree of $T$ rooted at vertex $v$ and let $C(v)$ denote the set of children of $v$.  Assume $T_i$ has been defined for some $i\ge 0$ and that $r$ is still the root of $T_i$. Traverse down $T_i$ from $r$ until encountering a vertex $v$ (if one exists) such that $|s(T_i,v)| > \ell$ and $|s(T_i,u)| \le \ell$ for all  $u\in C(v)$. Then $s(T_i,v)$ satisfies
\begin{align}\label{eq:t-right-size}
\ell < |s(T_i,v)| = 1 + \sum_{u\in C(v)}|s(T_i,u)| \le 1 + \Delta \ell.
\end{align}
If $v\neq r$, let $S_{i+1} = s(T_i, v)$ and $T_{i+1} = T_i - S_{i+1}$ and repeat for $i+1$. If $v=r$ or if no such vertex $v$ exists, then set $S_{i+1}=S_t = T_i$.
Each $S_i$ is connected by construction. Property \ref{t1} is satisfied by \eqref{eq:t-right-size}.  Property \ref{t2} follows since each $S_i$ is connected and thus  $\sum_{i=1}^t |E(G[S_i])|  \geq  \sum_{i=1}^t (|S_i| -1) = p - t$.

Finally, if $\ell= \floor{\sqrt{p}}$ we have
\[     (t-1)(\floor{\sqrt{p}} + 1) \le  \sum_{i=1}^t|S_i| = p \le t(1+ \Delta\sqrt{p})\]
and from each of $(t-1)(\floor{\sqrt{p}} + 1) \le p$ and $p \le t(1+ \Delta\sqrt{p})\leq t\sqrt{p}(1+\Delta)$, we derive the bounds on $t$ in \ref{t3}.
\end{proof}

The next tool we need is the following concentration inequality of McDiarmid \cite{M} (see also \cite{FK}).

\begin{lemma}[McDiarmid's ineqauality]\label{lem:mcd} Let $Z=Z(X_1, \ldots, X_N)$ be a random variable that depends on $N$ independent random variables $X_1, \ldots, X_N$.  Suppose that
\[|Z(X_1,\ldots, X_k, \ldots, X_N ) - Z(X_1, \ldots, X_k', \ldots, X_N)| \le c_k\]
for all $k=1,\ldots, N$ and $X_1,\ldots, X_n, X'_k$. Then for any $t\ge 0$ we have
\[\mathbb{P}\sqbs{Z\ge \mathbb{E}[Z]  + t} \le \exp\of{-\frac{t^2}{2\sum_{k\in [N]} c_k^2 }}.
\]
\end{lemma}

We are now ready to prove the main lemma.

\begin{proof}[Proof of Lemma \ref{partition}]
Apply Lemma \ref{treepartition} with $\ell = \floor{\sqrt{N}}$ to partition the components of $G$ into  $\frac{\sqrt{N}}{\Delta+1} \le t\le \sqrt{N}+1$ connected subgraphs $S_1, \ldots, S_t$ each of order at most $1+\Delta\sqrt{N}$.  There are at least $N-(t-1)\geq N-\sqrt{N}$ edges accounted for in these subgraphs.  Define $m=|E|-(N-\sqrt{N})$ to be an upper bound on the number of edges of $G$ which are not contained in these subgraphs.

We independently at random place each such connected subgraph in one of the sets $V_1, \dots, V_k, V_{k+1}$ with probabilities $\alpha_1, \dots, \alpha_1, \alpha_2$ respectively. Let $Z_i$ represent the number of vertices which land in the set $V_i$ for all $i\in [k+1]$.

Then $\mathbb{E}\sqbs{Z_1} = \dots= \mathbb{E}\sqbs{Z_k} = \alpha_1N$ and $\mathbb{E}\sqbs{Z_{k+1}} = \alpha_2N$.
Note that changing the position of one of $S_1, \ldots, S_t$ can change any of these variables by at most $1+\Delta\sqrt{N} \le N^{9/16}$.
Thus we may apply McDiarmid's inequality (Lemma \ref{lem:mcd}) and the union bound to conclude that the probability that for some $i\in [k]$, $Z_i$ exceeds $\alpha_1 N+N^{7/8}$ or $Z_{k+1}$ exceeds $\alpha_2 N + N^{7/8}$ is at most
\[(k+1)\cdot\exp\of{- \frac12\cdot\frac{N^{7/4}}{(\sqrt{N}+1)\cdot(N^{9/16})^2}   } = \exp\of{-\Omega(n^{1/8})}.\]
Thus at least $1 - e^{-\Omega(n^{1/8})}$ proportion of the partitions satisfy
\begin{align}\label{eq:parts-small}
|V_1|, \dots, |V_k| \le \alpha_1 N+N^{7/8}\,\, \textrm{ and }\,\, |V_{k+1}|\le \alpha_2 N + N^{7/8}.
\end{align}
Now, by linearity of expectation, the expected number of edges $\mu$ crossing the partition satisfies
\begin{align*}
\mu\le (1-k\alpha_1^2-\alpha_2^2)m.
\end{align*}
So there is a partition $V_1, \dots, V_{k}, V_{k+1}$ satisfying \eqref{eq:parts-small} with at most $(1-k\alpha_1^2-\alpha_2^2)m+1$ edges crossing the partition; otherwise we would have
\[ (1-k\alpha_1^2-\alpha_2^2)m\ge \mu \ge (1 - e^{-\Omega(n^{1/8})})((1-k\alpha_1^2-\alpha_2^2)m+1)>(1-k\alpha_1^2-\alpha_2^2)m,\]
a contradiction.

Let $S=\{v\in V(G): d(v)\leq 800k^3\}$ and note that $|S|>(1-\frac{1}{4k^3})N$; otherwise, there are at least $\frac{N}{4k^3}$ vertices of degree greater than $800k^3$ and we have that $$200N\geq 2|E| = \sum_{v\in V(G)} d(v)> \frac{N}{4k^3}(800k^3)=200N,$$ a contradiction.  If $|V_{k+1}|<\frac{N}{2k^3}$, we move vertices from $S\cap (V_1\cup \dots \cup V_k)$ to $V_{k+1}$ until $|V_1|, \dots, |V_k|\leq  n-3$; a total of at most $kN^{7/8}$ vertices.  If $|V_{k+1}|\geq \frac{N}{2k^3}$, we do the following: if $|V_i|>n-3$ for $i\in [k]$ or $|V_{k+1}|>N-k(n-3)$, there must exist $j\in [k]$ such that $|V_j|<n-3$ or $V_{k+1}<N-k(n-3)$, so we select a vertex from $V_i\cap S$ and we move it to $V_j$.  Because of the size of $|S|$, we can repeat this process until we have $|V_1|=\dots=|V_k|$ and $|V_{k+1}|=N-k(n-3)$.  The total number of vertices moved will be at most $kN^{7/8}$.  In either case, at the end of this process, the number of edges crossing the partition is at most $$(1-k\alpha_1^2-\alpha_2^2)m +1+ kN^{7/8}\cdot 800k^3 < (1-k\alpha_1^2-\alpha_2^2)(|E|-N) + N^{15/16}.$$
\end{proof}

\subsection{Extending Proposition \ref{3n/2}}\label{extend}

The following observations extend Proposition \ref{3n/2}.  We note that there is a similarity between this observation and the concept of the \emph{integrity} of a graph (see \cite{V}).

\begin{observation}\label{n/2-path}
If $G$ has a set $S$ of at most $\frac{n}{2}-1$ vertices such that every component of $G-S$ has no path of order $n$, then there exists a 2-coloring of the edges of $G$ such that every monochromatic path has order less than $n$.
\end{observation}

\begin{figure}[ht]
\begin{center}
\includegraphics[scale=1]{fig_obs.pdf}
\caption{Coloring the edges of $G$ in Observation \ref{n/2-path}}\label{obs31}
\end{center}
\end{figure}

\begin{proof}
We color all edges incident to $S$ with red and every other edge blue. Clearly there will be no blue path of order $n$. Any pair of consecutive vertices on a red path must contain at least one vertex of $S$. Thus the longest red path is of order less than $n$.
\end{proof}

We also note that there is a similarity between the following observation and the concept of the \emph{edge integrity} of a graph (see \cite{AHK}).

\begin{observation}\label{edgeint}
If $G$ has a subgraph $H$ such that $H$ has no path of order $n$ (in particular, if $H$ has at most $n-2$ edges) and every component of $(V(G), E(G)\setminus E(H))$ has order less than $n$, then there exists a 2-coloring of the edges of $G$ such that every monochromatic path has order less than $n$.
\end{observation}

\begin{proof}
Color the edges of $H$ with red and color the remaining edges blue.
\end{proof}

The following lemma says that if the number of vertices is not too much more than $3n/2$ and the number of edges of $G$ is small enough, we can essentially color $G$ in a way which resembles the coloring in Proposition \ref{3n/2}.

\begin{lemma}\label{close3n/2}
Let $0<\epsilon<\frac{1}{6}$, let $n$ be sufficiently large, and let $G=(V, E)$ be a graph with $\delta(G)\geq 3$ on $\frac{3}{2}(n-2)-2+\sigma n$ vertices where $0<\sigma\leq \frac{1}{6}-\epsilon$. Let $d$ be an integer such that $4\leq d\leq \min\{\frac{1/2-3\epsilon}{\sigma}+1, 100\}$ and $q$ be an integer such that $0\leq q\leq \frac{\epsilon n}{d+1}-5$.  If $H = (U,E')$ is a subgraph of $G$ with $|U| = |V| - q$ and  $$|E|\leq \left(\frac{3(d+1)+6\sigma}{4}-\epsilon\right)n,$$ then there exists a partition $\{X,Y,Z\}$ of $U$ such that
\begin{enumerate}
\item every vertex in $X$ has at most one neighbor in $Z$ and
\item $|Z|\leq n-3$, $|Y|\leq \frac{n-6}{2}-q$, and $|X|+|Y|\leq n-3$.
\end{enumerate}
Consequently, there exists a 2-coloring of $G$ such that every blue path has order at most $n-3$ and every red path has order at most $n-3$.
\end{lemma}

\begin{proof}
First note that there is a value of $d$ which satisfies  $4\leq d\leq \frac{1/2-3\epsilon}{\sigma}+1$ since $\sigma\leq \frac{1}{6}-\epsilon$ and there is a value of $q$ that satisfies $0\leq q\leq \frac{\epsilon n}{d+1}-5$ since $d\leq 100$ and $n$ is sufficiently large.
Let $X^*=\{v\in U: d_H(v)\leq d\}$.  We first show that $|X^*|$ is significantly larger than $\sigma n$.  Indeed, we have
\begin{align*}
\left(\frac{3(d+1)+6\sigma}{2}-2\epsilon\right)n\geq 2|E|\ge \sum_{v\in U}d_G(v)&\geq 3|X^*|+(d+1)(|U|-|X^*|)\\
&=(d+1)|U|-(d-2)|X^*|.
\end{align*}

Rearranging and using $|U|=\frac{3}{2}(n-2)-2-q+\sigma n$ and $q\leq \frac{\epsilon n}{d+1}-5$ gives
\begin{align*}
(d-2)|X^*|&\geq (d+1)(\frac{3}{2}(n-2)-2-q+\sigma n)-\left(\frac{3(d+1)+6\sigma}{2}-2\epsilon\right)n\\
&=(d-2)\sigma n+2\epsilon n-(d+1)(q+5)\geq (d-2)\sigma n+\epsilon n,
\end{align*}
and thus
\begin{equation}\label{X*}
|X^*|\geq (\sigma +\frac{\epsilon}{d-2})n\geq \floor{(\sigma +\frac{\epsilon}{d-1})n}\geq  \sigma n+1,
\end{equation}
where the last inequality holds since $d\leq 100$ and $n$ is sufficiently large.

Now let $X\subseteq X^*$ such that $|X|=\floor{(\sigma +\frac{\epsilon}{d-1})n}$, let $Y^*=N_H(X)\setminus X$, and note that
\begin{equation}\label{Y*}
|Y^*| \le d|X|.
\end{equation}

Since $N_H(X)\subseteq X\cup Y^*$ we would be done if $|Y^*| \le \frac{n-6}{2}-q$ by taking $Y=Y^*, Z = U\setminus (X\cup Y)$ (see Figure \ref{proof2i} for the coloring).  We now show that if $|Y^*| > \frac{n-6}{2}-q$, then we can move at least $|Y^*|-(\frac{n-6}{2}-q)$ vertices from $Y^*$ to $Z$.  We do this by showing that there exists an induced matching in the bipartite graph $H[X, Y^*]$ of size at least $|Y^*|-(\frac{n-6}{2}-q)$.

Let $Y_1=\{v\in Y^*: d_H(v, X)=1\}$ and $Y_2=\{v\in Y^*: d_H(v, X)\geq 2\}$.  We note that since every vertex in $X$ sends at most $d$ edges to $Y^*$, $H[X, Y^*]$ has an induced matching of size at least $|Y_1|/d$.  We have
\begin{align*}
d|X|\geq e_H(X, Y^*)\geq |Y_1|+2(|Y^*|-|Y_1|)=2|Y^*|-|Y_1|
\end{align*}
which implies
\begin{align*}\frac{|Y_1|}{d} \geq \frac{2|Y^*|}{d}-|X|= |Y^*| - \frac{d-2}{d}|Y^*| - |X|
&\stackrel{\eqref{Y*}}{\ge} |Y^*| - (d-1)|X|\\
&= |Y^*|-(d-1)\floor{(\sigma +\frac{\epsilon}{d-1})n}\\
&\geq |Y^*|-(\frac{1}{2}-2\epsilon)n\\
&\geq |Y^*|-(\frac{n-6}{2}-q),
\end{align*} where the penultimate inequality holds by the upper bound on $d$ and the ultimate inequality holds by the upper bound on $q$.

Let $X'=\{v\in X: N(v)\cap Y_1\neq \emptyset\}$ and choose a matching from $X'$ to $Y_1$ which saturates $X'$ (which must exist by the definition of $Y_1$ and $X'$) and let $f(X')$ be the vertices in $Y_1$ which are saturated by the matching.   Set $Y'=Y^*\setminus f(X')$.  By the above we have $|Y'|\leq \frac{n-6}{2}-q$.

\begin{figure}[ht]
\centering
\subfloat[Subfigure 1 list of figures text][Coloring the edges of $H$ if $Y^*$ is small enough]{
\includegraphics[scale=1]{fig_proof4i.pdf}
\label{proof2i}}
\hfill
\subfloat[Subfigure 1 list of figures text][Coloring the edges of $H$ after moving vertices from $Y^*$]{
\includegraphics[scale=1]{fig_proof4.pdf}
\label{proof2ii}}
\hfill
\subfloat[Subfigure 1 list of figures text][Coloring the edges of $G$]{
\includegraphics[scale=1]{fig_proof3i.pdf}
\label{proof2iii}}
\caption{Coloring the edges in Lemma \ref{close3n/2}}
\end{figure}

Let $Y\subseteq U\setminus X$ such that $Y'\subseteq Y$ and $|Y|=\floor{\frac{n-6}{2}-q}$.  Now let $Z=U\setminus (X\cup Y)$ and note that
\begin{align*}
|Z|=|U|-|X|-|Y|&= \frac{3}{2}(n-2)-2+\sigma n -q-|X|-|Y|\\
&\stackrel{\eqref{X*}}{\leq} \frac{3}{2}(n-2)-2 +\sigma n-q-(\sigma n+1)-(\frac{n-6}{2}-q) = n-3.
\end{align*}
Color all edges inside $X\cup Y$ blue, all edges inside $Z$ blue, and color all other edges red (see Figure \ref{proof2ii}). Clearly there is no blue path of order $n-2$ since $|X|+|Y|, |Z|\leq n-3$.  Since $|Y|\leq \frac{n-6}{2}-q$, the longest red path in the bipartite graph induced by $[Y,Z]$ has order at most $n-5-2q$ and since the red edges from $X$ to $Z$ form a matching, the longest red path overall in $H$ has order at most $n-3-2q$.

Finally, coloring all edges from $V_0=V\setminus U$ to $X\cup Y\cup V_0$ blue and all edges from $V_0$ to $Z$ red (see Figure \ref{proof2iii}), we have that the longest red path is at most $n-3-2q+2|V_0|=n-3$, and the longest blue path has order at most $n-3$ since $|X|+|Y|+|V_0|, |Z|\leq n-3$.
\end{proof}

\section{Two colors}\label{sec:2-col}

We are now ready to give a proof of Theorem \ref{thm:main-2-col}.  We note that if $G$ had bounded maximum degree, we would be able to directly apply Lemma \ref{partition} (or Lemma \ref{close3n/2}).  So dealing with the ``high degree'' vertices is the main challenge which remains.  We also note that the $\epsilon$ in the following proof can be taken to be as small as $\epsilon =n^{-\Theta(1)}$; however, for the sake of readability, we didn't try to optimize the value of $\epsilon$.

\begin{proof}[Proof of Theorem \ref{thm:main-2-col}]
Let $\epsilon>0$ and let $n_0$ be a sufficiently large integer (the value of which we don't explicitly compute, but we will point out which inequalities depend on $n$ being sufficiently large).
Let $G=(V,E)$ be a connected graph on $N$ vertices with at most $(3+\gamma-\epsilon)n$ edges, where $0\leq \gamma<3/2$ is to be chosen later (ultimately, we will choose $\gamma=3/4$, but we leave it undetermined for now because it helps to see where there is slack in certain parts of the proof).  By Lemma \ref{mindegree} it suffices to assume that $\delta(G)\geq 3$ and thus $N\leq 2|E|/3\leq (2+2\gamma/3-2\epsilon/3)n<3(n-2)$.  Since we are using Lemma \ref{mindegree}, our goal for the rest of the proof is to exhibit a 2-coloring of $G$ which has no monochromatic paths of order $n-2$.  So by Proposition \ref{3n/2} we may assume that $N\ge \frac{3}{2}(n-2)-1$.

Let $V_0=\{v\in V(G): d(v)>n^{1/32}\}$.  We have
$n^{1/32}|V_0|\leq 2|E|$ and thus since $n$ is sufficiently large,
\begin{equation}\label{V0}
|V_0|\leq 2(3+\gamma-\epsilon)n^{31/32}\leq \frac{\epsilon n}{200}.
\end{equation}
We say that a component $C$ of $G-V_0$ is \emph{small} if $|C|<n-2$, \emph{medium} if $n-2\leq |C|\leq \frac{3}{2}(n-2)-|V_0|-2$, and \emph{large} if $\frac{3}{2}(n-2)-|V_0|-1\leq |C|$.

Regarding the components of $G-V_0$, if there is at most one medium component $B$, and the rest of the components are small, select a set $S\subseteq B$ such that $|V_0|+|S|=\floor{\frac{n-2}{2}}-1$.  Note that every component of $G-(S\cup V_0)$ has order at most
$$\frac{3}{2}(n-2)-2-|V_0|-|S|\leq \frac{3}{2}(n-2)-2-\left(\frac{n-2}{2}-1\right)=n-3$$
and thus we are done by Observation \ref{n/2-path}.

Since $N<3(n-2)$, if there is more than one medium component, then there are exactly 2, call them $B_1$ and $B_2$ and the remaining components $C_1, \dots, C_t$ are small. Likewise, if there is one large component $A$, then there is exactly one and either we have a medium component $B$ and the remaining components $C_1, \dots, C_t$ are small, or there are no medium components and the remaining components $C_1, \dots, C_t$ are small.  Let $U=A$, $U=A\cup B$, or $U=B_1\cup B_2$ depending on the case and note that in any case $|U| \ge \frac{3}{2}(n-2)-1-|V_0|$.

\noindent
\textbf{Case 1} First suppose that $\frac{3}{2}(n-2)-1-|V_0|\leq |U| \leq 2(n-3)$. We parameterize this by introducing a variable $\sigma$ such that $|U|=\frac{3}{2}(n-2)-2-|V_0|+\sigma n$ where $0<\sigma< \frac{1}{2}+\frac{|V_0|}{n}$.

\textbf{Case 1.1} Suppose $0<\sigma\leq \frac{1}{6}-\epsilon$ (in this case $U$ must consist of one large component).  We apply Lemma \ref{close3n/2} to $G[U\cup V_0]$ with $\epsilon$ and $\sigma$ as given, $d=\min\{\floor{\frac{1/2-3\epsilon}{\sigma}+1}, 100\}$, and $q=|V_0|$.  Note that $q=|V_0|<\frac{\epsilon n}{200}\leq \frac{\epsilon n}{d+1}-5$.  Also note that in order to apply the lemma we must have $(3+\gamma-\epsilon)n\leq (3.9375-\epsilon)n\leq \left(\frac{3(d+1)+6\sigma}{4}-\epsilon\right)n$ (because of this, it would suffice to choose $d=4$ if $\sigma \ge 1/8$ and $d=5$ otherwise).

The lemma provides a partition of $U$ as $\{X,Y,Z\}$.  We color the edges exactly as in Lemma \ref{close3n/2} and additionally we color all edges from $V_0$ to the small components red and all edges inside the small components blue.  Since $|Y|+|V_0|\leq \frac{n-6}{2}$ the longest red path still has order at most $n-3$ (that is, the extra red edges from $V_0$ to the small components do not change the properties of the coloring from Lemma \ref{close3n/2}).

\begin{figure}[ht]
\begin{center}
\includegraphics[scale=1]{fig_proof3.pdf}
\caption{Coloring the edges of $G$ in Case 1.1}\label{proof}
\end{center}
\end{figure}

\textbf{Case 1.2}
Now we deal with the case where $\frac{1}{6}-\epsilon< \sigma< \frac{1}{2}+\frac{|V_0|}{n}$ (in this case $U$ must consist of one large component).
Apply Lemma \ref{partition}  (with $k=1$) to $G[U]$ to get a bipartition of $U$ into sets $U_1, U_2$ of order at most $n-3$ such that the number of edges crossing the partition is
\begin{equation*}
\left(1-\left(\frac{1}{1+(1/2+\sigma)}\right)^2-\left(\frac{1/2+\sigma}{1+(1/2+\sigma)}\right)^2\right)(3/2+
\gamma-\sigma-\epsilon)n+N^{15/16}<(1-\frac{\epsilon}{4})n,
\end{equation*}
where the last inequality holds provided $n$ is sufficiently large and
\begin{equation}\label{gammaupper}
\gamma\leq \frac{3/4+\sigma+3\sigma^2}{1+2\sigma}.
\end{equation}
Since we are assuming $0< \sigma< \frac{1}{2}+\frac{|V_0|}{n}$, we have $\frac{3/4+\sigma+3\sigma^2}{1+2\sigma}\geq 0.75$ with the minimum occurring when $\sigma= \frac{1}{6}-\epsilon$ (note that the minimum of $\frac{3/4+\sigma+3\sigma^2}{1+2\sigma}$ over the entire interval $0< \sigma< \frac{1}{2}+\frac{|V_0|}{n}$ is $\sqrt{3}-1\approx 0.732$ and occurs when $\sigma=\frac{2\sqrt{3}-3}{6}\approx 0.07735$).

Now color the edges inside the sets $U_1, U_2$ blue, the edges between the sets $U_1$, $U_2$ red, the edges inside $V_0$ blue, the remaining edges incident with $V_0$ red, and the edges inside the small components blue.  Note that since $|V_0|\leq \frac{\epsilon n}{200}$ and the number of edges between the sets $U_1$, $U_2$ is at most $(1-\frac{\epsilon}{4})n$, the longest blue path has order at most $(1-\frac{\epsilon}{4})n+2|V_0|\leq n-3$ (c.f. Observation \ref{edgeint}, thinking of $H$ as the graph induced by the red edges).

\begin{figure}[ht]
\begin{center}
\includegraphics[scale=1]{fig_proof.pdf}
\caption{Coloring the edges of $G$ in Case 1.2 and Case 2}\label{proof4}
\end{center}
\end{figure}

\noindent
\textbf{Case 2}
Now suppose that $|U| > 2(n-3)$. We parameterize this by introducing a variable $\tau$ and assuming that $|U| = (2+\tau)(n-3)$ where $0 < \tau \le 2\gamma/3<1$
(in this case $U$ either consists of one large component, one large and one medium component, or two medium components).  Apply Lemma \ref{partition} (with $k=2$) to $G[U]$ to get a tripartition of $U$ into sets $U_1, U_2, U_3$ of order at most $n-3$ such that the number of edges crossing the partition is at most
\begin{align*}
\left(1-2\left(\frac{1}{2+\tau}\right)^2-\left(\frac{\tau}{2+\tau}\right)^2\right)(1+\gamma-\tau-\epsilon)n+N^{15/16}<(1-\frac{\epsilon}{4})n,
\end{align*}
where the last inequality holds provided $n$ is sufficiently large and
\begin{equation*}
\gamma\leq \frac{1+\tau+5\tau^2/2}{1+2\tau}.
\end{equation*}
We have $\frac{1+\tau+5\tau^2/2}{1+2\tau}\geq \frac{3}{4}(\sqrt{5}-1)\approx 0.927$
with the minimum occurring when $\tau=\frac{3\sqrt{5}-5}{10}\approx 0.1708$.

Now color the edges as we did at the end of Case 1.2.
\end{proof}

\begin{figure}[ht!]
\begin{center}
\includegraphics[scale=6]{updatedfigure.png}
\caption{The value of $c$ on the $x$-axis represents $|U|=cn$.  For a given value of $c$, the curve shows the maximum number of edges $G$ can have so that our proof gives a 2-coloring of $G$ with no monochromatic $P_n$.  The blue curve corresponds to Case 1.1, the red curve to Case 1.2, and the green curve to Case 2.  Note that the minimum over the entire interval is $3.75$ and occurs when $c=5/3$.}\label{3.75}
\end{center}
\end{figure}

We note two things about the previous proof.  We originally dealt with the case $|U|=\frac{3}{2}(n-2)-1+\sigma n$ as a whole rather than splitting into the subcases $0< \sigma\lesssim 1/6$ and $1/6\lesssim \sigma\lesssim 1/2$.  Without the subcases, the bound we obtained in \eqref{gammaupper} was $\gamma\leq \sqrt{3}-1$ which gives an overall lower bound of $\hat{R}(P_n)\geq (2+\sqrt{3}-o(1))n\approx 3.732n$.  So by dealing with the subcases separately, we got an improvement of about $(0.018-o(1))n$.

If one were to attempt to improve the lower bound of $(3.75-o(1))n$, a good test case would be when $|U|\approx\frac{5n}{3}$, since this corresponds to the case where $|U|=\frac{3}{2}(n-2)-1+\sigma n$ and $\sigma\approx 1/6$ which is the bottleneck of the above proof (see Figure \ref{3.75}).

\section{More than two colors} \label{sec:r-col}

The following theorem implies the first part of Theorem \ref{thm:main-r-col}.  We simply use the $r=2$ case and induction.

\begin{proposition}\label{r=3}
For all $r\geq 2$ and sufficiently large $n$, if $G$ is a graph with at most $(\frac{(r-1)r}{2}+2.75-o(1))n$ edges, then there exists a $r$-coloring of the edges of $G$ such that every monochromatic path has order less than $n$.
\end{proposition}

\begin{proof}
Let $G=(V,E)$ be a connected graph.  For $r=2$, this holds by Theorem \ref{thm:main-2-col}.  So let $r\geq 3$ and suppose the result holds for all smaller $r$.  If $N\leq (r-1)(n-2)$, then we are done by Proposition \ref{3n/2_r}; so suppose $N\geq (r-1)(n-2)+1$.  Let $T$ be a spanning tree of $G$ and apply Lemma \ref{snip} to get a forest $F$ with no paths of order $n$ and at least $N-2r^2$ edges. Color the edges of the forest with color $r$.  The number of remaining edges is at most $(\frac{(r-1)r}{2}+2.75-o(1))n-(r-1)n-2r^2=(\frac{(r-2)(r-1)}{2}+2.75-o(1))n$ and thus we may apply induction to color the remaining edges with the remaining $r-1$ colors.
\end{proof}

\begin{remark}
The bound is Proposition \ref{r=3} is larger than the bound in Remark \ref{r^2-1} for $2\leq r\leq 6$, and for $r\geq 7$, the bound in Remark \ref{r^2-1} is larger.
\end{remark}

\begin{definition}
An affine plane of order $q$ is a $q$-uniform hypergraph on $q^2$ vertices (called points), with $q(q+1)$ edges (called lines) such that each pair of vertices is contained in exactly one edge.
\end{definition}

It is well known that an affine plane of order $q$ exists whenever $q$ is a prime power (and it is unknown whether there exists an affine plane of non-prime power order).  We collect two key properties of affine planes in the following proposition.

\begin{proposition}\label{affprop}
Let $q\geq 2$ such that there exists an affine plane of order $q$.  There exists a $q+1$-coloring of the edges of $K_{q^2}$ such that
\begin{enumerate}
\item\label{ap1} every color class (called a parallel class) consists of a collection of $q$ vertex disjoint $K_q$'s, and
\item\label{ap2} every vertex is contained in exactly one edge of each color and the union of these $q+1$ edges is all of $V(K_{q^2})$.
\end{enumerate}
\end{proposition}

The following theorem implies the second part of Theorem \ref{thm:main-r-col}.  We modify Krivelevich's proof \cite[Theorem 8]{K} in such a way that no color is ``wasted'' on the high degree vertices.  This improves the lower bound from $((r-2)^2-o(1))n$ to $((r-1)^2-o(1))n$.

\begin{proposition}\label{affine}
Let $r\ge 3$ and let $q\leq r-1$ be the largest integer such that an affine plane of order $q$ exists (effectively, let $q\leq r-1$ be the largest integer such that $q$ is a prime power) and suppose $n$ is sufficiently large. For all graphs $G$ with at most $q^2 n - 6q^4n^{0.9}  = (q^2-o(1)) n$ edges, there exists a $q+1$-coloring (which is an $r$-coloring) of the edges of $G$ such that every monochromatic path has order less than $n$.
\end{proposition}

\begin{proof}
Let $G=(V,E)$ be a graph with $|E|\leq q^2-6q^4n^{.9}$.  Let $V_0:=\set{v\in V(G)}{d(v) \ge n^{0.1}}$. Then $q^2 n \ge |E(G)| \ge \frac{1}{2}|V_0|n^{0.1}$ implies that
$|V_0| \le 2q^2n^{0.9}$.
Now randomly partition $V\setminus V_0$ into $q^2$ parts $V_1, \ldots V_{q^2}$ by placing each vertex into one of these sets independently with probability $1/q^2$.
 Let $L$ be a line of the affine plane $A_{q}$ on point set $[q^2]$. For each edge $e$ in $G[V\setminus V_0]$, we assign color $i$ to $e$ if the endpoints of $e$ are in sets $V_x$ and $V_y$ where the unique line containing $x$ and $y$ in $A_{q}$ is in the $i$'th parallel class of $A_{q}$. We color $e$ arbitrarily if both of its endpoints are in $V_x$ for some $x$.

For a line $L$ of $A_{q}$, define the random variable $X_L := |E\of{\bigcup_{x\in L}V_x}|$.
Then \[\mathbb{E}\sqbs{X_L}\le \frac{1}{q^2}\cdot |E(G)| \leq n - 6q^2n^{0.9}.\]
 Since every vertex of $V\setminus V_0$ has degree at most $n^{0.1}$, we have that moving any one vertex from $V_x$ to $V_y$ can change $X_L$ by at most $n^{0.1}.$ Thus we may apply McDiarmid's inequality (Lemma \ref{lem:mcd}) with $c_k = n^{0.1}$ for all $k$ to conclude that
 \[\mathbb{P}\sqbs{X_L \ge n-5q^2n^{0.9}} \le \exp\of{-\frac{(q^2n^{0.9})^2}{2|V\setminus V_0|\cdot (n^{0.1})^2}} = \exp\of{-\Omega(n^{0.6})}.\]
Where we used $|V\setminus V_0|\leq |E|\leq q^2n$ in the last inequality. Thus taking a union bound over all $(q+1)q$ lines $L$, we conclude that there exists a partition of $V\setminus V_0$ in which at most $n - 5q^2n^{0.9}$ edges lie inside $\bigcup_{x\in L}V_x$ for all lines $L$.
 Suppose $V_1, \ldots, V_{q^2}$ is such a partition.

Let $L_1, \dots, L_{q+1}$ be the lines from $A_q$ incident with the point $1$, one from each parallel class (which is possible by Proposition \ref{affprop}(ii)).  Note that for all $j\in [q^2]\setminus \{1\}$, $V_j$ intersects precisely one such $\bigcup_{x\in L_i}V_x$ for $i\in [q+1]$.  For each $i\in [q+1]$, we color the edges from $V_0$ to $\bigcup_{x\in L_i}V_x$ with color $i$ (coloring the edges from $V_0$ to $V_1$ arbitrarily).  Now every edge from $V_0$ to $V\setminus V_0$ has been colored and for each color $i\in [q+1]$, there exists a unique line $L_i$ such that $V_0$ sends edges of color $i$ to $\bigcup_{x\in L}V_x$.
Any path contained in  $V_0\cup \bigcup_{x\in L_i}V_x$ can have order at most \[\abs{E\of{\bigcup_{x\in L_i}V_x}} + 2|V_0| \le n - 5q^2n^{0.9} + 4q^2n^{0.9} < n\]
The sets $\bigcup_{x\in L}V_x$ where $L$ does not contain point 1 still contain less than $n-1$ edges and thus have no path of order $n$.
\end{proof}

\begin{remark}
The bound in Proposition \ref{affine} is larger than the bounds in Proposition \ref{r=3} and Remark \ref{r^2-1} for all $r\geq 4$.
\end{remark}

\section{Additional observations and conclusion}\label{sec:concl}

In this section we collect a few additional thoughts, none of which fit into into the main thread of the paper.  The four observations below quantify the intuitive notion that if $G$ is a graph having the property that every 2-coloring of the edges of $G$ contains a monochromatic $P_n$, then $G$ must be ``expansive'' in some sense.

For a graph $G=(V,E)$, let $S_V$ be the set of permutations of $V$. The \emph{bandwidth}, $\varphi$ of $G$ is defined as \[\varphi(G):=\min_{f\in S_V}\max_{uv\in E}|f(u) - f(v)|.\]

\begin{observation}
If $\varphi(G) \le \frac{n}{2}-1$, then there is a 2-coloring of the edges of $G$ such that every monochromatic path has order less than $n$.
\end{observation}

\begin{proof}
Order the vertex set to witness the minimum bandwidth, then split the vertices into sets $V_1, \dots, V_t$, with $|V_1|=\dots=|V_{t-1}|=\floor{\frac{n}{2}-1}$ and $|V_t|\leq n-1$.  For all odd $i\in [t]$, color the edges from $V_i$ to $V_i\cup V_{i+1}$ red, and for all even $j\in [t]$ color the edges from $V_j$ to $V_i\cup V_{j+1}$ blue.
\end{proof}

A \emph{depth first search} (DFS) tree (or \emph{normal} tree) $T$ rooted at $x$ in a graph $G$ is a subtree of $G$ such that for all $uv\in E(G)$ with  $u,v \in V(T)$, either $v$ is on the $x-v$ path in $T$ or $u$ is on the $x-u$ path in $T$.

For a connected subgraph $H$ of a graph $G$ and vertices $u,v\in V(H)$, let $d_H(u,v)$ be the length of the shortest path between $u$ and $v$ in $H$.  A \emph{breadth first search} (BFS) tree $T$ rooted at $x$ is a subtree of $G$ such that for all $v\in V(T)$,  $d_T(x, v)=d_G(x, v)$.  Such a tree has the property that for all $uv\in E(G)$ with $u,v\in V(T)$, $|d_T(x,u)-d_T(x,v)|\leq 1$.  The vertices at each fixed distance from the root are called the \emph{levels} of $T$.
It is well known that for every connected graph $G$ and every vertex $x\in V(G)$, there exists a spanning DFS tree $T$ rooted at $x$ and a spanning BFS tree rooted at $x$.

Using the notation for rooted trees from the proof of Lemma \ref{treepartition}, we have the following observation.
\begin{observation}
Let $G$ be a connected graph.  If there exists a vertex $x$ and a DFS tree $T$ rooted at $x$ so that every child $y\in C(x)$ satisfies $|S(T,y)|\le \frac{5n}{4}-2$, then there exists a 2-coloring of the edges of $G$ such that every monochromatic path has order less than $n$.
\end{observation}

\begin{proof}
For each sub-tree $S(T,y)$ where $y\in C(x)$, we partition the vertices of $S(T,y)$ into sets $A_y$ and $B_y$ where $|A_y|\le \frac n4 -1$, $y\in A_y$ and $|B_y|\le n-1$. Let $A=\{x\}\cup \bigcup_{y\in C(x)}A_y$ and $B=\bigcup_{y\in C(x)}B_y$.  We color the edges of $G$ within $B$ blue and the edges from $A$ to $A\cup B$ red.  Note that this is all the edges of $G$ since no edges go between $S(T,y)$ and $S(T,z)$ for $y,z\in C(x)$, $y\neq z$. Clearly there are no blue paths of order $n$. Any red path may intersect at most two of the sub-trees $S(T,y)$, $S(T,z)$ for $y,z\in C(x)$, $y\neq z$ and any such path must pass through $x$. For all $y\in C(x)$, the longest possible red path in $G[A_y\cup B_y]$ is of order at most $\frac n2 -1$ and so the longest red path in $G$ is of order at most $n-1$.
\end{proof}

\begin{observation}
Let $G$ be a connected graph.  If there exists a vertex $x$ and a BFS tree $T$ rooted at $x$ such that every pair of consecutive levels of $T$  have fewer than $n$ vertices, then there exists a 2-coloring of the edges of $G$ such that every monochromatic path has order less than $n$.
\end{observation}

\begin{proof}
For all $i\geq 0$, let $D_i=\{v: d_T(x, v)=i\}$.  For all $j\geq 0$, color the edges from $D_{2j}$ to $D_{2j}\cup  D_{2j+1}$ red and the edges from $D_{2j+1}$ to  $D_{2j+1}\cup D_{2j+2}$ blue.  By the property of BFS trees, this accounts for every edge in $G$.  Since every two consecutive levels contain fewer than $n$ vertices, there are no monochromatic paths of order $n$.
\end{proof}

The following observation was inspired by Figure 2 in both \cite{BKLL1} and \cite{BKLL2}.

\begin{observation}
If $G$ is a graph on $N$ vertices with $\alpha(G)\geq N-(n-2)$, then there exists a 2-coloring of the edges of $G$ such that every monochromatic path has order less than $n$.
\end{observation}

\begin{proof}
Let $S$ be an independent set of order at least $N-(n-2)$ and partition the vertices of $V(G)\setminus S$ into disjoint sets $X,Y$ with $|X|, |Y|\leq \frac{n}{2}-1$.  Color all edges incident with $X$ red and color all edges incident with $Y$ blue (so edges between $X$ and $Y$ can be either color).  The longest monochromatic path has order at most $2(\frac{n}{2}-1)+1=n-1$.
\end{proof}

Finally, we end with the following question which relates to the upper bound on the size-Ramsey number of a path.

\begin{question}
What is the largest monochromatic path one can find in an arbitrary 2-coloring of a $d$-regular graph on $n$ vertices?
\end{question}

For instance, suppose $d=5$ and there was an upper bound of $n/28$.  This would imply that 5-regular graphs on $28n$ vertices (having $70n$ edges) have a 2-coloring with no monochromatic $P_n$.  In other words, 5-regular graphs could never improve the current best \cite{DP2} upper bound $\hat{R}(P_n)\leq 74n$.


If you've always dreamed of going to Alaska, stop putting it off. And if you've been once and think you've crossed it off your list, think again. Alaska is a cruise destination unlike any other, showcasing a side of the United States that's rarely seen: still truly wild and open to endless adventure. The reasons to go -- and go now -- quickly outweigh the mundane cons of time and money. (You can make both!) If you're considering a first-time or repeat cruise to Alaska, the following are seven reasons to book sooner than later.
Take one of the most popular cruise destinations, and then squeeze its season into a third of the year; that's what you'll find when booking a cruise to Alaska. Although Alaska sailings are only available from mid-May to mid-September, high demand means there is room for plenty of ships to fill to capacity during those four summer months. Almost every oceangoing cruise line -- including luxury and expedition styles -- serve the 49th state in the U.S. of A, so there's more than a dozen lines to choose from.
Not sure which line to pick? We compare these 11 cruise ships in Alaska to give you an idea of which cruise line will offer the best Alaska experience for you. There's never been a better time for choice when it comes to exploring the Last Frontier; just make sure to book way in advance -- up to a year -- for your pick of the litter.
Many cruisers and tourists consider Alaska a bucket list destination, but why wait until the end of your life to experience such a spectacular place? That's the thought shared by the large numbers of families who travel there each year. Curious kids can immediately appreciate the grandiosity of watching a glacier calve or a whale breach. (In fact, dogsledding and whale-watching are included in our Best Shore Excursions for Kids.) Alaska meets the checklist of various activity levels and points of interest for all ages, as well as seasonality that falls during summer break. If you're looking for your next family trip with the kids and grandkids, you should start planning your Alaska cruise right now.
If you can reach out and touch an iceberg or come within feet of a brown bear fishing for salmon, we'd call that pretty close and personal. In Alaska, there's an abundance of glowing blue glaciers and wild animals patrolling the land, air and sea -- and you can see them from the comfort of your cruise ship. Don't believe us? Check out 10 Pictures of Alaska Cruise Wildlife.
Of course, encountering a moose on a hike or an otter by kayak is arguably more thrilling, but for the less adventurous or able, the wildlife viewing from the deck of your ship is equally satisfying. As a bonus, most if not all Alaska itineraries include scenic cruising days where your sole intention is to whip out the binoculars, grab a coffee and scour the horizon for any sign of a whale tail or bald eagle's white head. On Glacier Bay cruises, a national park ranger is required to board in those protected waters, and provide informative commentary about what you're looking at and what to look out for. Glaciers are receding at a rapid rate, so visit before the landscape is forever changed (or more protected from human visitation).
If you've heard anything from family, friends or forums about Alaska cruises, there's a strong chance that the elaborate excursions have come up. While the price ranges might be unlike other shore tours you've encountered before, the experiences are even more incredible. Anything involving flight doesn't come cheaply, but Alaska is one of the best places to grab an aerial view -- especially as planes are a regular mode of transportation in a state where many locales aren't accessible by road. Floatplanes to bear-viewing parks or helicopters to sled dog camps or tundras high above sea level leave an indelible cruise memory.
And for every tame salmon bake or city tour, there are once-in-a-lifetime activities like snorkeling in the frigid waters of Ketchikan or paddleboarding past Bear Glacier in Kenai Fjords National Park (Seward). If you take a cruise to Alaska, it's guaranteed that your albums won't be full of just another boring beach picture.
While a cruise is certainly a vacation highlight, Alaska can't be fully experienced in a week at sea. A few extra days on terra firma add to the awe with sights like Denali National Park or the Kenai River Valley, which aren't accessible by cruise ship. Whether you build your own Alaska cruise land tour or book a pre-arranged cruisetour with your cruise line, a trip out to the largest U.S. state (by far) is served well with a land tour before you board, after you disembark or even both. It will take many times over to see all there is to see across Alaska, which could fit New York inside its confines about 12 times. That's why, you better start exploring now -- there's many future Alaska cruises (and land tours) to book!
Alaska became a U.S. territory when Secretary of State William Seward purchased it from Russia in 1867. At the time, people thought this was a poor idea, and dubbed the transaction "Seward's Folly" -- can you imagine? Make up for the initial lack of enthusiasm in 2017, when Alaska celebrates its sesquicentennial or 150th anniversary as part of the United States. Sitka, the site of the original purchase, will certainly be recognizing the milestone with yearlong cultural events. And if you miss everything going on this year or next, make sure to plan on an Alaska cruise for 2019, when Alaskans can rejoice over 60 years of official statehood.


text:
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declare(strict_types=1);

namespace Sylius\Bundle\ThemeBundle\Translation\Provider\Loader;

use Symfony\Component\Translation\Loader\LoaderInterface;

/**
 * @author Kamil Kokot <kamil@kokot.me>
 */
final class TranslatorLoaderProvider implements TranslatorLoaderProviderInterface
{
    /**
     * @var LoaderInterface[]
     */
    private $loaders;

    /**
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     */
    public function __construct(array $loaders = [])
    {
        $this->loaders = $loaders;
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    /**
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    {
        return $this->loaders;
    }
}

redpajama_set_name: RedPajamaGithub


At least nothing is on fire, Eva thought with a welling of relief deep inside.
In fact, better than nothing being on fire, nothing looked damaged at all. Everything in the women's ward was just where she had left it. Outside, well, the prison was a fairly large place. Eva was mostly certain that everything was normal. Her blood wards gave her a view of the place in a manner reminiscent of her using clouds of blood to see and through that, she couldn't see anything amiss.
Neither could she sense the presence of Daru anywhere. Nor any sign of him physically being here through her sense of blood. If this were a trap, she would have expected some trail of breadcrumbs to point her in the right direction.
Then again, it was barely past dawn. Dew coated the grass and the sun was too weak to give off much heat. The note had said to come at nightfall.
Which meant that she might have a few hours to set up traps before they showed up.
First thing was to prepare her secondary blood wards around the prison. They were wards that Eva had set up after the Elysium Order had attacked. The nuns had sat outside of her wards, taking them down while being a safe distance away. Eva's secondary wards would lie dormant in a ring around the edges of her main ward, surrounding the prison. If someone walked up and tried to take down the main wards, the secondary wards would activate and hopefully ruin someone's day.
It probably wouldn't work against anyone who knew what they were doing. Someone would probably notice.
Hiding her wards better was one of the things that Eva was hoping to learn in next year's schooling.
But, delaying attackers for a few extra minutes might make all the difference in the world.
Turning back into the women's ward, Eva ran straight into Zoe. The professor immediately clasped her hands on Eva's shoulders, freezing her movement.
"Devon needs to be warned," Eva said. That was the second thing she needed to do. It would have been the first, but she didn't want to run from the women's ward all the way out to Devon's building only to have to run back to the women's ward.
Eva opened her mouth to retort, but found she had no real response. The women's ward, and the prison by extension, had always been safe. Well, almost always. Barring that one incident with the Elysium Order inquisitors. An incident that her secondary blood wards should mitigate in the future.
But she hadn't even considered that someone could have taken down her wards in the two or so hours since she was last here, set up traps, and waited.
Eva shuddered. There likely wouldn't have been anything left of her.
"Alright. I concede the point. But it is safe right now. My wards are still up and Devon is still around. Or, the carnivean is at least. I assume that he is with her.
"That's…" Zoe trailed off, rubbing her forehead. She pulled out her cellphone and started typing away. "I'm warning Wayne," she said.
Catherine was supposed to have met her at the dormitories along with Prax. That was obviously not going to happen now. She needed to get word to them.
But she also needed to do the wards and other traps.
Eva didn't wait for any further protests, questions, or comments from Zoe. She all but ran back into her room.
The hovering orb of blood that both powered her wards and held the blood from everyone keyed in had grown quite large since she had first created it. Just about everyone that Eva knew had wound up keyed in at some point. Though they needed to donate only a small vial's worth of blood, it all added up. A mixture of black and red, demon and human blood all swirled together.
Seeing the ward only reassured Eva that nothing untoward had happened either while she was hunting down Sawyer or during the few hours that she was gone rescuing Lucy. When people not keyed into the ward tried to walk around inside the area of effect, the orb would shrink. Not nearly so much as one of her shields, but enough to be noticeable.
And when she had finally gotten around to repairing the ward after the attack from the Elysium Order, there had been nothing left of the orb save for the blood used for keying people in.
Unfortunately, her secondary set of wards couldn't be activated from her room. They were essentially small yet overlapping bubbles set up around the walls of the prison.
Turning to her window, one inside her room that hadn't been obstructed by Serena, Eva blinked straight out to the nearest wall. It didn't take long for her to open up the small box that held the blood ward.
While the main ward had started out around the size of a basketball and had grown to match a small beach ball, the secondary wards were much smaller. About the size of a tennis ball. They had much less area to cover, but there were a lot more of them.
The container was just a simple plastic storage box. While it would have been nice to have Ylva make some solid void metal cubes with intricate locking mechanisms, Ylva had already offered Eva so many favors for free. Even though the favors suited Ylva's designs just as much as Eva's, asking for more seemed to be in poor taste. And, while she had money, even just a mundane lockbox was out of price range for the number that she needed.
So plastic boxes it was. Eva had added a few runes to lock them down, but that was more to keep the elements out than people. If someone managed to get this close to them, they would either be exploding from the main wards or find the runes to be less of a deterrent than a plastic tarp.
Opening the box, Eva activated the ward. It wouldn't actually cause death just yet, but it was ready to be fully turned on.
Unfortunately, that was only one of… a lot. Eva blinked straight to the next box after sealing the first and repeated the process.
She blinked to the next.
She only made it about halfway around before Devon stormed out. Zoe and the carnivean trailed just behind him, with Zoe wincing and rubbing her forehead. The reason for her headache quickly became apparent.
Just behind the carnivean was a certain familiar demon with waxy skin. The same one that he had summoned immediately before the Elysium Order had attacked.
Eva didn't bother to stand and greet them. She finished activating the ward that she had been working on before turning to regard the newly summoned demon.
Through her sense of demons, she had felt it pop into existence–or the mortal realm–somewhere around a third of the way through the wards. However, as before, she couldn't see any blood within the wax construct. Until she had looked at it with her eyes, all she had known was that a new demon had been summoned.
Whether or not it was the same one that he had summoned before, Eva couldn't say. The demon was female in form, made of black wax with green flames for hair and glowing red eyes. It looked like the same one, but Eva didn't know much about that species of demon. For all she knew, they all looked exactly the same.
It didn't help that she only vaguely remembered the demon. Being skewered by Sawyer's cursed blade had overshadowed pretty much every other event that day.
Eva turned her gaze away from the waxy demon's glowing eyes. The intense pain coursing through her mind was something that she definitely did recall.
Nowhere in sight was the Lord of Slaves or the non-euclidean demon.
Something that didn't fill Eva with confidence. These demon hunters thought that they could take on Zagan. A laughable prospect, but not something to take lightly anyway.
Arachne had been able to kill the carnivean without much trouble. The wax demon had a formidable ability, but it didn't look like it could take much punishment.
Eva would be feeling much safer if Devon had gone all out.
It was a good thing that Zoe had shown up. Going around activating all the wards wasn't a difficult task, but it was time consuming. If he was prepared earlier, all the better.
"Thought you said you weren't followed, girl," Devon called out as he neared, somehow managing to grumble under his breath and shout out over the short distance at the same time.
"Either I'll be alive and worry about the consequences later or I'll be dead and not care much," he said with a shrug.
Martina feels much the same way, Eva didn't bother saying. It wouldn't surprise her in the least if Devon didn't even remember who Martina was.
Eva glanced back to the wax demon. It was obviously dominated. The way it moved was stiff and unnatural. In comparison, the carnivean had her hands deep in the pockets of a sweatshirt and was glancing off to one side with an impotent scowl on her face.
The letter that was a torn, crumpled mess back in the women's ward.
Eva started, jumping slightly. Not so much at Zoe's suspicion, though the idea that the trap had already been set was a worrying one, but at the loud scoff from Devon.
Eva blinked away, straight to the next blood ward. She had no desire to get into yet another argument with Devon about Zagan. Especially not when half the prison was still unprotected from the secondary array.
After a few minutes of discussion between Zoe and Devon, Zoe teleported away. Probably to go find Zagan, though Ylva and Juliana were other possibilities.
Eva paid them no mind, watching Devon skulk back to his cell block as she rounded a corner. By the time she had finished traversing the perimeter of the prison, the sun had climbed the sky. Noon.
Still no sign of any demon hunters.
Martina Turner propped her elbows up on her desk and clasped her hands together just beneath her nose as she stared at the wall of her office.
She was not having a good day.
Most days weren't good, these days. Tolerable was about the best that she could hope for. Days where nothing happened, where Governor Anderson held his tongue on any reports that he may have, where investigative journalists weren't harping at her door about the sky—or worse, mundane journalists taking the sudden notoriety of Brakket City as an opportunity to question magic.
Those last ones were the worst. People should mind their own business. Especially the mundanes. It was getting to the point where she was thinking about sending Zagan to answer questions instead of Catherine.
Of course, it hadn't helped that she had personally had to chase away some of the journalists over the past week.
Catherine, her familiar, had gone missing for several days. As Martina had later found out, the succubus had been off gallivanting with the self-proclaimed diabolist. And not gallivanting in the manner normally associated with succubi. Rather, she had been researching some ritual circle nonsense.
Martina did not consider herself a slave driver in any sense of the phrase. Both her familiar and her contracted demons were free to go about whatever it was that they got up to in their spare time, so long as they did as she asked when she asked. Having Catherine just up and disappear for most of a week had Martina pulling hair from her head in large clumps.
Not literally, but the stress was there.
Just when she had reeled in the wayward succubus, the current situation sprung up.
Two of her security team murdered. Two demons missing.
Worse, the murders had not been behind closed doors. Not when both bodies had been lying out in the courtyard between the dormitory buildings in the early morning hours, ready for any students still around to wake up and spot them. It was a great deal more difficult to cover up public killings.
Two demon hunter related killings.
The only bright spot in the whole mess was that the hunters were not applying scorched earth tactics. They hadn't announced themselves in any manner, the school wasn't in flames, and the citizens were largely unaware of the possible danger posed by the hunters. Typically, if a community was suspected of willingly or knowingly harboring demons, all involved would be killed. Though exact amounts of survivors often depended on the temperaments of the specific hunters.
Martina wasn't sure what these particular hunters believed in, but she believed that a large portion of their passiveness—if kidnapping Lucy and Daru could count as passive—was because of Zagan. And the hel, to a lesser extent.
Martina did not miss the narrowed eyes and the clenched teeth on Zoe as she looked over Martina's shoulder.
Pausing for a moment, Martina leaned back in her chair. She half expected Zagan to cut in with his own observations. Or, more likely, for him to run off without her explicit consent.
He did neither, seemingly content to remain standing just behind Martina.
Martina nodded and waved the professor away. She had half a mind to insist on her staying around Brakket. Though they didn't often agree on things—especially things related to demons—Martina had to admit that Zoe was a talented mage. However, what she could do that Zagan couldn't was essentially nothing.
Zoe blinked, regarding Martina with an odd look in her eye before she finally nodded. "You as well," she said as she stood.
Martina just rolled her eyes as Zoe walked out of the room. She couldn't be quite sure what that last look had been, but it was almost as if Zoe had expected her to wish her death.
Obviously, Martina didn't want Zoe to die. Finding a replacement theory professor would be a nightmare. Especially if she was more or less alright with demons. Something of this scale would be near impossible to cover up. Half of her staff would probably be resigning before next year started.
As such, if Zoe and her little group could clean up these demon hunters quietly and away from Brakket, all the better for Martina.
As soon as the door had firmly shut behind the professor, Martina kicked back in her chair, propping her feet up on top of her desk. She pulled a bottle of Hellfire from her desk drawer and didn't even bother to pour it into a glass before drinking down half the bottle.
She slammed the bottle down on the desk. Breathing out a breath that felt like it was on fire, she glanced over her shoulder.
Zagan walked around her desk, seating himself in the chair that had just held Zoe.
"I agree with your assessment. They were trying to get me out of the way. A distraction while they cleaned up the rest of the demons in town, most likely. Powerful though I am, I am limited to a single location at a time.
Martina didn't so much as feel the slightest discomfort under his gaze. She had spent enough time around him for any danger he might have posed to have worn off. His words were another story. She buried her head into her hands, rubbing her eyes.
This entry was posted on Thursday, December 29th, 2016 at 12:05 am	and tagged with Devon, Eva, Martina, Zagan, Zoe and posted in Book 007.	You can follow any responses to this entry through the RSS 2.0 feed.


Description This stereograph is from the 'Scenes in the Indian Country' series at the treaty negotiations at Fort Laramie, Wyoming.
Notes Verso: [handwritten] No. 29. Bears Claw - a Crow Warror Mounted on his Mule.


Comedy sequels rarely enjoy the longevity, let alone the renown, of their predecessors. Not so in the case of Sister Act 2: Back in the Habit, an inspired follow-up that turns a fish-out-of-water tale into an underdog story with a funky fresh twist. Whoopi Goldberg puts on her inspirational teacher hat for the movie that courageously celebrates the true heroes of America's failing education system: show choirs.
Blythe Rudloff joins our mighty chorus to rap about the awesome power of musical performance. So if you want to be somebody and want to go somewhere, you gotta wake up and pay attention to this divine episode!


text:
void fb_nand_flash_write(const char *cmd, void *download_buffer,
			 unsigned int download_bytes);
void fb_nand_erase(const char *cmd);


Abby Hardy has just signed up. Say hello!
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Leonardo DiCaprio was an undeniably adorable toddler, but his childhood photos are more controversial than you'd expect.
Facebook page "History in Pictures" recently shared a cute photo of the Oscar winner (!) with his parents George DiCaprio and Irmelin Indenbirken from the 1970s and it has the Internet in an uproar.
The picture shows a very blond Leo in a pinstriped shirt, balancing on his lookalike mom's shoulder and his dad's hand. But instead of people focusing on the overall happiness exhibited in this family photo, many started to draw attention to his mother's unshaved armpits.
Commenters began attacking Indenbirken's hygiene, and subsequently, her appearance, calling her armpit hair "gross" and "disgusting." Funny how quickly people seem to forget that women had, and have, the right to decide if they want to shave or not.
We say, who cares? Indenbirken is gorgeous and glowing in the picture, just as she is at the Oscars with Leo today. For anyone who can't seem to get over her armpit hair, we say: Get a life.


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text:
Élie-Abel Carrière, né le  à May-en-Multien et mort le , est un horticulteur et un botaniste français.

Carrière
Il est chef jardinier des pépinières du Muséum national d'histoire naturelle de Paris. Il est le rédacteur en chef de la Revue horticole à partir de 1866 et exerce une profonde influence sur l'horticulture.

Spécialiste des conifères, il décrit de très nombreuses espèces ainsi que les nouveaux genres Tsuga, Keteleeria et Pseudotsuga. Sur ce sujet, il publie le Traité général des conifères ou description de toutes les espèces et variétés de ce genre aujourd'hui connues, avec leur synonymie, l'indication des procédés de culture et de multiplication qu'il convient de leur appliquer (1855, seconde édition largement revue 1867).

Il publie de très nombreux ouvrages sur la culture des arbres fruitiers ainsi que sur l'obtention des variétés horticoles. Il écrit aussi des livres de philosophie et de morale.

Il est membre de la Société nationale d'horticulture de France.

Bibliographie partielle
Jardin fruitier - Fruits à pépins - Poires (1845)
Pépinières (1855)
.
Entretiens familiers sur l'horticulture (1860)
Guide pratique du jardinier multiplicateur (1862)
Production et fixation des variétés dans les végétaux (1865)
Arbre généalogique du groupe pêcher (1867)
Description et classification des variétés de pêchers  et de brugnoniers (1867)
Encyclopédie horticole (1880)
Montreuil aux pêches-Historique et pratiques (1880)
Semis et mise à fruit des arbres fruitiers (1881)
Étude générale du genre Pommier (1883)
Pommiers microcarpes et pommiers d'agrément (S.D.)
La Vigne et Réfutations sur la culture de la vigne (S.D.)

L'Arbre fruitier haute tige. Manuel pratique d'arboriculture fruitière (1908)

Distinctions

Liens externes

Notes et références

Botaniste français du XIXe siècle
Horticulteur français
Naissance en juin 1818
Décès en août 1896
Décès à 78 ans
Chevalier de la Légion d'honneur

redpajama_set_name: RedPajamaWikipedia


I finally had the chance to finish this track (it seems like I've been working on it forever!). It's a dark brooding remix getting it's main guitar line from ugly by now's "wall drill." I also used bits of the vocal parts and some more of the guitar (but it's hard to recognize it). Pretend you're a spy on a mission.


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text:
(-) Philippines (11)
Philippines: "It Takes a Long Time for People to Rebuild Their Lives"
Over the past seven weeks, a Doctors Without Borders/Médecins Sans Frontières (MSF) team has run mobile clinics by boat to deliver medical and humanitarian aid to five islands south of Guiuan that ...
"Some Parts Are Still Devastated": Two Months in Tacloban, Philippines
Doctors Without Borders/Médecins Sans Frontières (MSF) doctor Natalie Roberts spent two months working in the Philippines, running MSF's inflatable hospital in Tacloban. Here, she describes her ...
Arizona Public Media: Philippines Recovery Includes Tucsonan Disaster Coordinator
Ibrahim Younis describes the devastation of Typhoon Haiyan in the Philippines and MSF's response. View external media
Arizona Public Media: Tucson Resident Helps Philippines Relief Efforts
MSF's Ibrahim Younis describes the aid response in the Philippines in the month after Typhoon Haiyan struck the country. View external media
Thomson Reuters: Doctors Race to Fight Preventable Tragedies in Typhoon-Hit Philippines
Johan von Schreeb, an MSF surgeon in Guiuan, the Philippines, speaks about the urgent need to treat injuries from the typhoon, which are at risk of serious and life-threatening infections. View ...
Media Coverage : Feb 10, 2014
NPR: Doctors Struggle To Provide Care After Typhoon
MSF's Natasha Reyes describes the destruction and medical needs in Guiuan, where Typhoon Haiyan first made landfall in the Philippines. View external media
New York Times: Access to Medical Care Improves in Storm-Ravaged Philippines
An MSF team had 80 patients within an hour of their arrival in Santa Fe, a typhoon-damaged inland town on the Philippines' Leyte Island. View external media
Anderson Cooper 360: Doctors Without Borders help typhoon survivors
MSF's Damien Moloney describes efforts to set up an inflatable field hospital in Tacloban, the Philippines, in the aftermath of Typhoon Haiyan. View external media
New York Times: Typhoon Aid Gridlock Paralyzes Epicenter of Damage
MSF's emergency teams in the Philippines are encountering immense needs and logistical difficulties in areas devastated by Typhoon Haiyan. View external media
NPR: In Typhoon-Ravaged Philippines, Aid Trickles In
MSF's emergency coordinator Natasha Reyes describes the destruction and injuries caused by Typhoon Haiyan in the Philippines. View external media

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text:
package ch.heigvd.quaris.api.dto;

import java.util.Objects;
import com.fasterxml.jackson.annotation.JsonProperty;
import com.fasterxml.jackson.annotation.JsonCreator;
import io.swagger.annotations.ApiModel;
import io.swagger.annotations.ApiModelProperty;
import org.joda.time.DateTime;

/**
 * EventDTO
 */
@javax.annotation.Generated(value = "class io.swagger.codegen.languages.SpringCodegen", date = "2017-01-25T16:31:01.726+01:00")

public class EventDTO   {
  private String type = null;

  private String identifier = null;

  private Object payload = null;

  private DateTime createdAt = null;

  private String application = null;

  public EventDTO type(String type) {
    this.type = type;
    return this;
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   /**
   * Event's type
   * @return type
  **/
  @ApiModelProperty(value = "Event's type")
  public String getType() {
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    return this;
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   /**
   * user identifier
   * @return identifier
  **/
  @ApiModelProperty(value = "user identifier")
  public String getIdentifier() {
    return identifier;
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  public void setIdentifier(String identifier) {
    this.identifier = identifier;
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    return this;
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   /**
   * Parameters to use in rules (optional)
   * @return payload
  **/
  @ApiModelProperty(value = "Parameters to use in rules (optional)")
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   /**
   * Get createdAt
   * @return createdAt
  **/
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  public DateTime getCreatedAt() {
    return createdAt;
  }

  public void setCreatedAt(DateTime createdAt) {
    this.createdAt = createdAt;
  }

  public EventDTO application(String application) {
    this.application = application;
    return this;
  }

   /**
   * Application name
   * @return application
  **/
  @ApiModelProperty(value = "Application name")
  public String getApplication() {
    return application;
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    return Objects.equals(this.type, eventDTO.type) &&
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        Objects.equals(this.payload, eventDTO.payload) &&
        Objects.equals(this.createdAt, eventDTO.createdAt) &&
        Objects.equals(this.application, eventDTO.application);
  }

  @Override
  public int hashCode() {
    return Objects.hash(type, identifier, payload, createdAt, application);
  }

  @Override
  public String toString() {
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    sb.append("class EventDTO {\n");

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}


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text:
Former EPL official Mark Clattenburg slammed over female referee comments
A former EPL official has caused a social media stir for his controversial comments regarding the challenges facing female referees.
Sandra Brobbey, The Sun
Kerr makes history in Matildas' 18-0 bloodbath
Goal scoring GOAT: Kerr claims Cahill's scalp
Border backflip has Glory boss 'flabbergasted'
Mark Clattenburg has come under fire for suggesting women referees have to choose between their jobs and having children.
The former Premier League ref reckons female match officials who get pregnant during their careers face a "difficult pathway".
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Clattenburg, who left England's top-flight in 2017 to manage referees in Saudi Arabia, was sharing his opinions with Ally McCoist, Alan Brazil and England rugby ace Joe Marler on talkSPORT.
And the Durham-born official, who went on to ref in China in February 2019, was asked for his thoughts on Sara Cox making history by becoming the first female referee to oversee a Premiership Rugby clash.
Clattenburg, 46, said: "We always had an assistant referee in the Premier League, Sian Massey (Ellis), and now we have a woman refereeing in the Football League, Rebecca Welch, so women are starting to develop in the men's game.
Clattenburg claimed that female officials who get pregnant in their refereeing career face a "difficult pathway". Picture: AFP Photo/Oli Scarff
"If you look at Uefa, for example, the French woman (Stephanie Frappart) refereed the Super Cup final, so Uefa are getting more and more women.
"The problem with women is, and certainly in refereeing in football, they have a difficult pathway if they get pregnant during their refereeing career – it can stop them a long way.
"So they have to make this choice: do they want to be pregnant or do they want to be referees?"
Clattenburg added: "They've also got to pass the men's fitness test.
"A lot of women struggle with the men's fitness test. If you want to be in the men's game, you have to meet those criteria.
"So, if they pass all this, and they choose the right path, I believe that women should be involved in the men's game, as well as men being involved in the women's game.
"If we look at the Women's Soccer League, for example, there's been some high-profile mistakes.
"Why not bring in the best referees, if you've got the best leagues."
Clattenburg's views got short shrift from some accounts on social media, as many were quick to slam the former Premier League whistleblower.
Mark Clattenburg says female referees have to choose between their careers & children.🤔Hmmm WHY? Maybe 50 years ago he'd have a point, but we've come a long way since then & most women juggle ALOT! It's exactly those sorts of outdated views that stop females progressing in sport
— Rebecca Williams 📺 (@RebeccaTVnews) September 30, 2021
Appalled at the comments Mark Clattenburg made (largely unchallenged) about female referees on talkSPORT this morning.
To suggest a woman can't pursue a certain career because they'd also like to have children is archaic. Can't believe we haven't moved past this kind of attitude https://t.co/PQgFlHEzNe
— Nancy Gillen (@Nancy_Gillen) September 30, 2021
On female referees: "Choose the right path" - wow Mark Clattenburg, what era are you living in?!!! "Pregnancy costs you 2/3 years of your life" This is exactly the type of ridiculous statement that has held us all back for so long and made many of us fearful of getting pregnant.
— Faye Carruthers (@FayeCarruthers) September 30, 2021
Does Mark Clattenburg realise there are international women's rugby players who play Test match rugby after giving birth? https://t.co/Fj05szaZNA
— Jessica Hayden (@_JessHayden) September 30, 2021
When quizzed on his opinion the former Prem ref, who oversaw the Euros 2016 final between Portugal and France, refused to alter his standpoint.
Clattenburg said: "Women have to make that sacrifice. Say if they get to a certain level in refereeing, and they want to get to that next level.
"If you become pregnant, it can cost you two or three years of your life.
"Once you lose that three years, somebody else takes your position, so that they move up.
"If you want to have babies, you've got to do it early in your career.
"If you start having babies later on in your career, you miss that opportunity and it's sad because there are some great women."
This article originally appeared in The Sun and has been republished with permission.
Australia's best-ever football player just became its greatest goalscorer, piling on five goals in a ridiculous rout of minnow Indonesia.
Australia's best-ever football player just became its greatest goalscorer, Sam Kerr piling on five goals to pass Tim Cahill in a ridiculous Asian Cup rout.
Perth Glory officials aren't happy with WA Premier Mark McGowan\s backflip on opening the WA borders next month.


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Welcome to day 21 of my 25 Days of Craftmas series! I'm super excited to share how to make this heat embossed Christmas apron. Not only do I love how it turned out, but I love that I was able to try something new and it worked! Now, I have all kinds of ideas in my head of how I can use heat embossing on other projects.
I've always been fascinated with heat embossing. I remember as a kid, watching my aunt do heat embossing with stamps. I thought it was amazing and have always wanted to try it But, I've never been much of a stamper, so I never invested in the tools. But then, I came across this post and knew it was time to give heat embossing a try. So, I bought my supplies, created a design, and made this adorable Christmas apron!
Open up the design in your software program. I sized my design to about 6″ x 4.85″ Add a box around the design to create a stencil. Then cut out the vinyl and weed out the design.
Apply transfer tape and then remove the vinyl backing.
Center the vinyl and apply it to the apron.
Use an application tool or scraper to rub over the vinyl.
Then remove the transfer tape.
Fill in the design with Fabric Mod Podge. You'll want to do this quickly so the Mod Podge doesn't have time to dry.
While the Mod Podge is still wet, quickly sprinkle embossing powder over the design. The quicker you can cover the design, the better.
Once it is all covered, press over the whole design to help the embossing powder stick to the Mod Podge. Let it dry for a few minutes.
Shake off the excess powder and save for later.
Carefully remove the vinyl stencil.
Use a heat gun to melt the embossing powder and watch the magic happen. I LOVE this part! I used this heat gun. Since I've never embossed before, I just found an inexpensive gun with good reviews and bought it.
It takes a little bit of time, but you'll be able to tell when it's done.
I wanted to test out how the embossing held up after washing. So, I took this picture before I threw the apron in the wash.
I washed the apron on a regular cycle and hung it to dry. This is what it looks like now, after washing. It still looks great!
I love the sparkle and the dimension the heat embossing gives to the apron. And I love how easy it was to do! What do you think? Would you give heat embossing a try?
Cute design, thanks. Does the heat embossing hold up in the wash? Same product used in paper craft?
I ran the apron through one regular cycle and hung it to dry. To me, it looked the same before and after washing. Yes, it's that Mod Podge.
What does the "fabric" mod podge do? Does it have to be fabric mod podge??
I just used fabric Mod Podge because that's what I read was used on a similar project. After reading the differences of the different Mod Podges, it sounds like you could probably use regular Mod Podge as well.
What a great idea! I am same… friends were into stamps and embossing especially with cards & scrapbooks but I was not interested in spending a fortune on stamps!
Did this require a special embossing powder? I saw fabric Modpoge…. Also, how do you think an everyday wear shirt might hold up?
I just used regular embossing powder. There was nothing special about it.
Wow, I love heat embossing stamps…now I'll have to give this a try.