text:
\section{Introduction}

The celebrated Max-Cut problem asks for the largest bipartite subgraph of a graph $G$, i.e.,
for a partition of the vertex set of $G$ into disjoint sets $V_1$ and $V_2$ so that
the number of edges of $G$ crossing $V_1$ and $V_2$ is maximal. This
problem has been the subject of extensive research, both from a largely
algorithmic perspective in computer science and from an extremal perspective in combinatorics.
Throughout, let $G$ denote a graph with $n$ vertices and $m$ edges with maximal cut of size $f(G)$.
The extremal version of Max-Cut problem asks to give bounds on $f(G)$ solely as a function of $m$ and $n$.
This question was first raised more than fifty years ago by Erd\H{o}s~\cite{E1} and has attracted a lot of attention
since then (see, e.g.,~\cite{E73, ER75, EFPS, AL96, SHE92, BS1, AL05, Su07, CFKS} and their references).

It is well known that every graph $G$ with $m$ edges has a cut of size at least $m/2$. To see this, consider a random
partition of vertices of the vertices $G$ into two parts $V_1, V_2$ and estimate the expected number of edges between $V_1$ and $V_2$. On the other hand, already in 1960's Erd\H{o}s~\cite{E1} observed that the constant $1/2$ cannot
be improved even if we consider very restricted families of graphs, e.g., graphs that contain no short cycles.
Therefore the main question, which has been studied by many researchers, is to estimate the error term $f(G)-m/2$, which we call {\it surplus},  for various families of graphs $G$.

The elementary bound $f(G) \geq m/2$  was improved by Edwards \cite{E73, E75} who showed that every graph with $m$ edges has a cut of size at least $\frac{m}{2}+ \frac{\sqrt {8m+1}-1}{8}$.
This result is easily seen to be tight in case $G$
is a complete graph on an odd number of vertices, that is, whenever $m=\binom{k}{2}$ for some odd integer $k$. Estimates on the
second error term for other values of $m$ can be found in~\cite{AH} and~\cite{BS1}.

Although the $\sqrt{m}$ error term is tight in general, it was observed by Erd\H{o}s and Lov\'asz \cite{ER75} that for triangle-free graph it can be improved to at least $m^{2/3+o(1)}$.
This naturally yiels a motivating question: what is the best surplus which can always be achieved if we assume that our family of graphs
is {\it $H$-free}, i.e., no graph contains a fixed graph $H$ as a subgraph.
It is not difficult to show (see, e.g.~\cite{AL03}) that for every fixed
graph $H$ there is some $\epsilon=\epsilon(H)>0$ such that $f(G) \geq \frac{m}{2}+\Omega(m^{1/2+\epsilon})$ for all $H$-free graphs with $m$ edges. However, the problem of
estimating the error term more precisely is not easy, even for relatively
simple graphs $H$. It is plausible to conjecture (see~\cite{AL05}) that for every fixed graph $H$ there is a constant $c_H$ such that every $H$-free graph $G$ with $m$ edges has a cut with surplus at least
$\Theta(m^{c_H})$, i.e., there is both a lower bound and an infinite sequence of example showing that exponent $c_H$ can not be improved.
This conjecture is very difficult. Even in the case $H=K_3$ determining the correct error term took almost twenty years. Following the works of~\cite{ER75, PT, SHE92}, Alon~\cite{AL96} proved that every $m$-edge triangle free graph has a cut with surplus of order $m^{4/5}$ and that this is tight up to constant factors. There are several other forbidden graphs $H$ for which we know quite accurately the error term for the extremal Max-Cut problem in $H$-free graphs.  For example, it was proved in~\cite{AL05}, that if $H=C_r$ for $r=4, 6, 10$ then  $c_H=\frac{r+1}{r+2}$. The answer is also known in the case when $H$ is a complete bipartite graph $K_{2,s}$ or $K_{3,s}$ (see~\cite{AL05} for details).

\paragraph{New approach to Max-Cut using semidefinite programming.}
Many extremal results for the Max-Cut problem rely on quite elaborate probabilistic arguments.
A well known example of such an argument is a proof by Shearer~\cite{SHE92} that if $G$ is a triangle-free
graph with $n$ vertices and $m$ edges, and if $d_1, d_2, \ldots, d_n$ are the degrees of its vertices, then
$f(G) \geq \frac{m}{2} + O(\sum_{i=1}^n \sqrt d_i)$. The proof is quite intricate and is based on first choosing a random cut
and then randomly redistributing some of the vertices, depending on how many their neighbors are on the same side as the chosen vertex in the initial cut.
Shearer's arguments were further extended, with more technically involved proofs, in~\cite{AL05} to show that the same lower bound remains valid for graphs $G$ with relatively sparse neighborhoods (i.e., graphs which locally have few triangles).

In this article we propose a different approach to give lower bounds on the $\MC$ of sparse $H$-free graphs using approximation by semidefinite programming (SDP). This approach is intuitive and computationally simple. The main idea was inspired by the celebrated approximation algorithm of
Goemans and Williamson~\cite{GW95} of the $\MC$: given a graph $G$ with $m$ edges, we first construct an explicit solution for the standard $\MC$ SDP relexation of $G$ which has value at least $(\frac{1}{2}+W)m$ for some positive surplus $W$.
We then apply a Goemans-Williamson randomized rounding, based on the sign of the scalar product with random unit vector, to extract a cut in $G$ whose surplus is within constant factor of $W$.
Using this approach we prove the following result.

\begin{theorem}
	\label{thm:sdp}
	Let $G = (V,E)$ be a graph with $n$ vertices and $m$ edges.
	For every $i \in [n]$, let $V_i$ be any subset of neighbours of vertex $i$ and $\varepsilon_i \leq \frac{1}{\sqrt{\vert V_i \vert}}$.
	Then,
	\begin{align}
	f(G) \geq \frac{m}{2} + \sum_{i=1}^n \frac{\varepsilon_i \vert V_i \vert }{4 \pi} - \sum_{(i,j) \in E} \frac{ \varepsilon_i \varepsilon_j |V_i\cap V_j|  }{2}.
	\end{align}
\end{theorem}

\noindent
This results implies the Shearer's bound \cite{SHE92}. To see this, set $V_i$ to the neighbors of $i$ and $\varepsilon_i=\frac{1}{\sqrt{d_i}}$ for all $i$. Then, if $G$ is triangle-free graph, then  $|V_i\cap V_j|=0$ for every pair of adjacent vertices $i,j$.

The fact that we apply Goemans-Williamson SDP rounding in this setting is perhaps surprising for a few reasons.
In general, our result obtains a surplus of $\Omega(W)$ from an SDP solution with surplus $W$, which is not possible in general.
The best cut that can be guaranteed from any kind of rounding of a Max-Cut SDP solution with value $(\frac{1}{2}+W)m$ is $(\frac{1}{2}+\Omega(\frac{W}{\log W}))m$ (see \cite{OW08}).
Furthermore, this is achieved using the RPR$^2$ rounding algorithm, not the Geomans-Williamson rounding algorithm.
Nevertheless, we show that our explicit Max-Cut solution has additional properties that circumvents these issues and permits a better analysis.

\paragraph{New lower bound for Max-Cut of triangle sparse graphs}

Using Theorem~\ref{thm:sdp}, we give a new result on the Max-Cut of triangle sparse graphs that is more convenient to use than previous similar results.
A graph $G$ is \emph{$d$-degenerate} if there exists an ordering of the vertices $1,\dots,n$ such that vertex $i$ has at most $d$ neighbors $j<i$.
Equivalently, a graph is $d$-degenerate if every induced subgraph has a vertex of degree at most $d$.
Degeneracy is a broader notion of graph sparseness than maximum degree: all maximum degree $d$ graphs are $d$-degenerate, but the star graph is 1-degenerate while having maximum degree $n-1$.
Theorem~\ref{thm:sdp} gives the following useful corollary on the Max-Cut of $d$-degenerate graphs.
\begin{corollary}
	\label{cor:sdp-2}
	Let $\varepsilon\le \frac{1}{\sqrt{d}}$.
	Let $G$ be a $d$-degenerate graph with $m$ edges and $t$ triangles.
	Then
	\begin{align}
	f(G)\ge \frac{m}{2} + \frac{\varepsilon m}{4\pi} - \frac{\varepsilon^2 t}{2}.
	\end{align}
\end{corollary}

\noindent
Indeed, let $1,\dots,n$ be an ordering of the vertices such that any $i$ has at most $d$ neighbors $j<i$, and let $V_i$ be this set of neighbors.
Let $\varepsilon_i=\varepsilon$ for all $i$.
In this way, $\sum_{i}^{} |V_i|$ counts every edge exactly once and $\sum_{(i,j)\in E}^{} |V_i\cap V_j|$ counts every triangle exactly once, and the result follows.
This shows that graphs with few triangles have cuts with surplus similar to triangle-free graphs.

This result is new and more convenient to use than existing results in this vein, because it relies only on the global count of the number of triangles, rather than a local triangle sparseness property assumed by prior results.
For example, it was shown that (using Lemma 3.3 of \cite{AL05}) a $d$-degenerate graph with a local triangle-sparseness property, namely that every large induced subgraph with a common neighbor is sparse, has Max-Cut at least $\frac{m}{2} + \Omega(\frac{m}{\sqrt{d}})$.
However, we can achieve the same result with only the guarantee that the global number of triangles is small.
In particular, when there are at most $O(m\sqrt{d})$ triangles, which is always the case with the local triangle-sparseness assumption above, setting $\varepsilon=\Theta(\frac{1}{\sqrt{d}})$ in Corollary~\ref{cor:sdp-2} gives that the Max-Cut is again at least $\frac{m}{2} + \Omega(\frac{m}{\sqrt{d}})$.

\paragraph{Corollary: Lower bounds for Max-Cut of $H$-free degenerate graphs.}
We illustrate usefulness of the above results by giving the following lower bound on the Max-Cut of $K_r$-free graphs.
\begin{theorem}
	\label{thm:kr-lb}
	Let $r \geq 3$.
	There exists a constant $c = c(r) >0$ such that, for all $K_r$-free $d$-degenerate graphs $G$ with $m$ edges,
	\begin{align}
	f(G) \ge \left(\frac{1}{2} +\frac{c}{d^{1-1/(2r-4)}}\right)m.
	\label{eq:kr-lb}
	\end{align}
\end{theorem}
\noindent
Lower bounds such as Theorem~\ref{thm:kr-lb} giving a surplus of the form $c\cdot \frac{m}{d^\alpha}$ are more fine-grained than those that depend only on the number of edges.
Accordingly, they are useful for obtaining lower bounds the Max-Cut independent of the degeneracy:
many tight Max-Cut lower bounds in $H$ free graphs of the form $\frac{m}{2}+cm^{\alpha}$ first establish that $f(G) \ge \frac{m}{2} + c\cdot \frac{m}{\sqrt{d}}$ for all $H$-free graphs, and then case-working on the degeneracy. \cite{AL05}

In the case of $r=4$ one can use our arguments together with Alon's result on Max-Cut in triangle-free graphs to improve Theorem~\ref{thm:kr-lb} further to $m/2+cm/d^{2/3}$.
While Theorem~\ref{thm:kr-lb} gives nontrivial bounds for $K_r$-free graphs, we believe that a stronger statement is true and propose the following conjecture.

\begin{conjecture}\label{conj:opt}
For any graph $H$, there exists a constant $c=c(H)>0$ such that, for all $H$-free $d$-degenerate graphs with $m\ge 1$ edges,
\begin{align}
	f(G)\ge \left(\frac{1}{2} + \frac{c}{\sqrt{d}}\right)m.
  \label{eq:conj_opt}
\end{align}
\end{conjecture}

Our Theorem~\ref{thm:sdp} implies this conjecture for various graphs $H$, e.g.,  $K_{2,s}, K_{3,s}, C_{r}$ and for any graph $H$ which contains a vertex whose deletion makes it acyclic.
This was already observed in~\cite{AL05} using the weaker, locally triangle-sparse form of Corollary~\ref{cor:sdp-2} described earlier.

Conjecture~\ref{conj:opt} provides a natural route to proving a closely related conjecture proposed by Alon, Bollob{\'a}s, Krivelevich, and Sudakov~\cite{AL03}.
\begin{conjecture}[\cite{AL03}]
\label{conj:abks}
For any graph $H$, there exists constants $\varepsilon=\varepsilon(H)>0$ and $c=c(H)>0$ such that, for all $H$-free graphs with $m\ge 1$ edges,
\begin{align}
  f(G)\ge \frac{m}{2} + cm^{3/4+\varepsilon}.
\end{align}
\end{conjecture}
Since every graph with $m$ edges is obviously $\sqrt{2m}$-degenerate, the Conjecture \ref{conj:opt} implies immediately a weaker form of Conjecture~\ref{conj:abks} with surplus of order $m^{3/4}$. With some extra technical work (see Appendix~\ref{sec:conj}) we can show that it actually implies the full conjecture, achieving a surplus of $m^{3/4+\varepsilon}$ for any graph $H$.
For many graphs $H$ for which Conjecture~\ref{conj:abks} is known, \eqref{eq:conj_opt} was implicitly established for $H$-free graphs \cite{AL05}, making Conjecture~\ref{conj:opt} a plausible stepping stone to Conjecture~\ref{conj:abks}.
As further evidence of the plausibility of Conjecture~\ref{conj:opt}, we show that Conjecture~\ref{conj:abks} implies a weaker form of Conjecture~\ref{conj:opt}, namely that any $H$-free graph has Max-Cut $\frac{m}{2} + cm \cdot d^{-5/7}$.
Using similar techniques, we can obtain nontrivial, unconditional results on the Max-Cut of $d$-degenerate $H$-free graphs for particular graphs $H$.
See Appendix~\ref{app:gen} for a table of results and proofs.

Conjecture~\ref{conj:opt}, if true, gives a surplus of $\Omega(\frac{m}{\sqrt{d}})$ that is optimal up to a multiplicative constant factor for every fixed graph $H$ which contains a cycle. To see this, consider an Erd\H{o}s-R\'enyi
random graph $G(n,p)$ with $p=n^{-1+\delta}$. Using standard Chernoff-type estimates, one can easily show that with high probability that this graph is $O(np)$-degenerate and its Max-Cut has size at most $\frac{1}{4}\binom{n}{2}p+O(n\sqrt{np})$. Moreover, if $\delta=\delta(H)>0$ is small enough, then with high probability $G(n,p)$ contains only very few copies of $H$ which can be destroyed by deleting few vertices, without changing the degeneracy and surplus of the Max-Cut (see Appendix~\ref{app:ub}).

\section{Lower bounds for Max-Cut using SDP}
\label{sec:general}
In this section we give a lower bound for $f(G)$ in graphs with few triangles, showing Theorem~\ref{thm:sdp}.
To prove this result, we make heavy use of the SDP relaxation of the $\MC$ problem, formulated below for a graph $G = (V, E)$:
\begin{align}
\text{maximize}\qquad & \sum_{(i,j)\in E}^{} \frac{1}{2}(1-\ab{v\ind{i},v\ind{j}})\nonumber\\
\text{subject to}\qquad & \|v\ind{i}\|^2=1\,\forall i\in V.
\label{eq:sdp}
\end{align}
We leverage the classical Goemans-Williamson \cite{GW95} rounding algorithm which that gives an integral solution from a vector solution to the $\MC$ SDP.

\begin{proof}[Proof of Theorem~\ref{thm:sdp}]
	For $i\in[n]$, define $\tilde v\ind{i}\in\mathbb{R}^n$ by
	\begin{align}
	\tilde v\ind{i}_j \ &= \ \left\{
	\begin{tabular}{ll}
	1 & $i=j$\\
	$-\varepsilon_i$ & $j \in V_i$\\
	0 & otherwise.\\
	\end{tabular}
	\right..
	\end{align}
	For $i\in[n]$, let $v\ind{i}\defeq \frac{\tilde v\ind{i}}{\|\tilde v\ind{i}\|}\in\mathbb{R}^n$.
	Then $1\le \|\tilde v\ind{i}\|\le 1 + \varepsilon_i^2 \vert V_i \vert \le 2$ for all $i$.
	For each edge $(i,j)$ with $i \in V_j$, we have
	\begin{align}
	v_i^{(i)} v_i^{(j)} = \frac{1}{\|\tilde v\ind{i}\|}\cdot \frac{-\varepsilon_j}{\|\tilde v\ind{j}\|} \le \frac{-\varepsilon_j}{4}.
	\end{align}
	For $k \in V_i \cap V_j$, we have $v_k^{(i)} v_k^{(j)}\le \varepsilon_i \varepsilon_j$.
	For $k\not \in \{i,j\}\cup (V_i \cap V_j)$, we have $v_k^{(i)} v_k^{(j)}=0$ as $v_k\ind{i} = 0$ or $v_k\ind{j} = 0$.
	Thus, for all edges $(i,j)$,
	\begin{align}
	\ab{v\ind{i},v\ind{j}}\le -\frac{\varepsilon_i}{4}\mathbbm{1}_{V_j}(i) -\frac{\varepsilon_j}{4}\mathbbm{1}_{V_i}(j) + |V_i\cap V_j| \varepsilon_i \varepsilon_j.
	\label{}
	\end{align}

	Here, $\mathbbm{1}_S(i)$ is 1 if $i\in S$ and 0 otherwise.
	Vectors $v\ind{1},\dots,v\ind{n}$ form a vector solution to the SDP~\eqref{eq:sdp}.
	We now round this solution using the Goemans-Williamson \cite{GW95} rounding algorithm.
	Let $w$ denote a uniformly random unit vector, $A=\{i\in[n]:\ab{v\ind{i},w}\ge 0\}$, and $B=[n]\setminus A$.
	Note that the angle between vectors $v\ind{i},v\ind{j}$ is equal to $\cos^{-1}(\ab{v\ind{i},v\ind{j}})$, so the probability an edge $(i,j)$ is cut is
	\begin{align}
	\Pr[(i,j)\text{ cut}]
	\ &= \  \frac{\cos^{-1}(\ab{v\ind{i},v\ind{j}})}{\pi} \nonumber\\
	\ &= \ \frac{1}{2} - \frac{\sin^{-1}(\ab{v\ind{i},v\ind{j}})}{\pi} \nonumber\\
	\ &\ge \ \frac{1}{2} - \frac{1}{\pi}\sin^{-1}\left(|V_i\cap V_j| \varepsilon_i \varepsilon_j -\frac{\varepsilon_i}{4}\mathbbm{1}_{V_j}(i) -\frac{\varepsilon_j}{4}\mathbbm{1}_{V_i}(j) \right) \nonumber\\
	\ &\ge \ \frac{1}{2} - \frac{1}{\pi} \cdot \left(
	\frac{\pi}{2}\cdot |V_i\cap V_j| \varepsilon_i \varepsilon_j -\frac{\varepsilon_i}{4}\mathbbm{1}_{V_j}(i) -\frac{\varepsilon_j}{4}\mathbbm{1}_{V_i}(j)
	\right) \nonumber\\
	\ &= \ \frac{1}{2} + \frac{\varepsilon_i}{4\pi}\mathbbm{1}_{V_j}(i) + \frac{\varepsilon_j}{4\pi}\mathbbm{1}_{V_i}(j) - \frac{|V_i\cap V_j| \varepsilon_i \varepsilon_j}{2}.\nonumber
	\end{align}
	In the last inequality, we used that, for $a,b\in[0,1]$, we have $\sin^{-1}(a-b)\le \frac{\pi}{2}a - b$.
	This is true as $\sin^{-1}(x)\le \frac{\pi}{2}x$ when $x$ is positive and $\sin^{-1}(x)\le x$ when $x$ is negative.
	Thus, the expected size of the cut given by $A\sqcup B$ is, by linearity of expectation,
	\begin{align}
	\sum_{(i,j)\in E}^{} \Pr[(i,j)\text{ cut}]
	\ &\ge \ \sum_{\substack{ (i,j) \in E \\ i < j}}
	\left(\frac{1}{2} + \frac{\varepsilon_i}{4}
	\mathbbm{1}_{V_j}(i)
	+ \frac{\varepsilon_j}{4}
	\mathbbm{1}_{V_i}(j)
	- \frac{|V_i\cap V_j| \varepsilon_i \varepsilon_j}{2} \right) \nonumber\\
	\ &= \  \frac{m}{2} + \sum_{i=1}^n \frac{ \vert V_i \vert \varepsilon_i }{4 \pi}- \sum_{(i,j) \in E} \frac{ |V_i\cap V_j| \varepsilon_i \varepsilon_j}{2}.
	\qedhere
	\end{align}
\end{proof}

In the proof of Theorem~\ref{thm:kr-lb} we use the following consequence of Corollary \ref{cor:sdp-2}.
\begin{corollary}
	\label{cor:lb-2}
	There exists an absolute constant $c>0$ such that the following holds.
	For all $d\ge 1$ and  $\varepsilon\le\frac{1}{\sqrt{d}}$, if a $d$-degenerate graph $G = (V, E)$ has $m$ edges and at most $\frac{m}{8\varepsilon}$ triangles then
	\begin{align}
	f(G) \ge \left(\frac{1}{2} + c \varepsilon\right)  \cdot m.
	\label{}
	\end{align}
\end{corollary}

 \section{Decomposition of degenerate graphs}

 In a graph $G=(V,E)$, let $n(G)$ and $m(G)$ denote the number of vertices and edges, respectively.
 For a vertex subset $V'\subset V$, let $G[V']$ denote the subgraph induced by $V'$.
 We show that $d$-degenerate graphs with few triangles have small subsets of neighborhoods with many edges.
 \begin{lemma}[]
 	\label{lem:lb-3}
 	Let $d\ge 1$ and $\varepsilon>0$, and let $G=(V,E)$ be a $d$-degenerate graph with at least $\frac{m(G)}{\varepsilon}$ triangles.
 	Then there exists a subset $V'$ of at most $d$ vertices with a common neighbor in $G$ such that the induced subgraph $G[V']$ has at least $\frac{|V'|}{\varepsilon}$ edges.
 \end{lemma}
 \begin{proof}
 	Since $G$ is $d$-degenerate, we fix an ordering $1,\dots,n$ of the vertices such that $d_<(i) \le d$ for all $i\in[n]$, where $d_<(i)$ denotes the number of neighbors $j<i$ of $i$. Then, if $t_<(i)$ denotes the number of triangles $\{i,j,k\}$ of $G$ where $j,k < i$, we have
 	\begin{align}
 	\sum_{i}^{} t_<(i)
 	\ = \ t(G)
 	\ \ge \ \frac{m(G)}{\varepsilon}
 	\ = \ \sum_{i=1}^{n} \frac{d_<(i)}{\varepsilon}.
 	\label{}
 	\end{align}
 	Hence, there must exist some $i$ such that $t_<(i) \ge \frac{d_<(i)}{\varepsilon}$.
 	Let $V'$ denote the neighbors of $i$ with index less than $i$.
 	By definition, the vertices of $V'$ have common neighbor $i$.
 	Additionally, $G[V']$ has at least $\frac{d_<(i)}{\varepsilon}$ edges and $d_<(i)\le d$ vertices, proving the lemma.
 \end{proof}

 We use this lemma to partition the vertices of any $d$-degenerate graph in a useful way.

 \begin{lemma}
 	\label{lem:lb-4}
 	Let $\varepsilon > 0$.
 	Let $G = (V, E)$ be a $d$-degenerate graph on $n$ vertices with $m$ edges.
 	Then there exists a partition $V_1,\dots,V_{k+1}$ of the vertex set $V$ with the following properties.
 	\begin{enumerate}
 		\item For $i=1,\dots,k$, the vertex subset $V_i$ has at most $d$ vertices and has a common neighbor, and the induced subgraph $G[V_i]$ has at least $\frac{|V_i|}{\varepsilon}$ edges.
 		\item The induced subgraph $G[V_{k+1}]$ has at most $\frac{m(G[V_{k+1}])}{\varepsilon}$ triangles.
 	\end{enumerate}
 	\label{lem:technical}
 \end{lemma}
 \begin{proof}
 	We construct the partition iteratively.
 	Let $V_0^*=V$.
 	For $i\ge 1$, we partition the vertex subset $V_{i-1}^*$ into $V_i\sqcup V_i^*$ as follows.
 	If $G[V_{i-1}^*]$ has at least $\frac{m(G[V_{i-1}^*])}{\varepsilon}$ triangles, then by applying Lemma~\ref{lem:lb-3} to the induced subgraph $G[V_{i-1}^*]$, there exists a vertex subset $V_i$ with a common neighbor in $V_{i-1}^*$ such that $|V_i|\le d$ and the induced subgraph $G[V_i]$ has at most $\frac{|V_i|}{\varepsilon}$ edges.
 	In this case, let $V_i^*\defeq V_{i-1}^*\setminus V_i$.
 	Let $k$ denote the maximum index such that $V_k^*$ is defined, and let $V_{k+1}\defeq V_k^*$.
 	By construction, $V_1,\dots,V_k$ satisfy the desired conditions.
 	By definition of $k$, the induced subgraph $G[V_k^*]$ has at most $\frac{m(G[V_k^*])}{\varepsilon}$ triangles, so for $V_{k+1}=V_k^*$, we obtain the desired result.
 \end{proof}

 \subsection{Large Max-Cut from decompositions}
 \label{s:framework}
 For a $d$-degenerate graph $G=(V,E)$, in a partition $V_1,\dots,V_{k+1}$ of $V$ given by Lemma~\ref{lem:lb-4}, the induced subgraph $G[V_{k+1}]$ has few triangles, and thus, by Corollary~\ref{cor:sdp-2}, has a cut with good surplus. This allows us to obtain the following technical result regarding the Max-Cut of $H$-free $d$-degenerate graphs.

 \begin{lemma}
 	\label{lem:general}
 	There exists an absolute constant $c>0$ such that the following holds.
 	Let $H$ be a graph and $H'$ be obtained by deleting any vertex of $H$.
 	Let $0<\varepsilon<\frac{1}{\sqrt{d}}$.
 	For any $H$-free $d$-degenerate graph $G=(V,E)$, one of the following holds:
 	\begin{itemize}
 		\item We have
 		\begin{align}
 		f(G) \ge \left( \frac{1}{2} + c\varepsilon \right)m.
 		\label{e:cond1}
 		\end{align}
 		\item There exist graphs $G_1,\dots,G_k$ such that five conditions hold: (i) graphs $G_i$ are $H'$-free for all $i$, (ii) $n(G_i)\le d$ for all $i$, (iii) $m(G_i) \ge \frac{n(G_i)}{8\varepsilon}$ for all $i$, (iv) $n(G_1)+\cdots+n(G_k)\ge \frac{m}{6d}$, and (v)
 		\begin{align}
 		f(G) \ge \frac{m(G)}{2} + \sum_{i=1}^{k} \left(f(G_i) - \frac{m(G_i)}{2}\right).
 		\label{e:cond2}
 		\end{align}
 	\end{itemize}
 \end{lemma}
 \begin{proof}
 	Let $c_1 < 1$ be the parameter given by Corollary~\ref{cor:lb-2}.
 	Let $c = \frac{c_1}{6}$.
 	Let $G = (V, E)$ be a $d$-degenerate $H$-free graph.
 	Applying Lemma~\ref{lem:technical} with parameter $8\varepsilon$, we can find a partition $V_1,\dots,V_{k+1}$ of the vertex set $V$ with the following properties.
 	\begin{enumerate}
 		\item For $i=1,\dots,k$, the vertex subset $V_i$ has at most $d$ vertices and has a common neighbor, and the induced subgraph $G[V_i]$ at least $\frac{|V_i|}{8\varepsilon}$ edges.
 		\item The subgraph $G[V_{k+1}]$ has at most $\frac{m(G[V_{k+1}])}{8\varepsilon}$ triangles.
 	\end{enumerate}
 	For $i=1,\dots,k+1$, let $G_i\defeq G[V_i]$ and let $m_i\defeq m(G_i)$.
 	For $i=1,\dots,k$, since $G$ is $H$-free and each $V_i$ is a subset of some vertex neighborhood in $G$, the graphs $G_i$ are $H'$-free.
 	For $i = 1, \ldots , k$, fix a maximal cut of $G_i$ with associated vertex partition $V_i = A_i \sqcup B_i$.
 	By the second property above, the graph $G_{k+1}$ has at most $\frac{m_{k+1}}{8\varepsilon}$ triangles.
 	Applying Corollary~\ref{cor:lb-2} with parameter $\varepsilon$, we can find a cut of $G_{k+1}$ of size at least $(\frac{1}{2} + c_1\varepsilon)m_{k+1}$ with associated vertex partition $V_{k+1} = A_{k+1} \sqcup B_{k+1}$.

 	We now construct a cut of $G$ by randomly combining the cuts obtained above for each $G_i$.
 	Independently, for each $i=1,\dots,k+1$, we add either $A_i$ or $B_i$ to vertex set $A$, each with probability $\frac12$.
 	Setting $B=V\setminus A$, gives a cut of $G$.
 	As $V_1,\dots,V_{k+1}$ partition $V$, each of the $m-(m_1+\cdots+m_{k+1})$ edges that is not in one of the induced graphs $G_1,\dots,G_{k+1}$ has exactly one endpoint in each of $A, B$ with probability $1/2$. This allows us to compute the expected size of the cut (a lower bound on $f(G)$ as there is some instantiation of this random process that achieves this expected size).
 	\begin{align}
 	\label{eq:general-1}
 	f(G) \ &\ge \  \frac{1}{2}(m - (m_1+\cdots+m_{k+1})) + \left( \frac{1}{2} + c_1\varepsilon \right) \cdot m_{k+1} + \sum_{i=1}^{k} f(G_i) \nonumber\\
 	\ &= \   \frac{m}{2} + c_1\varepsilon m_{k+1} + \sum_{i=1}^{k} \left( f(G_i) - \frac{m_i}{2} \right).
 	\end{align}
 	We bound~\eqref{eq:general-1} based on the distribution of edges in $G$ in $3$ cases:
 	\begin{itemize}
 		\item $m_{k+1}\ge \frac{m}{6}$.
 		Since $f(G_i)\ge \frac{m_i}{2}$ for all $i=1,\dots,k$, \eqref{e:cond1} holds:
 		\begin{align*}
 		f(G)
 		\ \ge \  \frac{m}{2} + c_1\varepsilon m_{k+1}
 		\ \ge \  \left(\frac{1}{2} + c\varepsilon\right)\cdot m.
 		\end{align*}
 		\item The number of edges between $V_1\cup\cdots\cup V_{k}$ and $V_{k+1}$ is at least $\frac{2m}{3}$.
 		Then, the cut given by vertex partition $V = A' \sqcup B'$ with $A'=V_1\cup\cdots\cup V_k$ and $B'=V_{k+1}$ has at least $\frac{2m}{3}$ edges, in which case $f(G)\ge \frac{2m}{3}> (\frac{1}{2} + \frac{c_1\varepsilon}{6})\cdot m$, so~\eqref{e:cond1} holds.
 		\item $G' = G[V_1\cup\cdots\cup V_k]$ has at least $\frac{m}{6}$ edges. We show \eqref{e:cond2} holds. By construction, for $i=1,\dots,k$, the graph $G_i$ is $H'$ free, has at most $d$ vertices, and has at least $\frac{m_i}{8\varepsilon}$ edges.
 		Since $G$ is $d$-degenerate, $G'$ is as well, so
 		\begin{align}
 		\frac{m}{6} \le m(G') \le d\cdot n(G') = d\cdot \sum_{i = 1}^k n(G_i),
 		\end{align}
 		Hence $n(G_1)+\cdots+n(G_k)\ge \frac{m}{6d}$.
 		Lastly, by \eqref{eq:general-1}, we have
 		\begin{align*}
 		f(G) \ge \frac{m}{2} + \sum_{i=1}^{k} \left(f(G_i) - \frac{m_i}{2}\right).
 		\end{align*}
 	\end{itemize}
 	This covers all possible cases, and in each case we showed either~\eqref{e:cond1} or~\eqref{e:cond2} holds.
 \end{proof}
 \begin{remark}
 	In Corollary~\ref{cor:lb-2} we can take $c = \frac{1}{60}$, and in Lemma~\ref{lem:general} we can take $c = \frac{1}{360}$.
 \end{remark}

 Lemma~\ref{lem:general} allows us to convert Max-Cut lower bounds on $H$-free graphs to Max-Cut lower bounds on $H$-free $d$-degenerate graphs.
 \begin{lemma}
 	\label{lem:general-2}
 	Let $H$ be a graph and $H'$ be obtained by deleting any vertex of $H$.
 	Suppose that there exists constants $a=a(H')\in[\frac{1}{2}, 1]$ and $c'=c'(H')>0$ such that for all $H'$-free graphs $G$ with $m'\ge 1$ edges, $f(G) \ge \frac{m'}{2} + c'\cdot (m')^{a}$.
 	Then there exists a constant $c=c(H)>0$ such that for all $H$-free $d$-degenerate graphs $G$ with $m\ge 1$ edges,
 	\begin{align*}
 	f(G) \ge \left(\frac{1}{2} + cd^{-\frac{2-a}{1+a}}\right)\cdot m.
 	\end{align*}
 \end{lemma}
 \begin{proof}
 	Let $c_2$ be the parameter in Lemma~\ref{lem:general}.
 	We may assume without loss of generality that $c'\le 1$.
 	Let $G$ be a $d$-degenerate $H$-free graph.
 	Let $\varepsilon\defeq c'd^{-\frac{2-a}{1+a}} < d^{-1/2}$ and $c\defeq \min(c'c_2, \frac{c'}{48})$.

 	Applying Lemma~\ref{lem:general} with parameter $\varepsilon$, either~\eqref{e:cond1} or~\eqref{e:cond2} holds.
 	If~\eqref{e:cond1} holds, then, as desired, $$f(G)\ge \left(\frac{1}{2} + c_2\varepsilon \right)m \ge \left(\frac{1}{2} + cd^{-\frac{2-a}{1+a}} \right)m.$$
 	Else~\eqref{e:cond2} holds. Let $G_1,\dots,G_{k+1}$ be the $H'$-free induced subgraphs satisfying the properties in Lemma~\ref{lem:general}, so that
 	\begin{align*}
 	f(G)
 	\ &\ge \  \frac{m}{2} + \sum_{i=1}^{k} \left(f(G_i) - \frac{m(G_i)}{2}\right)  \nonumber\\
 	\ &\ge \   \frac{m}{2} + \sum_{i=1}^{k} c'\cdot m(G_i)^{a}.
 	\end{align*}
 	For all $i$, we have
 	\begin{align*}
 	c' \cdot m(G_i)^{a} \overset{(*)}\ge \frac{c'\varepsilon}{8\varepsilon^{1+a}}\cdot n(G_i)^{a}  \overset{(**)}\ge \frac{\varepsilon d}{8(c')^{a}}\cdot n(G_i)  \overset{(+)}\ge \frac{\varepsilon d}{8}\cdot n(G_i),
 	\end{align*}
 	where $(*)$ follows since $m(G_i)\ge \frac{n(G_i)}{8\varepsilon}$, $(**)$ follows since $n(G_i)^{a-1}\ge d^{a-1}$ and $\varepsilon^{1+a} = (c')^{1+a}d^{a-2}$, and $(+)$ follows since $c' \le 1$.
 	Hence, as $n(G_1)+\cdots+n(G_k)\ge \frac{m}{6d}$, we have
 	\begin{align*}
 	f(G)
 	\ &\ge \   \frac{m}{2} + \varepsilon d \sum_{i=1}^{k}\frac{n(G_i)}{8}
 	\ \ge \   \frac{m}{2} + \frac{\varepsilon m}{48}
 	\ \ge \ \left(\frac{1}{2} + cd^{-\frac{2-a}{1+a}}\right)\cdot m,
 	\end{align*}
 	as desired.
 \end{proof}

 \section{Max-Cut in $K_r$-free graphs}
 \label{sec:kr}
 In this section we specialize Lemmas~\ref{lem:general} and~\ref{lem:general-2} to the case $H = K_r$ to
 prove Theorem~\ref{thm:kr-lb}.
 Let $\chi(G)$ denote the chromatic number of a graph $G$, the minimum number of colors needed to properly color the vertices of the graph so that no two adjacent vertices receive the same color. We first obtain a nontrivial upper bound on the chromatic number of a $K_r$-free graph $G$, giving an lower bound (Lemma~\ref{lem:kr-d}) on the Max-Cut of $K_r$-free graphs.
 This lower bound was implicit in \cite{AL03}, but we provide a proof for completeness. The lower bound on the Max-Cut of general $K_r$-free graphs enables us to apply Lemma~\ref{lem:general} to give a lower bound on the Max-Cut of $d$-degenerate $K_r$-free graphs per Theorem~\ref{thm:kr-lb}.
 The following well known lemma gives a lower bound on the Max-Cut using the chromatic number.
 \begin{lemma}(see e.g. Lemma 2.1 of \cite{AL03})\label{lem:mcchi}
 	Given a graph $G = (V, E)$ with $m$ edges and chromatic number $\chi(G) \le t$, we have $f(G) \ge (\frac12 + \frac{1}{2t}) m$.
 \end{lemma}
 \begin{proof}
 	Since $\chi(G) \le t$, we can decompose $V$ into independent subsets $V = V_1,\dots, V_t$.
 	Partition the subsets randomly into two parts containing $\floor{\frac{t}{2}}$ and $\ceil{\frac{t}{2}}$ subsets $V_i$, respectively, to obtain a cut.
 	The probability any edge is cut is $\frac{\floor{t/2}\cdot \ceil{t/2}}{\binom{t}{2}}\ge \frac{t+1}{2t}$, so the result follows from linearity of expectation.
 \end{proof}

 \begin{lemma}\label{lem:chikr}
 	Let $r\ge 3$ and $G = (V, E)$ be a $K_r$-free graph on $n$ vertices. Then, $$\chi(G) \le 4n^{(r-2)/(r-1)}.$$
 \end{lemma}
 \begin{proof}
 	We proceed by induction on $n$.
 	For $n\le 4^{r-1}$, the statement is trivial as the chromatic number is always at most the number of vertices.
 	Now assume $G=(V,E)$ has $n>4^{r-1}$ vertices and that $\chi(G)\le 4n_0^{(r-2)/(r-1)}$ for all $K_r$-free graphs on $n_0\le n-1$ vertices.
 	The off-diagonal Ramsey number $R(r,s)$ satisfies $R(r,s) \le \binom{r+s-2}{s-1} \le s^{r-1}$ \cite{ES35}.
 	Hence, $G$ has an independent set $I$ of size $s=\floor{n^{1/(r-1)}}$.
 	The induced subgraph $G[V\setminus I]$ is $K_r$-free and has fewer than $n$ vertices, so its chromatic number is at most $4(n-s)^{(r-2)/(r-1)}$.
 	Hence, $G$ has chromatic number at most
 	\begin{align}
 	1 + 4(n-s)^{(r-2)/(r-1)}
 	\ &= \   1 + 4n^{(r-2)/(r-1)}\left( 1-\frac{s}{n} \right)^{(r-2)/(r-1)} \nonumber\\
 	\ &\overset{(*)}\le \   1 + 4n^{(r-2)/(r-1)} - 4n^{(r-2)/(r-1)} \cdot \frac{s}{3n}
 	\ \overset{(**)}< \   4n^{(r-2)/(r-1)}
 	\label{}
 	\end{align}
 	In $(*)$, we used that $\frac{r-2}{r-1}\ge \frac{1}{2}$, that $\frac{s}{n} \le \frac{1}{4}$, and that $(1-x)^a\le 1 - \frac{x}{3}$ for $a\ge \frac{1}{2}$ and $x\le \frac{1}{4}$.
 	In $(**)$, we used that $s\ge 4$ and hence $\frac{3s}{4} < n^{1/(r-1)}$.
 	This completes the induction, completing the proof.
 \end{proof}

 \begin{rem}
 	The upper bound on the off-diagonal Ramsey number $R(r, k^{1/(r-1)})$ has an extra logarithmic factor which suggests that the upper bound on $\chi(G)$ of Lemma~\ref{lem:chikr} can be improved by a logarithmic factor with a more careful analysis.
 \end{rem}

 \begin{lemma}
 	\label{lem:kr-d}
 	If $G$ is a $K_r$-free graph with at most $n$ vertices and $m$ edges, then
 	\begin{align*}
 	f(G) \ge \left( \frac{1}{2} + \frac{1}{8n^{(r-2)/(r-1)}} \right) m
 	\end{align*}
 \end{lemma}
 \begin{proof}
 	This follows immediately via Lemma~\ref{lem:mcchi}
 	and Lemma~\ref{lem:chikr}.
 \end{proof}

 The above bounds allow us to prove Theorem~\ref{thm:kr-lb}.
 \begin{proof}[Proof of Theorem~\ref{thm:kr-lb}]
 	Let $G$ be a $d$-degenerate $K_r$-free graph and $\varepsilon=d^{-1 + \frac{1}{2r-4}}$.
 	Let $c_2$ be the parameter given by Lemma~\ref{lem:general}.
 	Let $c = \min(c_2,\frac{1}{388})$.

 	Applying Lemma~\ref{lem:general} with parameter $\varepsilon$, one of two properties hold.
 	If~\eqref{e:cond1} holds, then
 	\begin{align}
 	f(G)
 	\ \ge \ \left(\frac{1}{2} + c_2\varepsilon\right)m
 	\ &\ge \  \left(\frac{1}{2} + c d^{- 1 + \frac{1}{2r-4}}\right)m
 	\end{align}
 	as desired.
 	If~\eqref{e:cond2} holds, there exist graphs $G_1,\dots,G_k$ that are $K_{r-1}$-free with at most $d$ vertices such that $G_i$ has at least $\frac{n(G_i)}{8\varepsilon}$ edges, $n(G_1)+\cdots+n(G_k)\ge \frac{m}{6d}$, and
 	\begin{align*}
 	f(G)
 	\ &\ge \  \frac{m}{2} + \sum_{i=1}^{k} \left(f(G_i) - \frac{m(G_i)}{2}\right).
 	\end{align*}
 	For all $i$, we have
 	\begin{align*}
 	f(G_i) - \frac{m(G_i)}{2}
 	\ &\ge \  \frac{m(G_i)}{8n(G_i)^{(r-3)/(r-2)}}  \nonumber\\
 	\ &\ge \ \frac{n(G_i)}{64\varepsilon n(G_i)^{(r-3)/(r-2)}}
 	\ \ge \ \frac{n(G_i)}{64\varepsilon d^{(r-3)/(r-2)}}
 	\ = \ \frac{\varepsilon d n(G_i)}{64}.
 	\end{align*}
 	In the first inequality, we used Lemma~\ref{lem:kr-d}.
 	In the second inequality, we used that $m(G_i)\ge \frac{n(G_i)}{8\varepsilon}$.
 	In the third inequality, we used that $n(G_i)\le d$.
 	Hence, as $d(n(G_1)+\cdots+n(G_k))\ge \frac{m}{6}$, we have as desired that
 	\begin{align}
 	f(G)
 	\ &\ge \   \frac{m}{2} + \sum_{i=1}^{k}\frac{\varepsilon d n(G_i)}{64}
 	\ \ge \   \frac{m}{2} + \frac{\varepsilon m}{388}
 	\ \ge \ \left(\frac{1}{2} + cd^{-1 + \frac{1}{2r-4}}\right)\cdot m.
 	\qedhere
 	\end{align}
 \end{proof}

\begin{rem}
As we already mentioned in the introduction, we can improve the result of Theorem~\ref{thm:kr-lb} in the case that $r = 4$ using Lemma~\ref{lem:general-2} as follows.
Let $H=K_4$, and $H'=K_3$.
By a result of \cite{AL96}, there exists a constant $c'>0$ such that, for all triangle-free graphs $G$ with $m'\ge 1$ edges, we have $f(G)\ge \frac{m'}{2} + c'(m')^{4/5}$.
By Lemma~\ref{lem:general-2} with $H$ and $H'$ and $a=4/5$, there exists a constant $c>0$ such that any $K_4$-free $d$-degenerate graph $G$ with $m\ge 1$ edges satisfies
\begin{align}
f(G)  \ge    \left( \frac{1}{2} + cd^{-\frac{2-(4/5)}{1+(4/5)}} \right)\cdot m =  \left( \frac{1}{2} + cd^{-2/3} \right)\cdot m.
\end{align}
\end{rem}

\section{Concluding Remarks}
\label{sec:conclusion}
In this paper we presented an approach, based on semidefinite programming (SDP), to prove lower bounds on Max-Cut and used it to find large cuts in graphs with few triangles and in
$K_r$-free graphs. A closely related problem of interest is bounding the \textsf{Max-$t$-Cut} of a graph, i.e. the largest $t$-colorable ($t$-partite) subgraph of a given graph. Our results imply good lower bounds for this problem as well. Indeed, by taking a cut for a graph $G$ with $m$ edges and surplus $W$, one can produce a $t$-cut for $G$ of size
$\frac{t-1}{t}m +\Omega(W)$ as follows.  Let $A, B$ be the two parts of the original cut. If $t=2s$ is even, simply split randomly both $A, B$ into $s$ parts. If $t=2s+1$ is odd, then put every vertex of $A$ randomly in  the parts $1, \ldots, s$ with probability $2/(2s+1)$ and in the part $2s+1$ with probability $1/(2s+1)$. Similarly, put every vertex of $B$ randomly in  the parts $s+1, \ldots, 2s$ with probability $2/(2s+1)$ and in the part $2s+1$ with probability $1/(2s+1)$.
An easy computation (which we omit here) shows that the expected size of the resulting $t$-cut is
$\frac{t-1}{t}m +\Omega(W)$.

The main open question left by our work is Conjecture~\ref{conj:opt}. Proving this conjecture will require some major new ideas. Even showing that any $d$-degenerate $H$-free graph with $m$ edges has a cut with surplus at least
$m/d^{1-\delta}$ for some fixed $\delta$ (independent of $H$) is out of reach of current techniques.

\vspace{0.3cm}

\noindent
{\bf Acknowledgements.} \, The authors thank Jacob Fox and Matthew Kwan for helpful discussions and feedback.
The authors thank Joshua Brakensiek for pointing out an error in an earlier draft of this paper.
The authors thank Joshua Brakensiek and Yuval Wigderson for helpful comments on an earlier draft of the paper.

redpajama_set_name: RedPajamaArXiv


text:
Q: convert IE6 and 7 Hack into an IE6, 7 and 8 hack In my css code i have tose hack that i want to affect IE6/7
#topmenu li a.activa,
#topmenu li a.activa:hover{
    *background: url(../nImg/comunHomeSprite.png) no-repeat;
    *background-position: right -2169px;
    *float:left;
    *margin:0;
    *padding:0;
    *margin-left:10px;
}

Is there a simple way to convert them into IE8 also??

A: If you must use hacks, then read this: http://net.tutsplus.com/tutorials/html-css-techniques/quick-tip-how-to-target-ie6-ie7-and-ie8-uniquely-with-4-characters/
But note that CSS hacks are considered evil.
Conditional comments are a good alternative. They are easy to use, and guaranteed to work properly.
You can't put conditional comments directly into a stylesheet, but what you can do is define a class in your <body> tag using conditional comments, which you can then reference in the CSS:
Write your HTML <body> tag like this:
<!--[if IE 6]> <body class="ie6 ltie7 ltie8 ltie9"> <![endif]-->
<!--[if IE 7]> <body class="ie7 ltie8 ltie9"> <![endif]-->
<!--[if IE 8]> <body class="ie8 ltie9"> <![endif]-->
<!--[if IE 9]> <body class="ie9"> <![endif]-->
<!--[if (gt IE 9)|!(IE)]><!--> <body> <!--<![endif]-->

Then in your CSS, you can reference the relevant IE class in your selectors, and you'll have completely valid CSS code:
#topmenu li a.activa:hover {
   /*normal styles here*/
}

.ie8 #topmenu li a.activa:hover {
   /*IE8-specific styles here*/
}

Hope that helps.

A: Found this
selector {
    prop: value; /* real browsers */
    _prop: value; /* ie6 */
    *prop: value; /* ie6 ie7 */
    prop: value\9; /* ie8 */
}

don't know if this addresses IE9

A: /* IE8 */ @media \0screen { #topmenu li a.activa, @media \0screen #topmenu li a.activa:hover {
  background: url(../nImg/comunHomeSprite.png) no-repeat;
  background-position: right -2169px;
  float:left;
  margin:0;
  padding:0;
  margin-left:10px;
} }

I prefer this way personally as the attributes are still standard CSS and you need only change them once instead of however many different hacks you have. It can get a bit unwieldy otherwise.

redpajama_set_name: RedPajamaStackExchange


Katie has taken to wearing piggy tales just like her friend Lauren. She only cries a little when mommy puts them in. I am sure the crookedness of these would drive Aunt Michelle crazy, but luckily you can get away with quite a bit when you are two.
Katie started school today (Mother's Day Out). She goes Mondays and Wednesdays from 9:30am-1:30pm. She was so excited this morning. She wore the outfit that Geike bought her for school and her new running shoes so she can play on the playground. We packed her lunch and her backpack just like Grant. She ran into the building so excited we didn't even have a chance to get a picture outside. Mommy stayed for a few minutes, but then had to go. Katie was a little tearful, but Miss Rhonda was going to let her help pick out stickers for everybody so that was a nice distraction. We'll see how she is this afternoon!
Take the picture, silly man. I am busy.


The following is a brief excerpt from the new Starfleet Research study 'How to Leverage Business Intelligence to Improve Hotel Performance,' which is currently available for complimentary access. The research findings featured in the study are derived from surveys conducted earlier this year with more than 250 hotel executives and qualified staff.
The following is a brief excerpt from the new Starfleet Research study "How to Leverage Business Intelligence to Improve Hotel Performance," which is currently available for complimentary access. The research findings featured in the study are derived from surveys conducted earlier this year with more than 250 hotel executives and qualified staff.
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text:
Church gathers to worship outside DCH Regional Medical Center in Tuscaloosa
by: Drew Taylor
TUSCALOOSA, Ala. (WIAT) — Concerns over COVID-19 were not enough to keep people from gathering, albeit remotely, in the parking lot of DCH Regional Medical Center Sunday for a worship service.
The service was held by Church of the Highland Sunday afternoon, where worship music was played and prayer was held. Video of the event showed people standing by their cars, separate from one another, holding worship.
The service was held to pray for doctors, patients and medical workers dealing with coronavirus cases.
Those who took part in the service did not leave their cars.
Central AL Forecast: Gloomy Weather Returns Today With A Few Showers
by Art Franklin / Jan 18, 2021
BIRMINGHAM, Ala. (WIAT) -- The nephew of Dr. Martin Luther King Jr. joined Art Franklin on CBS 42 News at 10 p.m. to speak about the legacy of King from a family members perspective.
Watch the full interview in the video above.
by Malique Rankin / Jan 18, 2021
BIRMINGHAM, Ala. (WIAT) -- President-elect Joe Biden's $1.9 trillion dollar American rescue plan includes an increase to the federal minimum wage, bringing that wage to $15 an hour. The federal minimum wage has not been increased since 2009.
Small businesses in Alabama have had a tough year because of COVID-19. After closures, losses in sales, and a worsening economy, some are concerned an increased minimum wage will be a difficult transition.
by Landon Wexler / Jan 18, 2021
BIRMINGHAM, Ala. (WIAT) -- Aside from its unprecedented chaos, 2020 brought murals and a new sense of vibrancy to the streets of the Birmingham area.
The latest addition to that collection - a tribute to (from left to right) influential rapper Nipsey Hussle, NBA great Kobe Bryant, and award-winning actor Chadwick Boseman - all three of whom died in the past two years.

redpajama_set_name: RedPajamaCommonCrawl


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Imagine, if you will, a bunch of drunken college seniors, down in Mexico for spring break, ready to party one last time before the "real world" descends upon them. There is liquor. There is debauchery. There is infighting. There is more than one secret floating amongst the group. There is an evil presence released when they play a game of truth or dare in an abandoned mission which, unbeknownst to them, has a dark and mysterious past.
It is, in essence, the perfect setup for a stereotypical horror film, and that is exactly what "Blumhouse's Truth or Dare" is – a stereotypical horror film.
Directed by Jeff Wadlow and with a screenplay from Michael Reisz and Jillian Jacobs & Chris Roach & Wadlow, "Truth or Dare" has this group of kids forced into a deadly game of truth or dare. The rules are simple – play or die, tell the truth or die, do the dare or die.
Here is the thing about such a setup – it could, hypothetically, work. There is a reason for the stereotype and the opening of the game is suitably atmospheric. The students working out the rules is, similarly, intriguing. What doesn't work is anything else – the group themselves (the majority of whom are awful human beings), the film's resolution, the dialogue, the plotting, and the vast majority of the scares.
The college kids, as represented, are generally terrible people or, at the very least, exhibit obvious and fatal flaws. There's Markie (Violett Beane), who can't stop cheating on her bland and boring boyfriend, Lucas (Tyler Posey), even when he's 10 feet from her. There's Penelope (Sophia Taylor Ali), whose day-drinking is the stuff of jokes amongst the friends and who, despite living in an adult-free house with plenty of liquor, keeps a bottle in her dresser drawer. There's Tyson (Nolan Gerard Funk), who forges medical prescriptions and treats absolutely everyone in terrible fashion. There's Ronnie (Sam Lerner), whom no one likes because everything that comes out of his mouth is sexist and awful. There's Brad (Hayden Szeto), who is one of the exceptions to the rule but still written poorly since the film treats his sexuality as a plot device. And then there's the lead, Olivia (Lucy Hale), who is obviously too perfect to be true.
The only suitable conclusions about this group then is that these are, by and large, stereotyped individuals whom no one would want to spent time with, or an incredibly offensive depiction of how college students are viewed, or both. The supernatural villainy that forces the game of truth or dare upon the students is able to use their deficits against them, and that would work better if the individuals had more to them than those flaws. This is a film so focused on getting to the killing and revealing of horrible secrets that it doesn't bother to build anything else into most of the characters. They exist wholly to be used in the game, nothing more.
There are even opportunities when "Truth or Dare" could be more, but instead it shies away from them. There are serious conversations these students could have about who they are and why they've done the things in life that they've done—whether they're proud of them or not—and what it all means. The film opens up the questions via the "truth" option of the game, but then all too often chooses to shy away from the answers… answers that could actually turn the characters into three dimensional humans.
As the film speeds ahead and the reality behind the game is revealed, things go from bad to worse. Things progress to, and through, the climax with little regard for logic or anything that has come before. It is a conclusion included to impress and/or surprise the audience, and not built from the characters themselves.
Full of obvious, easy, jump scares, and just barely enough plot to get from one moment of the supernatural game to the next, "Truth or Dare" may be able to keep audiences in their seats for its entire length, but those watching will not walk away from it feeling as though the experienced something worthwhile. It is silly, forgettable, and a game better left unplayed.


STATE OF PLAY | ISSUE THREE a few things that are important in this business, and that is that every customer walks away happy with their expectations exceeded. " Exceeding expectations is where using a recruitment agency can be incredibly helpful. As well as saving you time, as they understand what you need and can articulate this in a well-written job description, they can help ensure potential staff are truly suited to the industry. If this is done by a company with a lot of experience, then the candidates to interview should be the best of the bunch, which makes the job of interviewing a whole lot easier. For both you and them. TOP TIPS TO RECRUIT AND RETAIN YOUR STAFF For any employer, the most critical asset that they have is their people. People are also every employer's most significant cost, says entrepreneur and managing director of recruitment specialist firm Chorus Executive, Christine Khor (above). Khor spoke at Tabcorp Gaming Solutions' (TGS) Big Picture Breakfast event in October last year and told the room that people are the number one issue in every business, and businesses in the hospitality and gaming sector are no exceptions. "If you think about your time as managers and leaders -- how much time did you spend worrying about who you're hiring, who you're firing, who you're training and how your staff are treating your customers? We need to focus on the right strategies to hire the right staff, as well as to retain them," she said. Khor, who also wrote Hire Love: How To Hire Passionate People To Make Greater Profit, shared her top recruitment strategies and poses some questions for employers to think about: • Research and spend the time to understand the talent you really need for the business. Be realistic about the requirements and have a few concise definitions of the role. • It's important to remember that recruitment today is different to what it was years ago. Today, candidates are assessing potential employers as much as the employers are evaluating them. According to Khor, the number one reason why staff leave is, you, the manager. "How does your place of work differ from the others? How do you, as an employer and as a business, attract these candidates? What's in it for them? Why are you great to work for? These are some of the questions employers need to think about," she said at the event. • Focus on your current staff -- your brand ambassadors. It's important to reflect on what they could be saying about your business in their social circles, and how they define you as an employer. If your current staff members are engaged, doing good work and being recognised, they are the best advocates for your customers as well as your future employees. • As you go through the hiring process, it's essential to be emphatic to the candidates. Finding a job is an arduous process. For a candidate, it is putting him or herself in front of people while understanding the high possibility of rejection. As you interview candidates, be present and not distracted by your other tasks. And if you do find them unfit for the role, let them down gracefully and respectfully. • Onboarding -- this isn't an induction process. Induction focuses on the formal and administrative part of the job and organisation such as OH&S (occupational health and safety) policies, sick leave policies and so on. Onboarding, however, is a socialisation process of finding out how to work together as a team. While managers are on hand to teach new staff the tasks, Khor recommends that employers pick a buddy for the new staff member to feel socially supported by the people in the business. According to Khor, if you onboard someone successfully in the first 90 days, you will get 60 percent higher productivity and retention rates.


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Home > Industry News > 1520 Strategic Forum opens in Sochi
1520 Strategic Forum opens in Sochi
President of Russian Railways (RZD) Vladimir Yakunin today opened the ninth 1520 strategic forum for Russian gauge railways.
The current political situation was recognised from the start. Jochen Eickholt, chief executive of Siemens Rail Systems Division felt it essential to maintain communication and focus on current projects. Siemens considered Russia to be a natural partner. Expecting the current situation in Crimea to de-escalate soon, the company has advised that sanctions would have no effect on its business with Russia.
Ruslan Grinberg, of the Russian Academy of Science felt that the forum had lost Ukraine as a result of the current 'regrettable situation'. Amongst the 600 or so forum participants there is no-one from Ukrainian Railways.
Topics being considered by the forum include improvements to rail transit corridors and high-speed rail.
Opening the forum, Yakunin stressed the importance of projects to enhance both the Trans Siberian and the more northerly Baikal Amur Mainline Railway which is celebrating its 40th year of construction this month.
In a later session, Professor Li Xin Chinese stressed China's intention to develop the "New Silk Road" route through Kazakstan in cooperation with all countries concerned.
Building on this theme Yakunin stressed the need to work together to promote infrastructure investment to promote international development and felt this was in contrast with the American market forces approach. He will no doubt expand on this when he meets foreign correspondents later today (June 19).
Report by David Shirres


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What happened to Magaly Rodríguez's life after the fame he achieved in Gran Hermano 2001: "I was scared to live in Buenos Aires"
Magaly Rodríguez in Big Brother
The second edition of big brother was perhaps the most complex. It began in August 2001 and ended on December 1, the eve of Argentina's social, political and economic explosion. But of course, inside the walls, for the participants – in complete isolation – everything was joy and happiness. On exit, it could be different: Roberto Parraby chance, he saw how his prize of 200 thousand dollars was trapped in the corralito.
Magaly Rodríguez participated in that season. And he still has fond memories of what he experienced in the reality show. He was 22 years old when he entered after winning the place in a casting that took place in his hometown, Mar del Plata. After the choice of the producers, he packed his bag, got on a microphone and started his journey to Buenos Aires, with all the excitement on him.
After the departure of GHhis first experience in the artistic environment was in the theater, being part of Onward my brave oneswork that was presented to Carlos Paz. He also did television, with recurring appearances on the shows of Marcelo Tinelli, Susana Gimenez y Maru Botanaamong others.
For Magaly, 2002 was a year of work at work, non-stop, living in a hotel. Until he got tired of it all. And he knocked on the door. Although she was happy with the fame she had acquired, she needed to return to everyday life with her family, in La Feliz. At that time his popularity, at least for the Buenos Aires media, was fading. What became of his life after that? He explains it to Teleshow.
Magaly Rodríguez, today
—What memory comes to mind when you think about your time in GH?
– Of absolute joy and happiness, no doubt. For me it was a beautiful and spectacular experience that I will never forget, but at the same time, I live it with nostalgia, I won't deny it. 22 years have passed, and every time a new edition of big brother I live it again. I have only words of thanks because it was something I enjoyed from start to finish.
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— Why did you sign up? What did you go looking for?
— I did it because my little sister didn't sign up. She has a very low profile and sent me to the casting. She was the one who found out they were doing it, but she told me to go and I didn't hesitate. I was 19 at the time. As for what I went in search of, of course I managed to experience what I wanted: to be chosen among the eight thousand people who signed up for Mar del Plata… I still don't know why I stayed, what they saw in me. Imagine being told to wear a leotard at the casting and wearing pants, told to dance and telling them I was embarrassed… The producers looked at me like they didn't understand what I was doing. I think that's why they chose me: to show me very different (laughs).
—How was your life in Mar del Plata before big brother?
—Like any girl of that age, she had a very normal life. I had been studying occupational therapy for three years, I was working and very much at home. My niece had just been born, I was smitten with her, so I spent a lot of family time. Honestly, I don't have much to say about what I was doing before GH, it was no different life than any other. He didn't go out much and stuff like that. That's why my entry into the house is very strange.
Magaly Rodríguez and the suitcase from Gran Hermano 2001, the one that was loaded with dreams
– And suddenly you were famous. How did you experience this exhibition?
– It was very crazy. I think about it and I can't believe it. You have to live it to understand it. It was nobody when I came in and when I came out, I was greeted even by the garbage collectors hanging from the truck. They knew me and called me from all sides. I didn't realize how crazy it was, because you quickly get used to it and don't sit around thinking about it. I remember that one day, just after I got out, they gave me a week to come to Mardel to see my family and I went to my hairdresser in the neighborhood, all very normal, but when I wanted to go I had to go and find my father because otherwise I wasn't going to be able to go out. He loaded the bike into the van and off we went. I don't know how many people were outside… When I finished there I dropped the exhibition card, what it really was. But I loved it, I always took that second to sign an autograph, it never bothered me, it wasn't a waste of time or anything like that.
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—She was one of the ones who worked the most in the middle at that time. Why was everything cut overnight?
– It's true, yes. I did circus, theater, television, and all the time I was getting offers. But I'll be honest with you: I was scared to live in Buenos Aires. I don't forget that one day I went to talk to Marcus Gorbanthe general producer of phone, and I told him: "I'm going back to Mar del Plata". He looked at me and said I was crazy. I know I had several things to do but I missed my family, my things, very much. At that time, when I put everything in the balance, the family could more than the proposals in a different city.
— When you decided to return to your city, were you in a relationship?
– She was alone and happy. I was experiencing all this madness which, I repeat, did not bother me: it fascinated me. I never denied that. But I was strange, and since I had nothing to tie me, I turned.
Magaly Rodríguez
– How is your life today?
—I was married for 14 years and raised a beautiful family. I have two sons aged 13 and eight who are my driving force, literally, and the loves of my life. I'm focused on it today. I'm not in a relationship. Single and no problems, I don't have time because I'm a 24-hour mom.
– Are you working?
— Yes, I've been working for a collection firm for a few years. However, I continue to maintain my hipon side and I also dedicate myself to a host of creative workshops: I teach macramé, painting, etc. This is more like a recreation, but one that I am passionate about. It's something I really like and I did it in my spare time.
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– Do people still recognize you?
— Yes, it still happens to me that they know me. Look, it's a rare thing, but people know me more now than when I was in big brother, at that time in the prime of life. It catches my eye because it should be the other way around, but it amuses me a lot. I like that the memory is still there. And if they ask me for a photo, of course I'm delighted.
Magaly Rodríguez is still in contact with all her former Gran Hermano colleagues, such as Silvina Luna
-Are you looking at the current edition? Do you see someone similar to you in the way you play?
— Yes, obviously I'm watching the screen and watching the debates, the galas, everything. I'm the number one fan. It's an entertaining program. It was before and it is now, in these times. I see the repercussions on social media and it's fabulous. I already have my favorites: for me Romi, Alfa, Marcos will reach the final and I thought Juli too, but well, they were all sent together and they eliminated her. Of all of them I think the winner will be Romi.
— Are you still in contact with your former colleagues?
— The vast majority of us are still in contact, we have a WhatsApp group and almost all of us are there. You can imagine that the group is clearly crazy… still 20 years later. And we are all still the same in our way of being. Time seems to have stopped.
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redpajama_set_name: RedPajamaGithub


The mechanical properties of the substrate upon which cells are cultured have been shown to influence a variety of cell properties including cell adhesion, spreading, protein expression and differentiation. The work presented here examines how the nonlinear mechanical properties of biopolymer gels affect the cellular responses to substrate stiffness. Cell spread area decreases with decreasing substrate stiffness when cells are cultured on linearly elastic polyacrylamide gels but display no spread area sensitivity when cultured on fibrin gels of various moduli. Fibrin gels, and other semiflexible biopolymer networks, exhibit strain stiffening, whereby the elastic modulus of the gel increases with increasing applied strain. Mechanosensitive cells and strain stiffening gels engage in a mechanical feedback loop with cells increasing their applied force and the gel modulus increasing as a result until the cells reach their maximum spread area. Cell applied forces locally induce anisotropy in an initially isotropic matrix providing a mechanism for cell/cell communication over a distance of ~5 cell lengths. This results in alignment of adjacent cells and formation of ring-like multicellular patterns. Finally, due in part to its mechanical properties, fibrin is an appealing scaffold for neural tissue repair. Initial animal studies confirm that salmon derived fibrin mitigates pain and inflammation after injury to the central nervous system.
Winer, Jessamine P., "Differential Cellular Response to Linear and Strain Stiffening Hydrogel Substrates" (2009). Publicly Accessible Penn Dissertations. 50.


After slogging through three tough years, General Motors climbed its way back to the top spot recently with news it had seen sales last year grow in every market worldwide.
, which runs through Sunday at the Oregon Convention Center.
The show features the usual vehicle lineup, as well as an expanded display of electric and hybrid vehicles. Smart Car lookalike Scion iQ will be available, too, for test drives on an indoor track.
, consumers age 19 to 31. A recent survey from consumer researcher Deloitte reported that 59 percent of GenYs prefer an "electrified vehicle," a category that includes hybrids, over any other option.
Around one out of four new vehicles sold this year -- and 40 percent over the next 10 years -- will be snapped up by GenY, a market segment that Deliotte estimates at nearly 80 million strong.
Industry experts say the nation has held on to their cars much longer than in the past, hoping to nurse along decrepit vehicles as they struggled to make it by on tightened budgets. They point to the record-high 10-year average age of cars on American roads.
The annual show represents metro-area dealers' shot at getting shoppers out browsing, test-driving and price-comparing. Dealers often experience an uptick in sales during the four-day show, although, they say, the serious buyers typically don't roll in until Sunday.


text:
Purdue is the second-lowest scoring team in the Big Ten and is at the bottom of the league standings in field-goal percentage (39 percent). It also is near the bottom of the league in defensive rebounds (11th, 26 per game). The Gophers are statistically superior, but the Boilermakers will have confidence from having beaten them earlier this season, and from their victory over Wisconsin.
Stollings said the Gophers learned plenty from those losses at Nebraska and Iowa, when they lost their poise in loud arenas. They have shown the ability to shake off bad games--and regroup after larger losses, such as Rachel Banham's injury.
"They're very resilient,'' Stollings said. "They understand they didn't play their best basketball (at Nebraska and Iowa). They know they're capable of much better. I think they have a renewed spirit. We'll hit the reset button going into the tournament. It's 0-0, and it's time for the Gophers to shine again.''
The folks who predict the NCAA tournament field have the Gophers firmly in the fold, including ESPN bracketologist Charlie Creme, who has them pegged as a No. 8 seed in the Albany bracket. But Stollings said she will rest easier with another victory or two. The Gophers have lost their opening game in the Big Ten tournament in five of the past seven years; last year, as a No. 6 seed, they beat Wisconsin before falling to No. 3 seed Nebraska.
"I don't know that we are secure in our belief (of making the NCAA tournament) yet,'' she said. "I know people are saying many things about it. I would love to go to the Big Ten tournament and see how big of a run we can make. Our goal would be to win the championship, obviously, but we're staying very focused game by game.
"We want to make sure we're in (the NCAA tournament). We don't want to sit on March 16 and wonder, 'Are we in, or not? Are we on the bubble?' We want to know we're in, and we're just waiting on seeding.''
The venue for this year's event is in Hoffman Estates, Ill., about 50 miles from downtown Chicago. It's the second consecutive year the tournament has been held here; it returns to Bankers Life Fieldhouse in Indianapolis next spring. And it's been pretty empty. The crowd for Wednesday night's opening games--featuring Wisconsin, Purdue, Penn State and Indiana--was announced as 3,471 in an 11,000-seat arena. The early session Thursday, with Michigan, Michigan State, Indiana and Rutgers, was announced as 4,941.
There were quite a few Gophers fans at the team hotel this afternoon, including a busload from New Richland, Minn., Carlie Wagner's hometown.
The Big Ten has been proudly noting that attendance during the regular season set a league record, with an announced total of 869,312 fans. Of course, this year's total includes two more teams. Maryland's average home attendance of 5,889 was fourth in the conference, and Rutgers' home average of 2,645 was 11th. The Gophers' home average of 4,460 was eighth.
March 2: Zahui named Big Ten player of the year by media
Gophers women's postgame: Blowout becomes a nail-biter
Rachel Blount writes about a wide variety of sports subjects, including Minnesota's women's basketball team. She also covers Olympic sports, Minnesota sports and social issues that intersect with the games we watch and play. She has been at the Star Tribune for 20 years, reporting on everything from hockey to horse racing to seven Olympic Games. An Iowa native, she holds degrees from the University of Notre Dame and the University of Missouri and is married to fellow sportswriter Pat Borzi.
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Gophers postgame: Little depth plus inexperience leads to loss for Stollings' team
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EIM Client will no longer time out due to large number of connections to the server.
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EIM will no longer show as none if it is an Active Directory field.
Export locator will use Active Directory for First Name, Last Name and Email Address.


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» Current students » Program admin » Program management » Document Certification and Translation
Document Certification and Translation
When you are required to submit an official document to the University (such as a medical certificate to accompany an application form), you may submit a certified copy of that document rather than the original document.
You should retain the original document as the University may request to view the document for verification purposes.
You must not submit photographs or videos with content that may be considered distressing or graphic. In addition, for any application where you are required to provide supporting evidence, photographs or videos are generally not considered appropriate evidence as their authenticity cannot be verified.
What is a certified copy?
A certified copy of a document is an unaltered copy of an original document that has been certified by an authorised person as being a true copy.
Please note: a scanned copy or a photocopy of a certified copy is not a certified copy of a document.
Who can certify documents within Australia?
Persons authorised to certify documents within Australia include:
a Justice of the Peace (currently registered in Australia);
a Commissioner for Declarations (currently registered in Australia);
a barrister or solicitor;
a police officer in charge of a Police Station or of the rank of Sergeant and above;
a Registrar or Deputy Registrar of a Court;
a medical practitioner (licensed or registered);
a pharmacist (licensed or registered);
a dentist (licensed or registered);
a nurse (licensed or registered);
an optometrist (licensed or registered);
a psychologist (licensed or registered);
a veterinary surgeon (licensed or registered);
a manager of a bank or credit union (other than managers of Bank Travel Centres);
an accountant who is a member of the Institute of Chartered Accountants in Australia or New Zealand, CPA Australia, the National Institute of Accountants or the Association of Taxation and Management Accountants; or
a permanent employee of the University with at least 5 or more years of continuous service.
Who can certify documents outside Australia?
Persons authorised to certify documents outside Australia include:
an authorised officer of an Australian overseas diplomatic mission;
an ANU officer; or
an official ANU offshore representative (Agent).
What does the certifier need to do?
When you ask someone to certify a copy of a document, you need to take both the original document and the copy with you as the authorised certifier must sight the original document and check that the copies are true and unchanged copies of the original documents.
The authorised certifier must certify every page of the documentation. Each page should include:
a stamp or statement that states "I certify this to be a true copy of the original document sighted by me";
the authorised certifier's full name and signature;
the autorised certifier's address and contact telephone number;
the authorised certifier's capacity to certify documents (e.g. their title and the institution they belong to, their profession or occupation. A Justice of the Peace should print their registration number and the state/territory in which they are registered); and
the date of certification.
When you are required to submit an official document to the University and the document is written in a language other than English, the document must be accompanied by a complete English translation.
The University accepts translations from the following organisations:
A translator accredited by the National Accreditation Authority for Translators & Interpreters (NAATI);
Multicultural New South Wales Language Services;
Interstate office of either the Ethnic Affairs Commission or the Department of Immigration and Border Protection;
Australian diplomatic missions.
Please note: each page of an authorised translation must bear the translator's signature, the date translated and the translator's accreditation details and/or the official stamp or seal of the translator's accrediting organisation.
Translations of official documents by a student are not accepted by the University.
Advisory information
Late withdrawal
Deferred examinations
Reducing your study load as an international student
Appeal an Academic Progress Committee Exclusion
Program leave - international coursework student
Change or correct your name, title, date of birth or gender
135 ANU (135 268)
Page owner: Division of Student Administration and Academic Services


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The 2020 Split Open was a professional tennis tournament played on clay courts. It was part of the 2020 ATP Challenger Tour. It took place in Split, Croatia between 28 September and 4 October 2020.

Singles main-draw entrants

Seeds

 Rankings are as of 21 September 2020.

Other entrants
The following players received wildcards into the singles main draw:
  Duje Ajduković
  Borna Gojo
  Nino Serdarušić

The following player received entry into the singles main draw as a special exempt:
  Tomás Martín Etcheverry

The following player received entry into the singles main draw using a protected ranking:
  Maximilian Marterer

The following players received entry from the qualifying draw:
  Marcelo Tomás Barrios Vera
  Altuğ Çelikbilek
  Malek Jaziri
  Peđa Krstin

Champions

Singles

 Francisco Cerúndolo def.  Pedro Sousa 4–6, 6–3, 7–6(7–4).

Doubles

 Treat Huey /  Nathaniel Lammons def.  André Göransson /  Hunter Reese 6–4, 7–6(7–3).

References

2020 ATP Challenger Tour
2020 in Croatian tennis
September 2020 sports events in Europe
October 2020 sports events in Europe
Split Open

redpajama_set_name: RedPajamaWikipedia


"I never tell people about my inside monster because I think they won't believe me. So he's here with me, getting stronger and wiser trying to lower me into his trap.
Picking at me, growing at my heart, memories all gone and extinct.
Will you help me get rid of him?
Last night I was about to give him what he wanted, but I told him no. He didn't like that one bit so he came up to me and started to break my ribs and gashing out my heart. I cried out in agony, I cried out to God, but the demon monster said that I'm all his no on else. I'm scared. I'm terrified. I don't know what to do. To say every time I see him. I fall deeper and deeper into his dungeon.
There's a monster inside me ripping at my intestines and breaking my ribs. This kind of monster shows no emotion, no empathy for me at all. he told me that I was worthless and ignorant. When I'm around people the monster really acts out. I see him in every dark corner of this room I lay in. I'm so scared but he's more terrified. He knows I can get rid of him. I just don't know how. Last night he was in my room telling me to fall with him to hell.
Creeping into a dark tunnel, after losing faith, I cry till my eyes are puffy.
Crying because there's no way out, having demons coming to steal me. My mind plays tricks everyday I fall deeper and deeper.
My good memories gone. Faded into darkness.
After he's done he leaves me to bleed out. Laughing at all of the pain I'm in. Ignoring like a fresh scent. I'm terrified whats going to happen next.
No is what I want to say but my mouth won't open.
Going with him might be a good idea after all.
At night I'm scared, the most because he's out lingering like smoke. Last night I had a dream, a nightmare. He crept in my room, got on top of me, and started to rip out my tongue so I wouldn't scream. He told me that God can't help me. He made me an offer. 1. He can take over my body. 2. Walk with him slowly to hell. I don't want to choose, thats what I told the monster. Then he called his demons and started breaking my bones and shattering my heart. I shrieked in pain and started to cry out to the universe. I hate the monster, I really do. No one can save me. No one can see my pain or even come close to my pain. I've been with the monster for a long time. Now its time to tame the monster.
Out of the heart and mind of a 15-year-old girl comes the truth that she faces, everyday.
I remember as a little girl, I loved watching movies on magical places, enchanted castles, true love, princess' and princes', that saving of souls. But I remember one movie, in particular, that scared me more than any of the others. A wicked woman turns into an evil dragon, and tries to destroy the innocence of one. That is what I picture when I read her piece. Not about a girl who will be saved by her one true love, her prince, and live happily ever after.
More of a girl who is a knight, saving herself, day after day, battle after battle with the dragon of a monster that she fights.
I don't know about you, but there are so many monsters that people fight that we are all oblivious too. Some are harsh and gripe and rip and tear at the innocent ones thoughts and heart and even action. Other monsters are softer. They creep in and stay a while, find a home within and then, eventually, move on.
But I do know that, when I hear of such stories, that I always hear of how they are battling through. No matter what is going on in their life, they find a way to push forward and win another day.


Rights of Writers: Can I Say That My Story Was Not Previously Published?
Well I would think that if you published an entire story on-line it might hinder your chances if they demand no on-line pubs. However if it's tiny excerpts (which in the final draft might be edited) I can't see how it would be considered publishing a whole story.
Your point is well taken. The portion of the story posted would clearly have a bearing on whether an editor would consider the posting commercially relevant to the acquisition of "first publication rights." This is a practical business calculation as much as (or more than) a question of legal interpretation of a contractual phrase.
great article! I'd love for you to have a column on my social network for writers. If you're interested, please let me know. Many writers will benefit from your work and get to know you.
This is such a helpful site! I was wondering if you could answer a question for me. I'm looking to send some of my poetry out to a few publications, but in December of last year, I "self-published" a chapbook of my work as a gift to friends and family. I only printed about 50 copies, and the book was not distributed via any other means but my own gift-giving. No money was exchanged and no one else is distributing my work. Are the poems printed in these gift chapbooks considered "published"? Are they off-limits for publications that are only interested in previously unpublished works? Feel free to either comment back, or send me an email at general_silliness@hotmail.com. Thanks!
I want to send in a story to a contest; and it says the prize is a certain amount of money and first publication rights; but after being published; I can 'add it to my collection.' Does that mean that I will never be able to have a publisher publish my story should I so desire if I actually win this contest? I would also like to know what the phrase 'add to my collection' means? Thank you very much for your time and iniative to aid budding writers.


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Highlights of Peru
With the rare combination of breathtaking Andean landscapes, abundant wildlife, rich history and exotic culture, Peru captivates visitors perhaps more than any other South American country.
On this tour we will explore 5 very diverse areas of Peru. We start and finish in the capital city of Lima, which features ruins from the Inca period, art galleries and museums, churches and cathedrals that date from the 16th century and other major historical sites.
Our next stop is the Paracas Nature Reserve (a UNESCO Natural World Heritage Site), dedicated to the conservation of local wildlife and to the history of pre-Inca culture. From here we board a small aircraft to see the famous Nazca Lines - huge line drawings on the coastal plain that pre-date flight by many centuries, yet they can only be seen from the air - a mystery that remains unsolved!
Then we head to Arequipa, surrounded by snow-capped volcanoes, thermal hot springs and the world's deepest canyons. From Arequipa we visit the Colca Canyon to marvel at the scenery and the flight of the Andean Condors.
Then we continue to Lake Titicaca and visit the man-made floating islands, where we discover an ancient way of life that has remained unchanged for centuries.
Our last stop is the lovely colonial city of Cusco, with an excursion to the fabled Machu Picchu. This magical and spectacular site, discovered less than 100 years ago, is located 13,000' above sea level in a deep valley of the towering Andes Mountains. It is considered one of the most well preserved and beautiful ancient cities on the planet.
The capital city, Lima
World-Renowned Nazca Lines
Arequipa, known as the "White City"
Colca Canyon - twice as deep as the Grand Canyon!
Lake Titicaca and the man-made floating islands
Machu Picchu - a UNESCO World Heritage Site
Cuzco - a UNESCO World Heritage Site
Inca ruins of Pachacamac, Andahuaylillas, Ollantaytambo and many more!
* This tour is suitable only for the fit and active traveller.
We depart for Lima on Air Canada Rouge's direct non-stop flight from Toronto.
We arrive in Lima and transfer to our hotel. Our afternoon city tour shows us the Old Town, Plaza Mayor (the main square), the Government Palace, the Cathedral and the House of El Oidor. We also enjoy a guided tour of the Rafael Larco Herrera Museum, built on the site of pre-Columbian pyramid houses. Tonight is our Welcome Dinner.
We drive along the ocean to Paracas, stopping en route at Pachacamac to visit the country's second largest Inca ruins - a sacred city dating to 900 B.C. featuring temples and other ancient structures.
Nazca Lines / Paracas
Today is one of the tour's highlights. We board a small aircraft for a flight over the world renowned Nazca Lines, which can only be seen from the air. We'll learn about the many theories regarding the mysterious Lines' origin and meaning. Afternoon at leisure.
Paracas / Lima
This morning we enjoy a boat tour to Ballestas Island. During our short sea crossing we can see the famous El Candelabro, a huge etching in the sandy hill - considered the beginning of the Nazca Lines. The island is home to huge colonies of birds, as well as sea lions, penguins and seals. Then we head back to Lima.
We fly to the "White City" of Arequipa, high in the Andes. On our city tour we see the extraordinary architecture of the Convent of Santa Catalina, the Main Square, the cathedral and the exhibit of the Sarita "Juanita" - a Virgin of the Sun who was sacrificed to the mountain spirits - recently discovered in a glacier in the Andes.
We proceed to the town of Chivay, located at the midpoint of the Colca Valley. Along the way see the surrounding volcanoes, a variety of flora and possibly some of the local wildlife.
Colca Canyon / Puno
We enter the Colca Canyon - twice as deep as the Grand Canyon! We proceed to the Cruz de Condors, a famed viewpoint to watch the condors gliding on the air currents which rise from the canyon - a spectacular scene with the river 1,200 meters below! We'll enjoy a boxed lunch, then continue our scenic drive to Puno.
We embark on a cultural journey as we head out across Lake Titicaca to explore the islands of Uros and Taquile, and learn about the indigenous people and traditions. These man-made floating islands were constructed entirely out of reeds. The friendly local people are happy to welcome you to their islands, and share their history, culture and traditions. The islands are untouched by the mainland's modernity. The local inhabitants wear traditional clothing and are known for their high-quality handicrafts. While the women make yarn and weave, knitting is actually done exclusively by men!
We transfer to Cuzco and our hotel in the Sacred Valley, former home of the Inca's wealthy families. En route we view the magnificent La Raya mountain range and visit Pukara - a ruin of the fortifications built by the natives of the Andes, and the Inca archeological site of Andahuaylillas with its beautiful baroque St. Peter Church.
Cuzco / Sacred Valley
We visit historic Moray to see the magnificent terraced circles built for planting experiments. Later we visit the Inca fortress of Ollantaytambo where steep stone staircases and terraces lead to an inner temple.
We take a train to the incredible mountaintop ruins of Machu Picchu. Our guided walking tour will illustrate what the city looked like centuries ago. It was once a flourishing ceremonial and agricultural site, and today it remains a beautiful jewel in a remarkable setting between two peaks of the Andes. Afternoon at leisure.
Machu Picchu / Cuzco
After further touring of mysterious Machu Picchu we take the Vista Dome train to Ollantaytambo. Overhead windows on the train enhance the views of the journey as we wind our way along the Urubamba Valley. Upon arrival we transfer back to Cuzco.
Cuzco / Lima / Depart for Home
This morning we first visit the Cathedral of Cuzco and Temple of the Sun - the most important temple in the Inca Empire. Then we proceed to the ancient fortress of Saqsayhuaman, and the picturesque San Blas - known as the "neighborhood of the artisans". After our tour we fly to Lima. This evening we enjoy our Farewell Dinner before heading to the airport for our direct flight to Toronto.
Arrival in Toronto, and connect to your flight home.