text:
\section{Introduction}
Let $X$ be a finite set and let ${\cal F} \subseteq 2^X$ be a family of the subsets of $X$. Let $a$ and $b$ be two positive integers. In the $(a:b)$ Maker-Breaker positional game $(X,{\cal F})$ two players, Maker and Breaker, take turns in claiming previously unclaimed elements of $X$, with Maker going first. In each turn, Maker claims $a$ unclaimed elements and then Breaker claims $b$ unclaimed elements of $X$. The game is played until all the elements of $X$ are claimed. Maker wins the game if she claims all the elements of some $F\subseteq {\cal F}$ by the end of the game. Otherwise, Breaker wins. If Maker can win against any strategy of Breaker then the game is said to be a \textit{Maker's win}. Otherwise, the game is said to be a \textit{Breaker's win}. The set $X$ is referred to as the \textit{board} of the game, while the elements of ${\cal F}$ are referred to as the \textit{winning sets}. The values $a$ and $b$ are called \textit{biases} of Maker, and Breaker, respectively. The most basic case of these games are \textit{unbiased games}, where $a=b=1$.

In this paper, we focus on Maker-Breaker \textit{graph games}, i.e., games where the board is the edge set of a given graph $G$.
In these games Maker's aim is to create a graph consisting only of edges claimed by her that contains some predefined graph theoretic structure. For example, in the \textit{$k$-clique game} (or sometimes abbreviated just as \textit{clique game} when the value of $k$ is not crucial), Maker's goal is to create a graph that contains a clique of order at least $k$. We denote this game by $(E(G),{\cal K}_k)$.

Here, we study a variant of the clique game - the \textit{$T$-tournament game} $(E(G),{\cal K}_{T})$. In the $T$-tournament game, introduced by Beck in~\cite{BeckBook}, the goal graph is a tournament $T$, a complete graph where each edge is directed. Before the game starts, the tournament $T$ is fixed, and Maker and Breaker in turns claim edges, and Maker also chooses one of the two possible \textit{orientations} whenever she claims an edge. If her graph contains a copy of the given tournament $T$ by the end of the game, Maker wins. Otherwise, Breaker does.

\noindent \textbf{Games on $\mathbf{K_n.}$} Very well-studied graph games are the ones where $G=K_n$ is the complete graph on $n$ vertices.
Erd\H{o}s and Selfridge~\cite{ES} initiated the study of the largest value of $k$, $k_c=k_c(n)$, such that Maker can win the $k_c$-clique game on $K_n$ and they were able to prove that $k_c\leq 2(1-o(1))\log_2 n$. Indeed, it turns out that in $(1:1)$ Maker-Breaker clique game on $K_n$, Maker has a strategy to occupy a clique of size $(2-o(1))\log_2(n)$, as shown by Beck~\cite{BeckBook}, and therefore $k_c=2(1-o(1))\log_2 n$ holds. The most interesting fact about this result is that it shows an intriguing relation between games and random graphs here, referred to as the {\em random graph intuition} or {\em probabilistic intuition}. To be precise,
if both players played randomly throughout the game, then Maker's graph would be distributed as a random graph with $n$ vertices
and $\lceil \frac{1}{2}\binom{n}{2} \rceil$ edges, which is well known to have clique size $(2-o(1))\log_2(n)$ with high probability, see e.g. \cite{AS}.
That is, for most values of $k$, a randomly played $k$-clique game on $K_n$ typically has the same winner as the deterministic game played by two intelligent players.

For the tournament game we can ask the same question, as initiated by Beck~\cite{BeckBook}. Motivated by the study of a randomly played $T$-tournament game, Beck conjectured that the largest value of $k$, $k_t=k_t(n)$, for which Maker can win in the $T$-tournament game, for any tournament $T$ on (at most) $k$ vertices, is of size $(1-o(1))\log_2 n$.
However, as the first author together with Gebauer and Liebenau ~\cite{CGL} showed, the truth is twice as large as the conjectured value, i.e., $k_t=(2-o(1))\log_2 n$. This, in particular, tells us two things. Opposite to the clique game, the tournament game does not satisfy the random graph intuition mentioned above. Secondly, since the two values, $k_c$ and $k_t$, are very close to each other, it does not make a big difference for Maker whether she needs to build a graph with or without orientations in an unbiased game on $K_n$.

In the following, we want to find out whether we have similar observations in case we fix $k$ to be a constant, while changing either the bias of Breaker or the board of the game. We start with biased games, in order to give more power to Breaker.
Chv\'{a}tal and Erd\H{o}s~\cite{ChEr} observed that Maker-Breaker games are \textit{bias monotone}, meaning that if the $(1:b)$ game $(X,{\cal F})$ is a Breaker's win, then the $(1:b+1)$ game is also a Breaker's win. Having this in mind, it thus becomes interesting to find the unique  \textit{threshold bias} $b_{{\cal F}}(n)=b_{{\cal F}}$, which is the largest non-negative integer such that for every $b\leq b_{{\cal F}}$ the $(1:b)$ game is a Maker's win. For the $k$-clique game on $K_n$, Bednarska and \L uczak~\cite{BL} showed that the threshold bias is $b_{{\cal K}_k}=\Theta(n^{\frac{2}{k+1}})$. Naturally, one may wonder what happens with the tournament game, and whether in this case orientations of the edges make things more complicated for Maker. We show that, for every tournament $T$ of order $k$, the threshold bias of the $T$-tournament game $(E(K_n),{\cal K}_{T})$ is of the same order as in the $k$-clique game.

\begin{proposition}
\label{biased_tournament}
Let $T$ be a tournament on $k\geq 3$ vertices, then
the threshold bias for the $T$-tournament game on $K_n$
is $b_{{\cal K}_{T}}=\Theta(n^{\frac{2}{k+1}})$.
\end{proposition}

\textbf{Games on random boards.} Another way to give Breaker more power in positional games is to play unbiased graph games on a random graph, as introduced by Stojakovi\'{c} and Szab\'{o}~\cite{SS}. The idea behind this approach is to make the board sparser before the game starts by randomly eliminating edges, so that some of the winning sets no longer exist. We look at the random graph model $\ensuremath{\mathcal{G}_{n,p}}$, which is obtained from the complete graph on $n$ vertices by removing each edge independently with probability $1-p$.

Now, if an unbiased game $(E(K_n),{\cal F})$ is a Maker's win, then we are curious about finding the \textit{threshold probability} $p_{{\cal F}}$ such that for $p=\omega (p_{{\cal F}})$ the game $(E(\ensuremath{\mathcal{G}_{n,p}}),{\cal F})$ is a Maker's win asymptotically almost surely (i.e.\ with probability tending to $1$ as $n$ tends to infinity and abbreviated \textit{a.a.s.\ }in the rest of the paper), and for $p=o(p_{{\cal F}})$, the game $(E(\ensuremath{\mathcal{G}_{n,p}}),{\cal F})$ is a.a.s.\ a Breaker's win.

When the $k$-clique game is played on $\ensuremath{\mathcal{G}_{n,p}}$, Stojakovi\'c and Szab\'o~\cite{SS} showed that for $k=3$, in the \textit{triangle} game, $p_{{\cal K}_3}=n^{-\frac{5}{9}}$ and for $k\geq 4$, it holds that $n^{-\frac{2}{k+1}-\varepsilon} \leq p_{{\cal K}_k}\leq n^{-\frac{2}{k+1}}$. M\"{u}ller and Stojakovi\'c~\cite{MS} recently proved that for all $k\geq 4$ the threshold probability is indeed $p_{{\cal K}_k} = n^{-\frac{2}{k+1}}$. This again underlines an intriguing relation between games and random graphs, again referred to as the \textit{probabilistic intuition}. Indeed, what we can observe here in case $k\geq 4$ (and also holds for several other natural graph games)
is that the threshold probability for Maker's win in the $(1:1)$ game $(E(\ensuremath{\mathcal{G}_{n,p}}),{\cal F})$ is of the same order of magnitude as the inverse of the threshold bias $b_{{\cal F}}$ in the $(1:b)$ game $(E(K_n),{\cal F})$. The triangle game is the only exception in this regard, as
here Maker a.a.s.\ can win also for probabilities below the so-called critical probability $1/b_{{\cal K}_3}$.

We show that the tournament game behaves similarly to the clique game when played on $\ensuremath{\mathcal{G}_{n,p}}$. So, even when played on a sparse graph $\ensuremath{\mathcal{G}_{n,p}}$, creating a graph with oriented edges is not much more difficult for Maker than creating a graph without oriented edges. For the tournaments on $k$ vertices, $k\geq 4$, we show the following, which also supports the probabilistic intuition.

\begin{proposition}\label{random_tournament}
Let $T$ be a tournament on $k\geq 4$ vertices, then
the threshold probability for winning the $T$-tournament game on
$\ensuremath{\mathcal{G}_{n,p}}$ is $n^{-\frac{2}{k+1}}.$
\end{proposition}

So again, for $k\geq 4$, the outcome of the game does not depend much on the choice of the tournament $T$ on $k$ vertices, i.e., on the way the edges of the goal tournament are oriented. However, our next theorem states that the tournament on three vertices behaves differently. In case $T$ is the acyclic triangle $T_A$, we obtain the same threshold probability as in the triangle game on $\ensuremath{\mathcal{G}_{n,p}}$. But, in case $T$ is the cyclic triangle $T_C$, the threshold probability is closer to the critical probability $1/b_{{\cal K}_3}$.

\begin{theorem}\label{T_3}
The threshold probability for winning the unbiased $T_A$-tournament game on
$\ensuremath{\mathcal{G}_{n,p}}$ is $p_{{\cal K}_{T_A}}=n^{-\frac{5}{9}}$, while for the unbiased $T_C$-tournament game
this threshold probability is $p_{{\cal K}_{T_C}}=n^{-\frac{8}{15}}.$
\end{theorem}

{\bf Notation and terminology.}
Our graph-theoretic notation is standard and follows that of~\cite{West}. In par\-ti\-cu\-lar, we
use the following. For a graph $G$, $V(G)$ and $E(G)$ denote its sets of vertices and edges
respectively, $v(G) = |V(G)|$ and $e(G) = |E(G)|$. For disjoint sets $A,B \subseteq V(G)$, let $E(A,B)$ denote the set of edges of $G$ with one endpoint in $A$ and one endpoint in $B$. Given two vertices, $x$ and $y$, an undirected edge is denoted by $xy$, while $(x,y)$ is a directed edge with orientation from vertex $x$ towards vertex $y$. If an edge is unclaimed by any of the players we call it \textit{free}. For a vertex $x \in V(G)$, $N(x) = \{u \in
G : \exists v \in S, uv \in E(G)\}$ denotes the set of neighbours of the vertex $x$ in $G$. We
let $d(x) = |N(x)|$ denote the degree of vertex $x$ in graph $G$. The minimum and maximum degrees of a graph $G$ are denoted by $\delta(G)$ and $\Delta(G)$ respectively. The density of a graph $G$ is defined as $d(G)=\frac{e(G)}{v(G)}$, while \textit{maximum density } is $m(G)=\max_{H\subseteq G} d(H)$.

Let $n,k \in\mathbb{N}$ be positive integers. Then with $T_{n,k}$ we denote the {\em Tur\'an graph}
with $n$ vertices and $k$ vertex classes.
That is, its vertex set $V(T_{n,k})=[n]$ comes with a partition $V(T_{n,k})=V_1\cup\ldots \cup V_k$ such that
$\Big| |V_i|- |V_j| \Big|\leq 1$ for all $1\leq i<j\leq k$, and such that its edge set is
$E(T_{n,k})=\{vw\ |\ v\in V_i,\ w\in V_j,\  1\leq i<j\leq k\}.$
Moreover, let $G$ be a graph on at most $k$ vertices, then we say that a subgraph $H\subseteq T_{n,k}$ is a {\em good copy}
of $G$ in $T_{n,k}$, if $G\cong H$ and $|V(H)\cap V_i|\leq 1$ for every $i\in [k]$.
Let $p\in [0,1]$ and moreover let $M\in [e(T_{n,r})]$. Then with ${\cal G}(T_{n,k},p)$ we denote the random graph model obtained from $T_{n,k}$ by deleting each edge of $T_{n,k}$ independently with probability $1-p$.
That is, ${\cal G}(T_{n,k},p)$ is the probability space of all subgraphs $G$ of $T_{n,k}$,
where the probability for a subgraph to be chosen is $p^{e(G)}(1-p)^{e(T_{n,k})-e(G)}$.
Similarly, with ${\cal G}(T_{n,k},M)$ we denote the probability space of
all subgraphs $G$ of $T_{n,k}$ with $M$ edges, together with the uniform distribution.

Let $Bin(n,p)$ denote the binomial distribution, i.e.\ the distribution of the number of successes
among $n$ independent experiments, where in each experiment we have success with probability~$p$.
Moreover, let us write $X\sim Bin(n,p)$ if $X$ is a random variable with distribution $Bin(n,p)$.

Finally, $W_k=(V,E)$ is called a {\em k-wheel}, if it is obtained from the cycle $C_k$
by adding one further vertex $z$ which is made adjacent to every vertex of $C_k$.
The special vertex $z$ is called the {\em center} of $C_k$.

Throughout the paper $\ln$ stands for the natural logarithm.

{\bf Organization of the paper.}
The rest of the paper is organized as follows. At first we collect some useful results in the Preliminaries. In Section~\ref{sec:prop} we prove Proposition~\ref{biased_tournament} and Proposition~\ref{random_tournament}. Finally, in Section~\ref{sec:triangle} we prove Theorem~\ref{T_3}.

\section{Preliminaries} \label{tg:preliminaries}

The following estimate is usually referred to as a Chernoff inequality \cite{JLR}.

\begin{lemma}[Theorem 2.1 in \cite{JLR}]\label{chernoff}
Let $X\sim Bin(n,p)$ and $\lambda = \mbox{\bf E}(X) = np$. Then for $t\geq 0$, it holds that
$Pr(X\geq \mbox{\bf E}(X) + t)\leq \exp\Big( -\frac{t^2}{2\lambda} + \frac{t^3}{6\lambda^2} \Big).$
\end{lemma}

As indicated above, we will consider the random graph models
${\cal G}(T_{n,k},p)$ and ${\cal G}(T_{n,k},M)$.
For this, we will make use of some general results about random sets.

Following \cite{JLR}, let $\Gamma$ be a set of size $N\in\mathbb{N}$.
For $p\in [0,1]$, we let $\Gamma_p$ denote the probability space of all subsets $A\subseteq \Gamma$,
where the probability of choosing $A$ is $p^{|A|}(1-p)^{|\Gamma\setminus A|}$.
Moreover, for $M\in [N]$, we let $\Gamma_M$ denote the probability space
of all subsets $A\subseteq \Gamma$ of size $M$,
together with the uniform distribution. In case we choose a random set $A$ according to the model
$\Gamma_p$, we shortly write $A\sim \Gamma_p$. Similarly, we write $A\sim \Gamma_M$,
when $A$ is chosen according to the uniform model $\Gamma_M$.

One important fact about the two models above is that in many cases they are closely related to each other
when $p\sim \frac{M}{N}$; see Section 1.4 in \cite{JLR}. In particular, we will make use of the following two
statements, which help us to transfer results from one model to the other.

\begin{lemma}[Pittel's Inequality, Equation (1.6) in \cite{JLR}]\label{random_pittel}
Let $\Gamma$ be a set of size $N$, let $M\in [N]$, and $p=\frac{M}{N} \in [0,1]$.
Let ${\cal P}$ be a family of subsets of $\Gamma$. Moreover, let $H_p\sim \Gamma_p$
and $H_M\sim \Gamma_M$, then
$$Pr(H_M\notin {\cal P})\leq 3\sqrt{M}\cdot Pr(H_p\notin {\cal P}).$$
\end{lemma}

\begin{lemma}[Corollary 1.16 (iii) in \cite{JLR}]\label{random_models}
Let $\Gamma$ be a set of size $N$ and let $M\in [N]$.
Let $\delta>0$ be such that $0\leq (1+ \delta)\frac{M}{N} \leq 1$, and let $p=(1+ \delta) \frac{M}{N}$.
Let ${\cal P}$ be a family of subsets of $\Gamma$. Moreover, let $H_p\sim \Gamma_p$
and $H_M\sim \Gamma_M$, then
$$Pr(H_M\in {\cal P})\rightarrow 1 \text{ implies }Pr(H_p\in {\cal P})\rightarrow 1.$$
\end{lemma}

Later we want to know whether a certain random graph contains a copy of a fixed graph
with high probability. In this regard, we make use of the following two theorems.

\begin{theorem}[Theorem 2.18 (ii) in \cite{JLR}]\label{random_bound1}
Let $\Gamma$ be a set, $p\in [0,1]$  and let $H\sim \Gamma_p$.
Let ${\cal S}$ be a family of subsets of $\Gamma$. Moreover, for every $A\in {\cal S}$
let $I_A$ be the indicator variable which is 1 if $A\subseteq H$, and $0$ otherwise.
Finally, let $X=\sum_{A\in S} I_A$ be the random variable counting the number of elements
of ${\cal S}$ that are contained in $H$. Then
$$Pr(X=0)\leq \exp \Big( - \frac{\mbox{\bf E}(X)^2}{\sum_{A\in {\cal S}} \sum_{\substack{B\in {\cal S}\\ A\cap B\neq \emptyset}} \mbox{\bf E}(I_AI_B)}\Big).$$

\end{theorem}

\begin{theorem}[Theorem 3.4 in \cite{JLR}]\label{threshold_subgraph}
Let $H$ be a graph, and let $X_H$ denote random variable counting the number of copies of $H$
in a random graph $G\sim \ensuremath{\mathcal{G}_{n,p}}$. Then, as $n$ tends to infinity, we have
\begin{align*}
Pr(X_H>0)\rightarrow
\begin{cases}
0 & \text{ if }p\ll n^{-\frac{1}{m(H)}}\\
1 & \text{ if }p\gg n^{-\frac{1}{m(H)}}.
\end{cases}
\end{align*}
\end{theorem}

\section{Most tournaments behave like cliques}\label{sec:prop}

The main idea for the proof of the propositions is as follows: Let $G$ be the graph on which the game is to be played.
Let $T$ be the goal tournament with vertices $v_1,\ldots, v_k$.
Then, before the game starts Maker splits the vertex set of $G$ into $k$ parts $V_1,\ldots,V_k$
with $\Big| |V_i|- |V_j| \Big|\leq 1$ for all $1\leq i<j\leq k$,
and she identifies each class $V_i$ with the vertex $v_i$ according to the following rule:
Whenever Maker claims an edge between some classes $V_i$ and $V_j$, she always chooses
the direction of this edge according to the direction of the edge $v_iv_j$ in $T$.
Because of this identification, it then remains to show that Maker has a strategy
for the usual Maker-Breaker game on $G$ to occupy a copy of $K_k$ with exactly one vertex in each $V_i$.

In order to show that Maker has such a strategy
for this game, we will make use of results from \cite{JLR},
and follow the proof ideas from \cite{BL, SS}.
As most parts are proven analogously to results in the aforementioned publications,
we rather keep our argument short and, whenever possible, we refer back to the known results.
At first, analogously to Theorem 3.9 in \cite{JLR}, we bound the probability that a random graph $G\sim {\cal G}(T_{n,k},p)$ does not contain a good copy of $K_k$.

\begin{claim}\label{cor_p}
Let $k\geq 3$ be a positive integer.
Then there is a constant $c_1=c_1(k)>0$ such that for every large enough $n$ the following is true:
If $n^{-\frac{2}{k+1}}\leq p\leq 4n^{-\frac{2}{k+1}}$
and if $X$ denotes the random variable counting the number
of good copies of $K_k$ in a random graph $G\sim {\cal G}(T_{n,k},p)$, then
$Pr(X=0)\leq \exp (-c_1n^2p).$
\end{claim}

{\bf Proof}
Let $G\sim {\cal G}(T_{n,k},p)$. Let $\cal S$ be the family of good copies of $K_k$ in $T_{n,k}$.
For each such copy $C_i\in {\cal S}$ let
$I_{C_i}$ be the indicator variable which is 1 if and only if $C_i\subseteq G$.
By Theorem~\ref{random_bound1},
$$Pr(X=0)\leq \exp\Big( - \frac{(\mathbb{E}(X))^2}{\sum_{C_1}\sum_{C_2:\ E(C_1)\cap E(C_2)\neq \emptyset} \mathbb{E}(I_{C_1}I_{C_2})}\Big).$$
The denominator in the above expression can be bounded from above by
\begin{align*}
\sum_{t=2}^k \sum_{C_1\in {\cal S}}\sum_{\substack{C_2\in {\cal S}:\\ \ C_1\cap C_2\cong K_t}} p^{2\binom{k}{2}-\binom{t}{2}}
	\leq &\ \sum_{t=2}^k n^{2k-t}p^{2\binom{k}{2}-\binom{t}{2}}\\
	= &\ \Theta(\mathbb{E}(X)^2)\cdot \sum_{t=2}^k n^{-t}p^{-\binom{t}{2}}\\
	= &\ \Theta(\mathbb{E}(X)^2\cdot n^{-2}p^{-1}) \sum_{t=2}^k \Big(n^{-1}p^{-\frac{t+1}{2}}\Big)^{t-2}\\
 = &\ \Theta(\mathbb{E}(X)^2\cdot n^{-2}p^{-1}),
\end{align*}
where in the last equality we use that $p=\Theta(n^{-\frac{2}{k+1}})$. Thus, the claim follows. \hfill $\Box$

\begin{corollary}\label{cor_M}
Let $k\geq 3$ be a positive integer.
Then there is a constant $c_1'=c_1'(k)>0$ such that for every large enough $n$ the following is true:
If $M=\lfloor n^{2-\frac{2}{k+1}}\rfloor$
and if $X'$ denotes the random variable counting the number
of good copies of $K_k$ in a random graph $G\sim {\cal G}(T_{n,k},M)$, then
$Pr(X'=0)\leq \exp (-c_1'M).$
\end{corollary}

\begin{proof}
Set $p=\frac{M}{e(T_{n,k})}$ and observe that  $n^{-\frac{2}{k+1}}\leq p\leq 4n^{-\frac{2}{k+1}}$.
The statement now follows by Claim \ref{cor_p} and Lemma \ref{random_pittel}.
\end{proof}

\begin{corollary}\label{cor_resilience}
Let $k\geq 3$ be a positive integer.
Then there is a constant $\delta=\delta(k)>0$ such that for every large enough $n$
and $M=2\lfloor n^{2-\frac{2}{k+1}}\rfloor$, a random graph $G\sim {\cal G}(T_{n,k},M)$
satisfies the following property a.a.s.:
Every subgraph of $G$ with at least $\lfloor (1-\delta)M \rfloor$
edges contains a good copy of $K_k$.
\end{corollary}

{\bf Proof}
We proceed analogously to \cite{BL}. Let $\delta>0$ such that $\delta-\delta \log(\delta)<c_1'/3$,
with $c_1'$ from Corollary~\ref{cor_M},
and count the number of pairs $(H,H')$ where $H$ is a subgraph of $T_{n,k}$ with $M$ edges
and where $H'\subseteq H$ is a subgraph with $\lfloor (1-\delta)M\rfloor$ edges that does not
contain a good copy of $K_k$. Then
using Corollary \ref{cor_M} (and simplifying the notation slightly by ignoring floor signs) we obtain that the number of such pairs is at most
\begin{align*}
\exp(-\frac{c_1'M}{2})\binom{e(T_{n,r})}{(1-\delta)M} \binom{e(T_{n,r})-(1-\delta)M}{\delta M}
\leq & \exp(-\frac{c_1'M}{2})\binom{M}{\delta M}\binom{e(T_{n,r})}{M}\\
 \leq & \exp\Big(-\frac{c_1'M}{2}+\delta M(1-\log(\delta))\Big) \binom{e(T_{n,r})}{M}\\
= & o(1)\binom{e(T_{n,r})}{M}. \text{\hspace{4.5cm}$\Box$}
\end{align*}

Using this last corollary, we can start proving the existence of Maker strategies. The following claim is an analogue statement to Theorem 19 in \cite{SS}, and thus its proof is analogous to \cite{SS}.

\begin{claim}\label{maker1}
Let $k\geq 3$ and $n$ be positive integers.
Then there is a constant $c_2=c_2(k)>0$ such that for every $M\geq c_2^{-1} n^{2-\frac{2}{k+1}}$, every $1\leq b\leq c_2Mn^{-2+\frac{2}{k+1}}$,
for a random graph $G\sim {\cal G}(T_{n,k},M)$ the following a.a.s.\ holds:
Maker has a strategy to occupy a good copy of $K_k$ in the $(1:b)$ Maker-Breaker game on $G$.
\end{claim}

\begin{proof}
Choose $\delta=\delta(G)$ according to Corollary \ref{cor_resilience} and let $c_2=\delta/10$.
Maker's strategy is as follows: in each of her moves she chooses an edge from $G$ uniformly at random
among all edges from $G$ that have not been claimed so far by herself. If she chooses an edge that is not claimed by Breaker so far,
she claims this edge. Otherwise, Maker declares her move as a failure and skips it.
Similar to \cite{SS}, we consider the first $M':=2\lfloor n^{2-\frac{2}{k+1}} \rfloor\leq \frac{\delta}{2}\cdot \frac{1}{b+1}M$
rounds of the game.
As only a $\frac{\delta}{2}$-fraction of all edges are claimed in these rounds, the probability for a failure is at most $\frac{\delta}{2}$
in each round. So, the number of failures can be ``upper bounded'' by a binomial random variable $X\sim Bin(M',\frac{\delta}{2})$,
which by Chernoff's inequality (Lemma~\ref{chernoff}) satisfies $Pr(X\geq 2\mbox{\bf E}(X))\leq \exp(-\frac{\mbox{\bf E}(X)}{3})=o(1)$.
That is, the number of failures will be at most $\delta M'$ a.a.s.\
Thus, Maker a.a.s.\ creates a graph $H\setminus R$ with $H\sim {\cal G}(T_{n,k},M')$ and $e(R)\leq \delta M'$, against
any strategy of Breaker,
which by Corollary \ref{cor_resilience} a.a.s.\ contains a good copy of $K_k$. Thus, a.a.s.\ Breaker cannot have a strategy to prevent good copies of $K_k$, and as either Maker or Breaker needs to have a winning strategy, the claim follows.
\end{proof}

\begin{corollary}\label{maker2}
Let $k\geq 3$ and $n$ be positive integers
Then there is a constant $c_3=c_3(k)>0$ such that for every $p\geq c_3n^{-\frac{2}{k+1}}$
and $G\sim {\cal G}(T_{n,k},p)$ the following a.a.s.\ holds:
Maker has a strategy to occupy a good copy of $K_k$ in the unbiased Maker-Breaker game on $G$.
\end{corollary}

\begin{proof}
The statement follows immediately from Corollary \ref{maker1} and Lemma \ref{random_models},
where we choose ${\cal P}$ to be the family of all graphs $G\subseteq T_{n,k}$
for which Maker has a strategy to occupy a good copy of $K_k$ in the unbiased Maker-Breaker game
on $E(G)$.
\end{proof}

Finally, we can prove the two propositions.

{\bf Proof of Proposition \ref{biased_tournament}.}
Let $T$ be the tournament, with $k\geq 3$ vertices, of which Maker aims to create a copy on $K_n$.
By Theorem 1 in \cite{BL}, we know that there is a constant $c>0$ such that for large enough $n$ and for every $b\geq cn^{\frac{2}{k+1}}$,
Breaker has a strategy to prevent cliques of order $k$. Using this strategy, Breaker
wins the $T$-tournament game on $K_n$.
Now, let $c_2=c_2(k)$ be given according to Claim \ref{maker1}, and let $M=e(T_{n,k})$, $b=0.25c_2n^{\frac{2}{k+1}}$. Then Claim \ref{maker1} implies that Maker has a strategy to occupy a good copy of $K_k$
in the $(1:b)$ Maker-Breaker game on $T_{n,k}$. But, as we argued earlier, this also gives Maker a strategy
for the $(1:b)$ $T$-tournament game on $K_n$.
\hfill $\Box$

{\bf Proof of Proposition \ref{random_tournament}.}
Let $T$ be the tournament, with $k\geq 4$ vertices, of which Maker aims to create a copy in an unbiased game on $G\sim \ensuremath{\mathcal{G}_{n,p}}$.
By Theorem 1.1 in \cite{MS}, we know that there is a constant $c>0$ such that for $p\leq cn^{-\frac{2}{k+1}}$,
Breaker a.a.s.\ has a strategy to block cliques of order $k$ in the unbiased Maker-Breaker game on $G$,
which again gives a winning strategy for Breaker in the $T$-tournament game on $G$.
Now, let $p\geq c_3n^{-\frac{2}{k+1}}$, with $c_3=c_3(k)$ from Corollary \ref{maker2}.
Before sampling the random graph $G\sim {\cal G}_{n,p}$ fix a partition $V_1\cup\ldots \cup V_k=[n]$ as before. Then, after sampling
$G\sim {\cal G}_{n,p}$, we know that the subgraph induced by those edges which intersect two different parts $V_i$ and $V_j$ is sampled like a random graph $F\sim {\cal G}(T_{n,k},p)$. According to Corollary \ref{maker2},
Maker a.a.s.\ has a strategy to occupy a good copy of $K_k$ in $F\subseteq G$,
and thus Maker a.a.s.\ has a strategy to create a copy $T$ in the unbiased tournament game on $G$.
\hfill $\Box$

\section{The triangle case}\label{sec:triangle}

In the following we prove {\bf Theorem \ref{T_3}}.

For the acyclic triangle $T_A$, the result can be obtained from \cite{SS} as follows: For $p\ll n^{-\frac{5}{9}}$
Breaker a.a.s.\ has a strategy to prevent triangles in the unbiased Maker-Breaker game on $G\sim \ensuremath{\mathcal{G}_{n,p}}$.
Applying such a strategy in the $T_A$-tournament game as Breaker obviously blocks acyclic triangles.
For $p\gg n^{-\frac{5}{9}}$ a.a.s.\ Maker has a strategy to gain an undirected triangle in the unbiased Maker-Breaker game on $G\sim\ensuremath{\mathcal{G}_{n,p}}$. In the $T_A$-game, Maker now can proceed as follows. She fixes an arbitrary ordering $\{v_1,\ldots, v_n\}$ of $V(G)$ before the game starts. Then she applies the mentioned strategy of Maker for gaining an undirected triangle, where she always chooses orientations from vertices of smaller index to vertices of larger index. This way, every triangle claimed by her will be an acyclic triangle, and thus she wins.

Thus, from now on, we can restrict the problem to the discussion of the cyclic triangle $T_C$. To show that $n^{-\frac{8}{15}}$ is the threshold probability for the existence of a winning strategy for Maker in the $T_C$-tournament  game on $G\sim\ensuremath{\mathcal{G}_{n,p}}$, we will study Maker's and Breaker's strategy separately.

\medskip

\begin{figure} [htbp]
\centering
\includegraphics[scale=0.8]{gnp_maker}
\caption{Graph $H$ without and with orientation.}
\label{graphH}
 \end{figure}

We start with {\bf Maker's strategy}. Let $p \gg n^{-\frac{8}{15}}$. Then, by Theorem \ref{threshold_subgraph}, a.a.s.\ $G\sim\ensuremath{\mathcal{G}_{n,p}}$
contains the graph $H$, presented in the left half of Figure \ref{graphH}, as $m(H)=\frac{15}{8}$. As indicated in the right half of the same figure, its edges can be oriented in such a way that each triangle has a cyclic orientation, and thus,
it is enough to prove that Maker has a strategy
to claim an undirected triangle in the unbiased Maker-Breaker game on $H$.
Her strategy is as follows. At first she claims the edge $e_1$, as indicated in the figure. By symmetry, we can assume
that afterwards Breaker claims an edge which is on the ``left side" of $e_1$. Then in the next moves, as long
as she cannot close a triangle,
Maker claims the edges $e_2$, $e_3$ and $e_4$, always forcing Breaker to block an edge
which could close a triangle, and Maker will surely be able to complete a triangle in the next round.

Now, let $p\ll n^{-\frac{8}{15}}$. We are going to show that a.a.s.\ there exists a {\bf Breaker's strategy} which blocks
copies of $T_C$, when playing on $G\sim \ensuremath{\mathcal{G}_{n,p}}$. We start with some preparations. Amongst others, we will consider {\em triangle collections},
as studied in \cite{SS}.

\begin{definition}\label{def:collections}
Let $G=(V,E)$ be some graph without isolated vertices. Further, let \linebreak
$T_G=(V_T,E_T)$ be the graph where
$V_T=\{H\subseteq G:\ H\cong K_3\}$ is the set of all triangles in $G$,
and $E_T=\{H_1H_2:\ E(H_1)\cap E(H_2)\neq \emptyset\}$
is the (binary) relation on $V_T$ of having a common edge. Then:
\begin{itemize}
\item $G$ is called {\em very basic} if $T_G$ is a subgraph of a copy of $K_3^+$ (triangle plus a pending edge),
or a subgraph of a copy of $P_k$ with $k\in\mathbb{N}$.
\item $G$ is called {\em basic} if there are distinct edges $e_1,e_2\in E(G)$
such that $T_{G-e_i}$ is very basic for both $i\in \{1,2\}$.
\item $G$ is a triangle collection if every edge of $G$ is contained in some triangle
and $T_G$ is connected.
\end{itemize}
If $G$ is a triangle collection we further call it a bunch (of triangles) if
we can find triangles $F_1,\ldots F_r\in V_T$ covering all edges of $G$
with the property that $|V(F_i)\setminus \cup_{j<i} V(F_j)|=1$
and $|E(F_i)\setminus \cup_{j<i} E(F_j)|\geq 2$ for every $i\in[r]$.
\end{definition}

\begin{figure} [htbp]
\centering
\includegraphics[scale=0.75]{basic}
\caption{Basic triangle collections.}
\label{basic}
 \end{figure}

Note that every collection on a given number $n$
of vertices, contains a bunch on the same number of vertices
with at least $2n-3$ edges.
Figure \ref{basic} shows some collections that are easily checked to be basic.
For each of the graphs, the edges $e_1$ and $e_2$ indicated in the figure satisfy the condition from the definition
of basic graphs. Moreover,
the following observation is easily verified.

\begin{observation}\label{reduction}
Let $G=(V,E)$. Maker has a strategy to create a triangle (a copy of $T_C$) on $G$
if and only if $G$ contains a collection $C$ such that she has a strategy to create a triangle (a copy of $T_C$)  on $C$.
\end{observation}

In the following we show now that Breaker can prevent Maker from occupying a triangle when playing
on basic graphs. This also ensures a winning strategy for Breaker in the corresponding
$T_C$-tournament game. We start with the following proposition.

\begin{proposition}\label{obs:basic}
Let $G=(V,E)$ be very basic, then Breaker can block every triangle
in the unbiased Maker-Breaker game on $E(G)$, even if Maker is allowed to claim two edges in the very first round.
\end{proposition}

\begin{proof}
Without loss of generality (abbreviated \textit{W.l.o.g.} in the rest of the paper) we can assume that $T_G\cong P_k$ for some $k$, or $T_G\cong K_3^+$,
with $T_G$ as given in Definition \ref{def:collections}.
We further can assume that Maker in the first round claims
two edges $f_1,f_2\in E(G)$ that participate in triangles of $G$.
If $T_G\cong P_k$ then observe that there is an ordering $F_1,\ldots, F_k$ of the elements in $T_G$,
such that $f_1\in E(F_1)$, and
$|V(F_i)\setminus \cup_{j<i} V(F_j)|=1$,
and $|E(F_i)\setminus \cup_{j<i} E(F_j)|=2$ for every $2\leq i\leq k$.
To see this one just has to start the sequence with a triangle $F_1$
containing $f_1$, and to extend the sequence along the path-like structure of $T_G$.
Finally, let $A_1:=E(F_1)\setminus \{f_1\}$
and $A_i:=E(F_i)\setminus \cup_{j<i} E(F_j)$ for every $i\in [k]\setminus\{1\}$.
These sets are pairwise disjoint, have cardinality $2$ and satisfy $A_i\subseteq E(F_i)$
for each $i\in [k]$. That is, Breaker can block triangles by an easy pairing strategy.
(In particular, for his first move, Breaker claims the unique edge $f$
for which there is an $i\in [k]$ with $A_i=\{f_2,f\}$.)
If $T_G\cong K_3^+$, then it can be shown that $G$ contains exactly four triangles
and that one can find an ordering $F_1,\ldots F_k$ (with $k=4$)
with the properties from the previous case. So, Breaker wins similarly.
\end{proof}

\begin{corollary}\label{cor:basic}
Let $G=(V,E)$ be basic, then Breaker can block every triangle
in the unbiased Maker-Breaker game on $E(G)$.
\end{corollary}

\begin{proof}
Let $e_1,e_2$ be the edges given by the definition of a basic graph.
Breaker's strategy is to claim $e_1$ or $e_2$ in the first round.
Afterwards, the game reduces to the graph $G-e_i$ for some $i\in [2]$,
where Maker claims 2 edges, before Breaker claims his first edge.
Now, since $G-e_i$ is very basic for both $i\in \{1,2\}$, Breaker then succeeds by the previous proposition.
\end{proof}

We further observe the following two statements which can be checked
by easy case distinctions.

\begin{observation}\label{K_4}
Breaker has a strategy to prevent cyclic triangles in an unbiased game on $E(K_4)$,
even if Maker is allowed to claim and orient two edges in her first turn.
\end{observation}

\begin{observation}\label{4-wheel}
Breaker has a strategy to prevent cyclic triangles in an unbiased game on $E(W_4)$,
even if Maker is allowed to claim and orient two edges in her first turn, as long as
not both edges are incident with the center vertex of $W_4$.
\end{observation}

Now, using the previous
statements we will show that for $p\ll n^{-\frac{8}{15}}$ a.a.s.\ every collection~$C$ in $G\sim G_{n,p}$
is such that Breaker has a strategy to prevent cyclic triangles in an unbiased game on $C$.
It follows then by Observation \ref{reduction} that a.a.s.\ Breaker wins on $G$.
To do so, we start with the following propositions, motivated by \cite{SS}, which helps to restrict
the set of collections we need to consider.

\begin{proposition}
\label{collection_density}
Let $p\ll n^{-\frac{8}{15}}$, then a.a.s.\ every triangle collection $C$ in $G\sim \ensuremath{\mathcal{G}_{n,p}}$.
 satisfies $m(C)<\frac{15}{8}$.
\end{proposition}

\begin{proof}
Each collection $C$ on at least 25 vertices contains a bunch $B$ on exactly 25 vertices with
$$d(B)=\frac{e(B)}{v(B)}\geq \frac{2v(B)-3}{v(B)}>\frac{15}{8}.$$
Since there are only finitely many such bunches and each of them a.a.s.\ does not appear
in $G$ according to Theorem \ref{threshold_subgraph}, together with the union bound
we obtain that a.a.s.\ each collection in $G$ lives on at most $25$
vertices. Since there are only finitely many collections with at most 25 vertices,
we also know by the same reason that a.a.s.\ each collection  in $G$ on at most 25 vertices
 needs to have maximum density smaller than $\frac{15}{8}$.
\end{proof}

\begin{proposition}\label{properties}
Let $C$ be a triangle collection with $m(C)<\frac{15}{8}$ such
that Maker has a strategy to create a cyclic triangle in an unbiased game on $C$,
but there is no such strategy for any collection $C'\subset C$.
Then the following properties hold:
\begin{enumerate}[(a)]
\item $5\leq v(C) \leq 7$,
\item $e(C)= 2v(C)-1$,
\item $\delta(C)\geq 3$,
\item $C$ is not basic.
\end{enumerate}
\end{proposition}

\begin{proof}
Property~(d) obviously holds, using Corollary \ref{cor:basic}. Moreover, (c) follows immediately. Indeed,
if there were a vertex $v$ with $d_C(v)\leq 2$,
then Breaker could prevent cycles on $C-v$ by the minimality condition on $C$,
and cycles containing $v$ by simply pairing the edges incident with $v$ (if there exist two such edges),
a contradiction. Furthermore, $v(C)\geq 5$ is needed,
according to Observation \ref{K_4}.
Now, let $B$ be a bunch contained in $C$ with $v(C)$ vertices, then $e(C)>e(B)$,
since $\delta(B)=2<\delta(C)$.
As such a bunch contains at least $2v(B)-3$ edges,
it follows that $e(C)\geq e(B)+1 \geq 2v(C)-2$. Furthermore
$e(C)\leq 2v(C)-1$, since otherwise $m(C)\geq 2$.
If $e(C)=2v(C)-1$, then together with $m(C)<\frac{15}{8}$, we deduce that $v(C)\leq 7$.
Otherwise, we have $e(C)=2v(C)-2$ and $e(C)=e(B)+1$. Analogously to the proof of Theorem 23 in \cite{SS}
it then follows that $C$ can only be a wheel; for completeness let us include the argument here:
Let $E(C)\setminus E(B)=\{v_1v_2\}$.
By the definition of a bunch, we can find triangles $F_1,\ldots F_r$ in $B$ covering all edges of $B$
with the property that $|V(F_i)\setminus \cup_{j<i} V(F_j)|=1$
and $|E(F_i)\setminus \cup_{j<i} E(F_j)|\geq 2$ for every $i\in[r]$.
As $e(B)=e(C)-1=2v(B)-3$ it then follows that $r=v(C)-2$
and $|E(F_i)\setminus \cup_{j<i} E(F_j)|= 2$ for every $i\in[r]\setminus\{1\}$,
as otherwise $e(B)> 3 + 2(r-1) = 2v(C)-3$, a contradiction. Thus,
for every $i \in[r]\setminus\{1\}$, $F_i$ needs to share exactly one edge with
$\cup_{j<i} F_j$. From this, we can conclude that $B$ needs to contain at least
two vertices of degree 2. However, as $\delta(C)\geq 3$
and $E(C)\setminus E(B)=\{v_1v_2\}$, we know that $v_1$ and $v_2$
must be the only vertices in $B$ of degree 2.
Now, by the definition of a triangle collection, $v_1v_2$
needs to be part of a triangle in $C$. Thus, there needs to be a vertex $v_3$
such that $v_1v_3,v_3v_2\in E(B)$. But this is only possible if $v_3$
belongs to every triangle $F_i$, $i\in[r]$, and thus, $C$ needs to be a wheel.
Now, to finish the proof, observe that Breaker can always prevent triangles
in an unbiased game on a wheel by a simple pairing strategy,
a contradiction to our assumption.
\end{proof}

So, the goal will be to show that there exists no collection $C$ which satisfies all the conditions
given in Proposition \ref{properties}.

\begin{figure} [htbp]
\centering
\includegraphics[scale=0.7]{collections}
\caption{Special collections.}
\label{special}
 \end{figure}

\begin{lemma}\label{collection}
If a collection $C$ satisfies (a) - (d) from Proposition \ref{properties}, then
either $C$  is isomorphic to $K_5^-$ ($K_5$ minus one edge) or $C$ is isomorphic to one of the graphs $S_i$, $1\leq i\leq 4$, given in Figure \ref{special}.
\end{lemma}

\begin{proof}
If $v(C)=5$, then $e(C)=9$, by Property (b), and the statement follows obviously. So, let $v(C)\neq 5$.
We will show now that a collection satisfying (a) - (c) either is isomorphic
to one of the collections $S_i$, or it is isomorphic to one
of the basic collections $A_i$ or $B_i$ from Figure \ref{basic}, thus contradicting Property (d).

Let us start with $v(C)=6$.
Assume first that $C$ contains a subgraph $H \cong K_4$ and let $\{x,y\}=V(C)\setminus V(H)$.
With $e(C)=11$ and $\delta(C)\geq 3$ we conclude
$xy\in E(C)$, and by the definition of a collection it follows that $x$ and $y$ have a common neighbour $v_1\in V(H)$. Because of (c),
we further have $xv_2\in E(C)$ for some $v_2\in V(H)\setminus \{v_1\}$. Now, if $yv_2\in E(C)$,
then $C\cong S_1$, otherwise by (c) we have $yv_3\in E(C)$ for some $v_3\in V(H)\setminus \{v_1,v_2\}$
and so $C\cong A_1$.\\
Assume then that $C$ does not contain a clique of order 4. We still find a subgraph $H'\subseteq C$
with four vertices $V(H')=\{v_1,v_2,v_3,v_4\}$ and five edges, say $v_1v_3\notin E(H)$.
Since $C$ is a triangle collection, there needs to be some $x\in V(C)\setminus V(H')$ that is part of the same triangle
as an edge $e$ from $H'$. Let $y$ be the unique vertex
in $V(C)\setminus(V(H')\cup \{x\})$.

Assume first that $e=v_2v_4$.  We know then that $\{x,v_1,v_3\}$
is an independent set in $C$, since otherwise we would have a 4-clique in $C$. By (b) and (c), it thus follows
that \linebreak $N(y)=\{x,v_1,v_3,v_i\}$ for some $i\in\{2,4\}$, which gives $C\cong A_2$.

Assume then that $e\neq v_2v_4$ and w.l.o.g.\ $e=v_3v_4$ by symmetry of $H'$.
If $v_1x\in E(C)$, it then follows that $d(y)=3$, since (b) and (c) need to hold;
moreover, $C[V(C)\setminus \{y\}]\cong W_4$ where $v_4$ represents the center of the wheel.
In case $v_4y\in E(C)$, we can only have $C\cong A_2$, as $C$ does not contain a 4-clique;
and in case $v_4y\notin E(C)$, we can assume that $N(y)=\{v_1,v_2,v_3\}$ (because of the symmetry of the 4-wheel),
which yields $C\cong A_3$.
If otherwise $v_1x\notin E(C)$,
then, since there is no 4-clique in $C$,
we immediately obtain $d(y)=4$ and $v_1,x\in N(y)$,
as $e(C)=11$ and $\delta(C)\geq 3$.
Moreover, $v_4\notin N(y)$, since we otherwise would
obtain a 4-clique, independently of the choice
of the fourth neighbour of $y$.
Thus, we conclude \linebreak $N(y)=\{v_1,v_2,v_3,x\}$
and $C\cong A_3$.

Now, let $v(C)=7$. We distinguish three cases.

{\bf Case 1.} Assume that $C$ contains a subgraph $H \cong K_4$. Let $\{x,y,z\}=V(C)\setminus V(H)=:V'$.
With $e(C)=13$ and $\delta(C)\geq 3$ it follows that $\{x,y,z\}$ is not an independent set, w.l.o.g.\ $xy\in E(C)$.
By the definition of a collection it further follows that $x$ and $y$ have a common neighbour -- the vertex $z$
or some vertex $v\in V(H)$.

Assume first that $z\in N(x)\cap N(y)$. By $\delta(C)\geq 3$ each vertex in $V'$ needs to have at least one neighbour in
$V(H)$. If there were a matching of size 3 between $V'$ and $V(H)$, then by (b), one of the matching edges
could not be part of a triangle, a contradiction.
If all the three vertices have a common neighbour in $V(H)$, then one easily deduces $C\cong S_2$.
Otherwise, by symmetry we can assume that there is a vertex $v_1\in V(H)$ such that
$v_1x,v_1y\in E(C)$ and $v_1z\notin E(C),$ and moreover, $v_2z\in E(C)$
for some $v_2\in V(H)\setminus \{v_1\}.$ Now, let $\{v_3,v_4\}=V(H)\setminus \{v_1,v_2\}$.
To ensure that $v_2z$ belongs to some triangle in $C$,
we finally need to have exactly one of the edges from $\{v_3z,v_4z,v_2x,v_2y\}$ to be an edge
in $C$. The first two edges however do not result in a triangle collection,
while for the other two edges we get $C\cong S_3.$

Assume then that $z\notin N(x)\cap N(y)$, but $v\in N(x)\cap N(y)$ for some $v\in V(H)$.
Because of (b) and (c), either $xz\in E(C)$ or $yz\in E(C)$, w.l.o.g.\ say $xz\in E(C)$ and  $yz\notin E(C)$.
As $\delta(C)\geq 3$, we then immediately get $yw\in E(C)$ for some
$w\in V(H)\setminus \{v\}$. Moreover, we then need two other edges incident with $z$ besides $xz$,
of which one is $zv$ to ensure that $xz$ belongs to a triangle. If the second edge is $zw$,
then $C\cong S_4$; otherwise $C\cong B_1$.

{\bf Case 2.} Assume that $C$ does not contain a clique of order 4, but there is some $H\subseteq C$
with $H\cong W_4$. Let $\{x,y\}=V(C)\setminus V(H)=:V'$ and
let $z$ be the unique vertex with $d_H(z)=4$. By (b) and (c), it follows that $xy\in E(C)$,
and since $C$ is a collection, there is a common neighbour of $x$ and $y$ in $V(H)$.

Assume first that $z\in N(x)\cap N(y)$. As $\delta(C)\geq 3$, both vertices $x$ and $y$
have another neighbour in $V(H)\setminus \{z\}$, however
there cannot be a second common neighbour, since there is no 4-clique in $C$.
One easily checks that $C\cong B_2$ or $C\cong B_3$ follows.

Assume then that $z\notin N(x)\cap N(y)$, but $v\in N(x)\cap N(y)$
for some $v\in V(H)\setminus \{z\}.$ If $xz\in E(C)$ (or $yz\in E(C)$),
we then need $yw\in E(C)$ (or $xw\in E(C)$) for some $w\in N_H(v)\setminus\{z\}$
to ensure that $e(C)=13$ and $\delta(C)\geq 3$ holds
while $C$ is a triangle collection. This gives $C\cong B_4$. Otherwise, we have $z\notin N(x)\cup N(y)$.
In this case, let $w'$ to be the unique vertex of $H$ not belonging to $N(v)\cup \{v\}$.
Then we also have $w'\notin N(x)\cup N(y)$. Indeed, if we had $yw'\in E(C)$ say,
then as $yw'$ needs to be part of some triangle and as $d(x)\geq 3$ and $e(C)=13$,
we would need $xw'\in E(C)$, in which case it is easily checked that $C$ is not a triangle collection.
So, we can assume that $xv_1\in E(C)$ for some $v_1\in V(H)\setminus \{v,w',z\}$,
and $yv_1\notin E(C)$, because $C$ does not have a 4-clique.
Finally, since $\delta(C)\geq 3$, we need $v_2y\in E(C)$
for the unique vertex $v_2\in V(H)\setminus \{v,w',z,v_1\}$, i.e.\ $C\cong B_5$.

{\bf Case 3.}  Finally assume that $C$ neither contains a 4-clique nor a 4-wheel.
It is easy to check that $C_0\subseteq C$ (with notation
of vertices as given in Figure \ref{subgraphs}), and by the assumption
of this case we further have $v_1v_3,v_1v_4,v_3v_5\notin E(C)$.
Since $C$ is a triangle collection, we find a vertex $x\in V':=V(C)\setminus V(C_0)$
which belongs to a triangle that also contains an edge $e\in E(C_0)$.
Let $\{y\}=V'\setminus\{x\}$.
By symmetry of $C_0$ we may assume that $e\in\{v_2v_5,v_4v_5,v_1v_5,v_1v_2\}$.

\begin{figure} [htbp]
\centering
\includegraphics[scale=0.6]{Ci}
\caption{Subgraphs.}
\label{subgraphs}
 \end{figure}

Assume first that $e=v_2v_5$ were possible, i.e.\ $C_1\subseteq C$.
Then by assumption of Case 3, every edge in $E(C)\setminus E(C_1)$ would need to be incident
with $y$. Because of (b) and (c) we then had that
$d(y)=4$ and $v_1y,v_3y,xy\in E(C)$. Since these three edges would need
to belong to triangles, we further would need $yv_2\in E(C)$,
which would create a 4-wheel on $V(C)\setminus\{v_3,v_4\}$ with center $v_2$,
in contradiction to the assumption.

So, as next assume that $e=v_4v_5$ were possible, i.e.\ $C_2\subseteq C$.
Then analogously every edge in $E(C)\setminus E(C_2)$ would need to be incident
with $y$, and $d(y)=4$ and $\{v_1,v_3,x\}\subseteq N(y)$, because of (b) and (c).
But then, independently of what the fourth
neighbour of $y$ is, one of the edges $v_1y,v_3y,xy$ could not belong to a triangle, again a contradiction.

As third, assume that $e=v_1v_5$, i.e.\ $C_3\subseteq C$.
By the assumption of Case 3,
every edge in $E(C)\setminus (E(C_3)\cup \{xv_3\})$ needs to be incident
with $y$.
If $xv_3\notin E(C)$, then we have $d(y)=4$ and $xy,v_3y\in E(C)$,
because of $e(C)=13$ and $\delta(C)\geq 3$. Depending on how the other two edges incident
with $y$ are chosen, we either obtain a contradiction by creating a 4-clique or a 4-wheel,
or we see that $C\cong B_6$.
So, let $xv_3\in E(C)$. Then $d(y)=3$, by
(b) and (c), and to have $xv_3$ in a triangle, we need
$yx,yv_3\in E(C)$. It follows that $C\cong B_6$, if $yv_1\in E(C)$ or
$yv_4\in E(C)$, or $C\cong B_7$, if $yv_2\in E(C)$ or $yv_5\in E(C)$.

As last, assume that $e=v_1v_2$, i.e.\ $C_4\subseteq C$. If $xv_3\in E(C)$
were possible, then we had $d(y)=3$ because of $e(C)=13$ and $\delta(C)\geq 3$.
But then, depending on the three edges incident with $y$, we would
get a 4-clique or a 4-wheel in $C$, or we would find an edge which is not contained in a triangle,
a contradiction. So, we can assume that $xv_3\notin E(C)$.
Then, by (b), (c) and the assumption of Case 3,
we deduce that $d(y)=4$ and $yx,yv_3\in E(C)$.
If $yv_2\in E(C)$ were also an edge of $C$, then for any choice of the fourth edge
incident with $y$, we would create a 4-clique or a 4-wheel in $C$.
That is, we can assume that $yv_2\notin E(C)$. But then we need
$v_1y,v_4y\in E(C)$ to ensure that $yx$ and $yv_3$ belong to triangles, which yields $C\cong B_7$.~
\end{proof}

\begin{lemma} \label{breaker_part}
For any collection given by Lemma \ref{collection}, Breaker has a strategy
to prevent cyclic triangles.
\end{lemma}

\begin{proof}
If $C\cong S_i$ for some $i$, note that $C$ is covered by two (not necessarily disjoint) graphs $C(1)$, $C(2)$, plus
at most one additional edge if $C\cong S_2$, where each of the $C(i)$ is isomorphic to $K_4$ or $W_4$.
Choose edges $a_1$ and $a_2$ as indicated in Figure \ref{special}.
In his first move, Breaker claims the edge $a_1$ if Maker did not orient it before;
otherwise he claims the edge $a_2$.
Afterwards, Breaker plays on $C(1)$ and $C(2)$ separately,
meaning: each time Maker orients an edge of $C(i)$, Breaker claims an edge of $C(i)$ if there remains one.
Now, using Proposition~\ref{obs:basic} and
Observation \ref{K_4}, Breaker can do this in a way such that he prevents cyclic triangles on each $C(i)$,
and therefore in $C$.

Finally, we need to look at the case when $C\cong K_5^-$. By an easy case analysis, it can be proven that Breaker has a strategy
to prevent cyclic triangles on $C$. We give a sketch in the following. Let $V(C)=X\cup Y$ with $X=\{v_1,v_2,v_3\}$ and $Y=\{v_4,v_5\}$, and let
$E(C)=\binom{X}{2}\cup \{xy:\, x\in X,\, y\in Y\}.$

\textbf{Case 1.} Maker orients an edge in $E(X,Y)$ in her first turn.

W.l.o.g.\ let $e=v_1v_4 \in E(X,Y)$ be the edge to which Maker gives an orientation in her first move.
Then Breaker's strategy is to delete the edge $v_1v_2$.
Note that $C-\{v_1v_2\}$ is isomorphic to the $4$-wheel $W_4$, here with center $v_3$, and Maker's first arc is not incident with $v_3$. Thus, Breaker can win by Observation \ref{4-wheel}.

\textbf{Case 2.} Maker orients an edge inside $E(X)$ in her first turn.

W.l.o.g.\ let Maker's first oriented edge be $(v_1,v_2)$. Then Breaker's first move will be to delete the edge $v_2v_4$.
Afterwards, Breaker's second move will depend on Makers second move, as follows:

If Maker orients $(v_1,v_3)$ or $(v_3,v_2)$ for her second move, then Breaker claims $v_2v_5$ and afterwards he wins by an easy pairing strategy, with the pairs $\{v_1v_4,v_3v_4\}$ and $\{v_1v_5,v_3v_5\}$.

If Maker for her second move chooses one of the arcs $(v_1,v_4)$, $(v_4,v_1)$, $(v_3,v_4)$, $(v_4,v_3)$, $(v_1,v_5)$, $(v_5,v_2)$, $(v_2,v_3)$ and $(v_3,v_5)$, then Breaker for his second move claims the edge $v_1v_3$.
As he claims $v_2v_4$ and $v_1v_3$ then, the only triplets on which Maker could create a triangle are $\{v_1,v_2,v_5\}$
and $\{v_2,v_3,v_5\}$. In either of the cases it is easy to check that from now on Breaker can prevent cyclic triangles.

If Maker for her second move chooses $(v_2,v_5)$ or $(v_5,v_3)$, then Breaker claims $v_1v_5$ for his second move. Afterwards there remain three triplets
on which Maker still could create a triangle, namely $\{v_1,v_3,v_4\}$, $\{v_1,v_2,v_3\}$ and $\{v_2,v_3,v_5\}$.
To block a triangle on $\{v_1,v_3,v_4\}$, Breaker can consider a pairing $\{v_1v_4,v_3v_4\}$. For the other two triplets
it is easy to check then that Breaker can prevent cyclic triangles, since the orientation which $v_2v_3$ needs, to create a cyclic triangle, is different for these two remaining triplets.

If Maker for her second move chooses $(v_3,v_1)$, then Breaker needs to claim $v_2v_3$. Afterwards there remain three triplets
on which Maker still could create a triangle, namely $\{v_1,v_3,v_4\}$, $\{v_1,v_2,v_5\}$ and $\{v_1,v_3,v_5\}$.
To block a triangle on $\{v_1,v_3,v_4\}$, Breaker can consider a pairing $\{v_1v_4,v_3v_4\}$. For the other two triplets
it again is easy to check that Breaker can prevent cyclic triangles, since the orientation which $v_1v_5$ needs, to create a cyclic triangle, is different for these two triplets.

Finally, if Maker for her second move chooses $(v_5,v_1)$, then Breaker needs to claim $v_2v_5$. Afterwards there remain three triplets
on which Maker still could create a triangle, namely $\{v_1,v_3,v_4\}$, $\{v_1,v_2,v_3\}$ and $\{v_1,v_3,v_5\}$.
To block a triangle on $\{v_1,v_3,v_4\}$, Breaker can consider a pairing $\{v_1v_4,v_3v_4\}$. For the other two triplets
it again is easy to check that Breaker can prevent cyclic triangles, since the orientation which $v_1v_3$ needs, to create a cyclic triangle, is different for these two triplets.
\end{proof}

To summarize, we have shown now that for $p\ll n^{-\frac{8}{15}}$, a.a.s.\ Breaker can prevent cyclic triangles
in the tournament game on $G\sim \ensuremath{\mathcal{G}_{n,p}}$. Indeed, by Proposition \ref{properties},
Lemma \ref{collection} and Lemma \ref{breaker_part},
we know that there exists no collection $C$ with $m(C)<\frac{15}{8}$
on which Maker has a strategy to create a copy of $T_C$.
By Proposition \ref{collection_density} we however know that for $p\ll n^{-\frac{8}{15}}$
a random graph $G\sim \ensuremath{\mathcal{G}_{n,p}}$ a.a.s.\ only contains such collections,
and using Observation \ref{reduction} we thus conclude that a.a.s.\
Maker does not have a winning strategy when playing on $G\sim \ensuremath{\mathcal{G}_{n,p}}$,
which at the same time guarantees a winning strategy for Breaker.\hfill $\Box$

redpajama_set_name: RedPajamaArXiv


Philonotis calcarea, copyright Michael Becker.
The subgenus Eu-Philonotis of the genus Philonotis mostly includes robust and erect mosses (Dixon 1924).
Characters (from Dixon 1924): Usually robust and erect; stems usually straight and regularly branched. Dioicous. Male flower discoid, terminal.


text:
publication date: Jan. 11, 2019
Clinical Roundup NCI's BRCA Exchange aggregates data on BRCA variants to inform understanding of risk
A global resource that includes data on thousands of inherited variants in the BRCA1 and BRCA2 genes is available to the public. The BRCA Exchange was created through the BRCA Challenge, a long-term demonstration project initiated by the Global Alliance for Genomics and Health to enhance sharing of BRCA1 and BRCA2 data.
The five-year BRCA Challenge project was funded in part by NCI and Cancer Moonshot. PLOS Genetics published the paper detailing development of the BRCA Exchange on Jan. 8.
The resource, available through a website and a new smartphone app, allows clinicians to review expert classifications of variants in these major cancer predisposition genes as part of their individual assessment of complex questions related to cancer prevention, screening, and intervention for high-risk patients.
Certain inherited variants in these genes can increase the risk of breast, ovarian, and other cancers by varying degrees, whereas others are not associated with disease. Clinicians and patients need to know whether a given variant is likely to be disease-associated and how likely a pathogenic variant is to cause cancer. Until now, the available data on the inherited variants in these genes were not aggregated in a comprehensive way.
The BRCA Exchange dataset is composed of information … Continue reading NCI's BRCA Exchange aggregates data on BRCA variants to inform understanding of risk
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redpajama_set_name: RedPajamaCommonCrawl


text:
News and Careers Advice
Keep up to date with the latest news and careers advice
How Engineering Has Become the Most Popular Role in Retail
21st April 2018 Uncategorised
When you think of retail nowadays, you tend to think of the online space. Retail is no longer all about traditional brick-and-mortar stores. That's not to say brick-and-mortar is going away, but they are closing at record numbers.
With retail evolving LinkedIn wanted to see how the talent needs of the retail industry are changing, especially in regards to sales and engineering talent. What LinkedIn found was that there was an emergency of engineering roles within retail, and a slight decline in sales roles.
Suddenly, software developers have become one of the most important jobs in retail, with the industry now firmly placing itself what is already a fierce market for tech talent.
Here are the most insights we found and what it means for employers and employees in retail and outside the industry.
Sales are still popular
With pretty much every other retail role declining in popularity, sales are still the biggest function in retail.
Sales talent makes up 29% of the workforce in retail, which is twice the size of the second-biggest function, operations. Engineering and IT talent are the third most common type of employee in retail, representing 9% of the industry, with support, marketing, and other roles rounding things out.
However, what is found is that the proportion of salespeople within retail is steadily shrinking as illustrated below:
The relative size of sales has shrunk from a peak of 33% of all retail jobs in 2013 down to 29% in 2017. At the same time, engineering and IT grew from 7% to 9% of employees in retail (but more on that later).
Engineering is on the rise
Despite the drop in the proportion of sales roles, other functions within retail are showing promising growth—particularly in engineering.
The demand for engineering and digital skills is growing quickly in retail. For example, the role of 'Software developer' is now one of the most common titles in the retail industry. Furthermore, The percent of retail employees in engineering roles has risen from 7% in 2013 to 9% in 2017. Sure that might not sound like a lot, but the speed of the increase is striking, especially when you look at job titles.
The most common skill for retail developers
The single most popular skill among retail developers was Java, a common programming language that's a is commonly used to build the retailers e-commerce stores.
JavaScript and HTML, the second- and fourth-most common skills, are fundamental to website development, while SQL. the third-most common skill is used to access and edit databases. Which is generally used to track inventory and customers.
It's obvious, everyone knows the retail industry is shifting. The move to online shopping and closure of many traditional stores is visible in our everyday lives. But the composition of the industry's talent is shifting, too. The number of sales associates, traditionally the backbone of retail, is shrinking, as many look for opportunities elsewhere.
This isn't a "death of retail" story, despite the decline in traditional retail roles. This showcases the shift from brick-and-mortar storefronts to online shopping, given the significant growth in tech roles, as retail becomes yet another industry competing for software developers and IT support.
About the author: Sharqua Abdullah is an Associate Insights Analyst at LinkedIn.
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Micronaut is presenting a portfolio collecting the works of Martin Oeggerli.
The acari-BOX contains a series of 12 mite portraits reproduced on archival Hahnemühle photo paper, including the five glorious pieces which have been awarded by IPA / Lucy awards in New York (2011). Each piece is carefully hand-made, numbered in editions, and includes Oeggerli's autograph.
The renowned scientific journal CELL has published a new cover produced by Micronaut.
The renowned scientific journal CELL has invited Micronaut to select images for a Cell Picture Show.
In parallel, they are presenting an interview with the artist.
The renowned scientific journal CELL has selected Micronauts iconic bacteria-still-life on their cover.
National Geographic Deutschland has published their new Cover story 'Das Leben in uns' (April 2018).
The renowned scientific journal NATURE has selected Micronauts iconic picture of a human cell placed on a cantilever as one of the top 10 scientific images of 2017.
On the occasion of their 200th anniversary (autumn 2017), the Natural Research Society in Basel (NGiB) proudly presents their brand new book 'natura obscura'. It includes articles of 200 scientists, including Nobel laureates, and celebrities. Illustrations by Micronaut and NASA.
The Swiss National TV broadcasts an artist portrait in their television show 'Kulturplatz' about Martin Oeggerli aka Micronaut. Wednesday 27th May, SRF 1, 10:25 p.m.
Proof is National Geographic's online photography journal, showcasing the voices of National Geographic photographers and editors to offer a behind-the-scenes look at the visual storytelling process… of Micronaut's brand new image series 'Seing Eye to Microscopic Eye'.
Martin Oeggerli experiments with Scanning Electron Microscope (SEM) – Technology to create art. By transforming his invisible discoveries into photorealistic sceneries, he is leading the observer to a place between reality and fiction. A series of hand-picked archival prints is now available on Oeggerli.com.
My beloved 'mite project' has just been published by National Geographic (new feature story called 'Mighty Mites', Feb 2015).
Mysteries of the Unseen World is nominated for the PANDA-Award in 3D/Science images at Wild Screen Festival, the world's largest and most prestigious international wildlife and environmental film festival.
The travelling exhibition "Microstructures I" displays 43 images of the invisible world of the microcosm. Rent it now!
Who is Who Basel elected Martin Oeggerli (the Micronaut) into the most exclusive club of the 200 most well known personalities in town.
The image that was selected displays the mouth brushes of a mosquito larva. With the delicate hairs it creates a water current that washes particles in its mouth.
This sliced open fly head by Micronaut gives you an idea of what can be expected from the upcoming National Geographic Society's documentary "Mysteries of the Unseen World" in 3D. Micronaut was responsible for microscopic special-effects that are a fascinating part of this educational movie.
Unique for this project was, that 3D-stereo IMAX quality moving image sequences were created using a scanning-electron-microscope (SEM). For the first time ever, by the way… !
Four years have passed since the Micronaut website has been developed… A lot has happened since then. The internet has become faster, monitor sizes and resolution have stepped up quite a bit and, depending on your equipment and personal preference you are used to navigate with your smartphone or iPad. We have tried our best to keep up and adapt over the years, but this time it was different – it is time for a change!
While conserving our most popular sections and functions, the new website version has otherwise been redesigned from scratch. The new version makes it easier to browse through the Gallery, lets you enjoy images at increased resolution, and automatically adapts to different monitor sizes. Hope you like it!
Micronaut's image of a Zebralongwing egg placed accurately on top of a vine has been elected into the 'Most Popular' section by the readers of the National Geographic Magazine. The image has been produced and published along with Micronaut's second feature article 'Exquisite Castaways' – National Geographic, September 2010.
Scientists from the School of Life Sciences FHNW and the University Basel report a new strategy for making synthetic nano-particulate materials capable of recognizing viruses is reported in Nature Communications, 20 Feb 2013.
On the left: Nano-particles with attached viruses under the scanning-electron-microscope: Scan: A. Cumbo, Artwork: Micronaut, 2013.
Continue reading: 'Microbes – Small, small world', published by National Geographic, January 2013.
On the left: The Nanomechanical Signature of Breast Cancer under the scanning-electron-microscope: SEM: E. Bieler, M. Plodinec, R. Lim, Artwork: Micronaut, 2013.
Read the full article and interview in 'Turning Point: Martin Oeggerli', Nature 487, 131-131(2012).
Micronaut's hand-coloured scanning-electron-micrograph 'Superhydrophobic Swimming Beauty I' has been awarded 'Best Scientific Cover Image 2012′.
Micronaut's images of butterfly eggs are awarded with the 'German Prize for Scientific Photography 2011′ (DP 2011). Initiated by the science magazine 'Bild der Wissenschaft', the prize is equipped with an illustrous tradition and honors the most striking images of research over the last 12 months.
Considered by some as the Oscars of Photography, the IPA represents the most comprehensive photography contest at the international level, and the largest to date. Needless to say Micronaut is extremely happy to announce that he is awarded with 1st prize in the 'Micro- (professional) category.
…even more, Micronaut has also been awarded with 2nd prize in the 'Micro- (professional) category.
Micronaut is proud to announce his contribution to the '4th Gwangju Design Biennale', on nomination by Ai Weiwei, the prominent Chinese artist and Artistic Director of the event.
Set to be one of the most compelling design events of the year, Asias oldest design festival is expecting more than half a million visitors and features installations of 130 selected artists.
Continue reading: 'Exquisite Castaways', published by National Geographic, September 2010.
Pollen, and ultimately flowers, led to explosive diversification in plants, turning a brown planet green and then red, yellow, white, orange, and all the rest… have a look at the fascinating structures and stunning diversity of pollen grains documented by Micronaut and published by National Geographic!
Continue reading: 'Love is in the Air', published by National Geographic, December 2009.
Micronaut's images of composite flowers are awarded with the 'German Prize for Scientific Photography 2009′ (DP 2009).
Swiss national TV, 13 March 2008. Invisibly small things grow mighty big in Oeggerli's highly magnified pictures.


text:
Morocco showcases the world's oldest jewelry
The Guardian Reporter
The Kingdom of Morocco commemorated Independence Day in style showcasing one of the world's oldest pieces of jewelry discovered at the famous Bismoun Cave in Essaouira, a port city on Morocco's Atlantic coast.
The famous Bismoun Cave archaeological discoveries are on show in Chellah space in Rabat in what authorities have described "as the unveiling of human history as a whole".
The exhibition, a brainchild of the Ministry of Culture, started on Thursday and will run for the next couple of days. Minister of Culture, Youth and Communication, Mr Mehdi Bensaid, and the research team witnessed the presentation at the exhibition, in the presence of Royal Advisor Andre Azoulay and ambassadors of various countries, who came to see the discoveries for the first time.
The search team, made up of Moroccan and foreign scientists, discovered 32 marine pieces made of a seashell called Tritia Gibbosula, formerly known as Nassarius Gibbosulus, making up one of the world's oldest pieces of jewellery so far.
The seashells were most likely used by humans as ornamental necklaces, which indicates a very ancient symbolic behaviour, according to researchers.
Bensaid said the Ministry of Culture arranged the event in celebration of Independence Day – 18 November – noting that this great discovery in Essaouira reflects the important history of human civilization's early years in North Africa, and precisely in Morocco.
He added that "such discoveries contribute to the unveiling of human history as a whole, and makes Moroccans aware of the ample treasures the Kingdom has to offer the world."
Royal Adviser Andre Azoulay described the moment as "significant," praising the Minister for choosing Chellah to mark the event. Azoulay also expressed gratitude to the scientists for their efforts and hard work, which spanned years.Azoulay added that "this discovery will contribute in the re-examination of historic facts, especially in regards to communication between different human groups of the era."
Abdeljalil Bouzoukar, professor of higher education at the National Institute of Archaeology and Heritage in Rabat, said that "this discovery carries great symbolism."
Bozukar added that "symbolic tools, unlike artefacts, can only be communicated through language," and therefore these symbols are linked to language, which suggests that we are at the cusp of language introduced in the history of human civilization.
The Moroccan researcher explained that such discoveries "highlight the beginning of the formation of collective identities," as the spread of such marine seashells dates back 35,000 years.
'Tanzania have about 500,000 heroin users'
'Govt enhancing women's economic empowerment through entrepreneurship'


07/24/2018 10:00 AM CDT - 07/31/2018 4:50 PM CDT CLOSED!
Closing: Time being of the essence, these sales shall be closed on or before August 30, 2018. The purchaser will pay for all closing costs associated with the transaction. Please refer to the Contract for Sale of Real Property, it is the document that governs the sale.


She's festive and ready for her next adventure. You can take this beauty on a tropical vacation or to dinner. The off-the-shoulder detail gives her a level of excitement that will make any girl feel beautiful. Off-the-shoulder top featuring pretty volume sleeves with cuff tie-up.


text:
Lying Under the Apple Tree
Author : Alice Munro
Publisher : Random House
Download PDF Lying Under the Apple Tree Book by Alice Munro Full eBook and published by Random House. This book was released on 2014-05-08 with total page 394 pages. Available in PDF, EPUB and Kindle. Book excerpt: 'Munro is still one of our most fearless explorers of the human being, as she descends, time and again, headlamp on full beam, pickaxe and butter-knife at the ready' The Times Spanning her last five collections and bringing together her finest work from the past fifteen years, this new selection of Alice Munro's stories infuses everyday lives with a wealth of nuance and insight. Beautifully observed and remarkably crafted, written with emotion and empathy, these stories are nothing short of perfection. A masterclass in the genre, from an author who deservedly lays claim to being one of the major fiction writers of our time.
Alice Munro's Miraculous Art
Author : Janice Fiamengo
Publisher : University of Ottawa Press
Download PDF Alice Munro's Miraculous Art Book by Janice Fiamengo Full eBook and published by University of Ottawa Press. This book was released on 2017-02-14 with total page 309 pages. Available in PDF, EPUB and Kindle. Book excerpt: Alice Munro's Miraculous Art is a collection of sixteen original essays on Nobel laureate Alice Munro's writings. The volume covers the entirety of Munro's career, from the first stories she published in the early 1950s as an undergraduate at the University of Western Ontario to her final books. It offers an enlightening range of approaches and interpretive strategies, and provides many new perspectives, reconsidered positions and analyses that will enhance the reading, teaching, and appreciation of Munro's remarkable—indeed miraculous—work. Following the editors' introduction—which surveys Munro's recurrent themes, explains the design of the book, and summarizes each contribution—Munro biographer Robert Thacker contributes a substantial bio-critical introduction to her career. The book is then divided into three sections, focusing on Munro's characteristic forms, themes, and most notable literary effects.
Alice Munro: Writing Her Lives
Author : Robert Thacker
Publisher : Emblem Editions
Download PDF Alice Munro: Writing Her Lives Book by Robert Thacker Full eBook and published by Emblem Editions. This book was released on 2011-05-03 with total page 616 pages. Available in PDF, EPUB and Kindle. Book excerpt: This is the book about one of the world's great authors, Alice Munro, which shows how her life and her stories intertwine. For almost thirty years Robert Thacker has been researching this book, steeping himself in Alice Munro's life and work, working with her co-operation to make it complete. The result is a feast of information for Alice Munro's admirers everywhere. By following "the parallel tracks" of Alice Munro's life and Alice Munro's texts, he gives a thorough and revealing account of both her life and work. "There is always a starting point in reality," she once said of her stories, and this book reveals just how often her stories spring from her life. The book is chronological, starting with her pioneer ancestors, but with special attention paid to her parents and to her early days growing up poor in Wingham. Then all of her life stages — the marriage to Jim Munro, the move to Vancouver, then to Victoria to start the bookstore, the three daughters, the divorce, the return to Huron County, and the new life with Gerry Fremlin — leading to the triumphs as, story by story, book by book, she gains fame around the world, until rumours of a Nobel Prize circulate . . .
The Cambridge Companion to Alice Munro
Author : David Staines
Publisher : Cambridge University Press
Download PDF The Cambridge Companion to Alice Munro Book by David Staines Full eBook and published by Cambridge University Press. This book was released on 2016-03-10 with total page 221 pages. Available in PDF, EPUB and Kindle. Book excerpt: This Companion is a complete introduction to the fictional and non-fictional writings of the Nobel Prize winner Alice Munro.
Publisher : Vintage
Download PDF Lying Under the Apple Tree Book by Alice Munro Full eBook and published by Vintage. This book was released on 2014 with total page 434 pages. Available in PDF, EPUB and Kindle. Book excerpt: 'Munro is still one of our most fearless explorers of the human being, as she descends, time and again, headlamp on full beam, pickaxe and butter-knife at the ready' The Times Spanning her last five collections and bringing together her finest work from the past fifteen years, this new selection of Alice Munro's stories infuses everyday lives with a wealth of nuance and insight. Beautifully observed and remarkably crafted, written with emotion and empathy, these stories are nothing short of perfection. A masterclass in the genre, from an author who deservedly lays claim to being one of the major fiction writers of our time.
The Fiction of Alice Munro
Author : Brad Hooper
Publisher : ABC-CLIO
Download PDF The Fiction of Alice Munro Book by Brad Hooper Full eBook and published by ABC-CLIO. This book was released on 2008 with total page 169 pages. Available in PDF, EPUB and Kindle. Book excerpt: A comprehensive, up-to-date analysis of Munro's fiction.
The View from Castle Rock
Download PDF The View from Castle Rock Book by Alice Munro Full eBook and published by Random House. This book was released on 2010-08-31 with total page 368 pages. Available in PDF, EPUB and Kindle. Book excerpt: WINNER OF THE NOBEL PRIZE IN LITERATURE The world's finest living short story writer turns to her family for inspiration; and what follows is a fictionalised, brilliantly imagined version of the past. From her ancestors' view from Edinburgh's Castle Rock in the eighteenth century to her parents' thwarted ambitions in Ontario, and her own awakening in 1950s Canada, Munro effortlessly weaves fact and myth to create an epic story of past and present, proving that fiction has much to tell us about life.
The Last Empire
Author : Nicolas Bradley
Publisher : ePressario Publishing Inc
Download PDF The Last Empire Book by Nicolas Bradley Full eBook and published by ePressario Publishing Inc. This book was released on 2011 with total page 199 pages. Available in PDF, EPUB and Kindle. Book excerpt:
Beautiful NOW, Beautiful ME, Beyond Guilt, Blame and Shame
Author : Larry V. Murphy
Download PDF Beautiful NOW, Beautiful ME, Beyond Guilt, Blame and Shame Book by Larry V. Murphy Full eBook and published by Lulu.com. This book was released on with total page 298 pages. Available in PDF, EPUB and Kindle. Book excerpt:
God's Dawning Light; Spiritual Awakening in a Dusty Old Farmhouse
Download PDF God's Dawning Light; Spiritual Awakening in a Dusty Old Farmhouse Book by Larry V. Murphy Full eBook and published by Lulu.com. This book was released on 2010-07-03 with total page 234 pages. Available in PDF, EPUB and Kindle. Book excerpt: This book describes a man's late life journey to Spiritual Awakeness and Unwavering Forgiveness. In fact, forgiveness changed the book's unfolding during the writing of it. The author began the book's writing with the desire to review a memory troubled life in a way that some degree of peace could be found within it. Forgiveness absolutely began to take over the book very early; forgiveness became the way of seeing the life rather than an end result. The author learned that forgiveness is not only cleansing and healing in retrospective, it completely changes the way we look at things. For the author, a lifetime previously perceived largely in ugliness was transformed into beauty. Prior unsavory scenes took their true places within auras of ignorance; eyes of forgiveness exposed ignorance as the true culprit in all unhappiness. Thus, things previously seen in contempt and blame became innocent of all charges. The author found total, Unwavering Forgiveness and Universal Love, gateways to God Consciousness.
Author : Ohio State Board of Agriculture
Download PDF Annual Report Book by Ohio State Board of Agriculture Full eBook and published by . This book was released on 1862 with total page 638 pages. Available in PDF, EPUB and Kindle. Book excerpt: Includes abstract of the Proceedings of the county agricultural societies.
Annual report of the Ohio State Board of Agriculture
Author : Ohio (State) Board of Agriculture
Download PDF Annual report of the Ohio State Board of Agriculture Book by Ohio (State) Board of Agriculture Full eBook and published by . This book was released on 1862 with total page 642 pages. Available in PDF, EPUB and Kindle. Book excerpt:
Download PDF St. Nicholas Book by Full eBook and published by . This book was released on 1883 with total page 996 pages. Available in PDF, EPUB and Kindle. Book excerpt:
Author : Mirosława Buchholtz
Download PDF Alice Munro Book by Mirosława Buchholtz Full eBook and published by Springer. This book was released on 2016-01-20 with total page 161 pages. Available in PDF, EPUB and Kindle. Book excerpt: The book offers a new approach to the study of Alice Munro's fiction. Its innovative quality consists in juxtaposing a variety of literary analyses of selected stories with two other ways of looking at her fiction: the perspectives of film adaptation and of pedagogy. The book is divided into three parts which mirror the key words in the title: understanding, adapting and teaching. Part One consists of four articles on various aspects of Munro's short fiction from a literary perspective. Part Two - four essays - addresses editing and film adaptations of Munro's stories (both television and feature films). Part Three consists of an essay on didactic aspects of Munro's fiction and of several interviews with teachers of Canadian literature who have included stories by Munro in their syllabi.
The Scouts of Stonewall
Author : Joseph A. Altsheler
Publisher : VM eBooks
Download PDF The Scouts of Stonewall Book by Joseph A. Altsheler Full eBook and published by VM eBooks. This book was released on 2016-06-29 with total page pages. Available in PDF, EPUB and Kindle. Book excerpt: A young officer in dingy Confederate gray rode slowly on a powerful bay horse through a forest of oak. It was a noble woodland, clear of undergrowth, the fine trees standing in rows, like those of a park. They were bare of leaves but the winter had been mild so far, and a carpet of short grass, yet green, covered the ground. To the rider's right flowed a small river of clear water, one of the beautiful streams of the great Virginia valleys.
The English and Scottish Popular Ballads
Author : Francis James Child
Download PDF The English and Scottish Popular Ballads Book by Francis James Child Full eBook and published by . This book was released on 1884 with total page 270 pages. Available in PDF, EPUB and Kindle. Book excerpt:
Download PDF The English and Scottish Popular Ballads Book by Francis James Child Full eBook and published by Cambridge University Press. This book was released on 2014-11-20 with total page 267 pages. Available in PDF, EPUB and Kindle. Book excerpt: Published 1882-98, this ten-part work by Harvard's first professor of English became an essential resource for scholars and folklorists.
Filthy Beautiful Lust (Filthy Beautiful Lies, #3)
The Hero's Guide to Storming the Castle (The League of Princes, #2)
Crooked Little Heart
His Forbidden Bride: 50 Loving States, West Virginia
Tote Mädchen lügen nicht
The Queen of Subtleties: A Novel of Anne Boleyn
Letters to a Secret Lover
Meet Me in Bombay
Faefever (Fever, #3)
Slan (Slan, #1)
Hellburner (The Company Wars, #5)
The Enemy (Jack Reacher, #8)
A Dark-Hunter Christmas (Dark-Hunter #3.6)
Beloved Vampire (Vampire Queen, #4)
The Spirit of Imagination (The Spirit of Imagination, #1)
Those Who Wish Me Dead
The Portal (An Alternative History Novel)
Planetary, Volume 1: All Over the World and Other Stories
The Practical Heart: Four Novellas
Last Stand (The Black Mage, #4)
The Day John Died
Bite Me (Pride, #9)
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Shadowmagic (Shadowmagic, #1)
Elvenblood (Halfblood Chronicles, #2)


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Like many fans packed into the Jeffersonville ArenA last November for Terry Harper's first wrestling show at the Southern Indiana venue, I was a little taken aback when I got my first look at Freight Train. He didn't move like other wrestlers. He didn't fight like other wrestlers. A couple of casual fans standing near me turned their noses up and left the building to smoke a cigarette until the match was over. It's a shame they did, because the longer I watched Freight Train do what he does best, the more intrigued I became.
I mentioned Freight Train didn't move like other wrestlers. Freight Train has mild autism spectrum disorder known as Aspergers Syndrome. One of the hallmarks of Aspergers is a singular focus on one or more key interests. For Charles Stevenson, the boy who grew up to become Freight Train, there were two things that drew his interest more than any other: trains and wrestling.
Rather than hinder Stevenson's dream of becoming a wrestler, his condition actually helped him to achieve his dream. Stevenson refused to take no for an answer when men told him he couldn't train. He refused to take no for an answer when people wouldn't book him. He ignored the people who laughed at him. He was persistent; he never gave up. And when 5 Dollar Wrestling opened a door, he made that dream come true.
Packed with interviews with friends and fans like like Colt Cabana, Mad Man Pondo, Crazy Mary Dobson, and even the notoriously old school Rip Rogers, "The Power of Freight Train" is a beautiful story not only of Charles Stevenson, but all the people whose lives have been touched along the way, especially Freight Train's good friend and mentor "Manscout" Jake Manning. It was Manning who gave Freight Train a chance with 5 Dollar Wrestling, and Manning, more than anyone else, has been inspired and transformed by his friendship with the gentle giant.
"The Power of Freight Train" is a story even non-wrestling fans can enjoy. It's an inspirational tale of a man who refused to give up on a dream and the hearts he changed along the way.
"The Power of Freight Train" is available only on High Spots.
Yes, jabroni, Mitchell Huff has more views than you. As a matter of fact at last count, he's crushing you.
Mitchell Huff is an OVW-trained talent, taught by the likes of Danny Davis and Rip Rogers.
Huff is no rookie. He started training years ago. He took several years off from the business, and it's a credit to his talent that as soon as he stepped back in, he became a sought after main event level player across the Midwest.
Huff is no beer bellied slouch either. Go see him in person. He's ripped, and he's worked hard to get there.
Mitchell's also become an in-demand trainer here in Southern Indiana. I'm sure he'd welcome you into his class, if you were man enough to try. I'm sure you wouldn't last.
Mitchell thought your video was funny. So did his friends. They know there's a reason why Mitchell is called the Chosen One.
Hope you got a lot of hits riding on Mitchell's coat tails. That's all you wanted, I'm sure, because you definitely don't belong in a ring with Mitchell Huff. He is the Chosen One, and you… well, you're just trying to get famous off someone else's hard work.
No, I will not repost your video here. But I will share Mitchell's. He is Tough Enough.
OVW HD – A new Indiegogo Campaign!
I saw him again almost two years later at OVW, the night I started work on Bluegrass Brawlers. He didn't wrestle that night, but I saw him a few times over the next couple of years. He had a great look, and he showed potential, but he was always in the mid-card, working underneath guys like Rob Terry and Jamin Olivencia. He was better each time I saw him, but he was always outshined by the main event players.
Wednesday night, I saw him again. He worked the main event against OVW champion Mohamed Ali Vaez. This was a completely different Ryan Howe than I had ever seen before. Same look, same gimmick, but there was a confidence and a swagger about him I hadn't seen before. Howe looked like he belonged in that main event. He looked ready for the next step. If history is any indication at OVW, he'll probably get it sooner rather than later.
That's the legacy of OVW. OVW has set the standard for wrestling schools for nearly 20 years. Cena, Orton, Lesnar, Batista, Punk, Ziggler, Cody Rhodes, Miz, Mizdow, Henry, Big Show, Beth Phoenix, Mickie James, Dinsmore, Conway, Shelton Benjamin, John Morrison, Lisa Marie Varon, Jamin Olivencia, Rockstar Spud. Over 100 students have gone on from OVW to work for WWE or TNA.
OVW just launched an Indiegogo campaign to upgrade their television equipment. OVW is the longest running wrestling television program in America outside of WWE (over 800 episodes!), and they're ready to step it up and go HD. This campaign will allow them to upgrade their studio, their cameras, and their editing equipment so they can continue to produce a top quality program while providing the best training for the business, from inside the ring to the editing room.
OVW television airs locally in Louisville, but it's also available to view online. OVW alums have shared with me how fans have come up to them in airports and venues around the country, fans who know them only from watching online. Most recently, OVW announcer Dean Hill told me he was approached by a fan in Seattle, Washington who watched OVW on TV!
Independent wrestling is growing in popularity once more, and OVW is positioning itself to take advantage of the changing tides. Check out the campaign on Indiegogo and the perks that are available – including and opportunity to train at the school. And by all means go to www.ovwrestling.com to check out their show for yourself!


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Talking Horses: BHA to act after Irish raiders dominate Cheltenham
blinellccom
The British Horseracing Authority have said that "the health of British jump racing is always on our radar" and that "work is well under way across several areas" to improve the competitiveness of British-trained National Hunt horses at the highest level, following the four-day Cheltenham Festival meeting where Irish-trained horses won 23 of the 28 races.
Tuesday's comment is the BHA's first public acknowledgement of the unprecedented dominance of Irish stables at the sport's showpiece event, despite the visitors fielding only 40% of the runners over the four days.
Willie Mullins and Henry de Bromhead saddled six winners apiece – one more than all British stables combined – while Denise Foster, the new licence holder at Gordon Elliott's stable following his recent suspension, sent out three. Twelve of the Festival's 14 Grade One contests went to Ireland, including all six of the non-novice Grade Ones over jumps, while 60% of places in the frame also went to Irish-trained runners.
Having congratulated "everyone who had a winner last week" and noted that this year's Festival "had lifted the nation's spirits with some spectacular racing after a hard winter of lockdown," a BHA spokesperson said that the Authority is "aware that, like us at the BHA, many parts of the British racing industry will be reflecting on what happened last week and what can be done to improve the competitiveness of British racing at the top level.
"However, this is about more than just four days in March. The health of British jump racing is always on our radar and work is well under way across several areas. The Jump Pattern Committee, for example, decided last year that there were elements of the programme that needed looking at, including the need to produce some more competitive races across the season. There are also important pieces of work across areas such as handicapping and safety and welfare which all feed into the longer term objective of building a strong, competitive jump racing industry in Britain."
The spokesperson also promised the BHA would "harness some of the strong views that currently prevail across the sport and engage further with our industry. These views will feed in to ongoing work and continue to help shape what steps the sport will take to support the future of British jump racing."
Competition from valuable alternative targets at the Punchestown Festival at the end of April could well mean that Irish-trained runners do not travel in similar strength to the Grand National meeting earlier at Aintree earlier in the month.
The significant overall levels of prize money on offer in Ireland, though, could also be part of the problem – from a British perspective, at least – in the first place, as the incentive for British-based owners such as Rich Ricci, a mainstay of the Willie Mullins stable, and Cheveley Park Stud to send their most promising horses to race in Ireland is obvious.
British racing's funding model, which involves a Levy on betting turnover on the sport, makes it difficult to compete on prize money with Ireland, where the government makes a direct contribution to the industry of around €70m (£60m) annually. A much less extensive programme – with fewer than 400 meetings as opposed to nearly 1,500 in Britain – also contributes to higher prizes per race.
Unless the BHA can find a way to lure many more of the brightest young jumping talents to spend their racing careers in Britain, an embarrassment in the West Country in March could soon be an almost annual event.
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Чемпіонат світу з трекових велоперегонів 1962 проходив з 24 по 28 серпня 1962 року в Мілані, Італія на велодромі Vigorelli. Вперше на чемпіонаті провели змагання з командної гонки переслідування. Усього розіграли 9 комплектів нагород — 7 у чоловіків та 2 у жінок.

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