text:
Fifteen sports and 21 disciplines will be contested at Minsk 2019.
Thirteen venues will host events at the second European Games.
The Opening and Closing Ceremonies of the second European Games will take place at the Dinamo Stadium, which will also host the athletics events.
Minsk is the capital and largest city of Belarus, situated on the Svislač and the Nyamiha rivers.
Minsk has a population of 1.98 million people.
The Minsk 2019 mascot is Lesik the Fox who represents friendship, development, harmony, discipline, determination, cheerfulness, invincibility and vigour!
Lesik is pretty easy to spot with his red hat, white and green shirt, black shorts, blue shoes and yellow fur. These same colours are used on the Minsk 2019 logo.
One-hundred-and-ninety-nine medal events will be contested at Minsk 2019; 91 men's events, 89 women's events and 19 mixed events.
In all, 4,022 competitors will take part over 10 days.
The European Games take place every four years, the first of which was held in Baku, Azerbaijan in 2015.
Eight sports taking place at the European Games will serve as qualifiers for the Tokyo 2020 Olympic Games.
Minsk was elected as the venue for the second European Games on October 21, 2016 at the General Assembly of the EOC.
Fifty One European National Olympic Committees will be represented in the Belarusian capital during the Games.
Belarusian athletes won 43 medals at the first European Games - 10 gold, 11 silver, and 22 bronze - and finished seventh on the medal table.
The second European Games will be held at the largest sports facilities in Minsk, such as stadium "Dinamo", "Minsk-Arena" and "Chizhovka-Arena".
More than 8,000 volunteers will help prepare and stage Minsk 2019.
Minsk 2019 volunteers are assigned to one of 20 roles ranging from spectator services, hospitality, venue management, medical services, transport and protocol assistants.
Belaruskali is a Premium Partner of Minsk 2019 whilst Yasna-Beltelecom, MTS, Fabrika Reklamy, Belarusbank, UNICEF and LUCH are National Partners.
Faroe Islands will be able to compete for the first time at the second European Games in archery, badminton, judo and table tennis.
The first buyer of a paper ticket was Minsk 2019 mascot Lesik, the baby fox.
The programme of the European Games also includes the European Judo Championships which will gather the best athletes in the Chizhovka Arena.
The second European Games will be held under the motto Bright Year, Bright You.
The logo of the second European Games has been inspired by the slogan "Follow Your Dream".
A Minsk Guest Card has been designed for tourists coming to the Belarusian capital and provides discounts and bonuses for museums, shopping centres, restaurants, and cafes.
Minsk is characterised by long avenues and neoclassical grandeur.
Flame of Peace
Mascot - Lesik
Minsk 2019 TV
The European Games: Day one of competition and Opening Ceremony
The European Games: Opening Ceremony
The European Games: Day two of competition and European Olympic Committees Extraordinary General Assembly
The European Games: Day three of competition
The European Games: Day four of competition
The European Games: Day five of competition
The European Games: Day six of competition
The European Games: Day seven of competition
The European Games: Day eight of competition
The European Games: Day nine of competition
The European Games: Final day of competition
The European Games: Closing Ceremony
Minsk 2019 European Games Medal Table
Belarus 24 16 29
Ukraine 16 17 18
Italy 13 15 13
Minsk 2019 European Games Results
Women's 10m air pistol Shooting Shotgun
Dorsa Arabshahi 240.1
Haniyeh Rostamiyan 238.7
Kim Min-jung 218.7
Seventy-eight athletes selected to represent Switzerland at Minsk 2019
By Nancy Gillen
Swiss Olympic has selected 78 athletes from 14 sports to compete in June's European Games in Minsk.
With 23 athletes already qualified, a further 55 have been picked during a second selection phase.
These have come from athletics, basketball 3x3, beach soccer, canoeing, cycling, wrestling and archery.
The Swiss beach soccer team secured a place at the Games after beating Germany and Portugal at the European qualification tournament in Salou, while 11 cyclists, including Théry Schir, a participant in the Rio 2016 Olympic Games, have been selected.
Switzerland are not sending their strongest teams in athletics or 3x3 basketball due to other competitions taking place at the same time.
It is hoped that the Games will be an opportunity for athletes to gain experience and prove their skills on the international stage, however.
Archer Iliana Deineko will represent Switzerland for a second time at the European Games ©World Archery
Archer Iliana Deineko has been selected to participate in the European Games for the second time, having featured in Baku in 2015.
Thirty-one women and 47 men will compete in total.
The Swiss delegation will meet at Swiss Olympic headquarters in Ittigen on June 12 to prepare, with the Games taking place from June 21 to 30.
"It is very difficult to predict how many medals we will get at such an event," said Ralph Stöckli, Swiss Olympic Chef de Mission.
"What is certain is that we have a motivated team with a great deal of potential, which also embodies the diversity of Swiss sports.
"I am really looking forward to accompanying this team and I hope that we can collect important experiences together and celebrate some successes."
Back to Minsk 2019 Home
May 2019: Drugs cheat Tsikhan carries European Games Flame
May 2019: European Games Torch Relay crosses border into Belarus
May 2019: Minsk 2019 Torch set to continue epic international journey by heading into Belarus
May 2019: Two sambists to help carry Minsk 2019 Flame to Belarus' capital
May 2019: Germany to send up to 150 athletes to Minsk 2019 European Games
Nancy Gillen
Follow @Nancy_Gillen
Nancy Gillen is a junior reporter at insidethegames.biz. She graduated from the University of Leeds with a first class honours degree in International History and Politics, and also completed a placement year at the University as communications assistant. Her first trip with insidethegames.biz was to Turin for the Men's Volleyball World Championships.
Read more of Nancy's articles
Follow @Nancy_Gillen on Twitter
Belarusian Rouble (BYN)
What the local time is now in Minsk


Discussion in 'GBAtemp & Scene News' started by Urza, May 31, 2008.
Update to ricco's jezzball clone. Changes include a new "stretch" controls mode, where instead of dragging the cursor, you tap where you want it to go, and swipe to deploy the cutters. Additional changes include being able to use X and Y to send out vertical and horizontal cutters respectively, and a new smiley theme for the balls. Download and full changelog below.
Interestingly enough, it looks almost exactly like the original I remember, can't wait to hop into this one.


text:
    <hawtio-pane class="qdr-overview" position="left" width="300">

    </hawtio-pane>

redpajama_set_name: RedPajamaGithub


The ceremonial opening and unveiling of the plaque officially commenced the activities of the King Sejong Institute. Pictured, UP Rector Jaroslav Miller and Seoung-Hyun Moon, Korean Ambassador to the Czech Republic.
Korean Ambassador Mr. Seoung-Hyun Moon, together with UP Rector Jaroslav Miller commenced the functioning of King Sejong Institute in the university premises. The new UP workplace has been offering courses in Korean as well as cultural events promoting Korean culture since mid-September. It has joined the ranks of more than one hundred institutes of the same name located all over the world.
The ceremony prepared for the opening of King Sejong Institute took place in the Ceremony Hall at the UP Rector's Office. It will be located in the premises at Křížkovského 8 and is the first of its kind in Moravia.
"Palacký University has Chinese, Japanese, and Indonesian studies, and I am immensely happy to announce that Korean studies have joined the family of Asian studies under the auspices of the Department of Asian Studies," said Rector Miller. Together with the ambassador of the Republic of Korea he unveiled the plaque of the institute on October 9.
According to Martin Kudláček, Vice-Rector for International Relations, the date is very significant for Korea. "It is a day celebrating the creation of Hangeul, the Korean writing system. It was created in the fifteenth century by King Sejong the Great in order to replace Chinese characters used until then in the country. Thus, he significantly contributed to spreading education across the entire Korean peninsula," he said.
The headquarters of King Sejong Institute was founded in Seoul, Korea, in 2007 and is largely financed by the Korean government, as well as 174 local King Sejong institutes promoting Korean language and culture located all over the world. All these institutes offer Korean courses which meet with increasing interest every year.
The Olomouc workplace was established thanks to the collaboration of the Department of Asian Studies at the UP Faculty of Arts and the Embassy of the Republic of Korea. They will work closely together, and apart from Korean courses, they will also organise cultural events which will introduce Korean culture to Olomouc citizens and students. In October, a Korean Culture Festival took place in the UP Armoury courtyard. Visitors could try wearing traditional hanbok clothes, have their names written in Hangeul, or play the traditional game Yut Nori. Film enthusiasts enjoyed watching several Korean movies for free in Premiere Cinemas.
Korean studies have a three-year history at Palacký University. According to David Uher, head of the Korean section at the Department of Asian Studies, its establishment was possible due to the efforts by Mi-Young Park. "She worked at UP since 2003 and during roughly ten years she accomplished a great deal in Korean studies for us. She was the first person in the Czech Republic to write a Czech-Korean dictionary. She was repeatedly granted the support of the prestigious Korea Foundation," Uher said, adding that the accreditation for a three-year Bachelor's study programme Korean in Economics at UP was approved already in 2015.
"We have met all the conditions of the accreditation commission only after an agreement with Kangnam University was signed and their professor could arrive in Olomouc to become the guarantor of the programme. Korean in Economics was launched thanks to the vital assistance by the Department of Applied Economics. This year in June our first alumni graduated," added David Uher.
Lack of interest is really not an issue. Out of the one hundred and seventy applicants, thirty-six succeeded in entrance examinations. During their studies they will get a hands-on experience in Korean companies, and some of them will even study in Korea via student exchange programmes. Alumni of this study programme are often hired by Korean companies, mainly in the Moravian-Silesian Region. They will soon be allowed to enrol in the following Master's study programme Korean Philology, which is already prepared for launching within Asian Studies. More information is available HERE and HERE.


text:
Max Mara The Cube
Max Mara The Cube is an integrated system of outerwear and accessories that has been around the "block" a few times. The project - which was first released in 2008 - boasts a clear-cut identity, which is further accentuated by the unmistakable "Cube" packaging.
"What I do when I take photos is I live in the moment. I create situations where something interesting will happen, and by allowing everyone to be part of the creativity you get spontaneous moments".
Malibu-based artist Tierney Gearon drew inspiration from her Californian surroundings for her second photographic project dedicated to Max Mara The Cube. Live in the moment: this is the principle Gearon follows in her shots, in which she recreates spontaneous moments that allow everyone to get involved and waits for something unexpected to happen.
Water-repellent technical cotton parka
The Max Mara The Cube's design is so innovative that it has been "boxed" into the display cases at the Berlin State Museums and New York's FIT, as a "cult object" in the "Fashion and Technology" exhibitions. In a bid to celebrate the research and innovation behind the Max Mara The Cube's design, we asked three photographers to provide us with their own visual interpretation of the project and provided them with just one prompt: life, cubed.
Water-repellent gabardine trench coat
Water-repellent reversible taffeta parka
Thirteen other photographers have previously interpreted the project: Erik Madigan Heck, Koto Bolofo, Rinko Kawauchi, Hellen Van Meene, Max Farago, Lorenzo Vitturi, Christopher Anderson, Tina Barney, Laurie Simmons, Marilyn Minter, Sam Taylor-Johnson, Stephen Shore and Alex Prager.
Introducing CAMELUXE
Powered by Imbotex Lab.
The innovative new project from Max Mara The Cube.

redpajama_set_name: RedPajamaCommonCrawl


text:
CONGRESS MUST TAKE ACTION TO CURE SAVINGS AND LOAN INDUSTRY MALAISE
Great Western Financial Corp. of Beverly Hills, Calif., which has acquired several troubled Florida savings and loan institutions in recent years, promises in its television commercials that "We'll always be there!"
From the perspective of depositors, that pledge rings true enough, because Great Western is a muscular giant in a thrift industry plagued by 97-pound weaklings.
But from the perspective of the troubled Federal Savings and Loan Insurance Corp. (FSLIC), there is considerable doubt that Great Western will always "be there."
Great Western has announced its intention of leaving FSLIC and switching to the Federal Deposit Insurance Corp. (FDIC), which insures the deposits of commercial banks and savings banks. The move makes good business sense -- and a $25 million annual saving in insurance premiums -- for Great Western.
There is some question whether Great Western can bail out of FSLIC, making it even more powerless to come up with the $50-billion-plus it needs to compensate depositors in insolvent thrifts. Congress, mindful of the danger to FSLIC if its stronger clients pull out, earlier prohibited transfers such as the one proposed by Great Western. But the prohibition expired in August.
Great Western believes it can finesse the problem by merging its savings and loan operations with a small, FDIC-insured savings bank it owns in Washington state.
Federal regulators, and possibly the courts, will have the final say. However they rule, the malaise of the thrift industry is not going to disappear, no matter how devoutly the Congress and the new administration might wish it.
Both presidential candidates are ignoring the problem and the banking industry is balking at paying a larger share of premiums to rescue depositors in the 500 to 1,000 S&Ls; on the endangered list.
Both banks and thrifts pay 1/12th of 1 percent of their deposits to the separate insurance funds. Four years ago, FSLIC more than doubled its premiums to cover the costs of closing the many institutions -- mostly in the Sunbelt -- driven bankrupt by the incompetent and corrupt management that flourished under inadequate regulation and congressional neglect.
Now, with even larger assessments looming, efficiently managed institutions such as Great Western are unfairly being asked to pay for the sins of their counterparts. It's no wonder they are fleeing FSLIC.
The new adminstration and Congress should begin to address the problem by equalizing insurance premiums paid by banks and thrifts -- raising the former and lowering the latter. That might stop the flight from FSLIC and add some -- but not nearly enough -- money to the bailout pot.
Ultimately, though, Congress almost certainly will have to reach beyond the banking industry and put the arm on the taxpayer to rebuild the nation's damaged financial system.


text:
\section{Introduction}
This is a story about treewidth, but it starts in the world of biology.
A phylogenetic tree on a set of leaf labels $X$
is a binary tree representing the evolution of $X$. These are studied extensively in computational biology~\cite{steel2016phylogeny}.
Given two such trees a natural aim is to quantify their topological dissimilarity~\cite{john2017shape}. Many such dissimilarity measures have been devised and they are often NP-hard to compute, stimulating the application of techniques from parameterized complexity~\cite{bulteau2019parameterized}.
Recently there has been a growing focus on treewidth.
This is because, if one takes two phylogenetic trees on $X$ and identifies leaves with the same label, we obtain an auxiliary graph structure known as the \emph{display graph}~\cite{bryant2006compatibility}. Crucially, the treewidth of this graph is often bounded by a function of the dissimilarity measure that we wish to compute~\cite{kelk2016monadic}. This has led to the use of Courcelle's Theorem within phylogenetics (see e.g. \cite{kelk2016monadic,janssen2018treewidth}) and explicit dynamic programs running over tree decompositions; see \cite{van2022embedding} and references therein.
In~\cite{kelk2017treewidth} the spin-off question was posed: is the treewidth of the display graph actually a meaningful measure of phylogenetic dissimilarity \emph{in itself} - as opposed to purely being a route to efficient algorithms?
A closely-related question was whether parameter-preserving reduction rules,  applied to two phylogenetic trees to shrink them in size, also preserve the treewidth of the display graph?
The well-known \emph{subtree reduction rule} is certainly treewidth preserving~\cite{kelk2017treewidth}.
However, the question remained whether the \emph{common chain reduction rule}~\cite{AllenSteel2001} is treewidth-preserving.
A common chain is, informally, a sequence of leaf labels $x_1, \ldots, x_k$ that has the same order in both trees.
Concretely, the question arose~\cite{kelk2017treewidth}: is it possible to truncate a common chain to \emph{constant length} such that the treewidth of the display graph is preserved?
Common chains form ladder-like structures in the display graph, i.e., this question is about how far ladders can be reduced in length without causing the treewidth to decrease.

In this article we answer this question affirmatively, and more generally. %
Namely, we do not restrict ourselves to display graphs, but consider arbitrary graphs.
A \emph{ladder $L$ of length} $k\geq 1$ of a graph $G$ is a $2 \times (k+1)$ grid graph such that $L$ induces (only) itself and that $L$ is only connected to the rest of the graph by its four cornerpoints.
First, we prove that a ladder $L$ can be reduced to length 4 without causing the treewidth to decrease, and that this is best possible: reducing to length 3 sometimes causes the treewidth to decrease.
We also show that if $tw(G) \geq 4$ then reduction to length 3 is safe and,  again,  best possible.
These tight examples are also shown to exist for higher treewidths.
Returning to phylogenetics, and thus when $G$ is a display graph, we leverage the extra structure in these graphs to show that common chains can be reduced to 4 leaf labels (and thus the underlying ladder to length 3) without altering the treewidth: this result is thus slightly stronger than on general~$G$.

Our proofs are based on first principles: we directly modify a tree decomposition to get what we need. In doing so we come across the problem that, unless otherwise brought under control, the set of bags that contain a given ladder vertex of $G$ can wind and twist through the tree decomposition in very pathological ways. Getting these \emph{snakes} under control is where much of the hard work and creativity lies, and is the inspiration for the title of this paper.

From a graph-theoretic perspective our results have the following significance.
First, it is standard folklore that shortening paths (i.e. suppressing vertices of degree 2) is treewidth-preserving, but there is seemingly little in the literature about shortening recursive structures that are slightly more complex than paths, such as ladders.
(Note that Sanders~\cite{sanders1996linear} did consider ladders, but only for recognizing graphs of treewidth at most 4, and in such a way that the reduction destroys the ladder topology).
Second, our results imply a new safe reduction rule for the computation of treewidth; a survey of other reduction rules for treewidth can be found in \cite{Abu-Khzam2022}.
Third, we were unable to find sufficiently precise  machinery, characterisations of treewidth or restricted classes of tree decomposition in the literature that would facilitate our results.
Perhaps most closely related to our ladders are the more general \emph{protrusions}: low treewidth subgraphs that ``hang'' from a small boundary~\cite[Ch.~15-16]{kernelization-book_2019}.
There are general (algorithmic) results~\cite{BodlaenderFLPST16} wherein one can safely cut out a protrusion and replace it with a graph of parameter-proportional size instead -- these are based on a problem having \emph{finite integer index}~\cite{BodlaenderF01}.
Such techniques might plausibly be used to prove that there is \emph{some} constant to which ladders might safely be shortened, but our tight bounds seem out of their reach.
Finally, the results imply that minimal forbidden minors for bounded treewidth cannot have long ladders.

The above context leads us to ask whether our results on ladders can be
\begin{enumerate*}[(i)] %
\item (elegantly) generalized to more complex structures than ladders;
\item can they be made constructive;
\item can recognition of such structures be performed efficiently?
\end{enumerate*}
We hope that our results will stimulate new research in this direction.

\section{Preliminaries}
We follow \cite{kelk2017treewidth} for notation.
A \emph{tree decomposition} of an undirected graph $G=(V,E)$ is a pair $(\mathcal{B}, \mathbb{T})$ where $\mathcal{B} = \{B_1, \dots ,B_q\}$, $B_i \subseteq V(G)$, is a multiset of \emph{bags}
and $\mathbb{T}$ is a tree whose $q$ nodes are in bijection with $\mathcal{B}$, and
\begin{itemize}
    \item[(tw1)] $\cup_{i=1}^q B_i = V(G)$;
    \item[(tw2)] $\forall e = \{ u,v \} \in E(G), \exists B_i \in \mathcal{B} \mbox{ s.t. } \{u,v\} \subseteq B_i$;
    \item[(tw3)] $\forall v \in V(G)$, all the bags $B_i$ that contain $v$ form a connected subtree of $\mathbb{T}$.
\end{itemize}

The \emph{width} of $(\mathcal{B}, \mathbb{T})$ is equal to $\max_{i=1}^q |B_i|-1$. The \emph{treewidth} of $G$, denoted $tw(G)$, is the smallest width among all tree decompositions of $G$.
Given a tree decomposition $\mathbb{T}$ of a graph $G$, we denote by $V(\mathbb{T})$ the (multi)set of its bags and
by $E(\mathbb{T})$ the set of its edges. Property (tw3) is also known as \emph{running intersection property}.
Without loss of generality, we consider only connected graphs $G$.

Note that subdividing an edge $\{u,v\}$ of $G$ with a new degree-2 vertex $uv$ does not change the treewidth of $G$.
In the other direction, suppression of degree-2 vertices is also treewidth preserving \emph{unless} it causes the only cycle in a graph to disappear (e.g. if $G$ is a triangle); unlike \cite{kelk2017treewidth} we will never encounter this boundary case.
An equivalent definition of treewidth is based on chordal graphs. Recall that a graph $G$ is chordal if every induced cycle in $G$ has exactly three vertices. The treewidth of $G$ is the minimum, ranging over \emph{all} chordal completions $c(G)$ of $G$ (we add edges until $G$ becomes chordal), of the  size of the maximum clique
in $c(G)$ minus one. Under this definition, each bag of a tree decomposition of $G$ naturally corresponds to a maximal clique in a chordal completion of $G$ \cite{Blair1993}.

We say that a graph $H$ is a \emph{minor} of another graph $G$ if $H$ can be obtained
from $G$ by deleting edges and vertices and by contracting edges.

A \emph{ladder $L$ of length $k \geq 1$} is a $2 \times (k+1)$ grid graph. A \emph{square} of $L$ is a set of vertices of $L$ that induce a 4-cycle in $L$. We call the endpoints of $L$, i.e., the degree-2 vertices of $L$, the \emph{cornerpoints} of $L$.
We say that a graph $G$ \emph{contains} $L$ if the following holds (see Fig.~\ref{fig:ladder} for illustration):
\begin{enumerate}
    \item The subgraph induced by vertices of $L$ is $L$ itself.
    \item Only cornerpoints of $L$ can be incident to an edge with an endpoint outside~$L$.
\end{enumerate}

Observe that a ladder of length $k$ is a minor of the ladder of length $(k+1)$. Treewidth is non-increasing under the action of taking minors, so reducing the length of a ladder in a graph cannot increase the treewidth of the graph.

Suppose $G$ contains a ladder $L$. We say that $L$ \emph{disconnects} $G$ if $L$ contains a square $\{u,v,w,x\}$ such
that the two horizontal edges of the square (following Fig.~\ref{fig:ladder}, these are the edges
$\{u,w\}$ and $\{v,x\}$) form an edge cut of the entire graph $G$.
Note that a square of $L$ has this property if and only if all squares of $L$ do.
Also, if we reduce the length of a ladder $L$ to obtain a shorter ladder $L'$, $L'$ disconnects $G$ if and only if $L$ does.
We recall a number of results from Section 5.2 of \cite{kelk2017treewidth}; these will form the starting point for our work.
\begin{lemma}[\cite{kelk2017treewidth}]
\label{lem:disconnecting}
 Suppose $G$ contains a disconnecting ladder $L$. %
 The ladder $L$ can be increased arbitrarily in length without increasing the treewidth of $G$.
\end{lemma}
For the more general %
case, the following weaker result is known.
\begin{lemma}[\cite{kelk2017treewidth}]
\label{lem:protoplusone}
Suppose $G$ has $tw(G) \geq 3$ and contains a ladder. %
If the ladder is increased arbitrarily in length, the treewidth of $G$ increases by at most one.
\end{lemma}
We now make the following (new) observation.
\begin{observation}
\label{obs:k4}
Suppose $G$ contains a ladder $L$ of length $2$ or longer. If $L$ is not disconnecting, then $tw(G) \geq 3$.
\end{observation}
\begin{hidden}
\begin{proof}
Let $a,b,u,v,w,x$ be the six vertices in the ladder $L$ with edges as shown in Fig.~\ref{fig:ladder}. Observe that $G$ contains a $K_4$ minor. Specifically,
take $\{a,b\}, u, v$ and $\{w,x\}$ as the four corners of the minor pre-contraction. The fact that $L$ is not disconnecting means that there is a path from $\{a,b\}$ to $\{w,x\}$ that leaves the ladder at one end and re-enters it via the other, inducing the minor edge between $\{a,b\}$ and $\{w,x\}$. $K_4$ is the (unique) forbidden minor for graphs of treewidth at most 2, so $tw(G)>2$.
\end{proof}
\end{hidden}

We can leverage Observation \ref{obs:k4} to reformulate Lemma \ref{lem:protoplusone} without the $tw(G) \geq 3$ assumption. However it then only applies to ladders of size at least two.
\begin{lemma}
\label{lem:plusone}
Suppose $G$ contains a ladder $L$ with length at least 2. %
If %
$L$ is increased arbitrarily in length, the treewidth of the graph increases by at most one.
\end{lemma}
\begin{hidden}
\begin{proof}
If $L$ is disconnecting then Lemma \ref{lem:disconnecting} applies. If $L$ is not disconnecting, then by Observation \ref{obs:k4} we have $tw(G) \geq 3$ and Lemma \ref{lem:protoplusone} applies.
\end{proof}
\end{hidden}
If we start from a sufficiently long ladder, can the ladder be increased in length without increasing the treewidth? Past research has the following partial result. %

\begin{theorem}[\cite{kelk2017treewidth}]
\label{thm:unbounded}
Let $G$ be a graph with $tw(G)=k$. There is a value $f(k)$ such that if
$G$ contains a ladder of length $f(k)$ or longer,
the ladder can be increased in length arbitrarily without altering (in particular: increasing) the treewidth.
\end{theorem}
Ideally we would like a single, universal value \emph{that does not depend
on} $k$. In this article we will show that such a single, universal constant does exist.
\section{Results}
We first consider graphs of treewidth at least 4; we later remove this restriction.
\begin{theorem}
\label{thm:main}
Let $G$ be a graph with $tw(G)\geq 4$. If $G$ has a ladder $L$ of length 3 or higher, the ladder can
be lengthened arbitrarily without changing  %
the treewidth.
\end{theorem}

\begin{proof}
Due to Lemma \ref{lem:disconnecting} we
can assume that $L$ is not disconnecting.  Our general strategy is to show that if $G$ contains the ladder $L$ shown in Fig.~\ref{fig:ladder}, we can insert an extra `rung' in the ladder without increasing the treewidth, thus obtaining a ladder with one extra square (see Fig.~\ref{fig:ladder2}).
The extension of the ladder by one square can then be iterated to obtain an arbitrary length ladder. %
\begin{figure}[t]
\begin{minipage}[b]{0.44\textwidth}
    \centering
    \includegraphics[scale=1.0]{figs/ladder}
    \caption{A ladder $L$ of length 3 with corner points $a,b,c,d$.
    }
    \label{fig:ladder}
\end{minipage}
\hfill
\begin{minipage}[b]{0.48\textwidth}
    \centering
    \includegraphics[scale=1.0]{figs/ladder2}
    \caption{Inserting a new edge $\{u',v'\}$ into ladder $L$ results in ladder $L'$ of length 4.}
    \label{fig:ladder2}
\end{minipage}
\end{figure}

Let $L$ be the ladder shown in Fig.~\ref{fig:ladder}, and assume that $G$ contains $L$.
Let $(\mathcal{B}, \mathbb{T})$ be a minimum-width tree decomposition for $G$.
We proceed with a case analysis. The cases are cumulative: we will assume that earlier cases do not hold.

\smallskip

\noindent
\textbf{Case 1.} \textbf{Suppose that $\mathcal{B}$ contains a bag $B$ such that all four vertices from one of the squares of $L$ are in $B$.}
Let $\{u,v,w,x\}$, say, be the square of $L$ contained in bag $B$, where the position of the vertices is as in Fig.~\ref{fig:ladder}. We prolong the ladder as in Fig.~\ref{fig:ladder2} and create a valid tree decomposition for the new graph as follows:
we introduce a new size-5 bag $B'=\{u',u,v,w,x\}$ which we attach pendant to $B$ in the tree decomposition, and a new size-5 bag $B''=\{u',v',v,w,x\}$ which we attach pendant to $B'$. Observe that this is a valid tree decomposition for the new graph. Due to the fact that $tw(G) \geq 4$, the treewidth does not increase, and the statement follows.
Note that in this construction $B''$ contains all four of $\{u',w,v',x\}$, which is a square of the new ladder, so the construction can be applied iteratively many times as desired to produce a ladder of arbitrary length.

\smallskip

\noindent
\textbf{Case 2.} \textbf{Suppose that $\mathcal{B}$ contains
a bag $B$ such that $|B \cap \{a,u,w,c\}| \geq 2$ and  $|B \cap \{b,v,x,d\}| \geq 2$.}
Let $h_1, h_2$ be two distinct vertices
from $B \cap \{a,u,w,c\}$ and $l_1, l_2$ be two distinct vertices
from $B \cap \{b,v,x,d\}$.

Observe that it is possible to partition the sequence $a, u, w, c$ into two disjoint intervals $H_1, H_2$, and the sequence $b,v,x,d$ into two disjoint intervals $L_1, L_2$ such that $h_1 \in H_1$, $h_2 \in H_2$, $l_1 \in L_1$ and $l_2 \in L_2$. If we contract the edges and vertices in each of $H_1, H_2, L_1, L_2$ we obtain a new graph $G'$ which is a minor of $G$. Note that $G'$ is similar to $G$ except that the ladder now has two fewer squares -- the three original squares have been replaced by a square whose corners correspond to $H_1, H_2, L_1, L_2$. This square might contain a diagonal but we simply delete this. We have $tw(G') \leq tw(G)$ because treewidth is non-increasing under taking minors. Now, by projecting the contraction operations onto $(\mathcal{B}, \mathbb{T})$ in the usual way\footnote{In every bag of the decomposition vertices from $H_1$ all receive the vertex label $H_1$, and similarly for the other subsets $H_2, L_1, L_2$.}, we obtain a tree
decomposition $(\mathcal{B}', \mathbb{T}')$ for $G'$ such that the width of $\mathbb{T}'$ is less than or equal to the width of $\mathbb{T}$. The bag in $(\mathcal{B}', \mathbb{T}')$ corresponding to $B$, let us call this $B'$, contains all four vertices $H_1, H_2, L_1, L_2$. Clearly, $\mathbb{T'}$ is a valid tree decomposition for $G'$. We distinguish two subcases.
\begin{enumerate}
\item If $\mathbb{T}'$ has width at
least 4, we can repeatedly apply the Case 1 transformation to $B'$ to produce an arbitrarily long ladder without raising the width of $\mathbb{T}'$. The resulting decomposition will thus have width no larger than $\mathbb{T}$. %
\item Suppose $\mathbb{T}'$ has width strictly less than 4, and thus strictly less than the width of $\mathbb{T}$. The width of $\mathbb{T}'$ is at least 3 because of the bag containing $H_1, H_2, L_1, L_2$. Case 1 introduces size-5 bags and can thus raise the width of the decomposition by at
most 1. Hence we again obtain a decomposition whose width is no larger than $\mathbb{T}$ for a graph with an arbitrarily long ladder.
\end{enumerate}
This concludes Case 2.  Moving on, any chordalization of $G$ must add the diagonal $\{w,v\}$ and/or the diagonal $\{u,x\}$.
Hence we can assume that there is a bag containing $\{u,w,v\}$ and another bag containing $\{v,w,x\}$. (If the other diagonal is added we can simply flip the labelling of the vertices in the horizontal axis i.e. $a \Leftrightarrow b, u \Leftrightarrow v$ and so on). As Case 1 does not hold we can assume that the bag containing $\{u,w,v\}$ is distinct %
from the bag containing $\{v,w,x\}$.

For the benefit of later cases we impose extra structure on our choice of minimum-width tree decomposition of $G$.
The \emph{distance} of decomposition $(\mathcal{B}, \mathbb{T})$ is the minimum, ranging over all pairs of bags $B_1, B_2$ such that $B_1$ contains $\{u,w,v\}$ and $B_2$ contains $\{v,w,x\}$, of the length of the path
in $\mathbb{T}$ from~$B_1$~to~$B_2$. \\
\\
\fbox{
\parbox{0.97\textwidth}{
\textbf{We henceforth let $(\mathcal{B}, \mathbb{T})$ be a minimum-width tree decomposition of $G$ such that, ranging over all minimum-width tree decompositions, the distance is minimized. Clearly such a tree decomposition exists.}}}\\
\\
Let $B_1, B_2$ be two bags from $\mathcal{B}$ with $\{u,w,v\} \subseteq B_1$, $\{v,w,x\} \subseteq B_2$ which achieve this minimum distance. Let $P$ be the path of bags from $B_1$ to $B_2$, including $B_1$ and $B_2$. We assume that $P$ is oriented left to right, with $B_1$ at the left end and $B_2$ on the right. As Case 2 does not hold, we obtain the following.

\begin{observation}
\label{obs:stuff_not_there}
$B_1$ does not contain $b$, $x$ or $d$, and $B_2$ does not contain $a$, $u$, $c$.
\end{observation}
\begin{hidden}
\begin{proof}
If this was not true then Case 2 would hold, contradiction.
\end{proof}
\end{hidden}

\noindent
\textbf{Case 3. $B_1$ and $B_2$ are adjacent in $P$.} Although this could be subsumed into a later case it introduces important machinery;
we therefore treat it separately.%

\emph{Subcase 3.1:} Suppose $a \in B_1$ (or, completely symmetrically, $d \in B_2$). Note that in this case all the edges in $G$ incident to $u$ are covered by $B_1$. Hence, we can
safely delete $u$ from all bags except $B_1$. Next, we create a new bag $B^{*} = \{a,u,w,v\}$ and attach it pendant
to $B_1$, and finally we replace $u$ with $x$ in $B_1$. It can be easily verified that this is a valid tree decomposition for $G$ and that the width is not increased, so it is still a minimum-width tree decomposition. However, $B_1$ is now a candidate for Case 2, and we are done. Note that replacing $u$ with $x$ in $B_1$ is only possible because $B_1$ is next to $B_2$ in $P$.

\emph{Subcase 3.2:} Suppose Subcase 3.1 does not hold. Then $a \not \in B_1$ (and, symmetrically, $d \not \in B_2$). Putting
all earlier insights together, we see $a, b, x, d \not \in B_1$ and $a,u,c,d \not \in B_2$. Observe that $a$, which is not in $B_2$, is not in any bag to the right of $B_2$. If it was, then the fact that some bag contains the edge $\{a,u\}$, and the running intersection property, entails that $B_2$ would contain at least one of $a$ and $u$, neither of which is permitted. Hence, if $a$ appears in bags other than $B_1$, they are all in the left part of the decomposition. Completely symmetrically, if $d$ is in bags other than $B_2$, they are all in the right part of the decomposition. Because of this, $b$ can only appear on the left of the decomposition (because the edge $\{a,b\}$ has to be covered) and $c$ can only be on the right of the decomposition (because of the edge $\{c,d\}$). Summarising, $B_1$ (respectively, $B_2$) does not contain $a$ or $b$ (respectively, $c$ or $d$) and all bags containing $a$ or $b$ (respectively, $c$ or $d$) are in the left (respectively, right) part of the decomposition. Note that $c \not \in B_1$. This is because edge $\{c,d\}$ has to be in some bag, and this must necessarily be to the right of $B_2$: but then running intersection puts at least one of $c,d$ in $B_2$, contradiction. Symmetrically, $b \not \in B_2$. So $a,b,c,d,x \not \in B_1$ and $a,b,c,d,u \not \in B_2$.

We now describe a construction that we will use extensively:  \emph{reeling in (the snakes) $a$ and $b$}. Observe that, due to coverage of the edge $\{a,u\}$, and running intersection, there is a simple path of bags $p_{ua}$ starting
at $B_1$ that all contain $u$ such that the endpoint of the path also contains $a$. The path will necessarily be entirely on the left of the decomposition. Due to coverage of the edge $\{b,v\}$ there is an analogously-defined simple path $p_{vb}$. (Note that $p_{ua}$ and $p_{vb}$
both exit $B_1$ via the same bag $B'$. If they exited via different bags, coverage of the
edge $\{a,b\}$ would force at least one of $a,b$ to be in $B_1$, yielding a contradiction). Now, in the bags along $p_{ua}$, except $B_1$, we relabel $u$ to be $a$, and in the bags
along $p_{vb}$, except $B_1$, we relabel $v$ to be $b$. This is no longer necessarily a valid
tree decomposition, because coverage of the edges $\{u,a\}$ and $\{v,b\}$ is no longer
guaranteed, but we shall address this in due course. Next we delete the vertices
$u,w,v$ from all bags on the left of the decomposition, except $B_1$; they will not be needed. (The only reason that $w$ would be in a bag on the left, would be to meet $c$, since $B_1$ and $B_2$ already cover the edges $\{u,w\}$ and $\{w,x\}$. But then, due to coverage of the edge $\{c,d\}$ and the fact that $d$ only appears on the right of the decomposition, running intersection would put at least one of $c,d$ in $B_1$, contradiction.) Observe that $B'$ contains $\{a,b\}$. We replace $B_1$ with 5 copies of itself, and place these bags in a path such that the leftmost copy is adjacent to $B'$, the rightmost copy is adjacent to $B_2$, and all other bags that were originally adjacent to $B_1$ can (arbitrarily) be made adjacent to the leftmost copy. In the 5 copied bags we replace $\{u,w,v\}$ respectively with: $\{a,u',b\}$, $\{u',b,v'\}$, $\{u',u,v'\}$, $\{v',u,v\}$ and $\{u,w,v\}$. It can be verified that this
is a valid tree decomposition for $G'$, and our construction did not inflate the treewidth - we either deleted vertices from bags or relabelled vertices that were already in bags - so we are done. The operation can easily be telescoped, if desired, to achieve
an arbitrarily long ladder.

\smallskip

\noindent
\textbf{Case 4. $P$ contains at least one bag other than $B_1$ and $B_2$.}
\begin{observation}
\label{obs:obs_v_w_everywhere}
All bags in $P$ contain $v,w$, by the running intersection property.
\end{observation}

We partition the bags of the decomposition into (i) $B_1$, (ii) bags \emph{left} of $B_1$, (iii) $B_2$, (iv) bags \emph{right} of $B_2$, (v) all other bags (which we call the \emph{interior}).

\begin{hidden}
\begin{itemize}
\item $B_1$,
\item bags \emph{left} of $B_1$,
\item $B_2$,
\item bags \emph{right} of $B_2$,
\item all other bags. These are exactly the inner bags of the path $P$ (i.e. all bags on $P$ excluding its endpoints $B_1$ and $B_2$), plus all bags that can be reached from an inner bag of $P$ without passing through $B_1$ or $B_2$. For this reason we call this group the \emph{interior}.
\end{itemize}
\end{hidden}

Recall that $b,d,x \not \in B_1$, $a,c,u \not \in B_2$ (because Case 2 does not hold).
\begin{observation}
\label{obs:no_u_no_x_in_interior_or_wrong_side}
No bag in the interior contains $u$ or $x$. $B_1$ does not contain $x$, and no bag on the left contains $x$. Symmetrically, $B_2$ does not contain $u$, and no bag on the right contains $u$.
\end{observation}
\begin{observation}
\label{obs:a_and_d_magnet}
At least one of the following is true: $a \in B_1$,  $a$ is in a bag on the left. Symmetrically, at least one of the following is true: $d \in B_2$, $d$ is in a bag on the right.
\end{observation}
Now, suppose $w$ is somewhere on the left. We will show that then either $w$ can be deleted from the bags on the
left, or Case 2 holds. A symmetrical analysis will hold if $v$ is somewhere on the right. Specifically, the only possible reason for $w$ to be on the left would be to cover the edge $\{w,c\}$ -- all other edges
incident to $w$ are already covered by $B_1$ and $B_2$. If no bags on the left contain $c$, we
can simply delete $w$ from all bags on the left. On the other hand, if some bag on the left
contains $c$, then $c \in B_1$, because: $d \not \in B_1$, the need to cover the edge $\{c,d\}$, the presence of $d$ on the other `side' of the decomposition (Observation \ref{obs:a_and_d_magnet}), and running intersection. So we have that $c,u,w,v \in B_1$. This bag already covers all edges
incident to $w$, except possibly the edge $\{w,x\}$. To address this, we replace $w$ everywhere
in the tree decomposition with $x$ - this is a legal tree decomposition because some bag contains $\{w,x\}$ - and then add a bag $B' = \{u,w,x,c\}$ pendant to $B$. This new bag serves to cover all edges incident to $w$.  But $B_1$ now contains $u,v,c,x$, so Case 2 applies, and we are done! Hence, we can assume that $w$ is nowhere on the left, and, symmetrically, that $v$ is nowhere on the right.
In fact, the above argument can, independently of $w$, be used to trigger Case 2 whenever $c \in B_1$ or $b \in B_2$.
So at this stage of the proof we know: $b,c,d,x \not \in B_1$ (and $c$ is not on the left) and $a,b,c,u \not \in B_2$ (and $b$ is not on the right).

\smallskip

\emph{Subcase 4.1:} Suppose $a \not \in B_1$. Then, $a$ must only be on the left. It cannot be in the interior (or on the right) because the edge $\{u,a\}$ must be covered, $a \not \in B_1$, $u \in B_1$, and $u$ is not in the interior (Observation \ref{obs:no_u_no_x_in_interior_or_wrong_side}).
Because $a$ is on the left, and
because some bag must contain the edge $\{a,b\}$, $b$ must also be on the left. In fact $b$ is only on the left. The presence of $b$ both on the left and in the interior (or on the right) would force $b$ into $B_1$ by running intersection, contradicting the fact that $b \not \in B_1$. So $a,b$ are only on the left. We are now in a situation similar to Subcase 3.2. We use the same \emph{reeling in $a$ and $b$} construction and we are done.

\smallskip

\emph{Subcase 4.2:} Suppose $a \in B_1$. Note that here $u$ has all its incident edges covered by $B_1$, so $u$ can be deleted from all other bags.
Recall that $b \not \in B_1$. Due to edge $\{b,v\}$ some bag must contain both $v$ and $b$. Suppose there is such a bag on the left. We attach a new bag $\{a,u,w,v\}$ pendant to $B_1$ and  delete $u$ from $B_1$. We put $x$ in $B_1$ and to ensure running intersection we replace $v$ with $x$ in all bags anywhere to the right of $B_1$. This is safe, because in the part of the decomposition right of $B_1$, $v$ only needs to meet $x$ (and not $b$, because $v$ meets $b$ on the left). Thus, $B_1$ now contains $\{a,v,w,x\}$ and Case 2 can be applied.

Hence, we conclude that $\{v,b\}$ is not in a bag on the left. Because of this $v$ can safely be deleted from all bags on the left. That is because any path $p_{vb}$ that starts at $B_1$ and finishes at a bag containing $b$ must go via the interior.
In fact, such a path must avoid $B_2$, and is thus entirely contained in the interior. It avoids $B_2$ because $a,b \not \in B_2$ and $\{v,b\}$ cannot be in a bag to the right: if it was, coverage of edge $\{a,b\}$, the fact that $a \in B_1$ and running intersection would mean that at least one of $a$ and $b$ is in $B_2$, yielding a contradiction.

The only case that remains is $a \in B_1$, $\{v,b\}$ is not in a bag on the left and thus $p_{vb}$ is in the interior.
By symmetry, we assume that $d \in B_2$, $\{w,c\}$ is not in a bag on the right and thus $p_{wc}$ is in the interior.
Consider any path $p_{ab}$ starting at $B_1$, defined in the now familiar way. Note that no bag on the left of $B_1$ can contain $b$. This is because $\{v,b\}$ is in a bag in the interior: hence if $b$ was also on the left, $b$ would then by running intersection be in $B_1$ and we would be in an earlier case. This means that $p_{ab}$ must go via the interior.
Suppose the following operation gives a valid tree decomposition: delete $u$ from $B_1$, attach a new bag $B^{*} = \{a,u,w,v\}$ pendant to $B_1$, and relabel all
occurrences of $a$ along the path $p_{ab}$ (except in $B_1$) with $b$. Then we are done, because we are back in Case 2. A symmetrical situation holds for the path $p_{dc}$.

\begin{figure}[t]
\centering
\includegraphics[scale=1]{figs/tricky_case}
\caption{%
Path $p_{ab}$ goes via the interior, but it cannot be relabelled to $b$ because it is used by other paths $p_{az}$ to some neighbour $z$ of $a$ that does \emph{not} lie on the ladder. %
}
\label{fig:trickycase}
\end{figure}

Assume therefore that this transformation does not give a valid tree decomposition. This is the most complicated case to deal with. It is depicted in Fig. \ref{fig:trickycase}. The issue here is that the path $p_{ab}$ (respectively, $p_{dc}$) necessarily goes via the interior, but cannot be relabelled with $b$ (respectively, $c$) because the path is also part of $p_{az}$ (respectively, $p_{dz}$) where
$z$ is some non-ladder vertex that is adjacent to $a$ (respectively $d$). We deal with this as follows. We argue that some bag in the decomposition \emph{must} contain $a,b,v$ (and possibly other vertices). Suppose this is not the case. By standard chordalization arguments, every chordalization adds at least one diagonal edge to every square of the ladder. If $\{a,b,v\}$ are not together in a bag, then this is because the corresponding chordalization did not add the diagonal $\{a,v\}$ to square $\{a,u,b,v\}$. Hence,
the chordalization must have added the diagonal $\{u,b\}$. This would in turn mean that some bag contains $\{a,u,b\}$. Such a bag must be on the interior, because this is the only place that $b$ can be found. However, no bags in the interior contain $u$ -- contradiction.

Hence, some bag $B'$ indeed contains $\{a,b,v\}$. Again, because $b$ is only on the interior, $B'$ must be in the interior. There could be multiple such bags, but this does not harm us. Let $B_{v\text{-done}}$ be the rightmost bag on the path $P$ that is part of a path, starting from $B_1$, from $a$ to some bag $B'$  containing $\{a,b,v\}$. Let $B_{a\text{-done}}$ be the rightmost bag on $P$ that contains $a$. Note that $B_{v\text{-done}}$ contains $a$ (because of
running intersection: $a \in B_1$ and $a \in B'$) and $v, w$ (because it lies on $P$). We also have $a,v,w \in B_{a\text{-done}}$. By construction, $B_{v\text{-done}}$ is either equal to $B_{a\text{-done}}$ or left of it. This is important because it means that the only reason $v$ might need to be in bags to the right of $B_{a\text{-done}}$ is to reach a bag containing $x$ (i.e. to cover the edge $\{v,x\}$) -- all other edges are already covered elsewhere in the decomposition; in particular, edge $\{v,b\}$ is covered by the bag $B'$ containing $\{a,b,v\}$. See Fig. \ref{fig:trickycase2} for clarification.
\begin{figure}[t]
\centering
\includegraphics[scale=1]{figs/tricky_case_2}
\caption{The bags $B'$, $B_{v\text{-done}}$ and $B_{a\text{-done}}$ illustrated. Note that $B_{a\text{-done}}$ cannot be the penultimate bag on the path $P$ from $B_1$ to $B_2$, due to the presence of $d$ in that bag.}
\label{fig:trickycase2}
\end{figure}
Recall that none of the bags on the path $P$ contain both $a$ and $d$. (If they did, there would be a bag containing $\{a,d,v,w\}$ and we would be in Case 2, done.) We also know that some path $p_{dc}$ goes via the interior and thus that the penultimate bag on $P$ (i.e. the one before $B_2$) thus definitely contains $d$. (To clarify: $c \not \in B_2$, $d \in B_2$, the edge $\{c,d\}$ must be covered, and $c$ is only in the
interior). Combining these insights tells us that this penultimate bag definitely does \emph{not} contain $a$, and hence $B_{a\text{-done}}$ is \emph{not} equal to the penultimate bag; it is further left. This fact is crucial. Consider $B_{a\text{-done}}$ and the bag immediately to its right on $P$. Between these two bags we insert a copy of $B_{a\text{-done}}$, call it $B_r$, remove $a$ from $B_r$ (i.e. forget it), and add the element $x$ to it instead. Finally, we switch
$v$ to $x$ in all bags on $P$ right of $B_r$, including $B_2$ itself, and delete $v$ from all bags in the tree decomposition that are anywhere to the right of $B_r$; there is no point having them there. It requires some careful checking but this is a valid (minimum-width) tree decomposition. Moreover, $B_r$ contains $w,v,x$. The fact that $B_{a\text{-done}}$ was not the penultimate bag of $P$, means that the length of the path from $B_1$ to $B_{r}$ is strictly less than the length of the path from $B_1$ to $B_2$: contradiction on the assumption that these were the closest bags containing $\{u,w,v\}$ and $\{w,v,x\}$ respectively. We are done.
\end{proof}

We now deal with the situation when the $tw(G) \geq 4$ assumption is removed.
\begin{hidden}
\begin{corollary}
\label{cor:mainAllTreewidth}
Let $G$ be an undirected graph of arbitrary treewidth. There is a universal constant $C'$ such that if there exists a
ladder $L$ of length length $C'$ or longer in $G$, the ladder can be increased arbitrarily in length without
altering (in particular: increasing) the treewidth.
\end{corollary}
\begin{proof}
If $L$ is disconnecting, a ladder of length 1 is already sufficient (Lemma \ref{lem:disconnecting}). So assume that $L$ is not disconnecting. From Observation \ref{obs:k4} we know $tw(G) \geq 3$. Now, recall the function $f$ from Theorem \ref{thm:unbounded}. We take $C' = \max(f(3), 3)$, where $f(3)$ indicates a starting length from which ladders can be extended in graphs of treewidth 3 without increasing the treewidth. The second term in the max operator is the constant 3 obtained from the statement of Theorem \ref{thm:main} and which holds for all graphs $G$ such that $tw(G) \geq 4$.
\end{proof}

The above corollary is somewhat unsatisfactory because, although the constant $f(3)$ does exist, it is much larger than the constant 3 from Theorem \ref{thm:main}. However, by carefully analysing the case when $tw(G) \leq 3$ we can produce a more satisfactory result that again works for all treewidths.
\end{hidden}
\begin{lemma}
\label{lem:main2}
If $G$ has a ladder $L$ of length 5 or longer, the ladder can be increased in length arbitrarily without
altering (in particular: increasing) the treewidth. This holds irrespective of the treewidth of $G$.
\end{lemma}
\begin{proof}
Let $L$ be a ladder of length 5 or longer.
We can assume that $L$ is not disconnecting and $tw(G) \leq 3$. We select the three most central squares and label these as in Fig. \ref{fig:ladder}. These are flanked on both sides by at least one other square. Hence, $a,b,c,d$ each has exactly one neighbour  outside the 3 squares, let us call these $a', b', c',d'$ respectively, where $\{a',b'\}$ is an edge and $\{c',d'\}$ is an edge. Now, $tw(G) = 3$ because $L$ is not disconnecting. The only part of the proof of Theorem \ref{thm:main} that does not work for $tw(G)=3$ is Case 1 and (indirectly) Case 2 because these create size-5 bags. We show that neither case can hold.

Consider Case 1.  Let $B$ be a bag containing one of the three most central squares $S$ of the ladder (these are the only squares to which Case 1 is ever applied).
A \emph{small} tree decomposition is one where no bag is a subset of another. If a tree decomposition is not small, then by running intersection it must contain two adjacent bags $B^{\dagger}, B^{\ddagger}$ such that $B^{\dagger} \subseteq B^{\ddagger}$. The two bags can then be safely merged into $B^{\ddagger}$. By repeating this a small tree decomposition can be obtained without raising the width of the original minimum-width decomposition. Furthermore, some bag $B$ will still exist containing $S$. If $B$ has five or more vertices we immediately have $tw(G) \geq 4$ and we are done. Otherwise, let $B'$ be any bag adjacent to $B$; such a bag must exist because $G$ has more than 4 vertices. Due to the smallness of the decomposition we have $B \not \subseteq B'$ and $B' \not \subseteq B$. Hence, $B \cap B' \subset B$ and $B \cap B' \subset B'$. A \emph{separator} is a subset of vertices whose deletion disconnects the graph. Now, $B \cap B'$ is by construction, and the definition of tree decompositions a separator of $G$. However, due to our use of the three central squares, $S$ is not a separator, and no subset of it is a separator either;
the inclusion of $a', b', c', d'$ and the edges $\{a',b'\}$ and $\{c',d\}$, alongside the fact that $L$ is not disconnecting, ensure this. This yields a contradiction. Hence Case 1 implies $tw(G) \geq 4$ i.e. it cannot happen when $tw(G)=3$.

We are left with Case 2. This case replaces the three centremost squares with a single square, and deletes any diagonals that this single
square might have, to obtain a new graph $G'$. We have $tw(G') \leq tw(G)$, by minors. Note that $tw(G') \geq 3$ because the shorter ladder in $G'$ (which has length at least 3) is still disconnecting. Hence, $tw(G') = tw(G) = 3$. The decomposition $\mathbb{T}'$ of $G'$ obtained by projecting the contraction operations onto the tree decomposition $\mathbb{T}$ of $G$, is a valid tree decomposition (as argued in Case 2) with
the property that the width of $\mathbb{T}'$ is less than or equal to the width of $\mathbb{T}$. $\mathbb{T}'$ cannot have width less than 3, so it must
have width 3. Hence it is a tree decomposition of $G'$ in which all bags have at most four vertices. We then transform $\mathbb{T}'$ into a small tree composition: this does not raise the width of the decomposition, and every bag prior to the transformation either survives or is absorbed into another. Consider
the bag $B'$ containing $H_1, H_2, L_1, L_2$. The presence of $a', b', c,', d'$ in $G'$ and the fact that the ladder in $G'$ is not disconnecting, means that $H_1, H_2, L_1, L_2$ is not a separator for $G'$, and neither is any subset of those four vertices. But the intersection of $B'$ with any neighbouring bag \emph{must} be a separator. Hence $B'$ must contain a fifth vertex, contradiction. So Case 2 cannot happen when $tw(G)=3$.
\end{proof}

We can, however, still do better. The following proof is in the appendix.

\begin{theorem}
\label{thm:main3}
If $G$ has a ladder $L$ of length 4 or longer, the ladder can be increased in length arbitrarily without
altering (in particular: increasing) the treewidth. This holds irrespective of the treewidth of $G$.
\end{theorem}
\begin{hidden}
\begin{proof}
As usual it is sufficient to consider the case $tw(G)=3$. The high-level idea is similar to the proof of Lemma \ref{lem:main2}; the extra complexity comes from the fact that we can only add a flanking square on one side of the ladder, not both. However, we can argue that the end of the ladder without the extra square, either (1) has a sufficient `connectivity' to simulate the effect of an extra flanking square, or (2) after decomposition into biconnected components and decomposition around
a certain clique separator, either $c$ or $d$ has degree-2, in which
case Lemma \ref{lem:pointed} applies. We defer the full details to the appendix.
\end{proof}
\end{hidden}
\begin{hidden}
\textbf{Theorem \ref{thm:main3}, restated.}
\emph{If $G$ has a ladder $L$ of length 5 or longer, the ladder can be reduced to length 4
without altering (in particular: lowering) the treewidth. This holds irrespective of the treewidth of $G$.}\\
\end{hidden}
\begin{hidden}
Finally, for graphs of slightly higher treewidth, we observe
that a slightly more aggressive shortening of the ladder is
possible.

\begin{observation}
\label{cor:hightw}
Let $G$ be a graph with treewidth 5 or higher. If $G$ has a ladder $L$ of length 4 or longer, the ladder can be reduced to length 3
without altering (in particular: lowering) the treewidth.
\end{observation}
\end{hidden}
\begin{hidden}
\section{Tightness}
\label{sec:tightness}
\end{hidden}
\noindent
\textbf{Tightness.} The constant 4 in the statement of Theorem \ref{thm:main3} is equal to the constant obtained for the `bottleneck' case $tw(G)=3$. An improved constant 3 for this case is not possible, as Fig. \ref{fig:prism} shows. It is natural to ask whether, when $tw(G)\geq 4$, we can start from ladders of length 2, rather than 3. This is
also not possible. See Fig. \ref{fig:tightstuff}.
In fact, we have examples up to treewidth 20. These
can be found at \textbf{\url{https://github.com/skelk2001/snakes\_and\_ladders}}.
We conjecture that this holds for all treewidths, but defer this to future work.\\
\\
\textbf{Implications for phylogenetics.} A phylogenetic tree is a binary tree whose leaves are bijectively labelled by a set $X$.  The \emph{display graph}
of $T_1, T_2$ on $X$ is obtained by identifying leaves with the same label. In \cite{kelk2017treewidth} it was asked whether \emph{common chains}
could be truncated to constant length without lowering the treewidth of the display graph. Theorem \ref{thm:main3} establishes that the answer
is \emph{yes}. In fact, due to the restricted structure of display graphs, we can prove a stronger result: truncation to
$4$ leaves (i.e. $3$ squares) is safe, and this is best possible. Theorem \ref{thm:preserve2} summarizes this. We provide full details in the appendix.

\begin{theorem}
\label{thm:preserve2}
Let $T_1, T_2$ be two unrooted binary phylogenetic trees on the same set of taxa $X$, where $|X| \geq 4$ and $T_1 \neq T_2$. Then exhaustive application of the subtree reduction and the common chain reduction (where common chains are reduced to 4 leaf labels) does not alter the treewidth of the display graph. This is best possible, because
there exist tree pairs where truncation of common chains to length 3 does reduce the treewidth of the display graph (see Fig \ref{fig:chains}).
\end{theorem}
\textbf{Acknowledgements.} We thank Hans Bodlaender and Bart Jansen  for insightful feedback. We also thank the members of our department for useful discussions.
\begin{figure}
\centering
\includegraphics[scale=0.8,page=1]{figs/square_prism.pdf}
\caption{A graph of treewidth 3 that contains a ladder with 3 squares, shown in red. Increasing the length of the ladder by 1 square increases the treewidth to 4.
}
\label{fig:prism}
\end{figure}

\begin{figure}
\centering
\hfill
\includegraphics[scale=0.8,page=1]{figs/tight_stuff.pdf}
\hfill
\includegraphics[scale=0.8,page=2]{figs/tight_stuff.pdf}
\hfill \
\caption{These graphs have treewidth 4 (left) and 5 (right) and each one has a length 2 ladder, induced by $\{1,2,3,4,5,6\}$.
In each case increasing the length of the ladder to length 3 increases the treewidth of the graph by one.}
\label{fig:tightstuff}
\end{figure}

\begin{figure}
\centering
\includegraphics[scale=0.8]{figs/chains3to4.pdf}
\caption{Lengthening the common chain $\{a,b,c\}$ to $\{a,b,c,d\}$ in these phylogenetic trees causes the treewidth of the display graph to increase. Equivalently: shortening common chains to 3 leaf labels is not guaranteed to preserve treewidth in the display graph.}
\label{fig:chains}
\end{figure}

\clearpage

\noindent
Ruben Meuwese was supported by the Dutch Research Council (NWO) KLEIN 1 grant \emph{Deep kernelization for phylogenetic discordance} (OCENW.KLEIN.305).

{\small
\bibliographystyle{splncs04}

redpajama_set_name: RedPajamaArXiv


text:
Cladosporium brachormium är en svampart som beskrevs av Berk. & Broome 1850. Cladosporium brachormium ingår i släktet Cladosporium och familjen Davidiellaceae.   Inga underarter finns listade i Catalogue of Life.

Källor

Sporsäcksvampar
brachormium

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text:
BioM: Tyrrell, Maureen Audrey (1957)
Surnames: Tyrell, Hubing, Newman, Smith, Omholt, Jones, Mandry, Wittenberg, Geis, Babka, Nelson, Fehrman, Jacob, Galstad, Hantke, Clark, Harris, Burns, Davis, Waterman, Vine, Okray, Kusiak, Lowe, Cater, Root, Looch, Hawkins, Spear
----Source: Clark County Press (Neillsville, Clark Co., WI.) September 5, 1957
Tyrrell, Maureen Audrey (Marriage - 24 August 1957)
The First Methodist Church of Wisconsin Rapids was the scene August 24, of the marriage of Miss Maureen Audrey Tyrrell and Bradley Hubing. The double ring ceremony was performed in the presence of 250 guests by the Rev. David A Spear. Parents of the bride are Mr. and Mrs. Harvey C. Tyrrell of route 1, Wisconsin Rapids. The groom's parents are Mr. and Mrs. Arthur Hubing of route 4, Neillsville.
The bride was attired in a white ballerina length gown of hand clipped lace fashioned with a billowing lace skirt rising to the trim lines of the bodice. The Sabrina neckline was finished with glittering sequin trim. Her imported illusion veil was caught to a tiny pearl embroidered contour crown, and her flowers were an arrangement of pink roses and feathered carnations.
Miss Helene Tyrrell, sister of the bride, was the bridesmaid, and Miss Nancy Newman, a cousin, was junior bridesmaid. Bridesmaids were Miss Artice Hubing, sister of the groom, Miss Lea Smith, Miss Barbara Omholt and Miss Marion Jones.
Wendell Hubing of Granton served his brother as bestman. Ushers were Paul Mandry and Charles Wittenberg, both of Wisconsin Rapids, James Geis of Neillsville, and Larry Babka of Sparta. Ringbearer was Kenneth Nelson, cousin of the bride. Little Miss Kathy Hubing, niece of the groom, was flower girl and carried a basket of pink and white pompons.
Miss Joyce Fehrman furnished organ music and accompanied the trio of Mrs. Donald Smith, and the Misses Lea and Nancy Smith, who sang, "Because" and "Whither Thou Goest."
The bride is a graduate of Lincoln High School in Wisconsin Rapids, and has attended Wisconsin State College in Eau Claire one year. She will attend Stout State College in Menomonie this year, where the groom is a senior. He is a graduate of Neillsville High School.
Guests included: Mr. and Mrs. Arthur Hubing, Mr. and Mrs. George Hubing, Mr. and Mrs. Wilfred Galstad, Mr. and Mrs. Charles Hubing and daughter, Luann, Mr. and Mrs. Robert Jacob and daughter, Charleen, and Mrs. Ruben Hantke and daughter, Pauline, of Neillsville; Mr. and Mrs. Arthur T. Clark, Grandparents of the bride, Mr. and Mrs. Douglas Harris and family, Mr. and Mrs. Elmer Burns and daughter, Judy of Mosinee; Mr. and Mrs. Truman Davis, Grandparents of the groom, Mr. and Mrs. Clark Waterman, Mr. and Mrs. Ernest Vine, Mr. and Mrs. Wendell Hubing and children of Granton; Mr. and Mrs. Ralph Okray and family of Madison; Mrs. Sigmund Kusiak of Dancy; Mr. and Mrs. William Lowe of Ripon; Mrs. Warren Cater and Mrs. Frank Root of Knowlton; Mr. and Mrs. John Looch and Mr. and Mrs. Garth Harris of Milwaukee; Mr. and Mrs. Richard Nelson of Great Lakes, Ill.; and Mrs. Warren Hawkins and Mr. and Mrs. Archie Clark and family of Chicago, Ill.


Speed drills don't work for everybody. Take it from someone who flunked every single Friday speed drill in fourth grade math…until my dear mother worked with me one-on-one to help me master the math facts. After that, I took off and aced every single drill. But the drilling didn't cement the facts in my mind…it only proved that the facts were already cemented.
But if your student is having trouble getting up to speed (ahem), and you're thinking you might need some fancy software or workbooks…maybe all you really need it the the right deck of cards.
Pyramath is an amazing game from I See Cards. It can be played alone or head-to-head (if your kiddos like a bit of competition---there's nothing quite like beating dear old Mom). So what's so special about this deck of cards? Why can't we just play "Math War" with a regular deck of 52?
These cards are a little special. In addition to having the proper distribution of the numbers 0-9 to play the game (and that is important in this game), they are also labeled with Roman numerals and the names of the numbers in English, Spanish, Chinese, Arabic, and French. Do my kiddos even notice that? Nope, totally lost on them. I could bring it to their attention, but as the point is to increase our calculating speed, it's not really a priority.
What really sets this game apart is the game itself. The game is fabulous. And highly addictive (try the free online version of Pyramath to see for yourself, you might even win your own deck of cards---you can read the rules here).
While most math card games seem to only concentrate on one operation or another at a time, in Pyramath you can use addition, subtraction, multiplication and even division to try to build your pyramid before your opponent. This is great, too, if you are working with kiddos of varying abilities---I could have my 1st grader play against my 5th grader, but using different operations.
Most math card games don't require much in the way of thinking or developing a strategy. In Math War (addition), for example, yeah, you can practice adding up the numbers, but who wins is really up to chance. There's no strategy whatsoever. In Pyramath, after playing a few games, your kiddos will very quickly figure out that it's possible to not be able to finish your pyramid. In fact, it's likely at least one of you won't finish if you are playing the 7-card version head-to-head…unless you play your cards right.
Engagement. Not mindless regurgitation. Love it.
It's fun to see the kiddos really giving their brains a workout. Even if it is to beat the pants off of me.
And the best part…this is not some $25 game with a million pieces to lose. Only $6.95!
Be sure to check out their other games, like Fractazmic and Prime Bomb.
You can read more reviews on this product at the TOS Homeschool Crew blog.
Disclosure: As a member of the TOS Homeschool Crew, I received a free deck of Pyramath cards for review purposes. I received no compensation. This review reflects my personal opinions and experiences.


text:
What to do when your spouse is unfaithful
Televangelist Pat Robertson recently offered advice to a woman struggling to regain trust in a cheating husband. His advice? Stop dwelling on his indiscretion; remind yourself why you love him; consider whether he is handsome and a good provider. His final comment - men do stray and the way to stop them is by creating a home environment that makes them want to stay put.
Regardless of what you think of Robertson's advice, his comments beg a larger question. When is enough, enough? Yes, most everyone experiences lapses in judgment or unintentionally disappoints their mate; and yes there is power in forgiveness. But there is also a saying that goes, don't allow yourself to be led away from the truth by what you would like to believe. So what do you do when your spouse is unfaithful?
When your spouse is unfaithful it isn't only what he says after disclosure that matters, what he does matters, too. In other words, what you see him do must be considered against what he says. Excuses cannot take the place of action. What kind of action? If the goal is to restore trust and rebuild the relationship, you should be able to see the work. Words are comforting only to a point.
Whether you ultimately decide to stay in the relationship or end it, tell yourself the truth about why you made your decision. What does your strongest, wisest self have to say about your choice?
Find someone you can talk with honestly about your feelings. Pretending one thing and feeling another is emotionally exhausting. You already have enough going on, you deserve to not play the fake game.
Repairing broken trust is often a long and difficult process. If you choose this route be clear about the way forward. How will you know you are making progress? Set up mile markers – for example, let's decide next steps after six months of counseling. And while we are in counseling we will also do x each week.
Be Wary of Why
It is inevitable that you will wonder why your spouse was unfaithful and how your relationship came to be in this difficult place. The questions are important but they do not warrant endless consideration. Give yourself permission to move to the next question – Now what? What do you want? What is driving your decisions? What is your motivation for going forward with whatever you decide? What will you take away from this experience?
When your spouse is unfaithful it is easy to beat yourself up and to feel ashamed. Don't. The only real shame is in thinking you will stay stuck here. You won't. With courage and candor you can choose the way forward.
Content copyright © 2021 by Barbara Gibson. All rights reserved.
This content was written by Barbara Gibson. If you wish to use this content in any manner, you need written permission. Contact Barbara Gibson for details.
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Whether you're Count Dracula himself, a Mummy or a little devil, if you're planning to have a scream this Halloween, spare a thought for those who might be dreading the night.
During the run up to and on Halloween Torbay Council's Community Safety Team and Devon and Cornwall Police will be reminding local shopkeepers of their right to refuse to sell eggs and flour to any person who they believe isn't planning on making a tasty cake, but instead intends to use the items for anti-social purposes.
Community Safety Sergeant Jeff Coe from Devon and Cornwall Police said "It's not our intention to spoil well spirited spooky fun this Halloween, but those choosing to avoid the celebrations should have their wishes respected. Anti-Social Behaviour and other disorder will not be tolerated in Torbay.
"On Halloween we ask all parents and guardians to check the whereabouts of their little monsters and to ensure that they are accompanied by a responsible adult when 'Trick or Treating'. Whilst there are lots of residents who are happy to take part in providing children with tasty treats, some residents including older people and those who may be more vulnerable are entitled not to be called upon. Please remind children that 'Trick or Treating' is a game, not a right, and if someone doesn't want to play that must be respected".
No Trick, No Treat, No Thanks posters can be downloaded from www.safercommunitiestorbay.org.uk/Halloween. A limited number of posters for those who do not have access to the internet or a printer will also be made available in Torbay libraries.


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You have requested the file: Name: 08 Beethoven - Fur Elise (Mauriat).mp3. Size: MB Uploaded: Last download: User image. Fur Elise Dubstep. By: parr Classical piano MIDI and MP3 sequences by Beethoven for free download. For Elise (). Part, Tempo, Duration, Date, Size, MIDI Format 0, MP3, OGG. Read reviews, compare customer ratings, see screenshots, and learn more about Piano with Songs. Download Piano with Songs and enjoy it on your iPhone. Hidden Citizens is a commercial music team that combines a wide array of genres and stylings to forge impactful, high-end music for media and advertising.


text:
Q: why fiona gives error while importing geopandas ? I am trying to import geopandas.
I have installed Fiona, gdal separately, and added path to osgeo before executing pip install geopandas.
However, when I try to import geopandas on my jupyternotebook, it gives me error saying below.
ImportError: cannot import name 'urlparse' from 'fiona.compat' (C:\Users\me\Anaconda3\lib\site-packages\fiona\compat.py)
Im using python3.7 with jupyternotebook anaconda3, and windows10.
Does anyone know why this error happens?

redpajama_set_name: RedPajamaStackExchange


text:
HATE CRIME: Cape Town Beautician Attacked for being Gay
July 11, 2016 Kuchu Times Editor
A young Western Cape man is still grappling in pain after being brutally targeted by two youths because of his sexuality.
Geno Jonkers, a beauty spa consultant from Crawford, said he fears for his life after being attacked last month at his uncle's home in Cape Town's Eerste River. The 20-year-old was visiting his family when he heard voices calling him outside. He walked out the house to the gate when he was grabbed from behind.
"He was choking me. I couldn't breath or call out for help. Then another one came around the corner and they dragged me across the road, behind a house, and they beat me there," he said.
"While attacking me the one said, 'Maak die moffie vrek' (Kill the faggot). They were banging my head against a concrete fence and the other one said, 'Ja, maak die ding dood, maak die moffie vrek' (Yes, kill the thing, kill the faggot)."
Jonkers believed that he was going to die: "My only thought while they were beating me was, 'What about my family?' I was worried about my mom."
He was eventually able to escape from the attackers and ran back inside his uncle's house. "They must have gotten tired of beating me and I just ran."The assailants continued to threaten their victim from outside the home for some time before they finally left.
Jonkers said he knew his attackers, who are believed to be in their teens. Earlier in the day he had interacted with them at a shop when they asked him for money.
Police spokesman Captain FC van Wyk told the Daily Voice that a 15-year-old was arrested and appeared in the Blue Downs Magistrates' Court on a charge of assault on June 24, but Jonkers claims that he was never informed and that the attacker is back home.
Jonkers before the attack
"I'm worried they will come after me again because the police haven't done anything about it. And today it is three weeks after the incident happened and the police still hasn't gotten back to me since I reported the case."
Jonkers explained that this was the first time he had been the victim of overt homophobia and that the experience has scarred him.
"I'm more scared because I don't know what will happen. It has changed me emotionally because I'm not the person I used to be. I used to be an outgoing person but I don't go out any longer. I'm just in the house."
When asked if he regrets reporting the case to the police, Jonkers responded: "A month ago one of my friends was murdered and justice hasn't been served in his case. At first I didn't want to report to the police but I thought to myself, 'It could happen to someone else tomorrow, so I have to take a stand'."
He added: "Something has to be done about it."
Triangle Project's Matthew Clayton said the organisation, which is one of the Love Not Hate campaign partners, has provided counselling to Jonkers.
Triangle has also forwarded details of the case to the Department of Justice's hate crime Rapid Response Unit and is waiting for a reply.
Sourced from Mamba Online
current south africa
Born this day


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Trubisky, at this early point his NFL career, has been critiqued as being more accurate on the move and or outside the pocket. also was limited the team's offseason program due to a hamstring issue after signing with the team May. 14 with musical act Seth . It's hard when you call into Chicago because it's loud there with people around him. Mondays through Fridays.
Their defense has allowed average of close to eight fantasy points per game to tight ends, and Olsen want to stick it to his former team. He was second the NHL among rookie defensemen goals and shots . Tuesday night San lost at home to Detroit, which was a big blow to the Stars if they want to grab the No.


While YouTube became popular by hosting unauthorized content, it's now a platform for self-expression and many companies use tools like Content ID to make money from the videos that include their content. Like many other online services, YouTube is protected by the DMCA, a United States copyright law which "creates a safe harbor for online service providers against copyright liability if they adhere to and qualify for certain prescribed safe harbor guidelines and promptly block access to allegedly infringing material (or remove such material from their systems) if they receive a notification claiming infringement from a copyright holder or the copyright holder's agent."
Even if YouTube's employees were aware that a lot of the videos weren't uploaded by the copyright owners, it was difficult to tell which videos should be removed.
This case shows that it's a bad idea to fight against those that love your work and want to promote it. Encouraging fans to be creative, learning from their ideas, finding which of your works is more popular and making money from ads - there's a lot to gain from being open-minded.
We are with you Google on this fight vs Viacom. It is about time there is a company with power to finally fight for the right of the internet.
I am scared where the modern net is going.
This way any internet service could be brought down and or sued very conveniently, just hire someone to upload illegal things on it.
"As YouTube's council recently pointed out, more or less any video is copyrighted. If you make a video, it's copyright to you upon time of creation. Few people actually bother to release their stuff in to the public domain, so the works remain copyrighted. Now that does NOT mean they can't be posted on Youtube. The holder of a copyright can determine how it is allowed to be used, including given away for free to anyone.
Now, as that applies to big media companies the problem is that they themselves, or their agents, like PR agencies, do indeed upload content to Youtube. So just because a work is uploaded that is owned by a big company, it doesn't mean there has been infringement. Perhaps the company themselves did the uploading. They don't always do it through some official account.
Wow, that's something new. I didn't even suspect Viacom to upload their own material to Youtube and then come up with accusations about piracy against Youtube and its users. This is sick. Are all media-companies that sick?
Viacom are f'ing retards! And so is the rest of those greedy bastards!
"you so its all about the money they don't care whats gonna happen to the net they just want money"
Sooner or later, bastards like Viacom are gonna get theirs and they will burn in hellfire for all eternity, guarantee!
I'm with you google. So f you viacom for f'ing ruining YouTube. You f'king took down my YouTube account which had nothing to do with you. First ebaumsworld.com next viacom. F&$@ you Viacom!
More man hours have been wasted watching youtube videos than have been wasted playing WoW. It is a plague like no other.
I honestly wish we would stop giving lazy fat a$$es more excuses to stay inside and continue being lazy fat a$$es.


text:
Stan Anderson 1933-2018
John Coyle
Stan Anderson, who managed Doncaster Rovers between February 1975 and November 1978 died on June 10 2018 at the age of 85.
Born in Horden, County Durham, Stan played for Sunderland from 1952 to 1963 making over 400 appearances for the Wearside club. Subsequently he moved on to Newcastle United and then Middlesbrough, where he was player-coach before becoming manager in 1966. A tough-tackling midfielder, Stan enjoyed the unique distinction of captaining all of the "Big Three" north-east clubs. While with Sunderland he was capped twice by England and he was also a member of the England squad at the 1962 World Cup finals in Chile.
Stan was in charge of Middlesbrough from April 1966 to April 1973, when he was replaced by Jack Charlton and then managed AEK Athens and (briefly) Queen's Park Rangers. In February 1975 he was appointed Rovers' manager, succeeding Maurice Setters. He arrived with Rovers second from bottom in Division Four: by the season's end the club was in a respectable 17th place after a run of nine wins in 16 games. The highlight was a 4-3 home win over Division Four leaders Mansfield Town, watched by a crowd of over 7,000. Just over a month before only 1,500 had watched a 1-1 home draw with Reading.
Anderson's reputation was enhanced in 1975-76 when his team reached the quarter-finals of the League Cup before losing at Tottenham. Anderson had sold winger Terry Curran to Nottingham Forest but had acquired goalkeeper Dennis Peacock and another wide man, Ian Miller, in part-exchange. He also snapped up veteran midfielder Chris Balderstone, but many of his players were those who had appeared in Maurice Setters' struggling sides. Despite playing some exciting football Rovers' inconsistency saw them finish 10th in a very strong Fourth Division.
For 1976-77, Anderson added another experienced player in the former Carlisle midfielder Joe Laidlaw. The manager's aim was to bring on some home grown talent, but the progress of the likes of Dave Wignall, Mark Jones, Ian McConville and David Binch was slow, to say the least. Anderson's key man was striker Peter Kitchen, who scored over 20 League goals in three successive seasons, but at the end of the season he joined Second Division Leyton Orient. Again, it had been a season of excitement but under-achievement: despite Kitchen's 23 goals and a 10 game unbeaten run between January and March 1977, Rovers could finish only 8th.
With the youngsters showing no signs of coming through, Anderson went for experience at the start of the 1977-78 season. After a poor start to the season Rovers enjoyed a revival up to Christmas, but thereafter they had a run of only two wins from 18 games. The poor form continued into the 1978-79 season, with only four victories from the first 18 games. After a home defeat by York City in November 1979, Anderson decided enough was enough. He resigned and took up an offer to work under Ian Greaves at Bolton Wanderers. When Greaves departed in February 1980, Anderson replaced him as Bolton manager but his own reign lasted only just over a year. Later he worked as a scout for Newcastle United.
Stan Anderson continued to live in the Doncaster area and was a frequent attendee at supporters' meetings and reunion days. He was regarded throughout the game as a gentleman and those Rovers fans who were fortunate enough to meet him remembered those encounters with great affection. The thoughts of everyone at Doncaster Rovers FC are with Stan's family.
STAN ANDERSON'S RECORD.
First match: 8th February 1975 v Torquay United (home) won 3-0.
Last match: 18th November 1978 v York City (home) lost 1-2.
Football League Record.
Played 172, Won 67, Drew 46, Lost 59, Winning percentage: 38.9%.
Overall Record.
Stan Anderson


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