There are times you want to prop your iPad up for hands-free use, and a good stand lets you do just that. While there are plenty of portable iPad stands out there, stationary (or desktop) stands tend to be a bit sturdier and more stable. (See all our iPad-stand reviews.) I took a look at eight stands designed for the desktop or floor to see how they compare.
Before you buy a stand, it's important to think about how and where exactly you plan to use your iPad. Are you looking for a solution for a business setting? Will you instead be using your iPad primarily in the kitchen, so you can follow a recipe or watch a video while you cook? Or are you looking for a couch companion, to relax with an e-book or movie? Each of these stands I tested will suit different needs, so keep your needs in mind as you read along.
The Arkon Tab-Stand (3 of 5 rating) comes in two sizes; the $90 Tab-Stand1 (also known as the "4" Extending iPad Floor Stand with Quick Release Holder"), designed to be set on the ground like a podium stand, and the $40 Tab-Stand2 (also known as the "10" iPad & iPad Mini Table Stand with Quick Release Holder"), intended for tabletop use. The design of both versions is the same—which size you choose is just a matter of how you plan to use your iPad.
The Arkon stands are made of plastic and metal, and they accommodate any size iPad, including the mini. (It also fits other brands of tablets.) However, the stands hold a full-sized iPad only in portrait mode. That said, the stand's cradle is quite adaptable, letting you keep your iPad in a non-bulky case; you can adjust the height of the cradle and iPad angle (from roughly vertical to 45 degrees from vertical). The stands' cradle doesn't block any ports or buttons.
The Arkon models are each the sort of stand that comes to mind when you think "iPad Stand." They're not exciting, but they're functional and versatile.
Some iPad stands have one very specific use case, and Heckler Design's $129 WindFall for iPad (4 of 5 rating) is one of those. It's designed for business settings, to let you use your iPad as a point-of-sale stand or a display for customers. You use the included hex wrench to lock your iPad securely in the stand, with an opening to allow the use of a credit-card dongle. The WindFall accommodates the second-, third-, and fourth-generation iPads, though only bare or with a thin decal/skin.
Like Heckler Design's @Rest iPad stand, the WindFall is constructed from a single, heavy piece of metal and is quite sturdy. The WindFall holds your iPad only in landscape orientation, and at just one angle (roughly 45 degrees from vertical). Your iPad's top and bottom buttons and ports are accessible, while the tablet's side switch and volume buttons are blocked. You can use the stand for showing videos, typing and tapping on the screen, or with an external keyboard, but whatever you use it for, your iPad isn't going anywhere—once it's in the WindFall, it's in—you'll need that hex wrench to remove it. Heckler Design sells various accessories you can use to lock the stand itself in place to prevent theft.
This isn't a stand for casual or home use. But if you're looking to use your iPad for customer interaction at your place of business, the WindFall might be the answer.
Mophie's $150 Powerstand (4.5 of 5 rating) is constructed primarily of aluminum, but with soft-rubber pads wherever the stand comes in contact with your iPad. You can easily switch between landscape and portrait orientation, and the stand adjusts to any angle from horizontal to completely upright. In addition, you can adjust the height of the Powerstand within a range of several inches to get the perfect viewing position.
The Powerstand is quite sturdy and stable—even on a bed or couch it's difficult to tip over. You can use it with an external keyboard, but its hinge is also sturdy enough for on-screen typing. The variety of angles lets you get the right angle for watching videos, surfing the Web, reading a book, or video chatting. The fit is very good, thanks to adapters that customize the fit for each iPad model. In other words, this is a very good all-purpose stand.
What makes the Powerstand unique, however, is that it also features a 30-pin dock connector in its cradle, along with a USB port in the base, so that you can charge and sync your iPad while it's in the stand. (This does mean, however, that the Powerstand works only with full-size iPads that use a 30-pin dock-connector port—it doesn't work with Lightning-connector iPads.) The included USB cable can plug into your computer or the included AC charger. Note that while Mophie is well known for its battery-focused accessories, the Powerstand does not have its own battery.
The Powerstand folds down for travel, though it's not especially lightweight—it's a desktop stand, through and through. It's main drawback is that you'll need to remove any iPad case in order for your iPad to fit in the Powerstand. Other than that, the Powerstand is solid.
XFlex's $100 XFlex Tablet Stand (2.5 of 5 rating) is composed of rubber, plastic, and metal. The base takes the form of a big rubber "X," the neck is flexible, and a suction cup at the top attaches any iPad (or any other device with a flat, smooth back). In my testing, the suction cup adhered securely to my iPad both bare and in a smooth, hard-plastic case.
You can position your iPad in landscape or portrait mode, and the angle is adjustable, though not so much that you could get the tablet flat enough to type comfortably on the screen. On the other hand, if you use a Bluetooth keyboard, the XFlex's positioning of your iPad about a foot off the desk or table makes for a more-ergonomic setup. The stand keeps all ports and buttons fully accessible.
Once you secure the stand itself, it does keep the iPad in place and makes for pleasant video-viewing or reading. That said, I wouldn't choose the XFlex for everyday use, due to its top-heavy nature. I'd be worried that the stand would tip over and my iPad would crash onto a hard surface. In addition, the fact that you depend on a suction cup makes me even more nervous, despite the suction cup working well in my testing.
The biggest advantage of this stand over some of the others I tested is that fact that it's compatible with virtually any brand of tablet. Its flexible neck also might allow you to use it in places that other stands wouldn't.
Twelve South's $80 HoverBar Adjustable Arm for iPad (3.5 of 5 rating) is advertised mainly as a way to use your iPad as a second screen for, or alongside, your computer, though it ends up being a bit more versatile than that. Designed primarily for use with Macs, you attach the HoverBar's clamp to the L-shaped stand of your iMac or Apple display display, or to the lip of your desk, so the stand takes up very little desk space while keeping your iPad conveniently close. You can also attach the stand to a counter to use your iPad in the kitchen while keeping it out of the way of splashes and food splatters.
The HoverBar is constructed of metal with the aforementioned clamp on one end and a plastic iPad frame on the other. You use an included hex wrench to lock the clamp into place—the HoverBar isn't meant to be moved from place to place.
When your iPad is in the HoverBar's frame, all buttons and ports are accessible. However, that frame requires a bare iPad—you can't use an iPad case with this stand. Apple's Smart Cover and similar screen covers are compatible, though when you open the cover, you must either either remove it from the iPad or let it dangle behind or down.
The HoverBar is limited to particular uses, but it performs well in those scenarios.
If you want simple, Bluelounge's $40 Mika (3.5 of 5 rating) is the way to go. This stand is just a curved piece of aluminum with black-plastic strips wherever your iPad contacts the stand. This design means that any iPad—or any tablet or any smartphone—can take advantage of the Mika, and you can even keep your iPad in a case as long as that case isn't terribly bulky.
The Mika can prop up your devices in both landscape and portrait orientation, though at only a single angle of about 60 degrees from horizontal. That angle is quite comfortable for reading or viewing, and the stand is secure enough for you to tap on the screen, but onscreen typing is awkward. If your iPad is in landscape orientation, all buttons and ports are accessible; of course, when the iPad is in portrait mode with the Home button at the bottom, the dock-connector or Lightning-connector port port is blocked by the stand.
You'll want to use the Mika on a hard, flat surface, because only gravity holds your iPad in place—if you use the stand on a soft surface such as a bed or a sofa, the iPad can easily tip out.
The Mika has no bells or whistles. It doesn't fold down like many smaller stands do, so despite its compact size, it's not really portable. But if you just want to prop up your iPad in your home or office, the Mika is an elegant option.
Octa's $100 TabletTail: Monkey Kit (3.5 of 5 rating) is a bit different. It's actually two pieces: a big suction cup and a long, flexible tail that's covered in plastic but that's obviously got quite a bit of metal on the inside. You can use the suction cup alone as a small tabletop stand, though I found that because it's a half-sphere, I had to place it just right to avoid the iPad tipping over when I typed on the screen.
Add the tail, and you can place your iPad anywhere: You can wrap the tail around a bar, wedge it between couch cushions, or coil it so that it stands alone on a desk. You might even be able to rig it up in the car for people in the back seat to watch a video. However, I found that while the stand worked well when wedged in the couch, it wasn't very sturdy as a standalone desk stand.
You can manipulate the Monkey Kit to display your iPad at essentially any angle, in either landscape or portrait orientation, and because it attaches to the back of your iPad, all buttons and ports remain fully accessible.
Since the suction cup attaches to the back of your device, it will work any iPad or other device larger than the suction cup itself. You can even use it with an iPad as long as the case is hard and smooth. When you attach the suction cup, you press a button to pump the air out. Over time, as air leaks in, that button will start to pop back out, so you need to press it occasionally to maintain solid suction. It's not hard to do, but you have to remember to do it—and if you don't, your iPad can fall off.
That aside, if you want a multipurpose stand that stands out of the crowd, the TabletTail: Monkey Kit is worth a look.
Among this batch of iPad stands, there's no clear "winner"—each has different strengths and weaknesses. If you're looking to use your iPad in a business setting, Heckler Design's WindFall is the way to go, as it makes for a nice iPad-based kiosk and it will help keep your iPad from wandering off. Bluelounge's Mika or Arkon's smaller Tab-Stand2 are more basic options for your desk. Arkon's larger Tab-Stand1 makes for a good lectern or music stand, while Twelve South's HoverBar is great for use at your desk if you don't have a lot of free desk space. For iPad use in the car or other non-traditional places, Octa's flexible-necked TabletTail: Monkey Kit may fit the bill.
Mophie's Powerstand comes the closest to being an all-purpose stand. It works at the office, in the kitchen, or on the couch, and its flexible positioning ensure that you can do just about anything with it. The fact that it doubles as a charge/sync station is a huge advantage. However, its relatively high price and the fact that you can't use it with an iPad case keep it from getting an unreserved recommendation.
Arkon's Tab-Stand models aren't exciting, but they're functional and versatile.
If you're looking to use your iPad for customer interaction at your place of business, the WindFall might be the answer.
Mophie's Powerstand is both a stand and an iPad charge/sync dock. It works well, though you'll need to remove any case your iPad might be wearing.
The XFlex Tablet Stand is versatile; however, with a heavier tablet, the setup isn't very stable.
If you're looking to let your iPad be a "second screen" in your computer setup, or a cooking companion in the kitchen, the HoverBar is a solid choice.
The Mika is as simple a tablet stand as you'll find. But it's also elegant and effective.
If you want a multipurpose stand that stands out of the crowd, the TabletTail: Monkey Kit is worth a look.


I've been thinking about how happiness affects your life. There are things in our lives that we carry for a long time. These can be memories of happy or sad events, thoughts of foods we like or dislike, and thoughts of opportunities taken and missed.
I remember a quote and this is a paraphrase, happiness is hard to recognize because unlike pain, happiness doesn't leave scars.
This should make us think about how happiness affects your life. I think we can have temporary happiness and lasting happiness based on where happiness comes from.
When I think of temporary happiness, I think of food (disclaimer: I always think about food). I can think of moments where I had great cheesecake or other desserts. I bet you can too.
Remember the feeling of happiness from eating that really good meal when you were on vacation? How everything about the food came together and at the moment there's joy in what you were experiencing.
Experiences like this can make us happy, but for how long? Maybe a little while but the experience is short-lived and maybe we remember it now and then but mostly we don't.
Should we depend on outside experiences as our source of happiness? These experiences can short-lived and so is the related happiness. So what does that leave us with?
Does happiness come from longer lasting experiences? If short lasting experiences have short lasting happiness, would the opposite be true?
Are longer lasting experiences result in longer lasting happiness? If so, how do we capture these events in a way happiness affects your life in a lasting way?
If longer lasting experiences result in longer-lasting happiness, can happiness be learned?
Sometimes there seem to be more questions than answers.
I think we can find where happiness and how happiness affects your life. We do this by finding things that match or overlap with our goals and personal vision. These are things, I think, have long-lasting effects.
By having this overlap, you can not only create and maintain a level of happiness but you can change and maintain it by evaluating yourself. There is a burden that's lifted when you're figuring out what you want from life.
Working towards that goal can, I think, improve our moods for the positive. This can be a possible contributor to bring happiness.
As usual, I see this line of thinking as something that can be applied to part of my life.
I'm hoping it's probably the same for you. If you skim or read these posts, you can see a trend of personal redemption from failure and exploration into personal growth.
I think how we define happiness for ourselves is part of that journey. We may have areas in our lives where we already have fairly lasting things (i.e. personal achievements or relationships) which help us remember that we are happy.
Relationships, I think, can be an important source of happiness and tend to be a fairly long term. Relationships aren't restricted to just people but can be applied to pets, they are in our lives just as much a people and we have a personal responsibility for them.
This brings me to the next thought, about what happiness means to me? Can happiness, as a form of satisfaction be considered vanity or ego? How do we know where to draw the line between ego and happiness?
Maybe one way to know is the amount of interference the happiness causes with other aspects of our lives. If there's minimal interference then it's probably alright.
This is until happiness becomes overwhelming in other aspects, this is when vanity or pride may be harmful to our concept of happiness.
Happiness for me is knowing that I have a direction for my life, I might not fully know this direction but I know there's one out there.
It's also a combination of things and experiences that have made my life richer. Not richer financially but full. I can be happy with what I've accomplished and happy with the combination of my experiences.
Take a moment to think about the long lasting things that are part of your life. The combination of memories and experiences that help you remember that happiness is always within your grasp. Start a note to yourself and write "happiness means to me" and start a list.
If you don't have a draft personal vision then let this be your starting point. Keep this list with you so you can see how your thoughts change day to day and over time.
Write as much down that comes to mind, you can sort through it later. Don't limit your options and only stick with the things I wrote about, explore for yourself.
Let me know if you find something interesting that I didn't discuss. I'm always interested in hearing new perspectives.
Take a moment to remember when you were happy. Did you realize you were happy at that moment? Is there a way to recreate that now?
Can happiness be learned? If so, can we learn to be happy by evaluating what we want out of life (goals and personal vision)? I think we can learn to be happy when we reevaluate what it means to be happy.
I think we often forget that we are happy so we should remind ourselves of this. We can enjoy the temporary happiness (my desserts) but not lose focus on the bigger things that make us happy.


Applicants who are considering attending medical schools in the Caribbean should be aware of the challenges they will face when it comes time to apply for U.S. residency programs. Every year, approximately 95 percent of senior medical students in the United States match into a residency program, compared to only 50 percent of U.S. citizens graduating from a foreign school.
When comparing potential Caribbean schools, it is crucial to ask where and in what specialties their graduates eventually match. Well-established schools such as Ross University, St. George's University, and the American University of the Caribbean have higher success rates.
It is important to note that the vast majority of Caribbean school graduates enter professions in primary care. While this provides a partial solution to the growing shortage of primary care physicians in the United States, students should understand that their chances of obtaining a residency in a more competitive specialty is much more limited compared to U.S. medical school graduates.
In 2012, 76 percent of the senior medical students at Ross University obtained residencies in primary care, including 43 percent in internal medicine, 26 percent in family medicine, and 7 percent in pediatrics. Recent residency placement data from St. George's University and the American University of the Caribbean demonstrate similar trends.


text:
\section{Introduction}
\label{sec:intro}

For a combinatorial structure $G$ and a set of parameters, we say that two disjoint substructures are \emph{twins} if those parameters coincide for each substructure.
This notion generalizes the pigeonhole principle and was investigated in case of sequences and graphs.  In the case of sequences it is known, see Axenovich, Person and Puzynina ~\cite{APP},  that any binary sequence
of length  $n$ contains two disjoint identical subsequences of length $n/2-o(n)$ each.
Twins are defined to be two disjoint  vertex subsets of the same size and the same multiset of pairwise distances were studied by   Albertson, Pach and Young~\cite{APY},  and by Axenovich and \"Ozkahya~\cite{AO}.
Edge-disjoint isomorphic subgraphs were also addressed in ~\cite{GR, LLS, AKS}.

Here, we concentrate on the following twin problem in graphs introduced by Caro and Yuster in~\cite{caroyuster}.
For a graph $G$, we call two disjoint subsets  of vertices \emph{twins} if  they have the same size and
induce subgraphs with the same number of edges.
Let $t(G)$ be the largest $k$ such that there are twins $A, B$ in $G$ with $|A|=|B|=k$.
Let $$t(n) = \min \{ t(G):  |V(G)|=n\}.$$ The best currently known bounds on $t(n)$ are given in the following theorem.

\begin{theorem}[Caro and Yuster~\cite{caroyuster}]\label{general-bounds}
There exists a positive constant $c$ such that $\sqrt{n} \leq t(n) \leq  n/2 - c\log\log n$.
\end{theorem}

 Ben-Eliezer and  Krivelevich,~\cite{BK},  proved that  $t(G(n,p))=\lfloor n/2\rfloor$ with high probability, where $G(n,p)$ is the Erd\H{o}s-Renyi random graph.
In addition, Caro and Yuster proved that in a sparse graph there are twins of size almost $n/2$:

\begin{thm}[Caro and Yuster~\cite{caroyuster}]\label{num-edges}
For every fixed $\alpha > 0$ and for every $\epsilon > 0$ there exists $N=N(\alpha, \epsilon)$
so that for all $n > N$, if $G$ is a graph on $n$ vertices and at most $n^{2-\alpha}$ edges
then $t(G)\geq (1-\epsilon)n/2$.
\end{thm}

In this paper, we do the following:
\begin{itemize}
   \item In Theorem~\ref{thm:size}, we improve the above result using a more general theorem on discrepancy for induced subgraphs.
   \item In Theorem~\ref{thm:exact}, we give several sufficient criteria under which the graph has \emph{perfect twins}; i.e., two twins spanning the whole graph.
   \item In Theorem~\ref{thm:forest}, we prove that all but at most two vertices of any forest could be split into two twins.
\end{itemize}

For disjoint vertex sets $A, B$   in a graph $G$, we denote by $e(A)$ the number of edges induced by $A$,
$e(A,B)$ the number of edges between $A$ and $B$.  {The} \emph{discrepancy} between $A$ and $B$ is ${\rm disc}(A,B) = |e(A) -e(B)|$.
Let $\lg n$  denote the logarithm of $n$ in base $2$.
The maximum and the minimum degree of a graph $G$ are denoted by $\Delta(G)$ and $\delta(G)$, respectively,
the set of vertices of degree $i$ is  $V_i$.  A pair of vertices is called \emph{consecutive} if the degrees of these vertices differ by exactly $1$ or $-1$.

\begin{thm}\label{thm:almosttwins}
   If $G$ is an $n$ vertex graph, $n\geq 16$, then there are vertex disjoint sets $A$ and $B$ such that
   \begin{itemize}
    \item ${\rm disc}(A,B) \leq 2\lg^2 n$ and $|A|=|B| \geq (n-2\lg n)/2$ or
    \item ${\rm disc}(A,B) \leq (\Delta(G)-\delta(G)+1)/2$ and $|A|=|B| = \lfloor n/2 \rfloor$.
   \end{itemize}
\end{thm}

\begin{theorem} \label{thm:size}
If $G$  is  a graph on  $n$ vertices and $e$ edges then
$t(G) \geq  \frac{n}{2} \left( 1- \frac{20 \sqrt{e} \lg n}{n}\right).$
\end{theorem}

This theorem implies, in particular, that any $n$-vertex graph on $o\left(n^2 / \lg^2n\right)$  edges  has twins of size $n/2 - o(n)$ each and
 that planar graphs have twins of size at least $n/2 - c \lg n$ each.

\begin{thm} \label{thm:exact}
 Let $G$ be  a graph on   $n$  vertices,  where $n$ is even. If one of the following conditions~\ref{enum:consecutive-degrees}-\ref{enum:disjoint-consecutive-pairs} holds then $t(G)=n/2$.
 \begin{enumerate}[label= \arabic*)]
 \item The degree sequence of $G$ forms a set of consecutive integers.\label{enum:consecutive-degrees}
 \item $|V_i|$ is even for each $i$.\label{enum:even-vertex-sets}
 \item $n\geq 90$ and  $|\{i: |V_i|\, \text{ is odd}\}|>n/2$.\label{enum:large-odd-set}
 \item There are at  least $\Delta(G)-\delta(G)$ disjoint consecutive pairs of vertices.\label{enum:disjoint-consecutive-pairs}
\end{enumerate}
\end{thm}

\begin{theorem}\label{thm:forest}
 If $G$ is a forest then $t(G) \geq \lceil n/2 \rceil - 1$.
\end{theorem}

 If $n$ is odd this is clearly best possible. For even $n$ this is attained, for example, by a star. Such a graph has no perfect twins.

The rest of the paper is organized as follows: In Section~\ref{sec:prelim}, we cite and prove some useful observations about degree sequences and include a short proof of Theorem~\ref{general-bounds} for completness. Sections~\ref{sec:almosttwins}--\ref{sec:forest}
contain the proofs of Theorems ~\ref{thm:almosttwins}--\ref{thm:forest}.

\section{Preliminary and known results}
\label{sec:prelim}

Here, we denote a degree of a vertex $v$ in a graph $G$ by $d(G, v)$ or simply $d(v)$.   For a set $A$ of  vertices,  $d(G,A)=d(A) = \sum_{a\in A} d(a)$.
The set of all $k$-element subsets of a set $X$ is denoted by $\binom{X}{k}$.
For a graph $G$ and a vertex set $W \subseteq V(G)$ the induced subgraph of $G$ on $V(G) \setminus W$ is denoted by $G - W$.

\begin{lemma}\label{degree-pairs}
Let $G$ be an $n$-vertex  graph  and $A, B$ disjoint vertex sets of equal size such that $V(G)=A\cup B$.
If $d(A) = d(B) $ then $e(A)= e(B)$.
If $A$ is a disjoint union of $A_1, \ldots, A_k$ and $B$ is a  disjoint union of $B_1, \ldots, B_k$,  such that $|A_i|=|B_i|$ and $|d(A_i)-d(B_i)|\leq \epsilon$, $i=1, \ldots, k$ then there are disjoint vertex  subsets $A', B'$ of size $n/2$ each such that ${\rm disc}(A',B') \leq \epsilon$.
\end{lemma}

\begin{proof}
 {First assume that $d(A) = d(B)$. Then} we have that $$e(A)-e(B) = (d(A)-e(A,B))/2 - (d(B)-e(A,B))/2 = (d(A)-d(B))/2 = 0.$$
 {Now let $A$ and $B$ be the disjoint union of $A_1,\ldots.A_k$ and $B_1,\ldots,B_k$, respectively, such that $|A_i| = |B_i|$ for each $i = 1,\ldots,k$.} Assume, without loss of generality, that $|d(A_1)-d(B_1)| \geq |d(A_2)-d(B_2)|\geq \cdots \geq |d(A_k)-d(B_k)|$.  Define, for each $i=1,\ldots,k$,
 \begin{equation}\nonumber
  C_i= \begin{cases}
   A_i & i \mbox{ is odd and } d(A_i)\geq d(B_i),\\
   B_i & \mbox{otherwise}.
  \end{cases}
 \end{equation}

 Let $C_i' = \{A_i, B_i\} - C_i$ and consider $A'=C_1 \cup \cdots \cup C_k$ and $B'=C_1'\cup \cdots \cup C_k'$.  Since $e(A')=(d(A')-e(A',B'))/2$ and $e(B')=(d(B')-e(A',B'))/2$, we have $2(e(A')-e(B'))= d(A')-d(B') = \sum_{i=1}^k (d(C_i)-d(C'_i)).$ Since this sum has terms non-increasing in absolute value with alternating signs and bounded in absolute value by $\epsilon$, the sum itself is at most $\epsilon$.
\end{proof}

We shall need the following number theoretic result.

\begin{lemma}\label{fixed-k}
 Let $n\geq 16$  and $k \geq \lg n$. Let $X$ be a multiset  $\{a_1, a_2, \ldots \}$  of  integers from $\{0, 1, \ldots, n-1\}$  such that for any $l\leq k$
 and for any disjoint index sets $\{i_1, \ldots, i_l\}$ and $\{j_1, \ldots, j_l\}$  the following holds:
 $a_{i_1}+a_{i_2}+\cdots+a_{i_l} \neq a_{j_1}+a_{j_2}+\cdots+a_{j_l}$.  Then $|X| < 2k$.
\end{lemma}

\begin{proof}Let $S$ be the set of all possible sums of $k$-element  sub-multi-sets of $X$. Since we assume that all of these are distinct, $|S|=\binom{|X|}{k}$.
On the other hand, there are at most $k(n-1)+1 < kn$ possible sums of $k$ elements from  $X$.
Thus, using a well-known consequence of Stirling's inequality,
$$ kn>|S|=\binom{|X|}{k}\geq\binom{2k}{k}\geq\frac{2^{2k}}{2\sqrt{k}} . $$
Consequently $n>2^{2k-1}/k^{3/2}$. This is a contradiction since $k\geq\lg n \geq 4$.
\end{proof}

Here, we say that two integers are \emph{almost equal} if they differ by $1, 0$, or $-1$.

\begin{thm}[Karolyi~\cite{karolyi}]\label{kar}
Let $X$ be a set of $m$ integers, each between $1$ and $2m-2$.
If $m\geq 89$, then one can partition $X$ into two sets, $X_1$ and $X_2$  of almost equal sizes such that
the sum of elements in $X_1$ is almost equal to the sum of elements in $X_2$.
\end{thm}

For positive integers $k,n$ ($k \leq n/2$), the {\em Kneser graph} $\mathcal{K}= KG(n,k)$ is the graph on the vertex set $\binom{[n]}{k}$ whose edge set consists of pairs of disjoint $k$-sets. Lov\'asz~\cite{lovasz} proved that the chromatic number of the Kneser graph $KG(n,k)$ is $n-2k+2$.
This result is a main tool for proving the lower bound on $t(n)$ used by Caro and Yuster~\cite{caroyuster}. Here we repeat it for completeness.

\begin{proof}[Outline of the  Proof of Theorem~\ref{general-bounds}]
For a graph $G$ with vertex set $[n]$,  we define a vertex coloring of $\mathcal{K}$
by letting the number of edges  induced by each $k$-subset of $V(G)$ be
the color of the corresponding vertex in $\mathcal{K}$.  The number of possible colors is most ${\binom{k}{2}}+1$.
For $k = \sqrt{n}$, we have ${\binom{k}{2}}+1<n-2k+2$. Thus the number of colors is less than the chromatic number of $\mathcal{K}$ and  there are two  {adjacent} vertices in $\mathcal{K}$ of the same color. Therefore there are two disjoint $k$-sets  of $G$ that induce the same number of edges.

To verify the upper bound, consider the disjoint union of cliques of odd orders $a_1, \ldots, a_m$, where $a_j>2(a_1^2+a_2^2+\cdots+a_{j-1}^2)$.
One can show that any pair of twins in this graph must omit at least $c\lg\lg n$ vertices for some constant $c$.
\end{proof}

\section{Proof of Theorem~\ref{thm:almosttwins}}
\label{sec:almosttwins}

Set  $k=\lg n$ and $G=(V,E)$. We shall find two large vertex sets $A$ and $B$ of $G$ of equal size and small discrepancy following the following procedure.\\

\noindent
\textbf{Step 1.} Choose disjoint sets $A_1,B_1\subseteq V$ such that $|A_1|=|B_1|\leq k$ and $d(A_1)=d(B_1)$.\\
\indent By Lemma~\ref{fixed-k} this is possible.

\noindent
\textbf{Step  $i$,  $i>1$.} Choose disjoint sets $A_i,B_i\subseteq V\setminus(A_1\cup B_1\cup\cdots\cup A_{i-1}\cup B_{i-1})$ such that $|A_i|=|B_i|\leq k$ and $d(A_i)=d(B_i)$.\\
\indent Again, by Lemma~\ref{fixed-k} this is possible as long as $|V\setminus (A_1\cup B_1\cup\cdots\cup A_{i-1}\cup B_{i-1})|\geq 2k$.\\

Assume that we have to stop after step $q$ and let $S=V\setminus (A_1\cup\cdots\cup A_q\cup B_1\cup\ldots\cup B_q)$ be the set of leftover vertices. So, $|S|<2k$. Consider the graph $G'=G-S$. Since $d(G,A_i)= d(G,B_i)$ and $V(G)\setminus V(G')=S$, we have that $|d(G',A_i)-d(G',B_i)|=|d(G,A_i)+e(A_i,S)-d(G,B_i)-e(B_i,S)|$. Thus,
$$ |d(G',A_i)-d(G',B_i)|\leq |d(G,A_i)-d(G,B_i)|+|e(A_i,S)-e(B_i,S)|=|e(A_i,S)-e(B_i,S)|\leq |A_i||S|\leq 2k^2.$$

Denoting $\epsilon_i =|d(G',A_i)-d(G',B_i)|$, we have that $\epsilon_i \leq |A_i||S|\leq 2k^2$.  Lemma~\ref{degree-pairs} applied to $G'$ and $A_1,\ldots,A_q,B_1,\ldots,B_q$ asserts the existence of disjoint vertex sets $A$, $B$ in $G$, $|A| = |B| = \sum_{i=1}^q |A_i|$, such that ${\rm disc}(A,B) \leq 2k^2 =2\lg^2 n$.\\

Next, we shall find two vertex sets $A$ and $B$ of $G$ such that $|A|=|B|=\lfloor n/2\rfloor$ and ${\rm disc}(A,B) \leq \frac{1}{2}(\Delta(G) - \delta(G) + 1)$. Let $G' = G$ if $n$ is even, and $G' = G \setminus v$ if $n$ is odd, where $v$ is any vertex of $G$. Thus $G'$ has $2 \lfloor n/2\rfloor$ vertices and $\Delta(G') - \delta(G') \leq \Delta(G) - \delta(G) + 1$. Let $A$, $B$ be two disjoint vertex sets in $G'$ with $|A|=|B|=\lfloor n/2\rfloor$ and minimum discrepancy. It remains to show that ${\rm disc}(A,B) \leq \frac{1}{2}(\Delta(G') - \delta(G'))$.

If $\Delta(G') = \delta(G')$, i.e.  $G'$ is regular,  then ${\rm disc}(A,B) = 0$ by Lemma~\ref{degree-pairs}. Hence we may assume $\Delta(G') - \delta(G') > 0$. Assume, without loss of generality, that ${\rm disc}(A,B) = e(A) - e(B)$. Since $\sum_{a \in A}d(G',a) = 2e(A) + e(A,B) > 2e(B) + e(A,B) = \sum_{b \in B} d(G',b)$ there is a vertex $a \in A$ and a vertex $b \in B$ with $d(G',a) > d(G',b)$. For $A' = (A \setminus a) \cup b$ and $B' = (B \setminus b) \cup a$ we have
$$e(A') - e(B') = e(A) - d(G',a) - e(B) + d(G',b) = {\rm disc}(A,B) - d(G',a) + d(G',b)< e(A)-e(B). $$
Due to the minimality of ${\rm disc}(A,B)$ it follows ${\rm disc}(A',B') = e(B') - e(A')$ and thus
$$ 2{\rm disc}(A,B) \leq {\rm disc}(A',B') + {\rm disc}(A,B) = d(G',a) - d(G',b) \leq \Delta(G') - \delta(G'),$$
which implies ${\rm disc}(A,B) \leq \frac{1}{2}(\Delta(G')-\delta(G'))$ as desired. \hfill $\Box$

\section{Proof of Theorem~\ref{thm:size}}
\label{sec:size}

This argument is very similar to the one used by Caro and Yuster. The only difference is that we are using stronger discrepancy results.
We shall remove a few vertices of large degree. In the remaining graph we find two sets $T$ and $S$,  such that   there are no edges between $T$ and
$S$,   $S$ is large enough, and $T$   induces a matching and independent vertices.  Finding ``almost'' twins guaranteed by Theorem~\ref{thm:almosttwins} in
$G[S]$, and using the edges and vertices of $T$ to balance those, gives us twins in $G$.\\

Let $G$ be an $n$-vertex graph on  $e$ edges.  We shall find large twins $A$ and $B$ of size at least  $\frac{n}{2} \left( 1- \frac{20\sqrt{e} \lg n}{n}\right)$ each.
To shorten the expressions we use $f = f(n,e) =  \frac{\sqrt{e} \lg n}{n}$, so we are looking for twins of size at least $\frac{n}{2} (1-20f)$ each. In particular we may assume that $n(1-20f) > 0$.
Assume that the size of the largest independent set in $G$ is less than $n(1- 20f)$, otherwise we find the desired twins as subsets of this independent set.
Assume also that  $e\geq 4$. \\

Consider the set $L= L(x)$ of vertices of degree at least $x= \frac{2e}{nf}$.
The number of edges incident to these vertices is at least $ |L|x/ 2$.
Since this number is at most the total number $e$ of edges,  $|L|\leq  \frac{2e}{x} = nf$.
Let $G'=G-L$, a graph with maximum degree less than $x$ and number of vertices at least $n(1-f)$.
Next we shall consider $G'$ only.

Let $l = 2\frac{(nf)^2}{e} = 2\lg^2 n$. For a vertex $v$, denote by $N[v]$ the set of vertices consisting of $v$ together with its neighbors in $G'$. For an edge $e_0=uv$, denote by $N[e_0]$ the set of vertices $\{u,v\}$ together with each of their neighbors in $G'$. We shall choose a set $T$ of $4l$ vertices from $V=V(G')$ inducing a matching $e_1, \ldots, e_l$ and $2l$ independent vertices $v_1, \ldots, v_{2l}$
 via a simple greedy procedure:\\

 \noindent
{\bf Step 1.}  Pick an edge $e_1$ with endpoints in $V$.\\
  \indent Since $|V| \geq n(1 - f)$, $V$ can not be an independent set, so the edge $e_1$ exists.\\
{\bf Step $i$, $1<i\leq l$.}  Pick an edge $e_i$ with endpoints in $V \setminus (N[e_1] \cup \cdots \cup  N[e_{i-1}])$.\\
  \indent  Since the size of  $V \setminus (N[e_1] \cup \cdots \cup N[e_{i-1}])$ is greater than $n(1-f) - 2(i-1)x > n(1-f) - 2lx = n(1-f) - 2\cdot 2\frac{(nf)^2}{e}\frac{2e}{nf} = n(1-9f) > 0$, this set is not an independent set and an edge $e_i$ can be selected for $i=1,\ldots,l$.\\

\noindent
{\bf Step $l+1$.}  Pick a vertex $v_1$ from $V \setminus (N[e_1] \cup \cdots \cup N[e_l])$.\\
{\bf Step $l+j$, $1<j\leq 2l$.}  Pick a vertex $v_j$ from $V \setminus (N[e_1] \cup \cdots \cup N[e_l] \cup N[v_1] \cup \cdots \cup N[v_{j-1}])$.\\
  \indent The fact that  this set is non-empty for each $j$ follows from the argumentation below.

Let $S$ be the last set $S= V \setminus (N[e_1] \cup \cdots \cup N[e_l] \cup N[v_1] \cup \cdots \cup N[v_{2l}])$.  Then $|S| >  n(1-f) - 4lx = n(1-17f) > 0$. Moreover no vertex of $S$ is adjacent to any vertex of $T$. Applying Theorem~\ref{thm:almosttwins} to $G[S]$ gives two disjoint subsets of $S$, $A$ and $B$, of the same size at  least $(|S|-2\lg |S|)/2$ and with discrepancy $\gamma$, $\gamma \leq 2\lg^2(|S|) \leq 2\lg^2 n \leq l$. Assume, without loss of generality, that $e(A)\geq e(B)$. We construct a set $A'$ by adding $2\gamma \leq 2l$ independent vertices from $T$ to $A$, and a set $B'$ by adding $\gamma \leq l$ edges from $T$ to $B$. Now, the discrepancy of $A'$ and $B'$ in $G$ is zero since there are no edges between $A\cup B$ and $T$. Thus $A'$ and $B'$ are twins in $G$ of size at least $(|S| - 2\lg |S|)/2 > \frac{n}{2}(1-17f) - \lg n = \frac{n}{2}(1-17f) - \frac{nf}{\sqrt{e}} \geq \frac{n}{2}(1-19f)$. \hfill $\Box$

\section{Proof of Theorem~\ref{thm:exact}}
\label{sec:exact}

 {Let $G$ be a graph on $n$ vertices, where $n$ is even. We shall show that if one of conditions~\ref{enum:consecutive-degrees}-\ref{enum:disjoint-consecutive-pairs} holds then $t(G)=n/2$, i.e., the vertex set of $G$ can be perfectly split into twins.}

 \begin{enumerate}
  \item[\ref{enum:consecutive-degrees}.]  {Assume that the degrees of $G$ form a set of consecutive integers. That is, for every integer $i$, $\delta(G)\leq i\leq\Delta(G)$, there exists a vertex of degree $i$.} Let $A$ and $B$ be sets that have equal cardinality that partition the vertex set and have the smallest discrepancy. If the discrepancy is nonzero, assume that $e(A)<e(B)$. In particular, we see that if $A$ has a vertex of degree $d$ then $B$ does not have a vertex of degree $d+1$ (otherwise swapping the two will give a partition with smaller discrepancy). Similarly, if $B$ has a vertex of degree $d$, then $A$ has no vertex of degree $d-1$. So, if $d$ is the smallest degree of a vertex in $A$ then $A$ contains all vertices of degrees $d+1, d+2, \ldots, \Delta(G)$. Similarly, if $d'$ is the highest degree of a vertex of $B$, then $B$ contains all vertices of degree $d'-1, d'-2, \ldots, \delta(G)$. Moreover $d'\leq d$. So, we have that the sum of the degrees in $A$ is strictly greater than the sum of
the degrees in $B$, a contradiction to the assumption that $e(A)<e(B)$.

  \item[\ref{enum:even-vertex-sets}.]  {Assume that the number of vertices of degree $i$ is even for each integer $i$, $\delta(G)\leq i\leq\Delta(G)$.} Observe using Lemma~\ref{degree-pairs} that it is sufficient to partition the vertex set into two parts of equal sizes such that the degrees of vertices in each part have the same sum. Since each degree appears an even number of times, we put half of the corresponding vertices in one part and half in another.

  \item[\ref{enum:large-odd-set}.]  {Assume that $n\geq 90$ and $|\{i: |V_i|\, \text{ is odd}\}|>n/2$.} For each degree $d$, put $\lfloor d/2 \rfloor$ vertices of degree $d$ in one part and $\lfloor d/2 \rfloor$ vertices of degree $d$ in another part. Let $S$ be the set of remaining vertices of distinct degrees. We shall split $S$ also using Theorem~\ref{kar}. Formally, for each degree $d$, let $V_d = A_d \cup B_d \cup S_d$, where $|A_d|=|B_d|$ and $|S_d| \leq 1$. Then $S = S_{\delta}\cup \cdots \cup S_{\Delta}$ is an even set of more than $n/2$ vertices of distinct degrees. Applying Theorem~\ref{kar} to the degrees of vertices in $S$, we can split $S$ in two parts, $U$ and $U'$ of equal sizes such that $|d(U)-d(U')|\leq 1$. Let $A = U\cup  \bigcup_{d=\delta}^{\Delta} A_d$ and $B = U'\cup \bigcup_{d=\delta}^{\Delta} B_d$. Since $d(V)$ is even and $d\left(\bigcup_{d=\delta}^{\Delta} A_d\right) = d\left(\bigcup_{d=\delta}^{\Delta} B_d\right)$, then $d(U)$ and $d(U')$ must have the same parity, and therefore
must be equal. So it must be the case that ${\rm disc}(A, B) =0$.

  \item[\ref{enum:disjoint-consecutive-pairs}.] {Assume that there are least $\Delta-\delta$ disjoint pairs of vertices whose degrees differ by exactly $1$.} With $x=\Delta-\delta$, there is a set of vertices $\{v_1,v_1',\ldots,v_x,v'_x\}$ such that $d(v_i)=d(v'_i)-1$. Partition $V(G) - \{v_1,v'_1,\ldots,v_x,v'_x\}$ into two sets $A$ and $B$, of equal cardinality with $0\leq d(A) - d(B) = x'$, $x'\leq x$.  Note that this could be done greedily
  by ordering the vertices in the non-increasing order of their degrees and alternately adding one vertex to $A$ and the next to $B$.
  Let $A'=A \cup \{v_1,\ldots,v_{x'}\}$ and $B'=B \cup \{v'_1,\ldots,v'_{x'}\}$. As a result, $d(A') = d(B')$. If $x'\neq x$, let $A''= A'\cup \{v_i: x\geq i>x', i \mbox{  is  even}\} \cup \{v'_i: x\geq i>x', i \mbox{  is  odd}\}$ and $B''= B' \cup \{v'_i: x\geq i>x', i \mbox{  is  even}\} \cup \{v_i: x\geq i>x', i \mbox{  is  odd}\}$. Consequently, $|d(A'') -d(B'')|\leq 1$. However, since the sum of the degrees of a graph is even and $V(G)= A''\cup B''$, we have that $d(A'')=d(B'')$. Therefore $A''$ and $B''$ are twins of size $n/2$ each. We note that this technique is
reminiscent of the one used by Ben-Eliezer and Krivelevich in~\cite{BK}.
 \end{enumerate}~ \hfill $\Box$

\section{Proof of Theorem~\ref{thm:forest}}
\label{sec:forest}

First, we observe that we may assume that no component of our forest consists of any isolated edges.  If it does, remove those edges and find the twins in the remaining graph and delete $0$, $1$ or $2$ vertices to get twins $A$, $B$.  We can add the endvertices of each isolated edge the twins -- one to $A$ and one to $B$  -- to obtain twins for the whole forest, apart from the at most $2$ vertices that were already deleted.

In order to proceed, we need to define a type of twins with additional structure. Let $G=(V, E)$ be a forest and $A$, $B$ be twins.  We say that $A\cup B$ is the set of \emph{colored} vertices and $G[A\cup B]$ is the \emph{colored graph}; the remaining vertices are \emph{uncolored}.
We define $A$ and $B$ to be \textbf{good twins} if
\begin{itemize}
 \item{}  each uncolored vertex has degree at most $1$ in $G$,
 \item{}  the set of colored vertices induces no isolated vertices,
 \item{}  all colored vertices except for a set $S\subseteq A$ of at  most two vertices  have all neighbors colored,
 \item{}  $S$ contains at most one leaf and at most one non-leaf of the colored graph,
 \item{}  if $S$ contains a non-leaf, $w$,  of the colored graph, then $w$ has at most one neighbor in $A$,
 \item{}  if  $S$ contains a non-leaf $w$ and  a leaf $v$ of the colored graph,     then $v$ has smaller number of uncolored neighbors than $w$.
\end{itemize}

Because each uncolored vertex has degree at most $1$ and there are no isolated edges, if $A$,$B$ are good twins then at least one vertex of each edge is colored. Therefore, the set of edges that have exactly one colored vertex come in at most two disjoint stars, each centered at a vertex in $S\subseteq A$. We refer to the middle and the right of Figure~\ref{fig:tree-example} for examples of good twins. The figure exemplifies one step in the inductive proof of existence of good twins (Theorem~\ref{thm:good_twins}).

\begin{figure}[htbp]
 \includegraphics{tree-example-2.pdf}
 \caption{Illustration of the proof of Theorem~\ref{thm:good_twins}. Left: A forest $G$ and a vertex $u$ adjacent to at least one leaf and at most one non-leaf. Middle: Good twins in the graph $G'$ obtained from $G$ by deleting all leaves adjacent to $u$. Right: Good twins in $G$ obtained by recoloring $u$ and $v$ and applying a $(u,v)$-move followed by a $(w,v)$-move -- c.f. Case~3.1.b. We use red circles to indicate members of $A$ and blue squares to indicate members of $B$.}
 \label{fig:tree-example}
\end{figure}

\begin{theorem}\label{thm:good_twins}
In every forest $G$ there exist good twins.
\end{theorem}
\begin{proof}

Let $G = (V,E)$ be a forest. We do induction on the number of edges of $G$. If $|E| = 0$, then $G$ consists of isolated vertices only and good twins of $G$ are given by $A = B = \emptyset$.

For the induction step let $u$ be any vertex in $G$ that is adjacent to at least one leaf of $G$ and to at most one non-leaf of $G$. We apply induction to the subgraph $G'$ of $G$ obtained by deleting all leaves adjacent to $u$ and obtain good twins $A$, $B$ of $G'$. Note that $u$ is either a leaf or an isolated vertex of $G'$.

What follows is a case analysis. In each case we will argue how to modify the good twins $A$, $B$ of $G'$ so as to obtain good twins of $G$. The modification of twins in each case is done in one or two steps called \textit{moves} that are defined for a pair of vertices.

Let $X$ and $Y$ be two sets of vertices, $x\in X$, $y\in Y$, for any vertex $v$, $L(v) = N(v) \setminus (X \cup Y)$. If $L(x)$ and $L(y)$ are disjoint, then an \textbf{$(x,y)$-move with respect to $X,Y$} (or an \textbf{$(x,y)$-move}, where the sets $X$ and $Y$ are understood) is an operation creating a pair of sets $X',Y'$ as follows:

\begin{itemize}
  \item Let $L'(x) \subseteq L(x)$ and $L'(y) \subseteq L(y)$ be sets with $|L'(x)| = |L'(y)| = \min\{|L(x)|,|L(y)|\}$.
  \item Let $X' = X\cup L'(y)$ and $Y' = Y\cup L'(x)$.
\end{itemize}

Note that if $X$ and $Y$ are twins and $L(x)$ and $L(y)$ contain only leaves of $G$, the $(x,y)$-move creates twins again. Moreover, all neighbors of at least one of $x,y$ are contained in the new twins $X' \cup Y'$. See Figure~\ref{fig:xy-move} for an illustration. We shall see an $(x,y)$-move as a map $f$ that takes two sets of vertices $X$, $Y$ and two vertices $x \in X$, $y \in Y$ and maps these to $X'$, $Y'$, defined as above, i.e., $f(x,y,X,Y) = (X',Y')$.

\begin{figure}[htbp]
 \includegraphics{xy-move.pdf}
 \caption{An $(x,y)$-move with respect to the sets $X$ of red vertices and the set $Y$ of blue vertices.}
 \label{fig:xy-move}
\end{figure}

Let $S \subseteq A$, $|S| \leq 2$, be the set of those vertices in $G'$ that have uncolored neighbors. Then in $G$ the set of non-leaf vertices with uncolored neighbors is given by $S \cup u$. We distinguish the cases $|S|=0$ (Case~1), $|S|=1$ (Case~2) and $|S|=2$ (Case~3), and the subcases $u \in A$, $u \in B$ and $u \notin A \cup B$. We apply $(x,y)$-moves to twins only, with $\{x,y\} \subseteq S \cup u$, each time decreasing the number of vertices that have uncolored neighbors by at least one. In most of the cases we end up with at most one vertex $x$ with uncolored neighbors, and to see that these twins are good twins it then suffices to check that $x$ is incident to at most one edge in $G[A]$ or $G[B]$. To be precise, we may have to swap the roles of $A$ and $B$ in the case where $x \in B$. However, due to simplicity of presentation we omit to mention this explicitly.

In many cases we do not apply $(x,y)$-moves immediately. Instead, we first color some uncolored vertices and/or recolor some leaves of the colored graph. In the first case, we always add to $A$ some vertex that has no neighbor in $A$ and to $B$ some  vertex that has no neighbor in $B$, i.e., we get twins again. If we recolor a leaf $x$, say $x \in A$, in the colored graph this has the following effects: $|A \setminus x| = |B \cup x| - 2$ and $e(A \setminus x) = e(B \cup x) - 1$. We repair this imbalance by either recoloring a leaf $y\in B$ in the colored graph or adding two uncolored vertices to $A$, one with no neighbor in $A$ and the other with exactly one neighbor in $A$. See Figure~\ref{fig:case2_4} and~\ref{fig:case2_5} for an example.

 \begin{itemize}
  \item \textit{Case 1 -- $|S|=0$:} If $u \notin A \cup B$, then $u$ is an isolated vertex of $G'$ and good twins of $G$ are given by $A \cup \{u\}$, $B \cup \{x\}$ for some vertex $x \in L(u)$. If $u \in A \cup B$ then $A$, $B$ are already good twins for $G$. In both cases, the new set $S$ is $\{u\}$.

\medskip

 \begin{figure}[htbp]
   \begin{center}
    \subfigure[Case~2.1 -- $u \in B$:\newline $(A',B') = f(w,u,A,B)$]{
     \includegraphics{case2_1.pdf}
     \label{fig:case2_1}
    }
    \\
    \hspace{2em}
    \subfigure[Case~2.2.a -- $u \in L(w)$:\newline $(A',B') = f(w,u,A \cup x,B \cup u)$]{
     \includegraphics{case2_2.pdf}
     \label{fig:case2_2}
    }
    \hspace{2em}
    \subfigure[Case~2.2.b -- $u$ has no colored neighbor:\newline $(A',B') = f(w,u,A \cup x, B \cup u)$]{
     \includegraphics{case2_3.pdf}
     \label{fig:case2_3}
    }
    \hspace{2em}
    \subfigure[Case~2.3.a -- $u \in A$, $|L(w)| \leq |L(u)|$ and its colored neighbor is in $A$:\newline $(A',B')=f(w,u,(A \setminus u)\cup \{x,y\}, B\cup u)$]{
     \includegraphics{case2_4.pdf}
     \label{fig:case2_4}
    }
    \hspace{2em}
    \subfigure[Case~2.3.b -- $u \in A$, $|L(w)| \leq |L(u)|$ and its colored neighbor is in $B$:\newline $(A',B')=f(w,u,(A \setminus u) \cup \{x,y\}, B\cup u)$]{
     \includegraphics{case2_5.pdf}
     \label{fig:case2_5}
    }
   \end{center}
   \caption{Illustration of Case~2 in the proof of Theorem~\ref{thm:good_twins}.}
   \label{fig:case2}
  \end{figure}

  \item \textit{Case 2 -- $|S|=1$:} Let $S = \{w\} \subseteq A$. We may assume, without loss of generality, that $w$ is a non-leaf of $G[A\cup B]$. Next, we distinguish several cases depending on whether $u \in A$, $u \in B$, or $u \notin A \cup B$. Figure~\ref{fig:case2} shows for each case how to obtain good twins for $G$ by (re)coloring some vertices and applying a $(w,u)$-move. The procedure is formally described as follows: \smallskip

      \begin{itemize}
        \item \textit{Case~2.1-- $u \in B$:} We perform a single $(w,u)$-move. \smallskip
        \item \textit{Case~2.2-- $u\not\in A\cup B$:} Either $u \in L(w)$ or $u$ was isolated in $G'$. In either case, we first add $u$ to $B$ and any $x \in L(u)$ to $A$. We obtain twins again, to which we perform a $(w,u)$-move. If $|L(w)|>|L(u)|$, then the set $S$ remains $\{w\}$. If $|L(w)|<|L(u)|$, then the roles of $A$ and $B$ are switched and $S$ becomes $\{u\}$. If $|L(w)|=|L(u)|$, then $S=\emptyset$. \smallskip
        \item \textit{Case~2.3 -- $u \in A$:} If $w$ is a non-leaf of $G'[A\cup B]$ and $|L(w)| > |L(u)|$, then $A$, $B$ are already good twins of $G$ and the new set $S$ is $\{w,u\}$. In fact, we may assume that $|L(w)| \leq |L(u)|$, since if $w$ is a leaf of $G'[A\cup B]$ and $|L(w)| > |L(u)|$, we can swap the roles of $w$ and $u$. First, we recolor $u$ (remove it from $A$ and add it to $B$), add some $x \in L(w)$ as well as some $y \in L(u)$ to $A$, and perform a $(w,u)$-move to the resulting twins. Figure~\ref{fig:case2_4} and~\ref{fig:case2_5} depicts the case that the colored neighbor of $u$ is in $A$ and $B$, respectively.\smallskip
      \end{itemize}
In Case~3 below when we recolor a vertex $x$ we do not distinguish the subcases whether the colored neighbor of $x$ is in $A$ or $B$, since the treatment is the same. We indicate in Figure~\ref{fig:case3} the first subcase only.

\medskip
 \begin{figure}[htbp]
   \begin{center}
    \subfigure[Case~3.1.a -- $u \in B$ and $|L(u)| \geq |L(v)|$:\newline $(A',B') = f(w,u,f(v,u,A,B))$]{
     \includegraphics{case3_1.pdf}
     \label{fig:case3_1A}
    }
    \hspace{1em}
    \subfigure[Case~3.1.b -- $u \in B$ and $|L(u)| < |L(v)|$:\newline $(A',B') = f(w,v,f(u,v,(A \setminus v) \cup u, (B \setminus u) \cup v))$]{
     \includegraphics{case3_2.pdf}
     \label{fig:case3_1B}
    }
    \hspace{2em}
    \subfigure[Case~3.2.a -- $u \in L(w)$, $|L(w)| \leq |L(u)| + |L(v)|$:\newline $(A',B') = f(w,v,f(w,u,(A \setminus v) \cup \{x,y,z\}, B \cup \{u,v\}))$\newline(The same is done if $u$ has no colored neighbor.)]{
     \includegraphics{case3_3A.pdf}
     \label{fig:case3_2A}
    }
    \hspace{2em}
    \subfigure[Case~3.2.b -- $u \in L(w)$, $|L(w)| > |L(u)| + |L(v)|$:\newline $(A',B') = f(w,u,A \cup x, B \cup u)$\newline(The same is done if $u$ has no colored neighbor.)]{
     \includegraphics{case3_3B.pdf}
     \label{fig:case3_2B}
    }
    \hspace{2em}
    \subfigure[Case~3.3.a -- $u \in L(v)$, $|L(u)| \geq |L(v)|$:\newline $(A',B') = f(w,u,f(v,u,A \cup x, B \cup u))$]{
     \includegraphics{case3_4.pdf}
     \label{fig:case3_3A}
    }
    \hspace{2em}
    \subfigure[Case~3.3.b -- $u \in L(v)$, $|L(u)| < |L(v)|$:\newline $(A',B') = f(w,v,f(u,v,(A \setminus v) \cup \{u,x\},B \cup v))$]{
     \includegraphics{case3_5.pdf}
     \label{fig:case3_3B}
    }
    \hspace{2em}
    \subfigure[Case~3.4 -- $u \in A$, $|L(u)| \geq |L(v)|$:\newline $(A',B') = f(w,u,f(v,u,(A \setminus u) \cup \{x,y\},B \cup u))$]{
     \includegraphics{case3_6.pdf}
     \label{fig:case3_4}
    }
   \end{center}
   \caption{Illustration of Case~3 in the proof of Theorem~\ref{thm:good_twins}.}
   \label{fig:case3}
  \end{figure}

  \item \textit{Case 3 -- $|S|=2$:} By the definition of good twins we have $S = \{v,w\} \subseteq A$, $v$ a leaf and $w$ a non-leaf of $G'[A\cup B]$, and $|L(v)| < |L(w)|$. We distinguish, respectively, whether $u\in B$, $u\in L(w)$, $u\in L(v)$ or $u\in A$. We further distinguish subcases depending on the sizes of $L(u)$, $L(v)$ and $L(w)$. The treatment for each case is shown in Figure~\ref{fig:case3}; the description follows. \smallskip
            \begin{itemize}
      \item \textit{Case~3.1.a-- $u \in B$ and $|L(u)| \geq |L(v)|$:} We perform a $(v,u)$-move and afterwards a $(w,u)$-move. Note that if $|L(u)|=|L(v)|$, i.e., $u$ has no uncolored neighbors after the $(v,u)$-move, the second move has no effect. \smallskip
      \item \textit{Case~3.1.b-- $u \in B$ and $|L(u)| < |L(v)|$:} In particular, this means $|L(u)| < |L(w)|$. Then we swap the colors of $u$ and $v$, i.e., remove $u$ from $B$ and add it to $A$ as well as removing $v$ from $A$ and adding it to $B$. Since both $u$ and $v$ are leaves in the colored graph, we obtain twins again. Then apply a $(u,v)$-move followed by a $(w,v)$-move. \smallskip
      \item \textit{Case~3.2.a-- either $u \in L(w)$ or $u$ was  isolated in $G'$ and $|L(w)| \leq |L(u)|+|L(v)|$:} We add $u$ to $B$ and some $x \in L(u)$ to $A$, obtaining twins again. If $|L(w)|<|L(u)|$, then perform a $(w,u)$-move followed by a $(v,u)$-move. If $|L(w)|\geq |L(u)|$, then, recolor $v$ and add some $y \in L(u)$ (if $|L(u)| = 1$ take some $y \in L(v)$ instead) and $z \in L(w)$ to $A$. To the resulting twins we apply a $(w,u)$-move followed by a $(w,v)$-move. \smallskip
      \item \textit{Case~3.2.b-- either $u \in L(w)$ or $u$ was uncolored and isolated in $G'$ and $|L(w)|>|L(u)|+|L(v)|$:} We again add $u$ to $B$ and some $x \in L(u)$ to $A$. To the resulting twins apply a $(w,u)$-move only. Note that in this case, both $v$ and $w$ still have uncolored neighbors. \smallskip
      \item \textit{Case~3.3.a-- $u \in L(v)$ and $|L(v)|\leq |L(u)|$:} We add $u$ to $B$ and some $x \in L(u)$ to $A$, and then perform a $(v,u)$-move followed by a $(w,u)$-move. \smallskip
      \item \textit{Case~3.3.b-- $u \in L(v)$ and $|L(v)| > |L(u)|$:} We recolor $v$ and compensate the imbalanced edge-count and vertex-count by adding $u$ as well as some $x\in L(u)$ to $A$. We perform a $(u,v)$-move followed by a $(w,v)$-move. \smallskip
      \item \textit{Case~3.4-- $u \in A$:}  Assume, without loss of generality, that $|L(u)| \geq |L(v)|$ (otherwise swap the roles of $u$ and $v$). We recolor $u$ and add some $x \in L(u)$ as well as some $y \in L(v)$ to $A$. This produces twins to which we perform a $(v,u)$-move and afterwards a $(w,u)$-move. \smallskip
      \end{itemize}
 \end{itemize}
 It is straightforward to check that our treatment in each of the above cases results in good twins $A',B'$ of $G$. The formal definition of $(A',B')$ in Case~2 and Case~3 is given in Figure~\ref{fig:case2} and Figure~\ref{fig:case3}, respectively.
\end{proof}

Having Theorem~\ref{thm:good_twins}, we can prove Theorem~\ref{thm:forest} quite easily.

\begin{proof}[Proof of Theorem~\ref{thm:forest}]
 Let $G = (V,E)$ be a forest. By Theorem~\ref{thm:good_twins}, $G$ admits good twins $A$, $B$. The vertices in $V \setminus (A\cup B)$, i.e., the uncolored vertices, come in three groups, each of which may be empty. The first group $S_1$ contains all isolated vertices of $G$. The second group $S_2$ consists of some leaves of $G$ that are all adjacent to a leaf $v\in A$ of $G[A\cup B]$. The third group $S_3$ consists of some leaves of $G$ that are all adjacent to a non-leaf $w \in A$ of $G[A\cup B]$. Moreover, if $S_3 \neq \emptyset$, then $|S_3| > |S_2|$.\smallskip

\begin{itemize}
  \item \textit{Case 1 -- $S_2 = S_3 = \emptyset$:} Partition $S_1$ into equal-sized subsets by removing at most one vertex, and add one subset to each of $A$, $B$. \smallskip

  \item \textit{Case 2 -- exactly one of $S_2$, $S_3$ is non-empty:}   We will delete the corresponding vertex $v$ or $w$ from $A$, say it's $w$. I.e., we will uncolor $w$ but color some of its neighbors.

  If $w$ has no neighbor in $A$, then uncolor $w$ and add all but at most one of the remaining uncolored vertices to $A$ and to $B$ evenly so that the resulting sets are twins of size $\lceil n/2\rceil-1$. So, we may assume that $w$ has a neighbor in $A$ (exactly one, by the condition of twins being good).

  It is easy to see that $G$ has a leaf in $B$. For example, contracting all edges in $G[A]$ and in $G[B]$ and removing the uncolored vertices in $S$ gives a forest $\tilde{G}$ whose bipartition classes are induced by $A$ and $B$ and are of equal size. Every leaf of $\tilde{G}$ in $B$ corresponds to at least one leaf of $G$ in $B$ and every component of $\tilde{G}$ with at least as many vertices in $B$ as in $A$ has a leaf in $B$.
  So let $u \in B$ be a leaf of $G$. Then uncolor $w$ and recolor $u$ to be in $A$. Add all but at most one of the remaining uncolored vertices to $A$ and to $B$ evenly so that the resulting sets are twins of size $\lceil n/2\rceil-1$.

   \item \textit{Case 3 -- $S_2 \neq \emptyset$ and $S_3 \neq \emptyset$:} If $w$ has a neighbor in $A$, then add some vertex $x \in S_2$ to $A$. If $w$ has no neighbor in $A$, then and some vertex $y \in S_3$ to $A$. Uncolor $w$. These are twins again.

       Next, add all vertices in $S_2$, except possibly $x$, to $B$, and add the same number of vertices from $S_3$ to $A$. (Note that $|S_3| > |S_2|$ and hence $|S_3 \setminus y| \geq |S_2|$.) In the resulting twins, no uncolored vertex has a colored neighbor and thus we can find twins of $G$ by removing at most one further vertex. In any case, the graph $G$ contains at most 2 uncolored vertices and the colors form twins.
  \end{itemize}
This completes the proof.
\end{proof}

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


Lingual braces were designed as an alternative to the traditional 'train track' braces we are all too familiar with. Lingual braces North London have the same look and structure of the traditional brace but are fitted behind the teeth so they cannot be seen.
Lingual braces or invisible are just as effective in straightening your teeth as other methods but have the great advantage of not imposing on your social situations.
The results of using lingual braces are fantastic and give you perfectly straight teeth. Lingual braces are one of the most common braces used by adults wanting to invest in their oral hygiene without a big impact on their lives because of the way they are hidden behind the teeth. Lingual braces are so subtle that no one will know you have braces in, as a result of this they are used among professionals whose day to day appearance is paramount to their lives and career.


text:
Dierenfeldt was patrolling on Interstate 635, northbound just past State Avenue, when he saw a weaving car, the driver unable to stay in the lanes.
He stopped the car knowing the driver needed to be saved, as well as any motorist or bystander who might have crossed his path. It turned out the driver was a father who had his daughter in the car. And she had been "praying on that Bible, she was so scared."
He thought about her, he said, as he prepared to open Saturday's ceremony by singing the national anthem.
Azeneth Rodriguez wasn't as blessed as the 12-year-old, who escaped harm.
She got into a car with a drunken driver on Oct. 26, 2014. No one stopped that driver, who collided with another vehicle on the Heart of America Bridge. Rodriguez was killed at the scene.
The driver was Rodriguez's boyfriend, said her sister, Ashley Rodriguez.
"She thought she could stop him," said Ashley Rodriguez, who said her sister was hopeful that she could stop him from driving recklessly after he left a bar. "She thought she could save his life and someone else's life."
Ashley Rodriguez came to walk with other friends and family and remember her sister.
MADD supporters know how many die — 308 in Missouri and Kansas alone in 2014, and 10,265 nationwide in 2015, a rate of 28 people killed every day.
They can only guess how many lives the Mothers Against Drunk Driving crusade has saved since 1980, when Candy Lightner started the effort after her 13-year-old daughter, Cari, was killed by a drunken driver while walking to a church carnival.
They know, as national board member Chris Mann told Saturday's crowd, that ignition interlock devices have stopped 1.6 million impaired driving attempts nationwide. The devices, championed by MADD and used by increasing numbers of jurisdictions, prevent a car from starting if the alcohol in a driver's breath exceeds a programmed level.
Some of the research cited by MADD projects that the group may have saved as many as 300,000 lives since 1980.
But people still die. Many of their names were on signs and stickers and shirts as the people who remember them walked to raise funds for the cause.
Joe Robertson: 816-234-4789, @robertsonkcstar
Shoes formed the number 308, representing the victims of drunk driving in Kansas and Missouri in 2014, at the Mothers Against Drunk Driving KC Heartland Chapter's second annual Walk Like MADD & MADD Dash Saturday at Theis Park. SUSAN PFANNMULLER Special to the Star
Ashley Rodriguez, (front right) and members of her extended family walked in honor of Ashley's sister Azeneth Rodriguez, who was killed in a drunk driving crash in 2014. Rodriguez was taking part in the Mothers Against Drunk Driving KC Heartland Chapter's second annual Walk Like MADD & MADD Dash Saturday at Theis Park. SUSAN PFANNMULLER Special to the Star
Co-founders of the Mothers Against Drunk Driving KC Heartland Chapter's annual Walk Like MADD & MADD Dash, Leann Lewis (left) and Annette Murray, spoke at the opening ceremony Saturday at Theis Park. SUSAN PFANNMULLER Special to the Star
Kansas City, Kansas police officer, Dustin Dierenfeldt, sang the national anthem at the Mothers Against Drunk Driving KC Heartland Chapter's second annual Walk Like MADD & MADD Dash Saturday at Theis Park. SUSAN PFANNMULLER Special to the Star
Taylor Hirth talks about the sentencing of man who raped Johnson County deputy
Johnson County DA happy with sentencing of man who raped a sheriff's deputy
Two men die in grain silo as Ohio firefighters rush to dig them out, officials say

redpajama_set_name: RedPajamaCommonCrawl


Our clients are Hollywood Studios, Distributors in the TV, Film and Music industry. Our team specializes in digital content strategy, distribution, monetization including collecting, reporting, distributing of all royalties and IP protection for content creators worldwide. Please let us if we can help.
Do you own exclusive rights to this content?
Yes. I own exclusive rights to this content.No. I do not own exclusive rights to this content.
What services you are looking for?


text:
Nissan sets goal to sell more than 1.5 million hydrogen-powered cars by the end of 2016
November 19, 2011	0 By Stephen Vagus
Japanese automaker Nissan has been making waves in the hydrogen fuel industry lately.
Earlier this month, the company unveiled a next-generation fuel cell that is twice as powerful as conventional models while being less expensive. While the company's previous ventures in alternative energy have revolved around plug-in vehicles, Nissan has recently adopted hydrogen as its focus. The company has high expectations for the fuel, declaring that it will sell more than 1.5 million hydrogen-powered vehicles by 2016.
The ambitious claim may sound like bravado, but Nissan's hydrogen-powered vehicles seem to be able to back up the claim. Powered by the company's new fuel cells, the vehicles can travel twice the distance of other hydrogen-powered cars without having to refuel. This means that the cars are less burdened by the lack of an adequate fuel infrastructure than their competitors.
While hydrogen is getting a great deal of attention from the company, Nissan is not abandoning their battery-powered cars.
The company believes that their popular plug-in vehicle, the Leaf, will continue seeing success, especially as upgrades to the lithium-ion battery system are being made.
Together, Nissan's battery and fuel cell cars are being called the Green Program, which essentially encompasses the entirety of the company's alternative energy ventures. Nissan hopes to be a leader in the transportation industry as hydrogen continues to gain momentum around the world.
CategoryBusiness Featured News Industry International Transportation
Tagsbattery powered cars electric cars hydrogen fuel hydrogen technology Nissan nissan Green Program plug in vehicles
Super skin material may unlock the future of alternative energy and medical progress
Ballard Power sees success in South Korea with innovative fuel cell system


When Wynand & Sarah's wedding day dawned gray and cloudy, we were initially a little worried that the weather might steal some of their joy. But we couldn't have been more wrong… especially if you have the dear Sarah twirling around in her dress, ready hours before the ceremony started! These two were literally rays of sunshine thought the clouds, full of smiles and willing to climb mountains for us, kiss in the rain and let the wind rummage thought their hair. It was such a blessing to see their love for one another shining through the clouds!
Despite the moody skies and plays of lightning, Wynand & Sarah's wedding was absolutely gorgeous and pulled at everyone's heartstings! We are so grateful that we were there to capture all of it, from the deeply spiritual ceremony, Sarah's little girl letter " to my future husband" that made it into Wynand's hands after being written many years ago and not to mention the most delightful reception and fabulous cake table.Here are a few of our favourite images from this one-of-a-kind wedding at the beautiful @Boshoek just outside Lydenburg.


A haze settles on Seattle-Hempfest Weekend is upon us!
Groovy times on the Seattle waterfront this weekend at Hempfest! We live near the festival so we just had to check it out on our walk home after shopping downtown at two of our favorite stores, World Market and Mud Bay. Thor was needing a big boy diaper so a trip to Mud Bay was required–he's one year old now and still prone to accidents so on goes the protection until such time as he can control his little self. But enough about Thor's incontinence and on to a more popular topic, Hempfest!
Hop aboard the Canna-bus and take a ride! Guaranteed you'll smoke a pipe!
Trying to avoid the line to herd people into security for bag check, we tried skirting Hempfest crowd to walk along the waterfront. Quickly stopped by a friendly Staffing Pro gal, her supervisor immediately intervened after taking one look at us and waved us by without further ado. Guess we didn't fit the Hempfest profile and he could instantly ascertain we were just wandering by on our way home with some purchases. Glad to keep on walking (my Fitbit is such a step whore–10,000 or more to be happy!) we quickly fell into the crowd walking as slow as humanly possible so they could smoke, eat and chat while strolling. Amazing how slow you can be when stoned I guess because I was weaving in and out like a madwoman just trying to move along. Seeing the food truck featuring fried PB&J and other fried delicacies was just info overload for their weed-laden brains. Squirrel!
"Is weed really legal to just smoke in public?" I innocently asked Thom. He explained that while it is not technically legal to light up in general, there certainly weren't enough cops in Seattle to enforce any type of law so unless there was trouble, it was pretty much a smokefest of weed in the park. CHILL! The Canna-bus was quite the attraction selling branded apparel and booking events on the weed party bus. Just had to pose for a picture with it. Bet they have to have lots of fridges and microwaves in that bus to crank out the munchies!
James just got a pt job with a security company and his first assignment was yesterday in the pouring rain and lightning checking bags at Hempfest for alcohol, which was NOT allowed but a small handful of weed was cool. Interesting standards but at least the crowd was subdued vs. hyped up and crazy. After the worst working day of his life, James crawled home for hot food and rest before another fun day keeping the peace at Hempfest. Jobs like these will help him stay focused on excelling at law school, which he starts next week. I'm super proud he persevered through a ten hour day of Hell. He is surely my son with that type of work ethic!
As the haze settles on Seattle, we are looking forward to seeing how Amsterdam embraces their pot culture by comparison to Seattle. We leave in four weeks for our Netherlands adventure and I cannot wait!


Published 04/25/2019 12:22:10 pm at 04/25/2019 12:22:10 pm in Aluminum Overhang.
aluminum overhang tapcons or wedge anchors to secure aluminum columns to concrete slab for patio overha roofing pool house remodeling decorating construction gutters for 35 metal roof overhang.
how to extend metal roof overhang on carport,wood overhang kits,metal roof overhang,extending metal roof overhang,how to clean aluminum overhang,metal patio overhang,paint aluminum overhang for door,wood overhang for door,metal patio overhang designs,hanging lights from an aluminum overhang,gutters for 3.5 metal roof overhang.


MRU Services can provide a range of services to support Television and Film Productions. We supply registered healthcare professionals, both Nurses and Paramedics, covering Unit Nurse activities, rigging / de-rigging, set building and Stunt cover, supporting crew and artistes throughout their activities on set.
Contact us to discuss the requirements of your production.
Equipped 4×4 response vehicle to support the Unit Nurse on location.
A wide range of equipment and medications, to support the Unit Nurse on location or at a studio.
We support artistes and crew in a proactive manner, maintaining the health of all, and minimising the risk of 'down time' through unexpected lack of cast or key personnel.
For a complete list of our previous productions go to our TV Production Support page.


Description: The Classic 250 MPPT Charge Controller has Arc Fault, Ground Fault, free web monitoring, a graphics panel and Solar, Hydro and Wind Modes. The Classic 250 has a maximum output current of 63 amps, an operating voltage of 250 volts and works with 12 to 72 volt battery systems. The FREE MidNite Solar Local Application enables you to monitor your Classics over your local network or the Internet. As with the whole Classic Line, the Classic 250 has MidNite's exclusive HyperVOC which extends VOC limits when needed. The Classic is the only MPPT charge controller that has Arc Fault Detection, making this controller the safest controller available.


Moncler Jackets Men past Half a century roughly Moncler provides sustained the position available for sale and can be a global quality brand that provides apparel and also accessories to be able to its precious customers. Moncler outdoor jackets, vests, neckties, boots, bags etc come in huge demand because of the unmatched high quality and classy designs. Moncler straight down jackets certainly are a rage within the world of fashion and they are in sought after. Heavy excellent skiing conditions and intense weather have caused that it is a must to defend canadian arctic jackets do-it-canadian arctic jacketsself adequately. Moncler outdoor jackets are lightweight and provide warmth hence ensuring any cozy winter to suit canadian arctic jackets needs. The tone-on-tone quilting stitches ensures virtually no broken seams for many years. Made with all of the special polyamide liner and authentic down completing, Moncler outdoor jackets are the better companion in the course of winters. Moncler straight down jackets make use of a double top zipper regarding added defense. Depending on canadian arctic jackets own tastes canadian arctic jackets can pick outdoor jackets in sound shades regarding brown, beige, moncler coats black or perhaps white or pick from some radiant colors which exist too. A highly liked amongst kids, Moncler Jackets outdoor jackets offers unique designs as well as unmatched top quality. Moncler shoes are slightly less popular than Moncler lower jackets yet are strongly suggested by customers whove purchased a single. They may be purchased in bright colors and so Moncler Jackets Women are light around the feet preserving canadian arctic jackets comfortable constantly. Moncler provides incorporated the most recent technology inside manufacturing the sandals in which ensure highest comfort in the feet even if worn for extended stays. The radiant hues certainly are a perfect summer months shade in which go beautifully along with canadian arctic jackets floral summer months canadian arctic jacketses and also completes canadian arctic jackets lifestyle. The good thing about most of these sandals is that they are incredibly resilient too. moncler sale Moncler shoes top is made of leather together with nylon along with the inner is going to be of cotton allowing adequate ventilation in canadian arctic jackets feet although wearing these. Moncler provides priced the sandals reasonably so that it is reasonably priced lv bags to become bought simply by its consumers. By washing these sandals canadian arctic jackets will not sweat at the same time. All canadian arctic jackets will require is any dry organic cotton cloth to completely clean the sandals and have them in the cool moisture-free spot for a dry. Extremely low priced in servicing, Moncler shoes are canadian arctic jackets better friend throughout summers. Aside from canadian arctic jackets winter assortment Moncler boasts smart natural cotton T-shirts, hand bags, boots as well as scarves to.
Moncler Coats meet canadian arctic jackets requirements. The boot styles look extremely smart as well as charming whenever teamed upwards with jeans plus a sleek jacket. Add on the scarf as well as snazzy bag to do the seem. Moncler is aware of the transforming needs and also requirements regarding its customers and thus designs the products to provide nothing nevertheless the best.


text:
D'acord amb la mitologia grega, Penteu (en grec antic Πενθεύς), va ser un rei de Tebes, fill d'Equíon, un dels Esparts, i d'Agave, una de les filles de Cadme.

La tradició més freqüent el fa hereu directe de Cadme, quan aquest, molt vell, va abdicar al seu favor. Una altra tradiació diu que entre Cadme i Penteu va regnar un fill de Cadme, Polidor, que Penteu va destronar.

La història de Penteu està relacionada amb el culte a Dionís. Com se sap, Dionís és un déu d'origen tebà, i per part de la seva mare, Sèmele, cosí de Penteu. Després de conquerir Àsia, el déu va decidir tornar a Tebes, la seva pàtria, i instaurar el seu culte, també per castigar les germanes de la seva mare, especialment Agave, per les calúmnies que havien llançat, un temps enrere, contra Sèmele. Després de passar Tràcia, Dionís va arribar a Tebes, on va fer embogir les dones, que van vestir-se com a bacants i van fugir a la muntanya a celebrar els misteris del déu. Però Penteu, tot i els advertiments de Cadme i de Tirèsias, es vol oposar a la propagació d'aquest culte violent. Es dirigeix a Dionís i el tracta de xerraire i d'impostor, tot i que el déu realitza alguns miracles, i el vol fer encadenar. El déu es deslliga de les seves cadenes i incendia el palau reial. El déu suggereix a Penteu que vagi en persona al mont Citeró, la muntanya de Tebes, per espiar les dones i ser testimoni dels excessos als quals es lliuren. Penteu escolta el consell i s'amaga darrere un pi. Però les dones el descobreixen, arrenquen el pi, i quan el tenen sota el seu poder, esquarteren el seu cos. Agave és la primera a agredir-lo, li agafa el cap i el clava a la punta d'un tirs. Torna a Tebes, passejant orgullosa el que ella creu que és el cap d'un lleó. Cadme la treu del seu error i Agave s'adona que ha matat el seu propi fill i no una fera.

Aquest mite va ser portat a escena per Eurípides i per Esquil. Va ser molt conegut en la literatura i l'art antics i se li atribuïa una significació religiosa, ja que Penteu era el prototip de l'impiu, castigat pel seu orgull.

Referències

Bibliografia
 Parramon i Blasco, Jordi: Diccionari de la mitologia grega i romana. Barcelona: Edicions 62, 1997, p. 171. (El Cangur / Diccionaris, núm. 209).

Reis de Tebes

redpajama_set_name: RedPajamaWikipedia


text:
Q: What is the usual meaning of the notation $ \| f \|_{C^k (A)}$? If $f \in C^k ( \mathbb R)$ and $A \subset \mathbb R$, what is the usual meaning of the notation below?
$$ \| f \|_{C^k (A)} $$

A: Assuming you meant $A\subseteq\mathbb{R}$, the notation usually means
$$\|f\|_{C^k(A)}=\|f\|+\|f'\|+\cdots+\|f^{(k)}\|$$
where $f^{(i)}$ is the $i$th derivative of $f$, and where for a function $g:A\to\mathbb{R}$, the notation $\|g\|$ refers to the sup norm:
$$\|g\|=\sup_{x\in A} \,|g(x)|$$
I've also seen the definition
$$\|f\|_{C^k(A)}=\max\left\{\|f\|,\|f'\|,\ldots,\|f^{(k)}\|\right\}$$
which (if I remember correctly) induces the same topology.

redpajama_set_name: RedPajamaStackExchange


All the comodities of a town meet the adventures nature preserves for you. And on top of this you get a breathtaking view of the Dolomites! That' s the deal.
Behind the mighty towers of the Cathedral in Brixen, the oldest town in Tyrol, the mountain Plose rises up with all its chances and offers of living and enjoying nature.
Today Brixen counts more than 21.000 inhabitants and is one of the biggest communities in South Tyrol (after Bolzano/Bozen and Merano/Meran). According to the census of 2011 73% of the inhabitants speak German as their mother language, 26% speak Italian and 1% speak an old Roman language, Ladin.
A town of culture and congresses Brixen attracts many visitors every year. The romantic old town with its magnificent cathedral and the unique square in front of it, the protected large-scale historic centre dating back to the Middle Ages, the porticos and lanes decorated with merlons, museum treasures, craft tradition and archaeological exhibits are a real treasure for those who love culture and arts and are definitely worth a visit. Thanks to climbing hall, fun swimming pool Aquarena and other sports facilities even sport freaks get a great deal.
Services and Tourism are the biggest employers with almost 68% of all employees, followed by Industry (19%), Crafts (7%) and Agriculture (6%).
Vacation in town or in the mountains? Why not both?
What does perfect holidays look like? There is only one thing that is sure: perfect holidays in South Tyrol feel and look different for anyone. In order to make individual wishes come true, you need chances to live your day the way you want to. Brixen Plose offers a whole lot of possibilities to spend your day exactly the way you enjoy it.
Active and challenging: The area Brixen Plose reaches from 560 m a.s.l. in Brixen up to a height of 2.542 m a.s.l. on the Plose. In between there are many chances to be active or to enjoy a quiet day in summer as well as in wintertime. With the sun shining on 300 days a year, the pleasure should be yours.
If you want to leave your car in the garage, please feel free to do so. Thanks to public transport you will always get where you need to go and you can easily explore the surroundings of Brixen and the Plose. The valley station of the gondola lift Plose is only 7 km away from Brixen.
This guestcard is the perfect proof of the connection between town an mountain. It offers you a whole range of advantages, as for example a free go and return trip with one of the gondola lifts Plose every day, or the free use of public transport all over South Tyrol. For more information please click here.


text:
// ------------------------------------------------------------------------------
// <auto-generated>
//     This code was generated by a tool.
//     Runtime Version: 15.0.0.0
//
//     Changes to this file may cause incorrect behavior and will be lost if
//     the code is regenerated.
// </auto-generated>
// ------------------------------------------------------------------------------
namespace King.Mapper.Generator.Templates
{
    using System.Linq;
    using System;

    /// <summary>
    /// Class to produce the template output
    /// </summary>

    #line 1 "C:\Users\jefkin\Documents\GitHub\King.Mapper.Generator\King.Mapper.Generator.Core\Templates\StoredProcedures.tt"
    [global::System.CodeDom.Compiler.GeneratedCodeAttribute("Microsoft.VisualStudio.TextTemplating", "15.0.0.0")]
    public partial class StoredProcedures : StoredProceduresBase
    {
#line hidden
        /// <summary>
        /// Create the template output
        /// </summary>
        public virtual string TransformText()
        {
            this.Write(@"// ------------------------------------------------------------------------------
// <auto-generated>
//     This code was generated by a tool.
//     Runtime Version: 1.0.2.7
//
//     Changes to this file may cause incorrect behavior and will be lost if
//     the code is regenerated.
// </auto-generated>
// ------------------------------------------------------------------------------
namespace King.Mapper.Generated.Sql
{
    using System;
    using System.Data;
    using King.Mapper.Data;

");

            #line 18 "C:\Users\jefkin\Documents\GitHub\King.Mapper.Generator\King.Mapper.Generator.Core\Templates\StoredProcedures.tt"

foreach (var proc in this.Manifest)
{

            #line default
            #line hidden
            this.Write("    /// <summary>\r\n    /// Class that Represents ");

            #line 23 "C:\Users\jefkin\Documents\GitHub\King.Mapper.Generator\King.Mapper.Generator.Core\Templates\StoredProcedures.tt"
            this.Write(this.ToStringHelper.ToStringWithCulture(proc.Preface));

            #line default
            #line hidden
            this.Write(".");

            #line 23 "C:\Users\jefkin\Documents\GitHub\King.Mapper.Generator\King.Mapper.Generator.Core\Templates\StoredProcedures.tt"
            this.Write(this.ToStringHelper.ToStringWithCulture(proc.Name));

            #line default
            #line hidden
            this.Write(" Stored Procedure\r\n    /// </summary>\r\n\tpublic partial class ");

            #line 25 "C:\Users\jefkin\Documents\GitHub\King.Mapper.Generator\King.Mapper.Generator.Core\Templates\StoredProcedures.tt"
            this.Write(this.ToStringHelper.ToStringWithCulture(proc.Preface+proc.Name));

            #line default
            #line hidden
            this.Write(" : IStoredProcedure\r\n\t{\r\n        /// <summary>\r\n        /// Gets Stored Proc name" +
                    " with Schema\r\n        /// </summary>\r\n\t\tpublic virtual string FullyQualifiedName" +
                    "()\r\n\t\t{\r\n\t\t\treturn \"[");

            #line 32 "C:\Users\jefkin\Documents\GitHub\King.Mapper.Generator\King.Mapper.Generator.Core\Templates\StoredProcedures.tt"
            this.Write(this.ToStringHelper.ToStringWithCulture(proc.Preface));

            #line default
            #line hidden
            this.Write("].[");

            #line 32 "C:\Users\jefkin\Documents\GitHub\King.Mapper.Generator\King.Mapper.Generator.Core\Templates\StoredProcedures.tt"
            this.Write(this.ToStringHelper.ToStringWithCulture(proc.Name));

            #line default
            #line hidden
            this.Write("]\";\r\n\t\t}\r\n\r\n");

            #line 35 "C:\Users\jefkin\Documents\GitHub\King.Mapper.Generator\King.Mapper.Generator.Core\Templates\StoredProcedures.tt"

if (null != proc.Variables && 0 < proc.Variables.Count())
{

            #line default
            #line hidden
            this.Write("\t\t#region Parameters\r\n");

            #line 40 "C:\Users\jefkin\Documents\GitHub\King.Mapper.Generator\King.Mapper.Generator.Core\Templates\StoredProcedures.tt"

foreach (var parameter in proc.Variables)
{
	var paramName = parameter.ParameterName.Replace("@", string.Empty);
	if (!string.IsNullOrWhiteSpace(paramName))
	{
		var dataTypeCSharp = DataTypeMappings.DataTypeCSharp(parameter.DataType);
		var dataTypeDb = DataTypeMappings.DataTypeDbType(parameter.DataType);

            #line default
            #line hidden
            this.Write("        /// <summary>\r\n        /// Gets or sets the parameter value\r\n        /// " +
                    "</summary>\r\n\t\t[DataMapper(\"@");

            #line 52 "C:\Users\jefkin\Documents\GitHub\King.Mapper.Generator\King.Mapper.Generator.Core\Templates\StoredProcedures.tt"
            this.Write(this.ToStringHelper.ToStringWithCulture(paramName));

            #line default
            #line hidden
            this.Write("\", ");

            #line 52 "C:\Users\jefkin\Documents\GitHub\King.Mapper.Generator\King.Mapper.Generator.Core\Templates\StoredProcedures.tt"
            this.Write(this.ToStringHelper.ToStringWithCulture(dataTypeDb));

            #line default
            #line hidden
            this.Write(")]\r\n\t\tpublic virtual ");

            #line 53 "C:\Users\jefkin\Documents\GitHub\King.Mapper.Generator\King.Mapper.Generator.Core\Templates\StoredProcedures.tt"
            this.Write(this.ToStringHelper.ToStringWithCulture(dataTypeCSharp));

            #line default
            #line hidden
            this.Write(" ");

            #line 53 "C:\Users\jefkin\Documents\GitHub\King.Mapper.Generator\King.Mapper.Generator.Core\Templates\StoredProcedures.tt"
            this.Write(this.ToStringHelper.ToStringWithCulture(paramName));

            #line default
            #line hidden
            this.Write("\r\n\t\t{\r\n\t\t\tget;\r\n\t\t\tset;\r\n\t\t}\r\n\r\n");

            #line 59 "C:\Users\jefkin\Documents\GitHub\King.Mapper.Generator\King.Mapper.Generator.Core\Templates\StoredProcedures.tt"

	}
}

            #line default
            #line hidden
            this.Write("\t\t#endregion\r\n");

            #line 64 "C:\Users\jefkin\Documents\GitHub\King.Mapper.Generator\King.Mapper.Generator.Core\Templates\StoredProcedures.tt"

}

            #line default
            #line hidden
            this.Write("\t}\r\n\r\n");

            #line 69 "C:\Users\jefkin\Documents\GitHub\King.Mapper.Generator\King.Mapper.Generator.Core\Templates\StoredProcedures.tt"

}

            #line default
            #line hidden
            this.Write("}");
            return this.GenerationEnvironment.ToString();
        }
    }

    #line default
    #line hidden
    #region Base class
    /// <summary>
    /// Base class for this transformation
    /// </summary>
    [global::System.CodeDom.Compiler.GeneratedCodeAttribute("Microsoft.VisualStudio.TextTemplating", "15.0.0.0")]
    public class StoredProceduresBase
    {
        #region Fields
        private global::System.Text.StringBuilder generationEnvironmentField;
        private global::System.CodeDom.Compiler.CompilerErrorCollection errorsField;
        private global::System.Collections.Generic.List<int> indentLengthsField;
        private string currentIndentField = "";
        private bool endsWithNewline;
        private global::System.Collections.Generic.IDictionary<string, object> sessionField;
        #endregion
        #region Properties
        /// <summary>
        /// The string builder that generation-time code is using to assemble generated output
        /// </summary>
        protected System.Text.StringBuilder GenerationEnvironment
        {
            get
            {
                if ((this.generationEnvironmentField == null))
                {
                    this.generationEnvironmentField = new global::System.Text.StringBuilder();
                }
                return this.generationEnvironmentField;
            }
            set
            {
                this.generationEnvironmentField = value;
            }
        }
        /// <summary>
        /// The error collection for the generation process
        /// </summary>
        public System.CodeDom.Compiler.CompilerErrorCollection Errors
        {
            get
            {
                if ((this.errorsField == null))
                {
                    this.errorsField = new global::System.CodeDom.Compiler.CompilerErrorCollection();
                }
                return this.errorsField;
            }
        }
        /// <summary>
        /// A list of the lengths of each indent that was added with PushIndent
        /// </summary>
        private System.Collections.Generic.List<int> indentLengths
        {
            get
            {
                if ((this.indentLengthsField == null))
                {
                    this.indentLengthsField = new global::System.Collections.Generic.List<int>();
                }
                return this.indentLengthsField;
            }
        }
        /// <summary>
        /// Gets the current indent we use when adding lines to the output
        /// </summary>
        public string CurrentIndent
        {
            get
            {
                return this.currentIndentField;
            }
        }
        /// <summary>
        /// Current transformation session
        /// </summary>
        public virtual global::System.Collections.Generic.IDictionary<string, object> Session
        {
            get
            {
                return this.sessionField;
            }
            set
            {
                this.sessionField = value;
            }
        }
        #endregion
        #region Transform-time helpers
        /// <summary>
        /// Write text directly into the generated output
        /// </summary>
        public void Write(string textToAppend)
        {
            if (string.IsNullOrEmpty(textToAppend))
            {
                return;
            }
            // If we're starting off, or if the previous text ended with a newline,
            // we have to append the current indent first.
            if (((this.GenerationEnvironment.Length == 0)
                        || this.endsWithNewline))
            {
                this.GenerationEnvironment.Append(this.currentIndentField);
                this.endsWithNewline = false;
            }
            // Check if the current text ends with a newline
            if (textToAppend.EndsWith(global::System.Environment.NewLine, global::System.StringComparison.CurrentCulture))
            {
                this.endsWithNewline = true;
            }
            // This is an optimization. If the current indent is "", then we don't have to do any
            // of the more complex stuff further down.
            if ((this.currentIndentField.Length == 0))
            {
                this.GenerationEnvironment.Append(textToAppend);
                return;
            }
            // Everywhere there is a newline in the text, add an indent after it
            textToAppend = textToAppend.Replace(global::System.Environment.NewLine, (global::System.Environment.NewLine + this.currentIndentField));
            // If the text ends with a newline, then we should strip off the indent added at the very end
            // because the appropriate indent will be added when the next time Write() is called
            if (this.endsWithNewline)
            {
                this.GenerationEnvironment.Append(textToAppend, 0, (textToAppend.Length - this.currentIndentField.Length));
            }
            else
            {
                this.GenerationEnvironment.Append(textToAppend);
            }
        }
        /// <summary>
        /// Write text directly into the generated output
        /// </summary>
        public void WriteLine(string textToAppend)
        {
            this.Write(textToAppend);
            this.GenerationEnvironment.AppendLine();
            this.endsWithNewline = true;
        }
        /// <summary>
        /// Write formatted text directly into the generated output
        /// </summary>
        public void Write(string format, params object[] args)
        {
            this.Write(string.Format(global::System.Globalization.CultureInfo.CurrentCulture, format, args));
        }
        /// <summary>
        /// Write formatted text directly into the generated output
        /// </summary>
        public void WriteLine(string format, params object[] args)
        {
            this.WriteLine(string.Format(global::System.Globalization.CultureInfo.CurrentCulture, format, args));
        }
        /// <summary>
        /// Raise an error
        /// </summary>
        public void Error(string message)
        {
            System.CodeDom.Compiler.CompilerError error = new global::System.CodeDom.Compiler.CompilerError();
            error.ErrorText = message;
            this.Errors.Add(error);
        }
        /// <summary>
        /// Raise a warning
        /// </summary>
        public void Warning(string message)
        {
            System.CodeDom.Compiler.CompilerError error = new global::System.CodeDom.Compiler.CompilerError();
            error.ErrorText = message;
            error.IsWarning = true;
            this.Errors.Add(error);
        }
        /// <summary>
        /// Increase the indent
        /// </summary>
        public void PushIndent(string indent)
        {
            if ((indent == null))
            {
                throw new global::System.ArgumentNullException("indent");
            }
            this.currentIndentField = (this.currentIndentField + indent);
            this.indentLengths.Add(indent.Length);
        }
        /// <summary>
        /// Remove the last indent that was added with PushIndent
        /// </summary>
        public string PopIndent()
        {
            string returnValue = "";
            if ((this.indentLengths.Count > 0))
            {
                int indentLength = this.indentLengths[(this.indentLengths.Count - 1)];
                this.indentLengths.RemoveAt((this.indentLengths.Count - 1));
                if ((indentLength > 0))
                {
                    returnValue = this.currentIndentField.Substring((this.currentIndentField.Length - indentLength));
                    this.currentIndentField = this.currentIndentField.Remove((this.currentIndentField.Length - indentLength));
                }
            }
            return returnValue;
        }
        /// <summary>
        /// Remove any indentation
        /// </summary>
        public void ClearIndent()
        {
            this.indentLengths.Clear();
            this.currentIndentField = "";
        }
        #endregion
        #region ToString Helpers
        /// <summary>
        /// Utility class to produce culture-oriented representation of an object as a string.
        /// </summary>
        public class ToStringInstanceHelper
        {
            private System.IFormatProvider formatProviderField  = global::System.Globalization.CultureInfo.InvariantCulture;
            /// <summary>
            /// Gets or sets format provider to be used by ToStringWithCulture method.
            /// </summary>
            public System.IFormatProvider FormatProvider
            {
                get
                {
                    return this.formatProviderField ;
                }
                set
                {
                    if ((value != null))
                    {
                        this.formatProviderField  = value;
                    }
                }
            }
            /// <summary>
            /// This is called from the compile/run appdomain to convert objects within an expression block to a string
            /// </summary>
            public string ToStringWithCulture(object objectToConvert)
            {
                if ((objectToConvert == null))
                {
                    throw new global::System.ArgumentNullException("objectToConvert");
                }
                System.Type t = objectToConvert.GetType();
                System.Reflection.MethodInfo method = t.GetMethod("ToString", new System.Type[] {
                            typeof(System.IFormatProvider)});
                if ((method == null))
                {
                    return objectToConvert.ToString();
                }
                else
                {
                    return ((string)(method.Invoke(objectToConvert, new object[] {
                                this.formatProviderField })));
                }
            }
        }
        private ToStringInstanceHelper toStringHelperField = new ToStringInstanceHelper();
        /// <summary>
        /// Helper to produce culture-oriented representation of an object as a string
        /// </summary>
        public ToStringInstanceHelper ToStringHelper
        {
            get
            {
                return this.toStringHelperField;
            }
        }
        #endregion
    }
    #endregion
}

redpajama_set_name: RedPajamaGithub


Girls on the Run (GOTR) inspires girls to be joyful, healthy, and confident using a fun, experience-based curriculum which creatively integrates running. Watershed is excited to host two middle school GOTR Heart & Sole teams on-site this fall!
Watershed's teams will meet Tuesday & Thursday afternoons from 5:00-6:30PM. The 10-week season kicks off the week of September 17th. If you're interested in coaching, please plan to attend GOTR Coaches Training (also at Watershed) on August 23rd. Express interest in coaching here.


text:
Rachel at the double for Belles
Women's Football News 15 Mar 2010
National Division leaders Sunderland, playing their first league match for three months, went down to a narrow 2-1 defeat away to Doncaster Rovers Belles.
The first half ended goalless and it wasn't until the last fifteen minutes that the goals finally came. Two well taken goals from Rachel Williams proved to be enough to give victory for the Belles, despite Sunderland captain Steph Bannon pulling one back in injury time. A disappointing day for the Wearsiders was completed as Bannon was shown a red card following a scuffle after her goal.
Chelsea kept up their good run of form by running out 4-1 winners over Blackburn Rovers, a result which takes them into third place. Claire Rafferty celebrated her England call up by giving the Blues a first half lead which was cancelled out just before the interval by an own goal from Sophie Perry. However after the break, Rebecca Jane put Chelsea back in front and they went on to seal the victory with two goals from in-form Dunia Susi.
Millwall Lionesses are looking set for a mid-table finish as they won 4-2 against Watford, whose worrying run continues. Emma Whitter put Millwall in front after eight minutes and Victoria Boardman added a second just ahead of half time, but Watford hit back to quickly level the scores after the break through Kate Longhurst and Charlotte Nicholson. The game looked to be heading for a draw until Millwall struck twice in the dying minutes with goals from Dionne Lennon and Meghan McKeag.


34 yr old Transport Engineer Kevin Harrold from Gravenhurst, has many pursuits that include paintball, brazil and vehicle racing. that included visiting Longobards in Italy. Places of the Power (- A.D.). If you loved this write-up and you would such as to obtain more information pertaining to Daniel Dantas kindly visit our own web-page.


text:
Q: Make all lower case and replace spaces with - I am trying to generate some html pages and I have:
$category="Category Name";
$zone="Zone Name";

I am trying to get $page="category-name-in-zone-name.html"

A: You're looking for the concept of 'slug creation':
function makeSlug($input)
{
    $interim = iconv('UTF-8', 'ASCII//TRANSLIT//IGNORE', $input);
    $interim = preg_replace("/[\"']+/i", '', $interim);
    $interim = preg_replace("/[^a-z0-9\\d]+/i", '-', $interim);
    return trim(strtolower($interim), ' -');
}
echo makeSlug("$category in $zone").'.html';

Demo here.