Why are Dell and HP struggling in the PC market? Smart phones and tablets are the latest trend! And quickly accessing BPM information on mobile devices is the next logical step. A while ago our product management team had to think about this market evolution.
SAP Services combine SAP experts, methodologies, tools and partners to accelerate implementations which helps you transferring knowledge to staff while continuously improving operations and therefore building long term success.
TransWare is an SAP® Services Partner delivering expertise for SAP Solution Manager® and SAP® NetWeaver® technologies with ASAP, Run SAP® and BPM methodologies.
We are proud to deliver SAP® technology and valid methodology to our valued clients to support the transformation of knowhow into business performance.
For the optimization of SAP® landscapes we offer reengineering and redocumentation of SAP Systems with our solution Profiling for SAP that uses SAP Solution Manager and web-based technologies.
As SAP® project experts we offer consulting for ASAP and BPM methodologies to improve the blueprinting of new and upgraded SAP® solutions, especially in the preparation and blueprint project phases, and in the latest ASAP methodology.
For SAP® Solution Manager we offer a set of qualified services and project support for the implementation of ITIL compliant IT processes with Run SAP® methodology.
Business-process driven testing is based on the methodology and algorithms of model-based testing using business process models to generate a set of valid test paths.
The table shows the value proposition of the model-based testing solution.
You have missed one of our successful webinars in May?
BPM-Xchange® is the leading enterprise management integration platform in terms of model, master and meta data interchange and transformation. It allows for a seamless linkage of management tools and repositories for EA, BPA, BPM(S), CMDB, GRC and others. Our platform implements a hub-and-spoke architecture with an enterprise service bus on the enterprise meta-level comparable to a TIBCO®-like BPM engine for transactional process execution.
Beyond that BPM-Xchange® utilizes and extends Microsoft's industry standards Visio® 2010 and SharePoint® Server 2010 to help your organization to significantly lower the costs for business process management.
Employing the new BPM-X Designer 2.7 add-in for rule-based process modeling in Visio® 2010 and using Microsoft® SharePoint® Server 2010 as a process repository and Microsoft SQL Server® Enterprise Reporting gives you striking business advantages. Your business users and process owners love easy-to-use office-based modeling.
BPM-Xchange® simply links the Microsoft® BPM environment with more complex and costly BPM tools like ARIS® and delivers considerable benefits for your organization.
Enterprise-wide reporting or: "BI meets BPM"


text:
Every year there are a few films which disappoint us. There are others which outright suck. When I started Unseen the idea was never to discuss the dregs of cinema except occasionally. However because the focus of Unseen shifted from highlighting just lost treasures to covering more recent films I've been forced to write about a lot more misses and turkeys than I used to.
Here then are my lists of the misses and turkeys I saw in 2019.
Disappointments-these films almost kinda could maybe been something
RED PENGUINS - The follow up to the director's RED ARMY is half a good film. The story of the Pittsburgh Penguins attempt to sponsor the Red Army team with the aim of siphoning off the players for the NHL. However like the plan it goes off the rails somewhere along the way and ever comes back.
LOS ULTIMOS FRIKIS- A film about a heavy metal band in Cuba that strangely has very little music.
A GIRL MISSING- a nurse's life is turned upside down when a relative kidnaps one a member of the family she works for. It's a jumbled story that is icy cold
ZOMBI CHILD weird mix of coming of age tale with a true life zombie tale goes banana shaped when it turns into the scariest horror film of the year in the last 15 minutes. WTF?
HENRY GLASSIE FIELD WORK- Portrait of the anthropologist has some stunning sections but Glassie is AWOL for most of the film. Instead we get footage of the people he is studying. Huh?
SADAKO- The restart of the Ring Franchise crashes and burns- despite having some moments
PLUCKED-The story of the theft of a Stradivarius violin is an hour too long making a compelling story a big WTF.
LOW TIDE- What should have been a great thriller misfires all over the place. Not only doesn't it know when it's set, it has so many story problems I put half my review in inviso textbecause I had to discuss it and didn't want to spoil if for those who are interested in seeing it. (And there is a good film possible with a better script)
TO THE ENDS OF THE EARTH- One of my most favorite directors in the world Kiyoshi Kurosawa makes a film in order to make a trip to Central Asia. Despite having great moments and a deeply moving version of an Edith Piaf classic, this film has no reason to exist.
WASP NETWORK- Olivier Assayas takes a ten part miniseries idea and chops it to two hours. It's a mess- so much so Assayas has been recutting the film
THE DRONE- a drone is possessed - beyond bad horror film that sadly isn't even funny.
DUMBO- mean and ugly live action rethink of the classic Disney tale will scar generations
JOKER- over rated and over hyped origin for Batman's greatest foe is just bad.
LIBERTE-ugly aristocrats have sex in a forest. Long, dull and stupid it's just utterly pointless and boring.
EL PADRE MEDICO-The story Padre Ferdinand Bendoraitis who was once considered second to Mother Theresa is so badly told I have no idea about anything his life or what the filmmakers were trying to say about it.
KNIVES AND SKIN- dreadful thriller is winning all sorts of awards- but I have yet to meet anyone who actually likes it. Frankly I don't know anyone who hasn't either walked out on it or turned off a screener.
TASTE OF SKY- Portrait of a restaurant in South America rambles all over the place and just bores.
YOU DON'T NOMI- Portrait of the film SHOWGIRLS is so inside that it will only means something if you have more than a working knowledge of SHOWGIRLS.
CLEMENTINE- beyond awful film about a woman obsessed with her ex-girlfriend who breaks into her summer house where she meets another young woman while nothing happens for 90 minutes.
YOUNG AHMED- the grooming of a young terrorist has no understanding of religious extremism except from the headlines- worse it wusses out in its WTF ending.
OVER 18- A way too simplistic film on the evils of porn.
WELCOME TO MARWEN- Robert Zemekis makes a film so bad you'll want to scrape your eyeballs with steel wool dipped in acid if you simply don't simply pluck them out of your head
Labels: year end 2019

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: How to update a value for a key in HashMap? I am having HashMap like this,
HashMap<String,Set<String>> map = new HashMap<String,Set<String>>();

I am trying to do before adding an element in map,

*

*Want to check whether the key exist or not, i can get it by using map.containsKey().

*If the key exist, i want check the size of Set respective to that key.

*If size <= 1  i want add an element in that set.

A: Sounds like this:
HashMap<String,Set<String>> map = new HashMap<String,Set<String>>();

Set<String> value = map.get("key");
if(value != null) {
    if(value.size() <= 1) {
        value.add("some value");
    }
} else {
    map.put("key", new HashSet<String>());
}

Now, either the last point was poorly worded (i.e. you want to update the Set associated with the key) or you really want to update the key itself, in which case you'd probably have to just remove it and add a new entry.

A: I wouldn't use containsKey and get as this means two lookups when you only need one.
private final Map<String,Set<String>> map = new HashMap<String,Set<String>>();

Set<String> set = map.get(key);
if(set != null && set.size() <= 1)
    set.add(some$value);

The only problem with this is that the value will always be null unless you set it somewhere so what you may want is
private final Map<String,Set<String>> map = new HashMap<String,Set<String>>();

Set<String> set = map.get(key);
if(value != null)
    map.put(key, set = new HashSet<String>());
if (set.size() <= 1)
    set.add(some$value);

It is unusual to have a set with a maximum size of 2. Is there any reason for this?

A: You could get the set from the map with map.get(String key).
Then test the size of the Set. If needed, add your element.
Now you can simply remove the old set from the map with map.remove(String key) and reinsert it with put(String, Set);

redpajama_set_name: RedPajamaStackExchange


We want you to be able to get the most from your inventions, creations, know-how, brand, designs and reputation. But protecting your intellectual property can be challenging in an ever-changing commercial and legal environment.
That's where we come in. We start by understanding your business objectives and establish a logical course of action. Whether you want to better exploit your IP, protect it from infringement or you need help with issues that might have arisen with external organisations, we can help. Our dedicated IT and IP specialists are right by your side through research and development, patent/know-how licensing and supply and contract manufacturing.
Intellectual property – who owns it?
As your business grows so too will your legal needs. That's why we can also support you with information technology, data privacy, cybersecurity and general commercial law solutions.


text:
13.07 Güstrow
27.07 Toruń
10.08 Vojens
28.09 Chorzów
INSIDE SEC
SEC Rules
SEC History
SEC Formula
SEC Girls
FIM Europe
One Sport
Vaclav Milik: I'm still learning
Vaclav Milik is probably the brightest star of Czech speedway, however, his last season couldn't be classified as a good one. This also touches his starts in TAURON Speedway Euro Championship.
Initially, it looked like, after getting bronze and silver medal, Milik can join the rivalry for gold, but what happened is that he couldn't have locked the spot for upcoming year in the series.
Vaclav Milik started rivalry in Euro Championship from two 7th positions and getting 8 points twice: in Gniezno and Guestrow, while only one year back, in the premiere rounds, he got himself into the final heat of the night.
This time, however, he was able to reach that only in the third round - in Daugavpils (Latvia), where he ended as 4th, right behind the podium.
While looking at his statistics, in each round, he was getting three points less than in 2017 and won only half of the heats than one year earlier. What's more, when we would not consider Josef Franc, who was failing constantly, it was Milik who finished his heats at the fourth position most often.
- To be hones, I had moments, when I really had no idea what was the issue. It just goes like that - ups and downs all the way. Season 2017 was quite a good one and this year, my form got worse. There's nothing left to be done that just draw some conclusions. Like I always say: I'm still learning. At the end of the season, I had some good moments. Maybe the key is to have fun with speedway, this pure joy, instead of keeping everything into professional frames - Czech rider wonders.
Breakthrough in Daugavpils had let Milik believe in his further progress and a positive ending of 2018 season. Ahead of the last round in Chorzów. Czech riders was only two points down to Antonio Lindbaeck and Mikkel Michelsen, classified ex-aequo at the fifth place. However, his performance at the Śląski Stadium got him rid of chances to stay in TAURON SEC in 2019 - only six points gathered got him the overall eight position. Even though, he started this evening with two points in the first heat and a win in his last, that was just not enough to catch up his loss in the mid-phase, when in three heats, he only got one point.
- Surely, I wanted to achieve more. Most importantly, I wanted to secure a spot in the TOP 5. On the other hand, it was my first time starting at the artificial track, so while considering this fact, it was not that bad. In the first phase of the meeting, we had been looking for proper setups. What's more, in these first heats, I was choosing some wrong lines, which cost me a lot. In the last heat of the night, I was able to race way better. I got a different trajectory and while racing near the inside line, I was way faster than with doing it in the middle of this track. We just started to adjust to this surface - "Vasek" reminisces.
Despite an underwhleming score, Milik's performance in Chorzów, opened him into a new experiences and got him lots of valuable notices. Rivalry at the big arenas, with support of a huge crowd, let him taste a totally different dimension of speedway, well known for years, from the Grand Prix series. These kinds of experiences make rider want to be able to participate in it once more.
- I really enjoyed it, even more with the beautiful Śląski Stadium, making such a great impression. And adding the fans' support - every rider would love to race in such conditions. Naturally, it was tough to adjust to this track, but if I would have a chance to be back here, I would be way smarter. Basic difference between artificial and natural tracks is that these firtst doesn't hold us that much and you're getting more holes and ruts. To be honest, thought, when I first saw this surface, I had thought it would've been worse than it actually was - Milik admitted.
Recently, there has been a lot of criticism over Milik. Not so good and unstable disposition found its reflection also in Polish PGE Ekstraliga starsts, where Czech rider decreased his average points with 0,3, comparing to 2017 and 0,6 when it comes to 2016. For some time, he couldn't have even been sure of a spot in Wrocław team for upcoming season. Ultimately, he got his chance for a breakthrough and that's exactly what he'll be trying to do also at the international arena. For a long time, it has been said that while being in shape, Czech rider, would be a great added value to Grand Prix series. Right now, however, Milik keeps his focus on Individual European Championship and settling the score in this rivalry.
- I would like to be back to TAURON SEC. I think that reaching the promotion through qualification rounds and further SEC Challenge could give me even bigger motivation to fight for the top goals and trying to reach for gold - Milik ended.
Tags: speedway Vaclav Milik speedwayeuro letsmakespeedway
Go to News section
Jepsen Jensen's solidity
Word that would perfectly describe Michael Jepsen Jensen is "solidity". Dane hasn't used to get under some level and it was the same when it comes to his starts in TAURON SEC 2019. Rider of Falubaz Zielona Góra had been in the game for medals for a long time, but ultimately he had to leave empty handed.
Bright shine for the future
For a long time, 2019 was a really tough season for Paweł Przedpełski. Polish rider couldn't be satisfied with both his starts in the league and individual meetings. Luckily, a successful ending occurred, which may give a bright shine for the future. One of the main reasons of thinking so is Pole's performance in Chorzów.
Great moment for Michelsen
Not everything had been going in his line when it comes to rivalry in TAURON SEC 2019. Mikkel Michelsen worked hard through the whole season and made his dream about the gold medal true. He seems like a perfect example of a great determination that can take you to the top.
French debut
French ace is a proof that you can link a good sport level with broadening speedway map. In 2019, David Bellego made it right on multiple fronts. And what can be regretted the most is the second part of TAURON SEC 2019...
Michelsen: Two last rounds were amazing
Mikkel Michelsen became the winner of the final round of TAURON SEC 2019, which was taken at the Silesian Stadium in Chorzów (Poland). Michelsen got 15 points, which made him equal Grigorii Laguta's score in the overall classification. In the additional heat, it was Dane, who defeated Russian and ultimately ended as the new European Champion.
Laguta: I am missing luck
Over 70 heats. For so long, Grigorii Laguta had been the leader of TAURON SEC 2019. Russian lost the gold medal in the very last, additional heat, where he had to accept Mikkel Michelsen's superiority.
M. Michelsen #155
G. Laguta #111
L. Madsen #30
K. Woryna #223
B. Smektała #115
M. Jensen #52
N. Pedersen #110
P. Przedpełski #323
J. Hampel #33
K. Huckenbeck #744
V. Milik #225
A. Lindbaeck #85
R. Lambert #505
A. Thomsen #11
D. Bellego #415
T. Lahti
A. Miedziński
F. Jakobsen
A. Lyager
M. Drabik
J. Miśkowiak
A. Kudryashov #191
P. Kildemand
K. Woelbert
I. Kopeć - Sobczyński
Madsen: I couldn't achieve more
Mikkel Michelsen is the new European Champion
First French rider at this level
Track in Chorzów with a license
Go to Photos section
TAURON SEC 2019 Round 4: Chorzów
TAURON SEC 2019 Round 3: Vojens
TAURON SEC 2019 Round 2: Toruń
TAURON SEC 2019 Round 1: Güstrow
GIRLS Section
Host Cities and Regions
Official Broadcaster
SPEEDWAY EURO CHAMPIONSHIP| © Copyright 2020 - All rights reserved | Privacy policy | Contact us | Term of use


text:
Q: Perl specific code The following program is in Perl.
cat "test... test... test..." | perl -e '$??s:;s:s;;$?::s;;=]=>%-{<-|}<&|`{;;y; -/:-@[-`{-};`-{/" -;;s;;$_;see'

Can somebody help me to understand how it works?

A: This bit of code's already been asked about on the Debian forums.
According to Lacek, the moderator on that thread, what the code originally did is rm -rf /, though they mention they've changed the version there so that people trying to figure out how it works don't delete their entire filesystem. There's also an explanation there of what the various parts of the Perl code do.
(Did you post this knowing what it did, or were you unaware of it?)
To quote Lacek's post on it:

Anyway, here is how the script works.
It is basically two regex substitutions and one transliteration.
  Piping anything into its standard input makes no difference, the perl
  code doesn't use its input in any way. If you split the long line on
  the boundaries of the expressions, you get this:
$??s:;s:s;;$?::
s;;=]=>%-{\\>%<-{;;
y; -/:-@[-`{-};`-{/" -;;
s;;$_;see

The first line is a condition which does nothing save makes the code
  look more difficult. If the previous command originated from the perl
  code wasn't successful, it does some substitutions on the standard
  input (which the program doesn't use, so effectively it substitutes
  the nothing). Since no previous command exists, $? is always 0, so the
  first line never gets executed.
The second line substitutes the
  standard input (the nothing) for seemingly meaningless garbage.
The third line is a transliteration operator. It defines 4 ranges, in
  which the characters gets substituted to the one range and the 4
  characters given in the transliteration replacement. I'd prefer not to
  write the whole transliteration table here, because it's a bit long.
  If you are really interested, just write the characters in the defined
  ranges (space to '/', ':' to '@', '[' to '(backtick)', and '{' to '}'), and
  write next to them the characters from the replacement range ('(backtick)' to
  '{'), and finally, write the remaining characters (/,", space and -)
  from the replacement pattern. When you have this table, you can see
  what character gets replaced to what.
The last line executes the
  resulting command by substituting the nothing with the resulted string
  (which is 'xterm'. Originally it was 'system"rm -rf /"', and is held
  in $_), evaluates the substitution as an expression and executes it.

(I've substituted 'backtick' for the actual backtick character here so that the code auto-formatting doesn't kick in.)


I was just surfing the internet the other day, only to find that there are quite a few sites with sloppy titles. I really condemn the use of sloppy headers in a website as this only helps in sabotaging and not in improving the SEO of the website.
Sidetrack! Wishing to finally up grade the bland look with my page. Thinking concerning the sweet style at http://smilesunlimited.ca? Sincerely a top dental professional based in Coquitlam if ever wanted within the general British Columbia territory. Write your feedback. Many thanks!
This is mainly because these sloppy headers tend to confuse search engine spiders. Sloppy headers usually spring up when using site builders on hosts like freeservers.com as while you build your website, you will be asked to save the website as you like, while you work on providing content for the website.
In the process, you may not include the non-visible area of the page which in turn makes search engines ignore the page, thinking it to be duplicate content. This problem not only exists with site builders, but also with do it yourself site software like Dreamweaver. The end result is a website with identical titles and Meta tags on all pages, and different visible content.
So to avoid this, I suggest your developing content and a title for your new web page, before heading on to the next page. With this practice, you will also be able to implement some effective SEO techniques while framing your header as well.
I suggest you head this bit of advice as most search engines tend to penalize pages that look identical to them. With this penalization, they don't include these pages in their search engine indexes at all.
Another bad effect of sloppy headers on web pages is a URL that looks very strange. These URLs will not have any of the keywords that you had used in your web content, and thus, will in no way be helpful in drawing any positive traffic to your website.
Thanks folks! I assured a buddy that we could certainly cite his dependable propane conversion company in Quebecs, http://www.ecopropane.ca for a future post. If you're researching to get a gas to propane conversion service around the general Quebec vicinity, they really are amazing.
Oh, naturally I will tell you the outline for this unique posting with thoughtfully granted via Ira with Interiortech. They are definitely an awesome sign companys. We really value a wonderful proposal!
http://greenguardpestcontrol.com - You trully are amazing human being.
http://www.appraisemuskoka.com - Call these guys regularly.


text:
\section{Introduction}

We answer in the affirmative a conjecture posed by Hegselmann and
Krause about the long-term behavior of opinions in a finite group of
individuals, some of them attracted to the truth, the so-called
\emph{truth seekers}.  Our contribution: Under mild assumptions, the
opinions of all truth seekers converge to the truth, despite being
distracted by individuals not attracted to the truth, the
\emph{ignorants}.

The underlying model for opinion dynamics is the
\emph{bounded-confidence model}: Opinions, which themselves are
represented by real numbers in the unit interval, are influenced by
the opinions of others by means of averaging, but only if not too far
away. This bounded-confidence model (formal definitions below) was
first suggested by Krause in 1997.  It received a considerable amount
of attention in the artificial societies and social simulation
community \cite{lorenz_survey,hegselmann_new,
  hegselmann2006,hegselmann2002,15.1021F,7291M}.

The concept of truth seekers was invented in 2006 by Hegselmann and
Krause~\cite{hegselmann2006}, along with a philosophical discussion
about the scientific context with respect to the notion of truth.  We
blind out the philosophical discussions here and focus on the
resulting dynamical system, governed by difference equations that we
find interesting in their own right.

The opinions of \emph{truth seekers} are not only attracted by
opinions of others; they are additionally attracted by a constant
number, the truth.  The resulting opinion is weighted average of the
result of the original bounded-confidence dynamics and the truth.
Individuals not attracted by the truth in this sense are
\emph{ignorants}. In their paper, Hegselmann and Krause show that if
all individuals are truth seekers -- no matter how small the weight
--, then (the opinions of) all the individuals converge to consensus
on the truth value.

The question we answer in this paper arises when some of the
individuals are ignorants, i.e.{}, the weight of the influence of the
truth is zero for them.  Numerous simulation experiments led
Hegselmann and Krause to the conjecture, that still the opinions of
all the truth seekers finally end up at the truth.  However, a proof
of this fact could not be found so far.  Evidence by simulation only,
however, bears the risk of numerical artefacts -- very much so in the
non-continuous bounded-confidence model.  Therefore, it is desirable
to provide mathematically rigid proofs of structural properties of
bounded-confidence dynamics.

Allthough the conjecture may seem self-understood at first glance
because of the contraction property of the system dynamics for truth
seekers, a second look on the situation reveals that the conjecture
and its confirmation in this paper are far from trivial: several
innocent-looking generalizations of the conjecture are actually false,
as we will show below in the technical parts of the paper.  Relying on
intuition only is dangerous.

Even in the affirmative cases, convergence turns out to be quite slow
in general and far from monotone.  The main difficulty is the
following: the convergence of truth seekers heavily depends on their
long-term influence on ignorants.  Depending on the configuration of
ignorants and the parameters of the system, there are arbitrarily many
iterations in which the truth seekers deviate from the truth.  The
crucial observation is that, during these iterations, the
configuration of ignorants is somehow ``improved'' because the truth
seekers attract them.

After all, the proof is elementary but extremely technical.  We
introduce some structures like the \emph{confidence graph}, that might
prove useful also in other contexts.  Other structures we need are
rather special, probably with limited use beyond this paper.  It
would, therefore, be desirable to find a more elegant proof, revealing
the reason why the conjecture is true.  For example: find a suitable
Lyapunov function.  The examples we give as we go along in the proof,
however, indicate that a certain amount of complexity has to be
captured by the arguments because the line between true and false
conjectures is extremely thin.

\section{Formal problem statement}
\label{sec:form-probl-stat}

Suppose there is a set $[n]:=\{1,\dots,n\}$ of individuals with
opinions $x_i(t)\in[0,1]$ at time $t$ for all $i\in[n]$,
$t\in\mathbb{N}$. The abstract truth is modeled as a constant over
time, denoted by $h\in[0,1]$. The opinion of an individual $i\in[n]$ is
influenced in a time step $t$ only by those individuals which have a
similar opinion, more precisely which have an opinion in the
\textit{confidence interval} of $x_i(t)$.

\begin{definition}
  For $x\in [0,1]$ and a parameter $\varepsilon\ge 0$ we define the
  \emph{confidence set of value} $x$ \emph{at time} $t$ as
  $$
    I_x^\varepsilon(t):=\{j\in[n]\mid |x-x_j(t)|\le \varepsilon\}.
  $$
  As a shorthand we define
  $I_i^\varepsilon(t):=I_{x_i(t)}^\varepsilon(t)$ for any $i\in[n]$.
\end{definition}
The update of the opinions is modeled as a weighted arithmetic mean of
opinions in the confidence set and a possible attraction towards the
truth.
\begin{definition}
  A \emph{weighted arithmetic mean symmetric bounded
    confidence opinion system} (WASBOCOS) is a tupel
  \begin{equation*}
    (n, h, \varepsilon, \alpha, \beta; \alpha_i(t), \beta_{ij}(t), x_i(0)),
  \end{equation*}
  where
  \begin{itemize}
  \item $n\in\mathbb{N}$ ist the number of \emph{individuals},
  \item $h\in[0,1]$ ist the \emph{truth},
  \item $\varepsilon\in[0,1]$ is the \emph{bounded confidence radius},
  \item $\alpha\in(0,1]$ is a lower bound for the weight of the truth for truth seekers,
  \item $\beta\in(0,\frac{1}{2}]$ is a lower bound for the weight of opinions in the bounded confidence interval,
  \item $\alpha_i(t)\in[\alpha,1]$ or $\alpha_i(t)=0$ for all
    $t\in\mathbb{N}$ is the actual weight of the truth for truth
    seeker~$i$ at time step~$t$,
  \item $\beta_{ij}(t)\in[\beta,1-\beta]$ with $\sum_{i=j}^n
    \beta_{ij} = 1$ for all $i\in[n]$ and for all $t\in\mathbb{N}$ is
    the weight of opinion~$j$ in the view of agent~$i$,
  \item $x_i(0)\in[0,1]$ is the starting opinion of Individual~$i$.
  \end{itemize}
  The \emph{bounded confidence dynamics} on such a system is defined
  by simultaneous updates of the opinions in the following form:
  \begin{equation}
    \label{eq_update}
    x_i(t+1):=\alpha_i(t)\cdot h+\Big(1-\alpha_i(t)\Big)\frac{\sum\limits_{j\in I_i^\varepsilon(t)}\beta_{ij}(t)x_j(t)}
    {\sum\limits_{j\in I_i^\varepsilon(t)}\beta_{ij}(t)}.
  \end{equation}
  \emph{Individuals} are members of the index set $[n]$.
  \emph{Truth seekers} are members of the set $K:=\{k\in[n]\mid
  \alpha_k(t)\ge \alpha\, \forall t\in\mathbb{N}\}$.  All other
  individuals, i.e.{}, those with $\alpha_i(t)=0$ for all $t$, are
  called \emph{ignorants}; their set is denoted by~$\overline{K}$.
\end{definition}

See Figure~\ref{fig:example-0} for a sketch of a typical set of
trajectories.  Remark: The term \emph{symmetric} in the notion of a
(WASBOCOS) refers to the confidence radius, not to the weights that
individuals assign to other individuals' opinions.
\begin{figure}
  \begin{center}
  \includegraphics[width=0.65\linewidth]{Example_6.pdf}

  \label{fig:example-0}
  \end{center}
\end{figure}
The main result we wish to prove is the following:
\begin{theorem}(Generalized Hegselmann-Krause Conjecture)
  \label{main_result}
  All truth seekers in an (WASBOCOS) $\Omega$ converge to the
  truth~$h$. Formally, for each $\gamma>0$ and each $\Omega$ there
  exists a $T(\gamma,\Omega)$ so that we have $|x_k(t)-h|<\gamma$ for
  all $k\in K$ and all $t\ge T(\gamma,\Omega)$.
\end{theorem}
Note that we use $\gamma>0$ in the statement of convergence instead
of~$\varepsilon>0$ because $\varepsilon$ is traditionally used for the
bounded confidence radius.

It is important that convergence is not just implied by the
contraction property of the dynamics with ignorants ignored.
Ignorants and where their opinions are make a huge difference (see
Figure~\ref{fig:example-1} for an example).
\begin{figure}
  \centering
  \includegraphics[width=0.65\linewidth]{Example_1}
  \caption{The only truth seeker is located at the bottom on the
    truth; it gets attracted away from the truth for quite some time;
    eventually, the ignorants either are ``converted'' or left
    behind. Thus, we cannot expect monotone convergence of the truth
    seeker farthest from the truth, i.p, reaching the truth once does
    not imply convergence to the truth.}
  \label{fig:example-1}
\end{figure}

It would be nice if one could derive a bound on the speed of
convergence, i.e., a bound on $T(\gamma,\Omega)$, in terms of the
structural parameters $\varepsilon$, $\alpha$, $\beta$, and $n$.
Unfortunately, this is not possible.  The speed of convergence is not
determined by the structural parameters alone.  This can be seen in
the following simple example.
\begin{example}
  \label{ex_interupted_convergent}
  Consider a (WASBOCOS) with truth $h=\varepsilon$, $\varepsilon>0$,
  $\alpha_1(t)=\alpha$, $\alpha_2(t)=0$, $\beta_{ij}(t)=\frac{1}{2}$,
  $\beta=\frac{1}{2}$, $x_1(0)=2\varepsilon$,
  $x_2(0)=\tilde{\varepsilon}$, where
  $\varepsilon>\tilde{\varepsilon}>0$. Let $T\in\mathbb{N}$ be the
  smallest integer so that
  $(1-\alpha)^T\varepsilon\le\tilde{\varepsilon}$. Then by induction
  we have $x_1(t)=\varepsilon+(1-\alpha)^t\varepsilon$ and
  $x_2(t)=x_2(0)=\tilde{\varepsilon}$ for all $t\le T$. So truth
  seeker $1$ seems to monotonically converge to the truth, but at time
  $T+1$ we have
  $x_1(T+1)=\alpha\varepsilon+\frac{1-\alpha}{2}\left(\varepsilon+(1-\alpha)^T\varepsilon+\tilde{\varepsilon}\right)
  \le \frac{1+\alpha}{2}\cdot\varepsilon+\tilde{\varepsilon}$.
\end{example}
See Figure~\ref{fig:example-2} for a sketch of the situation.
\begin{figure}
  \centering
  \includegraphics[width=0.65\linewidth]{Example_2}
  \caption{A sketch of interrupted convergence: a lonely truth seeker
    starting at $2\varepsilon$ seems to monotonically converge to the
    truth right away, but suddenly its confidence interval picks up
    the ignorant on the other side of the truth, and the truth seeker
    gets distracted.  However, finally the ignorant gets distracted
    himself, and convergence is eventually established.}
  \label{fig:example-2}
\end{figure}
Since we may choose $\tilde{\varepsilon}$ arbitrarily small, we find
the following: in general, we can not expect that
for every $\gamma>0$ there is a $T(\gamma,\varepsilon,\alpha,\beta,n)$
such that for all $t\ge T(\gamma,\varepsilon,\alpha,\beta,n)$ we have
$|x_k(t)-h|<\gamma$ for each truth seeker $k\in K$.  But we may have
\emph{interrupted convergence}: In a first phase, the truth seekers
come arbitrarily close to the truth in time only dependent on the
structural parameters; then, they may temporarily get distracted at
some point; finally, they converge to the truth in time depending only
on the structural parameters \emph{and the time of distraction}.  This
can be formalized as follows:
\begin{definition}
  \label{def_interupted_convergent}
  Given $\varepsilon$, $\alpha$, $\beta$, $n$, we say that truth
  seekers $k\in K$ are (1-fold) \emph{interrupted convergent to the
    truth}, if for each $\gamma > 0$ there exist two functions
  $T_1^s(\gamma,\varepsilon,\alpha,\beta,n)$ and
  $T_2^s(\gamma,\varepsilon,\alpha,\beta,n,T_1^e)$, so that for each
  (WASBOCOS) $\Omega$, with structural parameters $\varepsilon$,
  $\alpha$, $\beta$ and $n$, there exists an $T_1^e\in\mathbb{N}$
  satisfying
  \begin{eqnarray*}
    &&\forall k\in K,\,\forall t\in[T_1^s(\gamma,\varepsilon,\alpha,\beta,n),T_1^e]:\, |x_k(t)-h|<\gamma,\\
    &&\forall k\in K,\,\forall t\ge T_2^s(\gamma,\varepsilon,\alpha,\beta,n,T_1^e):\,|x_k(t)-h|<\gamma.
  \end{eqnarray*}
\end{definition}

Theorem \ref{main_result} is now a corollary of the following
substantially strengthened Theorem:
\begin{theorem}
  \label{thm_interupted_convergent}
  All truth seekers in an (WASBOCOS) $\Omega$ are (1-fold) interrupted
  convergent to the truth.
\end{theorem}

Originally Hegselmann and Krause considered the (WASBOCOS) model for
$\alpha_{i}(t)\in\{0,\alpha\}$ and $\beta_{ij}(t)=\frac{1}{n}$.  In
the case of complete absence of truth seekers they have already
proved, that the opinion of each individual converges, as can be
expected, not necessarily to the truth. In fact in general the
individuals form several clusters, where two individuals of different
clusters converge to different opinions.

We give an example without truth seekers where the individuals will
converge to five different clusters.

\begin{example}
  \label{ex_1}
  Consider a (WASBOCOS) with $\alpha_i(t)=0$ (no truth seekers),
  $\beta=\beta_{ij}(t)=\frac{1}{n}$, $n=12$; the values of~$\alpha$
  and~$h$ do not matter.  The starting positions are given by
  \begin{eqnarray*}
   &x_1(0)=x_2(0)=0,\,
   x_3(0)=\varepsilon,\,
   x_4(0)=2\varepsilon,\,
   x_5(0)=3\varepsilon,\,
   x_6(0)=x_7(0)=4\varepsilon,\,\\
   &x_8(0)=5\varepsilon,\,
   x_9(0)=6\varepsilon,\,
   x_{10}(0)=7\varepsilon,\,
   \text{and}\,
   x_{11}(0)=x_{12}(0)=8\varepsilon,
  \end{eqnarray*}
  see Figure \ref{fig_ex_1}.

  \medskip

  In Table \ref{table_ex_1} we give the complete dynamics of the
  opinions of all $12$ individuals over time until the opinion of
  every individual has converged. For brevity we write $x_i$ instead
  of $x_i(t)$. After three time steps, see Figure~\ref{fig:example-5}
  for the dynamics, we have reached a stable state, see Figure
  \ref{fig_ex_1_2} for the resulting positions of the individuals.
\end{example}

\begin{figure}[htp]
  \begin{center}
    \setlength{\unitlength}{1.0cm}
    \begin{picture}(8,1.4)
      \put(0,1){\circle{0.3}}
      \put(0,1.4){\circle{0.3}}
      \put(1,1){\circle{0.3}}
      \put(2,1){\circle{0.3}}
      \put(3,1){\circle{0.3}}
      \put(4,1){\circle{0.3}}
      \put(4,1.4){\circle{0.3}}
      \put(5,1){\circle{0.3}}
      \put(6,1){\circle{0.3}}
      \put(7,1){\circle{0.3}}
      \put(8,1){\circle{0.3}}
      \put(8,1.4){\circle{0.3}}
      \put(-1,0.5){\line(1,0){10}}
      \multiput(0,0)(1,0){9}{\put(0,0.4){\line(0,1){0.2}}}
      \put(-0.2,0){$0\varepsilon$}
      \put(0.8,0){$1\varepsilon$}
      \put(1.8,0){$2\varepsilon$}
      \put(2.8,0){$3\varepsilon$}
      \put(3.8,0){$4\varepsilon$}
      \put(4.8,0){$5\varepsilon$}
      \put(5.8,0){$6\varepsilon$}
      \put(6.8,0){$7\varepsilon$}
      \put(7.8,0){$8\varepsilon$}
    \end{picture}
    \caption{Starting positions of the individuals in Example \ref{ex_1}.}
    \label{fig_ex_1}
  \end{center}
\end{figure}

\begin{table}[htp]
  \begin{center}\renewcommand{\arraystretch}{1.5}
    \begin{tabular}{r@{\qquad}rrrrrrrrr}
      \toprule
      $t$ & $x_1=x_2$ & $x_3$ & $x_4$ & $x_5$ & $x_6=x_7$ & $x_8$ & $x_9$ & $x_{10}$ & $x_{11}=x_{12}$ \\
      \midrule
      0 & $0\varepsilon$ & $1\varepsilon$ & $2\varepsilon$ & $3\varepsilon$ & $4\varepsilon$ & $5\varepsilon$ &
      $6\varepsilon$ & $7\varepsilon$ & $8\varepsilon$ \\[0.9mm]
      1 & $\frac{1}{3}\varepsilon$ & $\frac{3}{4}\varepsilon$ & $2\varepsilon$ & $\frac{13}{4}\varepsilon$ & $4\varepsilon$
      & $\frac{19}{4}\varepsilon$ & $6\varepsilon$ & $\frac{29}{4}\varepsilon$ & $\frac{23}{3}\varepsilon$ \\[1.6mm]
      2 & $\frac{17}{36}\varepsilon$ & $\frac{17}{36}\varepsilon$ & $2\varepsilon$ & $\frac{15}{4}\varepsilon$ & $4\varepsilon$
      & $\frac{17}{4}\varepsilon$ & $6\varepsilon$ & $\frac{271}{36}\varepsilon$ & $\frac{271}{36}\varepsilon$ \\[1.6mm]
      3 & $\frac{17}{36}\varepsilon$ & $\frac{17}{36}\varepsilon$ & $2\varepsilon$ & $4\varepsilon$ & $4\varepsilon$ &
      $4\varepsilon$ & $6\varepsilon$ & $\frac{271}{36}\varepsilon$ & $\frac{271}{36}\varepsilon$\\
      \bottomrule
    \end{tabular}
    \medskip
    \caption{The dynamics of Example \ref{ex_1} in numbers.}
    \label{table_ex_1}
  \end{center}
\end{table}

\begin{figure}[htbp]
  \centering
  \includegraphics[width=0.65\linewidth]{Example_5}
  \caption{The dynamics in Example \ref{ex_1}.}
  \label{fig:example-5}
\end{figure}

\begin{figure}[!ht]
  \begin{center}
    \setlength{\unitlength}{1.0cm}
    \begin{picture}(8,2.6)
      \put(0.4722,1){\circle{0.3}}
      \put(0.4722,1.4){\circle{0.3}}
      \put(0.4722,1.8){\circle{0.3}}
      \put(2,1){\circle{0.3}}
      \put(4,1){\circle{0.3}}
      \put(4,1.4){\circle{0.3}}
      \put(4,1.8){\circle{0.3}}
      \put(4,2.2){\circle{0.3}}
      \put(6,1){\circle{0.3}}
      \put(7.5277,1){\circle{0.3}}
      \put(7.5277,1.4){\circle{0.3}}
      \put(7.5277,1.8){\circle{0.3}}
      \put(-1,0.5){\line(1,0){10}}
      \multiput(0,0)(1,0){9}{\put(0,0.4){\line(0,1){0.2}}}
      \put(-0.2,0){$0\varepsilon$}
      \put(0.8,0){$1\varepsilon$}
      \put(1.8,0){$2\varepsilon$}
      \put(2.8,0){$3\varepsilon$}
      \put(3.8,0){$4\varepsilon$}
      \put(4.8,0){$5\varepsilon$}
      \put(5.8,0){$6\varepsilon$}
      \put(6.8,0){$7\varepsilon$}
      \put(7.8,0){$8\varepsilon$}
    \end{picture}
    \caption{Final positions of the individuals in Example \ref{ex_1}.}
    \label{fig_ex_1_2}
  \end{center}
\end{figure}

We remark that for symmetric weights $\beta_{ij}(t)=\beta_{ji}(t)$ one
can easily show that in the absence of truth seekers the dynamics
becomes stable after a finite number of time steps. In the case of
asymmetric weights $\beta_{ij}(t)\neq\beta_{ji}(t)$ we only have
convergence, but need not reach a stable state after an arbitrary,
problem dependent, but finite number of time steps, as illustrated in
the following example.

\begin{example}
  \label{ex_2}
  Consider a (WASBOCOS) with $\alpha_i(t)=0$ (no truth seekers),
  $n=2$, $x_1(0)=0$, $x_2(0)=\varepsilon$,
  $\beta_{11}(t)=\frac{2}{3}$, $\beta_{12}(t)=\frac{1}{3} = \beta$,
  $\beta_{21}(t)=\frac{1}{2}$, $\beta_{22}(t)=\frac{1}{2}$; the values
  of $\alpha$ and~$h$ do not matter.

  \medskip

  One can easily verify, e.g.{}, by induction, that we have
  $$
    x_1(t)=\left(\frac{2}{5}-\frac{2}{5\cdot 6^t}\right)\cdot\varepsilon\text{ and }
    x_2(t)=\left(\frac{2}{5}+\frac{3}{5\cdot 6^t}\right)\cdot\varepsilon
  $$
  for all $t\in\mathbb{N}$. So we have $|x_1(t)-x_2(t)|=\frac{1}{6^t}\cdot\varepsilon>0$ but
  clearly the opinions of the two individuals converge to $\frac{2}{5}$.
\end{example}
All stated insights with the absence of truth seekers were known so
far. It becomes a bit more interesting if we allow truth seekers,
i.e.{}, if we consider a general (WASBOCOS).
\begin{example}
  \label{ex_3}
  Consider a (WASBOCOS) with $\alpha_1(t)=\alpha$, $\alpha_i(t)=0$ for $i\neq 1$, $\beta_{ij}(t)=\frac{1}{n}$,
  $h=\frac{1}{2}\varepsilon$, $x_1(0)=0$, and $x_i(0)=\varepsilon$ for $i\neq 1$. The opinion $u_t$ of the
  truth seeker $1$ at time
  $t$ and the opinion $v_t$ of the other ignorants at time $t>0$ are given by
  \begin{eqnarray*}
    u_t&=&\left[\frac{1}{2}-\alpha
    \left(\frac{1}{2}-\frac{1}{n}\right)\left(1-\frac{\alpha}{n}\right)^{t-1}\right]\varepsilon,\\
    v_t&=&\left[\frac{1}{2}+
    \left(\frac{1}{2}-\frac{1}{n}\right)\left(1-\frac{\alpha}{n}\right)^{t-1}\right]\varepsilon
  \end{eqnarray*}
  respectively. This can be verified, e.g.{}, by induction. We see
  that the opinions of the truth seekers, and here also those of the
  ignorants, converge to the truth $h=\frac{1}{2}\varepsilon$.

  Note the opinions of ignorants may in general fail to converge to
  the truth as one can see by adding some further ignorants with
  $\tilde{x}_i(0)=3\varepsilon$.
\end{example}
As our analytical investigation of the previous example was rather
technical, we also depict the situation for special values $n=6$ and
$\alpha=\frac{2}{3}$ in Figure \ref{fig_ex_3}.  We sketch the truth
seeker by a filled circle and the ignorants by an empty circle.

\begin{figure}[h]
  \begin{center}
    \includegraphics[width=0.65\linewidth]{Example_4}
    \caption{The dynamics in example \ref{ex_3}.}
    \label{fig_ex_3}
  \end{center}
\end{figure}

One can easily imagine more complicated configurations as in Example
\ref{ex_3} where one has little chance and willingness to describe the
situation analytically. Our result Theorem \ref{main_result} states
that -- whatever the parameters of a (WASBOCOS) are -- the opinions of the
truth seekers converge to the truth. This settles an open conjecture
of Hegselmann and Krause.

\section{The crucial objects}

To get a first impression of what we may expect in terms of
convergence we consider a lonely truth seeker, i.e.{}, $n=1$.

\begin{lemma}
  \label{lemma_lonely}
  For a lonely truth seeker $i=1$ we have
  $$
    |x_i(t+r)-h|\le |x_i(t)-h|\cdot(1-\alpha)^r.
  $$
\end{lemma}
\begin{proof}
  $$
    |x_i(t+1)-h|=|x_i(t)-h|\cdot(1-\alpha_i(t))\le |x_i(t)-h|\cdot(1-\alpha).
  $$
\end{proof}

Clearly this bound is tight. Similar to this very special situation of
a lonely truth seeker is the case $\varepsilon=0$, so that we now
assume $\varepsilon>0$ for the remaining part of this article.

In order to describe the states of the discrete time dynamical system
with more than one truth seeker, we look at the truth seekers with the
most extreme opinions.

\begin{definition}
  \label{def_lower_upper}
  We define $\tilde{u}(t)\in K$ as the lexicographically smallest
  truth seeker which fullfills $x_{\tilde{u}(t)}(t)\ge h$ and
  $x_{\tilde{u}(t)}(t)\ge x_k(t)$ for all $k\in K$. If there is no
  truth seeker with opinion greater or equal to the truth $h$ we set
  $\tilde{u}(t)=0$.  In order to avoid case distinctions, we define
  $x_0(t'):=h$ for all $t'\in\mathbb{N}$. Similar we define
  $\tilde{l}(t)$ as the lexicographically smallest truth seeker that
  fullfills $x_{\tilde{l}(t)}(t)\le h$ and $x_{\tilde{l}(t)}(t)\le
  x_k(t)$ for all $k\in K$. Again, we set $\tilde{l}(t)=0$ if
  there is no such truth seeker.
\end{definition}
Due to the \emph{symmetrical} -- one could say \textit{fair} --
definition of the confidence set, the confidence structure between the
individuals can be described as a simple graph with loops.
\begin{definition}
  The \emph{confidence graph} $\mathcal{G}(t)$ with vertex set $V(t)$
  and edge set $E(t)$, of a configuration $x(t)=(x_1(t) , \dots ,
  x_n(t))\in\mathbb{R}^n$ and the additional is defined as follows:
  \begin{eqnarray*}
    V(t):=[n]\cup\{0\},\\
    E(t):=\{\{i,j\} \in \tbinom{V}{2} \mid |x_i(t)-x_j(t)|\le\varepsilon\}.
  \end{eqnarray*}
  For $i\in V(t)$ let $C_i(t)$ be the set of vertices in the
  connectivity component of vertex $i$ in $\mathcal{G}(t)$.
\end{definition}

Because we want to keep track of the individuals which can influence
the truth seekers in the future, we give a further definition for
individuals, which is similar to Definition \ref{def_lower_upper} for
truth seekers.

\begin{definition}
  \label{def_extreme_individuals}
  We define $\hat{u}(t)\in C_{\tilde{u}(t)}(t)$ as the
  lexicographically smallest individual with $x_{\hat{u}(t)}(t)\ge
  x_c(t)\,\,\forall c\in C_{\tilde{u}(t)}(t)$ and $\hat{l}(t)\in
  C_{\tilde{l}(t)}(t)$ as the lexicographically smallest individual
  with $x_{\hat{l}(t)}(t)\le x_c(t)\,\,\forall c\in
  C_{\tilde{l}(t)}(t)$ for all $t\in\mathbb{N}$.
\end{definition}

The opinions of $\hat{u}(t)$ and $\hat{l}(t)$ form an interval
$[x_{\hat{l}(t)}(t),x_{\hat{u}(t)}(t)]$ called the \emph{hope
  interval} which is crucial for our further investigations. To prove
the main theorem we will show that the length of this hope interval
converges to zero.

\medskip

In Figure \ref{fig_def_borders}, we have depicted a configuration to
illustrate Definition \ref{def_lower_upper} and Definition
\ref{def_extreme_individuals}. In particular, we have $\tilde{l}=4$,
$\tilde{u}=9$, $\hat{l}=2$, and $\hat{u}=12$. Individual~$1$ is
\textit{lost} and not contained in the hope interval, because there is
no path in $\mathcal{G}$ from $1$ to $\tilde{l}=4$. So we already know
that the opinion of Individual~$1$ will not converge to the truth.

\begin{figure}[htp]
  \begin{center}
    \setlength{\unitlength}{1.0cm}
    \begin{picture}(6,3.0)
      \put(-1,1){\circle{0.3}}
      \put(1.7,1){\circle{0.3}}
      \put(2.3,1){\circle{0.3}}
      \put(3,1){\circle*{0.3}}
      \put(3.35,1){\circle{0.3}}
      \put(3.7,1){\circle*{0.3}}
      \put(4,1){\circle*{0.3}}
      \put(4.5,1){\circle{0.3}}
      \put(5,1){\circle*{0.3}}
      \put(5,1.4){\circle*{0.3}}
      \put(5.4,1){\circle{0.3}}
      \put(5.8,1){\circle{0.3}}
      \put(-1.1,1.7){$\downarrow$}
      \put(-1.1,2.1){$1$}
      \put(1.6,1.7){$\downarrow$}
      \put(1.6,2.1){$2$}
      \put(2.2,1.7){$\downarrow$}
      \put(2.2,2.1){$3$}
      \put(2.9,1.7){$\downarrow$}
      \put(2.9,2.1){$4$}
      \put(3.25,1.7){$\downarrow$}
      \put(3.25,2.1){$5$}
      \put(3.6,1.7){$\downarrow$}
      \put(3.6,2.1){$6$}
      \put(3.9,1.7){$\downarrow$}
      \put(3.9,2.1){$7$}
      \put(4.4,1.7){$\downarrow$}
      \put(4.4,2.1){$8$}
      \put(4.9,1.7){$\downarrow$}
      \put(4.9,2.1){$9$}
      \put(4.75,2.5){$10$}
      \put(5.3,1.7){$\downarrow$}
      \put(5.15,2.1){$11$}
      \put(5.7,1.7){$\downarrow$}
      \put(5.65,2.1){$12$}
      \put(-1,0.5){\line(1,0){8}}
      \multiput(0,0)(1,0){8}{\put(-1,0.4){\line(0,1){0.2}}}
      \put(-1.2,0){$0\varepsilon$}
      \put(-0.2,0){$\frac{1}{2}\varepsilon$}
      \put(0.8,0){$1\varepsilon$}
      \put(1.8,0){$\frac{3}{2}\varepsilon$}
      \put(4.25,-0.2){\line(0,1){1.5}}
      \put(4.15,1.7){$\downarrow$}
      \put(4.13,2.1){$\mathbf{h}$}
      \put(2.8,0){$2\varepsilon$}
      \put(3.8,0){$\frac{5}{2}\varepsilon$}
      \put(4.8,0){$3\varepsilon$}
      \put(5.8,0){$\frac{7}{2}\varepsilon$}
    \end{picture}
    \caption{Illustration of Definition \ref{def_lower_upper} and Definition \ref{def_extreme_individuals}.}
    \label{fig_def_borders}
  \end{center}
\end{figure}

In the configuration depicted in Figure \ref{fig_def_borders_2} we have $\tilde{l}=2$,
$\tilde{u}=0$, $\hat{l}=2$, and $\hat{u}=5$.

\begin{figure}[htp]
  \begin{center}
    \setlength{\unitlength}{1.0cm}
    \begin{picture}(6,3.0)
      \put(-1,1){\circle{0.3}}
      \put(1.7,1){\circle*{0.3}}
      \put(4.5,1){\circle{0.3}}
      \put(5.4,1){\circle{0.3}}
      \put(5.8,1){\circle{0.3}}
      \put(-1.1,1.7){$\downarrow$}
      \put(-1.1,2.1){$1$}
      \put(1.6,1.7){$\downarrow$}
      \put(1.6,2.1){$2$}
      \put(4.4,1.7){$\downarrow$}
      \put(4.4,2.1){$3$}
      \put(5.3,1.7){$\downarrow$}
      \put(5.31,2.1){$4$}
      \put(5.7,1.7){$\downarrow$}
      \put(5.71,2.1){$5$}
      \put(-1,0.5){\line(1,0){8}}
      \multiput(0,0)(1,0){8}{\put(-1,0.4){\line(0,1){0.2}}}
      \put(-1.2,0){$0\varepsilon$}
      \put(-0.2,0){$\frac{1}{2}\varepsilon$}
      \put(0.8,0){$1\varepsilon$}
      \put(1.8,0){$\frac{3}{2}\varepsilon$}
      \put(4.25,-0.2){\line(0,1){1.5}}
      \put(4.15,1.7){$\downarrow$}
      \put(4.13,2.1){$\mathbf{h}$}
      \put(2.8,0){$2\varepsilon$}
      \put(3.8,0){$\frac{5}{2}\varepsilon$}
      \put(4.8,0){$3\varepsilon$}
      \put(5.8,0){$\frac{7}{2}\varepsilon$}
    \end{picture}
    \caption{Illustration of a special case in Definition \ref{def_lower_upper}.}
    \label{fig_def_borders_2}
  \end{center}
\end{figure}

Note that the weights $\beta_{ij}$ may be assymmetric.  Thus, the
sequence of the opinions of the individuals may reorder during the
time steps. As an example, consider, e.g.{}, three ignorants with
starting positions $x_1(0)=1\varepsilon$,
$x_2(0)=\frac{3}{2}\varepsilon$, and $x_3(0)=2\varepsilon$. The
weights may be given as $\beta_{11}(0)=0.01$, $\beta_{12}(0)=0.01$,
$\beta_{13}(0)=0.98$, $\beta_{21}(0)=0.98$, $\beta_{22}(0)=0.01$,
$\beta_{23}(0)=0.01$, $\beta_{31}(0)=0.4$, $\beta_{32}(0)=0.4$, and
$\beta_{33}(0)=0.2$. After one time step the new opinions are given by
$x_1(1)=1.985\varepsilon$, $x_2(1)=1.015\varepsilon$, and
$x_3(1)=1.4\varepsilon$. We remark that it is possible to achieve
every ordering of the three opinions in one time step by choosing
suitable weights $\beta_{ij}$ in this example. Nevertheless we have
the following straight-forward lemma:

\begin{lemma}
  \label{lemma_interval_1}
  Let $i$ be an ignorant, $l\in I_i^\varepsilon(t)$ be an individual
  with smallest opinion and $u\in I_i^\varepsilon(t)$ be an individual
  with largest opinion then we have $x_i(t+1)\in[x_l(t),x_u(t)]$.
\end{lemma}
\begin{proof}
  This follows directly from the system dynamics in Equation
  (\ref{eq_update}).
\end{proof}
For truth seekers we have a similar lemma:
\begin{lemma}
  \label{lemma_interval_2}
  Let $i$ be a truth seeker, $l\in I_i^\varepsilon(t)$ be an
  individual with smallest opinion, and let $u\in I_i^\varepsilon(t)$
  be an individual with largest opinion. For $x_i(t)\le h$ we have
  $x_i(t+1)\in[x_l(t),\max(h,x_u(t))]$ and for $x_i(t)\ge h$ we have
  $x_i(t+1)\in[\min(h,x_l(t)),x_u(t)]$.
\end{lemma}
Our goal is to prove that the length of the hope interval converges to
zero. To this end, we show first that the length does not increase
after an iteration of Equation (\ref{eq_update}).
\begin{lemma}
  \label{lemma_decreasing}
  For all time steps $t\in\mathbb{N}$ we have $x_{\hat{u}(t+1)}(t+1)\le x_{\hat{u}(t)}(t)$ and
  $x_{\hat{l}(t+1)}(t+1)\ge x_{\hat{l}(t)}(t)$.
\end{lemma}
\begin{proof}
  We only prove the last inequality since the proof is symmetric for
  the first inequality. Due to Definition
  \ref{def_extreme_individuals}, we have $x_{\hat{l}(t+1)}(t+1)\le h$
  and $x_{\hat{l}(t)}(t)\le h$. By $\mathcal{L}(t)$ we denote the set
  of individuals with opinion strictly smaller than
  $x_{\hat{l}(t)}(t)$.  That is, $\mathcal{L}(t'):=\{i\in[n]\mid
  x_i(t')<x_{\hat{l}(t')}(t')\}$ for all $t'\ge t$. We remark that, by
  definition, $\mathcal{L}(t')$ does not contain a truth seeker. We
  set $\mathcal{U}(t'):=[n]\backslash\mathcal{L}(t')$; this set
  contains the remaining individuals.

  \medskip

  Let $u$ be an individual in $\mathcal{L}(t)$ with the largest
  opinion. By applying Lemma \ref{lemma_interval_1} we get
  $x_i(t+1)\le x_u(t)$ for all $i\in\mathcal{L}(t)$. Now let $l$
  (e.g.{}, $l=\hat{l}(t)$) be an individual in $\mathcal{U}(t)$ with
  smallest opinion then by applying Lemma \ref{lemma_interval_1} and
  Lemma \ref{lemma_interval_2} we receive $x_i(t+1)\ge x_l(t)$ for all
  $i\in\mathcal{U}(t)$. Thus, we have $\hat{l}(t+1)\in\mathcal{U}(t)$
  and so $x_{\hat{l}(t+1)}(t+1)\ge x_{\hat{l}(t)}(t)$ follows.
\end{proof}

In the remaining part of this article we prove that the length of the
hope interval $|x_{\hat{u}(t)}(t)-x_{\hat{l}(t)}(t)|$ converges (in
some special sense) to zero, as $t$ tends to infinity.

\section{Proof of the Generalized Hegselmann-Krause Conjecture}

One difficulty in the proof arises from the fact that convergence
happens in two phases: in a first phase, the hope interval becomes
sufficiently small so that the confidence graph is the complete graph.
Then, it may happen that truth seekers approaching the truth from one
side get distracted to the other side of the truth.  At that point,
however, the confidence structure is so simple that all individuals in
the hope interval converge to the truth.  Since all truth seekers are
in the hope interval at all times, this proves the theorem.  Where
exactly we split the phases is a technical decisison.

First, we show that after a finite number $T_1$ of time steps,
depending only on $n$, $\varepsilon$, $\alpha$, and $\beta$, the hope
interval $[x_{\hat{l}(T_1)}(T_1),x_{\hat{u}(T_1)}(T_1)]$ is contained
in the interval
$[h-\varepsilon-\frac{\varepsilon\alpha\beta}{12},h+\varepsilon+\frac{\varepsilon\alpha\beta}{12}]$. Therefore,
we introduce the following notion.
\begin{definition}
  A \emph{good iteration} is an iteration where for $1\le r\le 3$
  one of the following conditions is fullfilled:
  \begin{enumerate}
    \item[(1)] the number of individuals in the hope interval decreases,
    \item[(2)] the opinion of $\hat{l}(t+r)$ reaches or passes $h-\varepsilon-\frac{\varepsilon\alpha\beta}{12}$,
    \item[(3)] the opinion of  $\hat{u}(t+r)$ reaches or passes $h+\varepsilon+\frac{\varepsilon\alpha\beta}{12}$,
    \item[(4)] $|x_{\hat{u}(t+r)}(t+r)-x_{\hat{u}(t)}(t)|\ge\frac{\varepsilon\alpha\beta^2}{12}$,
    \item[(5)] $|x_{\hat{l}(t+r)}(t+r)-x_{\hat{l}(t)}(t)|\ge\frac{\varepsilon\alpha\beta^2}{12}$.
  \end{enumerate}
\end{definition}

Clearly, there is only a finite number of good iterations. We may
choose $T_1=3\cdot\left(n+2\cdot
  1+2\cdot\frac{12}{\varepsilon\alpha\beta^2}\right)$. We formulate
the next two lemmas only for the lower bound $x_{\hat{l}(t)}(t)$
because analog arguments hold for $x_{\hat{u}(t)}(t)$. As a shorthand
we define $d(i,j,t):=\left|x_i(t)-x_j(t)\right|$. For each point in
time $t$ we define the sets
\begin{eqnarray*}
  &&\mathcal{N}(t):=\left\{i\in[n]\mid d(\hat{l}(t),i,t)\in\Big[0,\frac{\varepsilon\alpha\beta}{12}
  \Big)\right\},\\
  &&\mathcal{M}(t):=\left\{i\in[n]\mid d(\hat{l}(t),i,t)\in\Big[\frac{\varepsilon\alpha\beta}{12},
  \varepsilon\Big]\right\},
  \text{ and}\\
  &&\mathcal{F}(t):=\left\{i\in[n]\mid x_i(t)-x_{\hat{l}(t)}(t)>\varepsilon\right\}.
\end{eqnarray*}

\begin{lemma}
  \label{lemma_middle}
  If $\mathcal{M}(t)\neq\emptyset$ then there is a good iteration after $1$ step.
\end{lemma}
\begin{proof}
  We assume that there is an individual $j\in\mathcal{M}(t)$, i.e.{},
  $d(\hat{l}(t),j,t)\in\left[\frac{\varepsilon\alpha\beta}{12},\varepsilon\right]$.
  For the evaluation of Equation (\ref{eq_update}) for elements of
  $\mathcal{N}(t)$, $\mathcal{M}(t)$, or $\mathcal{F}(t)$ we do not
  need to consider the opinion of individuals in
  $[n]\backslash(\mathcal{N}(t)\cup\mathcal{M}(t)\cup\mathcal{F}(t))$. Let
  $i$ be an element of $\mathcal{N}(t)$ with opinion
  $x_i(t)=x_{\hat{l}(t)}+\delta$, where
  $0\le\delta<\frac{\varepsilon\alpha\beta}{12}$.  Let us first assume
  that $i$ is an ignorant. Due to Individual~$j$ we have
  \begin{eqnarray*}
    x_i(t+1)&\ge&
    x_i(t)-\underset{\text{individuals in }\mathcal{N}(t)\backslash\{i\}}
    {\underbrace{\delta\left(1-2\beta\right)}}
    +\underset{i}{\underbrace{0\cdot\beta}}
    +\underset{j}{\underbrace{
    \left(\frac{\varepsilon\alpha\beta}{12}-\delta\right)\cdot\beta}}\\
    &\ge& x_{\hat{l}(t)}+\frac{\varepsilon\alpha\beta^2}{12}.
  \end{eqnarray*}
  For a truth seeker we similarly get
  \begin{eqnarray*}
    x_i(t+1)&\ge&x_i(t)+\alpha\varepsilon+(1-\alpha)
    \left(-\delta\left(1-2\beta\right)+\left(\frac{\varepsilon\alpha\beta}{12}
    -\delta\right)\cdot\beta\right)\\
    &\ge&x_{\hat{l}(t)}+\frac{\varepsilon\alpha\beta^2}{12}.
  \end{eqnarray*}
  Now let $i$ be an element of $\mathcal{M}(t)\cup\mathcal{F}(t)$ with $x_i(t)=x_{\hat{l}(t)}+\delta$ where
  $\delta\ge\frac{\varepsilon\alpha\beta}{12}$. In any case ($i$ being a truth seeker or an ignorant) we have
  $$
    x_i(t+1)\ge x_{\hat{l}(t)}+\delta-\underset{\text{individuals with smaller opinion than
    $i$}}{\underbrace{\delta(1-\beta)}}+\beta\cdot 0\ge
    x_{\hat{l}(t)}+\frac{\varepsilon\alpha\beta^2}{12}.
  $$
\end{proof}

\begin{lemma}
  If $x_{\hat{l}(t)} <
  h-\varepsilon-\frac{\varepsilon\alpha\beta}{12}$ then after at least
  $3$ time steps we have a good iteration.
\end{lemma}
\begin{proof}
  Due to Lemma \ref{lemma_middle} we can assume
  $\mathcal{M}(t)=\mathcal{M}(t+1)=\mathcal{M}(t+2)=\emptyset$. We can
  also assume
  \begin{eqnarray*}
    \left|x_{\hat{l}(t)}(t)-x_{\hat{l}(t+1)}(t+1)\right|&<&\frac{\varepsilon\alpha\beta^2}{12},\\
    \left|x_{\hat{l}(t+1)}(t+1)-x_{\hat{l}(t+2)}(t+2)\right|&<&\frac{\varepsilon\alpha\beta^2}{12},\text{ and}\\
    d\left(\hat{l}(t+1),0,t+1\right)&>&\varepsilon+\frac{\varepsilon\alpha\beta}{12}
  \end{eqnarray*}
  since otherwise we have a good iteration in at most $2$ time
  steps. At first we claim $\mathcal{N}(t+1)\cap K=\emptyset$.  If at
  time $t$ there is a truth seeker $i\in\mathcal{N}(t)\cap K$ then we
  have
  \begin{eqnarray*}
    x_i(t+1)&\ge& x_{\hat{l}(t)}(t)+\alpha\varepsilon-\frac{(1-\alpha)(1-\beta)\varepsilon\alpha\beta}{12}\\
    &\ge& x_{\hat{l}(t)}(t) +\frac{\varepsilon\alpha\beta^2}{12}+\frac{\varepsilon\alpha\beta}{12}\\
    &\ge& x_{\hat{l}(t+1)}(t+1) +\frac{\varepsilon\alpha\beta}{12}.
  \end{eqnarray*}
  So the only truth seekers that have a chance to move into the set
  $\mathcal{N}(t+1)$ could be those of the set $\mathcal{F}(t)$. So
  let truth seeker $i$ be in the set $\mathcal{F}(t)\cap K$, with
  $x_i(t)=x_{\hat{l}(t)}(t)+\delta$, where
  $\varepsilon<\delta<\varepsilon+\frac{\varepsilon\alpha\beta}{12}$.
  (Truth seekers where $\delta\ge
  \varepsilon+\frac{\varepsilon\alpha\beta}{12}$ are ruled out by
  Lemma \ref{lemma_interval_2}.) We have
  \begin{eqnarray*}
     x_i(t+1) &\ge& \underset{\le x_i(t)}{\underbrace{x_{\hat{l}(t)}(t)+\varepsilon}}-(1-\alpha)(1-2\beta)\\
              &\ge& x_{\hat{l}(t)}(t)+\varepsilon\alpha\\
              &\ge& x_{\hat{l}(t)}(t)+\frac{\varepsilon\alpha\beta^2}{12}+\frac{\varepsilon\alpha\beta}{12}\\
              &\ge& x_{\hat{l}(t+1)}(t+1)+\frac{\varepsilon\alpha\beta}{12}.
  \end{eqnarray*}
  Similarly, we can deduce $\mathcal{N}(t+2)\cap K=\emptyset$. Now we
  can assume that the individuals of $\mathcal{N}(t+1)$, who are all
  ignorants, are in the hope interval at time $t+1$, since otherwise
  we would have a good iteration after $1$ time step. So there exist
  individuals $i\in\mathcal{N}(t+1)$ and $j\in\mathcal{F}(t+1)$ with
  $\left|x_i(t+1)-x_j(t+1)\right|\le\varepsilon$. We set
  $x_i(t+1)=x_{\hat{l}(t+1)}(t+1)+\delta$, where $0\le
  \delta\le\frac{\varepsilon\alpha\beta}{12}$ and calculate
  \begin{eqnarray*}
    x_i(t+2) &\ge& x_i(t+1)-(1-2\beta)\delta+
    \underset{i}{\underbrace{\beta\cdot 0}}
    +\underset{j}{\underbrace{\beta\left(\varepsilon-\frac{\varepsilon\alpha\beta}{12}\right)}}\\
    &\ge& x_{\hat{l}(t+1)}(t+1)+\frac{\varepsilon\alpha\beta^2}{12}+\frac{\varepsilon\alpha\beta}{12}\\
    &\ge& x_{\hat{l}(t+2)}(t+2)+\frac{\varepsilon\alpha\beta}{12}.
  \end{eqnarray*}
  For the other direction we have
  \begin{eqnarray*}
    x_i(t+2) &\le& x_i(t+1)-\underset{\hat{l}(t+1)}{\underbrace{\beta\delta}}+(1-2\beta)\varepsilon\\
             &\le& x_{\hat{l}(t+1)}(t+1)+\frac{\varepsilon\alpha\beta}{12}+\varepsilon-2\beta\varepsilon\\
             &\le& x_{\hat{l}(t+1)}(t+1)+\frac{\varepsilon\alpha\beta^2}{12}+\varepsilon\\
             &\le& x_{\hat{l}(t+2)}(t+2)+\varepsilon.
  \end{eqnarray*}
  Thus, $i\in\mathcal{M}(t+2)$, which results in a good iteration in three time steps.
\end{proof}

Thus, we can conclude:
\begin{corollary}
  After a finite number $T_1(\varepsilon,n,\alpha,\beta)$ of steps we
  have $x_{\hat{l}(T_1)}(T_1)\ge h-
  \varepsilon-\frac{\varepsilon\alpha\beta}{12}$ and
  $x_{\hat{u}(T_1)}(T_1)\le
  h+\varepsilon+\frac{\varepsilon\alpha\beta}{12}$.
\end{corollary}

Due to Lemma \ref{lemma_lonely} there can not exist a general bound on
the convergence that does not depend on $\alpha$. We consider the two
side lengths $\ell_2(t):=|x_{\hat{u}(t)}(t)-h|$ and
$\ell_1(t):=|x_{\hat{l}(t)}(t)-h|$ of the hope interval. Clearly $\ell_1(t)$
and $\ell_2(t)$ are not increasing due to Lemma
\ref{lemma_decreasing}. For $t\ge T_1$ we have
$\ell_1(t),\ell_2(t)\le\varepsilon+\frac{\varepsilon\alpha\beta}{12}$

\begin{lemma}
  \label{lemma_epsilon_interval}
  If $\ell_1(t)+\ell_2(t)\le\varepsilon$ then we have
  $$
    (\ell_1(t+2)+\ell_2(t+2))\le (\ell_1(t)+\ell_2(t))\cdot\left(1-\frac{\alpha\beta}{2}\right).
  $$
\end{lemma}
\begin{proof}
  Let us assume, without loss of generality, that $\ell_1(t)\ge
  \ell_2(t)$. At first we consider the case $\ell_2(t)>0$. If $i$ is an
  ignorant with $x_i(t)=h-\ell_1(t)+\delta$ then we have
  \begin{eqnarray*}
    x_i(t+1) &\ge& h-\ell_1(t)+\delta -(1-2\beta)\delta+\beta(\ell_1(t)+\ell_2(t)-\delta)\\
    &\ge& h-(1-\beta)\ell_1(t).
  \end{eqnarray*}
  For a truth seeker $i$ with $x_i(t)=h-\ell_1(t)+\delta$ we have
  \begin{eqnarray*}
    x_i(t+1) &\ge& h-\ell_1(t)+\delta-\alpha(\delta-\ell_1(t))-(1-\alpha)(1-2\beta)\delta+\\
    &&(1-\alpha)\beta(\ell_1(t)+\ell_2(t)-\delta)\\
    &\ge&h-\ell_1(t)+\beta\delta(1-\alpha)+\alpha \ell_1(t)(1-\beta)+\beta \ell_2(t)(1-\alpha)+\beta \ell_1(t)\\
    &\ge&h-(1-\beta)\ell_1(t).
  \end{eqnarray*}
  Similarly we obtain $x_i(t+1)\le h+(1-\beta)\ell_2(t)$ in both cases.

  \medskip

  Next we consider the case $\ell_1(t)>\ell_2(t)=0$ and
  $l_i(t+1)>l_i(t)\cdot\left(1-\frac{\alpha}{2}\right)$.  Let $i$ be
  an arbitrary truth seeker with opinion $x_i(t)=h-\ell_1(t)+\delta$.  We
  have
  \begin{eqnarray*}
    x_i(t+1)&\ge& h-\ell_1(t)+\delta+\alpha(\ell_1(t)-\delta)-(1-\alpha)(1-\beta)\delta\\
    &\ge & h-\ell_1(t)+\alpha \ell_1(t).
  \end{eqnarray*}
  Thus, we have $x_i(t+1)\ge h-\ell_1(t+1)+\frac{\alpha}{2}\cdot
  \ell_1(t)$. If $j$ is an ignorant with $x_j(t+1)=h-\ell_1(t+1)+\delta$,
  then we have
  \begin{eqnarray*}
    x_j(t+2)&\ge& h-\ell_1(t+1)+\delta-(1-2\beta)\delta+\beta\left(\frac{\alpha}{2}\cdot \ell_1(t)-\delta\right)\\
    &\ge& h-\ell_1(t)+\frac{\alpha\beta \ell_1(t)}{2}.
  \end{eqnarray*}
  For an arbitrary truth seeker $j$ we have
  \begin{eqnarray*}
    x_i(t+2)&\ge& h-\ell_1(t+1)+\alpha \ell_1(t+1)\\
    &\ge& h-\ell_1(t)+\frac{\alpha\beta \ell_1(t)}{2}.
  \end{eqnarray*}

  \medskip

  \noindent
  Thus, in all cases we have $(\ell_1(t+2)+\ell_2(t+2))\le
  (\ell_1(t)+\ell_2(t))\cdot\left(1-\frac{\alpha\beta}{2}\right)$.
\end{proof}

This states that once the length of the hope interval becomes at most
$\varepsilon$ its lengths converges to zero.

\begin{lemma}
  \label{lemma_one_step}
  Let $t\ge T_1$.  If there exists an individual~$i$ with
  $\frac{\alpha\beta \ell_1(t)}{12}\le d(\hat{l}(t),i,t)\le\varepsilon$,
  then we have $\ell_1(t+1)\le \ell_1(t)\cdot
  \left(1-\frac{\alpha\beta^2}{12}\right)$. If there exists an
  individual $i$ with $\frac{\alpha\beta \ell_2(t)}{12}\le
  d(\hat{u}(t),i,t)\le\varepsilon$, then we have $\ell_2(t+1)\le
  \ell_2(t)\cdot \left(1-\frac{\alpha\beta^2}{12}\right)$.
\end{lemma}
\begin{proof}
  Due to symmetry it suffices to prove the first statement.  Let $j$
  be an ignorant with $x_j(t)=h-\ell_1(t)+\delta$, where $\delta\ge
  0$. We have
  \begin{eqnarray*}
     x_j(t+1) &\ge& h-\ell_1(t)+\delta-(1-2\beta)\delta+\underset{i}{\underbrace{\beta\left(
     \frac{\alpha\beta \ell_1(t)}{12}-\delta\right)}}\\
     &\ge& h-\left(1-\frac{\alpha\beta^2}{12}\right)\ell_1(t).
  \end{eqnarray*}
  For a truth seeker $j$ with $x_j(t)=h-\ell_1(t)+\delta$, $\delta\ge 0$
  we have
  \begin{eqnarray*}
    x_j(t+1) &\ge& h-\ell_1(t)+\delta+\alpha(\ell_1(t)-\delta)-(1-\alpha)(1-2\beta)\delta+\\
    &&\underset{i}{\underbrace{(1-\alpha)\beta\left(\frac{\alpha\beta \ell_1(t)}{12}-\delta\right)}}\\
    &\ge& h-\ell_1(t)+\beta\delta(1-\alpha)+\alpha \ell_1(t)\left(1-\frac{\alpha\beta^2}{12}\right)+\frac{\alpha\beta^2
    \ell_1(t)}{12}\\
    &\ge& h-\left(1-\frac{\alpha\beta^2}{12}\right)\ell_1(t).
  \end{eqnarray*}
\end{proof}

For transparency we introduce the following six sets:
\begin{eqnarray*}
  \mathcal{N}_1(t):&=&\left\{i\in[n]\mid d(\hat{l}(t),i,t)
  <\frac{\alpha\beta \ell_1(t)}{12}\right\},\\
  \mathcal{N}_2(t):&=&\left\{i\in[n]\mid d(\hat{u}(t),i,t)
  <\frac{\alpha\beta \ell_2(t)}{12}\right\},\\
  \mathcal{M}_1(t):&=&\left\{i\in[n]\mid \frac{\alpha\beta \ell_1(t)}{12}\le d(\hat{l}(t),i,t)\le\varepsilon\right\},\\
  \mathcal{M}_2(t):&=&\left\{i\in[n]\mid \frac{\alpha\beta \ell_2(t)}{12}\le d(\hat{u}(t),i,t)\le\varepsilon\right\},\\
  \mathcal{F}_1(t):&=&\left\{i\in[n]\mid d(\hat{l}(t),i,t)>\varepsilon,\,x_i(t)\le h+\ell_2(t)\right\},\\
  \mathcal{F}_2(t):&=&\left\{i\in[n]\mid d(\hat{u}(t),i,t)>\varepsilon,\,x_i(t)\ge h-\ell_1(t)\right\}.\\
\end{eqnarray*}

With this the individuals of the hope interval are partitioned into
$$
  \mathcal{N}_1(t)\cup\mathcal{M}_1(t)\cup\mathcal{F}_1(t)=\mathcal{N}_2(t)\cup\mathcal{M}_2(t)\cup\mathcal{F}_2(t).
$$

\begin{lemma}
  \label{lemma_two_step}
  If for $k\in\{1,2\}$ and $t\ge T_1$ there exists an ignorant $i\in
  \mathcal{N}_k(t)$ and an individual $j\in\mathcal{F}_k(t)$ with
  $|x_i(t)-x_j(t)|\le\varepsilon$ then $l_k(t+2)\le l_k(t)\cdot
  \left(1-\frac{\alpha\beta^2}{12}\right)$.
\end{lemma}
\begin{proof}
  If $l_k(t+1)>l_k(t)\cdot \left(1-\frac{\alpha\beta^2}{12}\right)$,
  then it is easy to check that the influence of Individual~$j$
  suffices to put ignorant $i$ in set $\mathcal{M}_k(t+1)$. In this
  case we can apply Lemma \ref{lemma_one_step}
\end{proof}

\begin{lemma}
  \label{lemma_no_near_truth_seeker}
  If $\mathcal{N}_k(t+1)\cap K\neq\emptyset$ and $t\ge T_1$ then
  $l_k(t+1)\le l_k(t)\cdot\left(1-\frac{\alpha}{2}\right)$ for
  $k\in\{1,2\}$.
\end{lemma}
\begin{proof}
  Due to symmetry it suffices to consider $k=1$. So let $i$ be a truth
  seeker with $i\in\mathcal{N}_1(t+1)$.  We set
  $x_i(t)=h-\ell_1(t)+\delta$ and calculate
  \begin{eqnarray*}
    x_i(t+1) &\ge& h-\ell_1(t)+\delta +\alpha(\ell_1(t)-\delta)-(1-\alpha)(1-\beta)\delta\\
    &\ge& h-(1-\alpha)\ell_1(t).
  \end{eqnarray*}
\end{proof}

\begin{lemma}
  \label{lemma_one_shrinks}
  For $t\ge T_1$ we have $l_k(t+3)\le l_k(t)\cdot
  \left(1-\frac{\alpha\beta^2}{12}\right)$ for at least one
  $k\in\{1,2\}$.
\end{lemma}
\begin{proof}
  Due to Lemma \ref{lemma_no_near_truth_seeker} we can assume
  $\mathcal{N}_k(t+1)\cap K=\emptyset$. At time $t+1$ there must be a
  truthseeker $i$. Without loss of generality, we assume $x_i(t)\le h$
  and $i=\tilde{l}(t+1)$. Due to Lemma \ref{lemma_one_step} we can
  assume $i\in\mathcal{F}_1(t+1)$. Now let $j_1$ be the ignorant with
  smallest opinion fulfilling $d(i,j_1,t+1)\le\varepsilon$. If
  $j_1\in\mathcal{N}_1(t+1)$ then we can apply Lemma
  \ref{lemma_two_step} with $j_1$ and $i$. Otherwise we let $j_2$ be
  the ignorant with smallest opinion fulfilling
  $d(j_1,j_2,t+1)\le\varepsilon$. So we have
  $d(j_2,i,t+1)>\varepsilon$ and $j_2\in\mathcal{N}_1(t+1)$. Thus, we
  can apply Lemma \ref{lemma_two_step} with $j_2$ and $j_1$.
\end{proof}

\begin{lemma}
  \label{lemma_greater_epsilon}
  If $l_k(t)>\varepsilon$ and $t\ge T_1$ then we have
  $l_k(t+3)-\varepsilon\le (l_k(t)-\varepsilon)\cdot
  \left(1-\frac{\alpha\beta^2}{12}\right)$ or
  $l_k(t+3)\le\varepsilon$.
\end{lemma}
\begin{proof}
  Due to Lemma \ref{lemma_no_near_truth_seeker}, we can assume
  $\mathcal{N}_k(t+1)\cap K=\emptyset$ and, due to Lemma
  \ref{lemma_one_step}, we can assume
  $\mathcal{M}_k(t+1)=\emptyset$. Due to symmetry, we only consider the
  case $k=1$.  Let $i\in\mathcal{N}_1(t+1)$ the ignorant with largest
  opinion $x_i(t+1)$, meaning that $d(\hat{l}(t+1),i,t+1)$ is
  maximal.

  If there exists an individual $j\in\mathcal{F}_1(t+1)$ with
  $d(i,j,t+1)\le\varepsilon$, then we can apply Lemma
  \ref{lemma_two_step}. If no such individual $j$ exists then we must
  have $d(i,0,t+1)\le\varepsilon$ or $\ell_1(t+1)=0$. So only the first
  case remains. We set $\delta=d(\hat{l}(t+1),i,t+1)\ge
  \varepsilon-\ell_1(t+1)$. Let $h\in\mathcal{N}_1(t+1)$ be an ignorant
  with $x_h(t+1)=x_{\hat{l}(t+1)}(t+1)+\mu$, where
  $0\le\mu\le\delta$. For time $t+2$ we get
  \begin{eqnarray*}
    x_h(t+2) &\ge& x_{\hat{l}(t+1)}(t+1)+\mu -(1-2\beta)\mu+\beta(\delta-\mu)\\
             &\ge& x_{\hat{l}(t+1)}(t+1)+\beta\delta\\
             &\ge& x_{\hat{l}(t+1)}(t+1)+\beta(\varepsilon-\ell_1(t+1)).
  \end{eqnarray*}
\end{proof}

From Lemma \ref{lemma_one_shrinks} and Lemma
\ref{lemma_greater_epsilon} we conclude:

\begin{corollary}
  \label{lemma_everything_small}
  There exists a finite number $T_2(\varepsilon,n,\alpha,\beta)$ so that we have
  $$
    \ell_1(t)+\ell_2(t)\le\varepsilon+\frac{\varepsilon\alpha^2\beta^3}{60},\quad\text{and}\quad
    \min(\ell_1(t),\ell_2(t))\le\frac{\varepsilon\alpha^2\beta^3}{60}
  $$
  for all $t\ge T_2(\varepsilon,n,\alpha,\beta)$.
\end{corollary}
We would like to remark that, e.g.{},
$T_2(\varepsilon,n,\alpha,\beta)=T_1(\varepsilon,n,\alpha,\beta)+\frac{36}{\alpha^2\beta^4}$
suffices.

\begin{lemma}
  \label{lemma_final}
  For each $t\ge T_2(\varepsilon,n,\alpha,\beta)$ we have
  $$
    \ell_1(t+3)+\ell_2(t+3)\le\varepsilon
  $$
  or
  $$
    d(k,0,t)\le\frac{\varepsilon\alpha^2\beta^3}{60}\cdot\left(1-\frac{\alpha\beta}{2}\right)^{\left\lfloor\frac{t-T_2}{2}\right\rfloor}
  $$
  for all truth seekers $k\in K$.
\end{lemma}
\begin{proof}
  Without loss of generality, we assume
  $\ell_1(t)\le\frac{\varepsilon\alpha^2\beta^3}{60}$ and prove the
  statement by induction on $t$. Due to Lemma \ref{lemma_one_step} and
  Lemma \ref{lemma_no_near_truth_seeker}, we can assume
  $\mathcal{M}_2(t+r)=\mathcal{N}_2(t+r)\cap K=\emptyset$ for
  $r\in\{0,1\}$ since otherwise we would have
  $\ell_1(t+3)+\ell_2(t+3)\le\varepsilon$. Thus, we have
  $d(k,0,t+r)\le\frac{\varepsilon\alpha^2\beta^3}{60}$ for all $k\in
  K$ and $K\subseteq \mathcal{F}_2(t+r)$.

  Due to Lemma \ref{lemma_one_step} for $r\in\{0,1\}$, the individuals
  in $\mathcal{F}_2(t+r)$ are not influenced by the individuals in
  $\mathcal{F}_2(t+r)$ since otherwise we would have
  $\ell_1(t+3)+\ell_2(t+3)\le\varepsilon$. Thus, we can apply Lemma
  \ref{lemma_epsilon_interval} for the individuals in
  $\mathcal{F}_2(t)$.
\end{proof}

From the previous lemmas we can conclude Theorem \ref{main_result} and
Theorem \ref{thm_interupted_convergent}.
\begin{proof}(Proof of Theorems~\ref{main_result} and~\ref{thm_interupted_convergent}.)
  After a finite time $T_2(\varepsilon,n,\alpha,\beta)$ we are in a
  \textit{nice} situation as described in Lemma
  \ref{lemma_everything_small}. If we have
  $\ell_1(T_2+3)+\ell_2(T_2+3)\le\varepsilon$ then we have an ordinary
  convergence of the truth seekers being described in Lemma
  \ref{lemma_epsilon_interval}. Otherwise we have
  $d(k,0,T_2)\le\frac{\varepsilon\alpha^2\beta^3}{60}$ for all truth
  seekers $k\in K$. Due to Lemma \ref{lemma_final} and Lemma
  \ref{lemma_epsilon_interval} either we have
$$
d(k,0,t)\le\frac{\varepsilon\alpha^2\beta^3}{60}\cdot\left(1-\frac{\alpha\beta}{2}\right)^{\left\lfloor\frac{t-T_2}{2}\right\rfloor}
$$
for all truth seekers $k\in K$ and all $t\ge T_2$, or there exists an
$S\in\mathbb{N}$, such that we have
\begin{enumerate}
\item[(1)]
  $d(k,0,t)\le\frac{\varepsilon\alpha^2\beta^3}{60}\cdot\left(1-\frac{\alpha\beta}{2}\right)^{\left\lfloor\frac{t-T_2}{2}\right\rfloor}$
  for all $T_2\le t\le S$,
\item[(2)]
  $d(k,0,t)\le\varepsilon\left(1-\frac{\alpha\beta}{2}\right)^{\left\lfloor\frac{t-S-3}{2}\right\rfloor}$
  for all $t\ge S+3$,
\end{enumerate}
for all $k\in K$.  The latter case is $1$-fold interrupted
convergence. Thus, the Hegselmann-Krause Conjecture is proven.
\end{proof}

\section{Remarks}

In this section we would like to generalize the Hegselmann-Krause
Conjecture and show up which requirements can not be weakened.

\begin{lemma}
  A finite number $n$ of individuals and symmetric confidence
  intervals are necessary for a convergence of the truth seekers.
\end{lemma}
\begin{proof}
  Infinitely many ignorants can clearly hinder a truth seeker in
  converging to the truth. If the confidence intervals are not
  symmetric then it is easy to design a situation where some ignorants
  are influencing a truth seeker which does not influence the
  ignorants, so that the truth seeker has no chance to converge to the
  truth.
\end{proof}

\begin{lemma}
  The condition $\beta_{ij}(t)\ge\beta>0$ is necessary for a
  convergence of the truth seekers.
\end{lemma}
\begin{proof}
  If we only require $\beta_{ij}(t)>0$, then we have the following
  example: $n=2$, $x_1(0)=1-\frac{1}{5}\varepsilon$,
  $x_2(0)=1-\varepsilon$, $\alpha_1(t)=\frac{1}{5}$, $\alpha_2(t)=0$,
  $\beta_{11}(t)=\left(\frac{1}{2}\right)^{t+1}$,
  $\beta_{12}(t)=1-\left(\frac{1}{2}\right)^{t+1}$,
  $\beta_{21}(t)=\left(\frac{1}{2}\right)^{t+1}$,
  $\beta_{22}(t)=1-\left(\frac{1}{2}\right)^{t+1}$, and $h=1$. By a
  straight forward calculation we find that
  $|x_1(t)-h|\ge\frac{1}{2}\varepsilon$ for $t\ge 1$.
\end{proof}

We remark that conditions like $\beta_{ij}(t)+\beta_{ij}(t+1)\ge
2\beta$ would also not force a convergence of the truth seekers in
general. One might consider an example consisting of two ignorants
with starting positions $h\pm\frac{7}{10}\varepsilon$ and a truth
seeker $k$ with starting position $h-\frac{1}{5}\varepsilon$. We may
choose suitable $\beta_{ij}(t)$ and $\alpha_i(t)$ so that we have
$|h-x_k(t)|\ge\frac{1}{5}\varepsilon$ for all $t$,
$h-x_k(t)\ge\frac{1}{5}\varepsilon$ for even $t$ and
$x_k(t)-h\ge\frac{1}{5}\varepsilon$ for odd $t$.

\medskip

For the next lemma we need a generalization of Definition
\ref{def_interupted_convergent}.

\begin{definition}
  Given $\varepsilon$, $\alpha$, $\beta$, $n$, we say that the truth
  seekers $k \in K$ are $r$-fold interrupted convergent, if for each
  $\gamma > 0$ there exists $r+1$ functions
  $T_i^s(\gamma,\varepsilon,\alpha,\beta,n,T_{i-1}^e)$,
  $i=1,\dots,r+1$, so that for each (WASBOCOS) $\Omega$ with structural
  parameters $\varepsilon$, $\alpha$, $\beta$ and $n$ there exist
  $T_i^e\in\mathbb{N}$, $i=1,\dots,r$ satisfying
  $$
    \forall k\in K,\,\forall t\in[T_i^s(\gamma,\varepsilon,\alpha,\beta,n,T_{i-1}^s),T_i^e]:\, |x_k(t)-h|<\gamma
  $$
  for $i=1,\dots,r$, where $T_0^e=0$, and
  $$
    \forall k\in K,\,\forall t\ge T_{r+1}^s(\gamma,\varepsilon,\alpha,\beta,n,T_r^e):\,|x_k(t)-h|<\gamma.
  $$
\end{definition}

\begin{lemma}
  The condition $\alpha_i(t)=0$ for all $i\in\overline{K}$ is
  necessary for Theorem \ref{thm_interupted_convergent}.  If it is
  dropped then the truth seekers are not ($|\overline{K}|-1$)-fold
  convergent in general.
\end{lemma}
\begin{proof}
  At first we remark that it clearly suffices to have $\alpha_i(t)=0$
  for all $i\in\overline{K}$ only for all $t\ge T$, where $T$ is a fix
  integer. W.l.o.g.{} we assume $T=0$ and consider the following
  example: $h=1$, $x_i(0)=1-2i\varepsilon$, $1\in K$, $1\neq
  i\in\overline{K}$, $\beta_{ij}(t)=\beta$, $\alpha_i(t)=\alpha$ for
  the truth seekers, and $\alpha_i(t)=0$ for the ignorants until we
  say otherwise. Let there be a given $\gamma>0$ being sufficiently
  small. There exists a time $T_1$ until $x_1(T_1)<1-\gamma$. Up to
  this time no other individual has changed its opinion. After time
  $T_1+1$ we suitably choose $\alpha_2(t)$ so that we have
  $\frac{1}{2}\varepsilon\le
  x_1(\tilde{T}_1)-x_2(\tilde{T}_1)\le\varepsilon$. So at time
  $\tilde{T}_1+1$ the convergence of truth seeker $1$ is interrupted
  the first time. After that we may arrange it that $x_1$ and $x_2$
  get an equal opinion and will never differ in there opinion in the
  future. Now there exists a time $T_2$ until
  $x_2(T_2)=x_1(T_2)<1-\gamma$ and we may apply our construction
  described above again. Thus, every ignorant $i\in\overline{K}$ may
  cause an interruption of the convergence of the truth seekers.
\end{proof}

\begin{conjecture}
  If we drop the condition $\alpha_i(t)=0$ for all $i\in\overline{K}$
  in Theorem \ref{thm_interupted_convergent} then we have
  ($|\overline{K}|$)-fold convergence of the truth seekers.
\end{conjecture}

The Hegselmann-Krause Conjecture might be generalized to opinions in
$\mathbb{R}^m$ instead of $\mathbb{R}$ when we use a norm instead of
$|\cdot|$ in the definition of the update formula. Using our approach
to prove this $m$-dimensional conjecture would become very technical,
so new ideas and tools are needed. We give an even stronger
conjecture:

\begin{conjecture}
  The $m$-dimensional generalized Hegselmann-Krause Conjecture holds
  and there exists a function $\phi(\Omega,\gamma)$ so that the truth
  seekers in an arbitrary generalized (WASBOCOS) $\Omega$ are
  $\phi(\Omega,\gamma)$-fold interrupted convergent in $\varepsilon$,
  $\alpha$, $\beta$, and $n$.
\end{conjecture}

redpajama_set_name: RedPajamaArXiv


text:
story from Business
Travel Booking Site CheapAir Is Switching Bitcoin Payment Processors
Wolfie Zhao
Apr 23, 2018 at 6:00 a.m. UTC Updated Apr 23, 2018 at 6:57 a.m. UTC
Online travel booking agency CheapAir.com said Friday that it will switch bitcoin payment processors following a service change by Coinbase.
CheapAir.com – which began accepting bitcoin back in 2013 in a first for the global travel industry – explained the move in a blog post, revealing it is eyeing a move to accept bitcoin payments through Bitpay.
"We were recently informed by our processing partner, Coinbase, that they will no longer support 'custodial' solutions for merchants, and are removing a number of the tools and features that we rely on to accept bitcoin from CheapAir.com shoppers," CheapAir wrote. "These changes are scheduled to occur in a matter of weeks."
According to CheapAir, the process that converts the bitcoin it receives from customers to fiat currencies is key to the platform's ability to accept the cryptocurrency, since traditional airlines and hotel services only take fiat currencies as payment.
The company also said that, as part of the shift, it's looking at accepting other cryptocurrencies, naming bitcoin cash, dash and litecoin as potential options. CheapAir announced in 2014 that it was accepting litecoin and dogecoin by way of GoCoin, another cryptocurrency payment processor.
However, while CheapAir said the integration with Bitpay has already completed, it foresees the adoption may not be as smooth as previously.
"Our one giant concern is that Bitpay does not support 'non-payment protocol wallets' (wallets that aren't BIP-70 compliant). So if you do not have a compatible wallet, you would have to get one and use it as an intermediate stage for your Bitcoin payment," the company said in the statement, going on to request that users send feedback on the Bitpay option:
"This is where we'd love your input."
Airplane image via Shutterstock
CoinbasePaymentsTravelCheapairMerchantsNewsUse Cases & VerticalsBusiness News
Mt. Gox Creditors Can Claim 90% of Bitcoin Left in Bankruptcy: Bloomberg
Kevin ReynoldsJan 15, 2021


Dr. Knut Scherhag, Professor Destinationsmanagment at Worms University, talks about the pedagogical approach "Project Work" from his point of view.
Andrea Menn, Head of International Office at Jade University, talks about the pedagogical approach "Project work" at her university.
Josef Timmerberg, Prof. Dr.-Ing. from Jade Hochschule talks about Project Work.
Yolande Tervelde, Lecturer at Windesheim University, talks about the importance of leadership whithin the pedagogical approach "Co-Creative learning".
05 WHAT IS CO-CREATIVE LEARNING?
Yolande Tervelde, Lecturer at Windesheim University, describes the pedagogical approach "Co-Creative learning".


The last reported accreditation for this school was done by the Accrediting Council for Independent Colleges and Schools. To see more and get consumer disclosures, please visit dec.edu.
The following is a list of areas of study available at Douglas Education Center. Programs that have a link will take you to the program's profile page where education requirements, career pathways, and other information is available.
Get more information about the programs offered by Douglas Education Center by clicking the request information to the right.
The overall cost students will pay at Douglas Education Center is approximately - after scholarships, grants and other forms of aid are taken into consideration. In order to calculate a more personalized cost check out their cost calculator here. There may be an application fee required, please check with Douglas Education Center for more information. The financial details shown on this page are an approximation. The actual costs may differ, please contact this school for more information.
The list of nearby schools has been selected by researching which schools offer similar programs to Douglas Education Center.


While I'm not surprised that VR can be used as a tool for empathy, I'm a little taken aback that a tool as immersive as VR needs to be used to teach people to be empathetic. What does this say about other modes of communication? What makes it so hard for people to put themselves aside to understand another persons position? I think learning from an "experience" is great, but I think it would be more fruitful if people could finally learn to be empathetic through listening.
Additionally if VR has the power to teach empathy in ways other media can't what other things is it capable of? Bias is something always lurking in the media and responsible journalism is something that is disappearing from the media. What kind of responsibility needs to be used when using VR?
I can't speak for VR makers or researchers, but I don't think the issue is so much about VR _needing_ to be used as a tool for empathy. Asking why it its hard for people to consider another's position is akin to asking why don't we all just get along? We don't, we know this, but just as there has always been conflict, there has also been efforts in conflict resolution.
Social justice advocates have used many forms of media in creative ways to enact change. http://beautifultrouble.org is a wonderful archive of such methods. There is a great post by Steve Duncombe on what he calls Ethical Spectacle. The idea of detournement, of using dominant forms of media for people rather than commerce. Immersive media will most likely be used for spectacle; but can it be used for Steven's idea of ethical spectacle?


text:
CONTACT: Mark McCormick, Director of Strategic Communications, 913-490-4113, mmccormick@aclukansas.org
OVERLAND PARK, KS - Bo Dana Rupert, is no longer "banished" from the state of Kansas as the result of a settlement reached Monday between the ACLU of Kansas and Montgomery County Attorney Larry Markle.
As a part of a previous plea agreement, which the ACLU argued was as archaic as it was unlawful, Rupert had agreed to leave Montgomery County for good. This settlement removes that probation stipulation and Markle has agreed not to prosecute Rupert for his absconder status, which Rupert received solely as a result of complying with the plea agreement's banishment terms.
"We hope attorneys on both sides will think twice before adding stipulations like this one to probation agreements in the future," said ACLU of Kansas Legal Director Lauren Bonds of the settlement. "This was an unusual situation, to say the least."
Mr. Rupert has agreed to dismiss his lawsuit against Larry Markle, County Attorney for Montgomery County, in exchange for an amended plea agreement. As a part of the resolution of the case, Rupert has clarified his defense attorney, Heath Lampson, prepared the problematic terms of the plea agreement, not Mr. Markle.
The ACLU of Kansas had argued that the banishment provision in Rupert's agreement had been all but eradicated by a compact between the States as early as 1935.
For almost a century, states "that have considered the question have unambiguously found that banishing a criminal defendant from an entire state is a form of illegal sentence that violates state public policy," the ACLU's previous filing read.
But when Rupert's lawyer told him, "Don't still be here tomorrow when the sun comes up," Rupert felt he had no choice but to accept the agreement.
This was the second case the ACLU of Kansas has filed against Markle. The affiliate along with the national ACLU, filed a June 2018 suit in the Kansas Supreme Court to force Markle to follow state law and begin notifying every defendant about Montgomery County diversion programs. That suit is ongoing.
About the ACLU of Kansas: The ACLU of Kansas is the statewide affiliate of the national American Civil Liberties Union. The ACLU of Kansas is dedicated to preserving and advancing the civil rights and legal freedoms guaranteed by the United States Constitution and the Bill of Rights. For more information, visit our website at www.aclukansas.org.


Botanical illustration - coloured - Cheiridopsis ? namaquensis. Dissected drawing of flower included plus dissected flower in attached capsule. Drawn from a collection made by R.H. Compton (s.n.. = sub-National Botanical Garden 14052/20), Matjesfontein [Matjiesfontein, Laingsburg District]. Strongly scented. sub-National Botanical Garden 1045/20 Compton = 538/20 Compton = 378/16 Pillans.
Botanical illustration - coloured - Cheiridopsis namaquensis. Dissected drawing of flower included. Drawn from a collection made by N.S. Pillans (5481), Namaqualand in October 1926. Flower open 1.30 p.m.
Botanical illustration - coloured - Cheiridopsis namaquensis. Dissected drawing of flower included. Drawn from collection sent by R.H. Compton, 15 August 1927. [No locality given - 1 and 2 planted on Bolus Herbarium Bank [at Kirstenbosch] 16/08/1927. Diameter of flower 4 cm. Flower deep yellow and open 11.30 a.m.


When you are facing DUI charges, you will probably want to hire a lawyer to protect your rights and help you get the best possible results. But how much are the attorneys fees?
There is no single answer to this question. A spectrum exists in terms of what attorneys charge. If you cannot afford to hire a lawyer, you can always contact the Public Defender to represent you for a nominal fee.
Mr. Lawson is an experienced and aggressive DUI attorney who has successfully tried dozens and dozens of DUI trials. His rates are based on a number of factors that are unique to each case. What your prior criminal history and driver's record look like, as well as the circumstances of the instant case make a difference. For example, if you had children in the car or there was an accident, it would be a more complicated case. If it is a felony DUI charge vs. a misdemeanor charge, that makes a difference as well.
In addition, what you are hiring a lawyer for makes a difference, most lawyers will charge less if you know you want to plea vs. if you want a jury trial.
Contact Mr. Lawson today to get an attorney fee quote and to discuss possible payment options.


text:
Rajasthan CM appeals for calm in 'Jodhpur's tradition of love, brotherhood'
Jaipur (Rajasthan) [India], May 3 (ANI): Calling the Jodhpur clashes between two groups "unfortunate", Rajasthan Chief Minister Ashok Gehlot instructed the district administration to maintain law and order situation at all costs while also appealing to the people to maintain peace and harmony.
Gehlot also appealed to the people to maintain peace and harmony in the district.
"It is an unfortunate incident that a tense atmosphere has prevailed due to the clashes between the two groups at Jalori gate in Jodhpur. The administration has been instructed to maintain peace and order at all costs," the chief minister tweeted.
He further called upon the people to help in maintaining the law and order situation in line with the "Jodhpur's tradition of love and brotherhood".
Jodhpur district administration temporarily suspended internet services in the district on Tuesday as a precautionary measure after clashes on Monday night, ahead of Eid-ul-Fitr and the Akshaya Tritiya festivals today.
According to an order issued by Jodhpur Divisional Commissioner Himanshu Gupta, all internet services have been suspended in Jodhpur beginning from 1 am today.
This follows after several people were injured in stone-pelting between two communities on Monday night after some miscreants replaced one community's religious flag at Balmukand Bissa at the Jalori Gate intersection circle with that of another. People from the other community objected to this move resulting in clashes.
During this, the crowd also uprooted loudspeakers installed in the area for the Eid prayers. Police reached the spot and intervened in the matter to disperse the crowd. (ANI)
Nothing for poor but "election speech": Mallikarjun Kharge on Union Budget
Budget, a step towards achieving 'Amrit Kaal' vision: Health Minister Mansukh Mandivya
Budget empowers 140 cr people, unlike an election budget where 'rewris' are distributed: Piyush Goyal
Akshay Kumar, Emraan Hashmi's recreated version 'Main Khiladi Tu Anari' is out
DC Studios Chiefs reveal 'Superman & Lois' series will conclude after "one or two more seasons"
Ranbir Kapoor, Shraddha's track 'Tere Pyaar Mein' from 'Tu Jhoothi Main Makkaar' is out
We will trust our processes, bring our strengths to table: Dubai Capitals' George Munsey
Gulf Giants camp set up for players to thrive and express themselves: Gerhard Erasmus
South Africa's Dane van Niekerk 'broken' after being left out of T20 World Cup squad
'I'm so bullish about this team,' Steve O'Keefe believes Australia will win BGT against India
Jaipur (Rajasthan) [India], May 3 (ANI): Calling the Jodhpur clashes between two groups "unfortunate", Rajasthan Chief Minister Ashok Gehlot instructed the district administration to maintain law and order situation at all costs while also appealing to the people to maintain peace and harmony.
"It is an unfortunate incident that a tense atmosphere has prevailed due to the clashes between the two groups at Jalori gate in Jodhpur. The administration has been instructed to maintain peace and order at all costs," the chief minister tweeted.
He further called upon the people to help in maintaining the law and order situation in line with the "Jodhpur's tradition of love and brotherhood".
This follows after several people were injured in stone-pelting between two communities on Monday night after some miscreants replaced one community's religious flag at Balmukand Bissa at the Jalori Gate intersection circle with that of another. People from the other community objected to this move resulting in clashes.


No picture book review this week does not mean I couldn't find one! Instead I've picked another form of storytelling I am passionate about – Perfect Films!
My parents asked if I could get a hold of this British classic from 1952 (IMDb), they remembered the PBS channel in NY used to show it. I found it too difficult, so I called on a film buff friend. Nowadays you can get a copy on ebay. Maybe you could then and I just couldn't figure it out! In any case it is a tender delight – don't pass it up next time you get a chance.
Directed by George More O'Ferrall, the script was adapted and produced by Anatole de Grunwald from a play by Wynyard Browne. Ralph Richardson plays the English clergyman who, in his devotion to his parishioners, seems to have sorely misunderstood, possibly even neglected, his own children, played by Celia Johnson, Margaret Leighton and Denholm Elliot. Over 80 minutes we see the family coming together to celebrate Christmas, the first Christmas since the mother has passed. The aunts, played by Margaret Halstan and Maureen Delaney, are my favorite characters, and I do hope you trouble yourself to find out why!


text:
NameDegree Offered
Popular Degrees
Largest Universities & Colleges
Online Colleges & Degrees
Online Only Colleges
Online Degrees by Career Type
cosmetology-schools
Home > Massachusetts > New England Hair Academy
New England Hair Academy
492-500 Main St, Malden, MA
New England Hair Academy is located in Malden, MA, and is a career school that focuses on cosmetology education. New England Hair Academy has been accredited by the Accrediting Commission of Career Schools and Colleges since 1979. The main program taught by New England Hair Academy is in cosmetology, which trains students with the skills and education needed to obtain licensure and seek out gainful employment after graduation. In addition, New England Hair Academy also offers a nail technology program, which provides training in nail care and a hair styling program, which specializes in the art of hair management.
New England Hair Academy graduates close to 160 students per year, with about 94% of students who start the program continue on to complete it.
New England Hair Academy offers student services including cosmetology career guidance to help students and job placement opportunities for students looking for work. State licensing procedures may vary, and as such it is suggested students lookup obtaining a license in cosmetology. Get all the information by visiting the school's official website, where you can see program details, financial aid & tuition, and disclosures.
The list below shows all the programs that may be offered at New England Hair Academy. Actual programs may differ from the list below, to get an official list of programs offered please visit the school's website or by phone at (781) 324-6799.
Cosmetology (1,000 Hours)
Cosmetology program takes on average eight months to complete. Students who complete this program may be eligible to pursue licensing.
Program Details:
Tuition costs are estimated to be $13,000*
The estimated cost of supplies that may be required is $900*
44 students completed this program in the previous reported year
The length of the cosmetology program is usually 8 months
Learn more about cosmetology programs by visiting our cosmetology schools profile page, which lists schools offering this program along with a career overview.
Barber (1,000 Hours)
The barber program offered usually takes about eight months to complete and may satisfy education requirements for licensing.
Cost of required supplies: $900*
60 students completed the barber program last year
Takes about 8 months to complete
(100 Hours)
program takes on average one month to complete. Students who complete this program may be eligible to pursue licensing.
Tuition costs are estimated to be $1,000*
The length of the program is usually 1 month
*Amounts for tuition, fees, and supplies are an estimate which are calculated based on historical data and should be solely used for informational purposes only.
**Repayment rates, earnings, and loan payments are estimates and should be treated as informational tools only, contact the school for official information.
Estimated Costs & Financial Aid
The items listed below are for informational purposes only and may not represent current tuition, fees, and expenses.
Note: Financial aid is only available for those who qualify, check with the school for details.
Financial Aid Details
The table below breaks down the types of financial aid, the percent of students receiving aid, and average amount of aid per student.
Percent of Students
Average Amount
Any Financial Aid 82% -
Receiving Grants & Scholarships 82% $5,427
Receiving Student Loans 22% $8,023
Other Cosmetology Schools Nearby
The list of nearby cosmetology schools below offer beauty and cosmetology programs similar to New England Hair Academy. Click the links below to get a detailed profile for each school.
Empire Beauty School - Malden in Malden, MA (0.5 miles away)
Catherine Hinds Institute of Esthetics in Woburn, MA (6 miles away)
Empire Beauty School - Boston in Boston, MA (12 miles away)
Massachusetts School of Barbering in Quincy, MA (13 miles away)
Mansfield Beauty Schools - Quincy in Quincy, MA (13 miles away)
EINE Inc in Tewksbury, MA (14 miles away)
Lowell Academy Hairstyling Institute in Lowell, MA (19 miles away)
Empire Beauty School - Framingham in Framingham, MA (20 miles away)
School Code: 212200
Financial Aid Page
FINANCIAL AID RATE
STUDENT:TEACHER RATIO
FINANCIAL AID AWARD AVG.
MatchCollege.com © 2020
MatchCollege is an independent, online repository of college information created to help you find your perfect college match.