text:
보이스 뉴스
우먼동아
中文 (簡体)
Government and KSOC should all be held accountable
Posted January. 12, 2019 07:46,
Updated January. 12, 2019 07:46
Government and KSOC should all be held accountable. January. 12, 2019 07:46. .
With allegations raised by Olympic short track champion Shim Suk-hee that she'd been sexually assaulted by her former coach, the ugly truth of the nation's sports community is being laid bare. The Korean Sport & Olympic Committee (KSOC) said that 136 cases of sexual violence were reported last year. Yeo Jun-hyung, head of the Solidarity for Young Skaters, said Thursday that he has "received reports on a half dozen suspected cases of sexual violence in skating, and in two cases, victims themselves confirmed that they had been sexually assaulted." The assault allegations revealed by the Olympic gold medalist has sent shock waves through the nation, and it is almost dreadful when we think of how many other athletes including young, aspiring skaters have been subjected to such serious crimes.
Individual sports organizations, the KSOC, and the Ministry of Culture, Sports and Tourism have practically neglected their duty to protect athletes. The stenographic records of the KSOC Athletes' Committee, which was held to determine disciplinary measures for the head coach of a semi-professional short track team who was convicted of sexual misconduct in February 2016, include quotes that leave one dumbfounded. "Please consider that it can be your younger or older brother that's coaching the team," the committee said. "Coaches or head coaches may touch athletes on the shoulder (when coaching)." The head coach was not permanently expelled but had his qualification suspended for three years.
Athletes' distrust in the KSOC is clearly evidenced by the fact that only four cases of sexual violence have been received by the Clean Sports Center that the nation's top sports body has operated since 2017. The Culture and Sports Ministry should be also held accountable. Even though it found out during last year's audit of the KSOC that the organization had disregarded reports of violence and sexual misconduct, the government merely demanded disciplinary measures, not taking further actions. No wonder that victims of sexual assault in the sports community have remained tightlipped in fear.
The sports community has a power structure that makes it difficult for victims to seek justice. Saying no to a coach's unjust demands can threaten their future career. Across sports, coaches, and senior and junior athletes are all in a closed society based on apprenticeship and obsessed with results. Under the current system, coaches, athletes, and their parents implicitly agree that even violence can be tolerated if they can get good results. In addition, camp and off-season training, which usually takes place in an isolated location, also makes is hard to protect athletes from violence. The Culture and Sports Ministry has said that it will launch a full investigation into sexual violence and expand the scope of crimes subject to permanent expulsion. Though belated, the government should revise the composition of sports organizations and improve training environments. Most of all, the sports community needs to reflect and correct itself unless it wants another competent athlete falls victim to crimes.
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Illicit actors – from terrorists to organized crime syndicates to corrupt foreign officials – take advantage of the global financial infrastructure to finance their unlawful activities and hide their ill-gotten gains.
The U.S. Intelligence Community has insight into these actors' identities, networks, and activities, but it has little ability to monitor actual financial transactions; private sector companies, particularly financial services firms, know how funds move around but often find it challenging to assess the identities or motivations of those involved in financial transactions. Effectively combating illicit financial activities thus requires greater collaboration between the public and private sectors.
INSA's Financial Threats Council works to strengthen public-private cooperation and information sharing regarding the broad range of threats faced by government, the financial services sector, and other industries, which include cyber security, money laundering, terrorist finance, transnational organized crime, corruption, and confidence in U.S. and global financial infrastructure.
Identify key problems and challenges associated with financial threats to our national security.
Shape public policy regarding financial threats that affect American businesses' interests and activities.
Develop specific new policy, legislation, intelligence, and law enforcement options to expose and counter these threats and the activities of the enablers who facilitate illicit financial activity.
Promote close collaboration between the private sector and the U.S. Government on efforts to identify financial vulnerabilities and establish tools and processes to counter efforts to exploit said vulnerabilities.


LOCKLINEAfter five years a son with identity issues returns to his father to claim the body of his stepbrother.
Abel, a thirty year old who finally integrated into society, knocks on the door of his father, who tried to show him a much more primal way of living in the past. What follows is a dual between two tormented characters who strongly test each other's faiths and beliefs.
Since junior high I, Beau Meevis, have been developing stories. I often wrote them with dark themes and while envisioning a movie as the end result. During my studies I've had the opportunity to make many new contacts with professionals as well as students and we finally all come together on this project. What started out as my final masters degree project has become something much bigger.
While writing I increasingly started to take inspiration from psychology and philosophy. As a result I began to look for a way to create characters that where much more distinguishable and often mentally contorted. This idea of a confrontation between two such characters came to me when I was reading about Nietzsche's Master-slave morality. The essence of master morality is nobility and masters themselves create their own morality. The slave morality is a reaction to the master morality and is based on re-sentiment. It opposes everything the master morality stands for.
In this short film these two extremes will have a face off on a rainy night, in a small warm house, deep in the woods.
A villa located in a natural monument in Brabant will serve as a backdrop for this movie. A part of the movie is set in the woods themselves. The coarse-looking environment provides a strong contrast to the modern interior of the villa. With that, the duality in the environments symbolizes one of the main themes in the film: human nature.
Professionals as well as students are working together to realize this movie. Two famous dutch actors, who we will announce in due time, will star in the lead roles of the film; known artist Paul van Dongen will create sketches, which serve as props in the film; and on set there are many more crew members who work in the field.
For the students in our midst, mainly Brabanders, this project is a great opportunity to work with professionals in their field of expertise. The soundtrack of the film is being composed by Laurens Hoynck van Papendrecht, a student of The Rock academy in Tilburg, for whom the project now serves as his Minor. Vince Donders, a student of the academy of arts in Den Bosch, is creating the biggest set piece of the film: a 7 feet tall statue. This way different disciplines of art work together in this production and that's one of the beautiful things of making movies.
Crew members from left to right: Beau Meevis (Director), Daan Bothof (DOP), Mayke Harkema (Producer, Kevin Schaduw (producer).
Up until now we managed to bring together half of our budget by doing various small jobs, but the clock is ticking and we still need the other half. That's where you come in, dear sir or madam. Support us so we can realize our vision; by doing so you will also help students to graduate or finish their internship. As our thanks we have created a great amount of rewards. For instance: receive concept art, an invitation to the set or tickets to the premiere.
So, do you like psychological thrillers where nobody's intentions are certain? Do you like suspense and dark themes? Support us! And we will see you at the premiere, July 2017.
This graduation project from Beau Meevis offers different young creators from the film- music- and visual sector the opportunity to develop themselves further in a special cooperation with professionals from the field. bkkc happily supports this.
If anyone does not provide for his own, and especially for those of his household, he has denied the faith and is worse than an unbeliever.
Hugo Koolschijn will play Henri in 'Verwanten'. Koolschijn played in several films like 'Soldier of Orange' and 'Accused' and in series as 'Fenna's Law' and 'Cellblok H'. Besides film and television Hugo is also an active member of Toneelgroep Amsterdam where he is a member since 1987.
Curently studying his master degree for Television and Film at the Luca school of Arts in Genk, Belgium. The short film 'Restitution' is part of his Master assignment. Previously he directed the short film 'Lily' and he took part in the Dutch and Belgium co-production 'Als de Dijken Breken' (High Tides).
Kevin graduated at the NHTV in Breda where he studied International Media and Entertainment management with the specialization production. Previously he worked on short films like 'Black premise' and 'Nasser'. 'Nasser' was shown on different big film festivals like 'BFI Flare' in London, TIFF Kids international Film Festival, in Toronto Canada and many other countries. Besides that he worked on many series for different Dutch networks. Series like, 'Als de Dijken Breken' (High Tides), de 'De Maatschap' (The Partnership) en 'Moordvrouw' (Fenna's Law).
Mayke graduated at the academy of Amsterdam in Media, Information and communication with a specialization in editorship and media production. Previously Mayke worked as a production intern for Zodiak Netherlands and editorial intern for SBS productions.
Currently a 3rd year student at the Dutch Film Academy in Amsterdam. He studies cinematography and took part in many Dutch film productions. As a freelancer he previously worked on 'Wiplala' (The Amazing Wiplala), 'Michiel de Ruyter' (The Admiral) and 'Flikken Maastricht'.
Liefs Marie-Louise N. en Floor N.
Succes met je film, Nerd!
Beau, veel succes, ik wil je wel met klem verzoeken, onze Bram, op geen enkele wijze in beeld te laten komen.
Top Beau! Ik hoop dat je het gaat halen.
Vic Vega cuts you: Special thanks on website and credits, Vimeo link and 2 première tickets.
Kyung Chul dismembers you: Special thanks on website and credits, Vimeo link, 4 première tickets, Digital script, DVD, special mention on Facebook and a booklet with stills and concept art of the movie.
John Doe prays with you: Associate producer credit, Vimeo link, 6 première tickets, Digital script, DVD, special mention on Facebook, a booklet with stills and concept art of the movie and a setvisit for 2 persons.
Norman Bates services you: Logo company in the credits, Vimeo link, 6 première tickets, Digital script, DVD, special mention on Facebook, a booklet with stills and concept art of the movie and a setvisit for 2 persons.
Hannibal Lecter eats you: We will contact you to discuss a fitting package!
32 supporters have contributed to this project!


text:
describe Frameit do
  describe Frameit::StringsParser do
    describe "parse" do
      it "raise error when file can't be found" do
        expect do
          Frameit::StringsParser.parse("./nothere")
        end.to raise_exception("Couldn't find strings file at path './nothere'")
      end

      it "raise error when file isn't a .strings file" do
        expect do
          Frameit::StringsParser.parse("./frameit/spec/fixtures/background.jpg")
        end.to raise_exception("Must be .strings file, only got './frameit/spec/fixtures/background.jpg'")
      end

      describe "successfully parsing" do
        it "parses a valid .strings file (utf16)" do
          translations = Frameit::StringsParser.parse("./frameit/spec/fixtures/translations.strings")

          expect(translations).to eq({
            "Cancel" => "Abbrechen",
            "OK" => "OK",
            "Multiple words working" => "einlangesdeutshceswort mit Abstand"
          })
        end

        it "parses a valid .strings file (utf8)" do
          translations = Frameit::StringsParser.parse("./frameit/spec/fixtures/translations.utf8.strings")

          expect(translations).to eq({
            "Cancel" => "Abbrechen",
            "OK" => "OK",
            "Multiple words working" => "einlangesdeutshceswort mit Abstand"
          })
        end

        it "parses a valid .strings file (utf8) with space in path" do
          translations = Frameit::StringsParser.parse("./frameit/spec/fixtures/translations file.utf8.strings")

          expect(translations).to eq({
            "Cancel" => "Abbrechen",
            "OK" => "OK",
            "Multiple words working" => "einlangesdeutshceswort mit Abstand"
          })
        end
      end

      describe "failure parsing" do
        it "logs a helpful message on a bad file" do
          expect(Frameit::UI).to receive(:error).with(/.*translations.bad.strings line 2:/)
          expect(Frameit::UI).to receive(:verbose).with(/undefined method .\[\]. for nil:NilClass/)
          expect(Frameit::UI).to receive(:error).with(/Empty parsing result for .*translations.bad.strings/)

          translations = Frameit::StringsParser.parse("./frameit/spec/fixtures/translations.bad.strings")
          expect(translations). to eq({})
        end

        it "explains that only UTF-8 and UTF-16 encoded are allowed" do
          expect(Frameit::UI).to receive(:error).with(/.*translations.utf32.strings.*UTF16/)

          translations = Frameit::StringsParser.parse("./frameit/spec/fixtures/translations.utf32.strings")
          expect(translations). to eq({})
        end
      end
    end

    describe "encoding_type" do
      let(:path) { "/Users/fastlane/directory/en-US/title.strings" }
      let(:path_with_space) { "/Users/fastlane/directory name/en-US/title.strings" }

      it "escapes path with no spaces" do
        expect(Fastlane::Helper).to receive(:backticks)
          .with("file --mime-encoding #{path}", print: false)
          .and_return("").exactly(1).times

        Frameit::StringsParser.encoding_type(path)
      end

      it "escapes path with spaces" do
        escaped_path = path_with_space.shellescape
        expect(Fastlane::Helper).to receive(:backticks)
          .with("file --mime-encoding #{escaped_path}", print: false)
          .and_return("").exactly(1).times

        Frameit::StringsParser.encoding_type(path_with_space)
      end
    end
  end
end

redpajama_set_name: RedPajamaGithub


text:
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CRESTWOOD CITIZENS ASSOCIATION
WASHINGTON, DC | ESTABLISHED 1941
Here is the most recent information on crime in our neighborhood.
CRIMES REPORTED IN PATROL SERVICE AREA 404*
From October 1 through 31, 2019
In PSA404 Total
Near Crestwood
In Crestwood
Subtotal Mon - Thu Fri - Sun
Burglary 5 1 1 1
Stolen Auto 9 4 1 1
Theft 33 6 1 1
Theft from Auto 21 3 6 3 3
Total 68 14 9 6 3
* Data generally reflect the crime "start" time reported to MPD (not the often later "report" time recorded by MPD); Crestwood reported 0 crimes which were not reported by MPD; details are at the internet site http://crimemap.dc.gov.
Crestwood Crimes By Block:
Theft from Auto (6)
1700 Allison St (Thu, 10/24, 1:00 pm)
1600 Shepherd St (Mon, 10/24, 3:30 pm)
1900 Shepherd St (Fri, 10/11, 2;16 pm)
1700 Varnum St (Fri, 10/4, 2;12 pm)
1600 Webster St (Fri, 10/11, 2:40 pm)
4400 Blagden Av (Thu, 10/8, 3:42 pm)
1600 Varnum St (Mon, 10/14, 7:12 pm)
1600 Decatur St (Thu, 10/24, 3:05 pm)0
Stolen Auto (1)
1600 Buchanan (Thu, 10/31, 10:31 pm)
Crestwood & Area Trends:
-- In Crestwood (16th St. & west), 9 crimes were reported in October, up from 3 reported in September, and above the 6-month average of 6.8.
-- Also in Crestwood, 6 of the 9 reported crimes was a Theft from Auto; all took place in the daylight hours.
-- Near Crestwood (14th St. & west to 16th St.), 14 crimes were reported in October; more than the 4 reported for September. Six were along 14th St.
-- In PSA 404, 68 crimes were reported in October, up from the 38 reported in September. Thefts more than doubled, from the 15 reported last month.
CRIMES REPORTED IN CRESTWOOD BY MONTH/YEAR
Click Underlined Number for Detail
Offense Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Tot# Tot %
Burglary 1
Robbery 1 1 1
Stolen Auto 1
Theft 1 1 4 2 1 1 1 2 1 1
Theft/auto 5 4 4 8
Total 2019 6 7 8 11 2 5 6 7 3 9
Assault 1
Burglary 1 1 2 1 1
Stolen Auto 2 2 1 2 1
Theft 1 1 1 1 2 2 1 2
Theft/auto 3 2 4 3 4 4 9 4 4 1
Total 2018 3 2 6 7 6 7 5 13 7 5 4
Burglary 3 2 1 3 2 2 5 3 21 25%
Robbery 1 1 2 2%
Stolen Auto 1 4 1 2 1 1 2 3 15 18%
2 1 1 2 2 6 1 1 20 25%
Theft/auto 1 1
2 5 1 3 2 2 4 2 1 24 29%
Other 1 1 1%
Total 2017 1 5 4 10 9 4 7 6 6 13 10 8 83 100%
Assault 0 0
Burglary 3 1 1 1 1 7 9
Homicide 0 0
Robbery 2 1 1 4 5
Stolen Auto 4 3 3 1 11 14
Theft 1 3 2 1 2 2 2 3 16 20
Theft/auto 3 3 3 3 5 6 3 1 6 4 2 2 41 51
Other 1 1 1
Total 2016 3 4 9 10 10 8 8 6 9 6 5 2 80 100
Offense Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec Tot#
Burglary 1 3 1 1 2 1 9 10
Robbery 1 2 1 1 1 6 7
Theft 2 1 1 1 3 6 1 1 2 18 20
Total 2015 6 1 4 8 11 13 13 8 5 4 6 9 88 100
Burglary 1 2 1 1 1 1 2 4 4 1 2 20 17
Robbery 1 2 4 3
Stolen Auto 1 1 1 1 1 3 2 1 1 12 10
Theft/auto 5 5 4 2 2 3 2 6 10 4 10 5 58 50
Total 2014 7 9 6 8 4 6 5 12 19 10 16 13 115 100
Burglary 1 1 1 1 1 2 2 3 3 15 17
Stolen Auto 1 1 2 1 3 8 9
Theft 1 1 1 5 1 1 2 1 4 2 4 23 27
Theft/auto 2 1 2 4 2 1 3 6 7 5 5 38 44
Other 0 0
Total 2013 5 4 4 9 4 4 7 8 15 9 10 7 86 100
Assault 1 1 1 3 3
Burglary 1 4 2 2 1 1 11 11
Stolen Auto 2 1 1 1 2 1 1 1 10 10
Theft 6 1 1 1 1 1 11 11
Theft/auto 4 8 3 2 2 3 9 10 6 5 3 5 60 59
Other 1 1 2 2
Total 2012 14 9 14 4 3 11 15 16 6 9 5 6 102 100
Assault 1 1 2 2
Burglary 3 6 1 1 8 1 1 21 17
Stolen Auto 2 2 1 2 1 2 1 11 9
Theft 1 5 1 3 2 4 4 2 1 4 27 22
Theft/auto 3 1 1 1 10 2 2 5 3 11 3 8 50 41
Total 2011 6 4 7 3 21 5 5 17 11 18 11 14 122 100
Assault 2 2 1 1 6 4
Stolen Auto 1 1 1 1 1 1 1 2 9 6
Theft 2 1 3 4 1 2 3 16 11
Theft/auto 2 4 5 6 9 4 6 5 41 27
Other 3 1 3 4 14 6 7 8 2 5 2 4 59 39
Total 2010 8 4 10 17 26 16 19 16 13 17 4 4 152 100
Assault 1 4 3 1 1 2 1 13 7
Robbery 2 1 1 3 1 1 2 11 6
Stolen Auto 1 1 1 3 3 3 2 1 1 1 17 10
Theft 2 2 1 4 1 3 1 14 8
Theft/auto 12 2 2 7 6 9 3 2 3 1 47 27
Other 14 5 1 4 3 2 3 8 3 4 49 28
Total 2009 35 6 7 20 20 26 18 10 15 6 8 5 176 100
Burglary 1 1 1 5 3 8 2 3 4 2 30 20
Robbery 2 1 1 1 5 3
Theft 1 3 2 3 3 2 14 9
Other 1 4 1 4 2 4 2 8 3 4 33 22
Total 2008 6 14 1 5 11 15 14 25 11 21 16 11 150 100
Assault 2 2 2
Theft/auto 4 2 1 2 1 2 8 3 4 25 29
Total 2007 8 2 6 7 7 5 6 4 11 13 7 9 85 100
Assault 1 1 2 1 1 6 6
Burglary 4 1 10 7 1 2 2 2 2 1 2 34 32
Total 2006 15 6 12 15 8 3 9 7 8 7 12 4 106 100
Robbery 3 1 1 1 1 1 7 8
Stolen Auto 11 3 4 4 1 5 1 4 3 3 39 42
Theft 1 1 2 1 1 1 1 1 1 10
Total 2005 17 8 8 8 2 6 11 7 8 2 6 8 91 100
This website and the Crestwood Citizens Association is supported by the dues of CCA members. Membership has its benefits including access to members-only resources and the knowledge that you are supporting a great neighborhood!
info@crestwooddc.org

redpajama_set_name: RedPajamaCommonCrawl


text:
\section{Introduction}
\label{sec:intro}
An autonomous mobile robot system, which is firstly introduced in \cite{SY99},
is a distributed system which consists of many mobile computational entities
(called \emph{robots}) with limited capabilities,
{\it e.g.,~} robots do not have identifiers, cannot distinguish other robots,
or cannot remember their any past actions.
The robots operate autonomously and cooperatively;
each robot observes the other robots (called \emph{Look}),
computes the destination based on the observation result (called \emph{Compute}),
and moves to the destination point (called \emph{Move}).
Each robot autonomously and cyclically performs the above three operations, \emph{Look}, \emph{Compute}, and \emph{Move},
to achieve the given common goal.
The literature \cite{gatherimpos,perspec13,distalg06,SY99} provide a formal discussion on the capabilities of the robots
for the distributed coordination ({\it e.g.,~} gathering, scattering, or pattern formation).
Since the introduction of the autonomous mobile robot system,
many related studies about the computational power of the system have been introduced
and many researchers are interested in clarifying the relationship between the capabilities of the robots
and solvability of the problems.

Generally, in \emph{Look}~operation,
each robot can observe all other robots (within its visibility range if it has a limited visibility range)
to compute the destination point to move.
In other words, each robot takes a snapshot that consists of all other robots' (relative) positions in its \emph{Look}~operation.
This implies that each robot temporarily remembers the positions of up to
$N-1$ robots,
where $N$ is the total number of robots.
From the more practical viewpoint,
we claim that a robot with low functionality may not have such large working memory.
That is, the main question we address in this paper is ``\emph{what occurs if a robot cannot observe some of other robots?}''.
More precisely, ``\emph{how many other robots should be observed to achieve the goals of the problems?}''.

\vspace{5pt}
\noindent
{\bf Related works:}
The gathering problem \cite{gatherfea},
which requires all the robots move to a common (non-predetermined) position,
is one of the basic and essential problems in autonomous mobile robot systems.
There are many studies about the gathering of the autonomous mobile robots;
M. Cieliebak \etal~presented the first algorithm to achieve the gathering from any arbitrary configuration \cite{gatheranyconf},
R. Klasing \etal~studied the gathering of mobile robots in one node of an anonymous unoriented ring \cite{gatherring},
G. D'Angelo \etal~introduced a gathering algorithm by robots without multiplicity detection on grids and trees \cite{gathergrid},
and many works for the gathering by robots with dynamic (or inaccurate) compass are also introduced
\cite{gatherinac3,gatherinac2,gatherinac4,gatherinac1}.
The capability of the robots deeply affects the solvability of the gathering problem, thus
some investigations about the required capability or impossibility are introduced \cite{gatherfea,gatherimpos}.
However, all of these works assume that each robot can observe all other robots
within its visibility range if there is no obstruction between the robots {\it e.g.,~} any opaque robot.

The works most related to this paper is those with the limited visibility range \cite{gathernulimited,gatherlimited,convlimited,circlelimited}.
The robots with the limited visibility cannot necessarily observe all robots, which is  similar to the defected view model we introduce.  But visibility is limited by distance in the model and thus all robots can be observed when they gather closely enough.
On the other hand, our model cannot guarantee such a full view of the robots.

The works for fault-tolerance \cite{gathermulfault,gatherfault,gatherweakmulfault} are also closely related to this paper.
The defected view in our model can be considered as a new type of faults in autonomous mobile robot systems.

\vspace{5pt}
\noindent
{\bf Contribution:}
To provide some answers for the above research questions,
we propose a new computational model
with restriction on the number of the robots that each robot can observe,
named the \emph{defected view model},
where each robot observes only $k$ other robots for $1 \leq k < N-1$.
The assumption of this model naturally arises
by considering some issues for robots with low functionality
such as
(1) each robot does not have enough working memory to store the entire observation result,
(2) each robot may miss some of observation results due to memory failure,
or (3) each robot fails to observe some of other robots by sensing failure.
It is obvious that when $k$ becomes the lower, the problem becomes the harder
(possibly impossible) to solve.
We consider two different defected view models
regarding which $k$ robots are observed:
the distance-based ($N$,$k$)-defected model
and the adversarial ($N$,$k$)-defected model.
In the former model, each robot observes the $k$ closest robots to its current position,
and in the latter one, $k$ robots which are observed by each robot are determined adversarially.

More precisely, the $k$ robots that each robot can observe are chosen from the robots located at points different from $r$'s current position.
Concerning $r$'s current position, $r$ can detect only whether another robot exists at the point or not (so called the \emph{weak multiplicity detection}).
Such an assumption that the robots at $r$'s current position are excluded from the candidates of the observed $k$ robots is motivated from the following observation:
each robot $r$ observes the robots at remote points and those at $r$'s current position by different ways usually.
Each robot observes the remote robots by, for example, a radar sensor or a vision senor,
but senses the other robots at the same point by, for example, a contact sensor.

As the first step of the gathering in the defected view model,
we investigate only the case of $k=N-2$.
The main contributions of this paper is as follows:
(1) we propose a gathering algorithm in the adversarial ($N$,$N-2$)-defected model for any $N \geq 5$,
(2) we present another algorithm to solve the gathering problem in the distance-based (4,2)-defected model,
and (3) we provide the impossibility result showing that there is no (deterministic) algorithm to solve the gathering problem
in the adversarial or distance-based
(3,1)-defected model.
Moreover, we present another impossibility result
in a naturally relaxed ($N$,$N-2$)-defected model where the observed $k$
robots can contain the robots at the observer's current position.
This impossibility result shows the necessity of the assumption that
the observed $N-2$ robots should be chosen from robots other than
those located at the observer's current position.

The rest of this paper is organized as follows:
Section \ref{sec:model} presents the system model (two defected view models) and problem definition;
Section \ref{sec:nn-alg} introduces an algorithm to solve the gathering problem in the adversarial ($N$, $N-2$)-defected model
for any $N \geq 5$;
Section \ref{sec:42alg} gives a gathering algorithm in the distance-based (4,2)-defected model;
Section \ref{sec:impos} shows two impossibility results
showing that there is no algorithm in adversarial or
distance-based (3,1)-defected model
and the relaxed adversarial ($N$,$N-2$)-defected model;
and Section \ref{sec:conclusion} concludes the paper and provides some open problems.

\section{Model and Problem Definition}
\label{sec:model}
\subsection{Robots}
Let $R=\{r_1, r_2, ..., r_N\}$ be the set of $N$ autonomous mobile robots deployed in a plane.
Robots are indistinguishable by their appearance ({\it i.e.,~} identical),
execute the same algorithm ({\it i.e.,~} uniform or homogeneous),
and have no memory ({\it i.e.,~} oblivious).
There is no geometrical agreement;
robots do not agree on any axis, the unit distance, nor chirality.

A point in the plane is called an \emph{occupied point} if there exists a robot at the point.
We allow two or more robots to occupy the same point at the same time.
We call a robot a \emph{single robot} if the point occupied by the robot has no other robot.  Otherwise, we call it an \emph{accompanied robot}.

Each robot cyclically performs the three operations, \emph{Look}, \emph{Compute}, and \emph{Move}:
{\bf (\emph{Look})}
a robot obtains the positions (based on its local coordinate system centered on itself)
of all other (visible) robots,
{\bf (\emph{Compute})}
a robot determines the destination according to the given algorithm based on the result of \emph{Look}~operation.
Since each robot has no memory, the result of \emph{Compute}~is determined only by the result of \emph{Look}~operation,
and
{\bf (\emph{Move})}
a robot moves to the destination computed in \emph{Compute}~operation.
We assume \emph{rigid movement} which ensures
each robot can reach the destination during its \emph{Move}~operation, {\it i.e.,~} a robot never stops before it reaches its destination.

\subsection{Schedule and Configuration}
We assume a fully-synchronous scheduler ($\mathsf{FSYNC}$):
all robots fully-synchronously perform their operations (\emph{Look}, \emph{Compute}, and \emph{Move}).
This means that all robots perform the same operation at the same time instant and duration.
We call the time duration
in which all robots perform the three operations (\emph{Look}, \emph{Compute}, and \emph{Move}) once
\emph{a round}.

Let \emph{configuration} $C_t$ be the set of the (global) coordinates of all robots at a given time $t$:
$C_t = \{(r_{1.x}^t,r_{1.y}^t), (r_{2.x}^t,r_{2.y}^t) \ldots$ $(r_{N.x}^t,r_{N.y}^t)\}$,
where $r_{i.x}^t$ (resp. $r_{i.y}^t$) is the $X$-coordinate
(resp. $Y$-coordinate) of robot $r_i$ at time $t$.
Note that no robot knows its global coordinate.
Configuration $C_t$ is changed into another configuration $C_{t+1}$
after one round ({\it i.e.,~} all robots execute the three operations once).

\subsection{Observation: Visibility Range and Multiplicity Detection}
We basically assume that every robot has unlimited visibility range,
{\it i.e.,~} any two robots can observe each other regardless of their distance, while we introduce in Definition \ref{def:model} the defect in the information obtained by \emph{Look}\ operation.
Moreover, we assume a \emph{weak multiplicity detection},
{\it i.e.,~} each robot cannot get the exact number of robots occupying the same point but can distinguish whether the point is occupied by one robot or by multiple robots.
This implies that when each robot observes any point,
it can distinguish the three cases: there is \emph{no robot}, \emph{one robot}, or \emph{two or more robots} at the point.

We consider a \emph{defected view} such that each robot may not observe all other robots.
We define the ($N$,$k$)-defected model, where $1 \leq k < N$ as follows:

\begin{definition}
\label{def:model}
{\bf (($N$,$k$)-defected model)}
Each robot $r$ can get from \emph{Look}\ operation the set of occupied points
(in its coordinate system) where $k$ robots not accompanied with $r$ are located
({\it i.e.,~} the $k$ robots contains no robot located at $r$'s current point).
When the number of robots not accompanied with $r$ is less than $k$,
all such robots are observed.
The weak multiplicity detection concerning the $k$ robots is assumed:
a point occupied by only one of the $k$ robots can be distinguished from that occupied by two or more of the $k$ robots.
Moreover, $r$ can distinguish whether $r$ is single or accompanied.
\end{definition}

Note that the ($N$,$N-1$)-defected model is equivalent to the commonly used model (with the weak multiplicity detection) where each robot can observe all robots.
The ($N$,$k$)-defected model has options depending on how the observed $k$ robots are chosen.
We consider the two options in this paper,
named \emph{adversarial ($N$,$k$)-defected model} and \emph{distance-based ($N$,$k$)-defected model}.
In the \emph{adversarial ($N$,$k$)-defected model},
$k$ robots observed by each robot are determined adversarially.
In the \emph{distance-based ($N$,$k$)-defected model},
each robot $r$ observes the $k$ closest robots to the $r$'s current point.
Tie breaks among the robots the same distance apart is determined in an arbitrary way.

\begin{figure}[b]
	\begin{center}
	    \includegraphics[scale=0.8]{exampleconf.pdf}
	\end{center}
    \caption{An example configuration by 8 robots}
    \label{fig:exconf}
\end{figure}

To help to understand, we explain the model using examples.
Figure \ref{fig:exconf} illustrates an example configuration by 8 robots; $R = \{r_1, r_2, \ldots, r_8\}$.
Robots $r_1$ and $r_2$ (resp. $r_4$, $r_5$ and $r_6$) are accompanied, and the other robots are single.
The dotted arrow between robots represents the distance between the points occupied by the robots.
Let $p_i$ denote the point occupied by robot $r_i$.
Now we explain the models as follows:

\begin{itemize}
    \item {\bf The adversarial (8,4)-defected model}. In this model, each robot observes 4 other robots chosen adversarially.
    Assume that robot $r_3$ observes 4 robots, $r_1$, $r_2$, $r_6$, and $r_8$. In this case, robot $r_3$ gets a set of points
    $P^{r_3} = \{p_1^*, p_3, p_6, p_8 \}$ including point $p_3$ occupied by $r_3$ itself,
    where $p_i^*$ denotes that $p_i$ is recognized to be occupied by two or more robots.

    Robot $r_3$ knows that two or more robots exist at point $p_1$ because both robots $r_1$ and $r_2$ are chosen,
    however, $r_3$ does not know that there is another robot at $p_6$ because it observes only $r_6$ at $p_6$.
    Robot $r_4$ (or $r_5$, $r_6$) observes 4 robots among 5 robots, $r_1$, $r_2$, $r_3$, $r_7$, and $r_8$.
    If robots $r_1$, $r_3$, $r_7$, and $r_8$ are chosen, $P^{r_4} = \{p_1, p_3, p_4^*, p_7, p_8\}$ holds, which means that
    $r_4$ observes all points, however, it does not know that another robot exists at $p_1$.
    Robot $r_4$ can know that point $p_4$ is occupied by another robot other than itself.
    If robot $r_4$ observes robots $r_1$, $r_2$, $r_7$, and $r_8$,
    $P^{r_4} = \{p_1^*, p_4^*, p_7, p_8\}$ holds, which means that
    robot $r_4$ knows there exist two or more robots at $p_1$, but it cannot observe point $p_3$ occupied by robot $r_3$.
    Notice that $r_5$ and $r_6$ located at $p_4$ are allowed to observe the set of points different from those observed by $r_4$.
    \item {\bf The distance-based (8,3)-defected model}. In this model, each robot observes 3 closest robots to itself.
    Robot $r_7$ observes 3 robots, $r_8$ (the closest one) and two robots among three robots at point $p_4$,
    thus $P^{r_7} = \{p_4^*, p_7, p_8\}$ always holds.
    Robot $r_4$ observes robot $r_3$ (the closest one)
    and two robots among 4 robots, $r_1$, $r_2$, $r_7$, and $r_8$, which are the same distance apart.
    Note that the observed robots are determined in an arbitrary way,
    thus in this case, $P^{r_4}$ becomes one among $\{p_1^*, p_3, p_4^*\}$, $\{p_1, p_3, p_4^*, p_7\}$, $\{p_1, p_3, p_4^*, p_8\}$,
    or $\{p_3, p_4^*, p_7, p_8\}$.
\end{itemize}

It is obvious that the adversarial ($N$,$k$)-defected model is weaker\footnote{Strictly speaking, we do not know the adversarial ($N$,$k$)-defected model is properly weaker than the distance-based one yet;
it is obvious that the adversarial ($N$,$k$)-defected model is NOT stronger
than the distance-based one.}
than the distance-based one, that is, any gathering algorithm for the adversarial ($N$,$k$)-defected model works correctly also in the distance-based ($N$,$k$)-defected model.

\subsection {Problem Definition: Gathering}
The \emph{Gathering} problem is to locate all robots at the same point.
We define the gathering problem as follows.

\begin{definition}
\textbf{The Gathering Problem}
Given a set of $N$ robots located at arbitrary points.
Algorithm $\mathcal{A}$ solves the gathering problem
if $\mathcal{A}$ satisfies all the following conditions:
(1)
algorithm $\mathcal{A}$ eventually reaches
a configuration such that no robot can move,
and (2) when the algorithm $\mathcal{A}$ terminates,
all the robots are located at the same point.
\end{definition}

\section{Algorithm in the Adversarial ($N$,$N-2$)-defected Model where $N \geq 5$}
\label{sec:nn-alg}
In this section, we introduce an algorithm to solve the gathering problem
in the adversarial ($N$, $N-2$)-defected model, where $N \geq 5$.

\begin{algorithm}[tbhp]
\caption{Gathering algorithm for robot $r_i$ in the adversarial $(N, N-2)$-defected model where $N\ge 5$}
\label{alg:nn-2gather}
{\bf functions:}
\begin{description}
\item[$~\cdot~ \mathsf{OPSET}()$] returns a set of points $\{p\:|\: p$ is occupied by $r_i$ or a point occupied by the robot that $r_i$ observed$\}$.
\item[$~\cdot~ \mathsf{isMulti}(p)$] returns $\mathsf{TRUE}$ if point $p$ is occupied by two or more robots that $r_i$ observed (weak multiplicity), otherwise $\mathsf{FALSE}$.
\end{description}
{\bf algorithm:}
\begin{algorithmic}[1]
\If{$\forall p \in \mathsf{OPSET}(): \mathsf{isMulti}(p) = \mathsf{TRUE}$}
\State{move to the center of the smallest enclosing circle of $\mathsf{OPSET}()$}
\ElsIf{($r_i$ is single) $\wedge$ $(\exists p \in \mathsf{OPSET}(): \mathsf{isMulti}(p) = \mathsf{TRUE})$}
\State{move to an arbitrary point $p \in \mathsf{OPSET}()$ such that $\mathsf{isMulti}(p) = \mathsf{TRUE}$}
\ElsIf{$\forall p \in \mathsf{OPSET}(): \mathsf{isMulti}(p) = \mathsf{FALSE}$}
\State{move to the center of the smallest enclosing circle of $\mathsf{OPSET}()$}
\EndIf \Comment{No action if ($r_i$ is accompanied) $\wedge$ $(\exists p \in \mathsf{OPSET}(): \mathsf{isMulti}(p) = \mathsf{FALSE})$}
\end{algorithmic}
\end{algorithm}

Algorithm \ref{alg:nn-2gather} presents an algorithm to achieve the gathering in the adversarial ($N$, $N-2$)-defected model where $N \ge 5$.
The algorithm adopts, as the destination of robot $r_i$, the center of the smallest enclosing circle (SEC) of the occupied points that $r_i$ observed in the \emph{Look}\ operation.
Before proving the correctness of the algorithm, we show some fundamental properties of the SEC of points in a plane.

\begin{proposition}
\label{prop:SEC}
Let $P$ be a set of $n$ distinct points in a plane and $C$ be the SEC of $P$.
The following properties hold.
\begin{enumerate}
    \item The SEC of $P$ is unique.
    \item Let $p \in P$ be any point (if exists) properly inside $C$, $C$ is the SEC of $P \setminus \{p\}$.
    \item When there exist three points $p_1, p_2, p_3 \in P$ on the boundary of $C$ that form an acute or right triangle, $C$ is the SEC of $\{p_1, p_2, p_3\}$.
    \item When three or more points in $P$ are on the boundary of $C$, there exist three points $p_1, p_2, p_3 \in P$ on the boundary of $C$ that form an acute or right triangle. \qed
  \end{enumerate}
\end{proposition}

A key property of the ($N$, $N-2$)-defected model,
which is used in the following proofs,
is that \emph{any accompanied robot can observe all the robots} (but only with the weak multiplicity detection).

\begin{lemma}
\label{lemma:3ac}
In the adversarial ($N$, $N-2$)-defected model ($N \ge 5$), Algorithm \ref{alg:nn-2gather} solves the gathering problem in two rounds from any configuration where there exist three or more accompanied robots.
\end{lemma}

\begin{proof}
When every robot is accompanied, each robot detects all the occupied points in the \emph{Look}\ operation and recognizes that each of them is occupied by multiple robots.  Every robot moves to the center of the SEC of all the occupied points (by lines 1 and 2 in Algorithm \ref{alg:nn-2gather}) and thus the gathering is achieved in one round.

When there exists a single robot $r$, every accompanied robot observes $r$ and does not move (see line 7 in Algorithm \ref{alg:nn-2gather}).  Every single robot misses at most one accompanied robot in its \emph{Look}\ operation and can detect at least one point occupied by multiple robots: a point occupied by three or more robots (if exists) or one of the two points each occupied by two robots.  Each single robot moves to one of such points (by lines 3 and 4 in Algorithm \ref{alg:nn-2gather}), which results in the configuration where every robot is accompanied.  Thus the gathering is achieved in the next round as shown above.
\end{proof}

\begin{lemma}
\label{lemma:2ac}
In the adversarial ($N$, $N-2$)-defected model ($N \ge 5$), Algorithm \ref{alg:nn-2gather} solves the gathering problem in two rounds from any configuration where there exist only two accompanied robots (that are at the same point).
\end{lemma}

\begin{proof}
Let $r_1$ and $r_2$ be the two accompanied robots.  Robots $r_1$ and $r_2$ observe all robots and recognize that single robots exist, which makes $r_1$ and $r_2$ stay at the current point.

Now consider actions of single robots.
A single robot $r$ misses one robot in its \emph{Look}\ operation, which implies that $r$ observes (a) both $r_1$ and $r_2$ or (b) only one of $r_1$ and $r_2$.
In case (a), $r$ moves to the point, say $p_a$, occupied by $r_1$ and $r_2$.
In case (b), $r$ moves to the center, say $p_b$, of the SEC of all the occupied points.
Thus after one round, all the robots are located at $p_a$ or $p_b$.  Note that $p_a$ is occupied by multiple robots including $r_1$ and $r_2$.

When $p_b$ is not occupied by any robot, the gathering is already achieved.
When $p_b$ is occupied by multiple robots, the robots at $p_b$ observe all the robots. Thus, all the robots move to the center of the SEC of $p_a$ and $p_b$ (or the midpoint of $p_a$ and $p_b$) in the next round (by lines 1 and 2 in Algorithm \ref{alg:nn-2gather}), which achieves the gathering.
When $p_b$ is occupied by only one robot $r$, $r$ detects that $p_a$ is occupied by multiple robots and moves to $p_a$ in the next round (by lines 3 and 4 in Algorithm \ref{alg:nn-2gather}) while the robots at $p_a$ recognize that $p_b$ is occupied by only one robot and does not move (see line 7 in Algorithm \ref{alg:nn-2gather}).  Thus, the gathering is achieved.
\end{proof}

\begin{lemma}
\label{lemma:0ac}
In the adversarial ($N$, $N-2$)-defected model ($N \ge 5$), Algorithm \ref{alg:nn-2gather} solves the gathering problem in three rounds from any configuration where all robots are single.
\end{lemma}

\begin{proof}
Each robot misses one robot in its \emph{Look}\ operation.
When there exist two robot $r_1$ and $r_2$ that miss the same robot, $r_1$ and $r_2$ get the same point set $\mathsf{OPSET}()$ and moves to the center of the SEC of $\mathsf{OPSET}()$ (by lines 5 and 6 in Algorithm \ref{alg:nn-2gather}).
From Lemmas \ref{lemma:3ac} and \ref{lemma:2ac}, two additional rounds are enough to achieve the gathering.

When no two robots miss the same robot, for any pair of two distinct robots $r_1$ and $r_2$, the robot missing $r_1$ is different from the robot missing $r_2$.  Let $C$ be the SEC of all the occupied $N$ points.  First, consider the case that two (or more) robots $r_a$ and $r_b$ are located properly inside $C$.
The SEC of $R\setminus \{r_a\}$ is also the SEC of $R\setminus \{r_b\}$ (that is $C$ from the second property of Proposition \ref{prop:SEC}), which implies that the two robots observing $R\setminus \{r_a\}$ and $R\setminus \{r_b\}$ move to the same point (or the center of the SEC).  From Lemmas \ref{lemma:3ac} and \ref{lemma:2ac}, two additional rounds are enough to achieve the gathering.

Second, consider the case that $N-1$ or $N$ robots are on the boundary of $C$.  From the last property of Proposition \ref{prop:SEC}, there exist three robots $r_1, r_2, r_3$ on the boundary of $C$ that form an acute or right triangle.  There exist two robots $r_4$ and $r_5$ other than $r_1, r_2, r_3$ from $N \ge 5$.  Both the robots observing $R\setminus \{r_4\}$ and $R\setminus \{r_5\}$ observe all of $r_1, r_2, r_3$. The third property of Proposition \ref{prop:SEC} implies that the two robots find the same SEC of $\mathsf{OPSET}()$ (or the SEC of $r_1, r_2, r_3$), which implies that they move to the same point (or the center of the SEC) (by lines 5 and 6 in Algorithm \ref{alg:nn-2gather}).
From Lemmas \ref{lemma:3ac} and \ref{lemma:2ac}, two additional rounds are enough to achieve the gathering.
\end{proof}

From Lemmas \ref{lemma:3ac}, \ref{lemma:2ac} and \ref{lemma:0ac}, the following theorem holds.

\begin{theorem}
In the adversarial ($N$, $N-2$)-defected model ($N \ge 5$), Algorithm \ref{alg:nn-2gather} solves the gathering problem in three rounds. \qed
\end{theorem}

Algorithm \ref{alg:nn-2gather} cannot solve the gathering problem for the case of $N=4$.
Assume that four robots, $R=\{r_0, r_1, r_2, r_3\}$.  Three robots $r_1, r_2$ and $r_3$ are deployed to form an equilateral triangle as Figure \ref{impos_eqtri} and $r_0$ is located at the center of the triangle ({\it i.e.,~} point $p_c$ in Figure \ref{impos_eqtri}).
Consider the case that $r_i$ observes $r_{(i+1)\ \text{mod}\ 3}$ and $r_{(i+2)\ \text{mod}\ 3}$ for each $i\ (0 \le i \le 3)$.
According to Algorithm \ref{alg:nn-2gather}, $r_0$ moves to the midpoint of $r_1$ and $r_2$, $r_1$ moves to $p_1$, $r_2$ moves to the midpoint of $r_2$ and $r_3$, and $r_3$ moves to the midpoint of $r_3$ and $r_1$.
In the resultant configuration, $r_0, r_2$ and $r_3$ form an equilateral triangle and $r_1$ is located at the center $p_1$ of the triangle, which shows by repeating the argument that the gathering is never achieved.

Thus we need another gathering algorithm for the adversarial (4, 2)-defected model,
however, we do not know whether the gathering problem in the adversarial (4,2)-defected model is solvable or not yet.
In the next section, we present an algorithm to solve the gathering problem
in the distance-based (4,2)-defected model.

\section{Algorithm in the Distance-based (4,2)-defected Model}
\label{sec:42alg}
Now we present an algorithm to solve the gathering problem
from any arbitrary initial configuration
in the distance-based (4,2)-defected model.

\noindent
{\bf Basic Strategy}.
In this model, the number of robots is 4 and each robot observes at most two other points occupied by some other robots
(three points in total including one occupied by itself).
In other words, the observation result of each robot
forms a triangle (by three points/robots) when every robot is single.
The strategy of the proposed algorithm is
to determine one unique point from the formed triangle.
Therefore, if any two robots observe the same three points occupied by the robots (including itself),
the two robots move to the same point according to the proposed algorithm.
If two or more robots are accompanied, the gathering can be achieved as the same manner
introduced in Algorithm \ref{alg:nn-2gather}.
Obviously, in this strategy, we have to consider the case so that all 4 robots observe the different triangles.
We resolve this problem by the geometrical property
(remind that each robot cannot observe the farthest robot from itself in
the distance-based defected model).

\begin{algorithm}[bthp]
\caption{Gathering algorithm for robot $r_i$ in the distance-based (4,2)-defected model}
\label{alg:42gather}
{\bf functions:}
\begin{description}
\item[$~\cdot~ \mathsf{OPSET}()$] returns a set of points $\{p\:|\: p$ is occupied by $r_i$ or a point occupied by the robot that $r_i$ observed$\}$.
\item[$~\cdot~ \mathsf{isMulti}(p)$] returns $\mathsf{TRUE}$ if point $p$ is occupied by two or more robots that $r_i$ observed (weak multiplicity), otherwise $\mathsf{FALSE}$.
\end{description}
{\bf algorithm:}
\begin{algorithmic}[1]
\If{$\forall p \in \mathsf{OPSET}(): \mathsf{isMulti}(p) = \mathsf{TRUE}$}
\State{move to the center of the smallest enclosing circle of $\mathsf{OPSET}()$}
\ElsIf{($r_i$ is single) $\wedge$ $(\exists p \in \mathsf{OPSET}(): \mathsf{isMulti}(p) = \mathsf{TRUE})$}
\State{move to an arbitrary point $p \in \mathsf{OPSET}()$ such that $\mathsf{isMulti}(p) = \mathsf{TRUE}$}
\ElsIf{$\forall p \in \mathsf{OPSET}(): \mathsf{isMulti}(p) = \mathsf{FALSE}$}
    \If{$\mathsf{OPSET}()$~forms an equilateral triangle}
    \State{move to the center of the triangle ({\it i.e.,~} incenter)} \Comment{Rule 1}
    \ElsIf{$\mathsf{OPSET}()$~forms an isosceles triangle}
    \State{move to the midpoint of the base of the triangle} \Comment{Rule 2}
    \Else \Comment{the other triangle or collinear three points}
    \State{move to the midpoint of the longest line} \Comment{Rule 3}
    \EndIf
\EndIf \Comment{No action if ($r_i$ is accompanied) $\wedge$ $(\exists p \in \mathsf{OPSET}(): \mathsf{isMulti}(p) = \mathsf{FALSE})$}
\end{algorithmic}
\end{algorithm}

Algorithm \ref{alg:42gather} presents an algorithm to achieve the gathering in the distance-based (4,2)-defected model.
Each robot which does not observe any accompanied robots executes one among three rules
(lines from 6 to 11 in Algorithm \ref{alg:42gather}).
Figure \ref{alg42rules} illustrates these three rules.
If a robot observes an equilateral triangle ({\it i.e.,~} the points observed by the robot form an equilateral triangle),
it moves to the center of the triangle (Figure \ref{alg42rules}(a)),
and if it observes an isosceles triangle, it moves to the midpoint of the base of the triangle (Figure \ref{alg42rules}(b)).
In the other case, it moves to the center point of the longest line of the triangle (Figure \ref{alg42rules}(c)).
It is obvious that if any two robots observe the same set of points ({\it i.e.,~} the same \emph{view}: $r_i.\mathsf{OPSET}() = r_j.\mathsf{OPSET}()$, where $i \neq j$),
the two robots move to the same point according to Algorithm \ref{alg:42gather}.
Hence the following lemma holds.

\begin{lemma}
\label{lem:sameview}
In any configuration, if two or more robots have the same view,
the robots move to the same point in one round by Algorithm \ref{alg:42gather}. \qed
\end{lemma}

\begin{figure}[b]
	\begin{center}
		\subfloat[Case of an equilateral triangle]{\includegraphics[scale=0.7]{rule1.pdf}}
        \hspace{10pt}
		\subfloat[Case of an isosceles triangle]{\includegraphics[scale=0.7]{rule2.pdf}}
        \hspace{10pt}
		\subfloat[The other case]{\includegraphics[scale=0.7]{rule3.pdf}}
	\end{center}
    \caption{Three rules in Algorithm \ref{alg:42gather}}
    \label{alg42rules}
\end{figure}

In Algorithm \ref{alg:42gather}, actions when a robot observes any accompanied robots (including itself)
are the exactly same as Algorithm \ref{alg:nn-2gather} (lines from 1 to 4 in both algorithms).
Therefore, by Lemma \ref{lemma:3ac}, the gathering is achieved in one round when there are three or more accompanied robots.
Now we consider the case that there are two accompanied robots
in the following lemma.

\begin{lemma}
\label{lemma:2acin4}
In the distance-based (4,2)-defected model,
Algorithm \ref{alg:42gather} solves the gathering problem in two rounds
from any configuration where there exist only two accompanied robots.
\end{lemma}

\begin{proof}
Let $r_1$ and $r_2$ be the two accompanied robots, and let $r_3$ and $r_4$ be the two single robots.
Robots $r_1$ and $r_2$ observe all robots and recognize that single robots exist, which makes $r_1$ and $r_2$ stay at the current point
(by line 13).
If robot $r_3$ (or $r_4$) observes both $r_1$ and $r_2$, it moves to the point of $r_1$ and $r_2$ (by lines 3 and 4),
this results in the configuration where there exist three or more accompanied robots and the gathering can be achieved by Lemma \ref{lemma:3ac}.
Now we assume that two robots $r_3$ and $r_4$ miss either $r_1$ or $r_2$ ({\it i.e.,~} they observe only one robot either $r_1$ or $r_2$).
This implies that robots $r_3$ and $r_4$ observe each other,
and these two robots observe the exactly same set of (three) points because robots $r_1$ and $r_2$ are at the same point (by lines from 6 to 11).
Therefore, robots $r_3$ and $r_4$ move to the same point in one round (by Lemma \ref{lem:sameview}) and the gathering is achieved in the next round.
\end{proof}

Even when all 4 robots are single, if two or more robots observe the same set of points,
the robots move to the same point (by Lemma \ref{lem:sameview}),
thus the gathering is achieved by Lemma \ref{lemma:3ac} and Lemma \ref{lemma:2acin4}.

Now we show that the gathering is eventually achieved in any configuration
where all 4 (single) robots have the different views ({\it i.e.,~} observe the different set of points).

\begin{lemma}
\label{lem:convex}
In the distance-based (4,2)-defected model,
if all robots have the different views,
the shape formed by the robots is a convex rectangle.
\end{lemma}

\begin{proof}
We prove the contraposition of the lemma:
if the robots do not form a convex rectangle, there exist two robots having the same view.

\begin{figure}[t]
	\begin{center}
	    \includegraphics[scale=0.7]{concave.pdf}
	\end{center}
    \caption{An example of a concave rectangle}
    \label{concave}
\end{figure}

Assume that the 4 robots, from $r_1$ to $r_4$, form a concave rectangle as Figure \ref{concave}
(Note that we can also assume that the robots form a triangle ({\it i.e.,~} three robots are collinear), it can be also proved in the same manner).
A concave rectangle has an interior angle which is larger than $180^{\circ}$,
so we assume robot $r_1$ is located at the point with such an angle as Figure \ref{concave}.
Let $e$ be the line $\overline{r_1r_2}$, either angle $\angle{r_2r_1r_4}$ or angle $\angle{r_2r_1r_3}$ is an obtuse angle ({\it i.e.,~} angle larger than $90^{\circ}$)
because interior angle $\angle{r_4r_1r_3}$ is larger than $180^{\circ}$.
Without loss of generality, we assume angle $\angle{r_2r_1r_3}$ is an obtuse angle (denoted by $\theta$).
Due to $\theta > 90^{\circ}$, $d$ is longer than $c$ and $e$ (see Figure \ref{concave}).
This implies that robot $r_3$ observes $r_1$ and robot $r_2$ also observes $r_1$ (because the farthest robot is missed in the distance-based defected model).
If angle $\angle{r_2r_1r_4}$ (denoted by $\theta'$) is also an obtuse angle, robot $r_4$ also observes $r_1$ by the same reason.
As a result, all robots observe $r_1$ (including $r_1$ itself) and the lemma holds
because there is a robot which has the same view with robot $r_1$ by the pigeonhole principle.
If angle $\angle{r_2r_1r_4}$ is an acute angle ({\it i.e.,~} angle smaller than $90^{\circ}$) or a right angle,
$\theta + \theta' < 270^\circ$ holds.
This means that an exterior angle $\angle{r_4r_1r_3}$ ({\it i.e.,~} $360^\circ - \theta - \theta'$) is an obtuse angle,
thus $b$ is shorter than $\overline{r_4r_3}$. Also in this case, robot $r_4$ observes $r_1$ and the lemma holds.
\end{proof}

\begin{lemma}
\label{lem:unob22}
Assume that all robots have the different views.
If robot $r_i$ cannot observe robot $r_j$ ({\it i.e.,~} robot $r_i$'s view does not include the point occupied by $r_j$),
$r_j$ cannot observe $r_i$ neither.
\end{lemma}

\begin{proof}
To help to explain, we introduce a directed graph $\vec{G} = (V,A)$ such that
$V = \{r_1, r_2, r_3, r_4\}$ and $(r_i, r_j) \in A$ if robot $r_i$ cannot observe $r_j$.
If all robots have the different views, there exist only two cases as Figure \ref{dag}.
And we show that there is no case as Figure \ref{dag}(a) to prove the lemma.

\begin{figure}[b]
 \begin{minipage}{0.6\hsize}
	\begin{center}
		\subfloat[]{\includegraphics[scale=0.7]{dag01.pdf}}
		\subfloat[]{\includegraphics[scale=0.7]{dag02.pdf}}
	\end{center}
    \caption{Directed graphs showing an unobserved robot}
    \label{dag}
 \end{minipage}
 \hfill
 \begin{minipage}{0.4\hsize}
	\begin{center}
	    \includegraphics[scale=0.7]{triangles.pdf}
	\end{center}
    \caption{Two isosceles triangles}
    \label{twotri}
 \end{minipage}
\end{figure}

Assume the case as Figure \ref{dag}(a):
robot $r_1$ cannot observe $r_2$, robot $r_2$ cannot observe $r_3$, and so on.
$\overline{r_1r_4} \leq \overline{r_1r_2}$ holds because robot $r_1$ cannot observe $r_2$.
For the same reason,
$\overline{r_1r_2} \leq \overline{r_2r_3}$, $\overline{r_2r_3} \leq \overline{r_3r_4}$,
and $\overline{r_3r_4} \leq \overline{r_1r_2}$ also hold.
Therefore, $\overline{r_1r_4} \leq \overline{r_1r_2} \leq \overline{r_2r_3} \leq \overline{r_3r_4} \leq \overline{r_1r_4}$ holds,
thus $\overline{r_1r_2} = \overline{r_2r_3} = \overline{r_3r_4} = \overline{r_1r_4}$ holds.
For simplicity, we assume that the length of $\overline{r_1r_2}$ is 1.

Now we consider the triangle $\triangle{r_1r_2r_3}$.
Due to $\overline{r_1r_2} = \overline{r_2r_3}$, triangle $\triangle{r_1r_2r_3}$ is an isosceles triangle (the base is $\overline{r_1r_3}$).
Similarly, triangle $\triangle{r_1r_3r_4}$ is also an isosceles triangle which has line $\overline{r_1r_3}$ as the base.
Line $\overline{r_1r_3}$ is the common base of these two isosceles triangles,
thus the locations of 4 robots are as Figure \ref{twotri}.

In Figure \ref{twotri}, we consider the lengths of two diagonal lines, $\overline{r_1r_3}$ and $\overline{r_2r_4}$.
By the assumption, robot $r_1$ cannot observe $r_2$, therefore, $\overline{r_1r_3} \leq 1$ holds because robot $r_1$ observes $r_3$.
As the same reason, $\overline{r_2r_4} \leq 1$ also holds.
However, both $\overline{r_1r_3} \leq 1$ and $\overline{r_2r_4} \leq 1$ cannot hold in this rhombus,
therefore, there is no case as Figure \ref{dag}(a) and the lemma holds.
\end{proof}

By Lemma \ref{lem:unob22}, if all robots have different views,
we have two disjoint pairs of robots such that robots in each pair cannot observe each other as in Figure \ref{dag}(b).
Now we discuss the location relations among the robots in this case by the following lemma.

\begin{lemma}
\label{lem:relation}
If all robots have different views in the distance-based (4,2)-defected model,
each of two robots which cannot be observed each other are
diagonally located on the formed convex rectangle.
\end{lemma}

\begin{proof}
We already proved that the robots form a convex rectangle if all robots have different views by Lemma \ref{lem:convex}.
Let $r_1$ and $r_2$ be two robots which do not observe each other,
and we assume for contradiction that $r_1$ and $r_2$ are not diagonally located ({\it i.e.,~} line $\overline{r_1r_2}$ is an edge of the convex rectangle).
For simplicity, we assume the length of $\overline{r_1r_2}$ is 1.

\begin{figure}[t]
	\begin{center}
	    \includegraphics[scale=0.5]{twocircles1.pdf}
	\end{center}
    \caption{Possible positions of robots $r_3$ and $r_4$}
    \label{twocircle1}
\end{figure}

Figure \ref{twocircle1} illustrates two circles, called $S_1$ and $S_2$, with radius 1
centered at $r_1$ and $r_2$ respectively.
Consider the position of robot $r_3$: robot $r_3$ should be located in area $S_1 \cap S_2$,
because both $r_1$ and $r_2$ observe $r_3$ (remind that $r_1$ and $r_2$ do not observe each other).
Locate $r_3$ in an arbitrary point in area $S_1 \cap S_2$.
Let $a$ be the length of longer one between $\overline{r_1r_3}$ and $\overline{r_2r_3}$ ({\it i.e.,~} $a = max(|\overline{r_1r_3}|,|\overline{r_2r_3}|)$),
here we assume $a$ is the length of $\overline{r_1r_3}$ without loss of generality.
Circles $C_1$, $C_2$, and $C_3$ present the circles with radius $a$ centered at $r_1$, $r_2$, and $r_3$ respectively.
By Lemma \ref{lem:unob22}, robots $r_3$ and $r_4$ cannot observe each other,
thus $|\overline{r_3r_4}| \geq a$ holds; robot $r_4$ should be located outside of $C_3$.
As a result, robot $r_4$ should be located in $(C_1 \cap C_2) - C_3$
which is presented as the shaded area in Figure \ref{twocircle1}.
In this case, robots $r_1$ and $r_2$ (resp. $r_3$ and $r_4$) are diagonally located on a convex rectangle,
which is a contradiction. Thus the lemma holds.
\end{proof}

Now we show that even when all single robots have different views,
two or more robots move to the same point by Algorithm \ref{alg:42gather}.
We consider the 6 lines derived by the combination of 4 robots (refer to Figure \ref{fig:6lines}).
We focus on the lengths of these 6 lines,
and the following corollary holds by Lemma \ref{lem:relation}.

\begin{figure}[b]
 \begin{minipage}{0.5\hsize}
  \begin{center}
   \includegraphics[scale=0.55]{6lines.pdf}
  \end{center}
  \caption{6 lines by 4 robots}
  \label{fig:6lines}
 \end{minipage}
 \hfill
 \begin{minipage}{0.5\hsize}
	\begin{center}
	    \includegraphics[scale=0.55]{1long.pdf}
	\end{center}
    \caption{Configuration with one longest line}
    \label{fig:1long}
 \end{minipage}
\end{figure}

\begin{corollary}
\label{col:longest}
Consider the 6 lines connecting distinct pairs of two robots.
If all robots are single and have different views,
there is no (side) line which is longer than any diagonal line.
\end{corollary}

It is worthwhile to mention that there can be at most 4 longest lines among 6 lines.
We focus on the number of longest lines and show that the algorithm works correctly in all cases.
By Corollary \ref{col:longest}, if there exist one or two longest lines, they are diagonal lines.
The following lemma holds.

\begin{lemma}
\label{lem:1long}
Assume that all robots are single and have different views in the distance-based (4,2)-defected model,
and consider the 6 lines connecting distinct pairs of two robots.
If there exist one or two longest lines, two or more robots become accompanied in one round.
\end{lemma}

\begin{proof}
Figure \ref{fig:1long} illustrates an example configuration including the only one longest line (as a diagonal line),
where the thick solid line represents the unique longest line.
Without loss of generality, we assume that line $\overline{r_2r_4}$ is the longest one.
From the assumption, $r_1$ and $r_3$ do not observe each other:
$r_1$ observes triangle $\triangle{r_1r_2r_4}$, and $r_3$ observes triangle $\triangle{r_2r_3r_4}$.
These two triangles are not equilateral triangles because line $\overline{r_2r_4}$ is the unique longest line.
Therefore, robots $r_1$ and $r_3$ move to the midpoint of line $\overline{r_2r_4}$ (by line 9 or 11).
If there are two longest lines, the both lines are diagonal lines by Corollary \ref{col:longest}
($\overline{r_1r_3}$ and $\overline{r_2r_4}$ in Figure \ref{fig:1long}).
However, this does not affect to the actions of robots $r_1$ and $r_3$; they move to the midpoint of line $\overline{r_2r_3}$.
Thus the lemma holds.
\end{proof}

Now we consider the case that
there is a side line whose length is the same as two diagonal lines;
there are three longest lines.

\begin{lemma}
\label{lem:3long}
Assume that all robots are single and have different views in the distance-based (4,2)-defected model,
and consider the 6 lines connecting distinct pairs of two robots.
If there are the three longest lines, two or more robots become accompanied in two rounds.
\end{lemma}

\begin{proof}
Figure \ref{fig:3long} illustrates the only configuration including three longest lines.
Three thick solid lines are the three longest lines.
Remind that robots $r_1$ and $r_3$ (or $r_2$ and $r_4$) cannot observe each other.
By Algorithm \ref{alg:42gather},
all robots move to the different points:
robot $r_1$ (resp. $r_2$) moves to the midpoint $p_1$ (resp. $p_2$) of line $\overline{r_1r_4}$ (resp. $\overline{r_2r_3}$) since $r_1$ (resp. $r_2$) observes an isosceles triangle $\triangle{r_1r_2r_4}$ (resp. $\triangle{r_1r_2r_3}$).
Robot $r_3$ (resp. $r_4$) moves to the midpoint $p_3$ (resp. $p_4$) of line $\overline{r_2r_4}$ (resp. $\overline{r_1r_3}$) that is the longest line of the observed triangle $\triangle{r_2r_3r_4}$ (resp. $\triangle{r_1r_3r_4}$).
In this case,
triangles $\triangle{r_2r_3r_4}$ and $\triangle{r_2p_2p_3}$ are similar,
the length of line $\overline{p_2p_3}$ is half of the length of line $\overline{r_3r_4}$,
and line $\overline{p_2p_3}$ and line $\overline{r_3r_4}$ are parallel.
Through the same argument for lines $\overline{p_1p_4}$ and $\overline{r_4r_3}$, we can show that the lengths of lines $\overline{p_1p_4}$ and  $\overline{p_2p_3}$ are the same
and these two lines are parallel.
This means that the rectangle formed in the next round is a parallelogram:
even if all robots have different views in this configuration,
two or more robots become accompanied in the next round
because diagonal line $\overline{p_1p_2}$ is the unique longest line (by Lemma \ref{lem:1long}).
\end{proof}

\begin{figure}[t]
 \begin{minipage}{0.5\hsize}
	\begin{center}
	    \includegraphics[scale=0.5]{3longb.pdf}
	\end{center}
    \caption{Configuration with three longest lines}
    \label{fig:3long}
 \end{minipage}
 \hfill
 \begin{minipage}{0.5\hsize}
	\begin{center}
	    \includegraphics[scale=0.46]{4long.pdf}
	\end{center}
    \caption{Configuration with four longest lines}
    \label{fig:4long}
 \end{minipage}
\end{figure}

\begin{lemma}
\label{lem:4long}
Assume that all robots are single and have different views in the distance-based (4,2)-defected model,
and consider the 6 lines connecting distinct pairs of two robots.
If there are four longest lines, two or more robots become accompanied in two rounds.
\end{lemma}

\begin{proof}
Figure \ref{fig:4long} illustrates the only configuration including four longest lines.
Four thick solid lines are the four longest lines.
By Algorithm \ref{alg:42gather},
all robots move to the different points:
robot $r_1$ (resp. $r_3$) moves to the midpoint $p_1$ (resp. $p_3$) of line $\overline{r_1r_4}$ (resp. $\overline{r_3r_4}$) since $r_1$ (resp. $r_3$) observes an isosceles triangle $\triangle{r_1r_2r_4}$ (resp. $\triangle{r_2r_3r_4}$).
Robot $r_2$ moves to the center $p_2$ of the equilateral triangle $\triangle{r_1r_2r_3}$ it observes,
and $r_4$ moves to the midpoint $p_4$ of the unique longest line $\overline{r_1r_3}$ it observes
(note that if $|\overline{r_1r_4}|=|\overline{r_3r_4}|$, triangle $\triangle{r_1r_3r_4}$ is an isosceles triangle,
however robot $r_4$ moves to the midpoint $p_4$ of the base line also in this case).
As a result, the four points, from $p_1$ to $p_4$, form a concave rectangle.
Hence, two or more robots become accompanied in the next round by Lemma \ref{lem:convex}.
\end{proof}

From Lemmas \ref{lemma:3ac}, \ref{lemma:2acin4}, \ref{lem:1long}, \ref{lem:3long} and \ref{lem:4long},
the following theorem holds.

\begin{theorem}
In the distance-based (4, 2)-defected model, Algorithm \ref{alg:42gather} solves the gathering problem in at most four rounds. \qed
\end{theorem}

\section{Impossibility Results}
\label{sec:impos}
In this section, we present two impossibility results for the gathering problem in the defected view model;
(1) there is no (deterministic) algorithm in the arbitrary or distance-based (3,1)-defected model,
and
(2) there is no (deterministic) algorithm in the relaxed adversarial ($N$,$N-2$)-defected model defined in Section \ref{sec:relaxed}.

\subsection{Impossibility in (3,1)-defected model}
By two gathering algorithms we introduced in the previous sections,
the gathering can be achieved in the adversarial (and thus also in the distance-based)
($N$,$N-2$)-defected model for $N \geq 5$,
and in the distance-based (4,2)-defected model.
These results bring us a problem to find an algorithm to solve the gathering problem
in the distance-based (or adversarial) (3,1)-defected model.
Here we show that there is no such algorithm.

\begin{theorem}
There is no (deterministic) algorithm to solve the gathering problem
in the distance-based or adversarial (3,1)-defected model.
\end{theorem}

\begin{figure}[t]
  \centering
		\includegraphics[scale=0.8]{impos_eqtri.pdf}
  \caption{Example for an unsolvable configuration in the distance-based (3,1)-defected model}
  \label{impos_eqtri}
\end{figure}

\begin{proof}
We prove this theorem by showing that there is no (deterministic) algorithm even in the distance-based (3,1)-defected model.
Note that the distance-based (3,1)-defected model is stronger than the adversarial one,
hence if the gathering cannot be achieved in the distance-based one, the gathering is also unsolvable in the adversarial one.
Assume that three robots, $R=\{r_1, r_2, r_3\}$, are arranged in an equilateral triangle as Figure \ref{impos_eqtri},
and robot $r_1$ (resp. $r_2$ and $r_3$) observes $r_2$ (resp. $r_3$ and $r_1$).
All robots do not agree on any geometrical agreement
({\it e.g.,~} direction, orientation, chirality, or unit distance),
thus we can assume that every robot $r_i$ considers the direction to the center of the triangle ($p_c$) from itself
({\it i.e.,~} $\overrightarrow{r_ip_c}$) as the positive direction of $X$-axis
in its local coordinate system.
Moreover, we also assume that all robots have the same chirality ({\it e.g.,~} clockwise) and the same unit distance.
This means that all robots have the exactly same result of \emph{Look}~operation.

Let $\mathcal{A}$ be an algorithm for gathering in the distance-based (3,1)-defected model.
In the above configuration,
all robots have the same observation results,
thus they execute the same behaviors according to $\mathcal{A}$
({\it i.e.,~} all robots move to the same $x$ and $y$ coordinates in their local coordinate systems).
This causes another configuration forming a different equilateral triangle, which shows
by repeating the argument that the robots cannot gather at the same point forever.
The only way to prevent the robots from forming another equilateral triangle is to move to point $p_c$,
{\it i.e.,~} each robot moves to the point located at $|\overline{r_ir_j}| / \sqrt{3}$
distance in the $30^\circ$ clockwise direction of the observed robot $r_j$.
However, if all robots agree on the opposite direction of chirality
(counter-clockwise in this case),
they move to the outside of triangle $\triangle{r_1r_2r_3}$
({\it i.e.,~} robot $r_1$ moves to point $p_c'$ instead of $p_c$).
As a result, the robots form another equilateral triangle.
\end{proof}

\subsection{Impossibility in the relaxed adversarial ($N$,$N-2$)-defected model}
\label{sec:relaxed}

The ($N$, $k$)-defected model assumes that $k$ robots observed by robot $r$ are chosen from the robots that are located at points other than $r$'s current position and that $r$ can detect whether it is single or accompanied.
Natural relaxation of the model is to choose the $k$ robots other than $r$ ({\it i.e.,~} robots at $r$'s current position can be chosen) and assume the weak multiplicity detection for the $k$ robots and $r$ itself.
We call the model with the relaxation the \emph{relaxed adversarial} ($N$,$k$)-defected model.
Notice that the key property of the ($N$, $N-2$)-defected model such that any accompanied robot can observe all the robots does not hold in the relaxed model.

The following theorem shows that the gathering is impossible (from some configuration) in the relaxed adversarial ($N$,$N-2$)-defected model.

\begin{theorem}
There is no (deterministic) algorithm to solve the gathering problem in the relaxed adversarial ($N$,$N-2$)-defected model.
\end{theorem}

\begin{proof}
Let $\mathcal{A}$ be a gathering algorithm in the relaxed adversarial ($N$,$N-2$)-defected model.
We consider only initial configurations where all robots are located at two points $p_1$ and $p_2$.

First, consider the initial configuration where $N-1$ robots are located at $p_1$ and one robot, say $r_1$, is located at $p_2$.  When the robots at $p_1$ do not observe $r_1$, they misunderstand that the gathering is already achieved and terminate.  To achieve the gathering, $r_1$ has to move to $p_1$.
This implies that $\mathcal{A}$ has the following action ({\bf Action 1}):
when a single robot $r$ observes only one occupied point other than $r$'s current point and recognizes that the point is occupied by multiple robots, $r$ has to move to the point.

Notice that {\bf Action 1} is sufficient to show that $\mathcal{A}$ cannot solve the gathering in the relaxed adversarial (4,2)-defected model.  Consider the initial configuration where two robots exist at both of $p_1$ and $p_2$ (four robots in total).  When the robots at $p_1$ (resp. $p_2$) observe only the two robots at $p_2$ (resp. $p_1$), the robots at $p_1$ (resp. $p_2$) move to $p_2$ (resp. $p_1$) by {\bf Action 1}.
At the resultant configuration, two robots exist at both of $p_1$ and $p_2$, which shows by repeating the argument that algorithm $\mathcal{A}$ cannot solve the gathering problem.

Second, consider the initial configuration where $N\ge 5$ and all robots recognize that both $p_1$ and $p_2$ are occupied by multiple robots, which can occur when a point is occupied by three or more robots and the other is occupied by two or more robots.
When all the robots at the same point observe the same set of robots (but still they recognize that both the points are occupied by multiple robots), the robots at the same point execute the same action ({\it i.e.,~} move to the same point).
Since algorithm $\mathcal{A}$ solves the gathering problem, all robots eventually have to move to the same point (precisely the midpoint of the two points occupied by robots) to achieve the gathering.
This implies that $\mathcal{A}$ has the following action ({\bf Action 2}):
when an accompanied robot $r$ observes only one occupied point other than $r$'s current point and recognizes that the point is occupied by multiple robots, $r$ has to move to the midpoint of the two points.

Finally, consider the initial configuration of $N\ (\ge 5)$ robots where two robots exist at $p_1$ and $N-2$ robots exist at $p_2$.
When each robot $r_1$ at $p_1$ observes only $N-2$ robots at $p_2$ (and recognizes itself as a single robot), $r_1$ moves to $p_2$ by {\bf Action 1}.
On the other hand, when each robot $r_2$ at $p_2$ observes the two robots at $p_1$ and $N-4$ robots (other than $r_2$) at $p_2$, $r_2$ moves to the midpoint of $p_1$ and $p_2$ by {\bf Action 2}.
At the resultant configuration, two robots exist at $p_2$ and $N-2$ robots exist at the midpoint of $p_1$ and $p_2$.
By repeating the argument, we can show that algorithm $\mathcal{A}$ cannot solve the gathering problem although all the robots converges at the same point ({\it i.e.,~} the distance between the two groups of robots becomes smaller and smaller but does not become zero).

Consequently, there is no gathering algorithm in the relaxed adversarial ($N$,$N-2$)-defected model.
\end{proof}

\section{Conclusion and Open Problems}
\label{sec:conclusion}
In this paper, we introduced a new computational model, the ($N$, $k$)-\emph{defected model},
where each robot cannot necessarily observe all other robots:
{\it i.e.,~} each robot observes at most $k$ other robots not located at its current position (where $k < N-1$).
We addressed the gathering problem, which is one of the basic problem in autonomous mobile robot systems,
in the ($N$, $N-2$)-defected model.
We proposed two gathering algorithms:
(1) an algorithm in the adversarial ($N$,$N-2$)-defected model that achieves the gathering in three rounds,
and (2) an algorithm in the distance-based (4,2)-defected model that achieves the gathering in four rounds.
Moreover, we showed that there is no (deterministic) algorithm in either the adversarial or distance-based (3,1)-defected model.
In the proposed model,
we assume that each robot $r$ observes $k$ other robots among the robots located at the different points
than the point occupied by $r$ itself.
The relaxation of this assumption, where $k$ robots are chosen among all other robots other than $r$,
can be considered, however, we proved that the gathering is unsolvable in this relaxed model.

The remaining problem we are most interested in is to clarify the solvability of the gathering problem
in the adversarial (4,2)-defected model.
Remind that the basic strategy of the proposed algorithm in the distance-based (4,2)-defected model is
to determine one unique point from the triangle formed by the observed set of points.
We call the algorithm using this strategy the \emph{set-based algorithm},
where each robot determines the destination referring to only the set of observed points:
for example, when a robot observes an isosceles triangle,
it always moves to the midpoint of the base, whether it is adjacent to the base or not,
{\it i.e.,~} we do not use the information of the (relative) position of the observing robot.
We can prove that \emph{there is no (deterministic) set-based algorithm to solve the gathering problem in adversarial (4,2)-defected model}.
This means that if there is a gathering algorithm in the adversarial (4,2)-defected model,
each robot uses its relative position in the set of observed points,
{\it e.g.,~} when a robot observes an isosceles triangle, the destination point changes
whether the robot is at a point incident to the base of the triangle or not.

An important future work is to find the minimum $k$ that allows a solution for the gathering problem in the adversarial or distance-based ($N$, $k$)-defected model.
In this paper, we considered only the gathering problem,
thus to challenge other problems under the ($N$, $k$)-defected model is another future work.

redpajama_set_name: RedPajamaArXiv


text:
The Making of a Theocracy: Government Funding and the Crisis of Compassion
Posted on November 9, 2013 November 9, 2013 by the uppity negro
This week, in the wake of a federal government shut down, a Congress with an approval rating uniformly in single digits, the GOP party with historical low approval numbers is still failing to see just how the uber-right wing arm of the party (namely the Tea Party faction) is concretizing their place on the wrong side of history. Despite government shut downs, and still not passing a budget and operating off of sequester funding, Congress has before it the Employee Non-Discrimination Act (ENDA) which was successfully passed with a bit of breathing room by the U.S. Senate earlier this week. The only thing before it and President Obama's signature is Speaker John Boehner (R-Oh.) to bring it before the House. Boehner repeatedly has said that he will not bring it to the floor of the House on the basis of frivolous lawsuits that may spawn from this law and also that most states and companies already have policies and laws in place that ban workplace discrimination on the basis of sexual orientation.
Opponents stayed largely silent until Thursday, when Sen. Daniel Coats (R-Ind.) said on the Senate floor that ENDA "diminishes" the religious freedom of organizations and employers who may feel compelled to hire people who hold religious views contrary to the organization.
"I oppose discrimination of any kind, and that includes discrimination, however, also, of individuals or institutions for their faith and values, which often gets lost and has been lost in this discussion," Coats said.
The conservative Family Research Council also warned that ENDA "would transform the workplace into an environment in which certain self-identifications and conduct must be given special privileges by employers" that might require people to suppress religious or moral views.
Those concerns are shared by Boehner, who reiterated his years-long opposition to the proposal in a statement this week. But supporters said Thursday that the House version of ENDA has at least 193 GOP co-sponsors and that they are pursuing dozens more. Congressional Democrats also suggested that socially conservative Republicans are increasingly out of step with most Americans.
Albert Mohler, President of Southern Baptist Seminary
I was prompted to write a response to this socially conservative way of thinking and I was spurred by the post "Render unto Caesar or unto God? Government Funding and the Crisis of Conscience" written by Southern Baptist Seminary President Albert Mohler. Mohler has been the de facto spokesperson for the [at times ultra] conservative Southern Baptist Convention. This is the same denominational organization that split in the 1840s specifically over the issue of slavery and took until the 1990s before an official apology was offered for their participation. They were noted for some semblance of progress when they elected their first black president in June 2012. Nevertheless, Mohler's media presence far outshadows elected officials of the organization, and no one seems to have a problem with that.
In his recent post Mohler meanders through the creation of the Louisville Baptist Orphan's Home established in the Reconstruction South and what it's original intent was and a 1986 pledge that was in keeping with "Christian principles." Mohler goes on to write that "That pledge is now very much in question as reports indicate that the ministry, now renamed Sunrise Children's Services, is poised to change its hiring policies to remove any barrier to homosexuals and lesbians working as employees of the ministry." The current president of the Sunrise Children's Services (the modern-day name), has now gone on record saying that the company, in accordance with modern practices will not discriminate against the hiring of those who are a part of the LGBT community. This in light of a lawsuit of the same nature filed in 1998.
Included in Smithwick's argument was his personal statement that he would "rather homosexuals see the love of God through us than be denied employment by us." He closed by offering the strange analogy of a missionary serving in Iran who wore a head covering out of respect for Muslims, apparently missing the point that no biblical command or biblical teaching is violated by wearing a head covering.
Kentucky Baptists were not alone in their shock over the Sunrise proposal. An attorney who had represented the terminated lesbian employee toldThe Courier-Journal (Louisville), "This is very surprising. They were very adamant that they wouldn't hire gays and lesbians." He is right, they wereadamant about the matter and, at least until the board votes later this week on Smithwick's proposal, they still are—at least officially.
All that can change in short order. Bill Smithwick is absolutely right about one aspect of this matter: there is every likelihood that governmental coercion on these issues is coming. The Employment Non-Discrimination Act (ENDA) is expected to pass in the U.S. Senate in coming days, and the Obama Administration has threatened to accomplish much the same by executive order. It is hard to imagine how an entity that describes itself as "not a church or a religious institution" can claim an exemption under such a legal mandate.
The choice faced by Sunrise, soon likely to be faced by a host of similar organizations, is to get smaller or get secular. The instant an organization takes government money it is transformed into an instrument of the state. What Caesar funds, Caesar controls. This is a hard lesson, and one likely soon to be learned by Christian institutions that have been taking government money and have grown dependent on those funds.
This will not end with children's homes. A good many Christian colleges and universities have grown dependent on funds flowing through federal student aid programs and similar forms of government funding. What happens when they face a similar choice? The math will not work in their favor. A hard choice will have to be made, and we will soon see who will stand on conviction and who will act to save their funding.
The question does not stop with funding. Soon after Britain passed antidiscrimination legislation like ENDA, Christian adoption agencies were basically put out of business. They were given a choice to sever ties with their churches or go out of business. In Massachusetts, the legalization of same-sex marriage meant the end of the adoption work done by Catholic Charities, since they could not and would not violate their convictions. In Illinois, the work of Catholic Charities in foster care and adoption came to an end in 2011, and the admired organization gave up millions in government funding because they would not violate their convictions.
The board of directors of Sunrise Children's Services faces a hard choice, but the choice is not just between several policy alternatives. They will decide to serve God or to serve Caesar. Paul Chitwood, executive director of the Kentucky Baptist Convention, urged Sunrise to step out in faith, even if it means losing massive funding. He urged the agency "to dramatically scale back its work in order to be faithful to Scripture and to model biblical values in front of hurting children." As for Kentucky Baptists, they will find a way to serve children and keep their convictions, assures Chitwood: "Either way, I am confident Kentucky Baptists will always minister to hurting children and will do so through a ministry with biblical values."
When asked about the payment of taxes, Jesus famously responded, "Render to Caesar the things that are Caesar's, and to God the things that are God's" (Mark 12:17). We dare not render to Caesar what belongs rightly and only to God.
I'm not so sure if this is a crisis of conscience or a crisis of compassion levied on behalf of those who so doggedly cling to their religious beliefs. One of the differentiations between a zealot and someone simply passionate about their beliefs is that a zealot will inflict or allow harm for their beliefs, whereas a passionate person will simply go down with a good fight and say "We'll get 'em next time." This blind approach to beliefs results in a "bad religion" that allows for basic human services to fall by the wayside simply because the Bible says is implemented to the point of exclusion rather than inclusion.
"Looking for a City" Patricia Wagner
I think what always gets missed in these discussion is the fact that this approach encourages a theocracy. This is not a battle between scriptures and which Jesus or Paul quotes we choose to engage today, not even does want to engage in an love ethic or an ethic of war, but actually that there are those who fundamentally believe "Thy will be done on earth…" as the zealous mandate to bring the Kingdom of God down here [from a real tangible heaven above, supposedly?] to earth. I agree that the movement of scriptures, specifically concluding with Revelation 21:2 where the New Jerusalem did come down from heaven to earth, does support this concept of "heaven on earth," it flies in the face of the precept of "separation of church and state."
When relying so heavily on the biblical "mandate" does to exclude persons from community and preclude helping those on the margins, I question what is the veracity of such a religious mandate. Those that agree with Mohler have created a homogenous community in which the fullness of humanity is not welcomed. This selected community is the same community that produces the Tea Party and allows them to engage in the divisive bombast that marks the likes of Sarah Palin to Ted Cruz. It's this same xenophobic conscience that allows Ted Cruz's father to tell Obama to go back to Kenya. It's this same concept that gives door to people fighting to "take back" the country; taking back the country from whom or from what has yet to be decisively answered.
Religious zeal of this nature has no room for a conscience. There is no crisis of conscience as Mohler blithely attempts to argue, he is quite clear how this game is to played out and for who's benefit. The underlying message, is that he's expecting the ultra-conservative religious groups to open up their pockets and subsidize the gap of what the end of government funding will leave as a result of ENDA potentially becoming law. The fictional vice president Sally Langston engaged in this similar dog-whistle strategy just this week on the latest episode of ABC's "Scandal" where the evangelical pastors and spokespersons whipped that faction into a frenzy and thus opened up pockets and checkbooks. There was never an crisis of conscience, but rather how to affect the bottom line: money.
Mohler writes this as an agent of Religion. Yes, Religion with a capital-R. Those of us who are in this system may claim that we are in the system to be agents of change, but as long as we collect paychecks twice a month or once a month from the system we are nothing more than errandboys and girls, working for the System (another proper noun). The System is a church staffmember, a pastorate, a professor at a school, a CEO of a religious-based charity whatever the case maybe; some perhaps are doing good work, but it's all a part of the System and we are all agents of Religion.
Systems inherently aren't good or bad. I love countering people's arguments when they say "All religion is bad," (yes, I still get people who speak in absolutes as though this is 1955) and I respond by saying "If there wasn't any organized religion, do you know how many homeless shelters, charity offices and other goodwill places wouldn't exist? Surely you don't expect the government to take up the slack?" Usually a blank stare follows. However, when a System is categorized by whom it excludes, rather than known for who it includes, it's at that moment that I begin to question the utter intent of why it was created.
Mohler states that the original intent in 1869 was to care for "orphan and destitute children," to which I ask, is this only doable when biblical mandates are fully achieved? It assumes that one's sexual orientation is the sole biblical mandate that stands in the way of doing this good work. Based on this logic, all fat people should be discriminated against from working there because of their greed, or that the majority of married people should be let go from their jobs because I'm sure at least one time they've looked at someone else wondering what it would be like to have sex with them.
I read Mohler's article, and most of what he's written with a grain of salt: I'm reading the works of a theotician. A theotician is a theological politician, merely working the System. Mohler must say what he's saying in order maintain his position and his social cache in the evangelical world. We need only look at the likes of Carlton Pearson to learn of the struggle and woes of one who was once in the System who decides to act against it.
The entire rhetoric of the evangelical church espoused in pulpits across America and even from media talking heads such as Mohler, all speak to creating a theocracy; it's about willingly participating in the System. It's actually something that has reached even into many average Christian households: when we talk about how bad things have gotten because prayer has been taken out of schools, we're likening ourselves to a theocracy. Every time we lament the ten commandments from being in front a public building, it's our sadness over the fact that this country isn't a theocracy. When we make those claims in our churches, in our public schoolhouses and in our personal homes, we are acting as agents of Religion.
I think it's the supreme example of xenophobia when Christians dislike another belief system to the point that they wish for legislation against inclusion. For me, Mohler's issues isn't the Christian's conscience, but rather the Christian asking just how far one's compassion will go. Obviously for Mohler and his ilk, that compassion stops when you have to work side-by-side a gay or transgendered person.
To push the envelope all the way, to be fully religious and to ultimately the engage the doctrine of the cross: does the atonement of Jesus' blood reach to the highest mountain and the lowest valley or has it been dammed up by religion's own shortcomings facilitated by our own crisis of compassion?
Tags: Albert Mohler, ENDA, EvangelicalCategories: Cultural Critique, Politics, Pop Culture, Religion
5 thoughts on "The Making of a Theocracy: Government Funding and the Crisis of Compassion"
If religious institutions were taxed, wouldn't that be 'rendering unto Caesar' as instructed?
dave bowen says:
Spirituality is a gift from the creator. Religion is a man made element,designed to guilt trip people,and control their thoughts and speech.You do not have to be religious to be spiritual. One has nothing to do with the other
volcanosunset says:
I agree with Mr. Bowen, spirituality has nothing to do with religion. I believe that the bible forewarns us about a broad path leading to destruction, is reffering to the path of religion, and not to the following Christ. They are separarte things. Yet the church, who clothes itself like a lamb, but speaks like a dragon?
They are, however, NOT mutually exclusive.
Response is to Dave Bowen re religion and spirituality.


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King Island Dairies
King Island Dairies were built on a century of dairying for which the island's climate makes it ideally suited. A co-operative butter factory operated at Loorana from 1902, and there were also private cheese factories on the island. Following difficulties in the 1970s, the butter factory was bought in 1978 by WJ Kirk who began to make a new range of products, including Mozzarella cheese and Emmenthal cheese curd. By 1984 it was clear that the company should fill niche markets rather than compete with the huge production of Tasmanian and mainland factories. Under new ownership, butter production ceased three years later, and the factory concentrated on producing specialty cheeses and other products such as triple cream which have been successfully marketed around Australia.
Further reading: J Cassidy, The dairy heritage of northern Tasmani, Launceston, 1995.
Jill Cassidy
Copyright 2006, Centre for Tasmanian Historical Studies


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