Well the awards season is over without much surprise, and I am pleased that A Separation won for Best Foreign Language film. For the first time since 2003 I agree with the Academy's choice.
It was also nice to see a French film dominate the major awards, even though I wasn't as impressed with it as much as most. However, the ceremony as always, was way too long and boring. I wish they would cut the unfunny fluff and let the winners have more time at the podium. Also why didn't they perform the nominated songs? Terrible decision.
The funniest line of the weekend was delivered on Friday by Antoine de Caunes, host of the César Awards, which was otherwise just as dull as the Oscars were. Just before the presentation of the in memoriam montage he quipped: "This year, we lost a giant of cinema. Megaupload."
For some quick blog news, I took a slight dip in page views compared to January, but still managed to pass the 250K mark. With some help from an increased interest in my coverage of the foreign language nominees, and also being selected as Lamb of the Month.
I've also been toying with the idea of adding a ranked blog roll widget to my sidebar. Similar to the LAMB's Leaderboard, to reward the blogs who send me the most referral traffic, except this one will be automatically updated. What do you guys think?
I caught up with quite a few 2011 releases this month, but there are still a bunch more I want to see, hopefully I can get to them in soon. In March, I'll also be continuing with my Czech New Wave series which I'm about half way through; and I've begun to source some films for my next major spotlight, I'm not yet ready to reveal what it will be, but the idea was inspired by this year's Berlinale.
Two young women, both named Marie, revolt against a degenerate, decayed and oppressive society, attacking symbols of wealth and bourgeois culture.
The daily grind for the cops of the Police Department's Juvenile Protection Unit is shaken up with the arrival of a photographer on a Ministry of the Interior assignment to document the unit.
A dramatization of one man's rescue of Jewish refugees in the Nazi-occupied Polish city of Lvov.
Eliezer and Uriel Shkolnik are father and son as well as rival professors in Talmudic Studies. When both men learn that Eliezer will be lauded for his work, their complicated relationship reaches a new peak.
An American writer in Rome is stalked by a serial killer obsessed with his latest book.
A young operata is stalked by a deranged fan bent on killing the people associated with her to claim her for himself.
When an elementary school teacher commits suicide, her class and the entire community is shaken to the core. No one wants to talk about it save her replacement, Bachir Lazhar, an Algerian immigrant seeking political asylum.
A musician is stalked by an unknown killer who's blackmailing him for an accidential killing of another stalker. But is everything what it appears to be?
A newspaper reporter and a retired, blind journalist try to solve a series of killings connected to a pharmaceutical company's experimental, top-secret research project and in so doing, both become targets of the killer.
A writer is stalked by a serial killer after witnessing a murder attempt on one woman's life.
I'm the LAMB of the Month!
Today I'm taking a break from my usual posting schedule to announce something that is quite exciting and a honor for me to be a part of. Just this past weekend, I was selected by Joel Burman—a fellow blogger and film producer who is also the recently appointed Shepherd of the LAMB (Large Association of Movie Blogs)—as the very first recipient of the LAMB of the Month Award, in recognition of some of my recent contributions to this great blogging community.
For those who have never heard of the LAMB, its the largest directory of movie blogs on the internet, open to any movie blog that's at least 3 months old, with approximately 1200 current members—though as I've recently discovered by paying a visit to almost every single one, sadly a number of them are now dead or dormant blogs—first created by Dylan from the blog Man I Love Films. I feel greatly honored for having been singled out of this large group of fine bloggers and chosen as January's LAMB of the Month.
We're now midway through the 62nd Berlin International Film Festival. With many of the first gems of 2012 already or about to be screened at the famed Berlinale Palast.
Let's have a closer look at some of the images and initial critical reception for the films in the 2012 Berlinale Competition Programme.
Joseph is an unemployed widower with a drinking problem, a man crippled by his own volatile temperament and furious anger. Hannah is a Christian charity shop worker, a respectable woman who seems wholesome and happy. When circumstance brings the pair together, Hannah appears as Joseph's guardian angel, tempering his fury and offering him warmth, kindness and acceptance. As their relationship develops, Hannah's own secrets are revealed - her husband is violent and abusive - and Joseph emerges as her unlikely savior.
A look at how the intense relationship between Carl Jung and Sigmund Freud gives birth to psychoanalysis.
Haunted by painful memories and increasing paranoia, a damaged woman struggles to re-assimilate with her family after fleeing an abusive cult.
The story of a terminally ill teenage girl who falls for a boy who likes to attend funerals and their encounters with the ghost of a Japanese kamikaze pilot.
When Emma meets Cassandra, they initiate a relationship filled with physical and psychological challenges. Emma does whatever it takes to master the rules of the game. Lines are crossed and the stakes get higher and higher. Despite this, Emma can't resist the intoxicating feeling of total control.
A look at the final days of an elegant, turn-of-the-century Paris brothel, the lives of the Madam, and the dozen girls who live and work there.
At a village railway station in occupied Czechoslovakia, a bumbling dispatcher's apprentice longs to liberate himself from his virginity. Oblivious to the war and the resistance that surrounds him, this young man embarks on a journey of sexual awakening and self-discovery, encountering a universe of frustration, eroticism, and adventure within his sleepy backwater depot.
Luca seduces girls and then leaves them in the hands of his friends at the Black Sea port of Constanta. But when he meets the beautiful and exuberant Veli, Luca falls in love... It's summer down the Danube river. The music is loud, the cars are fancy, the girls are tanned and Veli runs away from home to Luca's bed. First love has never felt more dangerous.
A farcical look at the Budapest subway system, the crazy ticket agents who earn their living there and the hostile citizens they deal with on a daily basis.
Roger Brown is an elite corporate headhunter who finances his lavish lifestyle by stealing art from his own clients, but trouble ensues when he targets a rare and valuable painting owned by a former mercenary.
In just a few short days, the 62nd Berlin International Film Festival (Internationale Filmfestspiele Berlin) or more commonly referred to as The Berlinale, one of the three major annual film festivals, opens with what promises to be another amazing lineup of films from all over the world.
First held in 1951, the festival has grown to be the largest publicly attended film festival in the world. According to their website there were 485,000 visitors and 300,000 tickets sold at last year's festival.
Films are screened in several categories, including the flagship Main Competition Programme, where this year 18 films will vie for the coveted Golden and Silver Bear prizes awarded by the International Jury. There's also the Panorama section, which focuses on art-house and auteur films from around the globe; the Forum section, covering avant garde, experimental, political minded, and other daring features and documentaries alike; and the Generation programme, which is dedicated to films for children and young people. With two sections titled Generation Kplus for younger audiences and Generation 14plus focusing on teens and above. In addition to those, there are special screenings of classics and important films, retrospectives, and homages to internationally renown film personalities, this year's Homage is dedicated to American actress Meryl Streep, and a special screening of her most recent film The Iron Lady will be held at the famed Berlinale Palast.


text:
Bukola Saraki's investigation is not personal – EFCC boss, Ibrahim Magu
Ibrahim Magu, the Acting Chairman, Economic and Financial Crimes Commission (EFCC), on Tuesday, said the ongoing investigation of the President of the Senate, Bukola Saraki, is not personal. He said this while receiving the award for Outstanding Leadership in Public Service by Euro Knowledge Group.
"No, no, it's not personal. I am not after anybody. Please, I don't want you to personalise this. It is not different from other investigations we are doing," he said.
When asked to comment on the statement by Saraki that he was being witch-hunted, Magu said "never. "You invite people who have committed a crime, people who are corrupt. "If you are not corrupt, you are not here. "We do our homework very well: sometimes we conduct preliminary investigations for up to six months and such persons won't even know we are investigating them," Magu said.
He described corruption as "a national disaster and mother of all evils. "If you remove disaster (corruption), you have a peaceful country." He said banditry, Boko Haram nuisance and other crimes were "caused by corruption and sustained by corruption. "It is been funded by stolen and laundered funds, that's why it is being sustained." The acting chairman, however, said that if corruption was tackled, banditry and kidnappings would be sorted out.
Photos from the CEO Nepas Group and His wife's Chi...
BBN star, Nina flaunts her hot body in new sexy ph...
Selena Gomez, Eva Longoria and other stars dazzle ...
Davido wasn't involved in getting Zlatan out of EF...
We are not planning to overthrow Buhari - Nigerian...
Bukola Saraki's investigation is not personal – EF...
New injury forces Serena Williams to withdraw from...
Here is Bobrisky's reply to Nigerians asking for t...

redpajama_set_name: RedPajamaCommonCrawl


text:
Q: Every time I run my code off of my phone, I get thrown out of the app and back on my computer it pops up with "Thread 1: Breakpoint 2.1" Here is the entire coded
import UIKit

class ViewController: UIViewController {

@IBOutlet weak var logo: UIImageView!

@IBOutlet weak var gameOver: UIImageView!

@IBAction func Play(_ sender: Any) { //What Happens After you click the "Play" button

    tapsValid = true
    ballChange = true
    self.gameOver.isHidden = true
    self.retry.isHidden = true
    self.Ball.isHidden = false
    self.logo.isHidden = true
    self.Play.isHidden = true
    self.scoreBoard.isHidden = true

    self.Ball.center.x = 178.0
    self.Ball.center.y = 390.0
    self.pillar.center = CGPoint(x: 175.0,y: 436.0)
    self.pillar2.center = CGPoint(x: 214.0,y: 407.0)

    timer = Timer.scheduledTimer(timeInterval: 0.045, target: self, selector: #selector(ViewController.movement), userInfo: nil, repeats: true)

    self.pillar3.center = pillarPlacement(x: pillar2.center.x, y: pillar2.center.y)
    self.pillar4.center = pillarPlacement(x: pillar3.center.x, y: pillar3.center.y)
    self.pillar5.center = pillarPlacement(x: pillar4.center.x, y: pillar4.center.y)
    self.pillar6.center = pillarPlacement(x: pillar5.center.x, y: pillar5.center.y)
    self.pillar7.center = pillarPlacement(x: pillar6.center.x, y: pillar6.center.y)
    self.pillar8.center = pillarPlacement(x: pillar7.center.x, y: pillar7.center.y)
    self.pillar9.center = pillarPlacement(x: pillar8.center.x, y: pillar8.center.y)
    self.pillar10.center = pillarPlacement(x: pillar9.center.x, y: pillar9.center.y)

    self.pillar.isHidden = false
    self.pillar2.isHidden = false
    self.pillar3.isHidden = false
    self.pillar4.isHidden = false
    self.pillar5.isHidden = false
    self.pillar6.isHidden = false
    self.pillar7.isHidden = false
    self.pillar8.isHidden = false
    self.pillar9.isHidden = false
    self.pillar10.isHidden = false
    self.pillarTop.isHidden = false
    self.pillarTop2.isHidden = false
    self.pillarTop3.isHidden = false
}
@IBAction func Retry(_ sender: Any) {

}

@IBOutlet var gameView: UIView!
@IBOutlet weak var pillar10: UIImageView!
@IBOutlet weak var pillar9: UIImageView!
@IBOutlet weak var pillar8: UIImageView!
@IBOutlet weak var pillar7: UIImageView!
@IBOutlet weak var pillar6: UIImageView!
@IBOutlet weak var pillar5: UIImageView!
@IBOutlet weak var pillar4: UIImageView!
@IBOutlet weak var pillar3: UIImageView!
@IBOutlet weak var pillar2: UIImageView!
@IBOutlet weak var pillar: UIImageView!
@IBOutlet weak var pillarTop3: UIImageView!
@IBOutlet weak var pillarTop2: UIImageView!
@IBOutlet weak var pillarTop: UIImageView!
@IBOutlet weak var Play: UIButton!
var timer = Timer()
var tapsValid:Bool?
var ballRight:Bool?
var ballChange:Bool?
var ballCenter: CGPoint?
var pillarCenter: CGPoint?
var pillar2Center: CGPoint?
var pillar3Center: CGPoint?
var pillar4Center: CGPoint?
var pillar5Center: CGPoint?
var pillar6Center: CGPoint?
var pillar7Center: CGPoint?
var pillar8Center: CGPoint?
var pillar9Center: CGPoint?
var pillar10Center: CGPoint?
var pillarTopCenter: CGPoint?
var pillarTop2Center: CGPoint?
var pillarTop3Center: CGPoint?

@IBOutlet weak var scoreBoard: UIImageView!
@IBOutlet weak var retry: UIButton!
@IBOutlet weak var Ball: UIImageView!
//Start Screen
override func viewDidLoad() {
    super.viewDidLoad()

    self.gameOver.isHidden = true
    self.retry.isHidden = true
    self.Ball.isHidden = true
    self.logo.isHidden = false
    self.Play.isHidden = false
    self.scoreBoard.isHidden = true
    self.Ball.isHidden = true
    self.pillar.isHidden = true
    self.pillar2.isHidden = true
    self.pillar3.isHidden = true
    self.pillar4.isHidden = true
    self.pillar5.isHidden = true
    self.pillar6.isHidden = true
    self.pillar7.isHidden = true
    self.pillar8.isHidden = true
    self.pillar9.isHidden = true
    self.pillar10.isHidden = true
    self.pillarTop.isHidden = true
    self.pillarTop2.isHidden = true
    self.pillarTop3.isHidden = true

}

override func touchesBegan(_ touches: Set<UITouch>, with event: UIEvent?) {
    if tapsValid == true
    {
        if ballRight == true
        {
            ballChange = false
        } else
        {
            ballChange = true
        }
    }
}

func movement()
{
    ballCenter = self.Ball.center
    pillarCenter = self.pillar.center
    pillar2Center = self.pillar2.center
    pillar3Center = self.pillar3.center
    pillar4Center = self.pillar4.center
    pillar5Center = self.pillar5.center
    pillar6Center = self.pillar6.center
    pillar7Center = self.pillar7.center
    pillar8Center = self.pillar8.center
    pillar9Center = self.pillar9.center
    pillar10Center = self.pillar10.center

    if ballChange == false
    {
        ballRight = false
    } else
    {
        ballRight = true
    }
    if ballRight == true
    {
        Ball.center.x += 6.5
        Ball.center.y -= 0.5
    }else
    {
        Ball.center.x -= 6.5
        Ball.center.y -= 0.5
    }

    Ball.center.y += 0.5
    pillar.center.y += 5.0
    pillar2.center.y += 5.0
    pillar3.center.y += 5.0
    pillar4.center.y += 5.0
    pillar5.center.y += 5.0
    pillar6.center.y += 5.0
    pillar7.center.y += 5.0
    pillar8.center.y += 5.0
    pillar9.center.y += 5.0
    pillar10.center.y += 5.0

    pillar.center = movePillarUp(floatx: pillar.center.x, floaty: pillar.center.y, pillarNumber: 1)
    pillar2.center = movePillarUp(floatx: pillar2.center.x, floaty: pillar2.center.y, pillarNumber: 2)
    pillar3.center = movePillarUp(floatx: pillar3.center.x, floaty: pillar3.center.y, pillarNumber: 3)
    pillar4.center = movePillarUp(floatx: pillar4.center.x, floaty: pillar4.center.y, pillarNumber: 4)
    pillar5.center = movePillarUp(floatx: pillar5.center.x, floaty: pillar5.center.y, pillarNumber: 5)
    pillar6.center = movePillarUp(floatx: pillar6.center.x, floaty: pillar6.center.y, pillarNumber: 6)
    pillar7.center = movePillarUp(floatx: pillar7.center.x, floaty: pillar7.center.y, pillarNumber: 7)
    pillar8.center = movePillarUp(floatx: pillar8.center.x, floaty: pillar8.center.y, pillarNumber: 8)
    pillar9.center = movePillarUp(floatx: pillar9.center.x, floaty: pillar9.center.y, pillarNumber: 9)
    pillar10.center = movePillarUp(floatx: pillar10.center.x, floaty: pillar10.center.y, pillarNumber: 10)

}
func movePillarUp(floatx: CGFloat, floaty: CGFloat, pillarNumber: Int) -> (CGPoint)
{

    var center = CGPoint(x: floatx,y: floaty)
    if checkPillarPosition(y: floaty) == true
    {
        switch pillarNumber
        {
        case 1:
            gameView.sendSubview(toBack: pillar)
            center = pillarPlacement(x: self.pillar10.center.x, y: self.pillar10.center.y)
            break
        case 2:
            gameView.sendSubview(toBack: pillar2)
            center = pillarPlacement(x: self.pillar.center.x, y: self.pillar.center.y)
            break
        case 3:
            gameView.sendSubview(toBack: pillar3)
            center = pillarPlacement(x: self.pillar2.center.x, y: self.pillar2.center.y)
            break
        case 4:
            gameView.sendSubview(toBack: pillar4)
            center = pillarPlacement(x: self.pillar3.center.x, y: self.pillar3.center.y)
            break
        case 5:
            gameView.sendSubview(toBack: pillar5)
            center = pillarPlacement(x: self.pillar4.center.x, y: self.pillar4.center.y)
            break
        case 6:
            gameView.sendSubview(toBack: pillar6)
            center = pillarPlacement(x: self.pillar5.center.x, y: self.pillar5.center.y)
            break
        case 7:
            gameView.sendSubview(toBack: pillar7)
            center = pillarPlacement(x: self.pillar6.center.x, y: self.pillar6.center.y)
            break
        case 8:
            gameView.sendSubview(toBack: pillar8)
            center = pillarPlacement(x: self.pillar7.center.x, y: self.pillar7.center.y)
            break
        case 9:
            gameView.sendSubview(toBack: pillar9)
            center = pillarPlacement(x: self.pillar8.center.x, y: self.pillar8.center.y)
            break
        case 10:
            gameView.sendSubview(toBack: pillar10)
            center = pillarPlacement(x: self.pillar9.center.x, y: self.pillar9.center.y)
            break
        default:
            break

        }

    }
    return(center)
}
func checkPillarPosition(y: CGFloat) -> (Bool)
{
    var low = false
    if y>720
    {
        low = true
    }
    return(low)
}

override func viewDidLayoutSubviews()
{
    if let newBallCenter = ballCenter{
        self.Ball.center = newBallCenter
    }
    if let newPillarCenter = pillarCenter{
        self.pillar.center = newPillarCenter
    }
    if let newPillar2Center = pillar2Center{
        self.pillar2.center = newPillar2Center
    }
    if let newPillar3Center = pillar3Center{
        self.pillar3.center = newPillar3Center
    }
    if let newPillar4Center = pillar4Center{
        self.pillar4.center = newPillar4Center
    }
    if let newPillar5Center = pillar5Center{
        self.pillar5.center = newPillar5Center
    }
    if let newPillar6Center = pillar6Center{
        self.pillar6.center = newPillar6Center
    }
    if let newPillar7Center = pillar7Center{
        self.pillar7.center = newPillar7Center
    }
    if let newPillar8Center = pillar8Center{
        self.pillar8.center = newPillar8Center
    }
    if let newPillar9Center = pillar9Center{
        self.pillar9.center = newPillar9Center
    }
    if let newPillar10Center = pillar10Center{
        self.pillar10.center = newPillar10Center
    }
    if let newPillarTopCenter = pillarTopCenter{
        self.pillarTop.center = newPillarTopCenter
    }
    if let newPillarTop2Center = pillarTop2Center{
        self.pillarTop2.center = newPillarTop2Center
    }
    if let newPillarTop3Center = pillarTop3Center{
        self.pillarTop3.center = newPillarTop3Center
    }
}
// This Function Places the Pillars
func pillarPlacement(x:CGFloat , y:CGFloat ) -> (CGPoint)
{
    var pillarNewX:CGFloat
    var pillarNewY:CGFloat

    var random:Int = Int(arc4random() % 2)

    if random == 1
    {
        pillarNewX = x + 39
        pillarNewY = y - 29

        if pillarNewX  >= 319
        {
            pillarNewX = x-40
            pillarNewY = y-30
        }
    }
    else
    {
        pillarNewX = x - 40
        pillarNewY = y - 30

        if pillarNewX  <= 17
        {
            pillarNewX = x+39
            pillarNewY = y-29
        }
    }
    var newPillarCenter = CGPoint(x: pillarNewX,y:  pillarNewY)
    return(newPillarCenter)

}

override func didReceiveMemoryWarning() {
    super.didReceiveMemoryWarning()
    // Dispose of any resources that can be recreated.
}

}
The problem pops up here
 func movement()
{
    ballCenter = self.Ball.center
    pillarCenter = self.pillar.center
    pillar2Center = self.pillar2.center
    pillar3Center = self.pillar3.center
    pillar4Center = self.pillar4.center
    pillar5Center = self.pillar5.center
    pillar6Center = self.pillar6.center
    pillar7Center = self.pillar7.center
    pillar8Center = self.pillar8.center
    pillar9Center = self.pillar9.center
    pillar10Center = self.pillar10.center

    if ballChange == false
    {
        ballRight = false
    } else
    {
        ballRight = true
    }
    if ballRight == true
    {
        Ball.center.x += 6.5
        Ball.center.y -= 0.5
    }else
    {
        Ball.center.x -= 6.5
        Ball.center.y -= 0.5
    }

    Ball.center.y += 0.5
    pillar.center.y += 5.0
    pillar2.center.y += 5.0
    pillar3.center.y += 5.0
    pillar4.center.y += 5.0
    pillar5.center.y += 5.0
    pillar6.center.y += 5.0
    pillar7.center.y += 5.0
    pillar8.center.y += 5.0
    pillar9.center.y += 5.0
    pillar10.center.y += 5.0

    pillar.center = movePillarUp(floatx: pillar.center.x, floaty: pillar.center.y, pillarNumber: 1)
    pillar2.center = movePillarUp(floatx: pillar2.center.x, floaty: pillar2.center.y, pillarNumber: 2)
    pillar3.center = movePillarUp(floatx: pillar3.center.x, floaty: pillar3.center.y, pillarNumber: 3)
    pillar4.center = movePillarUp(floatx: pillar4.center.x, floaty: pillar4.center.y, pillarNumber: 4)
    pillar5.center = movePillarUp(floatx: pillar5.center.x, floaty: pillar5.center.y, pillarNumber: 5)
    pillar6.center = movePillarUp(floatx: pillar6.center.x, floaty: pillar6.center.y, pillarNumber: 6)
    pillar7.center = movePillarUp(floatx: pillar7.center.x, floaty: pillar7.center.y, pillarNumber: 7)
    pillar8.center = movePillarUp(floatx: pillar8.center.x, floaty: pillar8.center.y, pillarNumber: 8)
    pillar9.center = movePillarUp(floatx: pillar9.center.x, floaty: pillar9.center.y, pillarNumber: 9)
    pillar10.center = movePillarUp(floatx: pillar10.center.x, floaty: pillar10.center.y, pillarNumber: 10)

}

Specifically this line
pillar9.center = movePillarUp(floatx: pillar9.center.x, floaty: pillar9.center.y, pillarNumber: 9)

Any help would be appreciated. Sorry If this question was already asked, but, I could not find it anywhere. I am new to Swift and XCode, so if you could go into some depth of what I am doing wrong, that would be lovely and extremely helpful. Thank you for all the help!

A: Aside from there being a breakpoint set (which should be checked first, of course!) you have a lot of repeated code which can be a great source of errors. Not having your entire project I can't test it all but I've simplified it quite a bit by using an array to hold your pillar UIImageView objects:
import UIKit

class ViewController: UIViewController {

  @IBOutlet weak var logo: UIImageView!

  @IBOutlet weak var gameOver: UIImageView!

  @IBAction func Play(_ sender: Any) { //What Happens After you click the "Play" button

    tapsValid = true
    ballChange = true
    self.gameOver.isHidden = true
    self.retry.isHidden = true
    self.Ball.isHidden = false
    self.logo.isHidden = true
    self.Play.isHidden = true
    self.scoreBoard.isHidden = true

    self.Ball.center.x = 178.0
    self.Ball.center.y = 390.0

    timer = Timer.scheduledTimer(timeInterval: 0.045, target: self, selector: #selector(ViewController.movement), userInfo: nil, repeats: true)

    // verify that we have more than 2 pillars before we use subscripts
    guard pillars.count > 2 else { return }
    self.pillars[0].isHidden = false
    self.pillars[0].center = CGPoint(x: 175.0,y: 436.0)
    self.pillars[1].center = CGPoint(x: 214.0,y: 407.0)
    var last = self.pillars[1]
    for pillar in self.pillars.suffix(from: 2) {
      last.isHidden = false
      pillar.center = pillarPlacement(center: last.center)
      last = pillar
    }
  }

  @IBAction func Retry(_ sender: Any) {
  }

  // outlet collection
  @IBOutlet var pillars: [UIImageView]!

  @IBOutlet var gameView: UIView!
  @IBOutlet weak var pillarTop3: UIImageView!
  @IBOutlet weak var pillarTop2: UIImageView!
  @IBOutlet weak var pillarTop: UIImageView!
  @IBOutlet weak var Play: UIButton!
  var timer = Timer()
  var tapsValid:Bool?
  var ballRight:Bool?
  var ballChange:Bool?
  var ballCenter: CGPoint?
  var pillarTopCenter: CGPoint?
  var pillarTop2Center: CGPoint?
  var pillarTop3Center: CGPoint?

  @IBOutlet weak var scoreBoard: UIImageView!
  @IBOutlet weak var retry: UIButton!
  @IBOutlet weak var Ball: UIImageView!
  //Start Screen
  override func viewDidLoad() {
    super.viewDidLoad()

    self.gameOver.isHidden = true
    self.retry.isHidden = true
    self.Ball.isHidden = true
    self.logo.isHidden = false
    self.Play.isHidden = false
    self.scoreBoard.isHidden = true
    self.Ball.isHidden = true
    self.pillarTop.isHidden = true
    self.pillarTop2.isHidden = true
    self.pillarTop3.isHidden = true

    // set the whole collection at once
    self.pillars.forEach { $0.isHidden = true }

  }

  override func touchesBegan(_ touches: Set<UITouch>, with event: UIEvent?) {
    if tapsValid == true
    {
      if ballRight == true
      {
        ballChange = false
      } else
      {
        ballChange = true
      }
    }
  }

  func movement()
  {
    ballCenter = self.Ball.center

    if ballChange == false
    {
      ballRight = false
    } else
    {
      ballRight = true
    }

    if ballRight == true
    {
      Ball.center.x += 6.5
      Ball.center.y -= 0.5
    } else
    {
      Ball.center.x -= 6.5
      Ball.center.y -= 0.5
    }

    Ball.center.y += 0.5

    // move all the pillars
    pillars.forEach(movePillarUp)
  }

  func movePillarUp(pillar: UIImageView)
  {
    pillar.center.y += 5.0
    if pillar.center.y > 720
    {
      gameView.sendSubview(toBack: pillar)
      pillar.center = pillarPlacement(center: pillar.center)
    }
  }

  override func viewDidLayoutSubviews()
  {
    if let newBallCenter = ballCenter{
      self.Ball.center = newBallCenter
    }
    if let newPillarTopCenter = pillarTopCenter{
      self.pillarTop.center = newPillarTopCenter
    }
    if let newPillarTop2Center = pillarTop2Center{
      self.pillarTop2.center = newPillarTop2Center
    }
    if let newPillarTop3Center = pillarTop3Center{
      self.pillarTop3.center = newPillarTop3Center
    }
  }

  // This Function Places the Pillars
  func pillarPlacement(center: CGPoint) -> CGPoint
  {
    var newCenter = center
    if Int(arc4random() % 2) == 1
    {
      newCenter.x += 39
      newCenter.y -= 29

      if newCenter.x >= 319
      {
        newCenter.x -= 40
        newCenter.y -= 30
      }
    }
    else
    {
      newCenter.x -= 40
      newCenter.y -= 30

      if newCenter.x  <= 17
      {
        newCenter.x += 39
        newCenter.y -= 29
      }
    }
    return newCenter
  }

  override func didReceiveMemoryWarning() {
    super.didReceiveMemoryWarning()
    // Dispose of any resources that can be recreated.
  }
}

You can connect to the pillars outlet collection the same way you do a normal outlet, it will appear under the heading "Outlet Connections". The order in which you add the connections will determine the index of each connection. Now all of your pillars can be under one collection and can be handled at once using loops and other methods. This will reduce code size, decrease the chance of errors, and allow for easier development.
Please test this code and see if it clears up your issue.

redpajama_set_name: RedPajamaStackExchange


text:
\section{Introduction}
\label{introduction}

Biological systems are known to be some of the most complex systems found in Nature. Lately, they have been
receiving great atention from the mathematical and physical sciences~\cite{murray}, that try to give
fundamental explanations to the phenomenology found in the field. One of the most important and changelling
problems in ecology is to understand the mechanisms that lead one population to extinction, interesting from
both the fundamental and applied point of view. Due to the complexity of the problem the most of the
approaches have been phenomelogical, like the use of stochastic differential equations~\cite{goel} or
reaction-diffusion equations~\cite{skellam}. Another consequence of this complexity is the lack of an
axiomatic theory describing this kind of systems. However, it has been noted that there is a deep connection between
ecological and reaction-diffusion processes~\cite{escudero};
a connection that may be exploted to get a more fundamental understanding
of ecology.

The problem under consideration is the so called critical patch size problem, already classic in the
mathematical literature. It was first considered by Skellam in his influencing article of
1951~\cite{skellam}, where he studied a population living in a finite refuge where the conditions are good
for life, but with hostile conditions outside. He showed that there is a critical size of the refuge such
that if the actual refuge is smaller than it the population get extincted. He used a reaction-diffusion equation
approach to that end, that considers a continuum population living in a continuum space. But since the
population is actually discrete and finite, we will use a different approach here, say, a master
equation approach, that considers a discrete population living in a discrete space. This
type of models have been
studied via Monte Carlo simulations and mean-field aproximations~\cite{weron1,weron2}, but here, instead, we will
perform exact analitical calculations. It is worth noting that Skellam's results have been improved recently in a
model that incorporates the internal fluctuations effects due to the discreteness of the population~\cite{escudero} (see also~\cite{doering} for a related problem on extinction).
This model, as well as Skellam's theory, deals with a continuum space and suppose every individual of the population punctual. The present model deals with a lattice, that may be interpreted as a correction to the infinitesimal size of the
individuals, while it represents the lost of the continuum space. This kind of analitical calculations are interesting
also as a guide for numerical simulations that take place, of course, on a lattice.

In this work we will get some insight into the extinction issue via two different models of reacting-diffusing particles:
in the first one the particles will obey the Fermi-Dirac statistics while in the second they will obey the Bose-Einstein
one. The biological interpretation is straightforward: the Fermi-Dirac statistics impose a maximum in the number of particles (individuals) per lattice site, i. e., a strict carrying capacity of the medium. In the case of Bose-Einstein statistics there is no such limit, even when some "crowded" realizations of the stochastic process could be highly unlikely.
It is easy to see that it is absolutely necessary to choose either statistics before studying a reaction-diffusion problem, be it numerical or theoretical.

At first look it seems that the statistics chosen has no relevance in this problem since the usual approach is to
consider the system near extinction~\cite{skellam,escudero,ludwig}. Indeed, the standard procedure is to linearize the
corresponding equation around the null population value, this is, to analize the low occupation number aproximation.
And since we are studying a system that seems to be controlled by its low occupation number state, this suggests that
the requirements on the occupation number (the particle statistics) will have no effect on the possible extinction event.
It is actually a question of theoretical interest if a system of reaction-diffusion bosons behave similarly to fermions in such systems dominated by a low occupation number state (see, for instance, the discussion in~\cite{cardy2}).
We will show that this is not the case. Actually, for the case of fermionic particles,
an arbitrarily well-adapted population will get extincted in finite time due
to a rare fluctuation event. This fact is very interesting also from the point of view of the stochastic
modelling, since rare fluctuations are some of worst understood issues that appear both in theory and
experiments, and whose understanding and control is very important for the pure science and applications
respectively~\cite{mc}. Also, the development of new techniques to study reaction-diffusion fermionic flows is a
very important subject that has proven itself very interesting for both points of view pure and
applied~\cite{mobilia1,mobilia2,mobilia3}.
In the case of bosonic particles
the situation changes completely and the population will be able to survive
even for the minimal refuge size. This implies that the importance of the statistics chosen for modeling the population
is complete.

In order to proof this statements we will use an interesting analogy with quantum mechanics, already common in the study of reaction-diffusion problems~\cite{cardy2,cardy} and that has proved itself useful when applied to biological systems~\cite{escudero,lopez}.

\section{The Fermionic Model}
\label{fermionic}

Our master equation will model a population of random walkers living in a finite lattice, that will be able
to reproduce, but will not neither compite for the nutrients nor die inside the refuge; death is only allowed
outside. This model can be thought as unreal, but we will study it because it overestimates the possibilites
of survival: if the population get extincted in this model, it will get extincted if we considered death
and/or competion. Every site of the lattice will be able to keep at most $N$ individuals, indicating the
finite resources of the medium to maintain a population. If we choose a low value of $N$, like $N=1$, we
are modelling a starving population, but because we can choose arbitrarily high values of $N$, we can model
arbitrarily well adapted populations.

Let us begin with the simplest case: A three sites one-dimensional chain undergoing the following series of
reactions. The first site (top to the left) allows diffusion to the right
\begin{equation}
A + \emptyset \to \emptyset + A
\end{equation}
at rate $D$, reproduction to the right
\begin{equation}
A + \emptyset \to A + A
\end{equation}
at rate $\alpha$, and death
\begin{equation}
A \to \emptyset
\end{equation}
at rate $\gamma$. The third site (top to the right) is symmetric to this one and allows death, reproduction
to the left, and diffusion to the left at the corresponding rates. The middle site allows diffusion and
reproduction to both the left and the right, but it is considered absolutely safe for the population, so no
death reaction is allowed. Additional reactions of death and competion can be added, but they will only
rend the extinction more likely. We will further consider a two-state chain: every site of the lattice
can be either empty or occupied by a single individual.
This system can be described via a master equation of the form:
\begin{equation}
\frac{dP_i(t)}{dt}=\sum_{j=1}^N [W(j \to i)P_j(t)-W(i \to j)P_i(t)],
\end{equation}
where $P_i(t)$ describes the probability of being in the state $\left| i \right>$ at time $t$, and
$W(j \to i)$ is the transition rate from the state $\left| j \right>$ to the state $\left| i \right>$.
It is worth pointing out that the algebraic properties of the master equation guarantee that its solution
consist of a linear combination of terms with a decaying exponential
time-dependence, and so will always show a stable approach to some steady
state~\cite{frensley,oppenheim}.

The
steady states of this equation are given by the condition $\frac{d\bar{P}_i(t)}{dt}=0$, and in our particular
case this condition reduces to the linear algebraic equation:
\begin{equation}
\label{steady}
\bar{\bar{M}} \cdot \vec{V} = \vec{0},
\end{equation}
where $\bar{\bar{M}}$ is the
$8 \times 8$~matrix
      \[ \left( \begin{array}{cccccccc}
     -2\gamma  & \alpha & 2\alpha & \alpha & 0 & 0 & 0 & 0 \\
     \gamma  & -T & D & 0 & \alpha & \alpha & 0 & 0 \\
     0 & D & -2T & D & 0 & 0 & 0 & 0 \\
     \gamma & 0 & D & -T & 0 & \alpha & \alpha & 0 \\
     0 & 0 & \gamma & 0 & -T & D & 0 & 0 \\
     0 & \gamma & 0 & \gamma & D & -2\alpha-2D & D & 0 \\
     0 & 0 & \gamma & 0 & 0 & D & -T & 0 \\
     0 & 0 & 0 & 0 & \gamma & 0 & \gamma & 0 \end{array} \right)\]
where $T=\alpha + D + \gamma$ and $\vec{V}$ is the vector
\begin{equation}
\left( \left| 111 \right>, \left| 110 \right>, \left| 101 \right>, \left| 011 \right>, \left| 100 \right>,
\left| 010 \right>, \left| 001 \right>, \left| 000 \right> \right)^t,
\end{equation}
that must fullfil the normalization condition
\begin{equation}
\label{normalization}
\left| 111 \right> + \left| 110 \right> + \left| 101 \right> + \left| 011 \right> + \left| 100 \right> +
\left| 010 \right> + \left| 001 \right> + \left| 000 \right> =1.
\end{equation}
In our notation $1$ stands for an occupied site and $0$ for an empty site, and $\left| ijk \right>$ for the
probability of being in such a state. The left hand side of Eq.(\ref{steady}) can be thought as a linear map;
this way our problem reduces to calculate its kernel. It is easy to see, by performing the matricial product
and solving the corresponding linear system, that, provided $\alpha>0$, $D>0$, and $\gamma>0$, the kernel
is generated by the linear span of the vector:
\begin{equation}
\left( 0,0,0,0,0,0,0,1 \right)^t.
\end{equation}
This is also the only element of the kernel that satifies the normalization condition
Eq.(\ref{normalization}), which means that it is the unique steady state
solution of the master equation. Physically,
this means that, no matter with which initial condition are we starting, the system will be in the state
$\left| 000 \right>$ in the infinite time limit. That is, \emph{extinction is certain} for long times.

This result is not surprising in view of the continuum space calculations of critical patch sizes, where one
can show that a patch shorter than the critical one leads to certain extinction, for a critical patch size
strictly greater than zero~\cite{skellam}.
In this discrete version of the problem, we have chosen the minimum possible
size of the refuge, so this result could be expected {\it a priori}. However, we would like to see what
happens in situations with different number of sites. This way we will see if it is possible to define
a critical patch number, that is, to determine what is the minimum number of sites being part of a refuge
that rends it effective to prevent an extinction in the infinite time limit.

Let us consider a one-dimensional two-state
finite lattice with $L+2$ sites. The first site (top to the left) allows
diffusion to the right, reproduction to the right and death at the rates defined below. The $(L+2)-th$ site
(top to the right) is defined as symmetric to the first one. The $L$ central sites allow reproduction and
diffusion both to the left and the right, but not death. Note that allowing only two sites to be out of the
refuge rends more difficult the extinction than if we consider more, since the individuals outside can only
diffuse inside. It is interesting to compare this with the continuum case, where it is considered an
infinite space out of the refuge at both boundaries.

In this case we can construct a master equation like in
the former one, and write the corresponding steady state condition generalizing Eq.(\ref{steady}). This will
be again a linear map given this time by a $2^{(L+2)} \times 2^{(L+2)}$~matrix.
Note that all the elements in the
$2^{(L+2)}-th$ column of this matrix are identically zero, reflecting the fact that
$\left| 0...0 \right>$ is an
absorbing state: once we arrive at this state, we will stay there forever. This is the mathematical
expression of the physical fact that no population can be created from nothing. This means that
$\left| 0...0 \right>$ always solves the master equation, something that should appear as obvious if we
realize that it can be chosen as initial condition, and therefore we will stay there for every $t>0$. For
$L=1$ we have also shown that the vacuum state is the unique steady state solution of the master equation,
implying that extinction is certain. We can establish therefore that the nonuniqueness of steady state
solutions of the master equation is a necessary condition to avoid extinction. This is equivalent to search
for situations in which the dimension of the kernel of the linear map has a dimension greater than or equal to
two. Provided that the determinant of the matrix is always zero (due to the fact that the smallest possible
kernel is one-dimensional), we are looking for situations in which the determinants of the minors of the
matrix are all zero. Fix $L>1$; in this case we have $2^{(L+2)} \times 2^{(L+2)}$ or less
algebraic equations of at most
$[(2^{(L+2)}-1) \times (2^{(L+2)}-1)]-th$ order,
which should be all zero in order to get a two-dimensional kernel. Suppose
that all this equations but one are identically zero, this way we arrive at a situation that is not
closer to extinction than the real one (we will consider the case of all equations identically zero below).
Fix $\alpha, D >0$; now we have a single algebraic equation in the variable $\gamma$ of at most
$[(2^{(L+2)}-1) \times (2^{(L+2)}-1)]-th$ order,
that has at most $(2^{(L+2)}-1) \times (2^{(L+2)}-1)$ solutions, as a consequence of
the fundamental theorem of algebra. Note that $\gamma = 0$ is a solution, since in this case we will arrive
certainly to the state $\left| 1...1 \right>$ in the infinite time limit, provided that the initial condition
is other than the vacuum state (in this case we can consider the null solution as ``unstable''). Suppose that
there are $N <(2^{(L+2)}-1) \times (2^{(L+2)}-1)$ solutions that are real and positive, and denote
$\gamma^*$ the smallest
of these solutions. For $\gamma = 0$ we have survival, as we have shown, and let us start now varying
continuously the value of $\gamma$ from zero to upper values. Because the set of solutions of an algebraic
equation is countable and finite, there is an open interval $(0,\gamma^*)$ of values of $\gamma$ that cause
certain extinction. That is, for values of $\gamma$ greater than zero but infinitely close to it, the system
is driven to extinction. As we vary continuously $\gamma$ we get certain extinction for an infinite number
of values till we arrive to $\gamma^*$, when we get again survival. This means that we get survival for
higher values of the death rate, while extinction is certain for lower values of it, something that is
absurd. We must necessarily conclude that there is not such $\gamma^*$, or what is the same, extinction is
certain for every $\gamma>0$. Note that in the case of more than one equation not identically zero, we can
reproduce the same argument defining $\gamma^*$ as the smallest common root of all the equations different
from zero.

Let us examine now the case in which all the equations are identically zero. Suppose that there is a value of
$L$, say $\bar{L}$, that fullfils this property. In this case, there should be at least another solution different from
the vacuum state, at which we should arrive if we start from the appropiate initial condition, no matter what
the values of the parameters are. Let fix $\gamma, D>0$, and $\alpha = 0$; in this case is easy to see that
we will get certain extinction in the infinite time limit for every initial condition, a contradiction. We
must necessarily conclude that there is not such $\bar{L}$, or what is the same, for every $L$ there is at
least one equation that is not identically zero. \emph{This means that, provided that
$\alpha, \gamma, D>0$, we get certain extinction in the limit $t \to \infty$}.

Generalizations of this problem are straightforward. Suppose that we have an hypercubic $d$-dimensional
lattice with $L$ sites per side. Suppose that we allow $N$ individuals per site, and diffusion and
reproduction to the first neighbours, and, in the case of $N>1$, also on-site reproduction. Suppose further
that we allow the death reaction to take place in just the boundary sites. A similar argument as the
above one will lead us again to the same conclusion: extinction is certain. Even in more complicated cases,
like when the occupation number depends on the lattice site $N=N(i)$, where this dependence may be
deterministic or stochastic (quenched disorder), we get the same result via an equivalent argument. We can
further claim that, because the probability of survival is identically zero in the steady state, all the
realizations of the stochastic process should reach the vacuum state in finite time.

The physical implications of this fact are very important. We can consider arbitrarily well adapted
populations by considering arbitrarily high (but bounded) values of $N$ and appropiate initial conditions.
We can further choose high values of the reproduction rate and low values of the death rate; no matter of
this the population will finally get extincted. This means that the dynamics of this system is dominated
by a rare fluctuation that always appears in finite time and kills the whole population.

\section{The Bosonic Model}
\label{bosonic}

In this section we will analize the same model but with bosonic particles for contrast. The Malthusian population under
consideration will undergo birth, $A \to A + A$, at rate $\sigma$, at every site, death, $A \to \emptyset$,
at rate $\gamma$, only at the boundary sites, and diffusing coupling between first neighbours.
To do the analysis we will need a very
different methodology. The master equation in this case reads
\begin{eqnarray}
\label{master1}
\nonumber
\frac{dP(\{n_i\};t)}{dt}=\sum_i \Big{(} D \sum_{\{e\}}[(n_e+1)P(...,n_i-1,n_e+1,...;t) \\
\nonumber
-n_iP(...,n_i,n_e,...;t)] \\
+\sigma[(n_i-1)P(...,n_i-1,...;t)-n_iP(...,n_i,...;t)] \Big{)},
\end{eqnarray}
where $\{e\}$ denotes the set of first neighbours of the site $i$ and
$i$ is not at the boundary. In the case $i$ is a boundary site the master equation is
\begin{eqnarray}
\label{master2}
\nonumber
\frac{dP(\{n_i\};t)}{dt}= \sum_i \Big{(} D \sum_{\{e\}}[(n_e+1)P(...,n_i-1,n_e+1,...;t) \\
\nonumber
-n_iP(...,n_i,n_e,...;t)]+ \\
\nonumber
\sigma[(n_i-1)P(...,n_i-1,...;t)-n_iP(...,n_i,...;t)]+ \\
\gamma[(n_i+1)P(...,n_i+1,...;t)-n_iP(...,n_i,...;t)] \Big{)}.
\end{eqnarray}
The analytical treatment of these equations comes as follows.
We can map this master equation description of the system into a quantum
field-theoretic problem. This connection was first proposed by Doi~\cite{doi} and
was deeply generalized in subsequent works~\cite{cardy}.
We can write this theory in terms of the second-quantized bosonic operators:
\begin{equation}
[a^\dag_i,a_j]=\delta_{ij}, \qquad [a_i,a_j]=0, \qquad [a^\dag_i,a^\dag_j]=0,
\qquad a_i \left| 0 \right>=0,
\end{equation}
whose effect is to create or to destroy particles at the corresponding lattice site:
\begin{equation}
a^\dag_i \left|...,n_i,...\right>=\left|...,n_i+1,...\right>,
\end{equation}
\begin{equation}
a_i \left|...,n_i,...\right>=n_i\left|...,n_i-1,...\right>,
\end{equation}
where we have defined the states as:
\begin{equation}
\left|\{n_i\}\right>=\prod_i (a_i^\dag)^{n_i} \left|0 \right>.
\end{equation}
Thus we can define the time-dependent state vector as:
\begin{equation}
\label{vector}
\left|\Phi(t) \right>=\sum_{\{n_i\}}P(\{n_i\};t)\left|\{n_i\} \right>,
\end{equation}
and claim that it obeys the imaginary time Schr\"{o}dinger equation
\begin{equation}
\label{schrodinger}
\frac{d}{dt}\left|\Phi(t)\right>=-H \left|\Phi(t) \right>,
\end{equation}
with the hamiltonian
\begin{equation}
\label{hamiltonian}
H=\sum_i \left(-D\sum_{\{e\}}a^\dag_i(a_e-a_i)
+\sigma [1-a^\dag_i]a^\dag_ia_i \right).
\end{equation}
Note that we recover Eq.(\ref{master1}) if we substitute Eq.(\ref{vector}) and Eq.(\ref{hamiltonian}) in
Eq.(\ref{schrodinger}). In the case of the linear chain, this second quantized theory is equivalent to the stochastic differential equation
\begin{equation}
da= D(a_{i+1}+a_{i-1}-2a_i)dt+\sigma a dt + \sqrt{2 \sigma a}dW(t),
\end{equation}
where $dW(t)$ denotes the increments of a Wiener process, and the correct interpretation of the equation is It\^{o}. This connection is explained in~\cite{cardy}, where it is also proved that the first moment of the stochastic process $a(t)$
is the first moment of the population. This implies
\begin{equation}
\frac{d\langle a_i \rangle}{dt}=D(\langle a_{i+1}\rangle +\langle a_{i-1}\rangle -2\langle a_i \rangle)+\sigma \langle a_i \rangle,
\end{equation}
in any site outside the boundary. If we repeat the calculation we get
\begin{equation}
\frac{d \langle a_i \rangle}{dt}=D(\langle a_e\rangle-\langle a_i \rangle)+(\sigma-\gamma) \langle a_i \rangle,
\end{equation}
when $i$ is at the boundary and $e$ denotes the first neighbour of $i$. This allows us to write the exact equation for the mean values of the population in the minimal chain
(L=3):
\begin{equation}
\frac{d\vec{a}}{dt}=\tilde{M} \cdot \vec{a},
\end{equation}
where $\vec{a}=(a_1,a_2,a_3)^t$ and $\tilde{M}$ is the $3 \times 3$~matrix
      \[ \left( \begin{array}{ccc}
     \sigma-\gamma-D & D & 0 \\
     D & \sigma-2D & D \\
     0 & D & \sigma-\gamma-D \end{array} \right)\]
We choose as initial condition $\vec{a}=(0,1,0)$, because is the minimal initial condition that keeps one particle under
the refuge. The solution can be calculated straightforwardly, but since the expresions are cumbersome we will only
report here the solution at the boundary, $a_b=a_1=a_3$,
\begin{eqnarray}
\nonumber
a_b(t)=-D\frac{\mathrm{exp}\left[\frac{1}{2}\left(-3D-\gamma-\sqrt{9D^2-2D\gamma+\gamma^2}+2\sigma\right)t\right]}
{\sqrt{9D^2-2D\gamma+\gamma^2}} \\
+D\frac{\mathrm{exp}\left[\frac{1}{2}\left(-3D-\gamma+\sqrt{9D^2-2D\gamma+\gamma^2}+2\sigma\right)t\right]}
{\sqrt{9D^2-2D\gamma+\gamma^2}}.
\end{eqnarray}
It is easy to see that the solution is increasing in time at some site if and only if it is increasing in time at
every site, thus it is enough to study the solution at one site, for instance, $a_b(t)$. And $a_b(t)$ is increasing
if and only if
\begin{equation}
\sigma \ge \frac{1}{2} \left(3D+\gamma-\sqrt{9D^2-2D\gamma+\gamma^2}\right).
\end{equation}
So we have seen that in the bosonic model the minimal chain allows survival for the appropiate parameter values.

\section{Conclusions}
\label{conclusions}

In this work, we have seen that the statistics of the reacting-diffusing particles modelling one ecological population
has a fundamental importance when we study the issue of extinction. It could be thought {\it a priori} that this would
not be the case since whenever we are close to extinction, due to the low number of surviving particles, restrictions
on the occupation number would not be important. We have shown that this is actually not the case, and that while
fermions are certainly driven to extinction for any refuge size, bosons can survive even in the minimal refuge.

We can also conclude that the risk of extinction can only be defined like a probability of
survival in a finite interval of time in the case of fermions. This definition has already been used in
biology~\cite{weron1,shaffer}, and it rules out the possibility of use those mathematical models
that suppose an absorbing boundary for enough high values of population. We have shown that
this supposition can not be assumed for enough long times, and we have put in a more rigorous footing
the definition first heuristically elucidated by the biologists.

But, in the case of bosons, the mathematical treatment for finite times is not necessary, because the infinite time
limit aproximation is enough to predict and separate surviving populations to those driven to extinction. Finally, as a remark, we would like to underline the methodological character of this work, that tries to clarify the mathematical
properties of the models commonly used to understand ecology, rather than give a realistic picture of the extinction
problem for one concrete population.

Very interesting mathematical problems
are still to be solved, for instance, to find the analytical expression of the time dependent solution of
the master equation in the case of the one-dimensional fermionic chain with $L+2$ sites.
This type of problem is usually
projected onto a spin chain problem expressed via Pauli operators, that turns out to be exactly integrable
in some cases~\cite{alcaraz}.

\section*{Acknowledgments}
This work has been partially supported by the Ministerio de Educaci\'{o}n y Cultura (Spain) through Grant
No. AP2001-2598, and by the Ministerio de Ciencia y Tecnolog\'{\i}a (Spain) through Project No. BFM2001-0291.

redpajama_set_name: RedPajamaArXiv


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Q: How to see or prove $L_p(\mathbb R^{n+1})= t^{-\frac{n+1}{p}}L_p(\mathbb R_+\times S^n ,\frac{dt}{t}d\phi)$ How to see or prove $L_p(\mathbb R^{n+1})= t^{-\frac{n+1}{p}}L_p(\mathbb R_+\times S^n ,\frac{dt}{t}d\phi)$. I know that i need to use change to polar coordinate because in all book says the same, $1<p<\infty$ and $S^n$ is the sphere unitary in $\mathbb R^n$.
I want to star taking one function $u$ in  $t^{-\frac{n+1}{p}}L_p(\mathbb R_+\times S^n ,\frac{dt}{t}d\phi)$ then $u(t,\phi)=t^{-\frac{n+1}{p}}\tilde{u}(t,\phi)$ with $\int_{0}^{\infty}\int_{S^n}|\tilde{u}(t,\phi)|^p\frac{dt}{t}d\phi<\infty$, how can transform this, what change of variable worth?
The idea is calculate
$$\int_{\mathbb R^{n+1}}|t^{-\frac{n+1}{p}}\tilde{u}(t,\phi)|^p dx= \int_{\mathbb R^{n+1}}|t^{-(n+1)}||\tilde{u}(t,\phi)|^p dx$$
i do not how is $dx$ and how transform the variable $(t,\phi)\in \mathbb R_{+} \times S^n$ in one variable $x\in R^{n+1}$, and perhaps apply Holder's inequality.
I was thinking in perhaps exist a stereographic representation  in $S^n$ as some day saw in calculus, with the pole $(0,0,1)$ and project one line between the pole and the points in the sphere unitary in $\mathbb R^3$ etc, please i want one hint, for me is important understand how to star please, thank you!!!


Want to transform your home into a style haven, but your lease is holding you back? There is a solution.
"A bit of creativity and a few adhesive hooks go a long way when it comes to revamping your rental home ," says the stylist for online homeware retailer The Home, Emma Blomfield.
Here she shares her top tips for updating your rental without breaking your lease.
If your landlord hasn't been kind enough to paint your walls a lovely crisp white, try creating a gallery wall of stretched canvases and framed prints to hide an ugly paint colour. 3M hooks are a renter's best friend and you can still achieve the full impact of a gallery frame wall with them so go nuts!
Invest in a large rug to cover rental carpet or scuffed floorboards. The largest rug you can afford will do wonders for your flooring and at the same time will help to zone and define the living space.
Is the kitchen looking a little tired and dated? Head to your local hardware store and purchase some new kitchen cupboard door knobs. Keep the old ones just in case you need to swap them back before moving out!
Use your chest of drawers, top of the buffet or coffee table to tell a visual story in your living room or bedroom. Place some trinkets, knickknacks and books from your travels your coffee table, or use the top of the buffet to rest a large piece of art that can't be drilled into the wall.
Need to add some reflective light into the space but can't hang heavy objects on the wall? Look for a floor length mirror that can lean against the wall or has a stand on the back so you can have it filling the entry way or in your bedroom to check yourself out while getting dressed of a morning.
All decor is available at thehome.com.au.


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Destinations Europe
8 Scandinavian Cruises for Any Time Frame
Terri Mapes
Terri Mapes specializes in giving first-time Scandinavia visitors practical tips for their travels.
Jillian Dara
Fact checked by Jillian Dara
Jillian Dara is a freelance journalist and fact-checker. Her work has appeared in Travel + Leisure, USA Today, Michelin Guides, Hemispheres, DuJour, and Forbes.
TripSavvy's fact-checking
Towering fjords, majestic waterfalls, UNESCO World Heritage Sites, and the Northern Lights dancing above the water at night: It's no wonder why Scandinavia is a popular cruise destination.
There are a number of cruises to choose from, ranging from three hours to three weeks, and while some only sail during summer, many are available throughout the year. Which one you choose will depend on the attractions that are important to you and how much you're willing to spend.
3 Hours: Norwegian Evening Cruise on the Oslo Fjord
Didier Marti/Getty Images
Finding it hard to squeeze a full-length Scandinavian cruise into your visit? You could always spend an evening cruising Norway's Oslo fjord on a traditional wooden sailing ship with Viator's three-hour sunset mini-cruise. You'll see the Akershus Fortress and Oslo Opera House, all for only $54 USD per person.
8 Hours: Icelandic Horse Riding and Whale Watching Cruise from Reykjavik
Michael Snell/Getty Images
Viator also offers a 8-hour tour from Reykjavik, Iceland, which includes a whale watching and horse riding experience through Iceland's open countryside. During the cruise, you may see minke whales, humpbacks and dolphins; your onboard guide will explain more about the diverse marine life.
4 Days: Scandinavian Capitals Cruise
Peter Adams/Getty Images
Explore the capital cities of Copenhagen, Denmark, and Oslo, Norway, on this four-day adventure through Scandinavia. Starting at $265 per adult, the itinerary includes two nights of hotel accommodations with an overnight ferry crossing from Copenhagen to Oslo. Authentic Scandinavia also offers capital tours from five to 18 days, which can include the capital cities of Helsinki, Stockholm, and Tallinn, along with the Russian cities of St. Petersburg and Moscow.
8 Days: Norwegian Fjords Cruise
Marco Wong/Getty Images
Your ship will depart from Copenhagen, Denmark, and make its way down the Norwegian coast, stopping in the scenic cities of Stavanger, Bergen, Alesund, and Geiranger before cruising back to Copenhagen. Experience Norway's Old-World charm, medieval and Art Noveau architecture, and natural beauty on this eight-day trip with Royal Caribbean, from $1898 USD per person.
Continue to 5 of 8 below.
8 Days: Scandinavia & Russia Cruise
Bruno Ehrs/Getty Images
You'll board Royal Caribbean's Jewel of the Seas in Copenhagen, Denmark, then set sail for Stockholm, Tallinn, St. Petersburg, and Helsinki before returning to Copenhagen. Highlights of the eight-day trip (from $1249 USD) include Sweden's medieval Old Town, the 13th-century castle where Estonia's parliament meets, architectural masterpieces in St. Petersburg, and the grand Helsinki Cathedral.
13 Days: In Search of the Northern Lights Cruise
Arctic-Images/Getty Images
If seeing the Northern Lights is on your bucket list, this Viking Ocean Cruises winter excursion to far-north Norway is a wonderful way to do so. This smaller vessel takes you to stops within the Arctic Circle: Bodø, Tromsø, Alta, Narvik and Bergen. Highlights include snowy landscapes, sled dog rides, stunning cathedrals, and of course the mesmerizing Northern Lights. This tour starts in London and ends in Bergen, Norway. Prices begin at $4,999 USD.
19 Days: Jewels of the Baltic & Viking Sagas
EschCollection/Getty Images
This 19-day cruise by Holland America starts in Amsterdam, The Netherlands, with stops in Berlin, Tallinn, St. Petersburg, Helsinki, multiple ports in Norway, and ends in Copenhagen, Denmark. Starting at $3,119 USD, the cruise showcases pristine fjords, Viking history, and an interesting lesson in the region's royalty.
20 Days: Norwegian Fjords & Baltic Beauty
Mark Hannaford/Getty Images
If you have the time, this 20-day Windstar Cruise is the trip of a lifetime. You'll depart from Edinburgh, Scotland, and pass through Bergen, Oslo, Copenhagen, Tallinn, Helsinki, St. Petersburg, and Stockholm, passing through fjords, evergreen forests, waterfalls, glaciers, and more. Highlights of this weeks-long adventure include kayaking through Alesund Archipelago, an overnight stay in St. Petersburg, and Klaipeda's Witches Hill. Prospective cruisers can request a quote online.
Watch Now: 10 New Bucket List Cruise Destinations
The Top 5 Cruise Destinations For Summer
Scandinavian Itinerary: Where to Visit and for How Long
Christmas in Scandinavia: Traditions, Events, and Foods
Scandinavia and the Nordic Region: Planning Your Trip
Nordic vs. Scandinavian: A Complete Guide to the Proper Use of the Terms
A Guide to Airports in Scandinavia
Maps of Northern European Countries Visited by Cruise Ships
The 8 Best Cruises for Solo Travelers of 2023
Top Tips for Planning a European Cruise Vacation
The 17 Best Countries to Visit in Europe
UNESCO World Heritage Sites in Scandinavia
How to Travel from Copenhagen to Oslo by Train, Bus, Car, Plane, and Ferry
The 7 Best Northern Lights Tours Around the World
The 9 Best Iceland Hotels of 2023
The 5 Best Cruises for Adults in 2023


It's Recipe Redux! Let's get out the slow cooker and make a big batch of beans for future meals. Get an easy recipe here.
Posted on July 21, 2017 by Erin Sills and filed under Recipe Redux, Recipe, Recipes and tagged dinner nutrition Nutrition entree.
#thereciperedux A fresh delicious recipe for Mini Rose Cakes.
Posted on June 21, 2017 by Erin Sills and filed under Recipe Redux, Recipes, Recipe and tagged sweets.
A recipe for a Mocktail of Sparkling Cardamom Grapefruit Juice.
Posted on May 22, 2017 by Erin Sills and filed under Recipe Redux, Recipe, Recipes and tagged recipe nutrition Nutrition beverage cardamom.
Don't toss those veggie scraps or chicken bones. Save, freeze and use to make broth instead. Get the details here.
Posted on April 22, 2017 by Erin Sills and filed under Recipe Redux, Recipes, Recipe and tagged Nutrition nutrition dinner entree turmeric.
A super easy dish for under $3 per serving.
Posted on January 22, 2017 by Erin Sills and filed under Recipe Redux, Recipes, Recipe and tagged dinner nutrition Nutrition recipe entree.
Posted on November 22, 2016 by Erin Sills and filed under Recipe Redux, Recipes, Recipe and tagged recipe beverage holiday cardamom.
#thereciperedux A Simple Stromboli Recipe!
Posted on September 28, 2016 by Erin Sills and filed under Recipe Redux, Recipes, Recipe and tagged recipe dinner nutrition Nutrition entree.


Following a ban on the cultivation of GM Maize in Mexico in 2013, the Campeche region of the country has now prohibited the growing of GM Soybeans following a two year court battle.
The Second District Court in Campeche ruled this week in favor of three Mayan communities from the Hopelchén township who had taken on the Ministry of Agriculture, Livestock , Rural Development, Fisheries and Food ( Sagarpa) and the Secretariat of Environment and Natural Resources ( SEMARNAT).
In a landmark ruling, the Court granted in favor of the Pac -chen and Cancabchen communities in Hopelchén and determined that the effects of the ruling it applied to all municipalities affected by the permit for GM soybean cultivation in Campeche. Sagarpa now has an obligation to ensure that no GM Soybeans are planted in the state.
The problem began on June 6, 2012 , when the Sagarpa , with the support of SEMARNAT permitted the release of Monsanto's Roundup-Ready GM Soybeans into the environment. The planting of 253 000 hectares was allowed in seven states of Mexico , including the municipalities of Campeche, Hopelchén , Tenabo , Calkiní , Escárcega , Carmen and Palisade .
In Campeche the Mayan authorities reacted angrily to the government authorization and Pac- Chen, Cancabchen and various associations of Campeche beekeepers decided to go to court to prevent the planting, because it affects the honey industry through GM 'pollution of production', which has resulted in the closure of some of the beekeepers' international markets.
The Mayan communities stated that the planting of GM Soybeans affected the traditional historical practices of the people (beekeeping) and that there was a violation of their right to a healthy environment through the overuse of herbicides and deforestation that GMOs encourage.
After nearly two years of litigation, the Second District Court supported the indigenous communities' claims.
In addition , the Directorate General of Environmental Impact and Risk of SEMARNAT , when giving the approval for the permit, was found to have violated the procedure laid down in its Rules of Procedure , by omitting three binding rulings Conabio , CONANP and INE, which advised against planting of genetically modified soy in marked polygons.
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On Nov. 6, voters will determine whether Michigan takes full advantage of a rare, golden opportunity to secure new investment, build new infrastructure, create jobs and enhance health and the environment for its citizens.
Proposal 3, aka "25 x 25," would increase Michigan's clean energy standard to require that 25 percent of our electricity come from renewable energy sources by 2025.
► First, our coal plants are among the country's oldest and many will soon need to close. Proposal 3 would allow us to replace old coal gradually with the cheapest new source of electricity available — wind energy. The Michigan Public Service Commission determined that onshore wind turbines produce energy at about one half the cost of a new coal plant. And while the costs of fossil fuels are increasing and volatile, "fuel-less" renewable energy technologies have long-term, predictable, fixed costs that protect us against that volatility.
► Second, clean energy technology is a trillion-dollar market with investment that is skyrocketing. Since April 2012, Wells Fargo, Goldman Sachs and Bank of America announced $120 billion of new investment in clean energy. Last year marked the first time investment in renewables exceeded global investment in coal and nuclear plants combined.
►Third, manufacturing clean energy technology hits our sweet spot. No one can compete with our strengths in advanced manufacturing, engineering and materials science, or our manufacturing-related research and development capacity. Michigan recently led all states in clean energy technology patents, fueling future opportunities as Michigan companies commercialize clean energy intellectual property.
If we allow ourselves to fall behind, Michigan risks surrendering market opportunities to other states and countries. Twenty states already have renewable energy standards of 20 percent or more and four have standards ranging from 30 percent to 50 percent. Sixteen other states have either enhanced or accelerated their renewable energy standards because they clearly see the benefits of clean energy development.
The number of states that derive 10 percent or more of their electricity from renewable energy doubled in the last year alone.
Clean energy should be of special concern to the people of northwest Lower Michigan. Our reliance on old energy generation comes with costs. More than half the man-made mercury deposited in our Great Lakes comes from coal-fired plants; bacterial action transforms that mercury into the more-dangerous methyl-mercury — a potent, bioaccumulating neurotoxin.
Perhaps because we have lived with that threat for so long, we don't fully appreciate that the fish advisories covering all our lakes not only concern our health, they limit the full economic benefits we could enjoy from our recreational and commercial fisheries.
With our manufacturing strengths, our tremendous research universities, the training capacity of our community colleges, and our ready and able workforce, we can fully capture the enormous benefits that clean energy brings while protecting the world's most magnificent natural resource endowment — our Great Lakes.
Vote "yes" on Nov. 6.
Stanley "Skip" Pruss is the former director of the Department of Energy, Labor and Economic Growth and Michigan's former Chief Energy Officer. He is currently a principal at 5 Lakes Energy LLC, a clean energy technology consulting firm. He lives in Northport.


Nestling at the bottom of the valley dominated by a rocky ridge, surrounded by vineyards, woodland and pastures, our village of Culles-les-Roches is situated on a fault line. This allows you to see the interesting geological particularities of our region. The very diverse rock formations clearly leave their mark on the hills, the vegetation, the way the land is cultivated or exploited and the style of local architecture.
We invite you to discover the secrets of 'terroir' and to marvel at the beauties of Nature. Not just the ground beneath your feet but also the aspect of the land, the topography, the exposure to the sun, the microclimate, composition of the soil, bed-rock and drainage. The 'terroir' has a vital impact on the quality and variety of wines grown on a particular vineyard.
Georama will enable you to visualise the landscape with a greater depth of understanding. The road of discovery begins here.
There are seven Information Panels along the path which explain and illustrate the geographical particularities that you are looking at. They will enable you to acquire a basic understanding, and greater appreciation of the Côte Chalonnaise. Follow the direction of the arrows on the itinerary over the 6km. You may also access the Panels by car or bicycle.
Walking tour: The circuit is approximately 6km in distance. At an average walking pace you should allow 2h30m to complete. Make sure you have sufficient water, snacks, etc. Also consider suitable protective clothing depending on the weather. Sun cream and insect repellent may also be needed. There is a short-cut which will avoid the Information Panels 3 and 4 and will reduce the distance to 3km. You can access those two Panels by car or some other time.
Visit by car or bike: If you don't have enough time or are unable to visit on foot, there is a parking space for one car near each Information Panel. Please use tarmac roads only. For vehicles transporting people of reduced mobility you may access the IPs via agricultural tracks but this is only at your discretion and responsibility.
Please take every precaution during your walk, especially if there are children under your responsibility. Be prepared for local traffic on the roads.
Do not wander off the designated itinerary in respect for the environment, and leave no evidence of your passage. Toilets are available near the Church (IP 5).