text:
Q: Highlight duplicate values in a multidimensional array I'm outputting a list of purchases, and I want to automatically highlight the presence of duplicate orders.
Here's what the array looks like. The first two orders are duplicate orders place by mistake. You'll notice that the orderid for each is different, while the email and userid remain the same. So the duplication will need to match on email and / or userid, but not on orderid.
array
  0 =>
    array
      'orderid' => string '2009091008261662'
      'email' => string 'john@example.com'
      'userid' => string '53'
array
  1 =>
    array
      'orderid' => string '2009091008261048'
      'email' => string 'john@example.com'
      'userid' => string '53'
array
  2 =>
    array
      'orderid' => string '2009091008262025'
      'email' => string 'fred@example.com'
      'userid' => string '103'
array
  3 =>
    array
      'orderid' => string '2009091008272082'
      'email' => string 'tom@example.com'
      'userid' => string '392'

How can I search for duplicate orders from the same person in a given array, in PHP?
I would like to output the above like so:
(pretend its in a table)
2009091008261662 - john@example.com - 53
2009091008261048 - john@example.com - 53
2009091008262025 - fred@example.com - 103
2009091008272082 - tom@example.com - 392
... so basically just highlight the two ( or more ) duplicates.

A: Assumes uniqueness based on userid value
<?php

$orders = array(
  array(
    'orderid' => '2009091008261662',
    'email' => 'john@example.com',
    'userid' => '53'
  ),
  array(
    'orderid' => '2009091008261048',
    'email' => 'john@example.com',
    'userid' => '53'
  ),
  array(
    'orderid' => '2009091008262025',
    'email' => 'fred@example.com',
    'userid' => '103'
  ),
  array(
    'orderid' => '2009091008272082',
    'email' => 'tom@example.com',
    'userid' => '392'
  ),
  array(
    'orderid' => '2009091008265555',
    'email' => 'john@example.com',
    'userid' => '53'
  )
);

$foundIds = array();
foreach ( $orders as $index => $order )
{
  if ( isset( $foundIds[$order['userid']] ) )
  {
    $orders[$index]['is_dupe'] = true;
    $orders[$foundIds[$order['userid']]]['is_dupe'] = true;
  } else {
    $orders[$index]['is_dupe'] = false;
  }
  $foundIds[$order['userid']] = $index;
}
?>

orderid	email
  <?php foreach ( $orders as $order ) { ?>
  <tr class="<?php echo $order['is_dupe'] ? 'dupe' : '' ?>">
    <?php echo $order['orderid']; ?>
    <?php echo $order['email']; ?>
    <?php echo $order['userid']; ?>

  <?php } ?>

A: Your best bet would be essentially "invert" the array into an associative one mapping values to keys from the original array:
$emails = array();
$userids = array();

foreach($inputarray as $key => $item) {
    if( isset($emails[$item['email']]) || isset($userids[$item['userid']]) ) {
        // This item has a duplicate email or userid as something already looked at!
        // $emails[$item['email']] or $userids[$item['userid']] has the key corresponding to the original location where it was seen.
        // $key has the key corresponding to the duplicate we just found.
    } else {
        $emails[$item['email']] = $key;
        $userids[$item['userid']] = $key;
    }
}

A: You could add a hash to the inner-array which represents the the array. Just loop through and compare the hashes.

A: This code works...
$array1[0]['orderid'] = '2009091008261662';
$array1[0]['email'] = 'john@example.com';
$array1[0]['userid'] = '53';
$array1[1]['orderid'] = '2009091008261662';
$array1[1]['email'] = 'john@example.com';
$array1[1]['userid'] = '53';
$array1[2]['orderid'] = '2009091008261662';
$array1[2]['email'] = 'john2@example.com';
$array1[2]['userid'] = '53';
$array1[3]['orderid'] = '209091008261662';
$array1[3]['email'] = 'joh3@example.com';
$array1[3]['userid'] = '53';
$array1[4]['orderid'] = '2001008261662';
$array1[4]['email'] = 'john@example.com';
$array1[4]['userid'] = '53';
$array1[5]['orderid'] = '20013344008261662';
$array1[5]['email'] = 'johnddd@example.com';
$array1[5]['userid'] = '53';
$array1[6]['orderid'] = '200133352008261662';
$array1[6]['email'] = 'johsdfgsdn@example.com';
$array1[6]['userid'] = '53';

$unique_array = array(); // Filtered array with no dupes
$email_array = array(); // Hash list
$order_array = array(); // Hash list
foreach($array1 as $i => $row) {

 if (array_key_exists($row['email'], $email_array)) {
  // This is a dupe based on email
  $array1[$i]['duplicate'] = 1;
  $array1[$email_array[$row['email']]]['duplicate'] = 1;
 }

 if (array_key_exists($row['orderid'], $order_array)) {
  // This is a dupe based on email
  $array1[$i]['duplicate'] = 1;
  $array1[$order_array[$row['orderid']]]['duplicate'] = 1;
 }
 $order_array[$row['orderid']] = $i;
 $email_array[$row['email']] = $i;
}
foreach($array1 as $i => $row) {
 if (!empty($row['duplicate'])) {
  echo "" . $row['orderid'] . $row['email'] . "\n";
  unset($row['duplicate']); // reset the array to original form
 } else {
  echo $row['orderid'] . $row['email'] . "\n";
 }
}

A: You'll need two passes of the orders array.  But it's really more simple than some have made it out to be:
$duplicateUserId = array();

// Mark user ID's with more than one order
foreach ( $orders as $order ) {
    $duplicateUserId[$order['userid']] = isset($duplicateUserId[$order['userid']]);
}

// Output each order
foreach ( $orders as $order ) {
    echo formatOrder($order, $duplicateUserId[$order['userid']]);
}

// Format the output of each order
function formatOrder($order, $isDuplicated) {
    // yadda yadda yadda
}

Assuming that $orders looks like
$orders = array(
  array(
    'orderid' => '2009091008261662',
    'email' => 'john@example.com',
    'userid' => '53'
  ),
  array(
    'orderid' => '2009091008261048',
    'email' => 'john@example.com',
    'userid' => '53'
  ),
  array(
    'orderid' => '2009091008262025',
    'email' => 'fred@example.com',
    'userid' => '103'
  ),
  array(
    'orderid' => '2009091008272082',
    'email' => 'tom@example.com',
    'userid' => '392'
  ),
  array(
    'orderid' => '2009091008265555',
    'email' => 'john@example.com',
    'userid' => '53'
  )
);

Also, it might be best to only match on userId since, presumably, users can change their emails and emails are unique to a single user.

A: Simple answer:
function hasDuplicate($arr,$email) {
  $count = 0;
  foreach ($arr as $row) {
     if ($row['email'] == $email) {
       $count++;
     }
  }
  return ($count >1);
}

redpajama_set_name: RedPajamaStackExchange


text:
Q: menu should hide when not being hovered over with timeout What I am trying to accomplish is having a menu that will show up when a "button" is clicked (the button and menu are in separate elements so the mouse is not hovering over the menu initially).
The menu should hide itself if it is not hovered over within the duration of the timeout (currently it does this, but only the first time it is clicked).
Also, if the element gets hovered over, I would like hide on mouseout and clear the timer and clicking the button again would reset the timeout (it is not resetting maybe?).
I have tried several incarnations and nothing I have tried has behaved correctly and am looking for advice. This his is what I am currently using:
var mytimer = window.setTimeout(function() {$('.menu-div').slideUp(function() {
        window.clearTimeout(this.mytimer);
})
}, 5000);

$().ready(function(){

    $('#menu-button').click(function () {
        $('.menu-div').slideDown();
        $(mytimer);
    });

    $('.menu-div').mouseenter(function() {
       window.clearTimeout(this.mytimer);
    });

    $('.menu-div').mouseout(function() {
        $('.menu-div').slideUp(200);
    });
});

A: I think you want something like this (just a guess, because you didn't give any HTML)
$(function(){
    var mytimer=0;
    $('#menu-button').click(function () {
        $('.menu-div').slideDown();
        mytimer = setTimeout(function(){
            $('.menu-div').slideUp();
        }, 5000);
    });

    $('.menu-div').mouseenter(function() {
        clearTimeout(mytimer);
    });

    $('.menu-div').mouseleave(function() {
        $('.menu-div').slideUp();
    });
});

DEMO.


text:
Q: iPhone 5 UI screen issues After lot of googling i am asking this question. I have an app, i want to make it compatible with the iPhone5. I added Default-568h@2x to my app which contains nearly 10views. So I changed views like this..
In viewDidLoad:
    if ([[UIScreen mainScreen] bounds].size.height >= 568) {
      self.view.frame = CGRectMake(0, 0, 320, 568);
    //iPhone5
    }
    else {
    }
window.autoresizingMask = UIViewAutoresizingFlexibleHeight | UIViewAutoresizingFlexibleWidth;

I have made this change in my didFinishLaunching with options:
window = [[UIWindow alloc] initWithFrame:[UIScreen mainScreen].bounds];

After doing like this I found Some issues..

*

*The bottom of the screen i.e 88px long is not touchable even the controls placed on it - for this i have set FullScreenAtLaunch checked in attributes of MainWindow.xib, but doesn`t worked

*I have textbox on view, when it is focussed keyboard is coming but on top of the Keyboard  view coming, how to get rid of that.

How to resolve this, any help would be greatly appreciated.
Thanks

A: This because in the line where you create your UIWindow in the App delegate class you will probably have something like this:
self.window = [[UIWindow alloc] initWithFrame:CGRectMake(0,0,320,480);

change it to something like:
self.window = [[UIWindow alloc] initWithFrame:[[UIScreen mainScreen] bounds]];

also where you now do:
if ([[UIScreen mainScreen] bounds].size.height >= 568) {
  self.view.frame = CGRectMake(0, 0, 320, 568);
//iPhone5
}
else {
}

You coud as well do:
self.view.frame = [[UIScreen mainScreen] bounds];

But most of the time this ins't needed since the view of the viewcontroller will take op the space correctly. If you have set the AutoResizeMasks in interface builder correctly you will not gave to set the frame in code.


federer and nadal after the wimbledon 2008 final, maybe the greatest in grand slam history.
yesterday, rafa nadal was eliminated in the quarterfinal of the australian open. he didn't seem to feel well, and the normally physically strong spaniard didn't resist and lost 0x3 sets to his countryman david ferrer. just now, roger federer suffered elimination at the hands of novak djokovic, the world's number 3. 0x3 sets, and this time, djokovic just beat him.
you might think, well, anyone can lose, even a nadal or a federer. why this sensacionalist talk about the end of an era? the answer: because although they do lose from time to time, it just doesn't happen too often. and normally not before a final.
almost six years ago, rafael nadal reached his first grand slam final, at his first attempt in roland garros. since then, there have been 24 grand slam finals; and wait for this: in 23 of them, either nadal or federer was in the final! and, more: 22 of those times, the grand slam winner was either roger federer (12 times) or rafael nadal (10 times). they actually played out 6 of those finals between themselves (nadal won 4), which seems even a low number for this kind of dominance.
exactly three years ago was the one time that neither one nor the other was playing on the second sunday. australian open 2008 had tsonga beat nadal 3×0 in the first semifinal, and djokovic beat federer 3×0 in the second. almost the same as this time around. so it might be just an exception, a "twice in a lifetime" thing. time will certainly tell, but it does seem possible that federer and nadal might start to share some of their glory with novak djokovic, who's been trying to enter their league for a while now.
there is another sign for that: this will be djokovic's fourth grand slam final in his career; he had two in a row in us open 2007 (lost to federer) and australian open 2008 (beat tsonga in the final); now he lost to nadal in the us open final last year, and might just win his second slam, again in australia. and, who knows, he might try for the grand slam this year, which neither nadal nor federer ever came close to winning. ok, that's a bit too much. but who knows?
by the way, it's now completed two years since the last roger federer x rafa nadal grand slam final, also down under. another sign for the end of an era?
notice that this is a personal list of how i perceived this year's sport events. the most intense sport moments of 2010, in chronological order.
anastasiya kuzmina from slovakia had won the first gold medal of the olympic games in female biathlon, just two seconds ahead of german star magdalena neuner in her first olympic games. but right in the beginning of the pursuit race, neuner passes her rival and increases her advantage after two error-free shootings to an incredible advantage. in the fourth and last shooting, neuner makes her second mistake and kuzmina has a perfect shooting, but it is not enough -12 seconds advantage at the end give the young bavarian her first ever olympic gold medal.
in the 12.5 km mass start, only 13 of the 30 starters make a mistake in the first shooting – and neuner is one of them. after her perfect second shooting, she was now in 8th place, still dreaming of a medal, but another mistake in her third shooting made it look tough. thanks to a perfect fourth shooting and her incredible pace on the slopes, magdalena neuner manages the incredible and wins her second gold medal – an even more exciting race than the one five days earlier.
in a cold, windy day, the people from fiore were all there, seemingly. it was the greatest day in football for this italian town, that had had european glory, but never in the main competition, the champions league. to eliminate bayern would be their biggest success ever, and the players knew it; bayern couldn't control the wind or the opponent, and early in the second half, fiorentina were up by 2×0. mark van bommel managed to score a lucky goal at the mark of 60 minutes, but the dream of extra time vanished only minutes later with jovetic's second strike against a desolate defense, 3×1. the italians looked like the certain winners now, but just a minute later, arjen robben silenced the crowd with an incredible masterpiece of a shot. with luck and robben, bayern were through to the quarterfinals.
what a semifinal match between schalke and bayern for the german cup. the kind of 0x0 that lifts the soul, good football and very nerve breaking. but even with the stress in the weeks before, and with only a few minutes to play in extra-time, arjen robben takes the ball, dribbles passed five people and concludes with a perfect shot – the goal of the year, and once again such a decisive one!
it didn't look good at all for the home team. rooney had scored an early goal, and bayern were trying to get the best possible result since then; they dominated the match, but weren't able to convert their advantage into goals; only ribéry was able to score a goal against the favorites and 2008 champs, and that goal only came in the 77th minute. in stoppage time, the 3-minute mark was coming close as mario gomez won a ball against rooney in midfield and somehow got it to the united box, where evra lost orientation for a second and olic scored the most exctiting goal of this life. incredible! 2×1 and munich now had a chance!
schalke against bayern – the decisive match for the title race in the bundesliga, where felix magath's team could've made a great leap towards the title with a simple victory over their pursuers. bayern, though, scored twice in the first half, conceded a goal and got a red card very early in the game. but they were able to play the match so well and united that schalke did not have one goal scoring chance in the second half, running away with the victory and, finally, with the leadership in the table.
lionel messi wrote his name in golden letters that evening. historians might call this the greatest performance ever by a football player, in history. against arsenal fc in the quarterfinals of the uefa champions league, messi's game was near perfection even without scoring a single goal; but he scored four, after arsenal had scored the first goal of the match. 4×1 final score, four messi goals in a perfect night for the argentinian, who marked his name as the best of the world – and applied for the "best of all time" category, although he'll need a world cup victory to state that claim.
what a day. manchester seemed to show that olic's last-second goal wouldn't matter at all. they demonstrated modern, fast and agressive football in the first 45 minutes, scoring three goals even without a good rooney (but with a great nani), and even olic's goal right before the break didn't seem to change that fact. in the second half, bayern were more present, but still found it difficult to score what would be a decisive 2×3. but they did: corner kick ribéry, volley from 20 meters by robben – perfect goal, a third all-decisive goal of the year for the incredible dutch! after that, bayern controlled the game and manchester had no answers. olic's goal from the first leg was worth gold, after all.
the last ten minutes of fc barcelona's semifinal second leg match against internazionale milan were some of the best sporting minutes of the year. inter had held 0x0 for a long time, and they could lose 0x1 and still advance to the final; but after piqué scored a beautiful goal, the nou camp became an inferno for the italian team (playing without a single italian player); attack after attack rolled onto julio cesar's goal, and some luck was involved, mixed with a lot of talent. in stoppage time, barcelona scored for the second time, which would see the holders of the title advancing to the final. but the goal was disallowed, and inter advanced. what a match!
what a match. bayern played almost perfect football in the german cup final against holders werder bremen. beautiful, efficient, dominating, fast. 4×0 was the final score, and the second title was in the sack for bayern; trully a great moment for louis van gaal and his football philosophy.
the uefa champions league final was too important to really be a fantastic game. bayern had the ball possession, inter the wit and experience; diego milito and wesley sneijder made the difference on the positive side, daniel van buyten and martin demichelis on the negative one. 2×0 to inter, in a game that seemed more like a 0x0.
in the most exciting championship race in history, race 7 had everything to be spectacular. mark webber and sebastian vettel led the pack with equal points but more wins for the australian, who had just won monaco. alonso had 75 points, only 3 points behind the duo, and button had 70, with felipe massa at 61. lewis hamilton, the most spectacular driver of the year thus far, wasn't even in the top five. and it was a battle between red bull and mclaren, between webber (pole man) and vettel, between hamilton and button. and between all four, who were racing on similar pace through the circuit during most of the race. vettel an edge faster than the leading team mate, and with hamilton pressing him. so he attemped a pass, encouraged by his mechanic – but webber didn't know of it, closed the door, and the two drivers crashed. vettel was out, webber still managed third, but the race was not over; in sight of the red bull drama, mclaren advised their drivers to "save fuel", in other words, hold positions. hamilton thought he was okay, but button still attacked him and passed with an incredible move. hamilton was determined not to let that happened and answered in specatular fashion. incredible! great race, and hamilton now emerged in third place in the standings – two places ahead of vettel, with mark webber still leading the pack.
a pretty good opening match, but what a goal by tchabalala, to erupt the stadium and the hearts of all african people. a singular moment in this sport year.
finally the world cup gets started in terms of quality. germany plays a perfect match and impresses the whole world with its incredible speed and light way of playing, a new kind of "beautiful game". 4×0 against australia, who didn't lose after that and had a good world cup in 2006. mesut özil, now at real madrid, appears to the world for the first time, even without scoring a goal – that's how well he played.
john isner against nicolas mahut. no one could have guessed that this might have been one of the most exciting moments in tennis history, but it was. round 1 in wimbledon, court 18. the match had started one day earlier, with four sets completed and a fifth set in need; darkness was the reason for the ending of the playing that day. basically what happened is that the american and the frenchman played the whole day and still didn't find a winner between them, due to the rule that you have to outscore your opponent by two games in a grand slam fifth set match, no tie breaks allowed. 59-59 was the astonishing score after that day, and for the second time, darkness impeded that match to be concluded. the match was concluded a day later, with an incredible 70-68 scoreline.
another american victory on the same day, and what a victory: landon donovan's deciding goal in stoppage time was one of the most deserving lucky strikes in this sports year and in any fifa world cup. the us had been denied a couple of decisive goals and would face elimination without the life-saving strike by the american soccer star.
the reigning champions are eliminated in last place of their group, after a gripping second half in the match against slovakia, who qualified for the knockout stage with an incredible 3×2 victory – many beautiful goals, lots of drama and an italian team playing like it should have during the world cup only in the last 10-15 minutes. the typical too litte too late.
asamoah gyan is a world cup star. he had a great group phase, but his decisive and beautiful goal against the us team kept african dreams alive, and it was in extra time after a 1×1 tie during the match. one could feel the whole continent united supporting the black stars.
arguably the best soccer match of the year. the first half was the best half of football i can remember watching, mostly because of the incredible performance of the german team in the first 35 minutes (and their 2×0 lead, more than deserving) and the great english performance after the 1×2, including the disallowed goal by frank lampard, the answer the the wembley-goal from 1966. germany shook, but didn't fall. england's pressure was on in the beginning of the second half, but after one of those incredible and fast moves, germany scored with müller. 3×1, and shortly thereafter, 4×1, again müller. a pitty that i watched that game by myself.
brazil had impressed me and the whole world with their focussed performances in the group phase. it was a different brazil, a more intelligent brazil, but a much less beautiful one. against chile, one could only come to the conclusion that this team could only beat itself, and that it wouldn't. so brazil were favorites against holland, not only by tradition. and the first half was a demonstration of the new brazilian way of playing: very firm in the defensive midfield, combative, efficient and cool. the 1×0 advantage could have easily been a 2×0, holland had not even seen julio cesar in the match. but the brazilians beat themselves; a goalkeeping error by julio cesar, very rare indeed, enabled an own goal by felipe melo later credited to free kicker wesley sneijder. melo saw the red card after the whole team had become nervous, and the game was now gone. the netherlands dominated the second half and won the game, 2×1. brazil was out, and the country didn't even care too much about it. a sign of how the public viewed the "new brazilian football".
what a match between ghana and uruguay, and what dramatic moments at the end. muntari had scored a beautiful goal right before the half time break, but forlan equaliezed in the beginning of the second half. ghana were stronger in he first half, uruguay had a better second half. but ghana was the team pressuring the south americans in extra time, and then it happened. after quite a lot of confusion in the uruguaian box, a defender saves from an shot, and a header is fished out of the goal by uruguay star attacker luiz suarez. red card, penalty, but no decisive 2×1 right before the penalty shootout. asamoah gyan could score it from the spot – but the ball hit the aluminium and went out. drama was not over, more penalties were one the program now. this time gyan scored, but two of his team mates didn't – africa cried, uruguay celebrated – mostly bad-boy suarez.
i'd watched argentina's 3×1 win over mexico and was not impressed. no tactics, only talent. a lot of talent, sure. but enough to beat germany's beautiful and efficient fast-paced way of playing the game? after 2 minutes, shooting-star and world cup golden boot winner thomas müller answered with the first goal. bastian schweinsteiger played the match of his life, eliminating lionel messi from the game and even organizing all attacking moves. he played it like a true quarter-back, and the runners scored another 3 goals. 4×0, and title favorites maradona were out. germany had convinced the world that they can actually play the most attractive football on the planet, not only the most efficient one.
germany couldn't play. spain didn't let them. it was as simple as that. and when germany finally started to have a little chance after after almost 70 minutes, the spaniards found their goal – after a corner, with a header. the deserving winner with perfect timing and somewhat boring, but extremely exact and perfect passing style. efficient enough for he germans not to get into the game at almost any time.
there were only two good things about the final: it was thrilling and the right team won. the football was pretty poor, the netherlands weren't even try to play football most of the time, which is totally contrary to the country's football history and philosophy. they played more german than the germans, and they could've won – robben missed the decisive goal twice. but spain was strong in the beginning and in extra-time, in the rest of the match they were only able o control the game. the goal by iniesta was deserving, the title as well; the best team won!
this has always been my favorite race (and i plan to watch it live next year), and it didn't disappoint; at least not in the spectacle, which had vettel trying to win the race. he was the fastest on the track, once again, and as it was getting dry, he tried to pass jenson button; but he crashed into him, provoking widespread criticism. lewis hamilton won the race and took over the lead of the championship after 13 races, ahead of mark webber. at this point, it seemed that the two of them would decide the title race, but things changed quite a lot after that.
a ryder cup final day on a monday was uncharacteristic, and it is not every time that the decision goes to the last hole. the young american zach johnson kept the us-hopes alive, and their great final day seemed to be able to produce a historical comeback after trailing by three points on sunday. at the end, greame mcdowell kept his cool under enormous pressure: the us open champ of 2010 beat hunter mahan and gave europe another title in this, the arguably greatest competition in golf.
fernando alonso had won the last two races and was back in the title race after a great end-of-summer run. japan was the last chance for sebastian vettel to still run for the title, with hamilton also in front of him and webber leading the standings. and he had a simply superb weekend (just like a year earlier) to claim a fantastic victory in the land of the rising sun. kubai kobayashi gave his home crowd an incredible show and made five passing moves, making it his own personal show and finishing in 7th. vettel had now passed hamilton, but alonso was still ahead of him – and so was leader and team mate webber.
the championship leader had held on to his lead without winning any of the previous four races, but this time, his nerves cost him the top spot. after a mistake, he crashed out and tried to take then-second placed fernando alonso with him, but actually hit nico rosberg. vettel now was the virtual leader, since he completely dominated the race; it would have been his first championship lead of the season, the sixth man to do so. but it was not to be, since his motor failed and alonso went on to win the race and become the runaway championship favorite. with only two races to go, vettel was a race victory (25 points) behind, webber only 11 points – a team order seemed the logical strategy for brazil and abu dhabi. but that also was not to be!
the podium was vettel, webber, alonso. which was curious in this order, since alonso remained on the top of the standings, only needing a second place in the emirates to win his 3rd title. but webber was denied the team order privilege, since his team mate still had slim chances himself. so vettel won an exciting brazil race, and went into the final race having to leave team mate webber and alonso behind him – and not only that!
vettel cries. he's done he impossible, and thanks the team for the opportunity to do so. he wins the race, webber finishes back in 8th (worst race of the year for him), alonso in 7th. schumacher has a frightening accident right at the beginning, ferrari ignores rosberg and petrov going to the box (i'd seen it all along!) during that safety-car phase, and their second mistake is to "mark" webber, who goes early into the pits after hitting one of his wheels. alonso goes to, worried to lose his spot in front of that red bull, and gets stuck behind petrov. vettel wins with luck and superb driving, after a year of some mistakes and lots of bad luck. and he cries. the youngest f1 champions ever, what a career so far!
what a year it had been for rafael nadal. although there was no nadal x federer final in a grand slam, in 2010 (the first time this had happened for quite a few years), nadal had won the last three slams and is going for a "false" grand slam in australia, to make it four in a row, but not in the same year. nadal finally met federer in the atp world tour finals, the inofficial world championships of tennis. and federer taught the spaniard a lesson of who still hasn't given up on the dream of coming back to become number one – and maybe even winning the grand slam itself. after this superb match, 2011 can come in the tennis world!
5- ivica olic scores the winning goal for bayern munich against manchester united in the first leg, seconds from stoppage time.
4- arjen robben scores a spectacular volley to eliminate manchester united in old trafford, making olic's goal count.
3- lukas podolski's 2×0 against england, a fantastic goal as an example of the new german game (great work by thomas müller) and virtually eliminating the until then astonished englishmen, traditional rivals.
2- sebastian vettel passing the finish line in abu dhabi and, half a minute later, realizing he's the world champion, breaking into tears.
1- arne friedrich's goal against argentina, eliminating the gauchos with the 3×0; bastian schweinsteiger's brilliant assist was the reason, and of course the tension breaking after another incredible performance of the best german team that ever played.
a love affair with a country... lena and germany.
… well, not really. but you gotta love that cute little blondie, with those blue eyes seeming to explode of joy and life. actually, germany is currently in love with two lenas. one is still a teenie, from north germany, brunette. the other is in her early twenties, blonde, from bavaria. and i think that the country's love will last longer with the latter.
you must understand that germans love wintersports. every weekend, there are hours of live events on tv, and not on some secondary channel. the two official tv stations ard and zdf can bring up to 16 hours per weekend, live! bobsled, alpine skiing, long skiing, ski jumping, nordic combination, speed skating, curling, the luge – you name it! but one sport has really made it to the center of attention: biathlon, that seemingly strange combination of cross-country skiing and shooting a gun. well, if you look at it, it's not so mysterious where a sport like that might have come from. and it's pretty damn exciting to watch. since germany has had pretty good success, it has become very popular over the last three decades.
and the popularity is still growing. thanks to "gold-lena". it's actually pretty simple: you are a star if you are successful in what you do, and if people are paying attention to what you do. like cristiano ronaldo in the sport of football. but you are loved if people like what you say and don't say in those interviews before and after a competition. and how you say it. this makes lena neuner a beloved star.
her sporting successes make people at least wanting to love her, anyway. magdalena neuner was 19 when she first competed in a biathlon world cup season, which is made out of racing weekends (similar to formula 1) with various competitions. she won her first race before turning 20, and in her first biathlon world championship (a competition only surpassed in importance by the winter olympics), she won three gold medals.
lena in shooting position during this year's olympic games.
germany had a new wunder girl. and, what made it even more appealing, she was short, very cute, always smiling with her face and her eyes, very natural in interviews and simply lovable. simple, from a small winter town, with no intention of becoming a star or a socialite. just the way germans love to like their stars. the last time they loved someone from the sports arena in a similar way must have been with steffi graf, who was arguably less cute!
what would the young woman do in her second season? well, she won the overall world cup (comparable to a f1 world championship after all the races), as the youngest woman to do so in history. and in the world championship (held every year), she won another three golds, upping her tally to six overall.
don't think for a moment that this is an easy feat; in biathlon, it's actually really tough to win races. the shooting is what adds the spice, with different sorts of punishments for errors (normally having to do one extra round for each shooting mistake). To win every weekend is basically impossible because of that, but to be amongst the best constantly is possible and a true sign of quality. lena's third season wasn't as successful as the first two, and she wasn't able to defend her title as the overall world cup champion. so she somewhat had a nice story going on for her before her first winter olympics, held in vancouver earlier this year. would she be able to come back? would she be able to get a few medals, maybe even a gold?
lena neuner would. and she did. two golds and a silver made her one of the big names in the games, and probably the biggest name in german sports, for the year. just to keep the comparison up, neuner's olympíc games are even more significant than vettel's first f1 title. she will probably win the vote for germany's sport woman of the year, after winning that for the first time in 2007; vettel is in the running for the men, but might not win it.
don't get me wrong. germany really likes vettel. he's young, straight-forward, has a good sense of humor and is down-to-earth, and very successful. but there is a slight difference: germany really loves lena neuner. because she has all of that, and she has that smile and those eyes.
neuner just started the defence of her 2010 world cup overall title, her second after 2008. she didn't race in the first weekend, because she had the flew; and germany missed her. now she's back, slowly finding her form. this weekend, she led the german team to win first place in the team event, and germany was reminded once again why they'd fallen in love with her only four winters ago.
this love affair will last.
there are no words to describe roger federer any more.
and to rafa nadal, it's kind of a blessing and a curse to be around at the same time. nadal is one of the best players the sport of tennis has ever seen. and yet, it will be tough for him to achieve what roger has achieved.
the third set was a demonstration of tennis. yet another one.
and roger was so happy to beat rafa. this is a fun rivalry. there is respect, irony and great will to beat each other. in the first number one against number two year ending duel, the number two taught the number one a lesson. just like a more experienced master tells the not-so new kid on the blog: "this is how you do it, friend".
and the world once again was awed.
exciting home victory for europe in the ryder cup!
mcdowall and mahan. it all came down to them. young american fowler had finished an incredible comeback in the 18th hole, halving a seemingly lost matchup against one of the molinaro brothers to give his team mate mahan the chance to win it for the usa. or, in this case, to tie it up. a 14×14 tie means the holders keep the trophy, those are the ancient ryder cup rules.
martin kaymer was one of the rookies. the us open winner won 2,5 points for europe!
but mahan was not up to it, and mcdowall was. europe wins with the slightest of margins, 14.5 x 13.5; this is their 9th win in the last 13 editions, which is a surprising feat, since generally europe is considered to be the underdog, if you only go by world rankings. tiger woods, phil mickelson – they were all there. but they couldn't do enough to beat europe and to retain a title for the first time since 1993, in this bi-annual event.
up to 1977, the ryder cup (first played as soon as 1927) was a competition between americans and brits. due to the second world war, only four editions were cancelled; every odd year besides those years have been ryder cup years in the golf calendar. up to 71, it was strictly inventors britain against pros usa. in 73, 75 and 77, the irish played with he brits, but even they failed to help: the usa only lost three times until 1977, and even when great britain became europe for the first time in 1979, there were still three defeats until he maiden victory in 1985. since then, it has been 9×4 for europe, and only one home match was lost (out of seven), but three away matches were won (out of six). europe's supremacy leads to their 9×7 overall lead since they entered the competition to join the brits in 1979, and it will be a question of honor for the americans to strike back in 2012.
the smart ones will have noticed that this year is not an odd year, but there was still ryder cup action. well, the second exeption to the rule was in 2001, where the ryder cup should've taken place, but didn't, for obvious reasons.
so now it's another 2-year wait until next time. and even for those who don't really like golf – ryder cup action is about as exciting as it gets, and it really is quite a great event and competition. young las vegan ricky fowler was able o shine, as well as old stars like lee westwood for europe.
europe won it with six rookies. and, at the end, it was current major holder macdowell who closed the deal.
timo boll is europe's best player and current world number two.
in january 2003, something quite remarkable happened. not to the great public in the world, or even the world of sports. but a remarkable feat anyway: a young 21-year-old german called timo boll had just risen to the number one position of the world rankings in table tennis. in a sport completely dominated by china, a german had risen to the spot top of the rankings, and whilst doing so, eliminating his chinese opponents in important tournaments.
boll had already won the european championship for the first time, which was quite an achievement since the more veteran players dominated the rankings and the games. currently, he is trying to win his fourth single title in a european championship – he's won this year's gold medals with the team and in the doubles tournament already.
on number 2 of the world and surrounded by five chinese players, timo has held his own in the last years, where he was never worse ranked than 6th since 2001.
timo still wants to win single medals at world championships and/or olympic games, a feat he has yet to achieve.
timo boll has had some injuries typical to table tennis players, mostly back injuries. those and other (emotional) factors have influenced his performance in the bi-anual world championships and in the olympic games, where timo still has no singles medals to showcase. in 2011, he'll try again in a world championship (in rotterdam), and in 2012, he'll go for gold in london. let's see if he can translate his great technique and his status as the unreachable european number one player into some even more significant medals.
he just started today, by winning his fourth singles gold medal in a european championship, beating his landsman bauer in he final.
this blog will have at least one post per day, about different topics. it's goal is to voice opinions and to gather some opinions from readers, to create a discussion around mostly football and formula 1, but also many other sports.


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Assad Will Gas Israel, Start WWIII
Assad has already decided that he will die in Syria. Unlike Gaddafi in Lybia, Assad has behind him the power of a religion, of the Alawite sect. Assad is not a man, he's a religious movement, and religious movements cannot be readily destroyed by force, in fact the very act of persecution often emboldens them.
The Alawites have their very own Samson Option, and that option involves Israel. In the past, Assad promised to use chemical weapons only against foreigners. That was not an idle threat, and Assad has spent two years planning a last ditch attempt at survival. Should the United States lead a kinetic coalition against Assad, then he will clearly see Gaddafi in the mirror.
Though one would presume that the rational thing to do when attacked by one individual is to not start attacks against another, thereby inviting more enemies, this logic does not apply to Israel. However, if Assad manages to successfully invite Israel into the fight, then he will have won the support of Iran even more. Assad needs to gas an Israeli center or two, his survival depends on bringing Israel into the fight. If I were Assad and knew that the US and the UK were bombing my country, I would gas an Israeli center; it's the intelligent thing for a psychopath to do. Assad has already successfully convinced many of his followers that it is his enemies using chemical weapons in order to discredit him.
Assad using chemical weapons recently, and just as UN inspectors arrived, was a calculated act to sow distrust in allegations that he would be so "stupid" as to use chemical weapons just as inspectors arrived, and his followers will not believe it. When Israel finds itself gassed, the Alawites will believe a conspiracy against their religion, and if they notice the Star of David in their land, then the world will plunge into a conflict that will bring about the death of millions.
Half a million Americans died during the Civil War; only 100,000 Syrians have died. Our history should teach us that interference in internal struggles can only bring about more doom than it will solve. Our interference will only hasten a bigger war, one that will destroy the US economy.
Let's recall 2003, and the daily press briefings of Mohammed Saeed al-Sahhaf, better known as Comical Ali, or Baghdad Bob. Comical Ali predicted: "They are trying to say that the Iraqi is easy to capture, in order to deceive the world that it is a picnic. One day, they [will] start facing bitter facts... The simple fact is this: they are foreigners inside a country which has rejected them. Therefore, these foreigners, wherever they go or travel, they will be rained down with bullets from everyone. Attacks by members of the resistance will only go up... The infidels are committing suicide by the hundreds on the gates of Baghdad."
The Atlantic summed up Bob's prediction about the suicide rate perfectly: "In 2003, this was probably Sahhaf's most quoted line. It was so conveniently ludicrous, so patently untrue, that commentators didn't have to do any research or devote any column inches to disproving it. A decade later, nobody jokes about military suicide. The Department of Veteran Affairs recently published the most comprehensive study of veteran suicides ever conducted, revealing that about 22 American vets committed suicide every day in 2010, the most recent year for which data was available. That's up from 18 per day in 2008. So Sahhaf was actually slightly wrong on this one. The United States really loses their servicemen and women to suicide not by the hundreds, but by the thousands. And not at the gates of Baghdad, but at home."
And yes, we plunder in foreign lands, but eventually the war comes home. This war promises to bring about not hundred, not thousands, but millions of Israeli and American lives. The political world as we know it is coming to an end.
By Unknown on Sunday, August 25, 2013

redpajama_set_name: RedPajamaCommonCrawl


Escape to the world of opera at Simonsig Estate for a memorable evening of fine food, wine and arias performed by the all-star South African cast of Madama Butterfly at Cuvée Restaurant on Thursday, 5 September 2013.
For this special evening in the heart of the Stellenbosch Winelands, Head Chef Lucas Carstens has designed a menu with flavours of the orient to reflect Puccini's operatic masterpiece. The overture of seared tuna with daikon salad followed by a green tea refresher will set the scene for the more dramatic Springbok loin served with oriental mushroom duxelle. For his grand finale, Lucas has chosen Mitsumame, a delicious Japanese sundae.
Hailing from various parts of South Africa the young and upcoming Madama Butterfly singers who will be entertaining Cuvée guests, include the award winning tenor, Tshepo Moagi, portraying the role of Pinkerton, mezzo soprano Nonhlanhla Yende as Kate Pinkerton, soprano Arline Jaftha as Butterfly and baritone Mandisinde Mbuyazwe as Sharpless.
Cuvée's Madama Butterfly dinner costs R500 per person and includes a glass of Simonsig wine with each course. Booking is essential and for your safety, let Scooter Angels be your designated driver at special rates to Cuvée guests. For reservations or more information contact Cuvée restaurant on +27.218884932 or send an email to cuvee@simonsig.co.za.
Opera lovers can enjoy the full production of Madama Butterfly produced by Christine Crouse at Artscape from 19 to 27 September 2013 or attend the special gala performance on Saturday, 21 September at 18h00. For more information visit www.capetownopera.co.zaa or book your tickets online at Computicket.


Has adding video to your nonprofit marketing plan seemed daunting? Well you may be worrying a bit too much.
Creating video for your nonprofit has become easier than ever. Forget about the big budget for fancy equipment and software—and start thinking about better SEO results and a higher viewership.
Watch our short In Kind video below as our Video Producer, Cole Jensen, explains more in depth why you need video in your nonprofit marketing plan – and how you can get started.
Video is one of the most powerful tools that a nonprofit can utilize. But video is usually the last thing a nonprofit thinks they can handle. With updating their website, staying relevant on social media, and keeping with the day to day of the nonprofit; jumping into video can seem daunting.
That's why I'm here. I'm going to walk you and your nonprofit through everything that you're going to need to begin using video. Cameras, audio, editing, software, hosting, utilizing video, shooting a video. In short 1 minute videos, we're going to cover it all, and make it extremely understandable.
And we'll start with why you need to start utilizing video.
The fact is, technology has now reached an intersection. Internet speeds now are fast enough to support good quality video without needing long buffering times. While the technology to shoot and edit high quality video has become affordable and uncomplicated.
So now, even nonprofits with the smallest budgets can make a video, and people can watch it without having to wait. Which is extremely important because you have 10 to 15 seconds to capture a viewer's attention.
And now lets throw some numbers at ya.
Comscore says 85% of internet viewers will watch online videos.
Well, I shot this on my phone.
So I hope you now get that adding video to your communication mix isn't out of your reach and is something that will help take your nonprofit to that next level.


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More Colette Cheyne: • Originals •
COLETTE CHEYNE
Colette Cheyne was born in Auckland, New Zealand in 1960. She started drawing things as a child, choosing her "favourite things" as subjects: a little ornament, a leaf or a feather, the Stubbs horses from her birthday cards or wild cats from a calendar. She studied art in high school and, after moving to Sydney, Australia in 1980, Colette attended a local art class where she began to experiment a little with paint and studied life drawing. Pencil remained her preferred medium, and upon her return to New Zealand in 1988, Colette began accepting commissions to draw children and pets. Her ability to portray something more than mere appearance made these sensitive portraits particularly successful. In 1992 Colette began painting in oils and sold the first of her animal paintings. She made the move to paint full time in 2000, and the next year won the prestigious best of show award in the international Wild Cat Art 2001 show for her painting, 'Serval'.
Colette's artwork has been featured in Sporting Classics and Wildlife Art magazines as an up and coming artist to watch. She is represented by galleries in New Zealand and the United States, and her oil paintings hang in private and public collections in New Zealand, the United States, and around the world.
Russell Fink Gallery P. O. Box 250 Lorton, VA 22199
Voice: 703-550-9699 Fax: 703-339-6150
email: info@RussellFinkGallery.com


I set up an appointment with the same attorney that my parents used for my brothers adoption. He is a very well respected adoption attorney and I knew he was who we were going to use.
We could go about our adoption two ways. The first is to have our home study done by social services and have them process our adoption as well. The second is to have social services do our home study and have an attorney handle the rest of the adoption.
I prefer to do the second option, so that is where we are.
At our first meeting our attorney explained the process to us. At birth, our attorney will open our adoption with our local court. He will come to the hospital when our baby is born and went over the form with us that the birth mom will sign. He will also have a custody order which will allow us to take our baby home from the hospital.
My main and all consuming question was: can they change their minds after signing? He explained that while yes, they can change their minds, they are required (in the State of Indiana) to provide proof to a court that it would be in the child's best interest to be returned to the birth parents. He went on to explain that they only have 30 days from the date they each sign to do so, and that in all the years he has been an adoption attorney he has never seen or heard of it happening. That helped my mind to rest easier.
I made the mistake of saying "the baby that she is giving up" while we were talking (the birth mom always says it). He stopped me mid-sentence and asked me to change it to "the baby she is choosing to place into a loving environment" or something similar to that. It takes the negativity out of my statement. Never thought of that.
We will have two in-home visits with social services after the baby is born just to make sure everything is working out well at home. When the baby is four months old, our attorney will prepare the proper paperwork to finalize our adoption. When our baby is six months old, our adoption will be finalized. Our adoption date should be sometime in August.
I have spent most of my working years employed by attorneys, so while I am not an attorney myself, I understand what he told us. This part does not make me nervous at all. It is the unknown that is social services that scares me. Step three, whenever it comes, will be our home study. Get ready for that.


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Telewest Restructures Bonds, But Bank Talks May Drag On
Hugo Dixon
Sept. 16, 2003 12:01 am ET
Telewest finally has agreed the terms of its debt-for-equity swap -- with its bondholders, at any rate. As expected, they will receive 98.5% of the equity of the restructured U.K. cable group in exchange for their £3.5 billion ($5.61 billion or €4.97 billion) of debt. Based on recent bond prices, that values the restructured equity at around £1.3 billion. The existing shareholders would be left with a stake of just £20 million -- a fraction of the £73 million they apparently still think their shares are worth.
Telewest is expected to retain its £2 billion of bank finance, which it is capable of servicing from its cash flow. Adding that to the restructured equity, the cable operator would emerge from its debt swap with an enterprise value of about £3.1 billion. That would place it on a multiple of 6.9 times forecast 2004 earnings before interest, taxes, depreciation and amortization.
This 5% premium to U.K. rival NTL doesn't look terribly generous. After all, Telewest is breaking even while NTL is burning cash. Moreover, Telewest will have the less onerous debt burden of the two once it has completed its restructuring. One might think this makes Telewest's bonds attractive. But the picture is not so straightforward. NTL's illiquid shares are tightly held by its former creditors, some of whom also have positions in Telewest bonds. They may be trying to prop up NTL's market value until Telewest establishes one of its own.
Telewest has cleared the biggest hurdle of its yearlong restructuring. It still needs its banks to sign up to the plan but hasn't said whether they have. The bondholders, moreover, are hardly helping the process. They reportedly are planning to domicile Telewest in the U.S. Their main aim in this would be to get a primary U.S. listing -- hoping the market values the group in line with U.S. peers, which trade at a premium to European rivals.
But there also would be a side effect: Telewest would be able to seek Chapter 11 bankruptcy protection if its bank debt became untenable. This probably won't be a deal-breaker for the banks: Their loans look secure enough. But the banks surely will demand a price for their assent. So the restructuring talks look set to drag on while the banks extract their concessions.
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Meghan Agosta has been a member of the Canadian Women's National Hockey Team since 2004. She played in four Olympic Winter Games, winning gold in Turin 2006, Vancouver 2010, Sochi 2014 and the silver medal in Pyongchang, South Korea 2018. With nine goals and six assists, she was the top scorer at the 2010 Olympic Winter Games and was voted Most Valuable Player and Best Forward as well as being named to the Media All-Star Team. Agosta has also been a part of eight world championship teams, winning gold in 2007 and 2012 to go with six silver medals.
After Sochi 2014, Agosta took a break from hockey to focus on her career in Law Enforcement. She graduated from the B.C. Justice Institute Police Academy, and is now a full-time member of the Vancouver Police Department. She returned to the National Team program in the fall of 2015, competing at the World Championships in 2016 and 2017. While playing for the Montreal Stars of the Canadian Women's Hockey League in 2012-13, Agosta became the first player to win the Angela James Bowl in back-to-back seasons as the CWHL's leading scorer. In 2011-12 she set the CWHL's single season scoring record.
Collegiately, Agosta has played for Mercyhurst University where she was named four times as a finalist for the Patty Kazmaier Award as the top women's NCAA player.
Without a CWHL team in Vancouver, Agosta keeps her skills sharp by practicing with the Valley West Hawks, Midget AAA team, playing for the Vancouver Police Centurions hockey team and playing on the over 30 Devils team. She also works with a skills coach out of UBC.
Dr. Jamie Bechtel is a managing partner at Kito Global and CEO of Kito Impact Foundation.
Jamie is a dynamic, innovative, and highly regarded impact expert with over 25 years of experience including work in over 35 countries. She has a knack for identifying and building non-traditional (unexpected) partnerships across sectors including private, public and local stakeholders. Jamie's ability to develop and pursue promising, novel approaches to complex conservation, development and humanitarian challenges is core to her success.
Prior to joining Kito Jamie was the CEO of New Course, an organization that works with women to find solutions to the interconnected problems of poverty and environmental degradation.
Prior to that, Jamie worked at Conservation International where she led the development of CI's signature marine program. Jamie served as an advisor to the Clinton Global Initiative for six years and was recently honored in Fast Company Magazine as 1 of 60 women heroes. She is an advisor to the United Nation's Framework Convention on Climate Change.
Jamie is a lecturer at University of Washington's School of Law. She has a Ph.D. from Boston University and a law degree from Boston College.
Paul K. Chappell is an international peace educator and serves as the Peace Literacy Director of the Nuclear Age Peace Foundation.
A graduate of the US Military Academy, West Point, Paul served with the US Army in Iraq. He left active duty as a Captain.
He is the author of the seven-book Road to Peace series about ending war, waging peace, the art of living, and our shared humanity. The first six published books in this series are Will War Ever End?, The End of War, Peaceful Revolution, The Art of Waging Peace, The Cosmic Ocean, and Soldiers of Peace.
Lecturing across the United States and internationally, he teaches courses and workshops on Peace Leadership and Peace Literacy.
Mr. Chappell grew up in Alabama, the son of a half-black and half-white father and a Korean mother. His father is a veteran of the Korean and Vietnam wars. Having grown up in a violent household, Paul has forged a new understanding of war and peace, rage and trauma, and vision, purpose, and hope. His website is www.peacefulrevolution.com.
Drew Kessler joined Rotary in 2001 at the age of 20. He has served the North Rockland Rotary club in all levels of leadership and became the youngest president in their 91-year history at the age 25.
He has been honored with the Distinguished Service Award and the Paul Harris Fellowship. Drew is a past District Governor of Rotary District 7210 which is made up of 1650 members in 57 clubs serving the 9 counties of the Hudson Valley. Currently he serves as the District's Rotary Foundation Chair, an Assistant Rotary Coordinator for Zone 32 that encompasses New England and is 1 of 8 members of the Young Past District Governor's Committee that serves the Rotary International President. He lives by the motto "Service above Self".
In 2008 Drew obtained his Real Estate license and became a Realtor to better serve the Realtor community. He was named President to the Rockland County Board of Realtors for the year of 2010. He most recently served as the 2015 President for the 10,000 member Hudson Gateway Association of Realtors which covers Rockland, Westchester, Orange and Putnam counties in New York. As a sign of his stature in the real estate community, Drew regularly appeared on Fox Business News from 2008-2013 for his expertise on the finance industry and the real estate market.
Drew resides in Hopewell Junction, New York with his wife Vicki and their five children Dilynn age 17, Joe age 16, Parker age 13, Jackson age 6 and Brandon age three.
John has been a Rotarian since 1988. He is currently a member of the Rotary Club of Mercer Island, Washington. He served as District Governor for District 5030 in 2010-11 and has been selected to serve as the Rotary International Director for Zones 25/26 for the years 2017-19.
In addition to his work with Rotary, John has completed two successful careers. He served 20 years in the U.S. Navy to retire as a Commander in the Supply Corps. His military skills translated to a 25- year career as a Senior Vice President with Costco Wholesale a $110B company with 180,000 employees worldwide. At Costco, John was responsible for Human Resource and Risk Management. John has been active in both corporate and community boards, including the Seattle Chamber of Commerce, The Advisory Board for the University of Washington Bothell, Mercer Island Presbyterian Church, Rotary First Harvest and NW Reinsurance.
23 years old Barbara Helena Nieviadonski was a Rotary Exchange student in Germany during 2013-2014. She was sponsored by District 4730 in Brazil. Upon returning to Brazil she got involved with Rotaract and Rotex, and the administration of the District Youth Exchange Commission.
While on exchange, Barbara realized how much she wanted to share the experience of being abroad. She wanted to really portray the program as accurately as possible, and demonstrate the life changing experience that an exchange year is. She knew that a Power Point presentation just wouldn't do. So she gathered her experience with drama classes and video editing to put together an unique presentation.
In an interesting and innovative narrative, Barbara tells the story of her exchange, an account of all the events that took place from her departure from Curitiba until her return.


Meet the evidence hunters risking it all to collect and document evidence of war crimes committed during Syria's war.
They film, take photographs and collect physical evidence in the hope that one day the perpetrators will be punished. They have recovered military orders, collected bomb fragments, taken toxic samples, filmed mutilated bodies and archived thousands of testimonies to make sure the trace of evidence does not disappear.
The story of these evidence-hunters is one of a Syria never told: a story of men and women who choose as their weapons of war their mobile phones, their pens and the penal code, to become witnesses for the prosecution against all abuses committed in Syria. In SYRIA – WITNESSES FOR THE PROSECUTION, we follow some of these unknown heroes who risk everything to ensure that one day justice will be served.
Garance Le Caisne is a free journalist and writes for "Le Journal du Dimanche" and "L'Obs". Since 1990 she reports on the Middle East. After several trips through shattered Syria she started the quest for "Caesar". It took her more than six months to find him and to gain his trust. In her book "Codename Ceasar" she writes about the discussions she had with the photographer. They talk about the atrocities that happen under the regime of Assad. With her book Garance Le Caisne became internationally known and received the "Geschwister-Scholl-Preis" in 2016.
Olivier Joulie works as a director and cinematographer for various TV broadcasters.


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On November 9, 2018, PCPCC hosted an Invite-Only Workshop during which primary care thought leaders from across the nation came together to discuss strategies to increase investment in primary care to improve healthcare quality and affordability.


text:
\section{Introduction}
The problem of classifying low-rank modular categories has its roots in the classification problem for rational conformal field theories going back to the 1980s (predating the definition of modular category \cite{Tur92}).  Currently, the most complete results are in \cite{RSW} where unitary modular categories of rank at most $4$ are classified.  More generally, Ostrik classified ribbon fusion categories of rank at most $3$ \cite{Ostrik2} and fusion categories of rank at most $2$ \cite{Ostrik1}.

By a generalized form of Ocneanu rigidity \cite[Prop. 2.31]{ENO} one may classify modular categories of a given rank up to finite ambiguity by classifying their Grothedieck semirings.  In \cite{HR} such an approach yielded a classification (up to Gr.-semirings) of modular categories of rank at most $5$ with the property that some  object is not isomorphic to its dual object.

Wang has conjectured that there are only finitely many inequivalent modular categories of each rank (see \cite[Conjecture 6.1]{RSW}), and the results mentioned above bear this out for rank at most $4$ and for rank at most $5$ in case some object is non-self-dual.  The most general class of modular categories for which Wang's conjecture has been verified is  for \emph{weakly integral} categories, that is, categories $\mathcal{C}$ with $\FPdim(\mathcal{C})\in\mathbb{N}$ \cite[Prop. 8.38]{ENO}.  The proof relies upon a classical result of Landau \cite{Landau} that the diophantine equation:
\begin{equation}\label{sumeq}
1=\sum_{i=1}^{n}\frac{1}{x_{i}}
\end{equation}
has finitely many solutions with $x_i\in\mathbb{N}$.   Such solutions are \emph{Egyptian fraction} representations of $1$ which are of independent interest in combinatorial number theory (see \cite[Section D11]{Guy} and \cite[Seq. A002966]{OEIS}).  Moreover the number of solutions to (\ref{sumeq}) is at least exponential in $n$ since $\frac{1}{x}=\frac{1}{x+1}+\frac{1}{x^2+x}$ so that each (non-constant) $n$-term Egyptian fraction representation of $1$ leads to at least $2$ $(n+1)$-term representation.  In \cite{HR} a computational approach to classifying \emph{integral} modular categories (that is, with $\FPdim(X)\in\mathbb{N}$ for all objects $X$) of  rank $n$ is suggested using two facts:
\begin{enumerate}
 \item \cite[Lemma 1.2]{EG}: $\sqrt{\frac{\FPdim(\mathcal{C})}{\FPdim(X_i)}}\in\mathbb{N}$ for all simple $X_i$
\item \cite{Curtiss}: $\dim(\mathcal{C})\leq u_n$ where $u_n$ is inductively defined by $u_1:=1$ and $u_k:=u_{k-1}(u_{k-1}+1)$.
\end{enumerate}
As $u_n$ is double exponential in $n$, a direct search for solutions by computer quickly becomes infeasible.

We attempt to circumvent this computational obstacle by finding interesting conditions on modular categories that imply integrality, and then use these conditions to simplify the classification problem for these classes of categories.  Specifically, we have:
\begin{thm}
Suppose $\mathcal{C}$ is a modular category such that either:
\begin{enumerate}
 \item[(a)] the twist matrix $T$ satisfies $T^N=I$ for $N\in\{2,3,4,6\}$ or
\item[(b)] the only simple object in $\mathcal{C}$ satisfying $X\cong X^*$ is the unit object $\mathbf{1}$
\end{enumerate}
then $\mathcal{C}$ is integral.
\end{thm}
Statement (a) is proved below in Theorem \ref{twistint} and was inspired by Davydov who posed the question for $N=2$ to the second author.  Statement (b) is \cite[Theorem 2.2]{HR} and follows from a Galois theory argument.

We are interested in classifying categories with one of these two properties (a) or (b).  We obtain a fairly explicit description of modular categories with property (a) in Theorem \ref{N2346}.  Modular categories with property (b) of rank at most $11$ are shown to be \emph{pointed} in Theorem \ref{mnsdthm}, that is, $\FPdim(X_i)=1$ for each simple object $X_i$.  While property (b) may seem to be a rare condition at first glance one can show (see Prop. \ref{equivconds}) that it is equivalent to the condition that $\FPdim(\mathcal{C})$ is odd.

The remainder of the paper is structured as follows: in Section \ref{prelim} we collect together some notation and useful facts about modular categories.  We address the classification problem for modular categories having property (a) or (b) in Sections \ref{lowordertwist} and \ref{MNSD} respectively, and give some perspectives and futher directions in Section \ref{conclusions}.

\subsection*{Acknowlegements}
We benefitted from discussions with K. Rusek, M. Papanikolas, Y. Sommerh\"auser and D. Naidu.  The second author was partially supported by NSA grant H98230-10-1-0215.

\section{Notation}\label{prelim}
A modular category $\mathcal{C}$ is a non-degenerate braided, balanced fusion category (see \cite{BK} or \cite{Rsurvey} for the complete axiomatic definition).  In this section we establish notation and describe some of the algebraic data and relations coming from the axioms of modular categories.

We shall typically adopt the notation and normalizations of \cite{NS} for the data of a modular category.  The fusion coefficients are $N_{i,j}^k:=\dim\Hom(X_i\otimes X_j,X_k)$ where $\mathbf{1}=X_0,\ldots,X_{n-1}$ are the (isomorphism classes of) simple objects the number of which ($n$) is called the \emph{rank} of $\mathcal{C}$.  The diagonal twist matrix $T_{ij}:=\delta_{ij}\theta_i$ has finite order and the $S$-matrix is normalized so that $S_{00}=1$.  We will denote by $d_i$ the dimension of the simple object $X_i$, i.e. $d_i=S_{i0}=\dim(X_i)$.  Defining the \textit{Gauss sums} by $p_+=\sum_i \theta_i (d_i)^2$ and $p_-=\overline{p_+}$ we have $(ST)^3=p_+S^2$, $S^2=p_+p_-C$ where $C_{ij}=\delta_{ij^*}$ is the (involutive) charge conjugation matrix, which commutes with $T$.  In particular $(S,T)$ give rise to a (projective) representation of the modular group $SL(2,\mathbb Z)$.  We define the fusion matrices $(N_i)_{jk}:=N_{ik}^j$ and denote by $\FPdim(X_i)$ the largest eigenvalue of $N_i$, i.e. the \emph{FP-dimension} of $X_i$.  The global dimension of $\mathcal{C}$ is $\dim(\mathcal{C})=\sum_i (d_i)^2$ and the global $FP$-dimension is $\FPdim(\mathcal{C}):=\sum_i \FPdim(X_i)^2$.  If $\FPdim(\mathcal{C})=\dim(\mathcal{C})$ then $\mathcal{C}$ is said to be \emph{pseudo-unitary}.  The \textit{Verlinde formula} relates the fusion coefficients to the $S$-matrix entries:
\begin{equation}\label{verlinde}
 N_{ij}^k=\sum_r\frac{S_{ir}S_{jr}\overline{S_{kr}}}{\dim(\mathcal{C})S_{0r}}.
\end{equation}
A pair of matrices $(S,T)$ satisfying the above relations such that the right-hand-side of (\ref{verlinde}) is a non-negative integer for all $i,j,k$ is called a \emph{modular datum} (\cite{Gan05}).  A modular category $\mathcal{C}$ with corresponding $S$ and $T$ matrices is called a \emph{categorification} of $(S,T)$.

A category $\mathcal{C}$ is called \emph{integral} if $\FPdim(X_i)\in\mathbb{N}$ for all simple objects $X_i$.  All categories encountered in this work will be integral and hence pseudo-unitary so that we may assume that $\FPdim(X_i)=d_i$ and $\FPdim(\mathcal{C})=\dim(\mathcal{C})$ by \cite[Prop. 8.23, 8.24]{ENO}.

The entries of $N_i$, $S$ and $T$ satisfy further relations:
\begin{equation}\label{GS}
 p_+p_-=\dim(\mathcal{C})
\end{equation}
and
\begin{equation}\label{twisteq}
  \theta_i\theta_jS_{ij}=\sum_k N_{i^*j}^k\theta_k d_k.
\end{equation}
Since $\dim$ is a character of the Grothendieck semiring of $\mathcal{C}$ we have:

\begin{equation}\label{chars}
\sum_k N_{ij}^k d_k=d_id_j.
\end{equation}

The (second) \textit{FS-indicator} is defined to be
 $$\nu_k=\frac{1}{\dim(\mathcal{C})}\sum_{i,j}N_{i,j}^kd_id_j\left(\frac{\theta_i}{\theta_j}\right)^2$$
 and satisfies:
\begin{equation}\label{fseq}
 \nu_k=\begin{cases} 0 & X_k\not\cong X_k^*\\ \pm 1 & X_k\cong X_k^*\end{cases}.
\end{equation}

More generally the $n$-th FS-indiactor $\nu_k^{(n)}$ may be defined for simple objects in modular categories in an analogous way (\cite[Theorem 7.5]{NS1}).  The quantity $\FSexp(\mathcal{C})$ is the smallest integer $m>0$ such that $\nu_k^{(m)}=d_k$ for all simple $k$.  For modular categories $\FSexp(\mathcal{C})$ coincides with the order of the $T$ matrix (\cite[Theorem 7.7]{NS1}).

\section{Low-order Twist Matrices}\label{lowordertwist}
In this section we study modular categories with twist matrix of order $2,3,4$ or $6$ of arbitrary rank.  Our first result is:
\begin{theorem}\label{twistint}
Suppose that $T$ is the twist matrix of a modular category $\mathcal{C}$ such that $T^N=I$ for $N\in\{2,3,4,6\}$.  Then $\mathcal{C}$ is integral.
\end{theorem}
\begin{proof} First we observe that by \cite[Prop. 5.7]{NS} the entries of the $S$-matrix for $\mathcal{C}$ must lie in $\mathbb Q\left(\theta_1,\ldots,\theta_{n-1}\right)=\mathbb Q\left(\zeta_N\right)$ where $\zeta_N$ is a primitive $N$th root of unity.  Since $s_{ij}$ are algebraic integers and $\varphi(N)\leq 2$ (Euler's totient $\varphi$), if $s_{ij}\in\mathbb{R}$ then $s_{ij}\in\mathbb Z$.  In particular some column of the $S$-matrix must be an integer multiple of the vector of $FP$-dimensions (since $\mathcal{C}$ is modular) and so $\FPdim(X_i)$ are rational integers.
\end{proof}

We now characterize modular categories with $T^N=I$ for $N \in \{2,3,4,6\}$ in the following:
\begin{theorem}\label{N2346} Suppose $\mathcal{C}$ is a modular category such that $T^N=I$.  Then:
\begin{enumerate}
\item[(a)] If $N=2$ then $\mathcal{C}$ is braided tensor equivalent to a subcategory of ${\rm Rep}(D^\omega G)$ where $G$ is an abelian $2$-group of exponent $2$ and $\dim(\mathcal{C})=2^{2s}$ (in particular, $\mathcal{C}$ is pointed),
\item[(b)] If $N=3$ then $\mathcal{C}$ is braided tensor equivalent to a subcategory of ${\rm Rep}(D^\omega G)$ where $G$ is a $3$-group of exponent $3$, and
    \item[(c)] If $N=4$ then $\mathcal{C}$ is braided tensor equivalent to a subcategory of ${\rm Rep}(D^\omega G)$ where $G$ is a $2$-group of exponent $2$ or $4$.
        \item[(d)] If $N=6$ then $\mathcal{C}$ is solvable and hence weakly integral.
\end{enumerate}
\end{theorem}
\begin{proof}

If  $T^2=I$ then each $X_k$ is self-dual since the entries of the $S$-matrix are real.   Computing the FS-indicator we have:
$$\nu_k=\frac{1}{\dim{\mathcal{C}}}\sum_{i,j} N_{ij}^kd_id_j\left(\frac{\theta_i}{\theta_j}\right)^2=\pm 1$$
for all $k$.  But $\left(\frac{\theta_i}{\theta_j}\right)^2=1$ by assumption and $\sum_iN_{ij}^kd_i=d_kd_j$ (as $N_{ij}^k$ is totally symmetric in the self-dual case) so we may simplify:

$$\nu_k=\frac{1}{\dim{\mathcal{C}}}\sum_{j} d_kd_j^2=d_k.$$
But $\nu_k=\pm 1$ and $d_k>0$ so this implies $d_k=1$ for all simple objects $X_k$.  We must then have $X_k^{\otimes 2}\cong \mathbf{1}$ as well.  Thus $\mathcal{C}$ is pointed and has the same fusion rules as the group $\mathbb Z_2^m$ where $2^m$ is both the rank and global dimension of $\mathcal{C}$.  Now since $p_+\in \mathbb{R}$ we have $p_+=p_-\in\mathbb Z$ so that by (\ref{GS}) we must have $\dim(\mathcal{C})=2^m=(p_+)^2$ and hence $m=2s$.  Since $\mathcal{C}$ is braided tensor equivalent to a subcategory of $Z(\mathcal{C})\cong{\rm Rep}(D^\omega G)$ we must have $G$ an elementary abelian $2$-group.

For (b) and (c) with $N=3$ (resp. $4$) we use \cite[Theorem 8.4]{NS1} to conclude that $\dim(\mathcal{C})=p^t$ where $p=3$ (resp. $2$).  In particular $\mathcal{C}$ is braided and nilpotent and hence group theoretical by \cite[Theorem 6.10]{DGNO1}. Thus we conclude that $\mathcal{C}$ is braided tensor equivalent to a subcategory of $Z(\mathcal{C})\cong{\rm Rep}(D^\omega G)$ where $G$ is a $p$-group and since $\FSexp(\mathcal{C})=\FSexp(Z(\mathcal{C}))=3$ (resp. $4$) by \cite[Corollary 7.8]{NS1} $G$ must have exponent $3$ (resp. $2$ or $4$ as the exponent of $G$ must divide that of ${\rm Rep}(D^\omega G)$).

For (d) with $N=6$ \cite[Theorem 8.4]{NS1} implies that $\dim(\mathcal{C})=2^a3^b$ with $a,b>0$ and so by \cite[Theorem 1.6]{ENO2} $\mathcal{C}$ is solvable, and in particular weakly group-theoretical.
\end{proof}

\begin{remark}\ \\
\begin{itemize}
 \item In principle, Theorem \ref{N2346}(a),(b),(c) can be used to completely classify modular categories with twist matrix of orders $2,3$ and $4$.
\item Theorem \ref{N2346}(d) is sharp in the sense that we cannot conclude $\mathcal{C}$ is group theoretical as \cite[Example 5.3(a)]{GNN} shows that non-group-theoretical modular categories with $T^6=I$ exist.
\item In \cite[Prop. 5.1]{Cuntz} an infinite family of modular data $(S,T)$ depending on an integer $k\geq 0$ is described with $S_{ij}\in\mathbb Z$ for all $i,j$ and $T^2=I$.  Theorem \ref{N2346}(a) shows that this family is not categorifiable for $k\geq 1$.
\end{itemize}
\end{remark}

\section{Integrality and Egyptian Fractions}\label{MNSD}
In this section we classify integral modular categories of rank at most $6$ and maximally non-self-dual modular categories of rank at most $11$.

If $\mathcal{C}$ is a (pseudo-unitary) integral modular category with dimensions of simple objects $d_1,d_2,\ldots,d_{n}=\dim(\mathbf{1})=1$ then by \cite[Lemma 1.2]{EG} the numbers $x_i=\dim(\mathcal{C})/(d_i)^2$ are \emph{integers} and so satisfy eqn. (\ref{sumeq}).  In \cite[Lemma 2.3]{HR} it is shown that after relabelling if necessary so that $x_1\leq x_2\leq\cdots\leq x_n=\dim(\mathcal{C})$ then:
$$k\leq x_k\leq u_k(n-i+1)\quad \mathrm{for all}\quad 1\leq k\leq n,$$
where $u_1=1$, $u_k=u_{k-1}(u_{k-1}+1)$.
These bounds rely upon two classical results in \cite{Landau} and \cite{Curtiss}.  Notice also that $x_{n}/x_i=(d_i)^2$ is a perfect square for each $i$.  The \emph{trivial solution} $x_1=\cdots=x_n=n$ corresponds to a \emph{pointed} modular category of rank $n$ as then $d_i=1$ for all $i$.  Our basic algorithm for finding $x_1\leq\cdots\leq x_n$ satisfying these conditions is as follows:
\begin{alg}\label{algorithm1} \ \\
\begin{enumerate}
\item form the set of pairs $S_2:=\{(j,j\cdot i^2):2\leq j\leq n, 1\leq i\leq \sqrt{u_n/j}\}$.  These are the possible pairs $(x_1,x_n)$.
\item having determined $S_k$ for some $2\leq k\leq n-2$ let $S_{k+1}$ be the set of $(k+1)$-tuples $(j_1,\ldots,j_{k-1},J,j_{k})$ such that
\begin{enumerate}
\item $(j_1,\ldots,j_k)\in S_k$,
\item $\max(j_{k-1},k)\leq J\leq \min(j_k,u_k(n-k+1))$,
\item $\sqrt{j_k/J}\in\mathbb Z$ and
\item $\frac{1}{J}+\sum_{i=1}^k \frac{1}{x_i}\leq 1$.
\end{enumerate}
\item the solution set $S_n$ is the set of $n$-tuples $(j_1,\ldots,j_{n-2},J,j_{n-1})$ where
\begin{enumerate}
\item $(j_1,\ldots,j_{n-1})\in S_{n-1}$,
\item $\max(j_{n-2},n-1)\leq J\leq \min(j_{n-1},2u_{n-1})$,
\item $\sqrt{j_{n-1}/J}\in\mathbb Z$ and
\item $\frac{1}{J}+\sum_{i=1}^{n-1} \frac{1}{x_i}=1$.
\end{enumerate}
\end{enumerate}
\end{alg}

For $n=5$ and $n=6$ the solutions are: $\{(5,5,5,5,5),(2,8,8,8,8)\}$ and $$\{(6,6,6,6,6,6),(3,3,12,12,12,12),(4,4,4,9,9,36)\}$$ respectively.  These computations were carried out in a few hours on a desktop computer using Maple, and we have:
\begin{theorem}
Any integral modular category of rank at most $6$ is pointed.
\end{theorem}
\begin{proof}
For rank at most $5$ this result is known, see: \cite{RSW} and \cite{HR}. For rank $6$ we need only show that no modular category may have simple dimensions $(1,1,1,1,2,2)$ or $(1,2,2,3,3,3)$ corresponding to the non-pointed solutions above.  The first case is eliminated by \cite[Prop. 4.11]{propF}: integral modular categories $\mathcal{C}$ with $\FPdim(\mathcal{C})=2^2 3$ must be pointed.  On the other hand, a fusion category of dimension $36$ must be solvable and contain a non-trivial invertible object by \cite[Prop. 8.2]{ENO2} which eliminates the second solution.
\end{proof}
We remark that there is a non-pointed integral modular category of rank $8$, namely ${\rm Rep}(D(S_3))$, the representation category of the double of the symmetric group $S_3$.  We expect that this is the smallest possible rank for non-pointed integral modular categories.
Unfortunately for $n=7$ one has $u_7\approx 10^{13}$ and attempts to implement Algorithm \ref{algorithm1} quickly overwhelms a computer's memory.
This motivates passing to a restricted class of modular categories that enjoy integrality.

With a view towards improving the upper bound $u_n$ above, suppose that $\mathcal{C}$ is a maximally non-self dual (MNSD) modular category, i.e. each non-trivial simple object $X_i$ satisfies $X_i\not\cong X_i^*$.  Then $\mathcal{C}$ is integral with rank $n=2k+1$ by \cite[Theorem 2.2]{HR}.  Since $d_i=\dim(X_i)=\dim(X_i^*)$ we may label the dimensions of simple objects: $d_1,d_1\ldots,d_k,d_k,d_{k+1}=\dim(\mathbf{1})=1$ so that $\FPdim(\mathcal{C})=1+2\sum_{i=1}^k d_i^2$ is odd.  After reordering the $k+1$ integers $x_i:=\dim(\mathcal{C})/d_i^2$ so that $x_1\leq x_2\leq\cdots\leq x_k\leq x_{k+1}=\dim(\mathcal{C})$ they satisfy the special case of eqn. (\ref{sumeq}):
\begin{equation}\label{newsumeq}
\frac{1}{x_{k+1}}+\sum_{i=1}^{k}\frac{2}{x_k}=1
\end{equation}
The maximal non-self-dual condition may seem somewhat exceptional, but it follows from  \cite[Theorem 2.2]{HR}, \cite[Corollary 8.2(ii)]{NS1} and \cite[Corollary 2.33]{DGNO2} that:
\begin{prop}\label{equivconds} Let $\mathcal{C}$ be a modular category of rank $n$.  The the following are equivalent:
\begin{enumerate}
 \item $\mathcal{C}$ is maximally non-self-dual
\item  $\FPdim(\mathcal{C})$ is odd
\item $\mathcal{C}$ is (monoidally) equivalent to ${\rm Rep}(A)$ where $A$ is an odd-dimensional, semisimple quasi-Hopf algebra.
\end{enumerate}
\end{prop}

The following generalization of \cite[Lemma 2.3]{HR} gives a linear improvement of the upper bounds in Algorithm \ref{algorithm1} for a restricted class of integral modular categories and includes the MNSD setting as the special case $\ell=2$:

\begin{lemma}
\label{new bounds 1}
Suppose that $\mathcal{C}$ is an integral modular category of rank $n$ and there exist $k$ (weakly decreasing) integers
$p_{1}\geq p_{2}\geq \cdots\geq p_{k}$ such that the non-trivial (isomorphisms classes of) simple objects can be partitioned into $k$ sets $P_1,\ldots,P_k$ such that $X\in P_i$ has $\dim(X)=p_i$ and $|P_i|=\ell$ for all $1\leq i\leq k$.
 Then:
\begin{itemize}
\item[(i)] The numbers $x_{i}:=\frac{\dim{\mathcal{C}}}{p_{i}^{2}}$ form a weakly increasing sequence of integers such that $\sum_{i=1}^{k}\frac{\ell}{x_{i}}+\frac{1}{x_{k+1}}=1$, and
\item[(ii)] $\ell i\leq x_{i}\leq\left(n+\ell-i\ell\right)\frac{A_{i}}{\ell}$ for $i\leq k$ and $(k+1)\leq x_{k+1}\leq\frac{A_{k+1}}{\ell}$ where $A_{1}=\ell$ and $A_{i}=A_{i-1}\left(A_{i-1}+1\right)$.
\end{itemize}
\end{lemma}
\begin{proof}
The proof of (i) proceeds exactly as in \cite[Lemma 2.3]{HR} so we focus on (ii).  First define $x_i^\prime=x_i$ for $1\leq i\leq k$ and $x_{k+1}^\prime=x_{k+1}/\ell$ so that:
\begin{equation}\label{sumwithell}
\sum_{i=1}^{k+1}\frac{1}{x_i^\prime}=\frac{1}{\ell}
\end{equation}
and $x_1^\prime\leq\cdots\leq x_{k+1}^\prime=\ell x_{k+1}$.  Observing that $n=k\ell+1$ we finds that (ii) becomes $\ell i\leq x_{i}^\prime\leq\left(n+\ell-i\ell\right)\frac{A_{i}}{\ell}$.
Since the $x_i^\prime$ are weakly increasing the lower bounds are clear: for if $\ell i>x_i^\prime$ for some $i$ then $\ell i> x_i^\prime\geq x_{i-1}^\prime\geq\cdots\geq x_1^\prime$ so $1/\ell=i\frac{1}{\ell i}>\sum_{j=1}^{i}\frac{1}{x_j^\prime}$, a contradiction.  For the upper bounds follow the same strategy as in \cite{HR}: we use Takenouchi's \cite{Ta} bound on the largest denominator $x_{k+1}^\prime$ in eqn. (\ref{sumwithell}) and Landau's \cite{Landau} estimate.  In particular, if
$$\sum_{i=1}^k\frac{1}{y_i}=r$$ with $y_1\leq y_2\leq\cdots\leq y_k$ then Landau's result says $y_i\leq (k-i+1)/r_{i-1}$ where $r_0=r$ and $r_i=r_{i-1}-1/y_i$.  Takenouchi's result says that the maximum denominator $y_i$
of a solution to $\frac{1}{\ell}=\sum_{i=1}^k\frac{1}{y_i}$ is $A_k$ where $A_1=\ell$ and $A_i:=A_{i-1}(A_{i-1}+1)$.  So in particular $1/r_{i-1}\leq A_{i}$ for all $i$ and the result follows.

\end{proof}

Lemma \ref{new bounds 1} provides a linear improvement on the bounds in \cite[Lemma 2.3]{HR}: one has roughly $\mathrm{rank}(\mathcal{C})/\ell$ variables as opposed to $\mathrm{rank}(\mathcal{C})$.
Taking the case $\ell=2$ one can modify Algorithm \ref{algorithm1} to search for possible MNSD modular categories in the obvious way.  As the bounds are still double exponential the best result we can achieve with our (modified) algorithm is:
\begin{theorem}\label{mnsdthm}
All maximally non-self dual modular categories of rank at most $11$ are pointed.
\end{theorem}
\begin{proof}  The only non-trivial solution the modified algorithm produces is in rank $11$, consisting of $9$ simple objects of dimension $1$ and two simple objects of dimension $3$.  To eliminate this possibility we use a result in \cite{GN}: any integral modular category $\mathcal{C}$ is faithfully graded by its universal grading group $U(\mathcal{C})$, which is isomorphic to the group of invertible objects in $\mathcal{C}$.  Since each component of a faithful grading must have the same $FP$-dimension we see that this is impossible as $|U(\mathcal{C})|=9$ and any component $\mathcal{C}_g$ of the grading containing a simple object $X$ with $\FPdim(X)=3$ must have $\FPdim(\mathcal{C}_g)\geq 9$.
\end{proof}

\begin{remark}
 An obvious source of MNSD modular categories are those of the form ${\rm Rep}(D^\omega G)$ where $|G|$ is odd (that is, the representation category of the twisted double of a finite group of odd order).
The smallest rank example of a non-pointed MNSD modular category we are aware of is the rank $25$ category ${\rm Rep}(D(\mathbb Z_3\ltimes\mathbb Z_7))$ of dimension $441$ where $\mathbb Z_3\ltimes\mathbb Z_7$ is a non-abelian semidirect product.  One might speculate that any MNSD modular category of rank at most $23$ is pointed.
\end{remark}

We close with the following related question that we find interesting:
\begin{question}
 Is there an odd dimensional modular category that is non-group-theoretical?
\end{question}

\section{Conclusions and Future Directions}\label{conclusions}
We have made significant progress towards classifiying modular categories with twist matrices of order $2,3,4$ or $6$.  It is not clear how to extend these techniques to other orders.  One might hope that one could generalize \cite[Theorem 8.4]{NS1} to non-integral categories to get some statement about the possible prime divisors of $\FPdim(\mathcal{C})$ in terms of those of the order of the twist matrix.  However, the relevant number field would no longer be $\mathbb Z$ so that other complications would arise particularly concerning units.

For rank greater than $11$ even the improved bounds for MNSD modular categories become too large for our computational techniques.  However, if one considers MNSD modular categories with:
\begin{enumerate}
 \item rank $r$
\item  $\FPdim(X_i)\leq M$, for each simple $X_i$ and
\item $|\{\FPdim(X_i):0\leq i\leq r-1\}|=t$  ($X_i$ simple).
\end{enumerate}
further headway can be made.  For example, taking $r\leq 23$, $M=45$ and $t\leq 6$ we set $k_i=|\{j:\FPdim(X_j)=d_i\}|$ with $d_0=1$ we obtain (using \cite[Lemma 1.2]{EG} and \cite[Prop. 8.15]{ENO}):
\begin{eqnarray*}
 &&k_0\equiv 1\mod{2}\\
&&k_i\equiv 0\mod{2}, (i\neq 0)\\
&&\sum_{i=0}^{t-1} k_i=r\\
&&\sum_{j\neq i} k_j(d_j)^2\equiv0\mod{(d_i)^2}\\
&&\sum_{j\neq 0} k_j(d_j)^2\equiv0 \mod{k_0}
\end{eqnarray*}
For any fixed $r\leq 23$ and subsets of $\{d: 3\leq d\leq 45,\/ d\in 2\mathbb Z+1\}$  of size at most $5$
we convert these congruences to a system of linear diophantine equations and then use the Smith normal form of the corresponding matrix to solve for the $k_i$.  Applying
 further classification theorems and \emph{ad hoc} techniques produces the following partial result:
\begin{theorem}
If $\mathcal{C}$ is a non-pointed maximally non-self dual modular category with $13\leq\text{rank}\left(\mathcal{C}\right)\leq23$ then either $\mathcal{C}$ has:
\begin{enumerate}
\item[(a)] $\FPdim(X_i)\geq 47$ for some simple $X_i$ or
\item[(b)]  $|\{\FPdim(X_i):0\leq i\leq m-1\}|\geq 7$
\end{enumerate}
\end{theorem}
This is further evidence that perhaps a non-pointed MNSD modular category of smallest rank is indeed the category of rank $25$ and dimension $441$ described above.

More generally, one can also obtain partial classification results for integral modular categories of small rank by bounding the $FP$-dimension further (below the double-exponential bound of $u_n$ in Algorithm \ref{algorithm1}).  For example, using a streamlined version of Algorithm \ref{algorithm1} provided by K. Rusek and some classification theorems we can prove:
\begin{theorem}
 If a non-pointed modular category $\mathcal{C}$ has rank $7$ then $\FPdim(\mathcal{C})>10^{5}$.
\end{theorem}

\bibliographystyle{amsxport}

redpajama_set_name: RedPajamaArXiv


text:
My Life Since the 2012 Sandy Hook Shooting: Sarah Clements's Story
"People always say, 'Nothing changed after Sandy Hook.' And that angers me, because I actually don't think this moment would be happening without the change that happened after Sandy Hook."
Shoes placed in front of the U.S. Capitol by activists last week, to protest those killed since Sandy HookEric Thayer / Reuters
Editor's Note: This is one in a series of conversations with those who have survived high-profile shootings or lost loved ones to them. The other interviews, as well as background about the series, can be found here.
On December 14, 2012, a gunman entered Sandy Hook Elementary and turned left, toward the main office and first-grade classrooms; had he turned right, he almost immediately would have encountered Sarah Clements's mother teaching in her second-grade classroom. This saved her mother's life, and the lives of her students, and the close call turned both mother and daughter into gun-violence-prevention activists.
That day, 20 children and eight adults, including the shooter and his mother, were killed. I spoke with Sarah, now a senior at Georgetown University, to learn about how she processed the shooting, and how she has been reacting to the national conversation around youth activism and gun violence after Parkland. The following conversation has been edited for length and clarity.
Steven Johnson: Can you walk me through the day of the shooting?
Sarah Clements: I was a junior in high school, in my first-period class, physics. Our principal came on the loudspeaker and said that we were going into lockdown, and it wasn't a drill. We realized that 30 minutes had passed, and maybe this wasn't actually a drill. I saw that I had a text from my dad that said there was a shooting at Sandy Hook.
The town was just an absolute mess. People were trying to figure out if their kids were alive, so there were mobs of parents at every school. We went home, and for the rest of that night, I think we all just sat in my parents' room and watched CNN to figure out who had been killed and who had lived. My mom was recounting what she went through, so that we knew. She always, even from that first day, wanted us to know what happened to her. I think her sharing her story immediately with us is something that affected me, and I got into activism. Of course we need to share our stories. Of course people need to know what that was like.
Johnson: And what was her story?
Clements: She was a second-grade teacher. When she first heard the shots from the hallway, she thought it was folding chairs falling. She walked into the hallway to check it out, and the janitor actually was sprinting by. That's when she knew, when it clicked with her. She immediately closed the door, and told her students to hide up against the walls where their coats were hanging.
Somehow the PA system in Sandy Hook was flipped on in the main office. Everybody in the school heard every single shot. So my mom tried to read her class picture books, and sing songs to try and block out the sound. Eventually it stopped, and there was knock on her door, and they said, "It's the police, it's the police!" They slid their badge under the doorway, and that's when she opened it.
Johnson: You mentioned how your feelings that day eventually turned into activism. How long did that process take?
Clements: I didn't start doing activism until about a month later. I saw a Facebook event for a march on Washington for gun control. I had never done something like that. My dad and I went down to D.C. with a group of maybe 100 people from Newtown, and there were about 6,000 people there. That was also the first time that I met other survivors from other instances of gun violence, and saw that this issue was far, far bigger than just Newtown.
Clements uses a cellphone with a picture of a Newtown victim on it to record the victim's family speaking at a rally in Hartford, Connecticut, two months after the shooting. (Jessica Hill / AP)
Johnson: And you focused specifically on youth activism.
Clements: A lot of people had different ideas of what healing looked like, and for me, it was taking action. But I felt this urgency: I was getting involved in these gun-violence-prevention groups, and [thinking], why weren't students' voices being lifted up? Why weren't they being centered as the people that are most affected, disproportionately, based on gun-violence rates around the country? That's something I recognized really early on, and advocated for. And that's one reason why I'm just so proud of these Parkland students. I'm grateful for this moment that we're in in terms of organizing. And I think it's going to be a real shift in how we talk about guns.
Johnson: How have you felt in the past, each time news of mass shootings would come up?
Clements: When shootings happen, my biggest emotion is just anger and rage that this keeps happening. A lot of people ask, "Do you go back to that day?" I don't think I do, and my mom says that she doesn't either, necessarily. How can another community be facing what my family had to face, what my community had to face? That feeling is so overwhelming and so suffocating, especially when you've been fighting for change for so long. The biggest reason I do this work is so that eventually, no family, no community has to face this terrible tragedy of gun violence. And when it keeps happening over and over again, it's extremely difficult to work through that rage, and to work through that impatience with the situation that we're in.
Johnson: And now feels different?
Clements: It does, it does. It's something that I didn't expect to see at this point in time. I think one [reason] is that my generation and Generation Z see gun violence as an issue that doesn't stand alone, that intersects with a lot of other issues that are happening in our communities and our country.
People always say, "Nothing changed after Sandy Hook." And that angers me, because I actually don't think that this moment would be happening without the change that happened after Sandy Hook. I think that there was a big culture shift, and a political shift—maybe not necessarily nationally, in terms of policy. I think after Sandy Hook, we forced the country to have a continued national dialogue. We built new organizations. That infrastructure of grassroots organizers in every state and in every community, and funding on these issues, and creating a leadership pipeline for young organizers—those things absolutely did not exist before Sandy Hook.
What did exist was [more] communities being affected calling for change, and many of those were communities of color that have been affected by gun violence disproportionately for years and years. What happened after Sandy Hook is that we, I think, came together and forced politicians to answer questions about guns. Small, simple things like that. And for five years, we've built that.
Steven Johnson is a former editorial fellow at The Atlantic.