"I was asked to write about how Islam allows men to beat their wives. Sorry, not true. Yes, I'm sure critics of Islam will quote random Qur'anic verses or ahadith but all are usually taken out of context. If a man does raise a finger to his wife, he is not allowed to leave a mark on her body … this is another way of the Qur'an saying; "Don't beat your wife, stupid".
"Now let's take a glance at some really interesting statistics, hmm. I can almost hear the words pot, kettle, black. According to the National Domestic Violence Hotline, four million American women experience a serious assault by a partner during an average 12-month period".
"On the average, more than three women are murdered by their husbands and boyfriends every day . . . that is nearly 5,500 women battered to death since 9/11″.
"Glossy magazines tell us as women that unless we are tall, slim and beautiful we will be unloved and unwanted. The pressure on teenage magazine readers to have a boyfriend is almost obscene".
The daily mail tried to get the answer of this question.
"In the West, we are stressed for super ficial reasons, like what clothes to wear. In Islam, everyone looks to a higher goal. Everything is done to please God. It was a completely different value system".
"the glitter of showbiz promised nothing new for me. I didn't want to pursue my professional career at any price. My orientation had changed. I wanted to integrate every aspect of my life into my new found value system".
"I am so grateful I found my escape route. This is the real me — I am happy to pray five times a day and take classes at the mosque. I am no longer a slave to a broken society and its expectations".
These revelations are strange for many ordinary Muslims of Pakistan like me. Because Muslims all over the world and especially in Pakistan have adopted an apologetic behavior. Bombardment of allegations, intense Islamo-phobic media campaign along with loss of innocent lives in war on terror brought us all in a defensive position The campaign against fundamental Islam is so intense that talking about Jihad can make you vulnerable to all sorts of problem from condemnation and criticism up to disappearance.
"I know women born Muslims who became disillusioned an d rebelled against it. When you dig deeper, it's not the faith they turned against, but the culture".
Whatever the reason may be, one thing is certain that many of us do not know our religion and message delivered by Islam.The message is so powerful that it can change you with in days. It brings blessings in the lives of followers. Those who have intact natural wisdom, are attracted by it. This message is so essential for all humanity that it should be delivered by any and all means. It can not be done without getting control of state.
We need a genuine Islamic leadership with a vision for Islamic Republic of Pakistan. The leadership which could eliminate all sorts of injustices, poverty and ignorance and could show the real picture of Islam by using state power in education sector and mass media.


text:
Q: Jquery Deferred object issue when chaining .then I'm having an issue with my returned data for the chained then being the 1st deferred data. The first example works:
api.getData().done(function(data){
  api.getData2().done(
    function(data2){
      $.log('success', data2);
    });
  });

But the second example SHOULD work, having data2 for the second .then(), but for some reason its the same as data1.
api.getData().then(function(data1){
  return api.getData2();
}).then(
function(data2){
  $.log('success', data2);
});

Any suggestions?

A: $.when will accept two asynchronous functions that returns a promise and execute the .then() function when both are done :
$.when( api.getData(), api.getData2() ).done(function(data, data2) {
    $.log('success', data2);
});

If for some reason (like needing the data) you need to perform getData() before getData2(), there really is no need for .then(), as your first example seems valid enough for that?

A: So After some research JQuery Deferred.Pipe I found out I can't chain .then, but instead need to use pipe like below (.pipe() is chainable):
api.getData().pipe(function(data1){
  return api.getData2();
}).then(
function(data2){
  $.log('success', data2);
});

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text:
«Бу́де ласка́вий дощ» або «Дощі́ випада́ють» () — короткий дванадцятистроковий вірш американської ліричної поетеси Сари Тісдейл, написаний в 1920 році.

Входить до збірки віршів «Flame and Shadow» («Полум'я і Тінь», розділ VIII, вірш 1).

Переклади

Також існують у різних варіантах аматорські та неофіційні переклади вірша.
Наприклад, такий:

 Прийде лагідний дощ, теплий дух від землі,
 І ширятимуть ластівки неба на тлі;
 У ставку жабин хор буде славити світ,
 З диких слив облітатиме ніжний їх цвіт;
 І зорянки, вдягнувши пухнастий вогонь,
 Нотоносець посядуть дротів осторонь.

 І не знатиме жоден війни - панни Зла,
 Що була тут колись, та в минуле пішла -
 Ні дерева, ні звір, ані птах... ніхто з них -
 Якщо людство щезло з теренів земних.

 І Весна, що повстане ізнову від сна,
 Пригадає навряд чи, що нас вже нема... (Вадим Ткач, Харків)

Також існує поширена, але невірна думка, ніби вірш належить перу самого Рея Бредбері.

У культурі
 За мотивами вірша Сари Тісдейл була написана глава з тією ж назвою в постапокаліптичному жанрі в збірнику «Марсіанські хроніки». У розділі вірш цитується повністю автоматичною системою догляду за будинком, яка вибрала його навмання зі свого архіву, так як місіс Маклеллан не відповіла, бо її вже не було в живих після ядерної війни.
 В однойменному мультфільмі за мотивами оповідання Рея Бредбері, знятому на студії Узбекфільм в 1984 році, скорочений варіант вірша (в перекладі Жданова) цитується голосом за кадром в наприкінці, поки йдуть титри.
 У постапокаліптичній відеогрі Fallout 3, в одній з локацій можна знайти покинутий будинок з роботом, який все ще вірно служить своїм давно загиблим господарям. За допомогою комп'ютерного терміналу, що знаходиться в підсобному приміщенні, роботу можна давати різні вказівки. Якщо його попросити почитати вірші дітям, робот прочитає «Буде ласкавий дощ» двом скелетам в ліжку.
 Існують пісні з цього вірша російською мовою. Один варіант склала А. Кудрявцева, він звучить у виконанні Олени Камбурової (альбом «Краплі Данського Короля»). Інший варіант належить казанському композитору і автору-виконавцю Сергію Бальцеру.
 Одне з маловідомих виконань пісні належить Василю Розумному - інженеру ЧАЕС, акторові і композитору. Її можна почути у документальному фільмі «Прип'ять. Поріг» 1988 на 37-й хвилині.
 За мотивами вірша написана симфонія Єфрема Підгайці «Буде ласкавий дощ». У фіналі симфонії звучить вірш у виконанні співачки.
 Також є і записаний на студії «МЕЛОДІЯ» пісенний варіант Олександра Градського.
 Перший рядок вірша послужив основою для однойменної пісні Віри Матвєєвої, однак зміст пісні не пов'язано з віршем.

Примітки

Посилання
 Переклад Михайла Рахунова
 Пісня "Ласкавий дощ", група "Смерш" (м.Харків), 1988 рік

Літературні твори за алфавітом
Вірші 1920

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For the best Southeastern Mexican zest, your first and only stop in Warner Robins must be La Parrilla Mexican Restaurant.
La Parrillia will be opening up its newest location in Warner Robins this summer. Construction for La Parillia has begun in the former Marco location on Watson. Stay tuned and we will continue to keep you updated as we hear more.
"Since 1994, La Parrilla has been serving up a Fiesta, with exciting and delicious menu items that are sure to enchant and fascinate our guests. Currently with 16 locations in Georgia, one in Dothan, Alabama and one in Greenville, South Carolina. We are a great place to go for ardent lovers of the Mexican Cuisine. The Chef and staff welcome our new and returning guests by preparing authentic Mexican entrees such as succulent enchiladas, juicy and tender fajitas, Quesadilla, burritos and nachos all cooked to perfection and served in a beautiful atmosphere just like in sunny Mexico. It is also La Parrilla's pleasure to now offer you our fantastic Catering Service. Our wonderful personnel will create an entertaining Latin Fiesta in your home or office. Whatever the occasion, our authentic Mexican favorites are always crowd pleasures and we serve it up in a festive style to spice up your event. It is always a Fiesta at La Parrilla!"


What you might not know about one of nature's oldest and most extraordinary events.
The annual migration of wildebeest from the plains of the Serengeti in Tanzania to the grasslands in Masai Mara, Kenya is perhaps one of the greatest natural phenomena to be witnessed. Not only is it the world's largest concentrated movement of wildlife, but also the longest in distance; the wildebeest travel over 1,800 miles on their on their 9 month journey.
1.The Great Migration isn't a singular event—it's a yearlong cycle, a voyage driven mainly by food and water availability. In November-December herds arrive on the plains of the Serengeti, where they'll remain through March, feeding on fresh vegetation. Slowly the masses being moving west across the grasslands and begin their migration North around April. May through August the herds continue traveling northward, occasionally being stalled at certain crossings. Around mid-August- September the wildebeest face their greatest obstacle: The Mara River Crossing, a truly fascinating spectacle. By October, they are reversing their journey, following the rains Southward, with more grace to reach the Southern Serengeti once again.
2.Wildebeest are not the only participants of this cyclical journey—the approximate 1.5million migrating wildebeest are joined by hundreds of thousands of zebras, along with gazelles and antelope.
3. Zebras and wildebeest have forged a natural friendship, their relationship being symbiotic and for this reason they never part ways. Although they feed off the same type of plant, the zebra and wildebeest actually eat different parts of the plant allowing them to graze amicably. The zebra also take on the role of protector and navigator—they keep the herd on course and also remain on the lookout for predators. In exchange, wildebeest help the zebra (and themselves) find water. The animal is known for its keen ability to sniff out water and so, the pack is able to find water nearly every day of the journey.
4. The period between January-March is when wildebeest birth their young. During this time, half a million wildebeest are born (yearly). In the month of February the birth rate is about 8,000 calves per day.
5. The Serengeti ecosystem is one of the oldest on the planet. The vegetation, climate and fauna have remained unchanged for over one million years.
6. To explain the wildebeest's odd appearance, ancient African legend holds that God created the wildebeest using leftover spare parts.
7. As an observer, the migration may appear chaotic and disorganized (especially at the river crossing), however, researchers have found wildebeest's operate using 'swarm intelligence,' which basically means they explore obstacles and generate solutions as one.
8. There are about 3,000 lions within the Serengeti that follow (and prey upon) the migrating herds.
9. Wildebeest are a noisy lot. They emit a constant, low moaning sound and when agitated or disturbed will snort violently.
10. Wildebeest are one of the only African antelope species to increase in population size; in 1960 there were an estimated 250,000 wildebeest and today there are over 1.5 million.


text:
Over 6,000 Myanmarese children studying in Mizoram schools
Aizawl, Sep 29 (PTI): More than 6,000 Myanmarese children, who have taken refuge in Mizoram along with their parents following a military coup in the neighbouring country last year, have been enrolled in various schools across the state, an official said.
Myanmar students in Government Thanchhuma High School
There is no concrete law prohibiting the Myanmarese children from enrolling in Mizoram schools but they could not be officially treated as 'refugee students' as the Centre did grant refugee status to them, School Education Department Director Dr H Lalthlangliana said.
Of the 6,195 such children, 5,221 were enrolled in government schools, 184 in government-aided institutes and 790 in private schools, he said.
Lalthlangliana said that Myanmar children do not face any language barrier as most of them belong to the Chin community. They share the same ancestry and culture as the Mizos of Mizoram and also speak the Mizo language.
In some cases, parents hired teachers for their children, the official said.
Lalthlangliana also said that 68 children of Myanmarese nationals have enrolled for class 10 board examinations under the Mizoram Board of School Education during the 2021-22 academic session.
Of them, 31 appeared in the examinations and all except three cleared the exam.
Recently, Mizoram University (MZU) and the Institute of Chin Affairs (ICA), a non-profit organisation, announced introduction of a diploma course in Burmese language and communication skills at the varsity.
Burmese is the official language of Myanmar and is spoken by around two-thirds of the population there.
The course will be offered by MZU jointly with International Development Research Centre (IDRC) from next month, Prof Lalnilawma, head of extension education and rural development department of MZU said.
"This course is recommended for students who wish to study Myanmar's complex, multi-layered society, and work with local and international agencies in Myanmar," ICA and Mizoram University said in a joint statement.
The course will initially be run for a period of 10 months (Semester I & II) beginning October.
As per official records, more than 30, 000 Myanmar refugees have taken shelter in different parts of Mizoram since the military coup in February last year.
The majority of them are living in relief camps, some in rented houses on their own while others are staying with their local relatives.
The government, in collaboration with village authorities and NGOs, have set up more than 150 relief camps across the state for them.
Mizoram shares a 510 km long international border with Myanmar.
Share:Over 6,000 Myanmarese children studying in Mizoram schools

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text:
Education Freedom Accounts Emerge as Top Priority in Upcoming Session
Posted to Education December 20, 2020 by Christopher Maidment
New Hampshire's GOP legislative majority got right to work after Organization Day, with the Fiscal Committee quickly approving a federal charter school expansion grant the day after the State Board of Education unanimously approved its first Learn Everywhere accredited program. Democrats had successfully blocked both these programs in the previous session, but the new GOP majority wasted no time getting them passed.
Republicans plan to continue that momentum in January with a bill to establish education freedom accounts (EFAs). Similar bills narrowly failed in the '17-'18 biennium, during which the NHGOP had a larger majority than they do today.
Legislative leaders aren't taking anything for granted this time around. Both the House and the Senate have filed versions of the bill, a move generally saved for high-priority legislation to ensure a backstop in case one fails. Senate Majority Leader Jeb Bradley (R-Wolfeboro) filed the upper chamber's version, while Acting Speaker Sherm Packard (R-Londonderry) took over the prime sponsor role of the House version after the death of Speaker Dick Hinch.
Governor Sununu isn't taking any chances either. In a virtual event last week with the Josiah Bartlett Center for Public Policy, Sununu took the opportunity to advocate for the EFAs. "You can sum all this up with: It's gotta be about outcomes for the kids, not outcomes for the system," the governor said. "We have to stop worrying about the system as much as the kids."
"This isn't about the traditional school choice battle," Sununu said. "If you're thinking about it that way, you're way behind."
Even with majorities in the House and Senate and control of the corner office, this Republican-led initiative is hardly a sure thing. In 2017-18, Republicans had to battle teacher's unions, fellow Republicans, and poor attendance.
Despite slimmer majorities, the wind may be at the back of the GOP this time around, partially due to the pandemic and what that meant for education services. Current polling shows that nearly 3-in-4 Granite Staters approve of education freedom accounts when the program is described to them.
Meanwhile, traditional public school attendance continues to decline in New Hampshire, a trend that's accelerated during the COVID-19 classroom shutdowns. "The reason is no mystery," according to The New York Times. "With public schools mostly shifting to remote or hybrid learning, parents are pulling their children out entirely, opting to keep them at home or looking for options that offer more in-person instruction."
As a result, Catholic school enrollment in the Granite State has soared, and evidence from across the country indicates more parents are choosing homeschool as well. Education Freedom Accounts are designed for families looking for an option other than the local, traditional classroom setting.
"People that traditionally weren't involved in this discussion are stepping up and saying, 'wait a minute, where is my money going? Why isn't my kid in school?" said Sununu, "and they're getting involved in this discussion about where their money — not our money, their money — is being spent. That's gonna raise the level of debate to where it needs to be."
The governor and legislators aren't fighting this battle alone, either. Respect New Hampshire, "an independent community dedicated to giving Granite Staters a voice in state and local government" according to their website, is hosting a virtual roundtable Tuesday centered on education freedom accounts with Rep. Glenn Cordelli (R-Tuftonboro) and Kate Baker Demers, the Executive Director of the Children's Scholarship Fund – NH. The group has also launched a petition drive to build support.
Americans for Prosperity New Hampshire, a conservative-libertarian political advocacy group, also plans to promote the legislation by engaging their grassroots activists.
"Education Freedom Accounts are a great step towards giving every child access to a personalized education that fits their unique needs and talents," Sarah Scott from AFP-NH tells NHJournal. "This pandemic has underscored that students and families need a more customized approach to education."
Challenges remain. Republicans only have a 212-187 majority in the House, and on any given day there are anywhere from a handful to several dozen legislators absent, not counting the GOP legislators that voted against the measure just two years ago, in 2018.
In their favor, though, is that only four of the sixteen GOP Reps that voted to send the bill to its polite death remain: Reps Dan Wolf (R-Newbury), Bob Elliot (R-Salem), Thomas Laware (R-Charlestown), and Joseph Guthrie (R-Hampstead).
Nashua Teachers Union Goes On the Attack as Data Show… Read More
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text:
IOI 2001
Return to IOI'2001 Home
by Don Piele, USA
Here is a question that stumps most Americans: "Name two cities in Finland.� After Helsinki, the mind goes blank. So when I told colleagues I was off to Tampere for the 13th International Olympiad in Informatics, July 14- 21, their response was, "Where?� Tampere, it turns out, is a thriving, clean city of 200,000 nestled between the banks of two large lakes, a two-hour bus ride north of Helsinki. Here, in a flat land filled with woods and lakes where nights never completely darken, would be our home for the next week. What follows is my trip report for IOI 2001.
The US delegation gathered at JFK airport in New York on Friday, July 13th for an evening flight aboard Finnair to Helsinki. Team members Steven Sivek from Virginia and Tom Widland from New Mexico were there. Team member Vladimir Novakovski would be joining us in Helsinski after competing in the International Physics Olympiad in Turkey and competitor Reid Barton would come over a day late after competing in the International Mathematics Olympiad in Washington D.C.
Team leader Rob Kolstad from Colorado, deputy leader Hal Burch from New Jersey, and myself (International Committee member) rounded out the normal USA delegation. We added five more to the delegation to study specific tasks that need to be done when we host the IOI in 2003: Greg Galperin and Brian Dean from Massachusetts, Tom Quayle from California, Jeff Polk from Colorado, and Lenny Klaver from Wisconsin. Kimberly Burch, Hal's new wife, was a guest.
Arrival in Helsinki
As soon as we cleared customs and entered the lobby area of the Helsinki airport we were met by two members of the welcoming committee in their blue IOI2001 T-shirts, patiently awaiting the arrival of all the teams.
We arrived in Tampere after a short bus ride, approximately twenty-two hours after leaving home. Team members were dropped off at the Villa Hotel and the delegation leaders were taken to the Ramada Hotel. Although it was tempting to succumb to our drowsiness and lie down for a short nap, we resisted. We�d learned that it�s best to suffer the sleep loss early in order to adjust quickly to the 8 hour jet lag. Delegations from 74 countries would arrive throughout the remainder of the day.
Our guide, Sini Riihijarn, met us at the registration desk after we had settled our accounts and had received IOI 2001 backpacks, filled with goodies. Backpacks in Finland came equipped with cell phone holders on the shoulder strap, since everyone in Finland seems to be born with a Nokia cell phone in their hand.
Next we were off to the cafeteria at the University of Tampere to greet other team leaders, who have become good friends over the years. Once someone becomes an IOI team leader, they often want to continue as long as possible. As a result, the annual IOI is getting to feel a bit like a class reunion.
The first day of an IOI is customarily devoted to getting used to the surroundings and overcoming jet lag. This was accomplished with an orienteering adventure in the city of Tampere. With the help of our guide, the team took off to visit numbered locations around the city and to answer a series of questions. At the Information Center, they discovered the population of Tampere (200,000) and the number of lakes within the city limits (200).
Practice Round
A new feature of this year�s competition was an automated grading system developed by Rob Kolstad. With his system, students can submit their source code over the Internet and get an immediate check that it compiles and correctly solves a simple case. In order to get used to this system it was necessary to schedule a practice round. After months of hard work, the system got its first real test under competition conditions and passed with flying colors. This was a watershed moment in the history of the IOI, made possible by switching to a Lenux environment for compiling and grading.
The opening ceremonies were held in the main auditorium of Tampere Hall on Sunday. Steel Pan Lovers opened the program followed by an Introduction from Jari Koivisto, Chair of IOI 2001, and a "Welcome to Finland� address by Mr. Jukka Sarjala, Director General of IOI2001. Folk dancing and music was interspersed with a few short remarks from retired Finnish gold medalist - Tero Karras; a Nokia representative - Tytti Varmavuo; and Jyrki Nummenmaa, Chair of the Scientific Committee.
A total of 509 people, including 272 contestants from 74 countries, watched the ceremony. Afterwards, the team leaders had a buffet dinner in the Fuuga Restaurant in Tampere Hall.
That evening the delegation leaders met to choose the problems for the first competition day. The problems were presented and accepted unanimously on the first try. Now the only thing left for many countries was translating the problems into their native language. Since the official language of the IOI is English and all contestants receive a copy in English, our delegation was able to skip this time consuming task.
First Day of Competition
On Monday morning, the contestants began the first five-hour competition. This gave the delegation leaders a break, and some headed back to the hotel for some sleep after an all night translation session. Five hours later, the first round was over and all problems had to be submitted to the grading system.
Rob had used his grading system at our training camp in June, where the programs from fifteen competitors were all graded within fifteen minutes after the contest had ended. Now the program had to work for 272 competitors. The processing load was distributed between 30 computers, so, barring unforeseen circumstances, it shouldn�t take much longer to grade all the IOI competitors.
There were two unforeseen circumstances: extremely large programs, and programs taking minutes rather than seconds to compile. The large programs were the result of students trying to do an abnormal amount of preprocessing. These were rare cases, but they had to be dealt with before the next round. It was decided by the general assembly to limit program size to one megabyte and compilation time to 30 seconds. Despite these first day problems, the results were ready by early afternoon, and every student could see exactly where his or her program succeeded and where it failed with the test data. Contestants got a detailed report of which data sets ran correctly and within the specified time limit. Points were automatically assigned and a total recorded.
Some questions were raised about the short time limits on one of the problems and about the accuracy of the timing process. The handling of these questions resulted in some ruffled feathers.
Excursion Day 1
The day between the two competition days is traditionally reserved for a nearby excursion. The skies where overcast and it looked like rain. Luckily one of the presents inside our backpack was a plastic raincoat. It would come in very handy on this trip. We walked down to the quay and boarded a passenger ferry for a short trip to Viikinsaari Island.
When we arrived, the rain was coming down in buckets and we made a mad dash for the only building in sight. There we stayed until the rains lifted and we were able to take a walk around the island. When the sun came out, some went on boat rides, a few swam, some played volley ball, others walked on stilts, and a few took a break dance lesson. The afternoon was beautiful, and everyone relaxed and had a good time enjoying nature.
Competition Day 2
This was pretty much a carbon copy of the first competition day. But the grading process was not so lucky. About everything imaginable that could go wrong did. Files were accidentally erased. Lightning struck and power was lost, trapping the programs in the computers. Rob, working non-stop for days, finally crashed too and went to bed. In the evening, while everyone else was sleeping, he was back at Tampere Hall with the power restored, rebooting, by hand, all 272 computers so the grading process could continue.
The new grading program advanced the IOI grading system by a quantum leap. But it came at a heavy cost to Rob, who was under constant pressure to keep everything working. He was not able to participate in the activities and excursions that make the IOI so enjoyable. This is something none of us anticipated and we don�t want to repeat.
The day after the last competition, the team leaders traveled to Ruovesi National Park for a 2.5 km hike in a pristine forest. We got information about the flora and fauna in the park along with a map and a box lunch. Meanwhile the contestants were getting their thrills and spills at the amusement park riding the Sky Flyer and the popular Tornado.
Once the cut off values for the gold silver and bronze medals had been determined, it is traditional to celebrate with a fancy dinner and some dancing. The venue this year was the hotel Rosendahl. Our delegation was color coordinated with polo shirts embroidered by Rob. The only problem was, Rob wasn�t there. He was too exhausted from the grading ordeal and needed rest.
On the afternoon of Friday the 20th, the closing ceremonies were held in the main auditorium of Tampere Hall. As we filed in we received a booklet prepared by Finland�s Scientific Committee, with all the problems used in the competition and the reserve problems worked out in complete detail. This was a wonderful idea and the first time it had been done. Often this material is available only a year after the competition is over. Another nice touch was when Jari Koivisto called all the guides to the stage, where they received the adulation of the audience for a job well done.
When the medal winners were announced, their names were displayed overhead on a large screen -- another idea which we will copy for 2003. There were 64 bronze medals and one went to Steven Sivek from the US.
A total of 45 silver medals were awarded, one going to Tom Widland
and another to Vladimir Novakovski.
Twenty gold medals were awarded and then it was time to announce the top three gold medals:
Third Place Ziqing Mao, China - 485 points,
Second Place Mihai Patrascu, Romania - 525 points,
First Place Reid Barton, USA - 580 points.
This was the biggest gap between first and second place in the history of the IOI. A very poised and humble young man received the applause from the audience and a new laptop computer from Mikrolog. It was the first time a contestant from the United States had won the top prize.
In addition to this victory, just a week earlier Reid had won the top gold medal at the International Mathematics Olympiad, making him the first competitor ever to win the top gold medal in two International Olympiads. Reid�s mother, Pam, who had planned to vacation in Finland with Reid after the IOI, was with us to witness this extraordinary achievement.
Finally the IOI flag was passed on to Ha-Jine Kimn of Korea, where the 14th IOI will be held next August.
Final Dinner
An elegant sit down dinner followed the ceremony. Finally it was the time to unload all those trinkets we had brought along as trading material. Reid handed out Daisy, a beenie baby cow, to everyone we could corral, including our guides. Yes, we had two guides, since Tuuli Rythy was supposed to be a guide for Mongolia but that team never showed up. Now the exodus would be non-stop until all 74 countries had left Tampere heading for home.
Thank you, Jari, Jryki, Tiina and the entire Finnish organization for your hard work and innovations that you brought to IOI 2001. It was another memorable IOI. Thank you, Rob, for your extraordinary efforts in creating the ultimate grading program. Now take a long rest.


text:
Los Melenchones est une localité de la commune d'Águilas (région de Murcie, Espagne), située à proximité de l'Autoroute AP-7 et de l' d'Águilas à Lorca.

Localité dans la Région de Murcie


The Midnight Mission discriminates against Homosexuals.
I changed my brown hair to blond.
The head honcho saw himself an "out."
He demanded me to change my hair back to brown or cut it off.
He kicked me out for not being a blond.
He used the blonde hair as "an excuse."
He discriminated me for being gay.
At least, I stood my ground.
Then listen to a low life jerk, Clancy Imislund, at Midnight Mission.
In closing, don't send any DONATIONS to the Midnight Mission.
They don't deserve any donations till they change their homophobia ways.


text:
Q: How to add mini_wishlist for django-oscar? I am working for e-commerce sites with django-oscar.
I have problem with wishlist. I want to add mini_whislist on navbar. It's looks like oscar's basket_quick.
I did try this one, but its only working on wishlists section.
{% if wishlists %}
   {% for wishlist in wishlists %}
       {{ wishlist.name }}
       {{ wishlist.lines.count }}
   {% endfor %}
{% else %}

    EMPTY
{% endif %}

How can I make it appear on every page?

A: create  context processor in customer app...
context_processors.py
def wishlists(request):
    if not (request.user and request.user.is_authenticated):
        return {}
    return {
        'wishlists': request.user.wishlists.all()
    }

and put that into settings file
TEMPLATE_CONTEXT_PROCESSORS = (

    'customer.context_processors.wishlists',
)

then you can use wishlists anywhere

A: One idea would be to put all the html you wrote for this feature into a HTML file and then use {% include 'your_file.html' %} in layout.html, it would not be a good idea to put in base.html since dashboard also extends the same base template.
Also, please note that you have to do the same in checkout/layout.html, because they have a separate layout for the checkout pages.


Marvin Minsky, a pioneer in the field of artificial intelligence who saw parallels in the functioning of the human brain and computers, has died aged 88.
Mr Minsky died at Brigham and Women's Hospital in Boston of a cerebral haemorrhage, the Massachusetts Institute of Technology (MIT) said.
The scientist viewed the brain as a machine whose functioning can be studied and replicated in a computer and considered how machines might be endowed with common sense.
He "helped create the vision of artificial intelligence as we know it today", said Daniela Rus, director of MIT's computer science and artificial intelligence laboratory.
Mr Minsky served in the US navy during the Second World War before earning mathematics degrees from Harvard and Princeton. He joined MIT's faculty in 1958.


We happened upon this place last year and liked it so much that we decided to try and find it again even though it had soft tip darts. Imagine our surprise to enter and find that the unthinkable had happened: the soft tip dart machine had been replaced by a steel tip board. Not only had it been replaced, it had been replaced with an eye to where the board should be and how it should be lit.
A very pleasant hour was spent playing darts, having a couple of beers and eating a basket of fried shrimp. The Lear Inn has a nice selection of beer (including locally brewed Great Lakes) and a menu of good, fast, very reasonably priced food. If you're in the area, drop by and support one of the few establishments actually bringing in steel tip and doing a good job of it.
Hi. I read your posting. My friend and I are road trippers, patronizers of small town bars, and always looking for steel tip. However, we live in Colorado, hence outside of the major metro regions of this state and those around it, steel sip boards are rare.


text:
Occidental College > Library > Special Collections > Online Exhibits > Miniature Books
Miniature Books Through the Years
Miniature Books by the Numbers
Miniature Books at Oxy
Resources & Acknowledgements
Visit Other Special Collections Exhibits
Miniature Books: Then and Now
An online exhibition of miniature books and how they have evolved throughout the years.
What is a Miniature Book?
A commonly asked question is what differentiates a miniature book from simply a small one. According to the Miniature Book Society, "a miniature book is usually considered to be one which is no more than three inches in size—height, width or thickness… Outside of the United States, books up to four inches are collected as miniature books." (www.mbs.org) The Occidental College Library's Special Collections uses this criteria to define miniature books for our own collection.
A Brief History of Miniature Books
Did you know that the first miniature book as we know it today was created in 1496 by Peter Schoffer, assistant to printer Johann Gutenberg? Early printers often had their apprentices make miniature books in order to test their skill at using small typefaces. Until the Industrial Revolution of the late 18th and 19th centuries, miniature books were primarily a work of convenience, designed for adults to slip into a pocket or strap to a girdle, or as a more accessible size for children. After the Industrial Revolution, paper became easier to make and books became easier to bind. As a result, miniature books were printed increasingly for novelty's sake. Creating miniature books today is also a book art form explored by artists and artisans in fine printing and binding. This exhibition, drawn from the Miniature Books in the Special Collections Dept, offers a look at the diverse types of these small books produced during the past 200 years.
Jessica Low '07
2007-2008 Library Fellow
Mellon Librarian Recruitment Program
" All the way home I kept remembering
The small book in my pocket. It was there. "
-Robert Frost , "A Fountain, a Bottle, A Donkey's Ears, and Some Books."
Page last edited by on 03/06/2013.
© 2008 Occidental College


Depart from the hotel, en rout we will stop at Quang Tri Province. From here, we will proceed to amazing Vinh Moc Tunnels. The tunnels remain unaltered and therefore offer an authenic view of the underground life. We will enter the tunnels and explore the subterranean passageways. Drive to Dong Hoi Town for dinner and overnight stay.
This morning, we will take a boat cruise on the Son River to visit Phong Nha Caves including Tien Son and Bi Ky Caves. After lunch, we will visit Hien Luong Bridge, Ben Hai River and Doc Mieu Fire Base on our way back to Hue City. Drop off at the hotel at the end of trip.


The Classic Phono TT-10 is a magnificent and stylish turntable with a true retro look. As the turntable is protected by a handy brown case that weighs less than 3 kg, you can always take it with you when travelling. So you always have the TT-10 and your favourite music to hand wherever you are.
The Classic Phono TT-10 is a magnificent and stylish turntable with a true retro look. As the turntable is protected by a handy brown case that weighs less than 3 kg, you can always take it with you when travelling. So you always have the TT-10 and your favorite music to hand wherever you are.
The turntable is a user-friendly unit without any unnecessary frills. Thanks to the built-in speakers, you can enjoy high-quality sound as it plays your favorite music. Take the case with you to a birthday party or reunion and surprise everybody with an extensive repertoire of old and new hits. Make the party truly unforgettable by contacting your friends and family beforehand and asking them to bring their favorite records as well.
The Classic Phono TT-10 features a ceramic stereo cartridge with integrated stylus. You can play your records at 3 different speeds: 33 RPM, 45 RPM and 78 RPM. The turntable is equipped with a belt drive, two built-in loudspeakers, an AUX input, an RCA jack and a line input for external speakers or other equipment.


text:
\section{Introduction}

\noindent
The Collatz conjecture is a long standing open conjecture in number theory. Among many other names it is often known as the conjecture for the $3x + 1$ problem, which concerns to an arithmetic procedure over integers. This conjecture is based on the Collatz function given by
\begin{equation}\label{eq:collatz}
  C(x) \;=\;
  \begin{cases}
    3x+1, & \text{if }\; x \equiv 1 \quad (\text{mod } 2) \\
    x/2,  & \text{if }\; x \equiv 0 \quad (\text{mod } 2)
  \end{cases}.
\end{equation}
Notice that the Collatz problem concerns to the dynamical behavior of the above map for any positive integer $x$.

\begin{conjecture}[Collatz conjecture]
Starting from any positive integer $x$, iterations of the function $C(x)$
will eventually reach the number $1$. Thus, iterations enter in a cycle,
taking successive values $\{1,\,4,\,2\}$.
\end{conjecture}

The problem has been addressed from several viewpoints along nearly a Century (see \eg  \cite{Lag10} for an overview), which come from approaches of number theory, dynamical systems,
ergodic theory, mathematical logic, and theory of computation as well as stochastic strategies. This paper consists of a numerical and dynamical hybrid analysis. The $3x + 1$ problem have been considered as a discrete map $C:\mathbb{Z} \to \mathbb{Z}$. The map here proposed maps an interval in the real numbers into itself, yet the given map is discontinuous almost everywhere.

In Sec.~\ref{sec:notation} we introduce a new binary function capturing the dynamics given by \eqref{eq:collatz}, yet for numbers in an interval in $\mathbb{R}$. In Sec.~\ref{sec:bin} this function rises up to an iterative map which allows to draw conclusions towards Collatz conjecture. Finally, a brief discussion of results can be found in Sec.~\ref{sec:conclusions}.

\section{Some useful notation and results}
\label{sec:notation}

\noindent
The Collatz function can be simplified into more tractable functions, a well-known case is the Everett function which considers a map for odd integers $(3x + 1)/2$. This function was also analyzed by Terras, \textit{cf.} \cite{Eve77,Ter76}. The iteration is optimized with the reduce Collatz function $R(x) = (3x + 1)/2^m$ (where $2^m$ is the larger power dividing $3x + 1$), \textit{cf.} \cite{Col17}.
Even though, this reduced map aims to simplify the Collatz procedure, it is yet not straight forward.

Let be $x$ an odd number, its \textbf{binary expression} is $(a_m a_{m-1} \cdots a_0)_2$ where $a_k \in \{0,\,1\}$ for $k = 1,\,2,\,\dots,\,m-1$, and $a_0 = 1$ and $a_m = 1$.
Notice that the reduced function satisfies that $R(x) = R(2^n x)$ for any $n \in \NN$.
However, for an even value of $R(x)$, the next iteration is divided by a power of two, which power is unknown \textit{a priori}. In order to remove this computation, let us take into account the binary expression of $y = 2^{-(m+1)}x$, \ie
$y = (0.a_m a_{m-1} \cdots a_0)_2$. Note that the 1 to 1 map from odd $x$ and $y$ is therefore given.
Let us define the map as
\begin{equation*}
M(x) = 2^{-m}x, \qquad \text{ with $\;m\;$ the smallest $\;\NN\,$ such that } \;x < 2^m\;
\text{ holds,}
\end{equation*}
hence $M(x) = y$ holds, and $M(2^n x) = M(x)$ is satisfied for all $x,\,n \in \NN$. Notice that $x = 1$ is related to $y = 2^{-1} = (0.1)_2$ by means of $M(x)$, odd integers $x$ are uniquely represented in the real interval $I = [1/2,\,1)$.

Now, we need to calculate
$3y + 2^{-(m+1)}$ for $y = (0.1 a_{m-1} a_{m-2} \cdots a_1 1)_2$,
which turns the problem out to calculate the \textit{total length} of the $a_k$
binary digits length. On the other hand, the benefit by iterating over
$I$, is that $3y + 2^{-(m+1)}$ lies on the interval $(3/2, 4]$.
Thus, in order to get a value in $I$, an iteration must be divided by $2$, $4$
or $8$.
For instance, $y = (0.1)_2$ uses $m = 0$ and
$3y + 1/2 = (1.1)_2 + (0.1)_2 = (10)_2 = 4$. Upon dividing by $8$
we get that $4/8 = (0.1)_2$, \ie, the fixed point $1$.
As a further matter, $y = (0.10101 \dots 01)_2$ yields to the same relation as
$3y + 2^{-(m+1)} = (1.1 \dots 11)_2 + (0.0 \dots 01)_2 = 4$, where $m+1$ is the
binary digits length of $y$.
For a practical use of additions in vertical form, see the sketch in Fig.~\ref{fig:add}; other uses of binary coding in number theory can be seen in \cite{Cas14}, for instance.

\begin{figure}[htb]
\centering
\includegraphics[width = 0.8\textwidth]{binary}
\caption{\small Binary addition example. On the left-handed panel, we notice that $(1)_2 + (1)_2 = (10)_2$; thus on vertical additions every $1$ added to $1$ places $0$ and carry $1$. On the right-handed panel the same numbers added on the left in both binary and decimal notations.}
\label{fig:add}
\end{figure}

\begin{definition
For a given $n \in \NN$, define $\ys_n = (0.1\{01\}^n)_2$, where $\{01\}^n$ stands
for $n$ repetitions of $01$.
Let be $\mN$ the set of all $\ys_n$ numbers as the set of \textbf{predecesor numbers} or simply \textbf{predecessors}.
\end{definition}

\begin{remark}\label{rem:predecessors}
Notice that $x$ defined as $2^{2n + 1}\ys_n = (1\{01\}^n)_2 = 2^{2n} +
2^{2(n-1)} + \cdots + 2^0$ which satisfies that $3x + 1 = 2^{2n+1} + 2^{2n} +
\cdots + 2^0 + 1 = 2^{2(n+1)}$. (Notice that $\ys_n \to 2/3$ as $n \to
\infty$.) Hence the Collatz function $C^{2n+3}(x) = 1$ and the reduced
form $R(x) = 1$ hold.
\end{remark}

In consequence, let us define the \textbf{binary function} by
\begin{equation}\label{eq:binary}
B(y) \;=\;
\begin{cases}
  1/2,                 & \text{if }\; y \in \mN \\
  (3y + 2^{-(m+1)})/2, & \text{if }\; y \in [1/2,\,2/3] \setminus \mN \\
  (3y + 2^{-(m+1)})/4, & \text{if }\; y \in (2/3,\,1)
\end{cases},
\end{equation}
for all $y \in I$, where $\ell = m+1$ is the \textit{total length}, \ie, the
``digits'' number of $y$ in binary notation.

\begin{remark}
A $y \in [1/2,\,1)$ without preimage $x \in \NN$ by $M(x)$ (\emph{e.g.} $y = 4/5$) will follow the map \eqref{map:circle} ahead, since in such a case $m$ goes to $\infty$.
\end{remark}

\begin{lemma}\label{lem:itself}
The binary function \eqref{eq:binary} maps the interval $I = [1/2,\,1)$ into itself.
\end{lemma}

\begin{proof}
The proof consists of three rigorous arguments based on binary adding (see Fig.~\ref{fig:add}):

(1) By definition and Remark~\ref{rem:predecessors}, the result holds for $y \in \mathcal{N}$.

(2) For $y \in [1/2,\,2/3] \setminus \mathcal{N}$, notice that
$y = (0.1 \{01\}^k 00 a_{m - 2k - 3} \cdots a_0)_2$ for $k \in \NN$, yet $m \geq 2k + 3$ holds since $y \notin \mathcal{N}$. Thus,
\begin{equation}
3y + 2^{-(m+1)} \;=\; (1.1\{11\}^k b_2 b_1 * \cdots * 1)_2 + (0.0\cdots01)_2,
\label{eq:N}
\end{equation}
where $*$ represents an unknown binary digit and both binary numbers at right have length $m+1$. Because of the $00$ in
the original $y$, we notice that the binary structure should satisfy $b_2 b_1
\neq 11$, thus by adding $2^{-(m+1)}$ expression \eqref{eq:N} cannot be greater
than $2$, yet is larger than $1$. Therefore, $(3y + 2^{-(m+1)})/2 \in I$ holds.

(3) For $y \in (2/3,\,1)$, its binary representation is
$y = (0.1 \{01\}^k 1 a_n \cdots a_0)_2$, for $k \in \NN$. Thus,
$$3y + 2^{-(m+1)} \;=\; (10.\{00\}^k * \cdots * 1)_2 + (0.0\cdots01)_2,$$
for $*$ any binary digit, which gives place to $3y + 2^{-(m+1)} \in
(2,\,4)$; in consequence, $(3y + 2^{-(m+1)})/4 \in I$.
\end{proof}

The equivalence between $B(y)$ and the reduced function $R(x)$ is given
multiplying by $2^{\pm\ell}$, where $\ell = m + 1$ is the length of both $x$
and $M(x)$ in binary form. Therefore, Remark~\ref{rem:predecessors} and
Lemma~\ref{lem:itself}, yields to the following result.

\begin{corollary}
For a given $x \in \NN$, $R(x) = 1$ if and only if $M(x) \in \mathcal{N}$ holds.
\end{corollary}

\begin{conjecture}[Binary Collatz conjecture]
For any $x \in \NN$, $y = M(x)$ belongs to $I$. Then, the iterative map $B^k(y) \to 1$ as $k \to \infty$. In other words, the binary function \eqref{eq:binary} has $1/2$ as a global atractor for the domain $I = [1/2,\,1)$ in binary notation.
\end{conjecture}

For a given $y_0 \in [1/2,\,1)$, we first notice that the behavior of the
length of $y_k = B^k(y_0)$ was successfully analyzed in \cite{BP08} for Baker's
map. This approach for the Collatz function remains nonetheless unclear. The numbers in
Table~\ref{table:initial} are the size of the initial length in binary
form against the maximal length found in several of runs considering $500$
random numbers of each particular length. The $200$ of these runs condensed
in Table~\ref{table:initial} give the computed maximum length and maximum
iterations in order to achieved the \textit{ground state} $1/2$. In
Table~\ref{table:initial}, we notice that these numerical simulations suggest
that in the binary form, the result may provide an useful guide to its proof.

\begin{table}[ht]
  \centering
    \caption{Computation of maximal length and maximum number of iterations for binary numbers in $I$ with fixed initial length.}\label{table:initial}
    \begin{tabular}{{r|rrrrrrr}}
Initial length & $50$ & $100$ & $500$ & $1,000$ & $5,000$ & $10,000$ & $50,000$
\\ \hline
Maximum length & $+14$ & $+13$ & $+13$ & $+12$ & $+13$ & $+12$ & $+12$
\\ \hline
Maximum stop time & $346$ & $449$ & $1,720$ & $3,136$ & $13,541$ &
$25,927$ & $124,514$
    \end{tabular}
\end{table}

\begin{remark}\label{rem:initial}
In Table~\ref{table:initial} the column for numbers with $50,000$ binary digits was obtained with ten runs of the routine iterating $500$ random numbers of this length. Nine out of ten runs returned the recorder numbers for \textit{Maximum length} and \textit{Maximum stop time}, $50,012$ and $124,514$ respectively.
In one case only, the result gives a maximum length of $+9$ binary digits and $118,914$ as maximum number of iterations before achieving $1/2$.
\end{remark}

Notice that the stopping time for the reduced function $R(x)$ and for the
binary function $B(M(x))$ are equal. Nevertheless,
Table~\ref{table:initial} and simulations suggest that the binary form has a regular
behaviour for the length in their binary expansion: a binary
length increasing of $14$ digits represents approximately $1.6 \times 10^{4}$
in decimal notation, thus a number in the first column of
Table~\ref{table:initial} may change from $2^{50}\approx 10^{15}$ to
$2^{64} \approx 2\times 10^{20}$.
The {\it hailstone numbers} introduced in \cite{Hay84} represent iterations where there is an abrupt increase of magnitude from, say $x_n$ to $C(x_n)$. In Fig.~\ref{fig:trayectory}, we may compare the smoothing from decimal expressions against their binary representations.

\begin{figure}[htb]
\centering
\includegraphics[width = \textwidth]{two}
\caption{\small Iterations of $x_0$ taken as $N := 63,728,127$ (left-hand panels), and $n_0 := 100 \lfloor \pi \cdot 10^{35} \rfloor$ (right-hand panels). On top panels the iterated $x_n \in \NN$; on the bottom panels, the respective $y_n = M(x_n)$ in binary: blank spaces for zeros and dots for ones binary bits, top lines stand for binary point.
Notice that $n_0$ has $38$ decimal digits, $119$ binary digits, and $255$ steps instead of $529$ steps with the original Collatz function, \cite{Lag10,Roos}.
It has a hailstone number $x_{13}$ with $35$ decimal digits and $116$ binary digits.
Notice that $N$ has $8$ decimal digits, $26$ binary digits; its hailstone number $x_{33} = 322,205,345,153$ has $12$ decimal digits and $39$ binary digits.}
\label{fig:trayectory}
\end{figure}

The growing thin of the string length is determinant in satisfying Collatz conjecture. Table~\ref{table:initial} suggests that for a length $m$, the longest representation has a maximum length of $m + 14$. Let us estimate the dynamics of the length for $y_0 = (0.1 a_{m-1} \cdots a_1 1)_2$ to $y_1 = B(y_0) = (0.1 b_{\tilde{m}-1} \cdots b_1 1)_2$. To do so, we count the ``head'' and ``tail'' behaviour: we define a \textbf{head} with the $n$ first binary digits, and a \textbf{tail} with the $n$ last binary digits.
For example, $(0.1001)_2$ is mapped to $(1.1100)_2$ by $3y + 2^{-(m+1)}$, now we count the length grow with the binary point as reference. We say that it increases by one at the beginning with a $1$ at the left of the binary point and decreases by two at the very end, because of the two last zeros and since $y_0$ originally has four binary digits at the right of the binary point. In so doing, we notice the head adds one to the total length and the tail subtracts two. In this case, the combination of head and tail behaviour is such that as a result of the iteration $y_1 = B(y_0)$, the total length is decreased by one.
Let us focus on $n = 3$ binary digits and notice that there are four possible configurations both for $3$-digit heads and $3$-digit tails.

\begin{theorem}[Heads and tails]\label{aff:h&t}
The head with three binary digits $h_1 = (0.100)_2$ adds one to the total length, the heads $h_2 = (0.101)_2$,  $h_3 = (0.110)_2$,  $h_4 = (0.111)_2$ add two to the total length by means of \eqref{eq:binary}. (A $h_2$ head may add only one digit.) The tail with three binary digits $t_1 = (\cdots 001)_2$ subtracts two to the total length, the tails $t_2 = (\cdots 011)_2$, $t_4 = (\cdots 111)_2$ subtract one to the total length and the tail $t_3 = (\cdots 101)_2$ subtracts at least three to the total length.
\end{theorem}

\begin{proof}
There are eight cases to consider. As above, $*$ stands for any unknown binary digit, notice we are using vertical adding as in Fig.~\ref{fig:add}. Notice that adding $2^{-(m+1)}$ remains implicit for the heads within the $*$, as can be seen by the binary notation. Notice that $3 = (11)_2 = (10)_2 + (1)_2$, so we get that
\begin{equation*}
h_1: \begin{array}{r@{} @{}c@{} @{}r@{} @{}r@{} @{}r@{} @{}r@{}}
     0&.&1&0&0&*\cdots \\
+ \; 1&.&0&0&*&*\cdots \\
\hline
{\bf 1}&.&1&*&*&*\cdots
\end{array}\;,
\;\;
h_2: \begin{array}{r@{} @{}c@{} @{}r@{} @{}r@{} @{}r@{} @{}r@{} @{}r@{}}
     0&.&1&0&1&a&*\cdots \\
+ \; 1&.&0&1&a&b&*\cdots \\
\hline
{\bf 10}&.&*&*&*&*&*\cdots
\end{array}\;,
\;\;
h_3: \begin{array}{r@{} @{}c@{} @{}r@{} @{}r@{} @{}r@{} @{}r@{}}
     0&.&1&1&0&*\cdots \\
+ \; 1&.&1&0&*&*\cdots \\
\hline
{\bf 10}&.&*&*&*&*\cdots
\end{array}\;,
\;\;
h_4: \begin{array}{r@{} @{}c@{} @{}r@{} @{}r@{} @{}r@{} @{}r@{}}
     0&.&1&1&1&*\cdots \\
+ \; 1&.&1&1&*&*\cdots \\
\hline
{\bf 10}&.&1&*&*&*\cdots
\end{array}\;.
\end{equation*}
Thus, $h_1$ adds one to the total length, and the others heads add less than three to the total length.
If for $h_2$, $a$ and $b$ are zero, then the sum is $(1.111*)_2$ and this head adds only one to the total length.

Similarly, for the tails we add $x$ in the first row and $(10)_2x + 2^{-(m+1)}$ on the second row.,
We obtain
\begin{equation*}
t_1: \begin{array}{@{}r @{\,}r@{} @{}r@{} @{}r@{} @{}r@{}}
     \cdots & *&0&0&1 \\
+ \; \cdots & 0&0&1&1 \\
\hline
\cdots&*&1&{\bf 0}&{\bf 0}
\end{array}\;,\quad
\quad
t_2: \begin{array}{@{}r @{\,}r@{} @{}r@{} @{}r@{} @{}r@{}}
     \cdots & *&0&1&1 \\
+ \; \cdots & 0&1&1&1 \\
\hline
\cdots & *&0&1&{\bf 0}
\end{array}\;,\quad
\quad
t_3: \begin{array}{@{}r @{\,}r@{} @{}r@{} @{}r@{} @{}r@{}}
     \cdots & *&1&0&1 \\
+ \; \cdots & 1&0&1&1 \\
\hline
\cdots & *&{\bf 0}&{\bf 0}&{\bf 0}
\end{array}\;,\quad
\quad
t_4: \begin{array}{@{}r @{\,}r@{} @{}r@{} @{}r@{} @{}r@{}}
     \cdots & *&1&1&1 \\
+ \; \cdots & 1&1&1&1 \\
\hline
\cdots & *&1&1&{\bf 0}
\end{array}\;.
\end{equation*}
By noticing that the binary point is fixed $m+1$ positions at the left of this adding, notice that $t_2$, $t_4$ subtract one to total length, $t_1$ subtracts two and $t_3$ substracts at least three to the total length.
\end{proof}

\begin{remark}\label{rem:total}
From Theorem~\ref{aff:h&t}, the contribution of all three digit head/tail possibilities to total length is depicted in the following table. Notice, that by adding all cells the total sum is less or equal than zero.
\newcolumntype{R}[1]{>{\raggedleft\let\newline\\\arraybackslash\hspace{0pt}}m{#1}}
\begin{center}
\begin{tabular}{ l | R{2.05cm} R{2.05cm} | r | R{2.05cm} }
Heads $\backslash$ Tails &
$t_1 = (\ddd 001)_2$ & $t_2 = (\ddd 011)_2$ & $t_3 = (\ddd 101)_2$ & $t_4 = (\ddd 111)_2$ \\
\hline
$h_1 = (0.100\ddd\!)_2$ &     $-1\qquad$ &      $0\qquad$ & \cellcolor{Gray} $\leq -2\qquad$ &      $0\qquad$ \\
\hline
$h_2 = (0.101\ddd\!)_2$ & \cellcolor{Gray} $\leq 0\qquad$ & \cellcolor{Gray} $\leq 1\qquad$ & \cellcolor{Gray} $\leq -1\qquad$ & \cellcolor{Gray} $\leq 1\qquad$ \\
\hline
$h_3 = (0.110\ddd\!)_2$ &      $0\qquad$ &      $1\qquad$ & \cellcolor{Gray} $\leq -1\qquad$ &      $1\qquad$ \\
$h_4 = (0.111\ddd\!)_2$ &      $0\qquad$ &      $1\qquad$ & \cellcolor{Gray} $\leq -1\qquad$ &      $1\qquad$
\end{tabular}
\end{center}
\end{remark}

As can be seen in Remarks~\ref{rem:initial} and \ref{rem:total} not only Collatz conjecture might be true, but they also point out to a novel way of analyzing a binary structure by each step iteration. In the following section, by taking into account a \textit{circle map} of the binary function \eqref{eq:binary} a visualization of the iterates of $B(y)$ is improved.

\section{Associated binary map and orbits in the iterative map}
\label{sec:bin}

\noindent
We now analyze the binary function \eqref{eq:binary} as an iterative map over the interval $I = [1/2,\,1)$ into itself. The one-dimensional dynamics is well understood when the map is a diffeomorphism or a homeomorphism (see {\it e.g.} \cite{dMvS93,MiTh88}).
Nevertheless, notice that the binary map $y_{n+1} = B(y_n)$ is highly discontinuous. For example, let be $y = (0.1011)_2$ and a number $z = (0.101011 \cdots 1)_2$ which is closed. Upon applying the binary function, we have $B(y) = (0.10001)_2$ and $B(z) =  (0.10000011 \cdots 1)_2$. Thus, the distance $|y - z| = 2^{-\ell}$ depends on the length $\ell$ of $z$, and can be bounded by any $\delta > 0$, however $|B(y) - B(z)| > 2^{-6}$ uniformly.

\subsection{A reduced circle map}
\noindent
By inspection, this discontinuous behaviour of \eqref{eq:binary} arises from both elements in $\mN$ and subtracting terms $2^{-(m+1)}$. Hence, we consider $B(y)$ as the perturbation of map
\begin{equation}
\label{map:circle}
\tB(y) \;=\;
\begin{cases}
  3y/2, & \text{if }\; y \in L \;:=\; [1/2,\,2/3) \\
  3y/4, & \text{if }\; y \in R \;:=\; [2/3,\,1)
\end{cases},
\end{equation}
which defines a \emph{circle homeomorphism} on $I$, see \cite{dMvS93}. This map has a unique critical value $c \in I$ such that $\tB(c) = 1/2$ where the map is discontinuous. This map may be extended to $\mathbb{R}$ by a \textit{lifting} for being a continuous function, with \emph{e.g.} $\tilde{H}(1/2) = 3/4$. Also, $\tB$ is continuous and strictly increasing in each component of $I\setminus\{c\}$, it is right-continuous at $c$, and $\tB$ maps both boundary points to a single point in the interior of $I$. Moreover, this map does not comprise fixed points.

An orbit of a point $y$ is given by the set of its iterations, say, $\mathcal{O}(y) = \{H^k(y)\}_{k \in \NN},$ where $\tB^k(y)$ stands for $k$-th iteration of $y$.
Upon taking into account this set, we notice that
\begin{equation}\label{eq:kmap}
\tB^k(y) \;=\;
\begin{cases}
  3^ky/2^{\mu(k)},   & \text{if }\; y \in [1/2,\,2^{\mu(k)}\!/3^k) \\
  3^ky/2^{\mu(k)+1}, & \text{if }\; y \in [2^{\mu(k)}\!/3^k,\,1)
\end{cases}, \;\text{ where }\; \mu(k) := \bigg\lfloor{k \frac{\ln 3}{\ln 2}}\bigg\rfloor.
\end{equation}
Thus, its critical point $c_k$ is $2^{\mu(k)}\!/3^k$, and we notice that there are no fixed points of $\tB^k$ with $k \in \NN$. As consequence there are no periodic points. In addition, it is \textit{topologically transitive} as the orbit of any open interval will cover eventually the whole interval $I$ when $k \to \infty$.
Now we define the inverse map $\tB^{-1}:I \to I$ such that $y \in \tB^{-1}(\tB(y))$ and $\tB^{-1}(J)$ is the set given for all values $z \in I$ such that $\tB(z) \in J$.

Therefore, $\tB(y)$ is a well defined circle map without chaotic behaviour. This clearly is in contradistinction to $B(y)$ as map. Nonetheless, we get information from $\tB(y)$ in order to investigate the veracity of Collatz conjecture.

\subsection{The binary map as a perturbation}
\noindent
As an illustration of $\tB(y)$ dynamics, let us choose as a motivation the set $T_5$ given by all numbers that can be expressed with binary length $\ell = 5$. This set is written with the first element $\tau_1 = 1/2$ and recursively with $\tau_{i+1} = \tau_i + 2^{-5}$. Thus, we have the representation $T_5 \;=\; \{\tau_1 = (0.10000)_2,\, \tau_2 = (0.10001)_2,\, \dots,\, \tau_{16} = (0.11111)_2\}$,
related with a disordered list of the odd numbers from $1$ to $31$ as $\tau_1 = M(1)$, $\tau_2 = M(17)$, $\tau_3 = M(9)$, $\tau_4 = M(19)$,
\dots, $\tau_{16} = M(31)$. Notice that $\tau_1 = (0.1)_2$, $\tau_5 = (0.101)_2$ and $\tau_6 = (0.10101)_2$ are in $\mN$, which have orbits $\mathcal{O}(\tau_i) = \{\tau_i,\,\tau_1,\,\dots\}$. Direct computations show that $B(\tau_i) \notin T_5$ for $i = 7,\,11,\,16$, however, their orbits also converge to $\tau_1$ for these cases,
\begin{eqnarray*}
 \mathcal{O}(\tau_7) &=& \{\tau_7   ,\,(100011)_2,\,(110101)_2,\,\tau_5,\,\tau_1,\,\dots\}, \\
 \mathcal{O}(\tau_{11}) &=& \{\tau_{11},\,(101001)_2,\,\tau_{16},\,\dots,\,\tau_1,\,\dots\},
\end{eqnarray*}
where the orbit of $\tau_{16}$ has $39$ iterations before $\tau_1$. In three occasions, the iteration $H^k(\tau_{16})$ has $12$ digits length.

To compute $\tB(y)$ instead of $B(y)$, an error can be estimated for $y \in J := I \setminus (T_5 \cup \mN)$, as we know that $\tB^k(y) \to 1/2$ for $y \in T_5 \cup \mN$. Notice that the calculated length for any $y \in J$ is $\ell \geq 6$, thus
$$\varepsilon_1 \;:=\; B(y) - \tB(y) \;\leq\; \left\{ \begin{array}{ll}
  2^{-7}, & \text{if }\; y \in L\setminus J \\
  2^{-8}, & \text{if }\; y \in R\setminus J \\
\end{array}\right\} \;\leq\; 2^{-7}.$$
As $B(y)$ is larger than $\tB(y)$, the absolute value is not included.

We define an error bound $\varepsilon_k := B^k(y) - \tB^k(y)$ for $k \in \NN$ and $y \in I$ on computing of $\tB(y)$. The orbit is well understood for $y \in I$ such that $B(y) \in T_5 \cup \mN$, so in each iteration we remove these relevant data from $I$. An inspection of plots for $B$ and $\tB$, see Fig.~\ref{fig:error}, shows that $y$ in $L = [1/2,\,2/3)$ satisfies $3y/2 = \tB(y) \approx B(y)$ and $y$ in $R = [2/3,\,1)$ satisfies $3y/4 = \tB(y) \approx B(y)$. Moreover, when $y \in L$ their images $B(y),\,\tB(y) \in R$, but the converse does not hold since $\tB(y) \in [1/2,\,3/4)$ for $y \in R$ although $B(y) \in [1/2,\,3/4 + \epsilon)$, where $0 \leq \epsilon \ll 1$.

The alternation of multiplying by $3/2$ or $3/4$ is the same in $\tB$ and $B$, once we remove conflicted data given by the numbers $\gamma_k$ ahead. In so doing, we notice that $\tB(y) \in L$ when $y \in [2/3,\,8/9)$, however for $2/3 \ll y < 8/9$, $B(y)$ may belong to $R$. As $8/9 = (0.\{111000\}^\infty)_2$ for an infinite repetition of the curly bracket, and $y = (0.\{111000\}^k0*\cdots*1)_2 < 8/9$ implies $B(y) \in L$,
we concern with three types of smaller numbers near to $8/9$ in binary form:
$$\alpha_k \;=\; (0.\{111000\}^k1)_2,\quad \beta_k \;=\; (0.\{111000\}^k11)_2,\quad \gamma_k \;=\; (0.\{111000\}^k111)_2,$$
where $k$ is the number of repetitions of the digits in curly brackets. From calculations: $B(\alpha_k) = (0.1\{01\}^{3k})_2 = \ys_{3k}$ and $B(\beta_k) = (0.1\{01\}^{3k+1})_2 = \ys_{3k+1}$, thus $B^2(\alpha_k) = B^2(\beta_k) = (0.1)_2$. Then $\alpha_k$, $\beta_k$ belong to the preimage of $\mN$ under $B$, see Remark~\ref{rem:predecessors}.
As a $\gamma_k$ satisfies $B(\gamma_k) > 2/3$ without a clear convergence to $\mN$, we denote by $\mC$ to the set of all $\gamma_k$.

Let $T_\ell$ to be a \textit{tested set} with length $\ell$: all natural numbers $x \leq 2^\ell$ give $y = M(x)$ with length $\ell$, the set is tested if the orbit of $y \in T_\ell$ contains $\tau_1$. Then, recursively define $J_1 := I \setminus (T_\ell \cup \mN \cup B^{-1}(\mN) \cup \mC)$ and $J_n := J_{n-1} \setminus B^{-1}(J_{n-1})$.

\begin{conjecture}\label{con:C}
For any $y \in \mC$, with preimage $x$ in $\NN$, there exists $k \in \NN$ such that $B^k(y) = 1/2$. Therefore, $\mC$ satisfies the Collatz conjecture.
\end{conjecture}

\begin{remark}
As for a number $z = n/3^k > 1/2$ with $n \in \NN$ we have that $H^k(z)$ has also a preimage $M(x)$ for some $x \in \NN$, then Conjecture~\ref{con:C} should hold for $z$.
\end{remark}

\begin{theorem}
The error bound $\varepsilon_n := B^n(y) - \tB^n(y)$ for any given $y$ in the subset $J_n \subset I$ defined above is
$$0 \leq \varepsilon_{2n} \leq 7 \left[\left(\frac{9}{8}\right)^{\!\!n} - 1\right] 2^{-\ell}, \qquad
0 \leq \varepsilon_{2n+1} \leq \left( \frac{15}{2} \left[\left(\frac{9}{8}\right)^{\!\!n} - 1\right] + \frac{1}{2}\right) 2^{-\ell},$$
for $\ell-1$ the smallest length of the elements of the tested set $T_\ell$.
\end{theorem}

\begin{proof}
In order to show this result, we highlight two facts. Firstly, notice that the difference between $B(y)$ and $\tB(y)$ for $y \in J_1$ is $2^{-(m+1)}/2^i$ where $m + 1 \geq \ell$ and $i$ is equal to $1$ or $2$, as $J_2$ neglects the elements that will be mapped into $J_1$, the length of any $y \in J_2$ satisfies that the length of $B(y)$ plus one is also larger than $\ell$. Secondly, $B^n(y)$ and $\tB^n(y)$ have the same power $i$ in the same step; both $B^{n-1}(y)$ and $\tB^{n-1}(y)$ are in the same subinterval $L$ or $R$ as we neglected the elements in $\mC$ and the preimages of $\mN$. Therefore, in the expansion of the difference between $B^n(y)$ and $\tB^n(y)$, we can rule out the $y$'s, \eg,
\begin{eqnarray*}
\varepsilon_3 &=& \frac{\displaystyle 3  \frac{\displaystyle 3 \frac{3y + 2^{-(m_1+1)}}{2^{i_{1}}} + 2^{-(m_2+1)}}{2^{i_2}} + 2^{-(m_3+1)}}{2^{i_3}} - \frac{3^3 y}{2^{\mu(3) + a}} \\
 &=& \frac{\displaystyle 3  \frac{\displaystyle 3 \frac{2^{-(m_1+1)}}{2^{i_1}} + 2^{-(m_2+1)}}{2^{i_2}} + 2^{-(m_3+1)}}{2^{i_3}},
\end{eqnarray*}
where $a$ is either $0$ or $1$, see Eq.~\eqref{eq:kmap}, however $i_1 + i_2 + i_3 = \mu(3) + a$ holds.

Now, notice that $2^{-(m+1)} \leq 2^{-\ell}$ in each iteration.
Recall that the powers $i_1,\,i_2,\,\cdots$ alternate from $2$ to $1$ or $2$ and from $1$ to $2$, thus the bound is enlarger when $i_n = 1$, $i_{n-1} = 2$, $i_{n-2} = 1$, $\cdots$.
These errors are bounded by setting all lengths $m+1$ as $\ell$ and the power laws alternating as $1,\,2,\cdots,\,1,\,2,\,1$ where the last one must be $1$. In other words, we split in even an odd iterations as follows
\begin{eqnarray*}
\varepsilon_{2n+1} &\leq& 2^{-1}(3\cdot 2^{-2}(\cdots(3\cdot2^{-2}(3\cdot2^{-1} + 1) + 1)\cdots) + 1) \, 2^{-\ell} \\
 &=& \left( \left[ \sum_{k=1}^n 3^{2k}2^{-(3k+1)} + 3^{2k-1}2^{-3k} \right] + \frac{1}{2} \right) 2^{-\ell} \\
 &=& \left( \frac{15}{2} \left[\left(\frac{9}{8}\right)^{\!\!n} - 1\right] + \frac{1}{2}\right) 2^{-\ell}, \\
\varepsilon_{2n}   &\leq& 2^{-1}(3\cdot 2^{-2}(\cdots(3\cdot2^{-1}(3\cdot2^{-2} + 1) + 1)\cdots) + 1) \, 2^{-\ell} \\
 &=& \left[ \sum_{k=1}^n 3^{2k-1}2^{-3k} + 3^{2(k-1)}2^{-3k+1} \right]2^{-\ell}
 \;=\; 7 \left[\left(\frac{9}{8}\right)^{\!\!n} - 1\right] 2^{-\ell},
\end{eqnarray*}
which prove the proposition.
\end{proof}

\begin{figure}[htb]
\centering
\includegraphics[width = 0.32\textwidth]{Error1}
\includegraphics[width = 0.32\textwidth]{Error2}
\includegraphics[width = 0.32\textwidth]{Error41}
\caption{\small Samples of $B^k(y)$ as perturbation of $\tB^k(y)$, shaded regions are potential places for the evaluations of the binary map. Left-hand and middle panels have the exact values of $B^k(\tau_i)$ for $\tau_i \in T_5$. (The error bound is twice for graphical purposes.) Here cases $k = 1,\,2,\,41$; the latter has $c_k$ near $1/2$; actually the next iteration runs far away the identity, since $c_{42} = 0.674352\dots$.}
\label{fig:error}
\end{figure}

Upon following these results, we notice that the binary map $B(y)$ do not have periodic points for $n \to \infty$. However, as $c_k$ can be close to $1/2$ as is shown in Fig.~\ref{fig:error}(b,c), the binary map may have periodic points of period $k$ if $1/2 + \varepsilon_k > c_k$ holds. In consequence we have the following result.

\begin{lemma}
\label{lem:tested}
Upon assuming a tested set $T_\ell$ consisting of all the binary expressions with length $\ell - 1$, and considering that the orbit of every $y \in T_\ell$ contains $\tau_1 = 1/2$, then there are no periodic orbits of $B$ of period $k < \ks$, where $\ks$ is the first number satisfying $1/2 + \varepsilon_{\ks} > c_{\ks}$.
\end{lemma}

\begin{remark}
The estimation in Lemma~\ref{lem:tested} shows that there are no periodic points in $\mC$. Notice that as the convergence test has been double checked for all $x \leq 2^{60}$ (\it{cf.} \cite{Roos}), thus we compute $\ks$ with these data showing that there are no periodic orbits of period $k < 600$. In this case, $\ell = 60$, and we have $c_{600} = 0.507858\cdots$ and $\varepsilon_{600} = 0.013460\cdots$ satisfying $c_{600} < 0.5 + \varepsilon_{600}$ for the first $k$.
\end{remark}

\section{Conclusions}
\label{sec:conclusions}

\noindent
This problem and its conjecture began as a curiosity of an arithmetical game, but it is in connection with various other areas of mathematics that have made it a respectable topic for mathematical research. Here we have presented a novel link with an one-dimensional dynamics which in turn has restricted the problem in showing that $\mC$ satisfies Collatz conjecture. This approach seems promising as we have proposed a strategy to remove an overseen difficulty which appears to reside in an inability to analyze the pseudorandom nature of successive iterates of $C(x)$. Here, its randomness can be controlled by the error bounds $\varepsilon_k$.

We have highlighted a path concerning the length in binary expressions which it seem to overpass the hailstone numbers given in \cite{Hay84} as well as a novel procedure that for a given set of tested numbers satisfying the conjecture, eliminates several periods for orbits. Hence, a proof for Conjecture~\ref{con:C} is needed to rigorously make this statement.

During the revision time of this manuscript, Hew's article \cite{Hew19} was published.
It shows that the binary map $B(y)$ in \eqref{eq:binary}  is actually a fractal under the metric of the dyadic rational numbers in $[1/2,\,1)$, \emph{i.e.},
the odd natural numbers via the $M(x)$ map. The incomplete analysis for string length of $y \in [1/2,\,1)$ under $B(y)$ is natural through Theorem~\ref{aff:h&t}: Hew
removes the ending repetition of $01$, therefore, instead of $y = (0.1* \cdots *11\{01\}^k)_2$, take $\hat{y} = (0.1* \cdots *11)_2$;
only the tail $t_3$ is prohibited. Then, for the $L$-domain, the $h_1$ and $h_2$ heads are possible, so by adding the allowed tails, the string length can grow or shrink.
For the $R$-domain, the $h_2$, $h_3$, $h_4$ heads are allowed and the string length grow or keep the same length.

\section*{Acknowledgments}
\noindent
I am grateful to V\a'ictor F. Bre\~na-Medina (\,ITAM\,) for detailed readings and corrections.
I am also grateful to the anonymous DCDS reviewer for his enlightening comments.
The author was partially supported by Asociaci\'on Mexicana de Cultura A.C.

redpajama_set_name: RedPajamaArXiv


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text:
Mount Morris Illinois Education Attainment Charts
This section of charts contains education data for Mount Morris Illinois based mainly on the latest year 2020 American Community Survey census data but also the survey from Common Core Data available for Public Elementary through Secondary Schools. In Figure 1, the percent of all people aged 25 years or older, who have either graduated from high school or completed the Graduate Equivalency Degree (GED) or some equivalent certification/credential. Mount Morris depicts it has a High School Grad or higher of 90% which is in the middle range of other places in the area. The city with the highest percent high school graduates or better in the area is Lost Nation CDP which depicts a high school graduates of 97% (8.1% larger).
In Figure 2, the percentage of people aged 25 years or older who have graduated from college/university with at least a bachelor's degree is provided. Note that the bachelor's degree is also called a four-year degree because it normally takes four years of full-time study to finish the course curriculum required to obtain the degree. This chart portrays the proportion of the population in this region who are college graduates with at least a bachelor degree or higher. In many ways, this analysis alongside the prior chart are very quick measures of the level of education in any particular area. Mount Morris shows it has a Bachelors Degree or higher of 27% which is at the top of all other places in the greater Mount Morris region.
Figure 4 shows the percentage of people 25 years of age or older who either have no schooling at all or dropped out of school before being able to complete high school. Additionally, these are people who also do not have a Graduate Equivalency Degree (GED) or some other high school level type credential/student achievement. This analysis, along with the last two charts, provides another high level review and comparative data on the Mount Morris Illinois educational level. Mount Morris depicts it has a School Dropout Rate of 10% which is in the mid range of other places in the local area. The city with the highest percent who dropped out of school in the area is Adeline which depicts a percent who dropped out of school of 14% (42.1% larger).
Figure 5 provides a more detailed look at the educational attainment for Mount Morris Illinois. This chart provides the proportion of people aged 25 years of age or older and what was their level of educational attainment. The chart provides 5 broad categories including: No Education/No School, Some High School, High School or equivalent, Some college or Associates Degree, and Bachelors Degree or higher. Mount Morris has the smallest proportion of percent of people with high school degree, ged, or equivalent at 25% of the total. Second, it has the largest proportion of percent of people with some college or an associates degree at 27% of the total and is ranked #1.
The next chart shows a break down of people who have received a bachelor's degree or higher advanced degree generally in a campus learning environment. In particular, Figure 6, provides the proportional breakdown of all the people who have received a postsecondary education along with what the level of advanced degree that was obtained. Note that these categories do not include any type professional development type activities such as those related to maintaining professional credentials in workshop lessons. Note Professional Degree includes medical, dental, lawyers, etc. Mount Morris has one of the largest proportions of percent of people with an associate degree at 47% of the total and is ranked #3. Only #2 Oregon (51%), and #1 Leaf River (53%) are larger. Second, it has the largest proportion of percent of people with a bachelors degree at 37% of the total and is ranked #1.
The chart in Figure 7 shows the broad area of academic concentration or the discipline for people who have received a bachelors degree. This high level classification is essentially the field of study for which a degree was obtained. Mount Morris has the percentage of percent of people with a science or engineering degree the second smallest in terms of percent of people with a science or engineering degree of all the other places in the local area at 6.5% of the total. Second, it has less than most other places in the greater region when sorted by percent of people with an arts, humanities, or other degree at 15.7% of the total. Third, it has the largest proportion of percent of people with a business degree at 40.9% of the total and is ranked #1.
The next chart (Figure 8) provides a more detailed deep dive on the category of major degree obtained for people aged 25 years or older who earned a bachelor's degree or higher. Specifically, this frequency distribution details out what the major field of study was the degree obtained. Mount Morris has one of the largest proportions of percent of people with a degree in psychology at 9.5% of the total and is ranked #2. The only larger city being Oregon with 11.4%. Second, it has the smallest proportion of percent of people with a degree in social sciences at 1.0% of the total. Third, it has less than most other places in the local area when ranked by percent of people with a degree in science and engineering related fields at 15.7% of the total. Also, it has the largest proportion of percent of people with a degree in business at 40.9% of the total and is ranked #1.
Figure 9 provides comparative data between the places in the greater Mount Morris Illinois region for broad educational attainment. This analysis uses provides five broad education attainment categories including: No Education/No School, Some High School, High School or equivalent, Some college or Associates Degree, and Bachelors Degree or higher. Mount Morris has the smallest proportion of percent of people with high school (or ged) at 24.9% of the total. Second, it has the largest proportion of percent of people with some college or an associates degree at 27.4% of the total and is ranked #1.
A more detailed frequency distribution of educational attainment is provided in Figure 10. In particular this illustration breaks out the highest levels of university educational opportunities beyond the four year college degree. Included in the breakout are the relative proportion of masters degrees, PhD/Doctorate/Doctorial holders, and professional degrees such as medicine, dentistry, lawyers, etc. Mount Morris has the smallest proportion of percent of people with some college but not a graduate at 3.7% of the total. Second, it has the largest proportion of percent of people with a bachelors degree at 11.6% of the total and is ranked #1. The next exhibit (Figure 11) provides detailed cross tabulation analysis that provides education success data broken out or cross tabulated by age group. Please note that the columns add to 100% and you must use the pagination buttons at the bottom of the table to see all the rows. Figure 12 is a cross tabulation analysis that shows large educational success categories and is broken out or cross tabulated by racial group. Please note that the columns add to 100% and you must use the pagination buttons at the bottom of the table to see all the rows.The final cross tabulation analysis is provided in Figure 13 and shows education success broken out by gender. Please note that the columns add to 100% and you must use the pagination buttons at the bottom of the table to see all the rows.
Mount Morris Illinois School Enrollment Charts
The next section of chart resources look at school enrollment by a variety of educational institutions and are categorized into a number of other groupings. Figure 14 provides the overall school enrollment by broad range of school age/level groupings. Mount Morris has the percentage of percent of children in kindergarten in the mid range of other places in the local area at 21.0% of the total. Second, it has the smallest proportion of percent of children in grade 1 to 4 at 19.5% of the total. Third, it has the largest proportion of percent of people in undergraduate colleges at 15.9% of the total and is ranked #1.
Figure 15 provides a simple high level comparison of the proportion of students that are enrolled in public schools versus students enrolled in private schools in the Mount Morris Illinois region. Mount Morris indicates percent enrolled in public schools very much bigger as the percent enrolled in private schools.
The next comparison provided in Figure 16 shows the proportion of students enrolled in private schools versus students enrolled public schools for students enrolled in Kindergarten through 8th grade (grade 8) or (K through 8). Mount Morris depicts percent of children enrolled in public k-8 grades very much bigger as the percent of children enrolled in private k-8 grades.
Figure 17 provides a comparative analysis all students enrolled in a public high school versus those students who are enrolled in private high school schools. Mount Morris depicts percent enrolled in a public high school considerably bigger as the percent enrolled in a private high school.
The next chart (Figure 18) provides a similar analysis to the last chart but provides the data for public post secondary education versus private higher education colleges. Mount Morris depicts percent enrolled in a public college considerably bigger as the percent enrolled in a private college.
Figure 19 shows another view of students enrolled in public or private schools-postsecondary schools or colleges/universities. This analysis shows the proportion for all graduate and professional schools (such as law or medical school) and whether those students are enrolled in private or public institutions. Mount Morris depicts percent enrolled in a public gradudate or professional school 6.4% larger as the percent enrolled in a private gradudate or professional school.
The next chart in this series of resources shown in Figure 20 looks at the total number of students enrolled in any educational institution for each place in the greater Mount Morris region. (Total enrollment in this case includes all students from preschool all the way through students enrolled in graduate school.) Mount Morris depicts it has a Total Enrolled of 711 which is the second most total enrolled of all the places in the greater Mount Morris region. The city with the highest total population enrolled in school in the area is Oregon which shows a total population enrolled in school of 865 (21.7% larger). Comparing total population enrolled in school to the United States average of 81,084,866, Mount Morris is only about 0.0% the size. Also, compared to the state of Illinois, total population enrolled in school of 3,228,360, Mount Morris is only about 0.0% the size.
Mount Morris Illinois Area Schools Charts
Figure 23 lists all the schools in the area along with the school district, county location and other program information/credential such as if they are a public charter school or private charter school or magnet school. Some of the Area Schools are: David L Rahn Jr High School, Oregon Elem Sch, Oregon High School, Polo Comm High School, and Aplington Middle School. The next illustration in Figure 24 shows the total child school enrollment for all grades (through 12th grade) at the school shown using NCES data (Common Core of Data, Public Elementary-Secondary School Universe Survey.) Looking at Enrollment for Area Schools we find that Oregon Elem Sch ranks the largest with a value of 779 enrolled students. The next largest values are for: Mary Morgan Elem Sch (660), Byron High School 9-12 (507), Oregon High School (492), and Forreston Jr/Sr High Sch (464). The difference between the highest value (Oregon Elem Sch) and the next highest (Mary Morgan Elem Sch) is that the enrolled students is about 18.0% larger.
Figure 25 show the ratio of the number of students to the number of teachers in the classroom. A good student to teacher ratio should be low because it indicates that there are less students for any one teacher to educate in a class and generally a better learning environment, better success and optimal teaching excellence. Teachers includes all educational staff such as special education teachers and any other educator. Note that distance learning (online learning/remote learning) is not included in these values. Looking at Student to Teacher Ratio for Area Schools we find that Chana Educ Center/Rock River ranks the largest with a value of 2.2 student to teacher ratio. The next largest values are for: Nachusa Campus School (4.1), Polo Comm High School (6.7), Highland Elem School (9.7), and Aplington Middle School (10.2). The difference between the lowest value (Chana Educ Center/Rock River) and the next lowest (Nachusa Campus School) is that the student to teacher ratio is about twice as large.
Mount Morris, Illinois Education Data
Figure 1: Mount Morris, IL At least High School Education
Figure 2: Mount Morris, IL Bachelors Degree or Better Education
Figure 4: Mount Morris, IL School Dropout Rate
Figure 5: Mount Morris, IL Education Attainment Breakdown
Figure 7: Mount Morris, IL Bachelors Degrees Field of Study
Figure 8: Mount Morris, IL Bachelors Degree Obtained
Figure 9: Mount Morris, IL Education Attainment by Level Comparison (Age 25+)
Figure 10: Mount Morris, IL Education Attainment Detailed Comparison (Age 25+)
Figure 11: Mount Morris, IL Detailed Education Attainment Breakout by Age Group (Age 18+)
Figure 12: Mount Morris, IL Detailed Education Attainment Breakout by Race (Age 25+)
Figure 13: Mount Morris, IL Detailed Male and Female breakdown of Educational Attainment
Figure 14: Mount Morris, IL School Enrollment by Aggregate Categories
Figure 15: Mount Morris, IL Overall Public vs. Private School Enrollment
Figure 16: Mount Morris, IL Public vs. Private K-8 School Enrollment
Figure 17: Mount Morris, IL Public vs. Private High School Enrollment
Figure 18: Mount Morris, IL Public vs. Private College Enrollment
Figure 19: Mount Morris, IL Public vs. Private Graduate or Professional School Enrollment
Figure 20: Mount Morris, IL Total Enrolled in Schools
Figure 22: Mount Morris, IL Public vs. Private Preschool
Figure 23: List of Schools in the Mount Morris, IL Area (2013)
Figure 24: Mount Morris, IL School Enrollment (2013)
Figure 26: Mount Morris, IL School Racial Mix (2013)
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LevelUp Will End Your Business' Credit Card Processing Fees. Really.
Merchants shell out $50 billion a year to the companies that process your credit card payments. One company, LevelUp, is dropping its fees altogether. Here's how they're getting away with it.
By Christina Chaey 3 minute Read
Credit card processing fees are such a common part of the consumer-merchant retail experience that we hardly give them any thought. So we tend to forget that it costs merchants money every time we pay with our credit cards, or that the extra cost is then passed along to us in the form of a higher price.
Starting today, LevelUp, a service from Cambridge-based SCVNGR that lets you make mobile payments via a QR code linked to your debit or credit card, is eliminating processing fees for its merchants, making it the first payments company to nix charges that cost merchants $50 billion each year. SCVNGR's Chief Ninja Seth Priebatsch refers to the concept as "Interchange Zero," and says it's the next step toward building an ecosystem in which payment companies' worth is measured only by how much value they can provide to their merchants beyond simply helping them make transactions.
"Right now, the competition is over who's the most entrenched or who's the cheapest," Priebatsch tells Fast Company. "But eventually, who gets to move money will be about who can create the best value to add to the business, which isn't a bad thing at all."
The logic behind Interchange Zero is simple: Decades ago, processing fees were a necessary evil because they covered material costs to lay the physical lines that connected merchants to payment networks and develop anti-fraud technology. Then innovators in the mobile payments sector started to crop up, from PayPal to Square, and figured out how to lower processing costs even further. But the payment companies kept interchange rates between 1 and 3 percent, meaning they continued to profit without providing extra value to merchants. That's where Interchange Zero comes in.
But eliminating transaction fees also happens to zap one of LevelUp's major revenue sources, because it's no longer charging merchants a flat 2% transaction fee on the $2 million it currently processes each month. So how can LevelUp conceivably expect to stay afloat? Priebatsch says the answer is LevelUp's suite of merchant campaigns, including customer acquisition and loyalty programs. Now, when a merchant uses a LevelUp campaign–say a store offers $10 for every $100 you spend–LevelUp will make 35 cents for every dollar of credit redeemed through that campaign. That way, a large part of LevelUp's revenue is its ability to help its merchants make more sales.
Priebatsch's hope is that others in the mobile payments industry will follow suit, eventually making the idea of charging merchants to move money a thing of the past. Of course, if the idea of Interchange Zero does catch on, LevelUp will have to create more sophisticated campaigns to keep competitive. For example, Square recently added an in-app punchcard feature to its Pay With Square app; Starbucks just launched mobile payments on Android. Currently, 66% of LevelUp customers return within 30 days. That's impressive, but "66 isn't 100," Priebatsch says, which is why LevelUp is constantly devising new solutions for merchants that push the boundaries of a typical rewards program. Right now, they're testing a campaign that offers special incentives to drive customers to stores when it's raining, when sales tend to suffer.
LevelUp is also partnering with banks to create custom loyalty programs for certain cardholders. Sovereign Bank noticed its cardholders using LevelUp were active purchasers, but few in numbers. So last month, LevelUp ran a campaign that gave Sovereign cardholders $5 in free credit as an incentive to link their card. And it expects to have similar partnerships with about two to three dozen banks by the end of the year, Priebatsch says.
The question is how many businesses will be interested in adopting the setup LevelUp offers. It's a natural fit for small businesses, which make up the majority of LevelUp's 3,000-strong merchant network, but you're not likely to start waving QR codes around at Walmart or Costco anytime soon. But Priebatsch says LevelUp is about to sign its first retailer with more than 500 stores, and they now serve more transactions at a given partner business than American Express in a 30-day period.
"Now we're able to start approaching everyone but the top merchants in the country and have serious conversations with them," he says. "Of course, not all of them will say yes."
[Image: Flickr user WSDOT]
Christina is an associate editor at Fast Company, where she writes about technology, social media, and business.
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A police officer was in critical condition last night after she was injured in a bizarre accident yesterday in Brooklyn.
Investigators say that at 2:11 p.m., the unnamed off-duty officer parked her black sedan at Whitney Avenue and Knapp Street in Gerritsen Beach. But the sedan began to roll after the officer got out. She tried to get back inside to apply the brakes, but got pinned between the sedan and another parked car.
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PlayStation Experience 2017 will begin on December 9; to also host Capcom Cup 2017
By Vishal Laul Neowin @vishal1082 · Aug 17, 2017 12:56 EDT with 0 comments
PlayStation Experience (PSX), the annual Sony event where game developers and publishers reveal their up-and-coming games for the PlayStation 4, is returning for its fourth year, this December.
This year's event will once again be taking place at the Anaheim Convention Center, just a few miles off the city of Los Angeles in California.
The two-day event will begin on the 9th of December, but Sony will also host a special event on the evening of Friday the 8th, where fans will be able to "get up-close and personal with game developers and PlayStation personalities."
Unfortunately, not everyone will be able to attend the special event; the price for a regular two-day ticket is $75, and does not include access to the special event – that is exclusive to the discounted $65 'early bird' ticket, which is available only in limited quantities.
Sony will also – once again – host the Capcom Cup 2017 tournament at the event, with a special stage dedicated to Street Fighter V where competitions will be taking place for an entire day on the 10th – the last day of the event. The players will battle for a prize pool of more than $250,000, with the winner being crowned as the '2017 Street Fighter V World Champion'; proceeds from the 2017 Capcom Pro DLC go towards increasing the tournament's prize pool.
Those who wish to attend the event can purchase the tickets over here; for those who may not be in attendance – it would perhaps be wise to keep a look-out for the PSX press conference on the first day of the event.
At PSX last year, we had our first look at the second installation for The Last of Us; it will be interesting to see what Sony and other game publishers have up their sleeves for 2018.
Source: PlayStation Blog
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