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As Teens Return To School After Holidays, Stress Levels Are Highest Ever; 58% Report Knowing Someone Who Has 'Considered Self-Harm'
Tigray Hospital In Urgent Need Of Assistance With Food And Medical Supplies
Controlling Inflation Requires A Better Policy Mix, Not Business Scapegoating
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Voting Rights And Public Opinion
Rep. Emmer Makes Huge Statement On Central Bank Digital Currency
Feb 15, 2021, 11:32am EST |
Anonymity More Important Than Ever For Charitable Giving
Carrie Lukas
Governor Kristi Noem came under attack last week, supposedly for advancing legislation to protect "dark money." That's the term used to make anonymous giving to charities sound sinister.
But Americans ought to appreciate now, more than ever, why anonymity is so important. Today, people routinely lose jobs, positions of influence, and face other forms or social sanction for comments or associations they make, even those that are decades old. To take one recent example, Cara Dumaplin, the woman running Taking Cara Babies, a popular sleep-training service, was exposed as having made a political donation to the Trump campaign. The result? Her copyrighted materials were made public and the social media mob pushed to have her "canceled."
Woke progressive may think that Dumaplin deserved it for supporting President Trump. But can they be confident that the donations they make today won't become cause for cancellation in the future?
One can easily imagine how giving to a seemingly uncontroversial nonprofit (say, one that supports medical research) could become controversial in the future, if someone associated with that charity later says something controversial. When Mozilla CEO Brendan Eich donated to California's Proposition 8 in 2008, he may have considered himself in good company — after all, back then Barrack Obama was opposed to gay marriage. But six years later, Eich was forced to resign when the contribution became public.
Today, some politicians want to destroy freedom of association by outlawing anonymous giving to policy-minded non-profits. Yet, these are exactly the types of charities where anonymity is desperately needed. Imagine, for example, a donor who wishes to support Planned Parenthood but is a member of a particular faith community or lives in a conservative area. A woman who wants to support an organization backing legal protections against female genital mutilation, honor killings, and arranged marriages, but whose family and community supports these traditions. Or a donor who wants to give to a fund for fallen police officers but worries that their charity will wind up in the crosshairs of Black Lives Matter, or even lead to charges of racism.
The question is, how many donors will decide that giving isn't worth the risk?
Anonymous support for controversial causes has been critical throughout our history. Philanthropist Olivia Sage publicized her charitable gifts to universities, hospitals, and other organizations, but insisted that her gifts to suffrage organizations be kept anonymous. In the mid-20th century, the NAACP fought and won in the Supreme Court to keep its membership its secret. The Supreme Court ruled that Alabama could not compel the civil rights champion to reveal the names and addresses of its members, recognizing that doing so would expose them "to economic reprisal, loss of employment, threat of physical coercion, and other manifestations of public hostility" and thereby restrain "their right to freedom of association." These are exactly the threats that donors face today if stripped of anonymity.
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Donors also want anonymity for other, very legitimate reasons: Someone giving millions of dollars to one charitable cause may not want to explain why other charities that had sought her support did not receive it. Big givers often do not want to face endless asks from other causes.
Americans are the most charitable people on Earth, together giving away $428 billion in 2018. These charitable dollars make possible a wide variety of activities, supporting everything from environmental causes to animal shelters, churches and other religious institutions, educational efforts, and caring for the poor and infirm. Americans shouldn't have to worry about the potential fallout from having their charitable giving exposed.
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Diy Home Alarm Systems - With out even giving it much thought . we use the doors within the home alot. New doorways may either be an important addition, or even perhaps a hindrance, dependent on the dimensions, shape, materials, how well they're created, and also how they fit. Instead of installing cheap doors that are overweight or do not close or open properly, why don't you place in time effort to buy fit ones that do not cause you any problems. It's certainly rewarding to achieve that. When we require only a little extra moment to look at each of the choices, materials, colours, and styles of the home's outside doors, they are able to say a lot regarding our dwelling and also its particular characteristics. Let us face it they are first item company sees when they encounter. Needless to say, a great sturdy doorway maybe not only looks pleasing, but it holds up to continuous use, the elements, and even retains us protected against break-in attempts.
The materials and finish of the do or are all nice, but aren't quite as crucial as how properly the do or is assembled. Glass doors, say, may be manufactured of exceptionally sturdy shatter-proof glass. Some wood doorways, on the flip side, can be quite thin and flimsy. When picking a doorway made of wood, make it obviously a dense hardwood, such as walnut, cherry, walnut, cherry or walnut. The thickness and width of the door's railings and stiles are critical aspects also. And don't forget about correct weatherstripping to keep the elements. For colder weather places, insulated doorways of alloy or fiberglass certainly are recommended. An adequately engineered alloy or fiberglass door provides increased protection against frigid winter months and blistering winds, using an inside tier of sustainable insulation. Without such a well insulated doorway, frost and moisture can creep into the door-frame and home, potentially leading to structural damage over time.
Now, let's look in the 4 principal kinds of inside doors you are able to purchase for your dwelling. Included in these are folding, skip . pocket, and conventional hinged doors. A folding doorway operates somewhat like your paper fan, that literally folds and unfolds. They truly are typically used for large closets that have limited shelving, or at other restricted places every time a conventional entrance do or would not swing open fully. A pocket door is used for cabinets, also, or different such areas such as storage. Such a magnificent doorway is normally wrapped with wheels into a track, at the door moves forth and back, slipping right into a specific pit or"pocket" spot whenever available. Bypass doors similarly proceed together the same kind of track system, and are useful for the exact purposes. The difference is there are at least two doors on separate tracks that move behind or facing the different doors though available.
In short, good preparation and preparation would be the key - obtaining the appropriate doors first time indicates not having to devote a ton of money on repairs or replacements, but not to mention that the time that it takes to put in them. Get every one of the specific dimensions you need before setup to guarantee a proper and efficient fit. Focusing to the door's specifications, and visual appeal, will help you to get the best doors for the house which you may appreciate for years in the future!
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The odds are you've reserved the right to suspend employees accused of misconduct whilst you investigate the allegations against them, either within your disciplinary policy or the contract of employment.
If you've not had cause to actually consider the implications of invoking this, now is as good a time as any.
A recent case in the High Court gives a stark warning that suspension should not be a knee-jerk reaction to an allegation of misconduct, particularly where the employee is of professional standing.
Whilst suspension is not a disciplinary sanction, the court reiterated that suspension is not a 'neutral' act. It carries with it connotations and potentially damages an employee's reputation, regardless of what may be contained within the most well-crafted disciplinary policy.
Despite an express right to suspend, an employee who feels that this has been exercised in breach of the implied duty of mutual trust and confidence might well feel aggrieved, leading to potential claims.
We're certainly not saying you cannot suspend an employee. It's often the best thing to do but let's assume a worst case scenario that you end up in front of a court or tribunal. The judge will want to see the evidence to determine whether you acted reasonably, so having your ducks in a row is crucial.
Do you have the right to suspend?
What are the allegations and are they substantive? Speak to the employee about the allegations and consider what they say.
Is there a potential danger to other employees, clients, customers or the public or is the investigation likely to be prejudiced by the employee remaining on normal duties?
Are there any alternatives: a move to another role, supervision, working from another location?
Confirm in writing, keep records and keep suspension under review.
If you require help with potential misconduct issues, including suspending employees, anyone within the Employment team will be pleased to advise.


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Cetronia Ambulance Corps recognizes Ricky Menendez as a "STAR" Paramedic
Featured Image: Ricky Menendez, Cetronia Ambulance Corps; PA Stars of Life Award
The 13th Annual Pennsylvania Star of Life Event: Recognizes those who serve others in EMS
Allentown, Pa., September 9, 2019 – The 13th Annual Pennsylvania Star of Life Event was held on Saturday, September 7 at Northern Valley EMS in Coplay, Pa. The event is hosted annually by the Ambulance Association of PA's – Pennsylvania EMS Provider Foundation which provides education grants to qualified applicants for the development of emergency medical providers. The PA Stars of Life Event was designed to allow EMS providers to recognize their star employee(s) during an evening celebration which honors them for their dedication and contributions in EMS.
This year, Cetronia Ambulance Corps chose Ricky Menendez as its Star employee for many reasons. Ricky, 38, resides in Whitehall, Pa. and started his working career in construction in his hometown of Brooklyn, NY. His passion for the EMS industry came from a co-worker on the job site who had been a medic on the side. Ricky chose to follow this path and attended EMT school in New York and hasn't turned back since. "I consider myself an adrenaline junkie and I love helping people. Being put into emergency situations makes your thinking skills go into action and I thrive on that type of pressure. Cetronia Ambulance Corps has given me the opportunity to greatly expand on the experience needed to be the best Paramedic I can be. They have allowed me to help train new Medics and EMTs which greatly expands on my management skills that will hopefully lead to even more opportunities for me in the future," said Menendez.
Ricky joined Cetronia Ambulance Corps in June 2006 as an Emergency Medical Technician. He later became a Paramedic in March 2016 and continued participating in employee groups as well as special teams. Ricky also serves as a part-time Paramedic for the City of Allentown.
Ricky has contributed significantly to the growth of advanced life support in the Commonwealth of Pennsylvania. He is a member of the Cetronia Ambulance Corps Mobile Stroke Unit team. This unit was the first of its kind in the state of Pennsylvania, allowing the crew to start high-level diagnostics onboard through the use of CTI and video and telehealth capabilities. Ricky happily received special training through Lehigh Valley Hospital in order to operate the advanced equipment used.
Ricky also volunteered to be part of community Paramedicine. He received additional training to do this and during the past year he has been going out to interact and treat patients in their home as part of the trial program.
Ricky joined the Strike and Bike Teams early in his career and has been deployed twice for Strike Team Operations. He took it upon himself to spearhead a fund-drive to raise money to help the victims affected by the flooding of Hurricane Harvey in Texas, which led to a hard earned donation to the Salvation Army's relief efforts.
Cetronia Ambulance Corps has also assigned Ricky to the Rescue Task Force Team and he has been doing active shooter training to prepare for such tragedies.
Ricky also serves as a field training officer and has received very positive feedback on his role as a preceptor. "He goes above and beyond daily helping new recruits develop into exceptional providers, and thoroughly enjoys coming to work each day. One of Ricky's treasured memories is working with the Miracle League in providing support for their youth program, which gives you a little insight into the type of compassionate employee Ricky has been," said Chris Peischl, Director of Operations for Cetronia Ambulance Corps.
"The PA Stars of Life program has been an excellent way to recruit and retain employees in a world where many EMS agencies face challenges doing just that. We couldn't be more proud of Ricky for his hard work and dedication to our organization," said Larry Wiersch, CEO for Cetronia Ambulance Corps.
About Cetronia Ambulance Corps:
Cetronia Ambulance Corps is a nonprofit 501c3 organization. Serving our community since 1955, we proudly serve Lynn, South Whitehall, Upper Macungie, Weisenberg and Whitehall townships, as well as portions of Lowhill, Salisbury, Greenwich and Albany Townships, and the Borough of Coplay in Eastern Pennsylvania. Our mission is to be the leader in emergency medical services, transportation and community health resources.
Cetronia Ambulance Corps is one of the largest providers of ambulance services in Eastern Pennsylvania; employing 135 full- and part-time career associates, 20 active volunteers and boasting a total fleet of 41 vehicles, including 19 ambulances, 20 para-transit vehicles, a supervisor's vehicle, and a special services unit. As the regional leader in High Performance EMS, Cetronia Ambulance Corps fields over 60,000 calls annually, covering an emergency service area that spans across 130 square miles, touching more than 226,000 people, and a non-emergency coverage area that encompasses five counties. Cetronia Ambulance Corps is proud to support the residents of the Greater Lehigh Valley and Pocono Mountains and remains dedicated to community involvement, outreach and education in the communities it serves.
Information and images provided to TVL by:
Kim Blichar
Director of Marketing & Public Relations
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Boeing Boeing, directed by Jake Van Hoorn, premiers at Corn Stock Theatre in Upper Bradley Park this Friday! The second show (and only play) of Corn Stock's 65th Summer Season is sure to be the perfect date night for you and your special someone. Come out early and enjoy dinner at Qdoba's booth before the show.
It's the 1960s in Paris: bachelor Bernard (played by Jeff Craig) couldn't be happier – a few gorgeous stewardesses, three to be exact, all engaged to him without knowing about each other. How does Bernard manage this? Gabriella (Dani West, making her CST debut) in a blue uniform from Alitalia (Italy), Gretchen (Autumn Vandeven, another CST debut) in yellow from Lufthansa (Germany), and Gloria (Anna Oxborrow) in red from TWA (America) all have different layovers in their flight schedules making scheduling a breeze. Plus Bernard's more than helpful housekeeper, Berthe (Trish Ballard), who always remembers to change out the photos. Things get bumpy as Bernard's old schoolmate Robert (Zachary Robertson) arrives in Paris right as the new and speedier Boeing jet is introduced. When this new Boeing jet brings all three of Bernard's fiancées to Paris, and his apartment, at the same time and catastrophe (and comedy) looms.
Boeing Boeing is a French farce written by Marc Camoletti and adapted for the English-speaking stage by Beverly Cross and Francis Evans, licensed with permission by Samuel French. A film of the same name premiered in 1965 featuring Tony Curtis as Bernard and Jerry Lewis as Robert.
"When I first read the script it really reminded me of the Cary Grant comedies from back in the day and I fell in love with it. Farces are fun to watch in general and that got me thinking that I'd only ever seen farces on a proscenium stage – so what would a farce look like under the tent?" says director, Jake Van Hoorn.
Boeing Boeing lands in the tent from Friday, June 22nd to Saturday, June 30th with all performances at 7:30pm. Tickets available now by visiting www.cornstocktheatre.com or by calling the box office at (309) 676-2196. Remember, come early and enjoy dinner provided by Qdoba, you'll want the sustenance before burning off the calories from laughter!


The biggest mistake that homeowners can do is not taking security measure when it comes to their gardens. If you are like most people, the likelihood is that you do not think that it is necessary, not knowing that in fact it is. Let us take the UK as an example. A great many people have had their gardening equipment stolen. As you can imagine, their households were the victims of thieves. The garden is your personal sanctuary, and you should not let others vandalise it. But what can you do? There are many things you can do to deter garden thieves. Here are some examples.
Out of the wish to give your front yard the wow factor, you have put fancy plants and ornaments. There is nothing wrong with that, only that these plants and ornaments catch the attention of thieves. When they look at the greens and decorations, the only thing they see is a source of money. Whether or not you have a fence, you should never place anything that stands out in the front garden. There are other things you can do to beautify this area of the home.
When someone asks you "How Secure is Your Garden?", you want to be able to respond positively. However, if you leave any ladders outside the home, you will certainly not be able to. As a rule, thieves look for items that can help them get into your home. It does not matter if you have all the doors locked if thieves manage to get their hands on a ladder. They will break one of the windows and enter anyway. Thefts are opportunistic, so you have to make sure that you are not careless. We do not live in the old times when you could leave the front door open at night, without having to worry that something bad might happen. Times have changed.
Installing a gravel driveway is one of the best things that you can do. You are probably wondering why. Well, the explanation is simple: you will be able to hear footsteps. Simply put, you will not be taken by surprise if someone attempts to steal items from the garden or even break into your home. A gravel drive is very affordable, not to mention that it is a beautiful addition to your home.
You can take as many measures as you like because it is almost impossible to deter thieves. Taking into consideration this aspect, it is useful to have a security camera. The camera will capture the fact of the wrongdoers and provide you proof for the police. What is more, when they see the security camera, the thieves will be less tempted to break the law. A security camera is the answer to your problem. The thing is that you have to place the camera strategically. So, go shopping right away.


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Kristen Martell
The Stage at St Andrew's6036 Coburg Road, Halifax NS, Canada
22apr7:30 pm9:30 pmKristen MartellEvent Type MusicThe Stage at St Andrew's, 6036 Coburg Road, Halifax NS, Canada Event Organized ByThe Stage at St Andrew's
Kristen Martell is an acoustic folk/pop artist from the shores of Nova Scotia. Her hazy, sun-dappled songs explore the self and the way we walk through life. With her debut
Kristen Martell is an acoustic folk/pop artist from the shores of Nova Scotia. Her hazy, sun-dappled songs explore the self and the way we walk through life. With her debut album 'Coming Home' (April 2020), nominated for the 2021 ECMA Inspirational Recording of the year and two Music Nova Scotia Awards, Kristen Martell is set to release new collaborative music with award-winning producer Daniel Ledwell and award-winning artist Gabrielle Papillon. She is not to be missed.
The Stage at St Andrew's
6036 Coburg Road, Halifax NS, Canada
The Stage at St Andrew's
The Underground Comedy Railroad begins its 2023 tour in Dartmouth January 31, 2023
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It is a Dark and Dirty World for Ted Simmons January 23, 2023
Randy Glynn to step down as Live Art Dance's artistic director January 20, 2023
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Jackson Fairfax-Perry named 2b theatre company's emerging artist-in-residence January 16, 2023
Discover other performing arts events in Halifax in our Calendar of Events.


To make a valid claim, you must complete and submit an online claim, write your name and auto generated tracking claim code on the back of the original tax invoice, and send via mail to "Brother Cash Back Promotion", PO Box 6246 BAULKHAM HILLS BC NSW 1755 within 30 days of the purchase date. All claims must be registered and original valid tax invoice/receipt must be received by the Brother Cash Back team within 30 days of the purchase date.
1. When does the promotion start and finish?
To be eligible for Cash Back you must purchase a Brother eligible machine between 1st December 2018 to 31st March 2019.
2. How much will I get back?
3. Where do I find the serial number?
The serial number is located on the box your product came in. The serial number is 9 alpha-numeric characters long.
The alpha numeric serial number can also be found on the back of all of the eligible printers. Please note that the serial number on the back of the machine will appear as part of a much longer code most commonly starting with e66 or e67. Only the last 9 alpha-numeric characters will need to be noted to complete the registration.
Not accepting your serial number?
Please make sure you haven't accidently used the letter 'o' instead of the number '0'. The '/' inside the '0' means that it's a number. Also, please make sure you have removed any spaces and that you are capturing the correct 9 alphanumeric characters as shown above. If you continue to experience problems please call 1300 761 785.
4. Can I make a cash back claim if I bought my product from Officeworks?
No. Officeworks are a non-participating reseller. Products purchased from this reseller are not eligible for cash back.
Eligible Retailers are Harvey Norman, The Good Guys, Joyce Mayne, Domayne and participating business to business resellers.
5. What do I need to do with my receipt/tax invoice?
Make a copy of the tax invoice for your personal records and post the original tax invoice. Receipts/Tax Invoices can be posted to: "Brother Cash Back Promotion" PO Box 6246, Baulkham Hills BC, NSW 1755. Remember to write your unique reference code and your full name on the back of your tax invoice/receipt. All valid tax invoices/receipts must be received within 30 days of the purchase date.
6. When will I receive my cash back?
Please allow approximately 5 working days from receipt and validation of correct documents for your claim to be processed.
7. Who can I call if I need to speak to a person regarding my claim?
8. Do I need to send in my original tax invoice?
Yes. Photocopies will not be accepted. No hand written tax invoice/receipts permitted.
9. Is this offer valid on all Brother Printers, Labellers and Consumables?
No. The Cash back offer is only valid on the products listed on the home page of this website.
10. Can I apply if I live outside of Australia or have purchased a Brother product abroad?
No. The offer is only open to residents of Australia who have purchased it within the country.
11. How will I know you have received my posted tax invoice?
Please allow up to two weeks from time of posting for us to process your claim. When your claim has been processed you will receive an email confirming payment or requesting more information. You can also track your claim online by using the unique code you were given when you registered.
12. If I return the product under any returns policy, must I also return the cash back?
13. What does EFT mean?
EFT stands for an Electronic Funds Transfer which simply means that the cash back will be deposited into your nominated bank account.


I tried to watch a video from Camtasia on my pc, but it doesn't run in windows 10. We made some tests at Windows 7 and the video run. Anyone know what can be happening?
Are you referring to a .trec file that is not being viewed from within the Camtasia Editor. But on a windows 10 computer? Because you con only watch a .trec video in the Editor.
Other than that, could you be more specific. Provide some screen shots or more information?
i think i found the fix.
I was search your problem at this forum.
Try it,and give us feedback about it.


Since winning his Pulitzer prize for criticism, Hilton Als has risen more visibly to the role of public intellectual, one that he plays particularly well: droll, genial, grand in the old-fashioned style, with occasional asides redolent of famous writers in less media-trained times. ("Derek," he says casually of the late poet and Nobel laureate Derek Walcott, a writer whose work he admires, "was a terrible person; awful man"). He is also, perhaps, instinctively at home in areas of apparent contradiction, the bon vivant who, as he puts it in White Girls, his most recent book of essays, believes in "half living life so I can get down to really living it by writing about it".
Did it strike him as a shortcoming? "I thought maybe I'm not smart enough. But I just don't think I was interested enough in the stories in that way. I was much more interested in what I felt about something. The beginning would be interesting, when I had that moment of interiority, then the minute I had to be quoting all the time …" He makes a snoring sound.
Als's writing was sharp and elliptical, and would continue to be so. He began contributing to the New Yorker in 1989, joining the staff five years later and becoming the theatre critic in 2002. Part of the joy of reading Als is that one can never predict which way he will go. On Lena Dunham, a few years ago, when Girls was criticised for being all-white: "Dunham is accurately describing the ways in which, once things get sexual in her world, and girls become women, the universe gets polarised, segregated – her female characters are looking for white male validation, which is their right." On Bob Fosse: "His strongest work is vibrant with a sense of right and wrong – or, more specifically, of why doing wrong can feel, to the corrupted soul, so damningly right." On Amy Schumer, "putting herself down in the language of schoolboy misogyny", he writes of "the limits of perversity, and how, at times, the comic has to circumnavigate her own intelligence in order to make the character 'Amy Schumer', self-described 'trash from Long Island', come to life".
• White Girls is published by Penguin. To order a copy for £8.49 (RRP £9.99) go to guardianbookshop.com or call 0330 333 6846. Free UK p&p over £10, online orders only. Phone orders min p&p of £1.99.


To determine if you have glaucoma, your eye doctor will perform an eye evaluation. Your visual acuity will be measured. Additionally, your refraction, or best vision with glasses, may also be checked. Your eye pressure will also be measured carefully, and the thickness of your corneas may also be checked, as thinner corneas are a risk factor for glaucoma. Your pupils will also be tested, which help the doctor determine if your optic nerves are working properly. Your peripheral vision will likely be checked by counting fingers with the edge of your vision. You may also have a computer check your peripheral vision during an automated visual field examination. Your doctor will perform a slit lamp examination to examine the eye and the lens. The trabecular drainage system of your eye may be examined with a special contact lens, called a gonioscopy lens. Your eyes may be dilated to help the doctor view the lens and the back part of the eye and examine the optic nerve carefully. A retinal exam may also be performed during the dilated examination. Photographs of your optic nerves or computer scans of the optic nerve shape may also be taken.
As glaucoma progresses to moderate or advanced disease, optic nerve fibers are destroyed, causing the optic nerve to look increasingly "cupped," as if a larger and larger ice cream scooper is used to remove a pit from its center. Your doctor may also be able to see the loss of small sections of retinal nerves around the optic nerve, or small areas of bleeding on the optic nerve. Additionally, if angle closure is present, your doctor may be able to see iris or scar tissue physically blocking the trabecular drain.


As we move into a new year, creating resolutions, making promises and generally turning our attention towards the bright and shiny future, I would like to pause. As I find myself Not Quite Twenty-Thirteen Ready, I would like to hang out for a moment or two in what for me is a liminal space.
We are so busy with our Hellos.
We move quickly from Howdy to the next wonderful thing which also gets a Hey There from us and yet our often unsaid Good-byes are important, too. Have you read Franny & Zooey? Check out the last page — Salinger puts together the best good-bye in fiction (feel free to share a better one with me if you disagree). I'll quote it at the end of this post.
As a couples counselor I focus my work on bringing two people closer together, helping them see more fully this most amazing person they chose as a partner, lover, spouse, friend. I aim for greater connection — for Hello and Who Are You rather than Good-bye We Are Done Here. And yet, there are times when the work I do with clients is to help them take leave of each other whether temporarily or not.
We do not learn how to do this well in our culture of day trading and drive through coffee shops. We are taught to move quickly through divorce or break-up or even death. We should move on, buck up, get back out there (whatever that means!). We do not learn how to honor an ending by lovingly letting the other person go, by allowing the pain of loss to fill us up and yet trusting we will make it through. We are not taught to study how we got from beginning to end so that we might learn more about ourselves.
Even smaller good-byes are rushed through as if inconsequential. We dash off to work, smartphones in hand, texting through breakfast, moving towards our first meeting and in the meantime we have left the person we love most in the world without a kiss, a hug, an acknowledgement of parting.
Good-byes are often the Best Thing For Us in theory and yet in practice they are so wrenching. I understand why we want to rush through them to get away from that experience. Pain is to be avoided – hence the amygdala's famous fight, flight or freeze function – we fear pain so we move quickly to avoid it. And yet without hanging out in the good-bye we also miss honoring a connection, a relationship, a love that we hold. By avoiding good-bye we miss acknowledging our attachment to others which is in the long term a lonelier and therefore more painful way to live.


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text:
Q: Nested loops for menus (basic c++) I'm running into errors where they say that a, q, h, etc aren't declared variables, but I'm not sure how to fix this. Here is an example of my code.
I'm also getting an error where it doesn't recognize userinp inside the while loop, but I'm not sure how to transfer the variable storage from outside the while loop to inside the while loop without declaring a global variable, which I want to avoid.
Last question: I want to be able to have any key immediately return to the main menu after each menu item, but I'm not sure what to use for that.
Any help is greatly appreciated!
//Runs a program with a menu that the user can navigate through different options with via text input

#include <iostream>
#include <iomanip>
#include <algorithm>
#include <ctype.h>
#include <string>
using namespace std;

int main()
{
    int userinp;
    cout<<"Here is the menu:" << endl;
    cout<<"Help(H)      addIntegers(A)      subDoubles(D)           Quit(Q)";

    cin >> userinp;
    userinp = tolower(userinp);
    while userinp != q //loop while no q is input
    {
        if userinp == h
        {
            cout <<"This is the help menu. Upon returning to the main menu, input A or a to add 2 intergers." << endl;
            cout <<"Input D or d to subtract 2 doubles. Input Q or q to quit.";
        }
        if userinp == a
        {
            int a, b, result;
            cout <<"Enter two integers:";
            cin << a << b;
            result = a + b;
            cout << "The sum of " << a << " + " << b << " = " << result;
        }
    }
}

A: Try this:
while(userinp != 'q') {

And this:
if(userinp == 'h')

And this:
if(userinp == 'a')

etc., and you probably want to declare userinp as a char rather than an int.
If you want to compare a literal character, you must use single quotes.  If you want to compare a literal string, you need double quotes.  The only compare literal numbers without using quotes.
You also have an error with your input.  Your >> should be <<. Use >> with cin and use << with cout.

A: problems with code:
u expect userinp to be a charachter but u have declared as integer
2)conditions for if should be within brackets

redpajama_set_name: RedPajamaStackExchange


Discussion in 'Sequencers & Software Tools' started by BundyRoy, Aug 28, 2014.
I've got 3 songs done now for Christmas so I thought I would have a crack at animating a face using papagayo. I've had a look at the program and I just want to make sure I have the process correct.
Then I use Bobo's Papagayo to LOR software to assign channels to the phonemes and import it into LOR as a clipboard.
I gather there are a few tricks like ensuring you use 25 FPS and when working with RGB I need to create a track with groups for each of the Phonemes.
Is this the whole process or is this now redundant and a new way has been invented. I see on Bobo's website and videos that he uses version 1.3.5.3 but when I did a google search I see Papagayo is up to version 2. Should I use the latest version or 1.3.5.3.
I also saw on Bobo's webpage a link to Sean's papagayo. What is the difference? I gather that Sean is in the process of incorporating this into xlights. This has not been been released yet has it.
Any tips people have on animating faces will be greatly appreciated. Thanks.
I'm not sure how I started this thread in community projects. I meant to start it in Aussie LOR - General discussions. Feel free to move it to another section that is better suited. I tried to remove it and restart it but it wouldn't let me. Thanks.
All in a Days work mate.
You may want to check out a thread that Sean from Nutcracker did which explains a bit about Papagyo and the plans for it to be put into xLights/Nutcracker.
I can see it in the new releases of xlight but I don't personally have a need for it and therefore haven't checked it out.
I think you will find that the big difference between Bob's and Sean's will be that one is basically designed for LOR and the other will export to most software thru xlights.
This has been imported into Vixen 3, and is now a native effect. Currently in the Dev Build, and will be in the 3.1.0 release which is due "real soon now".
Thanks Davrus. I plan on looking at other sequencers next year but will stick with LOR this year. Got to get my moneys worth out of it for starters!
Had a little more investigation into papagayo. Version 2, is a beta version of a re-write. Downloaded it and it doesn't seem to allow multiple voices so I went back to version 1.3.5.3 as I needed two voices. Seans papagayo on itsmebobo's page seems to be just a self installing version of the link Bob has to papagayo version 1.3.5.3.
I have now done the papagayo part of the skit I'm using (the kids recorded it). Habbosrus, kindly said he didn't mind if I kind of copied his face animation sequence. Hopefully 650km is far enough apart that not too many people will see both our shows.
Now tomorrow just need to work through getting it into LOR. Fingers crossed.
I have used Papagayo extensively in the past. While it does allow you to accomplish the task, it is extremely temperamental and buggy and it can cost a lot of time working around these problems. Version 1.2 is the older version and has the most bugs, version 1.3.5.3 is newer and has had some of the bugs addressed, but still has some. I have viewed Bobs tutorials, and he explains the process really well. I remember downloading version 2 a while ago and it wouldn't even run on my machine. If your are not familiar with any other sequencers, using the process laid out by Bob would likely be the route that would take the least amount of time. If you are familiar with other sequencers, then you have some other options.
Vixen3 has implemented most of the features of Papagayo natively, including the Lyric converter, and allows you to import work previously performed in Papagayo. If you sequence using Vixen3, your work can be exported into Comma delimited or Falcon Pi Player format. Presently, xLights will NOT allow you to convert either format to LOR but Sean has indicated interest in picking up support for this format. I can work with him, if needed, to pick up this support. You would then use xLights to covert the work back to LOR or whatever format you want.
xLights now allows you to do facial animations, I have not used this specifically, but Sean has some great tutorial videos out there. Search the forums and you will find them. You should be able to animate there and then export to LOR format.
HLS has also picked up support for facial animation and can do Lyric conversions. I have used it, and it is not a bad way to go if you are familiar with how to use HLS. Same thing, you could do the work there and then export.
Any of these sequencers will require you to overcome a learning curve. If you decide to us papagayo this year, remember to hold on to your native .pgo files so you can convert to the new sequencer in the future.
[SIZE=78%])[/SIZE]I've got 3 songs done now for Christmas so I thought I would have a crack at animating a face using papagayo. I've had a look at the program and I just want to make sure I have the process correct.
Thanks for the reply Ed. The only sequencers I have used so far is LOR and xlights. I have also used some of Bob's programs. This will be enough for me this year so it looks like papagayo it is. I figure if I try and learn too many of them at once I won't remember any of them. The other side of it is there is only so many video tutorials I can watch without going crazy. I really appreciate them and couldn't do it without them but sometimes they get you get frustrated having to watch so many. Maybe I am just a slow learner.
Anyway, thanks for the feedback. I look forward to giving some of the other sequencing options a crack later on. Some of the things people can do on some of them blows my mind.
Found one of those bugs you mentioned Ed. I changed the FPS to 25 to suit Bobs software. Then I did all the work and saved it. Once I reopened the file later on the FPS was 20. Not good. I then found out you have to click on the 25 FPS tab twice to make it change to 25. Lesson learnt. It is now 25FPS and I have to redo it all (you know you have changed it to 25FPS when the numbers on the frames go up in multiples of 25 not 20).
One other thing I had trouble with was editing the .pgo file. In my original work I had allocated some words to the wrong voice. I remembered that Sean had said in his tutorial that the .pgo file was a text file and you could change things in there. I thought beauty, I'll just change it in there from one voice to the other and that will stop me from having to do the whole lot again.
I opened the .pgo file in wordpad (after creating a backup of the original). Moved all the required voice commands from one voice to the other. Then changed the list of lyrics at the top of each voice section (|bits between the lines|), then changed the number of lyrics (eg 14) at the top each voice section and resaved the file.
When I tried to reopen the file in papagayo it wouldn't open. Nothing happens, no messages. Is there something I missed or should I use a different editor. When I closed papagayo down it said to look in papagayo.exe.log for the error but I can't find that file.
Anyway as it turns out I had to change the FPS anyway and that required me to hit the breakdown option, so I've done it all again in the meantime anyway. I was just interested in case I need to edit it again.
Just in case anyone else is as dumb as me. If you're having trouble finding how to export the .dat file from papagayo you need to switch back from the xsheet tab to the breakdown tab and use the export voices button found there. Spent ages looking through the web before I discovered that. I new it had to be simple because no one else was posting about having issues with it. Just me!


Submitted by mstod00 on Mon, 03/02/2009 - 7:10pm.
Payless Foods Inc. of Los Angeles, California is a small 9-store chain of grocery stores in predominately Hispanic neighborhoods. The POI file includes the stores, the warehouse, and the Corporate Office. Original store #6 is currently closed, but is listed for historical purposes only. The Internet URL for the company is: http://www.paylessfoods.com/ .


text:
\section{Introduction}
Considering comparative probability orders from the combinatorial viewpoint, Maclagan \cite{DM}  introduced  the concept of a flippable pair of subsets.  This concept appears to be very central for the theory as it has nice algebraic and geometric characterisations. Algebraically, comparisons of subsets in flippable pairs correspond to irreducible vectors in the discrete cone associated with the comparative probability order, i.e., those vectors that cannot be split into the sum of two other vectors of the cone \cite{PF1,CCS}. Geometrically, a representable comparative probability order corresponds to a polytope in a certain arrangement of hyperplanes and flippable pairs correspond (with one exception) to those facets of the polytope which are also facets of one of the neighboring polytopes.

Christian et al \cite{CCS} showed that in any minimal set of comparisons that define a representable comparative probability order all pairs of subsets in those comparisons are flippable.

Maclagan formulated a number of very interesting questions (see \cite[p. 295]{DM}).  In particular, she asked  how many flippable pairs a comparative probability order on $n$ atoms may have. In this paper we  show that a representable comparative probability order may have up to $F_{n+1}$ flippable pairs, which is the $(n+1)$th Fibonacci number.  We conjecture that this lower bound on the maximal number of flippable pairs is sharp. This conjecture was put forward by one of us and we call it Searles' conjecture. We provide an upper bound on the maximal number of flippable pairs in a representable comparative probability order that is not too far from $F_{n+1}$.

Section 2 contains preliminary results and formulates Maclagan's problem. In Sections 3 and 4 we discuss Searles' conjecture in relation to Maclagan's problem and prove the aforementioned lower and upper bounds. Section 5 concludes by stating several open problems.
\section{Preliminaries}
\label{prels}

{\bf 2.1. Comparative Probability Orders.}
Given a (weak) order, that is, a reflexive, complete and transitive binary relation, $\preceq$ on a set $A$, the symbols $\prec$ and $\sim$ will, as usual, denote the corresponding (strict)  linear order and indifference relation, respectively.

\begin{definition}
Let $X$ be a finite set. A linear order $\preceq$ on $2^X$ is called a
{\em comparative probability order} on $X$ if $\emptyset\prec A$ for every
nonempty subset $A$ of $X$, and $\preceq$ satisfies de Finetti's axiom, namely
\begin{equation}
\label{deFeq}
A\preceq B \ \Longleftrightarrow \ A\cup C\,\preceq\, B\cup C ,
\end{equation}
for all $A,B,C\in 2^X$
such that $(A\cup B)\cap C=\emptyset$.
\end{definition}

As in \cite{PF1,PF2} at this stage of investigation we preclude indifferences between sets.
For convenience, we will further suppose that $X=[n]=\{1,2,\ldots,
n\}$ and denote the set of all comparative probability orders on
$2^{[n]}$ by $\mbox{$\cal P$}_n$.

If we have a probability measure ${\bf p}=(\row pn)$ on $X$,
where $p_i$ is the probability of $i$, then we know the
probability $p(A)$ of every event $A$ which is given by $p(A)=\sum_{i\in A}p_i$.
We may now define an order $\preceq_{\bf p}$ on $2^X$ by
\[
A\preceq_{\bf p} B \quad \mbox{if and only if}\quad  p(A)\le p(B).
\]
If the probabilities of all events are
different, then $\preceq_{\bf p}$ is a comparative probability order on
$X$. Any such order is
called {\em (additively) representable}.  The set of representable
orders is denoted by $\mbox{$\cal L$}_n$. It is known \cite{KPS} that $\mbox{$\cal L$}_n$ is
strictly contained in $\mbox{$\cal P$}_n$ for all $n\ge 5$.

Since a representable comparative probability order does not have a unique probability measure representing it but a class of them, any representable comparative probability order can be viewed as a credal set (a closed and convex set of probability measures, see, e.g.,  \cite{L1}) of a very special type. We will return to this interpretation slightly later.\par

As in \cite{PF1,PF2}, it is often
convenient to assume that
$
1\prec 2\prec \ldots \prec n.
$
This reduces the number of possible orders under consideration by a factor of $n!$. The set of all comparative probability orders on $[n]$ that satisfy this condition will be denoted by $\mbox{$\cal P$}_n^*$, and the set of all such representable comparative probability orders on $[n]$ will be denoted by~$\mbox{$\cal L$}_n^*$.\par

We can also define a representable comparative probability order by any vector of positive utilities ${\bf u}=(\row un)$ by
\[
A\preceq_{\bf u} B \quad \mbox{if and only if}\quad  \sum_{i\in A}u_i\le \sum_{i\in B}u_i.
\]
We do not get anything new since this will be the order $\preceq_{\bf p}$ for the measure ${\bf p}=\frac{1}{S}{\bf u}$, where $S=\sum_{i=1}^nu_i$. However, sometimes it is convenient to have the coordinates of ${\bf u}$ integers. In this case we will call $u(A)=\sum_{i\in A}u_i$ the {\em utility} of $A$.

Kraft et al \cite{KPS} gave necessary and sufficient conditions for a comparative probability order to be representable. They are not so easy to formulate and they have appeared in the literature in various forms (see, e.g., \cite{DS,PF1}). The easiest way to formulate them is through the concept of a {\em trading transform} introduced in \cite{TZ}.

\begin{definition}
A sequence of subsets $(\row Ak; \row Bk)$ of $[n]$ of even length $2k$ is said to be a trading transform of length $k$ if for every $i\in [n]$
$$
\left|\{j\mid i\in A_j\}\right|=\left|\{j\mid i\in B_j\}\right|.
$$
In other words, sets $\row Ak$ can be converted into $\row Bk$ by rearranging their elements.
\end{definition}

Now the result of \cite{KPS} can be reformulated as follows.

\begin{theorem}[Kraft-Pratt-Seidenberg]
\label{KPStheorem}
A comparative probability order $\preceq $ is representable if and only if for no $k$ there exist pairs $A_i\prec B_i$, $i=1,2,\ldots, k$ such that
$
(\row Ak; \row Bk)
$
is a trading transform of length~$k$.
\end{theorem}

\par \medskip

{\bf 2.2. Discrete Cones.}
To every linear order $\preceq$~$\in \mbox{$\cal P$}_n^*$, there corresponds a {\em discrete cone} $C(\preceq)$ in $T^n$, where $T=\{-1,0,1\}$ (as defined in \cite{AK,PF1}).

\begin{definition}
A subset ${\cal C}\subseteq T^n$ is said to be a discrete cone if the
following properties hold:
\begin{enumerate}
\item[{\rm D1.}] $\{ {\bf e}_1, {\bf e}_2,\ldots, {\bf
e}_{n}
\}\subseteq {\cal C}$,
where $\{{\bf e}_1,\ldots,{\bf e}_n\}$ is the standard basis of $\mathbb{R}^n$,
\item[{\rm D2.}]  for every ${\bf x}\in T^n$, exactly one vector of the set $\{-{\bf x},{\bf x}\}$ belongs to $\cal C$,
\item[{\rm D3.}] ${\bf x}+{\bf y}\in C$ whenever ${\bf x},{\bf y}\in {\cal C}$ and
${\bf x}+{\bf y}\in T^n$.
\end{enumerate}
\end{definition}
We note that in \cite{PF1} Fishburn requires ${\bf 0}\notin {\cal C}$  because
his orders are anti-reflexive. In our case, condition D2 implies
${\bf 0}\in {\cal C}$.
\medskip

For each subset $A\subseteq X$ we define the characteristic vector
$\chi_A$ of this subset by setting $\chi_A(i)=1$, if $ i\in A$, and $\chi_A(i)=0$, if $ i\notin A$.
Given a comparative probability order $\preceq $ on $X$,
we define the characteristic vector
$\chi(A,B)=\chi_B -\chi_A\in
T^n$  for every possible pair $(A,B) $ such that $A\preceq B$.
The set of all characteristic
vectors $\chi(A,B)$  is denoted
by $C(\preceq)$. The two axioms of comparative probability guarantee
that $C(\preceq)$ is a discrete cone (see \cite[Lemma~2.1]{PF1}).

\iffalse
We can define a restricted sum for vectors in a discrete cone $\cal C$. Let ${\bf
u},{\bf v}\in {\cal C}$. Then
\[
{\bf u}\oplus {\bf v}=
\left\{
\begin{array}{cl} {\bf u}+{\bf v}& \text{if   ${\bf u}+{\bf v}\in T^n
$},    \\\
\text{undefined} &  \text{if   ${\bf u}+{\bf v}\notin T^n $}.  \end{array}
\right.
\]
This is equivalent to transitivity of the respective comparative probability order \cite{CCS}.

\begin{definition}
We say that the cone ${\cal C}$ is  {\em weakly generated} by vectors $\brow vk$ if
every nonzero vector ${\bf c}\in {\cal C}$ can be expressed as a
restricted sum of $\brow vk$, in which each generating vector can be used as
many times as needed. We denote this by ${\cal C}={<}\brow vk{>_w}$.
\end{definition}
\fi
\medskip

{\bf 2.3. Critical and flippable pairs.} Not all relations $A\prec B$ for pairs of subsets $(A,B)$ in a comparative probability order are equally informative. Some of these may be implied by others through transitivity or de Finetti's axiom. This is certainly true for any pair consisting of two nonadjacent sets or two sets with nonempty intersection.

\begin{definition}
Let $A$ and $B$ be disjoint subsets of $[n]$. The pair $(A,B)$ is said to be {\em critical} for $\preceq $ if $A\prec B$ and $(A,B)$ are adjacent, i.e., there is no $C\subseteq [n]$ for which $A\prec C\prec B$.
\end{definition}

In the above definition we follow Fishburn \cite{PF3}, while Maclagan \cite{DM} calls such pairs {\em primitive}.

It is known \cite{DM} that, if $\preceq$ and $\preceq'$ are distinct comparative probability orders, then there exists a critical pair $(A,B)$ for $\preceq$ such that $B \prec' A$.
This shows that critical pairs are of interest due to the fact that they define orders. But there are even more interesting pairs.

\begin{definition}
A critical pair $(A,B)$ is said to be {\em flippable\/} for $\preceq $ if  for every $D\subseteq [n]$, disjoint from $A\cup B$, the pair $(A\cup D,B\cup D)$ is adjacent in $\preceq$.
\end{definition}

We note that the set of flippable pairs is not empty, since the central pair of any comparative probability order is flippable \cite{KPS}. Indeed, this consists of a certain set $A$ and its complement  $A^c=X\setminus A$, and there is no $D$ which has empty intersection with both of these sets. It is not known whether this can be the {\em only\/} flippable pair of the order.\par
\medskip

Suppose now that a pair $(A,B)$ is flippable for a comparative probability order $\preceq  $, and $A\ne \emptyset$. Then reversing each comparison $A\cup D \prec B\cup D$ (to $B\cup D \prec A\cup D$), we will obtain a new comparative probability order $\preceq  '$, since the de Finetti axiom (\ref{deFeq}) will still be satisfied.  We say that $\preceq ' $ is obtained from $\preceq  $ by {\em flipping over\/} $A\prec B$. The orders $\preceq $ and $\preceq '$ are called {\em flip-related}. This flip relation turns $\mbox{$\cal P$}_n$ into a graph which we will denote ${\cal G}_n$.

\begin{definition}
An element ${\bf w}$ of the cone ${\cal C}$ is said to be {\em reducible\/} if there exist two other vectors
${\bf u}, {\bf v}\in {\cal C}$ such that ${\bf w}={\bf u} + {\bf v}$, and {\em irreducible\/}  otherwise. The set of all irreducible elements of ${\cal C}$ will be denoted as $\text{Irr}({\cal C})$.
\end{definition}

\begin{theorem}[\cite{DM,CCS}]
\label{flippable-irreducible}
A pair $(A,B)$ of disjoint subsets is flippable for $\preceq $ if and only if the corresponding characteristic vector  $\chi (A,B)$ is irreducible in ${\cal C}(\preceq )$.
So the cardinality $|\text{Irr}({\cal C}(\preceq))|$ is the total number of flippable pairs in $\preceq$.
\end{theorem}

Flippable pairs uniquely define a representable order but this does not hold for nonrepresentable orders \cite{CCS}.
\medskip

As we know, the flip relation turns $\mbox{$\cal P$}_n$ into a graph ${\cal G}_n$. Let $\preceq $ and $\preceq '$ be two comparative probability orders which are connected by an edge in this graph (and so are flip-related). We say that $\preceq $ and $\preceq '$ are  {\em friendly\/}  if they are either both representable or both nonrepresentable. \par\medskip

{\bf 2.4. Geometric Representation of Representable  Orders and Maclagan's Problem.}
Let $A,B\subseteq [n]$ be disjoint subsets, of which at least one
is nonempty. Let $H(A,B)$ be a hyperplane  consisting of all points ${\bf x}\in \mathbb{R}^n$ satisfying the equation
\begin{equation*}
\sum_{a\in A}x_a-\sum_{b\in B}x_b=0.
\end{equation*}
We denote the corresponding
hyperplane arrangement by ${\cal A}_n$.
Also let $J$ be the hyperplane $x_1+x_2+\ldots
+x_n=1,$ and let ${\cal H}_n={\cal A}_n^J$ be the induced hyperplane arrangement. Fine and Gill \cite{FG} showed that the regions of ${\cal H}_n$ in the
positive orthant $\mathbb{R}^n_+$ of $\mathbb{R}^n$ correspond to representable orders from ${\mbox{$\cal P$}}_n$.

Now we can see what is special in the credal sets that correspond to comparative probability orders. They are not only convex, as credal sets must be, but they are in fact interiors of polytopes.
When in the future we refer to a region of this hyperplane arrangement we will refer to the polytope which is the closure of that region. This will invite no confusion.

\begin{problem}[Maclagan \cite{DM}]
\label{prob1}
What is an upper bound for the number of representable neighbors for a representable comparative probability order on $n$ atoms? In other words, how many facets can regions of ${\cal H}_n$ have?
\end{problem}
The maximal number of facets of regions of ${\cal H}_n$ we will call the $n$th {\em Maclagan number} and denote $M(n)$, while the maximal number of flippable pairs for a representable order on $n$ atoms will be denoted $m(n)$. In this paper we provide bounds on these functions, some of which we suspect to be sharp.
  It is clearly sufficient to solve Maclagan's problem (Problem~\ref{prob1}) for comparative probability orders in $\mbox{$\cal L$}_n^{*}$. \par\medskip

The main combinatorial  tool for calculating or estimating $M(n)$ is the following semi-obvious proposition.

\begin{proposition}[\cite{DM,CCS}]
\label{mainprop}
Let $\preceq $ be a representable comparative probability order, and let $P$ be the corresponding convex polytope, which is a region of the hyperplane arrangement ${\cal H}_n$. Then the number of  facets of $P$ equals the number of representable comparative probability  orders that are flip-related to $\preceq$ (plus one if the pair $\emptyset \prec 1$ is flippable).
\end{proposition}

\begin{corollary}
\label{Mandm}
$M(n)\le m(n)$.
\end{corollary}

\begin{proof}
From the proposition it follows that $M(n)$ cannot be greater that $m(n)$. However theoretically it can be smaller since not all flips of the representable comparative probability order that has the maximal number of flips may be friendly.
\end{proof}

It is worth noting that the minimal number of facets of a region in ${\cal H}_n$ is known and equal to $n$ \cite{CS,CCS}.

\section{The Lower Bound}

\iffalse
Let us summarise what is known about the cardinality of $|\text{Irr}({\cal C})|$ in the following. For $n=5$ and $n=6$ a solution was found computationally,
using Proposition~\ref{mainprop}.

\begin{theorem}[\cite{CS,CCS}]
Let $\preceq $ be a comparative probability order on $2^X$ with $|X|=n$, and ${\cal C}$ be the corresponding discrete cone. Then
\begin{itemize}
\item  if $\preceq $ is representable, then the set of all irreducible elements ${\rm Irr}({\cal C})$ weakly generates $\cal C$ and $|{\rm Irr}({\cal C})|\ge n$, while
\item  if $\preceq $ is nonrepresentable, then the set of all irreducible elements ${\rm Irr}({\cal C})$ may not generate $\cal C$ and it may be that $|{\rm Irr}({\cal C})|< n$.
\end{itemize}
\end{theorem}
\fi

It is known that $M(3)=m(3)=3$ and $M(4)=m(4)=5$ \cite{CCS}.
Computations in {\sc Magma\/} {\cite{CCS}} show that $5\le |{\rm Irr}({\cal C})|\le 8$ for $n=5$ and $5\le |{\rm Irr}({\cal C})|\le 13$ for $n=6$
with all intermediate values being attainable for both values of $n$.
It was also observed that for $n=5$ and $n=6$, all comparative probability orders with the largest possible number of flips (namely 8 for $n=5$, and 13 for $n=6$) are representable, and all of their flips are friendly. This means that $M(5)=m(5)=8$ and $M(6)=m(6)=13$

Searles noticed that the four known values are  Fibonacci numbers, i.e., belong to the sequence defined by $F_1=F_2=1$ and $F_{n+2}=F_{n+1}+F_n$. He conjectured that \par
\medskip

\noindent {\bf Conjecture (Searles, 2007)}
The maximal number of facets of regions of $\mbox{$\cal H$}_n$ is equal to the maximal cardinality of $\text{Irr}({\cal C(\preceq)})$ for  $\preceq \in \mbox{$\cal L$}^{*}_n$, and equal to the Fibonacci number $F_{n+1}$ or, alternatively, $M(n)=m(n)=F_{n+1}$.\par\medskip

The first part of this conjecture will be proved if we show that for some representative comparative probability order $\preceq $, for which  $|\text{Irr}({\cal C(\preceq)})|$ is maximal, all flips of $\preceq $ are friendly. The existence of such an order was checked for all $n\le 12$. \par\medskip

In this section we prove that $M(n)\ge F_{n+1}$. To this end we prove

\begin{theorem}
\label{Dtheorem}
In ${\cal P}_n$ there exists a representable comparative probability order which (a) has $ F_{n+1}$ flippable pairs and (b)  whose flips are all friendly.
\end{theorem}

The proof will be split into several observations. Let us introduce the following notation first. Let ${\bf u}=(\row un)$ be a vector such that $0<u_1<\ldots<u_n$ and $q>0$ be a number such that $u_j<q<u_{j+1}$ for some $j=0,1,2,\ldots,n$ (we assume that $u_0 = 0$ and $u_{n+1}=\infty$). In this case we set $({\bf u},q)$ to be the vector of $\mathbb{R}^{n+1}$ such that
\[
({\bf u},q)=(u_1,\ldots,u_j,q,u_{j+1},\ldots, u_n).
\]
We also denote ${\bf \ell}_n=(1,2,4,\ldots,2^{n-1})$ and $2{\bf \ell}_n=(2,4,8,\ldots,2^{n})$. We start with an easy and well-known observation.

\begin{proposition}
\label{pr2}
$\preceq_{{\bf\ell}_n}$ is the lexicographic order on $2^{[n]}$.  The utilities of subsets from $2^{[n]}$ cover the whole range of integers between 0 and $2^n-1$ and the utilities of any two consecutive  subsets in it differ by~$1$.
\end{proposition}

\begin{proof} This is equivalent to every natural number possessing a unique binary representation. We leave the verification to the reader.\end{proof}

\begin{proposition}
\label{pr3}
Let $q$ be an odd positive integer smaller than $2^{n}$ and ${\bf m}=(2{\bf\ell}_n,q)$. Consider the order $\preceq_{\bf m}$ on $2^{[n+1]}$. Then the difference between the utilities of any two consecutive subsets in this order is not greater than~$2$.
\end{proposition}

\begin{proof}
Suppose $2^{j-1} \leq q<2^{j}$, that is, $q$ is the utility of $j$ in $\preceq_{\bf m}$.  By Proposition~\ref{pr2} the utilities of the subsets from $[n+1]\setminus \{j\}$ cover the range of even values from 0 to $2^{n+1}-2$.  Suppose $B$ is a subset in $\preceq_{\bf m}$, where $B \neq \emptyset$. If $u(B) \leq 2^{n+1}-2$, then by Proposition 2 there exists a subset $A$ such that $0 < u(B) - u(A) \leq 2$. If $u(B)> 2^{n+1}-2$, then we must have $j \in B$, and since $u(j) < 2^n$, $B'=B \setminus \{j\} \neq \emptyset$. As $j \notin B'$ we have $u(B') \leq 2^{n+1}-2$, and so by Proposition 2 there exists $A' \subseteq [n+1] \setminus \{j\}$ such that $A' \prec_m  B'$ and $u(B')-u(A') =2$. Then adding $j$ to both subsets we obtain $u(B) - u(A) = 2$ for $A = A' \cup \{j\}$. Therefore, for any nonempty $B$ in $\preceq_{\bf m}$, there exists a subset $A$ such that $0 < u(B) - u(A) \leq 2$, and so for any adjacent pair $(C,D)$ of subsets, we have $0 < u(D) - u(C) \leq 2$.
\end{proof}

Let us denote by $\mbox{$\cal S$}_{n+1}$ the class of orders on $X=\{1,2,\ldots,n+1\}$ of type $\preceq_{\bf m}$, where ${\bf m}=(2{\bf\ell}_n,q)$ for some odd $0 < q< 2^{n}$. And let $j$ denote the number such that $2^{j-1} \leq q<2^{j}$. Obviously, $j<n+1$. By $\row u{n+1}$ we will denote the respective utilities of elements of $X$, that is ${\bf m}=(\row u{n+1})$.

\begin{proposition}
\label{pr4}
From the position at which the subset $\{j\}$ appears in the order $\preceq_{\bf m}$ until the position after which all subsets contain $j$, subsets not containing $j$ alternate with those containing $j$, with the difference in utilities for any two consecutive terms being $1$.
\end{proposition}

\begin{proof}
All subsets not containing $j$ have even utility and all those containing $j$ have odd utility. If we consider these two sequences separately, by Proposition~\ref{pr2} the difference of utilities of neighboring terms in each sequence will be equal to 2. Hence they have to alternate in $\preceq_{\bf m}$.
\end{proof}

\begin{lemma}
\label{lemma_abc}
Let  $\preceq_{\bf m}$ be an order from the class $\mbox{$\cal S$}_{n+1}$ and let $(A,B)$ be a critical pair for $\preceq_{\bf m}$. Then the following conditions are equivalent:
\begin{enumerate}
\item[\rm (a)] $(A,B)$ is flippable;
\item[\rm (b)] either $A$ or $B$ contains $j$ but not both;
\item[\rm (c)] $u(B)-u(A)=1$.
\end{enumerate}
\end{lemma}

\begin{proof}
(a) $\Longrightarrow $ (b): Suppose $(A,B)$ is flippable. As $(A,B)$ is critical, it is impossible for $A$ and $B$ each to contain $j$ as $A\cap B=\emptyset$. We only have to prove that it is impossible for both of them not to contain $j$. If $j\notin A$ and $j\notin B$, then $u(A)+2=u(B)$. Since the pair is critical, by Proposition~\ref{pr4} both $A$ and $B$ appear in the order earlier than $\{j\}$. Hence $u(A)+2=u(B) < u(j)$. Then, in particular, $u(A)<u(B)<u(n+1)=2^n$, hence neither $A$ nor $B$ contains $n+1$. But then for $A'=A\cup \{n+1\}$ and $B'=B\cup \{n+1\}$ we have $u(j)<u(A')<u(B')$. Both $A'$ and $B'$ do not contain $j$, hence they are in the alternating part of the order, and since $u(B')-u(A')=2$, they cannot be consecutive terms. As $(A,B)$ is flippable, this is impossible, which proves that either $A$ or $B$ contains~$j$.

(b) $\Longrightarrow $ (c): This follows from Proposition~\ref{pr4}.

(c) $\Longrightarrow $ (a):  This is true not only for orders from our class, but also for all orders defined by integer utility vectors. Indeed, if $u(B)-u(A)=1$, then for any $C\cap (A\cup B)=\emptyset$ we have  $u(B\cup C)-u(A\cup C)=1$,  and so $A\cup C$ and $B\cup C$ are consecutive.
\end{proof}

Up to now, the values of $q$ and $j$ did not matter. Now we will try to maximise the number of flippable pairs in $\preceq_{\bf m}$, so we will need to choose them carefully. It should come as no surprise that the optimal choice of $j$ and $q$ will depend on $n$, so we will talk about $j_n$ and $q_n$ now. For the rest of the proof we will set
\begin{equation}
\label{numberq}
j_n=n-1,\qquad q_n=\frac{(-1)^{n+1}+2^n}{3}.
\end{equation}
An equivalent way of defining $q_n$ would be by the recurrence relation
\begin{equation}
\label{recrel}
q_n=q_{n-1}+2q_{n-2}
\end{equation}
with the initial values $q_3=3$, $q_4=5$. We also note:
\begin{proposition}
\label{pr5}
$
q_n\equiv 2+(-1)^{n+1} \pmod4.
$
\end{proposition}

\begin{proof}
Easy induction using (\ref{recrel}).
\end{proof}

Let us now consider a flippable pair $(A,B)$ for $\preceq_{\bf m}$, where ${\bf m}=(2{\bf\ell}_n,q_n)$. Since $j_n=n-1$, we have either $A=A'\cup\{n-1\}$ or $B=B'\cup\{n-1\}$ but not both. In the first case, $(A',B)$ is a pair of nonintersecting subsets from the lexicographic order induced by $2{\bf\ell}_n$ on $ [n+1]\setminus \{n-1\}$ with $u(B)-u(A')=q_n+1$. In the second, $(B',A)$ is a pair of nonintersecting subsets from the same lexicographic order with $u(A)-u(B')=q_n-1$.

As $ [n+1]\setminus \{n-1\}$ can be identified with $[n]$, we let $g_n$ be the number of pairs $(A,B)$ in the lexicographic order $\preceq_{2{\bf \ell}_n}$ on $n$ atoms with $u(B)-u(A)=q_n+1$, and let $h_n$ be the number of pairs $(A,B)$ in the same order with $u(B)-u(A)=q_n-1$. What we have proved is the following:

\begin{lemma}
\label{g_n+h_n}
Let ${\bf m}=(2{\bf\ell}_n,q_n)$. Then the number of flippable pairs in  $\preceq_{\bf m}$ is $g_n+h_n$.
\end{lemma}

This reduces our calculations to a rather understandable lexicographic order $\preceq_{2{\bf \ell}_n}$.\par\medskip

For convenience we will  denote $q_n^+=q_n+1$ and $q_n^-=q_n-1$. We note that Proposition~\ref{pr5} implies
\begin{proposition}
\label{pr6}
$
q_n^-\equiv 1+(-1)^{n+1} \pmod4$, and  $q_n^+\equiv 3+(-1)^{n+1} \pmod4.
$
In particular, if $n$ is even, $q^-_n\equiv 0 \pmod4$ and $q^+_n\equiv 2 \pmod4$ and if $n$ is odd, $q^-_n\equiv 2 \pmod4$ and $q^+_n\equiv 0 \pmod4$.
\end{proposition}

A direct calculation also shows that the following equations hold:

 \begin{proposition}
\label{pr7}
\begin{eqnarray}
\label{-2-}
q_{n+1}^-&=&2q^-_n \qquad\ \  \text{for all odd $n\ge 3$},\\
\label{-3-}
q_{n+1}^-&=&2q^-_n+2 \quad \text{for all even $n\ge 4$},\\
\label{-4-}
q_{n+1}^+&=&2q^+_n-2 \quad \text{for all odd $n\ge 3$},\\
\label{-5-}
q_{n+1}^+&=&2q^+_n \qquad\ \  \text{for all even $n\ge 4$}.
\end{eqnarray}
\end{proposition}

\begin{lemma}
\label{ghrecrel}
 The following recurrence relations hold:  for any odd  $n\ge 3$
\begin{eqnarray*}
g_{n+1}=g_n+h_n,\qquad
h_{n+1}=h_n,
\end{eqnarray*}
and for any even $n\ge 4$
\begin{eqnarray*}
g_{n+1}=g_n,\qquad
h_{n+1}=g_n+h_n.
\end{eqnarray*}
\end{lemma}

\begin{proof}
Firstly we assume that $n$ is odd. Then $n+1$ is even. We know from (\ref{-2-}) that $q_{n+1}^-=2q^-_n$. Given any nonintersecting pair $(A, B)$ of subsets in $[n]$, with $A \prec_{2{\bf \ell}_n} B$ and $u(B)-u(A)=q^-_n$, we may shift both subsets to the right, replacing each element $i$ in them with the element $i+1$, to obtain a nonintersecting pair $(\overline{A},\overline{B})$ of subsets in $[n+1]$, where $\overline{A}$ precedes $\overline{B}$ in $\preceq_{2{\bf \ell}_{n+1}}$.  This procedure of shifting doubles the difference in utilities, so $u(\overline{B})-u(\overline{A})=2q^{-}_n=q_{n+1}^{-}$. This proves $h_{n+1}\ge h_n$. Moreover, by (\ref{-2-}) and Proposition~\ref{pr6}, $q^-_{n+1}\equiv 0 \pmod4$, hence no nonintersecting pair $(C, D)$ in $\preceq_{2{\bf \ell}_{n+1}}$ with difference of utilities $q^-_{n+1}$ can include~$1$, either in $C$ or in $D$, as $u_1 = 2$. Therefore $C=\overline{A}$ and $D=\overline{B}$ for some nonintersecting pair $(A, B)$ in $[n]$ with $u(B)-u(A)=q^-_n$. This shows $h_{n+1}= h_n$.

Let $(A,B)$ be one of the $h_n = h_{n+1}$ nonintersecting pairs of subsets of $[n+1]$ with $u(B)-u(A)=q^-_{n+1}$ as above. As before, since  $q^-_{n+1}\equiv 0 \pmod4$, neither of the sets contain~$1$. We can use these pairs to construct the same number of nonintersecting pairs  of $\preceq_{2{\bf \ell}_{n+1}}$ with utility difference $q^+_{n+1}=q^-_{n+1}+2$. Indeed, adding~$1$ to $B$ will create a pair $(A,B\cup \{1\})$ with a utility difference $q^-_{n+1}+2=q^+_{n+1}$. We can also use (\ref{-4-}) and a shifting technique to create another $g_n$ nonintersecting pairs with utility difference $q^+_{n+1}$. Indeed, if $(A,B)$ is one of the $g_n$ nonintersecting pairs in $\preceq_{2{\bf \ell}_{n}}$ with utility difference $q^+_{n}$, then the pair $(\{1\}\cup \overline{A}, \overline{B})$ will be nonintersecting in $\preceq_{2{\bf \ell}_{n+1}}$ with utility difference $2q^+_n-2=q^+_{n+1}$. We observe that the $h_{n+1}$ pairs $(C,D)$ constructed in the first method all have $1\in D$ while the $g_n$ pairs $(C,D)$ constructed in the second method all have $1 \in C$, and so the two methods never construct the same pair. Thus $g_{n+1}\ge g_n+h_n$.

Now, let $(C,D)$ be any nonintersecting pair in $\preceq_{2{\bf \ell}_{n+1}}$ with  utility difference $u(D)-u(C)=q^+_{n+1}$. As $n+1$ is even,  Proposition~\ref{pr6} gives $q^+_{n+1}\equiv 2\pmod4$. This implies that either $1\in C$ or $1\in D$. Now as above, we can show that $(C,D)$ can be obtained as $(\{1\}\cup \overline{A},\overline{B})$ or $(A,\{1\}\cup B)$ by the second or the first method, respectively. Thus $g_{n+1}= g_n+h_n$.

For even $n$, the statement can be proved similarly, using the other two equations in Proposition~\ref{pr7} and congruences in Proposition~\ref{pr6}.
\end{proof}

\noindent{\it Proof of Theorem~\ref{Dtheorem} (a).}
Let us consider the case $n=3$. We have $q_3=3$, so $q_3^-=2$ and $q_3^+=4$. We have three nonintersecting pairs in $\preceq_{2{\bf \ell}_3}$ with utility difference two, namely $\emptyset \preceq_{2{\bf \ell}_3} \{1\}$, $\{1\}\preceq_{2{\bf \ell}_3} \{2\}$, and $\{1,2\}\preceq_{2{\bf \ell}_3} \{3\}$, and two nonintersecting pairs with utility difference four, namely, $\emptyset \preceq_{2{\bf \ell}_3} \{2\}$ and $\{2\}\preceq_{2{\bf \ell}_3} \{3\}$. Thus $g_3=2$ and $h_3=3$. Alternatively, we may say that $(g_3, h_3)=(F_3, F_4)$. It is also easy to check that $(g_4, h_4)=(5,3)=(F_5, F_4)$.  A simple induction argument with the use of Lemma~\ref{ghrecrel} now shows that $(g_n,h_n)=(F_n,F_{n+1})$ for odd $n$ and $(g_n,h_n)=(F_{n+1},F_n)$ for even $n$.
By Lemma~\ref{g_n+h_n} we find that the number of flippable pairs of $\preceq_{\bf m}$ is
\[
g_n+h_n=F_{n+1}+F_n=F_{n+2}.
\]
It remains to notice that $\preceq_{\bf m}$ is in ${\mathcal G}_{n+1}$.
\par\medskip

\noindent{\it Proof of Theorem~\ref{Dtheorem} (b).}
Let $\preceq$ be obtained from $\preceq_{\bf m}$ by a flip. Assume it was the pair $B \prec_{\bf m} A$ in $\preceq_{\bf m}$ which was flipped, so in $\preceq$ we have $A \prec B$. Assume $\preceq$ is not representable. By Theorem~\ref{KPStheorem} there must exist a trading transform  $(\row Ak; \row Bk)$ such that $A_i\prec B_i$ for $i=1,\ldots, k$. For each $i$ we may assume that $A_i \cap B_i = \emptyset$ since otherwise we could remove the intersection for each pair and obtain another trading transform with empty intersections.

Since each element of $[n]$ appears in the sequence $\row Ak$ exactly as many times as in $\row Bk$, for the weight function $u$ of $\preceq_{\bf m}$, we must have $\sum_{i=1}^k u(A_i) = \sum_{i=1}^k u(B_i)$. However the only nonintersecting pair $C \prec D$ in $\preceq$ with $u(C) \geq u(D)$ is the flipped pair $A \prec B$, and furthermore we know from Lemma~\ref{lemma_abc} that $u(A) - u(B) =1$ and for every other pair $(A_i,B_i)$ different from $(A,B)$, $u(A_i) - u(B_i) \leq -1$. Hence for  $\sum_{i=1}^k u(A_i) = \sum_{i=1}^k u(B_i)$ to hold at least half of the pairs $A_i \prec B_i$ must be the pair $A \prec B$. Without loss of generality assume that $A_i \prec B_i$ is the pair $A \prec B$ for $i = 1,2,...,r$ with $r \geq \frac{k}{2}$.

Let $j \in A$ be any element of $A$. Then $j$ appears $r$ times in the the sequence $\row Ar$ and no times in the sequence $\row Br$. Hence it must appear $r$ times in $(B_{r+1},\ldots, B_k)$, but $r \geq \frac{k}{2}$ and $j$ can appear at most once in each $B_i$ and so we must have $r=\frac{k}{2}$, $j \in B_i$ and $u(A_i)-u(B_i) = -1$ for $i=r+1,...,k$. But $j$ was an arbitrary element of $A$, so $A \subseteq B_i$ for $i=r+1,...,k$. The same argument shows that $B \subseteq A_i$ for $i=r+1,...,k$. But if $A_i = B \cup C_i$ and $B_i = A \cup D_i$ with $C_i$, $D_i$, $A$ and $B$ all disjoint for $i=r+1,...,k$ then
\[
u(A_i)-u(B_i) =u(B \cup C_i) - u(A \cup D_i) = u(B) + u(C_i) - u(A) - u(D_i) = -1.
\]
But $u(A) - u(B) = 1$ and so $u(C_i) = u(D_i)$. Since $\preceq_{\bf m}$ is a linear order, this implies $C_i = D_i = \emptyset$ and so $B \prec  A$ which gives the desired contradiction.

\section{The Upper Bound}

We now present a result giving an upper bound on the number of flippable pairs in any comparative probability order, representable or not. This will give us an upper bound for $m(n)$ and hence for $M(n)$. The basic result is in the following lemma which estimates the number of flippable pairs from above.

\begin{lemma}
\label{ub}
Let $\preceq$ be a comparative probability order on $n$ atoms. If $s$ is any positive integer such that $\sum_{i=0}^s2^{i}\binom{n}{i}\geq2^{n}-1$,
then $|\text{Irr}({\cal C(\preceq)})| \le \sum_{i=0}^s\binom{n}{i}$.
\end{lemma}

\begin{proof}
We first prove that if $A\prec B$ and $E\prec F$ are two distinct flippable pairs then $A\cup B\neq E\cup F$. Let $A\prec B$ be a flippable pair and consider $\preceq$ restricted to the subsets of $D=A\cup B$ and call this order $\preceq^{\prime}$. Clearly $\preceq^{\prime}$ is a comparative probability order:
\[
\emptyset=D_{1}\prec^{\prime}D_{2}\prec^{\prime}D_{3}\prec^{\prime}\ldots\prec^{\prime}D_{2^{r}-1}\prec^{\prime}D_{2^{r}}=D
\]
where $r=|D|$, $D_{i}\subseteq D$ and $D_{i}\prec^{\prime}D_{j}\Longleftrightarrow D_{i}\prec D_{j}$.
Because $A$ and $B$ were adjacent in $\preceq$, they will also be adjacent in $\preceq^{\prime}$, and since $A$ and $B$ are complements in $D$, they must be the central pair of $\preceq^{\prime}$, i.e. $A=D_{2^{r-1}}$ and $B=D_{2^{r-1}+1}$. However if $E\prec F$ was also a flippable pair with $E\cup F=D$, then it must also be the central pair of $\preceq^{\prime}$, and hence $A=E$ and $B=F$.

We now look at $\preceq$:
\[
\emptyset=A_{1}\prec A_{2}\prec A_{3}\prec\ldots\prec A_{2^{n}-1}\prec A_{2^{n}}=[n].
\]

Call the gap between two adjacent subsets an {\em adjacency}. There are a total of $2^{n}-1$ adjacencies, one for each $\prec$ sign in the order above. Consider a flippable pair $A\prec B$ in $\preceq$ and let $r=|(A\cup B)^c|$, which the size of the complement of $A\cup B$. From the definition of flippable pairs, every pair of the form $A\cup C\prec B\cup C$, where $C\subseteq (A\cup B)^c$, is adjacent. Let $C_{1},C_{2},\ldots,C_{2^{r}}$ be the subsets of $(A\cup B)^c$. Every pair $A\cup C_{i}\prec B\cup C_{i}$ will take up an adjacency, and hence the flippable pair $A\prec B$ will take up exactly $2^{r}$ adjacencies.

Hence we know that for every $r$ at most $\binom{n}{r}$ flippable pairs take up exactly $2^{r}$ adjacencies. This is because if there were more than $\binom{n}{r}$ such flippable pairs then by the pigeonhole principle two of the flippable pairs $A_{1}\prec B_{1}$ and $A_{2}\prec B_{2}$ will have $(A_{1}\cup B_{1})^c=(A_{2}\cup B_{2})^c$ and so must be the same pair. Hence there can be at most $\binom{n}{0}=1$ flippable pair that takes up $1$ adjacency, at most $\binom{n}{1}$ flippable pairs that take up $2^1$ adjacencies, at most $\binom{n}{2}$ flippable pairs that take up $2^2$ adjacencies, etc. But we have only $2^{n}-1$ adjacencies, so if we choose $s$ such that $\sum_{i=0}^s2^{i}\binom{n}{i}\geq2^{n}-1$ then  the number of flippable pairs cannot exceed $\sum_{i=0}^s \binom{n}{i}$.
\end{proof}

While the result is true for any such $s$, to maximize the strength of the upper bound we clearly wish to take the smallest value of $s$ possible. We will further need the binary entropy function ${H(\lambda)=-\lambda\log\lambda-(1-\lambda)\log(1-\lambda)}$ where the logarithms are of base 2. Using known approximations to the binomial coefficient we obtain the following:

\begin{corollary}
\label{c1}
Let $\lambda$ be the solution to the equation $\lambda+H(\lambda)=1$ and $\lambda<c<\frac{1}{2}$. Then
$m(n)\leq2^{H(c)n}$ for sufficiently large $n$. In particular, for any $\lambda<c<\frac{1}{2}$ we have $m(n)=O\left(2^{H(c)n}\right)$.
\end{corollary}

\begin{proof}
Let $\lambda<c<\frac{1}{2}$ and $\lambda < c' < c$ (e.g. $c' = \frac{\lambda + c}{2}$). Then $c'+H(c')>1$ since $H(x)$ is strictly increasing for $0<x<\frac{1}{2}$, and for sufficiently large $n$ it holds that $\lfloor cn \rfloor > \lceil c'n \rceil$. Hence
\[
\underset{i=0}{\overset{\lfloor cn \rfloor}{\sum}}2^{i}\binom{n}{i} > 2^{\lceil c'n \rceil}\binom{n}{\lceil c'n \rceil} \geq 2^{c'n}\frac{\sqrt{\pi}}{2}\frac{1}{\sqrt{2\pi nc'(1-c')}} 2^{H(c')n} > 2^n
\]
where the second inequality is obtained from \cite[p. 466]{PW} and the last inequality holds for sufficiently large $n$. So by Lemma~\ref{ub}, we have for sufficiently large $n$
\[
m(n)\leq\underset{i=0}{\overset{\lfloor cn \rfloor}{\sum}}\binom{n}{i}\leq c^{-cn}(1-c)^{-(1-c)n}=2^{H(c)n}
\]
where the second inequality is also obtained from \cite[p. 468]{PW}.
\end{proof}

\begin{example}
\label{upeg}
Take $c=0.25$ and consider $s=\lfloor{cn}\rfloor$. It can be checked that for $n \geq 102$ and $c' = c - \frac{1}{102} \leq \frac{s}{n}$ we have $\lceil c'n \rceil \leq s$ and the following inequalities hold
\[
\underset{i=0}{\overset{s}{\sum}}2^{i}\binom{n}{i} > 2^{\lceil c'n \rceil}\binom{n}{\lceil c'n \rceil} \geq 2^{c'n}\frac{\sqrt{\pi}}{2}\frac{1}{\sqrt{2\pi n c' (1-c')}}2^{H(c')n} > 2^n.
\]
Hence by Lemma~\ref{ub} it holds that
\[
m(n)\leq\underset{i=0}{\overset{\lfloor cn \rfloor}{\sum}}\binom{n}{i}\leq 2^{H(c)n}.
\]
Here $2^{H(c)} < 1.7548$.
Along with Theorem~\ref{Dtheorem} and standard bounds on the Fibonacci sequence, we have the following bounds for $n \geq 102$:
\[
F_{n+1} = \left[ \frac{\phi^{n+1}}{\sqrt{5}}\right] \leq m(n) \leq 1.7548^n,
\]
where $\phi \approx 1.6180$ is the golden ratio and $[x]$ is the closest integer to $x$.
\end{example}

Clearly in this example the exponent of 2 in the upper bound of $m(n)$ can be brought arbitrarily close to $H(\lambda)$ for sufficiently large $n$. As $2^{H(\lambda)} \approx 1.7087$, this gives the rough bounds $1.6180^n < M(n) \leq m(n) < 1.7087^n$ up to constant factors.

\begin{corollary}
For sufficiently large $n$
\[
1.6180^n < M(n) < 1.7087^n
\]
up to constant factors.
\end{corollary}

\section{Further Research}

We would like to know, of course, if Searles' conjecture is  true. Or at least, we would like to reduce the gap between the current bounds for $M(n)$ further.
There are some interesting questions that are not directly related to  Searles' conjecture but nevertheless  interesting.  One of them is the question of connectedness of
${\cal G}_n$. Since the subgraph of representable comparative probability orders is clearly connected, this question shows that we do not really understand much about
nonrepresentable orders. In particular, this question will be answered in the affirmative if we could show that any order is connected to a representable order by a series of flips. A similar
question is to find the minimum value of  $ |{\rm Irr}({\cal C})|$ in ${\cal G}_n$. For representable orders this minimum is $n$ but for nonrepresentable orders we cannot even say if it is possible that the central pair is the only flippable pair in the order. Maclagan also emphasised these questions \cite{DM}.

redpajama_set_name: RedPajamaArXiv


Hate and unforgiveness keeps you enslaved long after the ones you hate are in the ground.
Joseph Richardson, a New York millionaire lived and died in a house only five feet wide. It was called the "Spite House" and it deserved its name.
Owning the narrow lot of land on which it was built, Mr. Richardson wished to sell it to the neighboring property owners. They would not pay him what he asked, and so he put up this house, which disfigured the block—and then condemned himself to a life of discomfort in it.


Let's just jump right in to the Holiday spirit! In this one the guys take a peak at The Santa Clause which they consider to be a Holiday classic. Also, 2 movies to avoid, "Favorite Holiday Movies", and Rodney tries to stump Chet with Star Wars trivia with the help of a listener!


26 ed. A woman kept screaming, and Alice Tomlinson knew with vague pleasure that it was her girl Anne That was nice, she thought without malice: Anne was possibly screaming for her The air was thick. Alice wondered about breathing. Was it like freezing, when one grew drowsy, according to reports, and then snoozed off ? Would she have to choke a bit? She hated choking. The bed she was sitting on began to tip. Plainly, the floor beneath it was weakening. She thought, This is inter- esting- I am going to be tossed into the fires of Hell in spite of all my ef- forts to stay clear! She wished she had brought a nightcap of brandy to blaze with her. She felt somewhat light- headed in the thinning oxygen but quite clear in mind. The door crashed as if it were pa- per, and her son-in-law Hubert lunged into the room His beard and hair were gone, and she realized that he had a fine skull. How stupid of her not to have seen it before! He grabbed her nght hand, and pulled her off the teetering bed to the more solid floor near the hallway, which was as dark with smoke as a pit. Behind them the whole end of the house fen into the basement, so lately riddled with fiddle squeaks. She smiled delightedly. "Come on!" Hub e r t cried. He pulled her up. They were in the air. Below, the house blazed and there was an ambulance alongside some fire trucks, with a lot of light and bustle. "Come on!" She felt her hand strongly and warmly held in his, and they circled over the house as it crumbled into a red glowing heap. The ambulance drove off. Two of the fire trucks stayed near- by. NeIghbors went home, shaking their heads, and Hubert and Alice stayed lazily above them. She laughed a little, as lightly as snowbells. At first, she felt somewhat timorous or giddy, high over the hot turmoIl below. There had been a scary roar when the structure collapsed, but the hand that grasped hers was so strong and she felt such compassion and strength coming from it into her own bones that she held on firmly as they rose higher and higher. Never had she known such a warm lOVIng clasp. It told whatever she had always looked for. She felt innocent and undemanding, which she had al- ways meant to be anyway. And Anne would come along later. -M. F. K. FISHER . '}- .. > - Ü .ry .., .. :" .-\.': .. e A . . ,,;: c . ,( .Q ...: . .....:..( .... . . . / '. '. .' .". \. . ". .. . . . "" I " ". Y>-: f\' - " .'^" ....>>. . 1 . . '. \ . . , ;i.J t>> * , ).. 71 \, "" -f : : ,: i: ; '.' .,..' . . . .' :'.æ . "s' . f ' -<-' r < > . . \ :- - : ^/ " ... . '. ð 1. ; : } ^ ^ Lj illw " '1 :; ." '. t' . . , __ ,: J ;' " " ", . -It. , if .., \ Ii A ßD ". IHYO "" SfNE Tt E , . \ "foL " ,-- .:1#\ \' : .4., .J . Ii ., :. -- Y' , e,- of,,) <&+ '* "' r \ . . ^ t:t \:) ð""' '. ,,'>'. , . " '.,.. --- ...... . JM - ! iN :. ").. '""'. -c- 1! It ; J - I : . 11 ," ..) '<< -- .. BOOTH.


text:
ERFA welcomes the proposal for 2021 as the European Year of Rail
ERFA welcomes the Proposal of the European Commission to designate 2021 as the European Year of Rail. ERFA believes that a European Year of Rail should serve as an opportunity to finally deliver a legal framework that meets the needs of rail freight.
ERFA strongly believes that the proposal for a European Year of Rail is well aligned with the European Commissions proposed "Climate Law" which proposes setting a binding target of reaching net-zero carbon by 2050. It is impossible to achieve this objective if transport emissions are not addressed.
In 2017, 27% of all greenhouse gas emissions came from transport. Railways accounted for 0.5% of these emissions and clearly represent the best means by which to bring EU transport emissions under control. Modal shift is therefore essential.
However, for modal shift to be achieved, especially in relation to freight, an ambitious revision of the existing legal framework and financial mechanisms are required. The European Year of Rail should serve as the moment to highlight the benefits of rail, but also to openly discuss how the sector can further contribute to Europe's economic and climate related goals.
Dirk Stahl, ERFA President, said: "ERFA strongly supports the European Commission proposal to designate 2021 as the European Year of Rail. Rail has a critical role to play in ensuring that Europe can meet its 2050 objective of net-zero carbon emissions. The European Year of Rail must serve as the opportunity to finally begin work towards the creation of a competitive single European railway area that meets the needs of the rail freight industry".
Conor Feighan, ERFA Secretary General, stated: "The proposed European Year of Rail is well timed given that 2021 will be an important moment for the future of rail freight. 2021 will see the tabling of some key legal texts, such as the revision of the Rail Freight Corridors Regulation, which will play a pivotal role in determining whether rail freight can reach its true potential. ERFA is ready to contribute to the European Year of Rail in whatever way possible. The European Year of Rail should signal a new ambition for a European Single Railway Area and a departure from various national approaches and business as usual".
20200310 PR - ERFA welcomes the proposal for 2021 as the European Year of Rail ( 20200310 Press Release - ERFA welcomes the proposal for 2021 as the European Year of Rail-1583839221.pdf )


At present, with the continuous emergence of infrastructure projects, machine-made sand production is in full swing across the country. Faced with the demand of gravel market and the increasingly stringent environmental supervision, it is very important to do a good job in equipment selection and the efficient and environmentally friendly design of production lines. This article takes the semi-dry sand making process as an example, for everyone to talk about the equipment selection and process design points in the system sand production line for reference.
The coarse-breaking equipment used for aggregate production is generally a gyratory crusher, jaw crusher and impact crusher. If the production capacity is required to be higher than 1000t/h, it is recommended to use a gyration.
Medium-breaking equipment generally has a cone crusher and impact crusher to choose from, sand making machine is often used in the final level of sand making and shaping. The cone crusher is mainly suitable for crushing rocks with high hardness and high abrasive index, and the product has less powdery material. Impact crusher is mainly suitable for crushing rock with moderate or low abrasive index, and has more powdery material. If the demand for powdered material is poor, a standard cone crusher can be used for the middle smash and a short-head cone crusher for the fines.
Stone screening equipment generally uses a linear shaker or a circular shaker. Linear vibrating screen processing capacity is small, suitable for small particle size screening, small production use, can be used for water filter. Circular vibrating screen has high efficiency and large processing capacity. It is suitable for large-size particle size screening and large-scale production. When the circular vibrating screen is equipped with a pressure water spraying system, it can be used as a stone washing equipment. By means of hydraulic washing, the stones are tumbling and vibrating on the screen surface, and it is easier to wash away the fine mud adhering to the surface. For 0-4.75mm products in the mud and stone powder can generally choose spiral sand washing machine or bucket wheel washing machine after washing to obtain machine-made sand.
The mud-containing wastewater after washing the gravel aggregates needs to be treated in order to achieve zero discharge of wastewater from the production line. After examining the wastewater treatment equipment, a system consisting of a cyclone unit, a thickener and a filter press was selected to treat the waste water and mud, which can avoid the direct discharge of muddy wastewater into the sedimentation tank, resulting in a large area. The disadvantage of the sedimentation tank is difficult to clean.
The intermediate material stack can adjust the overall system capacity of the aggregate production line. When the rough section is overhauled, the downstream aggregate-made section will not stop production. When the mine can only be produced in whiteshift production, the downstream aggregate-making section can still be produced in two shifts, and the number of equipment can be reduced. Through the production class system adjustment system production capacity, reduced investment, to play the advantages of the system equipment.
Any design of the production line is a very good design if it can minimize construction investment and operating costs. Adapting to local conditions and making full use of topographical elevations for process planning can not only reduce civil investment and equipment investment, but also reduce production and operating costs.
Fine pre-screening design before the break. If the raw material contains clay, it is generally difficult to remove it by sieving. Designing pre-screening not only screens out soil-containing wastes smaller than a certain particle size by sieving when the raw material has a high soil content, but also allows aggregates smaller than the required particle size to pass through the medium and fine-break , And then increase the system yield and reduce the production of powdered material.
The finished product sieve is sprayed with a single layer shaker. The purpose of the first-order vibrating screen spray for the semi-dry sand aggregate production line is to remove part of the soil and to make the materials that are difficult to wash clean as early as possible, so that the mud-containing material can be soaked in advance. The secondary and tertiary finished vibrating screens are designed as a single layer, and the purpose of spray washing is to make each material product have the opportunity of direct contact with fresh water, thereby improving the cleanliness of aggregate products.
For sludge-containing wastewater, due to the large amount of mud sand and large particle size, it is not suitable to use the slurry pump to convey, and it can only flow through the water tank to the sand washing machine. In this link, the distance of the flow of the waste water and the angle of the sink should be fully considered. The sand washer workshop should be placed on the lower stage of the shaker workshop as much as possible. The trough is made into a U-shaped groove to reduce friction. A filter screen should be installed at the outlet of the sand washer and the outlet of the cyclone unit to prevent the moisture content of the material entering the finished product yard from being too large and affecting product sales.
After the sand-washed sludge-containing waste water enters the condensate pool, the position of the dosing point should be reasonable, and the dosing amount should be appropriate to improve the effect of the thickener system. Before the filter press should be set to buffer the mixing pool to meet the concentrator after the continuous operation of the slurry pump and press filter intermittent work requirements. The choice of filter press capacity must fully consider the time required for each job and the moisture content of the cake.
The design point of this link is to set at least one or two sets of weighing scales in addition to the appropriate discharge equipment for each batch, in order to adjust the reasonable gradation of sandstone, to meet the needs of customers for different grades of sand aggregate.
In the design, we generally adopt a policy of prevention and prevention. The dust points should be reduced as much as possible in each process before the middle and fine crushing, and the well-sealed conveying and feeding equipment should be used. Careful attention should be paid to the shape, angle, etc. of the material pipe connected to the equipment, and the drop should be minimized. All dust emission points adopt high efficient dust collectors with high dust collection efficiency and reliable technology to ensure that dust emission concentrations are lower than national standards. The storage of intermediate materials and soil-containing wastes uses closed sheds and spray dust removal to reduce the impact of dust on the surrounding environment.
Control of noise Select equipment with low noise as much as possible from the equipment selection, and arrange the noise source equipment in a closed factory building and green outside the workshop to block the noise transmission.
The wastewater is mainly domestic wastewater and sand washing wastewater. After the domestic wastewater is treated with pretreatment and other waste water is transferred to the wastewater treatment plant of the plant, it can be reused for the production line after reaching the standard. The wastewater generated by the screening and washing is treated with sand and water and processed by the waste cement processing system and recycled as production water. The mud cake after sludge treatment and the waste soil produced by the pre-screening are used for the reclamation of the cushion in the mine to achieve zero discharge of waste water and mud. This article mainly introduces the equipment selection and design points of the semi-dry sand production line. In the actual production, the sand aggregate production process and equipment configuration are various. The specific design should be adapted to local conditions and tailored to the production requirements.
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