A recent report from the Office of the Minnesota Secretary of the State found that even though more than twice as many absentee ballots were cast in 2012 than in 2010, fewer were rejected in last November's general election.
Of the 273,835 absentee ballots cast in 2012, 2,478, or 0.9 percent, were rejected. By comparison, 133,899 absentee ballots were cast, but 2,691, or 2 percent, were rejected in 2010.
The report noted that changes made to Minnesota's absentee-balloting process in 2010 and 2011 may have played a role in this decrease in rejections. Absentee ballot boards were created to review mail-voting envelopes after 2010. Soon after arrival, the envelopes are now checked to see if they meet legal requirements and can be accepted. If an error is found, local election officials must notify voters and send them a replacement ballot.


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Prices on site antidenuvo.com and on our other sites may vary. Before buying please tell us that you have found us on this site antidenuvo.com for geting games by lowest price.
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Published on March 6, 2019. Painting Rooms For Creative Hobby. Best 25 girls room paint ideas on pinterest paint girls, creative diy faux coral collection made with the help of. 17 best images about scale modelling hobby rooms & work. Mtac Los Angeles.
Hobby Room Looking Left Flickr Photo Sharing!


You guys liked the last sparring instruction video that we put up. Here are 7 more common sparring mistakes and tips, condensed into 5 minutes.
Break the Double Collar Tie | Namsaknoi Muay Thai - Duration: 41 seconds.
Break the Cross Face | Namsaknoi Muay Thai - Duration: 28 seconds.
Counter the Jab | Namsaknoi Muay Thai - Duration: 51 seconds.
Feint Setups for Knees & Elbows | Namsaknoi Muay Thai - Duration: 59 seconds.
Master the Roundhouse Kick with this Drill | Namsaknoi Muay Thai - Duration: 22 seconds.
Knee like Lamnammoon | Namsaknoi Muay Thai - Duration: 2 minutes, 37 seconds.
Lamnammoon The Vampire of Knees | Namsaknoi Muay Thai - Duration: 40 seconds.
How To Secure a Tight Head Control in the Clinch | Namsaknoi Muay Thai - Duration: 51 seconds.
Muay Thai Sparring : Common Mistakes + Tips | Namsaknoi Muay Thai - Duration: 5 minutes, 9 seconds.
Escape from the Collar Tie | Namsaknoi Muay Thai - Duration: 42 seconds.
What you have been doing wrong when you spar | Namsaknoi Muay Thai - Duration: 2 minutes, 11 seconds.
Elbow Strike in the Clinch | Namsaknoi Muay Thai - Duration: 42 seconds.
Namsaknoi vs Saenchai - Full Fight | Namsaknoi Muay Thai - Duration: 14 minutes.
Namsaknoi vs Attachai - Full Fight | Namsaknoi Muay Thai - Duration: 15 minutes.
Namsaknoi vs Rambo-Jiu - Full Fight | Namsaknoi Muay Thai - Duration: 23 minutes.
Namsaknoi vs Baiphet - Full Fight | Namsaknoi Muay Thai - Duration: 11 minutes.
Namsaknoi vs Khunsuek Sit Porramate - Full Fight | Namsaknoi Muay Thai - Duration: 16 minutes.
Namsaknoi vs Kaolan Kaovichit - Full Fight | Namsaknoi Muay Thai - Duration: 17 minutes.
Namsaknoi vs Kochasan Singklongsee - Full Fight | Namsaknoi Muay Thai - Duration: 16 minutes.
Namsaknoi vs Lamnammoon Sor Sumalee - Full Fight | Namsaknoi Muay Thai - Duration: 9 minutes, 54 seconds.
Fight Log: Kaito v. Kongkiat Rematch | Namsaknoi Muay Thai - Duration: 6 minutes, 55 seconds.
Team Namsaknoi wins 3-0 in Samui! - Duration: 4 minutes, 26 seconds.
9 Dec Fight Log: Paul wins via low kick TKO! - Duration: 3 minutes, 31 seconds.
26 June Fight Log: Kaito won by unanimous decision in Okinawa! - Duration: 4 minutes, 20 seconds.
Wai Kru Ram Muay | Namsaknoi Muay Thai - Duration: 61 seconds.
Koh Phangan Viewpoint Run | Namsaknoi Muay Thai - Duration: 47 seconds.
A Muay Thai Style Send Off | Namsaknoi Muay Thai - Duration: 34 seconds.
A Short Drive in Koh Phangan | Namsaknoi Muay Thai - Duration: 56 seconds.
Muay Thai Style Recovery Massage | Namsaknoi Muay Thai - Duration: 61 seconds.
Hello from NSN! - Duration: 67 seconds.
Buakaw vs. Namsaknoi: EXHIBITION fight in Phuket - Duration: 9 minutes, 54 seconds.
Thongchai Tor Silachai vs Namsaknoi Yudthagarngamtorn Muay Thai Lumpini - Duration: 10 minutes.
Namsaknoi: Centerline and Elbow Chains Into Knockout - Duration: 5 minutes, 29 seconds.
Namsaknoi: Exploiting Tendencies - Duration: 3 minutes, 21 seconds.
Nuengpichit vs. Namsaknoi - Duration: 14 minutes.
Satoshi Kobayashi vs Namsaknoi - Duration: 9 minutes, 29 seconds.
Numsaknoi Yudthagarngamtorn vs Kotchasarn Singklongsi - Duration: 13 minutes.
Best of Namsaknoi the emperor - Duration: 8 minutes, 6 seconds.
Namsaknoi Yuthkankunthon vs Craig O'Flynn @ KOMA - Duration: 10 minutes.
Ghost of Kiatsingnoi | Namsaknoi Muay Thai - Duration: 2 minutes, 42 seconds.
Stories of a Fighter | Namsaknoi Muay Thai - Duration: 103 seconds.
The Emperor of Muay Thai: Namsaknoi - Duration: 19 minutes.
Evolve Fighter Profile: Namsaknoi Yudthagarngamtorn - Duration: 7 minutes, 55 seconds.


text:
The next wave of Mets prospects is about to make an impact
By Albert DiCostanzo
Francisco Alvarez, 2021 SiriusXM All-Star Futures Game / Dustin Bradford/GettyImages
Washington Nationals v New York Mets / Mark Brown/GettyImages
2017: Ronny Mauricio, Mark Vientos
To date, only one Mets draft choice from 2017 has made it to the major leagues. That would be first round pick David Peterson, who quickly emerged as a solid left-handed starter in 2020 before struggling in 2021, both in terms of performance and avoiding injury.
Two other names from that year – one from the draft and another international signing – could debut with the Mets as early as this season. The key questions for both, however, are where they would play and if they'd even take the field in a Mets uniform.
Mark Vientos, whom the Mets selected out of high school in the second round, spent his first two professional seasons at Rookie level in Kingsport before climbing up the farm system. The 6'4'' third baseman played most of 2021 at Double-A Binghamton before moving up to Triple-A Syracuse late in the year on the strength of a breakout power season – 25 home runs and 43 extra-base hits across only 310 total at-bats after just 27 home runs in his first three minor league seasons combined.
The Mets' big splash internationally that year was Ronny Mauricio, a 17-year-old shortstop out of the Dominican Republic. Like Vientos, Mauricio's climb up the system has been slow given his age, but his potential has been consistently recognized, ranking in the top 100 of both Major League Baseball's and Baseball America's annual prospect lists each of the past two years. His 2021 season with High-A Brooklyn wasn't particularly noteworthy on paper (.242/.291/.449 slash line), but he did earn a promotion to Binghamton and, per scouts, has continued impressing with his athleticism.
The biggest roadblocks for both pertain to each one's primary position. Mauricio is likely not the Mets' shortstop of the future with Francisco Lindor under contract for the next decade, but at 6'3" with a strong arm, could find a home at third base or the outfield, where he's played some games in the minors. Vientos, meanwhile, has competition to landing the third base job in the form of another big-time prospect in the Mets' system (more on him later), which could ultimately make Vientos expendable. Regardless, both are high value prospects who could help in some way soon.

redpajama_set_name: RedPajamaCommonCrawl


text:
Creating a List in Scala
Scala Collections
Scala's List is an immutable collection data structure implemented using a linked list. In this tutorial, we'll learn about different ways to create and populate Lists in Scala using the help of some simple examples.
We'll also briefly look into the internal implementation and how it affects performance aspects during the build process.
2. List's Performance Characteristics
Before we start looking at the APIs, it's important to have an understanding of the performance characteristics. Appending an element to a List is a linear time complexity O(n) operation. However, prepending is a much faster operation, happening in constant time complexity O(1).
This property makes it suitable for Last-In-First-Out (LIFO) operations, as appending and removing elements from the head position is very quick. Most other operations, like random element access, follow linear time complexity O(n).
3. List Creation Methods
The Scala library provides us with plenty of options to create and populate Lists. There are methods in a functional style, object-oriented style, and there are many utility methods defined in its companion object. On top of that, there are conversion methods defined in other collection types to convert its data into a List.
We'll go through each one of them in detail.
3.1. The cons (::) Operator
One of the easiest ways to create a List is to use the "::" (also known as cons) operator and Nil, which is an empty singleton List object.
The intended way of building a List is by prepending a new element to it. The:: operator is particularly useful in this case, as it helps to build the collection by appending elements at the head position. Let's start by looking at an example:
val vehiclesList:List[String] = "Truck" :: "Car" :: "Bike" :: Nil
The cons operator is peculiar in its order of execution, as it's executed from right to left instead of the general left to right. Therefore, the above expression can be re-written as:
val vehiclesList:List[String] = Nil.::("Bike").::("Car").::("Truck")
In Scala, there's a general rule for the operators ending with a colon (:). Those expressions are evaluated from right to left. Therefore, our cons operator is also evaluated from right to left. This property is significant in the case of building Lists, as prepending an element to a List is much faster than appending at the end.
We can see that the cons operator requires two arguments, the element we are pretending and the list itself, or a Nil in case it's the first element. This style of programming is commonly found in the Lisp programming language and also generally in functional programming.
3.2. List's apply Method
Another common way of creating a List is to call its apply method. This is a simple API call, however, behind the scenes, it's a lengthy process happening in the background. The apply method creates a new Array of elements using the received varargs input.
Then the Array is converted to a Seq, and finally, the Seq is converted to List by calling the Seq.toList method on it.
Let's look at an example:
val vehiclesList:List[String] = List.apply("Truck", "Car", "Bike")
The apply function in Scala has a syntactic shortcut, that is, we can invoke it using its object-name as a function:
List("Truck", "Car", "Bike")
This is similar to the constructor method calls in Java classes.
3.3. The List.range Method
The List's companion object has an inbuilt range method to help us quickly create a list of continuously increasing integers. It takes a start element and an end element as parameters. Optionally, we can provide the step parameter also as a third parameter. By default, the step is taken as one.
val numList:List[Int] = List.range(1, 5)
println(numList)  // prints List(1, 2, 3, 4)
3.4. List.fill Method
Sometimes, we may want to create a list of elements with some default initial values. In such cases, we can invoke the fill method available in the List object.
Let's create a list of elements with an initial default value of an empty string:
val filledList:List[String] = List.fill(3)("")
println(filledList)  // prints List("","","")
3.5. toList Method
Many collection types in Scala have a toList method to convert their data to a List. Let's convert an Array of elements to a List using this function:
val convertedList:List[String] = Array("Truck", "Car", "Bike").toList
This method is especially useful when we have to deal with frequently changing data. In such cases, a mutable collection like ListBuffer is more suitable to build the collection, and once finished, we'd call the toList method to convert it to an immutable List.
3.6. List.tabulate Method
The List object has another useful function, tabulate. It helps us to create a list by applying a user-specific function to a series of numbers, starting at zero.
Let's assume we need to create a list of numbers in their binary representations:
val tabulatedList:List[String] = List.tabulate(3) (n => n.toBinaryString)
println(tabulatedList)  // prints List("0", "1", "10")
We can see the first three integers in their binary format.
In this tutorial, we've learned several methods to create and populate a List in Scala. Also, we've seen the performance characteristics of List during an append and prepend operation. Then we learned how the performance characteristics influence the creation methods.
As usual, the full source code can be found over on GitHub.
Scala Basics


ABB Home and Building Automation - How many users can operate the free@home system at the same time?
There is no limit on how many users can operate the free@home system at the same time. However, it is recommended to operate the system with only 4 users at the same time.
The reason for the limit is that, depending on the changes made, several users operating at the same time can lead to losses in performance (the changes take longer to implement).


text:
Shared Destiny. Shared Responsibility.
Prevention & Cures
Medical Advances
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How to drive in a downpour and get through a pandemic
by Richard Tucker | May. 28, 2020
By Memorial Day it was, finally, time to pick up the candy hearts.
During my son's freshman year at Vanderbilt University, I sent him a letter or a package every few weeks. Not so much to cheer him up; he loved the university and enjoyed being there. More to cheer myself up, and remind myself that I still mattered to his life.
Around the first of March, I spotted two-week-old Valentine candy hearts on super-duper sale at a CVS. No doubt they were 95 percent off by that point. Who wants hearts that say "I Love You" when it's already March, the Roman month of war? I dashed off a letter to my son, packed it in with the candy, and sent it off to Nashville.
Then they closed his school.
On March 9 Vanderbilt had a report of exposure to the coronavirus. The administration planned to shut down for a week or so, but my wife correctly predicted there would be no further classes on campus this spring. She booked my son on a flight out the next day. He hadn't gone back to Nashville since.
Zoom replaced office hours; online learning replaced lecture halls. The candy hearts sat, unopened, on his desk.
WHAT YOU NEED TO KNOW ABOUT CORONAVIRUS RIGHT NOW
AS STATES REOPEN, SOME SEE CORONAVIRUS CASES SPIKE
EXPERTS: 90% OF CORONAVIRUS DEATHS COULD HAVE BEEN AVOIDED
HERE'S WHEN IT'S SAFE FOR YOUR STATE TO REOPEN
HERE ARE THE 6 WAYS THE CORONAVIRUS PANDEMIC COULD END
In late May, the school announced its next step. Students should return at prearranged times to clear out of their dorm rooms. That's how we ended up driving to Nashville on Memorial Day weekend.
It was eerily quiet when we arrived at his dorm. When we'd moved him in back in August, the campus had been a madhouse. Teams of students unpacking cars and helping freshmen move in. Thousands of sad parents milling around. Now, we were about the only people on campus. We parked, illegally, in the roundabout by his dorm and headed in.
A gallon of milk from the fridge quickly went down the drain. A bag of apples ended up in the front seat for the drive home.
I unzipped the suitcase we'd brought, and realized there was a smaller suitcase inside. When I unzipped that one, I found a third. Because I didn't want to get stuck in an Abbot and Costello routine, I didn't open that third suitcase to find more and more and more suitcases. I just set it aside.
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My son didn't have much stuff. His clothing fit in the large suitcase, his schoolwork and books into the second one. His bedding went into the dirty clothes hamper. Some posters, a small keyboard, toiletries. Boom. Within an hour we'd packed everything from the room neatly into our car.
"Ready to roll?" I asked.
"Sure, but it's warmer than I thought. I shouldn't have worn a long sleeve shirt," he answered. We were both sweating from the trips up and down from the fifth floor.
"Well, you have about 100 short sleeve shirts in the trunk," I noted. It took just minutes to pull one out of the suitcase and we headed toward home.
The drive back is a metaphor.
For hours and hours, we rolled through the country with no problems. The sun was shining. I listened to podcasts, he tapped into music on his phone. It could have been 2019, except that we also spent some time plotting our family's future in this changing world.
Suddenly, after seven hours on the road, as I was preparing to have him drive again, came a challenge. A downpour of biblical proportions. I slowed but didn't stop.
Gripping the wheel and concentrating on the stripes in the middle of the road, I steered carefully for more than hour, able to see just well enough to keep us, like our nation, moving forward. While I kept a white-knuckle grip on the wheel, I noted that my son was playing on his DS, apparently oblivious to the storm.
And I realized: this is the difficulty of being an adult.
"Adulting" means, simply, that there are people depending on you. You cannot afford to let them down even when it's all too easy to do so. I suddenly had a memory. I was a child, riding home through a snowstorm on New York Route 17, an interstate-like highway. That trip meant hours of high-stress driving for my father. As a child, though, I had no idea the pressure he was under. Now I understand.
Also on my mind is the fact that, just a week before this trip, I'd lost my job. I'm laid off from a trade association because of COVID-19. It was nothing I'd done, they told me. They just needed to make some cuts in this new economic environment. I have no idea when or whether I'll land another gig. That makes for some white knuckles as well as gnawed fingernails.
What will the world look like after Covid-19? The thought is as scary to me as a downpour in I-81. However, we cannot avoid the coming crisis. We can only steer through it as best we can, keeping our eyes on the road and hoping for the best.
My family is depending on me, just as your family is depending on you. As Americans, we'll all get through this together.
As for us, well, we ate my son's candy hearts on the drive back. When I dropped them in the mail, I'd never expected to see them again. Yet despite the reminder, the world of Valentine's Day 2020 no longer exists.
Our family is luckier than most. My son had a normal high school graduation, and more than half of a normal freshman year. He can expect to see the Vandy campus again in good times.
As for the rest of us: Yes, there's a downpour. But we need to keep our eyes on the road and keep moving forward. Adulting remains difficult. But it's more important than ever right now.
Richard Tucker blogs at evenbetterwords.com. He's looking for a job in communications.
BREAKING NEWS ABOUT THE CORONAVIRUS PANDEMIC
WHO: THERE'S NO EVIDENCE WEARING A MASK WILL PROTECT YOU FROM CORONAVIRUS
FAUCI PREDICTS ANOTHER CORONAVIRUS OUTBREAK IN THE FALL WITH A 'VERY DIFFERENT' OUTCOME
MORE THAN 1000 TEST POSITIVE FOR CORONAVIRUS AT TYSON MEAT PLANT THE DAY IT REOPENS
TEXAS REPORTS SINGLE-DAY HIGH IN CORONAVIRUS DEATHS TWO WEEKS AFTER REOPENING
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Environment Here are the top national parks in the country, recommended by...
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Longevity Prescriptions of antipsychotic drugs for people with dementia...
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IBU provisionally suspends nine Kazakhstan athletes linked to criminal doping investigations
Nine Kazakhstan biathletes have been provisionally suspended in connection with criminal doping investigations in Austria and Italy, the International Biathlon Union (IBU) announced today.
The nine competitors were among the 10 athletes from the nation detained during last year's IBU World Championships in Hochfilzen.
The athletes involved now have the opportunity to provide the IBU with an "adequate explanation" of the alleged anti-doping rule violation.
The IBU will then decide if any further action should be taken after they have received the clarification from the nine biathletes.
The list of competitors given provisional sanctions today include Asian Games gold medallists Olga Poltoranina and Yan Savitsky, double Winter Universiade gold medallist Galina Vishnevskaya and Alina Raikova, who claimed the individual title at the 2015 Winter Universiade.
Anton Pantov, a double Winter Universiade medallist, Darya Klimina, Maxim Braun, Anna Kistanova and Vassiliy Potkorytov are the other athletes implicated.
In a statement, the IBU confirmed their provisional suspensions were linked to investigations being carried out in Austria and Italy.
It marks the latest development in the ongoing situation concerning the Kazakhstan biathlon team.
Kazakhstan's biathlon team doctor Khossilbek Tagayev was provisionally suspended in March after the Italian police found banned substance prednisolone in his hotel room.
It came during searches of the accommodation of Kazakhstan's biathlon team during a World Cup event in Antholz in South Tyrol in January, conducted by the Italian police.
Galina Vishnevskaya has also been provisionally suspended by the IBU ©Getty Images
The raid was directly linked to the probe conducted by Austrian authorities at the Hochfilzen World Championships, where several medical products were seized and 10 athletes were detained.
Later, the Kazakh athletes implicated took tests to check for doping but they all produced negative results.
The previous month, in East Tyrol, suspicion had loomed over the Kazakh team after a local resident reported seeing passengers in several minibuses disposing of large cardboard boxes at a petrol station.
The police recovered them and it was discovered the boxes contained disposable materials such as syringes and ampoules for injections.
There were also handwritten notes indicative of doping offences.
The boxes also contained various accreditations for IBU events and were transferred to the Anti-Doping Agency of Austria.
Analysis found documents and records which showed that the contents of the boxes belonged to the national team of Kazakhstan.
The World Anti-Doping Agency are also investigating and have met with Olle Dahlin, elected IBU President in September.
March 2018: IBU provisionally suspend Kazakhstan team doctor after banned substance found in hotel room
January 2018: Kazakhstan biathlon team have accommodation and vehicles searched by Italian police
February 2017: Kazakh biathletes hail redemption after doping suspicions at World Championships
February 2017: Detained Kazakh biathletes all return negative doping results
February 2017: Police raid Kazakh national team hotel at IBU World Championships amid doping suspicions
Liam Morgan Senior chief reporter
Follow @LMorgan21
Since joining insidethegames.biz, in 2015 Liam Morgan has covered a variety of international multi-sport events and conferences, including the Pyeongchang 2018 Winter Olympics, the Gold Coast 2018 Commonwealth Games and the Lillehammer 2016 Winter Youth Olympics. He also reported from the 2017 IOC Session in Lima and three editions of the FIFA Congress. He graduated from Southampton Solent University in 2014 with a BA First Class honours degree in Sports Journalism.
Contact Liam
Read more of Liam's articles
Follow @LMorgan21 on Twitter


Hi sorry for the preview instead of a full article. I just wanted to tell you hopes that are going on for this site and you guys deserve to know what's coming.
Anyway I was digressing a little… I told people of this a while ago that I would do a movie series. Well that movie series is the Left Behind movie trilogy! While not every movie is totally great (the first film decided not to share a message. Why? Just because people that didn't know the Lord would see it?? Did they not understand the reason for the movie?) but it is still better probably than the Nic Cage version. Note: I bought that version at a 7-11 and STILL have not watched it. I just hope it's worth the 3.99 and not nixing the 3, haha! Just kidding. Was an exaggeration but I don't remember the price, anyway….!
I will soon post a blog article on blessings. I rarely do drafts before posting online but it started that way for this case (also happened for "Tight and Secure" too!) so hope to finish it.
And there is one more tri-book review. This one under a completely different theme. Well kind of like the ones about the fiction thriller and the true story autobiography of the young woman who ended up knowing Christ, like that!
A close family of mine has just had a terrible tragedy. I have grown up with them almost all my life and a main member died a traumatic way. They need prayers of comfort and love revealed during this time.
Well these two days I feel like have been as long as a year! A time of growth and tragedy, a time to try to move on and mature, just so much going on.
It started on Wednesday. My brothers were planning on going to a bar in Philly and perform for Open Mic. I was getting ready to tell someone I knew from the past about how well I was doing when I was devastated by news.
The death. No one saw it coming. I worried and sent a message then called the family member that made the post. We cried as we talked. Then I told members I was planning to see them as soon as I could the next day… it was horrible to not see them, even panic because they were going through this and I felt like I could only do this stupid show!
I thought I was helpless with this music thing and had to leave it up to God for the moment. Boy, that was so tough.
After knocking some sense into some people about what had happened (haha, just kidding guys, hope you're not reading this!) they wanted to comfort me. They were worried like crazy about an entertainment show at a place far from here that they had hoped to go to and then found out about the death.
They realized immediately how serious it was.
While I was gathering my bearings I made them some breakfast. Was surprised I fed myself but I had to get some nourishment.
Making coffee then toast with some peanut butter, honey and sunflower seeds. I also gave the two people (one my bro!) a hard-boiled egg each since they hadn't eaten yet.
After a slow, agonizing few hours we eventually got to do a quick rundown of our songs.
Then we headed out to get to the Septa travelling, just the two bros and I!
Getting there it was so awesome to see in person a guy there leading the Open Mic (and he also was playing songs for the people there with his well-known band). I hope he will be alright and care about him a lot.
We got to perform and again we had sound issues. Oh well… I have got to learn to get right into the mic, almost kiss it when I sing (haha!).
Anyway awesome bands played. Indie music is pretty awesome. I thought one big highlight was FLNG. They played awesome! Also I got to talk to Rob McCall. He was doing some great stuff and after it was all over we walked most of the way toward our Subway destination and chatted with him.
I told him that my one bro and I had this band years ago. We just until recently had never even played a note, hahaha!
Then I told him stuff as to why I wasn't able to go to all Open Mics and he said, that is great. I was involved in ministry at my church.
He saw us off and we said good bye. Then we headed home. Oh yeah just because I was being silly I thought before we got there to write on my knuckles, a "tattoo" as it were.
People that know me know what the subject was. Rabbits!
I put on the eight fingers "I heart bunny" then an outline of a bunny face. I am weird, if you don't know that you don't know me, hahahaha!
Anyway I got home, ate and went to bed.
Then I got to see the family the next day….
Well it's just crazy how drastic life can be at times. None of us can get used to it but we always have to adjust. It's sometimes hard, very hard but you and I can find the strength to carry on if you have faith. Thank God if you know Him!
Just saw this and I think it's perfect for anyone who needs tips on crafting good films. Because of this video I got an idea on a way to punch up a script I hadn't put down yet. He says about avoiding cliché like having the first scene be the character waking up. This can be a good idea if used right (if it's integral to the story) but if it is only because it's the beginning it shouldn't have to happen.
Before I show this video I want to give you my tips (some might already be in the vid but still) if you want read it. One. If you want to make a good film story or transfer a short-story you've written into one you must show visuals and make them a little more important than dialogue scenes. People love dialogue but if yours isn't that interesting or goes on too long people will not stay invested in anything else you have to say. Two. Avoid putting in scenes that are not important. Because pacing is key to keeping the rhythm of a film and if you don't keep it well-balanced again you will lose your audience.
Okay it's all I can think of right now. I only think of this stuff because of hours of watching behind-the-scenes features of films and also hearing audio commentaries. If you haven't done that before and you want advice, it is great stuff to check out!
Wow, everything I said is in some way in the video. This has loads more tips though so check it out!!!
I plan on making at least a film review (maybe even covering a trilogy!) so keep your eyes peeled for that.
Hope Christians everywhere will want to or encourage this type of wonderful new tradition in any way they can. It is a spiritual alternative to the Satanic traditions that happen the 31st of every normal October.
and wait expectantly in the day.
Let the Lord just lead the way….!
I have noticed some heavy spiritual attack within my church community. People want to live for God and there was some chaos and heartbreak going on this past week. I hope that you even while you may be hurting that you will always feel encouraged by the One or know that Jesus loves you still. Jesus loves even you now matter what you are going or have gone through.
This is an original song that I hope you will enjoy reading the lyrics of and may you feel encouraged to know of hope and love.
With Jesus… will He fail on you? The answer is no.
With Jesus He knows the way because He is your King.


text:
May's Resignation means Brexit is Back to Square One
Posted By Ed Moya On 05/24/2019 @ 5:20 pm In AUD,CAD,Commodities,Crypto,Currency,Dean's FX,Economic Exposure,EUR,FX,GBP,Indices,News,USD,Yield and Indices | No Comments
One word perfectly sums up the events of this week. Disappointing. For some it began with the uninspiring conclusion of Game of Thrones, long-term stock investors grimaced at the lack of progress with the US-China trade war, and FX traders will see the British pound remain in limbo after Prime Minister May admitted defeat in failing to deliver a Brexit deal.
Theresa May's watch will end on June 7th and the Brexit process basically goes back to square one. The short-term fate of the British pound will depend on the final candidates preference for type of Brexit (soft/hard) and if they appear capable of negotiating across party lines to secure the votes in getting a deal pushed through Parliament.
Brexit Aftermath: New PM to decide on hard or soft Brexit
Trade talks need to resume before irreparable damage to confidence
Oil geopolitical risks to heat up this summer
European politics remain front and center as Nationalist and far-right parties are expected to make progress in gaining European Parliamentary seats. The result from the elections will likely deliver further friction within the bloc and more hurdles on integration. The trade war has delivered a major blow to risk appetite with all three major US indexes posting weekly losses, as well big monthly declines. Both the Chinese and Americans appear poised to ramp up domestic support/stimulus in the event of a prolonged standoff. Someone will blink and lead the return to the negotiating table, both sides are motivated, but we may need to see more market carnage before that happens. Key upcoming events include a rate decision from the Bank of Canada, the annual budget release for New Zealand, US preliminary Q1 GDP reading, China's Manufacturing PMI reading and Canadian GDP.
PM Boris Johnson unlikely
Theresa May's announcement means Brexit is right back to where we were right after the Referendum. The Brexit outcomes are still plentiful: We could see the Brexit deadline extended beyond Halloween, a general election is very possible or a second referendum asking for a no-deal or to remain.
Uncertainty will remain in place until we learn more about who will take over the negotiations and that process will take almost two months. The current favorite to replace Theresa May is former Foreign Secretary Boris Johnson, a hardliner that would suggest a greater risk for a no-deal Brexit. History however is not on Mr. Johnson's side, as he was the odds on favorite last time.
The British pound is likely to remain volatile and the risks could be to the upside following the recent precipitous drop. The markets will likely expect an orderly exit, and that should prove supportive for cable.
Dollar bottom in place
The latest reports on housing and durable goods is not pointing an optimistic outlook for the US economy. US economic data points still outperform the other advanced economies, but its strength is heavily priced in FX markets, which could mean we may have just seen a key bottom put in place for the dollar.
While the financial markets have been hit with a wave of risk aversion that has sent bond market yields plummeting. Despite the recent optimism from the Fed and their minutes, risks to the downside are growing and the Fed may need to deliver another dovish message in the coming months that cement rate cut bets in the nearer future. Current expectations are at 55% for a rate cut at the September meeting, but that could increase as growth has been dealt a strong blow from the US-China trade war and cautious tones from this past earnings season.
Crude prices have been battered by global demand concerns, rising stockpiles in the US and fading expectations that OPEC and their allies will unanimously be on board and deliver an extension of production cuts this year. The markets have been closely paying attention to the trade war between two largest economies and the lack of progress is likely to put an unexpected dent in demand forecasts for the second quarter.
The wildcard for oil remains geopolitical risks as tensions remain high between the US and Iran. The US appears set on showing Iran they are prepared for an escalation in the region. President Trump is sending 1,500 troops to assure freedom of navigation in the area. While both sides appear not interested in starting a war, tensions are flaring up and the risks for some conflict are growing.
Spare capacity and output at risk should support for a stabilization in prices in the short-term, but if we see another multi-million barrel build with US inventories, oil could see the bearish correction continue.
Gold prices remain in a quagmire as wave of risk aversion that has sent bond market yields plummeting has yielded minuscule gains for the safe-haven. The last six weeks of yellow metal trading has seen alternative bullish/bearish trading that for the most part has been kept in a tight $1,270 to $1,300 range. The deflationary conditions globally have hurt the inflationary pressure reason for owning gold. With US equities hovering within 5% of their record highs, gold is likely to have difficulty breaking out. Gold may need the Fed to confirm what markets are pricing in already and signal rate cuts are coming. That however may not happen anytime soon, if we see a resumption of constructive trade talks between the US and China.
Some past Bitcoin skeptics, such as JP Morgan and Facebook are becoming believers in cryptocurrencies. The recent surge with Bitcoin was supported on the continued progress with mainstream commerce adoption. First, we saw JP Morgan create the first US bank-backed cryptocurrency, then we saw Fidelity launch plans for institutional bitcoin trading, and E-Trade intends to offer cryptocurrency trading on their platforms. The financial community is seeing more companies want to take advantage of the digital currency and the latest to join is Facebook. The social media giant aims at competing with banks in offering secure and cheaper ways of sending money.
Bitcoin is meteoric rise immediately was followed by a tulip-mania crash. The current rally is seeing price trade around the $8,000 level and further mainstream acceptance could prove vital for the bullish momentum to continue. To the upside the psychological $10,000 level provides key resistance, while $6,500 remains major support.
Monday, May 27th
US Holiday
Tuesday, May 28th
1:45am ET CHF Q1 GDP q/q
4:00am ET EUR Eurozone M3 Money Supply y/y
5:00am ET EUR Eurozone Confidence data
9:00am ET USD FHFA House Price Index m/m
10:00am ET USD CB Consumer Confidence
5:00pm ET NZD RBNZ Financial Stability Report
2:45am ET EUR France Q1 GDP q/q
2:45am ET EUR France CPI m/m
3:30am ET SEK Q1 GDP q/q
10:00am ET CAD Bank of Canada (BOC) Interest Rate Decision
10:00am ET USD Richmond Fed Manufacturing Index
1:00pm ET MXN Banxico Inflation Report
9:00pm ET NZD ANZ Business Confidence
9:30pm ET NZD Private Capital Expenditure q/q
2:00am ET ZAR M3 Money Supply y/y
8:00am ET BRL Brazil GDP q/q
8:30am ET USD Q1 Preliminary GDP Annualized q/q (2nd reading)
8:30am ET USD Initial Jobless Claims
11:00am ET DOE US Crude Oil Inventories
7:50pm ET JPY Preliminary Industrial Production m/m
9:00pm ET CNY Manufacturing PMI
Friday, May 31st
1:00am ET JPY Housing Starts y/y
3:00am ET CZK Preliminary Q1 GDP q/q
4:30am ET GBP Mortgage Approvals
8:00am ET EUR Germany CPI m/m
8:00am ET ZAR Trade Balance
8:30am ET USD PCE Deflator m/m
8:30am ET USD Core PCE Price Index
8:30am ET USD Personal Spending m/m
8:30am ET CAD GDP m/m
9:45am ET USD Chicago PMI
10:00am ET USD Final University of Michigan Sentiment
Ed Moya [4]
Senior Market Analyst at OANDA
With more than 20 years' trading experience, Ed Moya is a market analyst with OANDA, producing up-to-the-minute fundamental analysis of geo-political events and monetary policies in the US, Europe, the Middle East and North Africa. Over the course of his career, he has worked with some of the world's leading forex brokerages and research departments including Global Forex Trading, FX Solutions and Trading Advantage. Most recently he worked with TradeTheNews.com, where he provided market analysis on economic data and corporate news. Based in New York, Ed is a regular guest on several major financial television networks including BNN, CNBC, Fox Business, and Bloomberg. He is often quoted in leading print and online publications such as the Wall Street Journal and the Washington Post. He holds a BA in Economics from Rutgers University. Follow Ed on Twitter @edjmoya
@https://twitter.com/edjmoya [5]
Follow @https://twitter.com/edjmoya [6]
Latest posts by Ed Moya (see all [4])
US Open – Stocks rally on earnings and on optimism China will contain virus, Oil slides, Gold remains steady [7] - 01/22/2020
BNN Bloomberg with OANDA discussing the US stock market [8] - 01/21/2020
US Open – Davos, Wuhan Virus pressures stocks and yields, Dax touches record high, Oil's supply problem, Gold's range, Bitcoin eases after CBDC think tank formed [9] - 01/21/2020
URL to article: https://www.marketpulse.com/20190524/mays-resignation-means-brexit-back-square-one/
[2] Twitter: #ts-fab-twitter-below-776376
[4] Ed Moya: https://www.marketpulse.com/author/ed-moya/
[5] @https://twitter.com/edjmoya: https://www.marketpulse.comhttp://twitter.com/https://twitter.com/edjmoya
[6] Follow @https://twitter.com/edjmoya: https://twitter.com/https://twitter.com/edjmoya
[7] US Open – Stocks rally on earnings and on optimism China will contain virus, Oil slides, Gold remains steady: https://www.marketpulse.com/20200122/us-open-stocks-rally-earnings-optimism-china-will-contain-virus-oil-slides-gold-remains-steady/
[8] BNN Bloomberg with OANDA discussing the US stock market: https://www.marketpulse.com/20200121/bnn-bloomberg-with-oanda-discussing-the-us-stock-market/
[9] US Open – Davos, Wuhan Virus pressures stocks and yields, Dax touches record high, Oil's supply problem, Gold's range, Bitcoin eases after CBDC think tank formed: https://www.marketpulse.com/20200121/us-open-davos-wuhan-virus-pressures-stocks-yields-dax-touches-record-high-oils-supply-problem-golds-range-bitcoin-eases-cbdc-think-tank-formed/


It is a very good school.
The school is good and the teachers are very nice.
not too bad, but not good. Teachers do not add value to the students. Many people go to tutor shops. Some go to good secondary schools, but many go to second or third tier schools.


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This entry was posted in detox, Raw Food, vegan food and tagged 5-day cleanse johannesburg, detox, healthy food delivery, Raw Food, raw vegan detox. Bookmark the permalink.


text:
Q: Cucumber hooks assistance I have the a feature file that has 5 scenarios within. What I've been googling without success is how to make each scenario start and close a new browser. What I want is to use the same instance of the browser I created when the first scenario is run.
@Before
public void setUp(){
    SeleniumDriver.chromeDriverSetUp();
    SeleniumDriver.getDriver().navigate().to(PropertiesManager.getInstance().getConfig(EPropertiesNames.BASE_URL));
    SeleniumDriver.getDriver().manage().timeouts().implicitlyWait(10, TimeUnit.SECONDS);
}

@After
public void tearDown(){
    SeleniumDriver.getDriver().quit();
}

The code above is are my BaseSteps for my features. I'm new to cucumber so any assistance would be appreciated.

A: As I understand you're looking for smthn like @BeforeSuite and @AfterSuite, i.e. methods that are going to be executed only once per execution, unfortunately there is no such thing in cucumber (in stadard library at least). However, you could consider using shutdown hooks to execute some code after your java app is about to shutdown, but I would strongly recommend to avoid doing so.
There are 2 other recommendations that I could give you:

*

*If for some reason your tests rely on each other - consider to try and make them independent.

*There are other BDD tools that actually have things you're looking for, one them is, for example, Gauge

redpajama_set_name: RedPajamaStackExchange


Right now I am more or less desperately trying to make ends meet for my upcoming trip on the last week in August. Top priorities are the making of a new, superultralight cuben tent and a personal flotation device using my water bottles. As I hinted in an earlier posting I am taking my packraft into what is probably the most inaccessible part of the Swedish mountains: I am going west of Virihaure.
The photo above is from an unforgettable backpacking trip when I was only seventeen and for the first time looked out over what has often been called "the most beautiful lake in Sweden". Virihaure is a big lake in the Laponia Heritage area, Padjelanta NP, close to the border of Norway. The sami summer camp at Staloluokta is seen in the photo and on the eastern shore of Virihaure there are trails and cabins. On the western shore there is nothing, and very often, no one.
The reason for the inaccessability is that Virihaure is part of a chain of some other big lakes and also guarded by some pretty ferocious rivers that have taken the lives of sevaral forders, even among the reindeer keepers in the area. So few people visit west of Virihaure and those that do usually come from Norway or let themselves be ferried over the lake by the Sami.
I have never been west of Virihaure but it struck me about a month ago that thanks to the packraft this area now would be open to a free ranging exploration of the kind that I like.
Wow, sounds fantastic. Virihaure is the most beautiful lake i've ever seen, the coral blue water and the snowcovered peaks of the norweigan mountains on the other side...Looking forward to reading about it when you come back.
On a personal note, trekked from abisko to nikkaluokta in sneakers this summer, and climbed the Kebnekaise. Will never buy another pair of hiking boots, and my nineyearold daughter was very jealous, and wants a pair of mesh sneakers for next year.
Yes, I am really looking forward to this trip. And I am really happy about your daughter and the mesh shoes. I hiked for a couple of days with my 11-year old son recently and have found that even if kids initially do not like to get their feet wet, once it has happened it is no big deal.
After being recently introduced to packrafting by Hendriks blog entry, I decided to buy a packraft myself. I had a vision of how I would use it, and that vision was to go west of Virihaure as I have wanted to get there since I first visited Sarek some five years ago. What a coincidence!
Yeah, I think I wrote in some of my articles on packrafting (you can search the blog)that one interesting thing with packrafting is that you can do hikes from a new perspective even in country you know well.You are sort of liberated from watery restraints. The Virihaure area is typical of an area with watery restraints, so I am not really surprised that other people get the same idea when you enter a packraft into the equation.
Looking forward to hearing about it when you get back as I've always been a kayaker. Developing my whitewater skills these days but sea kayaking in Abel Tasman, NZ is what started it all. Packrafting & open canoeing is something that truly enthrals me so really fancy doing some in the spring/summer next year.
Jörgen, this sounds like a wonderful trip, I cannot wait till I get the chance to visit northern sweden again. I also want to know more about the new shelter. Where did you get the cuben from and what colour is it?
Just got back, and yes it was a wonderful trip. The last couple of half days I walked the Nordkalotten trail at the end/beginning and ending up in Sulitelma. Truly a wonderful area that I think you should look into.
I'll do some write ups on the trip ASAP, but the Cuben was bought at www.extremtextil.de and I also used their tape. Seemed to work very well. But more on that later. Unfortuntately they were 40 cm short of the amount of olive green Cuben that I really wanted, so I bought the ordinary translucent.


text:
\section{Introduction}

In 1975, Szemer\'{e}di proved the following famous result about arithmetic progressions (see \cite{szemeredi75})

\begin{theorem}
Let $\epsilon > 0$ and let $k$ be a positive integer. Then there exists a $N_0 = N_0(k, \epsilon)$ such that whenever $N \ge N_0$ and $A \subset [N]$ has $|A| \ge \epsilon N$, then $A$ contains a non-trivial arithmetic progression of length $k$
\end{theorem}

Here we are using $[N]$ to denote the set $\{1, 2, \dots, N\}$. There are several ways one could try and strengthen this result. In \cite{bergelsonhostkra05}, Bergelson, Host and Kra conjecture that, instead of finding just one arithmetic progression, one can find a large number of arithmetic progressions all with the same common difference. More specifically, they make the following conjecture.

\begin{conjecture}\label{BHK}
Let $0 < \alpha < 1$, let $\epsilon > 0$ and let $k = 3$ or $4$. Then there exists an $N_0$ such that for all $N \ge N_0$ and any $A \subset [N]$ with $|A| \ge \alpha N$ there must exist a $d \neq 0$ such that the set $A$ contains at least $(\alpha^k - \epsilon)N$ distinct arithmetic progressions of length $k$ and with common difference $d$.
\end{conjecture}

Note that if we choose $A$ at random by putting each element of $[N]$ in $A$ independently with probability $\alpha$, then each arithmetic progression is in $A$ with probability $\alpha^k$. It follows that the bound $(\alpha^k - \epsilon)N$ is best possible. In the same paper, Bergelson, Host and Kra also provide a counter-example showing that the conjecture is false when $k=5$. Conjecture \ref{BHK} is now a theorem. Indeed, in 2005 Green \cite{ARLoriginal} proved the $k=3$ case of the above conjecture, by first  proving an arithmetic regularity lemma. This arithmetic regularity lemma is a kind of analogue of Szemer\'{e}di's graph regularity lemma, in the setting of abelian groups instead of graphs. In 2010 Green and Tao \cite{ARLadvanced} proved a more powerful arithmetic regularity lemma, and used it to prove Conjecture~\ref{BHK} for $k=4$ as well as $k=3$.

A different way to generalise Szemer\'{e}di's theorem on arithmetic progressions is to look for higher-dimensional patterns in subsets of $[N]^d$. This is known as the Multidimensional Szemer\'{e}di Theorem. The Corners Theorem is a special case of this.

\begin{theorem}\label{cornerstheorem} [Corners Theorem]
Let $\epsilon > 0$. There exists an $N_0=N_0(\epsilon)$ such that for any $N \ge N_0$ and any $A \subset [N]^2$ with $|A| \ge \epsilon N^2$, there exist $x, y \in [N]$ and an integer $d \neq 0$ such that $(x, y), (x+d, y)$ and $(x, y+d)$ are all in $A$. The triple of points $((x, y), (x+d, y), (x, y+d))$ is called a corner.
\end{theorem}

The Corners Theorem was first proved by Ajtai and Szemer\'{e}di in 1974 \cite{ajtaiszemeredi74}. Now, several proofs exist, see for example \cite{shkredov06} and \cite{furstenberg81}. In light of the Bergelson-Host-Kra conjecture, it is natural to make the following conjecture.

\begin{conjecture}\label{benshope}
Let $0<\alpha<1$ and $\epsilon>0$. There exists an $N_0$ such that for any $N \ge N_0$ and any $A \subset [N]^2$ with $|A| \ge \alpha N^2$, there is some integer $d \neq 0$ such that there are at least $(\alpha^3-\epsilon)N^2$ corners with that given $d$ all of whose vertices lie in $A$.
\end{conjecture}

We will show that this conjecture is in fact false. Indeed, there is a construction in which there are at most $C\alpha^{3.13}N^2$ corners for any given $d \neq 0$, where $C$ is an absolute constant. Chu \cite{chu11} considered an ergodic theory analogue of the above problem, and proved that the exponent 3 is not right in that setting. Our construction is broadly the same as Chu's, however her construction does not traslate directly to our setting and we need to use concentration of measure arguments. Our construction is also an improvement on Chu's in that we get a better bound on the exponent of $\alpha$.

We will be working with abelian groups, rather than the set $[N]$, but one can easily deduce the result for $[N]$ by embedding it into $\mathbb{Z}/N\mathbb{Z}$ in the usual way.

For a family of abelian groups $(G_n)_{n=1}^\infty$ with $|G_n|$ tending to infinity, we define the threshold density for corners $M(\alpha)$ as follows.

$M(\alpha)$ is the largest $M \in [0,1]$ such that for any $\epsilon > 0$ there is an $N_0$ such that for all groups $G$ in our family with $|G| = N \ge N_0$, and for all subsets $A \subset G \times G$ of density at least $\alpha$, there is a $d \in G \setminus \{ 0 \}$ with $|S_d|>(M - \epsilon) N^2$.

Over the course of this paper, we relate the problem of finding the threshold density for corners to the following problem about random variables.

Let $f$ be a piecewise constant function on $[0,1]^3$, taking values in $[0,1]$. By piecewise constant, we mean that there is a positive integer $m$ and real numbers $0 = x_0 \le x_1 \le \dots \le x_m = 1$, $0 = y_0 \le y_1 \le \dots \le y_m = 1$ and $0 = z_0 \le z_1 \le \dots \le z_m = 1$ such that $f$ is constant on $[x_i, x_{i+1}) \times [y_j, y_{j+1}) \times [z_k,z_{k+1})$ for each $0 \le i,j,k \le m-1$. We define
\[ T(f) := \int_0^1 \int_0^1 \int_0^1 \left( \int_0^1 f(x, y, z') \mathrm{d} z' \right) \left( \int_0^1 f(x, y', z) \mathrm{d} y' \right) \left( \int_0^1 f(x', y, z) \mathrm{d} x' \right) \mathrm{d} x \mathrm{d} y \mathrm{d} z\]
and
\[ m(\alpha) := \inf_f T(f) \]
where the infimum is taken over all piecewise constant $f: [0,1]^3 \to [0,1]$ with
\[ \int_0^1 \int_0^1 \int_0^1 f(x, y, z) \mathrm{d} x \mathrm{d} y \mathrm{d} z = \alpha. \]
Note that if we let $X, Y$ and $Z$ be independent $U[0,1]$ random variables, and setting $f = f(X, Y, Z)$, we have
\[ T(f) = \mathbb{E}(\mathbb{E}(f|X, Y)\mathbb{E}(f|X, Z)\mathbb{E}(f|Y, Z)) \]
and
\[ m(\alpha) = \inf_f T(f) \]
where the infimum is taken over all piecewise constant $f$ with $\mathbb{E}(f) = \alpha$.

Our main results are the following:

\begin{theorem}\label{pt1}
For the family of all abelian groups, $M(\alpha) \le m(\alpha)$.
\end{theorem}
\begin{theorem}\label{pt2}
Let $m':[0,1] \to [0,1]$ be any convex function of $\alpha$ satisfying $m'(\alpha) \le m(\alpha)$ for all $\alpha \in [0,1]$. For the family $G_n = \mathbb{F}_2^n$, $M(\alpha) \ge m'(\alpha)$.
\end{theorem}

Note that if $m(\alpha)$ is itself convex, then we may take $m'=m$ and combine the two theorems to get $M(\alpha) = m(\alpha)$ for $G_n = \mathbb{F}_2^n$.
In Section~\ref{construction} we prove Theorem~\ref{pt1}, using a random construction.
In section~\ref{regularity} we use regularity techniques to prove Theorem~\ref{pt2}.  The method generalises to the case $G_n = \mathbb{F}_p^n$, with $p$ any fixed prime.
Finally in section~\ref{bounds} we prove the bounds $\alpha^4 \le m(\alpha) \le C\alpha^{3.13}$ for an absolute constant $C$, and so deduce that $M(\alpha) \le C\alpha^{3.13}$ in the general case, and $M(\alpha) \ge \alpha^4$ in the finite field case.

We will work with an alternative formulation of the corners problem, which is more natural in some respects. Instead of working in $G \times G$, we work in the subset $P \subset G^3$ given by $P =\{(x, y, z) \in G^3 | x+y+z = 0\}$, and a corner is a triple of the form $(x+d, y, z), (x, y+d, z), (x, y, z+d)$. This is clearly seen to be equivalent to the original formulation, by forgetting one of the coordinates. We use this formulation because it treats the three vertices of the corner equally, whereas in the original formulation, the vertex $(x, y)$ was special.

We will also need the following Lemma, which is essentially giving us an alternative version of the random variable problem.
\begin{lemma}\label{discrete}
Given a piecewise constant $f : [0, 1]^3 \to [0, 1]$, we can construct independent random variables $X', Y'$ and $Z'$ which each take only finitely many values, and a function $f'=f'(X', Y', Z')$ taking values in $[0,1]$ with $\mathbb{E}(f') = \mathbb{E}(f)$ and $T(f') = T(f)$.

Conversely, given a trio of random variables $X', Y', Z'$ each taking only finitely many values, and a function $f' = f'(X', Y', Z')$ taking values in $[0,1]$, we may construct a piecewise constant $f : [0,1]^3 \to [0,1]$ with $\mathbb{E}(f) = \mathbb{E}(f')$ and $T(f) = T(f')$.
\end{lemma}
\begin{proof}
\underline{First part:} Suppose we are given such a function $f$. Let $m$, a positive integer, and $0 = x_0 \le x_1 \le \dots \le x_m = 1$, $0 = y_0 \le y_1 \le \dots \le y_m = 1$ and $0 = z_0 \le z_1 \le \dots \le z_m = 1$ be such that $f$ is constant on $[x_i, x_{i+1}) \times [y_j, y_{j+1}) \times [z_k,z_{k+1})$ for each $0 \le i,j,k \le m-1$. Given $X, Y, Z \sim U[0,1]$ independent we define new random variables $X', Y'$ and $Z'$ taking values in $[m]$ by setting $X'$ to be the $i$ such that $X \in [x_{i-1}, x_i)$, $Y'$ to be the $j$ such that $Y \in [y_{j-1}, y_j)$ and $Z'$ to be the $k$ such that $Z \in [z_{k-1}, z_k)$. The independence of $X', Y'$ and $Z'$ follows from the indepence of $X, Y$ and $Z$. By the piecewise constant assumption, the value of the function $f$ is determined by the values of $X', Y'$ and $Z'$, so there exists a function $f' : [m]^3 \to [0,1]$ such that $f(X, Y, Z) = f'(X', Y', Z')$. The first part follows easily.

\underline{Second part:} Now suppose we are given indepent random variables $X', Y'$ and $Z'$, each taking only finitely many values. Without loss of generality, we may assume that $X', Y'$ and $Z'$ take values in $[m]$ for some positive integer $m$. This is because the actual values taken by $X', Y'$ and $Z'$ are unimportant, so we can simply relable them. We then define real numbers $0 = x_0 \le x_1 \le \dots \le x_m = 1$, $0 = y_0 \le y_1 \le \dots \le y_m = 1$ and $0 = z_0 \le z_1 \le \dots \le z_m = 1$ by setting $x_i = \mathbb{P}(X' \le i), y_i = \mathbb{P}(Y' \le i)$ and $z_i = \mathbb{P}(Z' \le i)$. Then $\mathbb{P}(X' = i) = x_i - x_{i-1} = \mathbb{P}(X \in [x_{i-1}, x_i))$ so we may couple $X$ with $X'$ in the same way as we did in the first part. The same applies, of course, for $Y$ with $Y'$ and $Z$ with $Z'$. We can then construct a piecewise constant $f : [0,1]^3 \to [0,1]$ with $f(X, Y, Z) = f'(X', Y', Z')$, and so the second part follows.
\end{proof}
\section{Construction}\label{construction}
In what follows, we will need some properties of the functions $m$ and $M$, which we will establish now.

\begin{lemma}\label{continuity}
Let $m' : [0,1] \to [0,1]$ be convex with $m'(\alpha) \le m(\alpha)$ for all $\alpha \in [0,1]$. $M$, $m$ and $m'$ are all increasing functions of $\alpha$, and $m$ is uniformly continuous.
\end{lemma}
\begin{proof} $M$ and $m$ are trivially increasing. $m(0) = 0$ so $m'(0) = 0$, which implies $m'$ is increasing by convexity. So we focus on showing that $m$ is uniformly continuous.
Given any piecewise constant function $f: [0,1]^3 \to [0,1]$, we can form a new function $f_\epsilon$ by setting $f_\epsilon = \epsilon + (1-\epsilon)f$. This function will have $\mathbb{E}(f_\epsilon) = \epsilon + (1-\epsilon)\mathbb{E}(f) = \mathbb{E}(f) + \epsilon(1-\mathbb{E}(f))$, and
\begin{align*}
T(f_\epsilon) & = \mathbb{E}(\mathbb{E}(f_\epsilon|X, Y)\mathbb{E}(f_\epsilon|X, Z)\mathbb{E}(f_\epsilon|Y, Z)) \\
& \le \mathbb{E}((\mathbb{E}(f|X, Y)+\epsilon)(\mathbb{E}(f|X, Z)+\epsilon)(\mathbb{E}(f|Y, Z)+\epsilon)) \\
& \le T(f) + 3\epsilon + 3 \epsilon^2 + \epsilon^3.
\end{align*}
The last inequality above is obtained by expanding and using the trivial bound $f \le 1$. By taking $f$ to have $\mathbb{E}(f) = \alpha$ and $T(f)$ approaching $m(\alpha)$, we deduce $m(\alpha) \le m(\alpha + \epsilon(1-\alpha)) \le m(\alpha) + 3\epsilon + 3 \epsilon^2 + \epsilon^3$, which implies continuity of $m$ at all points except from 1. To show continuity at 1, we use the following inequality:
\[T(f) \ge \mathbb{E}(\mathbb{E}(f|X, Y)+\mathbb{E}(f|X, Z)+\mathbb{E}(f|Y, Z)-2) = 3\mathbb{E}(f)-2\]
so $m(\alpha) \ge 3\alpha-2$, and $m$ is therefore continuous at 1. $m$ is compactly supported and continuous, therefore it is uniformly continuous.
\end{proof}

We will use a probabilistic construction to prove Theorem~\ref{pt1}. We let $\epsilon_1, \epsilon_2, \epsilon_3 > 0$ and assume $N$ is sufficiently large in terms of $\epsilon_1, \epsilon_2$ and $\epsilon_3$. We take a piecewise constant function $f: [0,1]^3 \to [0,1]$ with $E(f)= \alpha$ and  $T(f)= m_1$, where $m_1 < m(\alpha) + \epsilon_1$ (this is possible by the definition of $m$). We then construct our subset as follows: for each $g \in G$, we take independent $U[0,1]$ random variables $X_g, Y_g, Z_g$. We then include each point $(x, y, z) \in P$ in our set $A$ independently with probability $f(X_x, Y_y, Z_z)$. Because $\mathbb{E}(f) = \alpha$, this means our subset $A$ will have expected size $\alpha N^2$.

Now, let $(x+d, y, z), (x, y+d, z), (x, y, z+d)$ be an arbitrary corner (with $d \neq 0$). The probability that this corner is contained in $A$ is
\[ \mathbb{E}(f(X_{x+d},Y_y, Z_z)f(X_x, Y_{y+d}, Z_z)f(X_x, Y_y, Z_{z+d})) .\]
The random variables $X_{x+d}, Y_{y+d}, Z_{z+d}$ are each independent of all the other random variables appearing here, so the above probability is in fact
\[ \mathbb{E}(\mathbb{E}(f|X_x, Y_y)\mathbb{E}(f|X_x, Z_z)\mathbb{E}(f|Y_y, Z_z))=T(f)=m_1 .\]
We would like to argue that the size of $A$ must, with high probability, be close to its expected value, and similarly for the size of the set of corners of given $d$.

\subsection{Size of $A$}
We say a subset $Q \subset P$ is independent if for each pair of distinct points $ p_1 =(x_1, y_1, z_1), p_2 =(x_2, y_2, z_2)$ in $Q$, $x_1 \neq x_2, y_1 \neq y_2$ and $z_1 \neq  z_2$. This condition is equivalent to the set of events $\{ p \in A | p \in Q\}$ being mutually independent. Now, let $Q$ be an independent set of size at least $N/3$. It is easy to see such a set must exist. Indeed, any maximal independent set has size at least $N/3$. Also note that, for any $p \in P$, $Q+p$ will also be an independent set.

Now, for any $p \in P$ the size of $(Q+p) \cap A$ is the sum of $|Q|\ge N/3$ independent Bernoulli random variables, so it is distributed as the binomial distribution $B(|Q|, \alpha)$. By concentration of measure arguments (see, for example, Hoeffding \cite{hoeffding63}) we know that for binomial distributions $X \sim B(n, p)$ the tail probabilities $\mathbb{P}(|X-np|/n > c)$ decrease exponentially with $n$. Therefore $\mathbb{P}((Q+p) \cap A < (\alpha - \epsilon_2)|Q|) < 1/3N^2$ for $N$ sufficiently large. So, with probability at least $2/3$, $|(Q+p) \cap A|>(\alpha - \epsilon_2)|Q|$ for all $p \in P$. By averaging over $p \in P$, we have $|A|>(\alpha - \epsilon_2)N^2$.
\subsection{Size of $S_d$}
We follow a similar method here, with a couple of modifications. For a given $d \neq 0$, we can consider $S_d$, the set of corners of size $d$, to be a subset of $P_d = \{(x, y, z) \in G^3 | x+y+z+d = 0\}$, by setting $(x, y, z) \in S_d$ if the corner $(x+d, y, z), (x, y+d, z), (x, y, z+d)$ is in $A$. We now define a subset $Q \subset P_d$ to be independent if for each pair of distinct points $ p_1 =(x_1, y_1, z_1), p_2 =(x_2, y_2, z_2)$ in $Q$, $x_1 \notin \{x_2, x_2+d, x_2-d \} $ and similarly for the other coordinates. This condition ensures that the set of events $\{ p \in S_d | p \in Q\}$ are mutually independent. For each $d$, we can take an independent $Q_d \subset P_d$ with $|Q_d| \ge N/9$, and for all $p \in P$, $Q_d+p$ will also be independent.

Since for all $p \in P$ the size of $(Q_d+p) \cap S_d$ is the sum of $|Q_d|\ge N/9$ independent Bernoulli random variables, it is binomially distributed and therefore $\mathbb{P}((Q_d+p) \cap S_d > (m_1 + \epsilon_3)|Q_d|) < 1/3N^3$ for $N$ sufficiently large. So, for each $d$, with probability at least $1-1/3N$, we have $|(Q_d+p) \cap S_d|<(m_1 + \epsilon_3)|Q_d|$ for all $p \in P$. By averaging over $p \in P$, we have $|S_d|<(m_1 + \epsilon_3)N^2$. This happens simultaneously for every $d \in G\setminus \{0\}$ with probability at least $2/3$.

Putting the two estimates together, we see that with probability at least $1/3$, we have that both $|A|>(\alpha - \epsilon_2)N^2$ and $|S_d|<(m_1 + \epsilon_3)N^2<(m(\alpha) + \epsilon_1 + \epsilon_3))N^2$. So, if we set $\alpha' = \alpha - \epsilon_2$, we have $M(\alpha') \le m(\alpha' + \epsilon_2) + \epsilon_1 + \epsilon_3$. By taking $\epsilon_1, \epsilon_2$ and $\epsilon_3$ to be sufficiently small, and using the fact that $m$ is continuous, we get that $M(\alpha') \le m(\alpha')$, as required.
\section{Regularity}\label{regularity}
Throughout this section, we set $G$ to be the group $\mathbb{F}_2^n$.
We define a boxing of $G$ to be the following data:
A subgroup $W \subset \mathbb{F}_2^n$, an integer $m$, and for each coset $V = W^3 + (x, y, z)$ with $(x, y, z) \in P$, partitions of each of $W+x, W+y, W+z$ into $m$ sets: $B_1, B_2, \dots, B_m$ $C_1, C_2, \dots, C_m$ and $D_1, D_2, \dots, D_m$. These cosets $V$ will be refered to as ``outer boxes''. There are $4^{\text{codim}(W)}$ of them, and they form a partition of $P + W^3$.

We will write $(B_i \times C_j)_P$ for the set $\{ (x, y, z) \in P | x \in B_i, y \in C_j \}$. Note that this set can be identified with $B_i \times C_j$ by forgetting the third coordinate. Similarly we write $(B_i \times D_j)_P$ for the set $\{ (x, y, z) \in P | x \in B_i, z \in D_j \}$ and $(C_i \times D_j)_P$ for the set $\{ (x, y, z) \in P | y \in C_i, z \in D_j \}$. We will refer to sets of this type as ``inner boxes''. For each outer box $V$, these inner boxes give us three partitions of the set $P \cap V$.

We will need some Fourier analysis. Given an abelian group $W$, let $W^* = \hom (W, S^1)$ be the dual group and $(g)^\wedge$ and $\hat{g}$ both denote the Fourier transform of a function $g$, defined as $\hat{g}(r) = \mathbb{E}_{x \in W} g(x)r(x) $ for $r \in W^*$. When $S$ is a subset of the group $W$, we will also use $S$ to denote the indicator function of the set $S$. If $B_i \subset W+x$, we define Fourier coefficients for the set $B_i$ to be the values of $\widehat{B_i - x}$ (and the same for $C_i$ and $D_i$). The size of these Fourier coefficients is independent of the choice of $x$.

We will also need the notion of quasirandomness from graph theory. This is the notion used in Szemer\' edi's graph regularity lemma \cite{szemeredi76}. By identifying $(B_i \times C_j)_P$ with $B_i \times C_j$, $A$ restricted to the inner box gives us a bipartite graph between $B_i$ and $C_j$. We say that $A$ is $\epsilon$-quasirandom if for any $B' \subset B_i$ and $C' \subset C_i$ with $|B'| \ge \epsilon|B_i|$ and $|C'| \ge \epsilon|C_i|$, the density of $A$ on $B' \times C'$ differs from the density of $A$ on $B_i \times C_j$ by at most $\epsilon$.

A boxing will be called regular if it satisfies the following two regularity conditions:
\begin{itemize}
\item [(1)] For all but a proportion of at most $\epsilon$ of the outer boxes $V$, all of the associated sets $B_i, C_j, D_k$ are $\epsilon/m^3$ uniform, in the sense that all of their non-trivial Fourier coefficients are bounded in size by $\epsilon/m^3$.
\item [(2)] For all but a proportion of at most $\epsilon$ of the outer boxes $V$, if we choose $(x, y, z) \in P \cap V$ uniformly at random, then for each of the three inner boxes containing $(x, y, z)$ (i.e. $(B_i \times C_j)_P$, $(B_i \times D_k)_P$, and $(C_j \times D_k)_P$ for some $i, j, k$), the probability that the set $A$ is $\epsilon$-quasirandom in the inner box is at least $1-\epsilon$.
\end{itemize}
Given a regular boxing, we will call an outer box $V$ regular if it is not one of the exceptional boxes for (1) or (2).

The main result of this section is:

\begin{theorem}\label{regularitylemma} [Regularity Lemma]
Given $\epsilon>0$, there exists an $n_1$ and $m_1$ such that if $n \ge n_1$, and $A \subset G = \mathbb{F}_2^n$, then there exists a regular boxing where $W$ has codimension at most $n_1$ and the parameter $m$ is at most $m_1$.
\end{theorem}

In the proof that follows, we will not obtain any effective bounds on $n_1$ or $m_1$. The method of proof goes via an energy increment argument, similar to that used in the proof of Green's regularity lemma \cite{ARLoriginal}. However, since we have two different regularity conditions to satisfy, we will be using two energies, one associated with each type of regularity. We define these energies now.

Let $\delta(B_i) := |B_i|/|W|$ be the density of $B_i$ inside $W+x$ and define $\delta(C_i), \delta(D_i)$ similarly. Let $\delta_{B_i, C_j} := |A \cap (B_i \times C_j)_P|/|B_i||C_j|$ be the density of $A$ in the inner box $(B_i \times C_j)_P$, and define $\delta_{B_i, D_j}, \delta_{C_i, D_j}$ similarly.
\[ E_1 = \frac{1}{4^{\text{codim}(W)}}\sum_{\text{cosets} V}\sum_{i=1}^{m}\frac{(\delta(B_i)^2+\delta(C_i)^2+\delta(D_i)^2)}{3} \]

\[ E_2 = \frac{1}{4^{\text{codim}(W)}}\sum_{\text{cosets} V}\sum_{i,j=1}^{m}\frac{1}{3}\big(\delta(B_i)\delta(C_j)\delta^2_{B_i, C_j}+ \]
\[\delta(B_i)\delta(D_j)\delta^2_{B_i, D_j}+\delta(C_i)\delta(D_j)\delta^2_{C_i, D_j}\big) \]

Note that $E_1$ and $E_2$ take values in $[0,1]$, so they cannot be incremented indefinitely.

We will show that:

\begin{itemize}
\item [(A)] If regularity condition (1) fails to hold, we may refine the boxing so that $E_1$ is incremented by some amount depending on $\epsilon$ and $m$ only, without increasing $m$ or decreasing $E_2$.

\item [(B)] If regularity condition (2) fails to hold, we may refine the boxing so that $E_2$ is incremented by an amount depending on $\epsilon$ only.
\end{itemize}

These two statements together imply that, starting with any boxing, we may refine the boxing until we get a regular boxing. Indeed, we have the following algorithm:

\begin{itemize}
\item[i.] If (2) doesn't hold, then apply (B) until it does hold.
\item[ii.] If (2) holds but (1) doesn't, apply (A) until (2) does hold. Then, if (1) doesn't hold, go back to i. Otherwise, both conditions hold and so we are done.
\end{itemize}

This must terminate because the number of times $E_2$ can be incremented in total is bounded, and the number of times $E_1$ can be incremented in a row is also bounded (with the bound depending on the parameter $m$).

\subsection{Refinement (A)}

We assume regularity condition (1) fails to hold. Suppose that $V$ is an outer box, and that some $B_i, C_i$ or $D_i$ associated with $V$ is not $\epsilon/m^3$-uniform. This will happen to a proportion of at least $\epsilon$ of the outer boxes. Without loss of generality, we may assume $B_1$ is non-uniform.

$B_1 \subset W + x$, so $B_1 - x \subset W$. The non-uniformity means that we can find some $\xi \in W^* \setminus \{0\}$ such that $|\widehat{(B_1 - x)}(\xi)| \ge \epsilon/m^3$. Set $W'$ to be the kernel of $\xi$. This will be a codimension-1 subspace of $W$, and so we can write $W = W' \cup (W' + v)$ for some $v$.

Each of the $B_i$ can be decomposed as $B_i = B'_i \cup B''_i$ where $B'_i = B_i \cap (W'+x)$ and $B''_i = B_i \cap (W' + v +x)$ (and similarly for $C_i$ and $D_i$). This gives us a way of refining our boxing by replacing the original subspace $W$ by a codimension 1 subspace $W'$. For each of the four cosets of $W'^3$ in the original coset $V$, we get the three new partitions by restricting the old partitions to the relevant coset of $W'$, so the new $B_1, \dots, B_m$ associated to a coset of $W'^3$ will either be $ B'_1, \dots, B'_m$ or $B''_1, \dots, B''_m$. Now we inspect what happens to $E_1$. If we let $\delta(B'_i) = |B'_i|/|W'|$ and $\delta(B''_i) = |B''_i|/|W'|$, then each term $\delta(B_i)^2$ will be replaced by

\begin{align*} \frac{1}{4}(2\delta(B'_i)^2+2\delta(B''_i)^2) & = \left( \frac{\delta(B'_i) + \delta(B''_i)}{2}\right)^2 + \left( \frac{\delta(B'_i) - \delta(B''_i)}{2}\right)^2 \\
& =
 \delta(B_i)^2 + \left( \frac{\delta(B'_i) - \delta(B''_i)}{2}\right)^2 ,
\end{align*}
and similarly for $\delta(C_i)^2$ and $\delta(D_i)^2$.

Since we haven't yet used any special properties of $W'$, this shows that $E_1$ does not decrease when we refine the boxing by replacing $W$ with any codimension~1 subspace. By iterating, this means that $E_1$ will not decrease when we replace $W$ by any subspace and restrict the sets $B_i, C_j$ and $D_k$ in the natural way to get a boxing.
Now, $|\widehat{(B_1 - x)}(\xi)| \ge \epsilon/m^3$ means that $|\delta(B'_1) - \delta(B''_1)| \ge \epsilon/m^3$, so replacing $W$ with $W'$ will increment $E_1$ by at least

\[ \frac{1}{4^{\text{codim}(W)}} \frac{\epsilon^2}{12m^6} .\]

Note that this increment can also be achieved by replacing $W$ with any subspace of $W'$. This increment still depends on $\text{codim}(W)$, so it is not enough on its own. However, we can apply the same method simultaneously for each outer box $V$ which does not satisfy the first regularity condition. There are at least $\epsilon 4^{\text{codim}(W)}$ such outer boxes. For each outer box $V$ which fails (1), we get a codimension~1 subspace $U_V$ of $W$. We set $T = \cap_V U_V$. The refinement we will take is the refinement obtained by replacing $W$ with $T$ and restricting the sets $B_i, C_j$ and $D_k$ in the natural way to get a boxing. Since $T$ is a subspace of $U_V$ for all outer boxes $V$ which fail (1), $E_1$ will be incremented by at least
\[\frac{\epsilon^3}{12m^6} ,\]
which depends only on $m$ and $\epsilon$, as required. Also note that the parameter $m$ does not change when we apply this refinement. However, we still need to check that $E_2$ has not been decreased.

To this end, the following lemma comes in useful.
\begin{lemma}\label{incrementformula}
Suppose we are given a set $S$, a subset $M \subset S$ with $|M| = \mu |S|$, and a partition $S_1, S_2, \dots, S_t$ of $S$. Then
\[ \sum_{i=1}^t \frac{|S_i|}{|S|} \left( \frac{|M \cap S_i|}{|S_i|} \right)^2 \ge \mu^2 .\]
If moreover there exists an $i$ with $|S_i| \ge c_1 |S|$, $\left| |M \cap S_i|/|S_i| - \mu \right| \ge c_2$, then
\[ \sum_{i=1}^t \frac{|S_i|}{|S|} \left( \frac{|M \cap S_i|}{|S_i|} \right)^2 \ge \mu^2 + c_1c_2^2 \]
\end{lemma}
\begin{proof}
Let $X$ be a random variable defined as follows: take $p \in S$ uniformly at random, and set $X = |M \cap S_i|/|S_i|$ where $i$ is such that $S_i \ni p$. Then we have
\[ \mathbb{E}(X) = \sum_{i=1}^t \frac{|S_i|}{|S|} \frac{|M \cap S_i|}{|S_i|} = \sum_{i=1}^t \frac{|M \cap S_i|}{|S|} = \mu \]
and
\[ \mathbb{E}(X^2) = \sum_{i=1}^t \frac{|S_i|}{|S|} \left( \frac{|M \cap S_i|}{|S_i|} \right)^2 .\]
So
\[ \sum_{i=1}^t \frac{|S_i|}{|S|} \left( \frac{|M \cap S_i|}{|S_i|} \right)^2 = \mu^2 + \operatorname{Var}(X)  \]
The first part follows from non-negativity of variance. In the second part, we may assume $|S_1| \ge c_1 |S|$, $||A \cap S_1|/|S_1| - \mu| \ge c_2$. We have
\[ \operatorname{Var}(X) \ge c_2^2 \mathbb{P}( |X-\mathbb{E}(X)|\ge c_2) \ge c_2^2 \mathbb{P}(p \in S_1) \ge c_1c_2^2, \]
which implies the result.
\end{proof}

Now let us look at $E_2$. $E_2$ is simply the weighted sum of the square of the density of $A$ in each inner box. One can check that in the refinement described above, each old inner box is replaced by a family of new inner boxes which partition it. So, by setting $S$ to be the old inner box, $M$ to be $A \cap S$ and $S_1, \dots S_t$ to be the relevant new inner boxes in Lemma~\ref{incrementformula}, we can see that $E_2$ is not decreased in the process.

\subsection{Refinement (B)}

If regularity condition (2) fails to hold, we will use a different kind of refinement. The subspace $W$ and the outer boxes will stay the same, but we will refine the three partitions $B_1, \dots, B_m$, $C_1, \dots C_m$ and $D_1, \dots D_m$. The first part of Lemma~\ref{incrementformula} tells us that $E_2$ will not decrease when we do this. Let $V$ be an outer box which does not satisfy (2), and let $(B_i \times C_j)_P$ be an inner box which is not $\epsilon$-quasirandom. Then we can find $B' \subset B_i$, $C' \subset C_j$ with $|B'| \ge \epsilon |B_i|$, $|C'| \ge \epsilon|C_j|$ with $|\delta_{B' \times C'}-\delta_{B_i \times C_j}| \ge \epsilon$. So if we refine our partitions by replacing $B_i$ with $B'$ and $B_i \setminus B'$ and replacing $C_j$ with $C'$ and $C_j \setminus C'$, we can apply the second part of Lemma~\ref{incrementformula} using

\[ S = (B_i \times C_j)_P, M = A \cap S,\]\[S_1 = (B' \times C')_P, S_2 = (B' \times C_j \setminus C')_P, S_3 = (B_i \setminus B' \times C')_P, S_4 = (B_i\setminus B' \times C_j\setminus C' )_P ,\]

Since $|S_1 \ge \epsilon^2 |S|$ and $\left||M \cap S_1|/|S_1|  - \mu \right| \ge \epsilon $, $E_2$ will be incremented by at least
\[ \frac{1}{4^{\text{codim}(W)}} \frac{\delta(B_i) \delta(C_j) \epsilon^4}{3} .\]

We can apply the same process simultaneously to all of the inner boxes inside $V$ which are not $\epsilon$-quasirandom (note that such boxes make up a proportion of at least $\epsilon$ of $V$'s inner boxes), and also do the same for each outer box $V$ which fails (2) (of which there are at least $\epsilon 4^{\text{codim}(W)}$). This gives us a total increment in $E_2$ of at least $\epsilon^6/3$. Note that, if we wish to ensure that the parameter $m$ is the same throughout the boxing, we may simply add empty sets to some of the partitions until they all have equal size.

We can now deduce that a regular boxing exists. To get the quantitative result of Theorem~\ref{regularitylemma}, we note that in each of the refinements, the parameters $\text{codim}(W)$ and $m$ increase in a controlled way.

\subsection{Counting}

Now that we have proved Theorem~\ref{regularitylemma}, we will use it to count corners. Let $m'(\alpha)$ be a convex function with $m'(\alpha) \le m(\alpha)$ for all $\alpha \in [0,1]$. Let $G = \mathbb{F}_2^n$ be sufficiently large so that we can apply the regularity lemma (Theorem~\ref{regularitylemma}) with $\epsilon$. Suppose $A \subset P$ has size $|A| \ge \alpha N^2$. We take an $\epsilon$-regular boxing for $A$ (this gives us the data $W \le G$, together with three partitions $B_1, \dots, B_m$, $C_1, \dots, C_m$ and $D_1, \dots, D_m$ for every outer box $V = W^3 + p$ with $p \in P$). Throughout what follows, we will use $o(1)$ to denote some quantities that tend to 0  as $\epsilon \to 0$. The main result of this section is the following:

\begin{claim}\label{countclaim}
The number of corners lying in $A$ with $d \in W$ is at least $(m'(\alpha) + o(1))N^2|W|$.
\end{claim}

Note that for a corner, the condition $d\in W$ is equivalent to requiring the three vertices of the corner to be in the same outer box. By the pigeonhole principle, it follows from  the above claim that there is some $d \in W \setminus \{0\}$ with $|S_d| \ge (m'(\alpha) + o(1) + O(|W|^{-1}))N^2$. Note that, for a fixed $\epsilon$, $|W|$ tends to infinity with $N$, so the $O(|W|^{-1})$ term can be made as small as we like. Therefore $M(\alpha) \ge m'(\alpha)$, which is Theorem~\ref{pt2}.

Now we set about proving Claim~\ref{countclaim}. For each outer box $V$, we define $\alpha(V)$ to be the density of $A$ inside $V$, $\alpha(V) = |V \cap A|/|V \cap P|$. The average value of $\alpha(V)$ over all $V$ is at least $\alpha$. We will show that if $V$ is a regular outer box, then the number of corners in $V$ is at least $(m(\alpha(V) + O(\epsilon)) + o(1))|W|^3$ (this is the content of Claim~\ref{1box} below). Since $m(\alpha)$ is uniformly continuous, $(m(\alpha(V) + O(\epsilon)) + o(1))|W|^3 = (m(\alpha(V))+ o(1))|W|^3 \ge (m'(\alpha(V))+ o(1))|W|^3$. Then, we can use the convexity of $m'$ to take the sum over all $V$, and so the number of corners with $d \in W$ is at least $ (m'(\alpha) + o(1))N^2|W|$ . Note that here we use the fact that the proportion of irregular boxes is $o(1)$, so the error that comes from those boxes is $o(1)$.

So, it is sufficient to prove the following claim:

\begin{claim}\label{1box}
If $V$ is a regular outer box, then there are at least $(T(f) + o(1))|W|^3$ corners in $V$, where $f$ is some function of independent random variables $X, Y$ and $Z$, where $X, Y$ and $Z$ each take finitely many values, $f$ takes values in $[0,1]$ and $\mathbb{E}(f) = \alpha(V) + O(\epsilon)$.
\end{claim}

By Lemma~\ref{discrete}, in the above claim, we may equivalently have assumed $f$ was piecewise constant $:[0,1]^3 \to [0,1]$ and $X, Y$ and $Z$ where $U[0,1]$.

\begin{proof}[Proof of Claim \ref{1box}]
Let $V$ be a regular outer box. Without loss of generality, we may assume $V = W^3$.
The corners whose vertices lie in $V \cap P$ are in bijection with the elements of $V$ as follows: $(x, y, z)$ correponds to the corner $K(x, y, z) := \{(x+d, y, z), (x, y+d, z), (x, y, z+d) \}$ where $d = -(x+y+z)$. For each $1 \le i, j, k \le m$, there are $|B_i||C_j||D_k|$ corners $K(x, y, z)$ with $(x, y, z) \in B_i \times C_j \times D_k$. The three vertices of these corners will lie in the inner boxes $(C_j \times D_k)_P, (B_i \times D_k)_P$ and $(B_i \times C_j)_P$ respectively. We may form a tripartite graph with vertex sets $B_i, C_j$ and $D_k$, by adding an edge between $x \in B_i$ and $y \in C_j$ if $(x, y, -x-y) \in A$, and similarly for the other two pairs. Triangles in this graph correspond to corners lying in $A$. Now, if the three inner boxes $(B_i \times C_j)_P, (B_i \times D_k)_P$ and $(C_j \times D_k)_P$ are all $\epsilon$-quasirandom, then we may apply the counting lemma from the proof of the triangle removal lemma to conclude that the number of such triangles is $|B_i||C_j||D_k|(\delta_{B_i, C_j}\delta_{B_i, D_k}\delta_{C_j, D_k} + o(1))$ (The statement of this counting lemma is Proposition~1.2 in \cite{ARLoriginal}. The triangle removal lemma, or rather a weaker form of it, was first proved by Ruzsa and Szemer\'{e}di in 1976 \cite{ruzsaszemeredi78}). Without quasirandomness, we still know the number of corners is at most $|B_i||C_j||D_k|$. Because $V$ is a regular box, the total size of all the non-quasirandom inner boxes is at most $3\epsilon|W|^2$, so the total of $|B_i||C_j||D_k|$ over all $i, j, k$ for which some inner box is not $\epsilon$-quasirandom is at most $3 \epsilon |W|^3$

Therefore, the number of corners of $V$ with vertices lying in $A$ is

\begin{equation} \label{eq:noofcorners}
\sum_{1 \le i, j, k \le m} |B_i||C_j||D_k|\delta_{B_i, C_j}\delta_{B_i, D_k}\delta_{C_j, D_k} + o(1)|W|^3 .
\end{equation}

Now we define the random variables $X, Y, Z$ and the function $f$. For any $(x, y, z) \in W^3$, define $B((x, y, z))$ to be the $B_i$ that contains $x$, $C((x, y, z))$ to be the $C_j \ni y$ and $D((x, y, z))$ the $D_k \ni z$. Let $p \in W^3$ be chosen uniformly at random, and let $X=B(p), Y=C(p), Z=D(p)$. $X, Y$ and $Z$ are independent random variables, each taking only finitely many values. Define
\[ f' = \frac{|A \cap X \times Y \times Z||W|}{|X||Y||Z|} \]
and
\[ f = \min(f', 1) .\]

One can easily check $\mathbb{E}(f') = \alpha(V)$. $\mathbb{E}(f' |X, Y) = |A \cap (X \times Y)_P|/|X||Y| = \delta_{X, Y}$ Similarly $\mathbb{E}(f' | X, Z) = \delta_{X, Z}$ and $\mathbb{E}(f' | Y, Z) = \delta_{Y, Z}$. Putting this together, we get

\[ T(f') = \sum_{1 \le i, j, k \le m} \frac{|B_i||C_j||D_k|\delta_{B_i, C_j}\delta_{B_i, D_k}\delta_{C_j, D_k}}{|W|^3} .\]

Comparing this with equation (\ref{eq:noofcorners}), we see that the number of corners is $(T(f')+o(1))|W|^3$. The problem with $f'$, though, is that it may take values greater than 1. We need to bound $f'$. Below, we will be using convolutions of functions on $W$, which are defined as follows: $g * h (y) := \mathbb{E}_{x \in W} g(x)h(y-x)$.

\begin{align*}
|A \cap B_i \times C_j \times D_k| &\le |P \cap B_i \times C_j \times D_k| \\
&= |\{(x, y, z) \in B_i \times C_j \times D_k | x+y+z = 0\}| \\ &=  |W|^2 B_i * C_j * D_k (0) \\ &= |W|^2 \sum_{r \in W^*} (B_i * C_j * D_k)^\wedge(r) \\ &= |W|^2 \sum_{r \in W^*} \hat{B}_i(r) \hat{C}_j(r) \hat{D}_k(r)
\end{align*}

We split the above sum into a ``main'' term, and error terms, which can be bounded using the uniformity assumption, plus applications of the Cauchy-Schwarz inequality and Parseval's identity.

\[\sum_{r \in W^*} \hat{B}_i(r) \hat{C}_j(r) \hat{D}_k(r) = \delta(B_i)\delta(C_j)\delta(D_k) + \sum_{r \in W^* \setminus \{0\}} \hat{B}_i(r) \hat{C}_j(r) \hat{D}_k(r) \]
\[ \le \delta(B_i)\delta(C_j)\delta(D_k) + \max_{r \in W^* \setminus \{0\}} |\hat{B}_i(r)| \left|\sum_{r \in W^* \setminus \{0\}} \hat{C}_j(r) \hat{D}_k(r) \right|\]
\[ \le \delta(B_i)\delta(C_j)\delta(D_k) + \frac{\epsilon}{m^3} \left(\sum_{r \in W^*} |\hat{C}_j(r)|^2 \right)^{1/2} \left(\sum_{r \in W^*} |\hat{D}_k(r)|^2 \right)^{1/2}\]
\[ \le \delta(B_i)\delta(C_j)\delta(D_k) + \frac{\epsilon}{m^3} \delta(C_j)^{1/2} \delta(D_k)^{1/2} \le  \delta(B_i)\delta(C_j)\delta(D_k) + \frac{\epsilon}{m^3} \]

We deduce that

\[ f'(B_i, C_j, D_k) \le \frac{|W|^3(\delta(B_i)\delta(C_j)\delta(D_k) + \epsilon/m^3)}{|B_i||C_j||D_k|} = 1 + \frac{\epsilon}{m^3 \delta(B_i) \delta(C_j) \delta(D_k)}\]
and hence
\[ \mathbb{E}(f' - f) \le \sum_{1 \le i, j, k, \le m} \frac{\epsilon}{m^3} \le \epsilon .\]

Therefore $\mathbb{E}(f) = \mathbb{E}(f') + O(\epsilon) = \alpha(V) + O(\epsilon)$. Using the fact that $\mathbb{E}(f' | X, Y) = \delta_{X, Y} \le 1, \mathbb{E}(f' | X, Z) = \delta_{X, Z} \le 1$ and $\mathbb{E}(f' | Y, Z) = \delta_{Y, Z} \le 1$, we also have
\begin{align*}
T(f') - T(f) & = \mathbb{E}(\mathbb{E}(f'-f|X, Y)\mathbb{E}(f'|X, Z)\mathbb{E}(f'|Y, Z)) \\
& + \mathbb{E}(\mathbb{E}(f|X, Y)\mathbb{E}(f'-f|X, Z)\mathbb{E}(f'|Y, Z)) \\
& + \mathbb{E}(\mathbb{E}(f|X, Y)\mathbb{E}(f|X, Z)\mathbb{E}(f'-f|Y, Z)) \\
& \le \mathbb{E}(\mathbb{E}(f'-f|X, Y)) + \mathbb{E}(\mathbb{E}(f'-f|X, Z)) + \mathbb{E}(\mathbb{E}(f'-f|Y, Z)) \\
& = 3 \mathbb{E}(f'-f) \le 3\epsilon
\end{align*}
So $T(f) = T(f') + o(1)$. Combining this with previous results, we get that the number of corners is $(T(f)+o(1))|W|^3$, where $\mathbb{E}(f)=\alpha(V) + O(\epsilon)$, as required.
\end{proof}

This completes the proof of Claim~\ref{1box}, and therefore also of Claim~\ref{countclaim} and Theorem~\ref{pt2}.

\section{Bounds}\label{bounds}

In this section, we prove some bounds on $m(\alpha)$, and hence get some bounds on $M(\alpha)$, the number of corners we can expect to find with the same $d$.

Recall that $m(\alpha)$ was defined to be the infimum of
\[ T(f) := \mathbb{E}(\mathbb{E}(f|X, Y)\mathbb{E}(f|X, Z)\mathbb{E}(f|Y, Z)) \]
over all possible piecewise constant functions $f:[0,1]^3 \to [0,1]$ with $\mathbb{E}(f) = \alpha$.

\subsection{Upper bound}
In section 5 of \cite{chu11}, Chu proves that for all $c > 0 $ there exists an $\alpha$ such that $m(\alpha) < c\alpha^3$. Her construction in fact shows that $m(\alpha) < C\alpha^{3.086}$ for some constant $C>1$. Her method involves taking a simple example of a function $f$ with $T(f) < \mathbb{E}(f)^3$, and then taking tensor products. We use the same method, except that our starting function is a bit simpler and we get a better bound.

Let $X, Y, Z \sim U[0,1]$ be independent. We define the piecewise constant function $g$ as follows:
\[ g(X, Y, Z) = \begin{cases}
    0       & \quad \text{if } (X,Y,Z) \in [0, 1/2)^3 \\
            & \quad \text{or}  (X,Y,Z) \in [1/2, 1)^3 \\
    1       & \quad \text{otherwise.}
  \end{cases}
  \]
A simple calculation shows us that $\mathbb{E}(g) = 3/4$ and
\[ T(g) = \frac{13}{32} = \left(\frac{3}{4}\right)^3\left(\frac{26}{27}\right) = \left(\frac{3}{4}\right)^{3+c} ,\]
where \[ c:= \frac{\log(26/27)}{\log(3/4)} = 0.131 \dots\]
So this shows that $m(\alpha) \le \alpha^{3+c}$ for $\alpha = 3/4$. We extend this using the construction detailed below.

We construct a new piecewise constant function $f^{\otimes n}$ from an old piecewise constant function $f$ essentially by taking the $n$-fold tensor power. However, in order to define the tensor power, we need to go to the alternative formulation given by Lemma~\ref{discrete}.
Given random variables $X, Y, Z \sim U[0,1]$, and $f : [0,1]^3 \to [0,1]$, we apply the first part of Lemma~\ref{discrete} to obtain independent random variables $X', Y', Z'$, each taking only finietly many values, and a function $f'=f'(X', Y', Z')$ such that $\mathbb{E}(f')=\mathbb{E}(f)$ and $T(f') = T(f)$. We then define $X'^{\otimes n} := (X'_1, X'_2, \dots , X'_n)$ where $X'_1, X'_2, \dots X'_n$ are independent $\sim X'$, and define $Y'^{\otimes n}, Z'^{\otimes n}$ similarly. Then define
\[ f'^{\otimes n} := \prod_{i=1}^n f'(X'_i, Y'_i, Z'_i) .\]
$f'^{\otimes n}$ is a function of $X'^{\otimes n}, Y'^{\otimes n}$ and $Z'^{\otimes n}$, and $X'^{\otimes n}, Y'^{\otimes n}$ and $Z'^{\otimes n}$ each take finitely many values.
Therefore, we can apply the second part of Lemma~\ref{discrete} to get a function $f^{\otimes n} : [0,1]^3 \to [0,1]$ with $\mathbb{E}(f^{\otimes n}) = \mathbb{E}(f'^{\otimes n}) = \mathbb{E}(f')^n = \mathbb{E}(f)^n$ and $T(f^{\otimes n}) = T(f'^{\otimes n}) = T(f')^n = T(f)^n$.

\begin{theorem} For all $\alpha \in [0,1]$, we have
\[ m(\alpha) \le \frac{27}{26}\alpha^{3+c} \]
\end{theorem}
\begin{proof} If $\alpha = 0$ this theorem is trivial, so assume $\alpha > 0$. We may write $\alpha = \beta (3/4)^k$ for some integer $k$ and $\beta \in (3/4, 1]$. Then we take $f$ to be the function $\beta g^{\otimes k}$. $\mathbb{E}(\beta g^{\otimes k}) = \beta \mathbb{E}(g)^k = \beta (3/4)^k = \alpha$, and $T(\beta g^{\otimes k}) = \beta^3 T(g^{\otimes k}) = \beta^3 (3/4)^{k(3+c)} \le (27/26) \beta^{3+c} (3/4)^{k(3+c)} = (27/26) \alpha^{3+c}$ as required.
\end{proof}
\subsection{Lower bound}

In this subsection, we will prove the lower bound $m(\alpha) \ge \alpha^4$. For convenience, we will write $f_3 = \mathbb{E}(f|X, Y), f_2= \mathbb{E}(f|X, Z) $ and $ f_1=\mathbb{E}(f|Y, Z)$.

Firstly, note that $\mathbb{E}(f \mathbf{1}(f_i < \alpha/4)) \le \alpha/4$ for $i = 1, 2, 3$ (where $\mathbf{1}$ denotes the indicator function of an event), because by the law of total expectation
\begin{align*}
\mathbb{E}(f \mathbf{1}(f_1 < \alpha/4)) & = \mathbb{E}(\mathbb{E}(f \mathbf{1}(f_1 < \alpha/4)|Y, Z)) \\
& = \mathbb{E}(\mathbb{E}(f| Y, Z) \mathbf{1}(f_1 < \alpha/4))) \\
&= \mathbb{E}(f_1 \mathbf{1}(f_1 < \alpha/4)) \\
& \le \alpha/4
\end{align*}
And similarly for the others. Therefore
\[ \mathbb{E}(f \mathbf{1}(f_1 \ge \alpha/4)\mathbf{1}(f_2 \ge \alpha/4)\mathbf{1}(f_3 \ge \alpha/4)) \ge \mathbb{E}(f) - \sum_{i=1}^3  \mathbb{E}(f \mathbf{1}(f_i < \alpha/4)) \ge \alpha/4 \]
Therefore the event $A = (f_1 \ge \alpha/4)\wedge(f_2 \ge \alpha/4)\wedge(f_3 \ge \alpha/4)$ has probability at least $\alpha/4$.

So we have, for any $f$

\[ T(f) = \mathbb{E}(f_1f_2f_3) \ge \mathbb{E}(f_1f_2f_3|A)\mathbb{P}(A) \ge (\alpha/4)^3(\alpha/4) = \frac{\alpha^4}{256} .\]

Now suppose for contradiction some function $h$ with $\mathbb{E}(h)=\alpha$ has $T(h)<\alpha^4$. So $T(h)=\alpha^4/C$, some $C>1$. Pick an integer $k$ such that $C^k > 256$. Then $\mathbb{E}(h^{\otimes k}) = \alpha^k$ but $T(h^{\otimes k}) = \alpha^{4k}/C^k < \alpha^{4k}/256$, contradicting the above inequality for $f = h^{\otimes k}$. Therefore $m(\alpha) \ge \alpha^4$.

\subsection{Conclusion}
Now that we have bounds on $m(\alpha)$, we can combine them with Theorem~\ref{pt1} and Theoram~\ref{pt2} to get bounds on $M(\alpha)$.

For the family of all abelian groups, we have $M(\alpha) \le (27/26) \alpha^{3+c}$, where $c = 0.131 \dots$. Meanwhile, since $\alpha^4$ is a convex function of $\alpha$, we can deduce that for the family $G_n = \mathbb{F}_2^n$, $M(\alpha) \ge \alpha^4$. This last bound also holds for the family $G_n = \mathbb{F}_p^n$ for any prime $p$.

It would be natural to ask what is the correct exponent $\log(m(\alpha))/\log(\alpha))$. So far we only know that this exponent lies somewhere between $3.131$ and $4$. The author is continuing to study this question and hopes to report on further advances in due course.

\emph{Acknowledgements.} I am grateful to Ben Green for plenty of helpful comments and discussion.

\bibliographystyle{plain}

redpajama_set_name: RedPajamaArXiv


text:
Home Crime News Authorities recover more than $300,000 in stolen property at Williams residence
Authorities recover more than $300,000 in stolen property at Williams residence
By Brian Pearson
An employee of a Fresno based construction company used a concealed GPS tracker to locate a stolen trailer in Williams, and ultimately helped authorities to discover more than $300,000 in stolen property at a Williams residence.
Authorities are now looking for James Thomas Vaughn, 50, of Williams – the man they suspect of stealing the property, which was taken from locations in several different cities across Northern California.
"We, and several other agencies from across the state, are seeking a warrant for his arrest," said Williams Police Sgt. Daniel Mata.
At around 8:05 AM on Friday, the construction company employee contacted the Williams Police Department in regards to a stolen trailer located in the 900 block of Larch Drive in Williams. According to the department, the employee tracked the vehicle to that location using a GPS device concealed in the stolen trailer, which was taken overnight from a PG&E substation in the Yuba City area. The trailer and the tools inside it have a value in excess of $100,000, the Williams Police Department said.
After locating the trailer, the Williams Police Department – in conjunction with investigators from the Colusa County Task Force, the Colusa County Sheriff's Office, the Colusa County District Attorney's Office – executed a search warrant at the residence on the 900 block of Larch Drive.
Sgt. Mata said that Vaughn is known to live at that residence, where investigators found an additional $200,000 worth of stolen property, including two enclosed trailers, a Harley Davidson motorcycle, a tow-behind generator, and a large amount of construction tools and supplies. Investigators also found evidence and property from recent thefts at the Love's Travel Stop and Dollar General Construction sites in Williams, the Williams Police Department said in a press release.
The department added that the investigation is ongoing at this time. Anyone with further information is being asked to contact the department at (530) 473-2661.■
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