text:
Inbox: What happens on the field is all that matters
It's a game of touches, not carries
Wes Hodkiewicz
packers.com digital reporter
Evan Siegle, packers.com
WR Davante Adams
Brad from Crofton, MD
Not to nitpick, but I believe the rodent art work referred to behind Wes is a gray squirrel, not a chipmunk. (I'm a highly trained wildlife biologist, but a Packer fan first and foremost!)
That is the fine work of the incredibly talented Katie Bobber. I bought it for my wife for Christmas. She loves chipmunks and squirrels, so I'll leave it up to you all to decide. It's like the blue dress of Insider Inbox. I swiped it from our living room and put it up in my makeshift office because it makes me happy. And we need to do whatever it takes to make ourselves happy right now. Good morning!
Mike from Austin, TX
I think I finally get it. Rather than being a follower in the evolution of NFL offensive strategy (the shiny object right now is "win the middle of the field"), Matt LaFleur wants to be that leader with new strategic concepts. It's likely crystallized only in his head right now, but we've drafted to build for that vision, which we all don't get just yet. Serenity now.
We can filibuster and debate the merits of this year's draft but Matt LaFleur is the man responsible for putting the 2020 team on the field. We all see the box of the puzzle but not every piece inside of it. LaFleur sounds energized by the group of players Brian Gutekunst is handing him. At the end of the day, that's what counts.
Ken from Boynton Beach, FL
Don't you think using the stat 75% of all Aaron Rodgers' TD passes were to receivers selected in Rounds 2 and 3 of the draft is more intriguing than using 274 TD passes to receivers selected in first three rounds since only one was to a first-rounder? One would think first-round receivers are more valuable, maybe to some QB but not to A-Rod. Just saying.
The Packers have drafted receivers in the second round better than almost every other NFL team over the past 15 years. And some intern in Bristol came up with a dumb stat graphic to make that look like it was a bad thing. But hey, retweets and likes over reality and facts, right? What happens on the field is all that matters.
Eric from Montgomery, IL
I've been reading stories on packers.com and elsewhere about A.J. Dillon and I'll say I'm getting excited to see him on the field. He seems to be in the Adrian Peterson mold, but of course I'm not projecting that type of success on the rookie. Thoughts?
I don't want to get into the comparison game with AJ Dillon, mainly because I don't think there are many running backs out there like him. Carrying 5% body fat at 247 pounds is ridiculous. He's a monster and a physical specimen. I also like how much pride he takes in his pass pro. Dillon has to earn his place but he has the makings of a complete NFL running back.
Matt from Minneapolis, MN
Both of our starting guards have a lot of flexibility with what positions they can play. Billy Turner has played guard and tackle, and the report on Elgton Jenkins in college was that he could play any position on the offensive line. With Bryan Bulaga gone, and Corey Linsley and David Bakhtiari in contract years, what position do you think Turner and Jenkins will be playing long-term?
I think both start at guard in 2020 and beyond but Turner has the versatility to fill a swing tackle role if injuries arise. The rest depends on how Rick Wagner performs and the development of all these young offensive linemen.
John from Liskeard, UK
I've read articles stating that this year's supplementary draft could be loaded. Would this be due to the possibility of no college football? If this is the case, how feasible is it that the Packers will participate? I assume this would also devalue next year's draft.
I've asked the same question myself for that very reason. It's been 22 years since the Packers last participated in the supplemental draft. They historically don't like sacrificing future draft capital, so that's something Gutekunst would have to consider if a big class of supplemental prospects forms.
Mike from Fort Wayne, IN
Do you think that when the players eventually get back that the new normal practices will be categorized as: practice in shorts/helmets, padded, and iPad practices?
I can just imagine Knute Rockne pulling his team together and declaring, "OK men, pull out your Microsoft Surfaces. It's time for a social-distancing practice." Or Vince Lombardi, banging on his tablet because it froze, "What the hell is going on in there?"
Venny from Montgomery, AL
I find myself getting excited about the potential of the offense this season. I just checked the stats of the 2003 Packers that featured a run-heavy game with Ahman Green, Tony Fisher, and Najeh Davenport. There were six games that year, where they rushed for 200 yards or more. We may see similar results with Jones, Williams and Dillon. League defense these days prep for the spread attack. We ebb, they flow.
Tony Fisher, one heck of a back and an even better player/alumni specialist. He had his own radio show. A true legend – then, now and forever. What was your question again?
Adam from Toronto, Ontario
I really enjoyed the piece on Dillon, well done. I was very happy with the pick although I thought he could have been a 3-4 rounder, but I'll trust the experts. I really don't understand all the negativity suggesting "Rodgers" wasn't provided any weapons. Looks like we will have a dynamic back field. Is it reasonable to expect 400 carries combined for Jones and Dillon?
It's a game of touches, not carries. And I expect Jones to touch the ball a whole heck of a lot in 2020. Dillon will get the reps he earns, with Jamaal Williams still a big part of this offense, as well.
Jason from Madison, WI
With all the hoopla around the Pack not drafting a WR, what about Reggie Begelton? I kind of feel his signing was like an extra draft pick. What is your opinion on him making the team?
Still an intriguing signing. Still a great backstory. Still a young receiver with a chance to show what he can do.
Don from Riverton, UT
In reading about the signing of free agent Treyvon Hester, it occurs to me that I often hear about signings days before BG announces them and they show up on Packers.com, while at other times I don't hear a peep until they're already official. What accounts for the disparities?
It depends on when the player signs/agrees to terms and whether the media catches wind of it. We announce "official" signings. However, five years ago, I was like the other reporters scratching the ground for news. That's the gig.
Grant from Charlotte, NC
So do you suppose we will keep four running backs this year – Jones, Ervin, Williams, and Dillion – on the game-day roster? Also, perhaps seeing three on the field at the same time, split one out wide with motion and two in the backfield? Just daydreaming while working here!
You guys know the Packers don't have a set number of spots labeled for (insert position here) on the 53- or game-day rosters, though I guess you now must keep eight O-linemen active. So they'll keep as many running backs as they see fit, but I will say Ervin's capabilities as a returner likely supersede whatever position is next to his name.
Nic from London, UK
Maybe Rodgers is a little annoyed and I'm sure he won't wave the white flag to his successor, but from how he has carried himself for 15 years, I also suspect he's a good enough person not to ruin this entire experience for a 21-year-old kid who didn't ask for any of it. He's never shown himself to be anything other than decent even if he might be opinionated and proud. Is it ever difficult for players to bond with such big age gaps? They're often not far off from parent and child.
I am going to say this once, in a full-throated voice with all of my conviction. I've covered Aaron Rodgers for the eight years – through collarbone injuries, an MVP award, big wins, tough losses, a coaching change and so much more. I expect Rodgers will be a model mentor for Love, while doing everything in his power to win championships and keep the youngster holding a clipboard. That's it. The rest is conjecture. Rodgers has done a fine job of relating to younger players and I expect Love to be no different.
Elliot from Eugene, OR
How well would you handle mentoring someone brought in specifically to take your job from you? Same question to the fans. Would be interesting to see their answers...
But am I doing what I'm doing and making what I'm making right now, or am I a two-time MVP quarterback who would be snatched up in seconds if he became a free agent? Listen, I work well with my co-workers. I've learned a lot from Spoff, Pete Dougherty, Rob Demovsky and Mike Vandermause and I hope they all learned a thing or two from me, too. That's teamwork. That's called being a professional.
Michael from Dover, PA
Ron Wolf traded a first-round pick for a second-round backup QB…and the wolves howled in angst. Ted Thompson took a QB while a first-ballot Hall of Famer was entrenched as the starter…and the wolves howled in angst. The Chiefs traded up to get a QB loaded with question marks while they had a No. 1 overall pick already at QB…and the wolves howled in angst. Stop the howling and Love the pick.
Go big or go home, right? It's called being a general manager, not a fan manager.
Tom from Iron River, WI
In the third linebacker conversation, I am anxious to see how Ty Summers looks this year. Last preseason he was all over the field making tackles. Any insight on how he progressed last season to see more field exposure this year?
Don't sleep on Summers. He led the special-teams unit in plays last years and he'll be right there with Oren Burks, Kamal Martin and Curtis Bolton in that battle to play alongside Christian Kirksey.
Ross from Roswell, GA
Gents, with the virtual offseason underway it got me wondering, when is the virtual schedule release supposed to happen? Do you think they will have contingencies built in?
League spokesman Brian McCarthy tweeted earlier this week the plan is still to release the schedule before May 9 (next Saturday). That's the extent of what I know.
Tory from Sammamish, WA
I just read the Titans aren't picking up the fifth-year option on Corey Davis. He was the fifth overall pick in the 2017 draft and accumulated a total of 142 receptions, 1,867 yards and six touchdowns. I thought all those top picks were sure things. It took the Titans four years to figure out this one wasn't. Maybe folks should relax and trust the process.
Draft picks are lottery tickets, not promises of greatness. I don't care if it's No. 1, No. 5 or Mr. Irrelevant – nothing is guaranteed.
Robert from Verona, WI
Watching the Bears add older guys such as Jimmy Graham and Ginn Jr. make me wonder what their philosophy is for building their roster. They clearly aren't "one player away" from a championship, so is the addition of some older players likely driven by the mentoring value for younger players, or simply that the GM thinks these signings improve their team?
Respectfully, I'm not sure what the Bears are doing or how it'll all play out. I do know Mitchell Trubisky was a lot more effective when Trey Burton was healthy in 2018, so the investment into tight ends makes sense in that regard.
Clipton from Pasadena, CA
I just read where Dallas assigned Cee Dee Lamb No. 88. Do you interpret that to mean they have no plans to sign Ty Montgomery? I prefer my ban not to exceed a month.
Talk to you on May 31.
Ryan from Columbus, WI
I am really tired of this narrative, but do you think we could get as many Super Bowl wins as the Falcons did after they pushed their chips in and drafted Julio Jones? It's a foolproof strategy! Love the work you guys do!
Ha. Thank you, Ryan. OK, stay safe everyone. Catch y'all next week.
Inbox: Neither team is going to blink
What's most crucial is the collective
Inbox: This is what you play for
The Packers have to start fast and finish strong against the 49ers
Inbox: They'll need their best
They shouldn't be surprised by anything
Inbox: Nobody deserved it more than De'Vondre Campbell
Welcome to the season of excitement
Inbox: This team is as tested as any
The stats themselves indicate the game is drastically different
Inbox: That win was both a statement and a turning point
The Packers' defense is different when both Kenny Clark and De'Vondre Campbell are available
Inbox: That best game is still out there this time
Now they have to stay on that path
Inbox: The urgency is there and so too is the focus
Be who you are and capture the trophy the seed says you're capable of capturing
Inbox: Packers have an interesting call to make
That says it all
Inbox: It's last call before playing for it all
Sunday could be a great opportunity for second-year QB Jordan Love
Inbox: He was the perfect guy at the perfect time in the perfect scheme
The Packers' trio is rock solid
Inbox: Aaron Rodgers deserves to be the NFL MVP…again
Allen Lazard is going to be a critical part of a championship run


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Q: How to avoid loop in this code I am sending prices to customers (10000+) but below code has loop that causes delays in the process for customers waiting for calculations.
PriceVisibleForCustomer = Price + CustomerMargin
Price - changing every 300ms - sent from central store, not related to customer instance
CustomerMargn - some plus or minus amount that is resulting from customer agreement/segment/administrator decision etc. It doesnt change during customer http session, I can keep it in memory
Customer - he takes part in the process after he logs in, he should see rapidly changing prices of 8 products.
Maybe I need some more technology ? I have Spring 3/4, Java, Weblogic and i could create even separate webapp for this task for providing calculated prices.
I thought about threads in Java but 10000+ customers would mean too many threads wouldnt it ? How to change this code? Maybe I should change architecture but how?
/**
     * Sends prices to customer. This method is called very often (300ms) as prices are changing in real time.
     * Customer should see prices also each 300ms
     * @param productId - id of a product that prices will be calculated
     * @param productIdToPriceMap
     * @param customerIdToMarginMap - this map is changed every time customer logs in or logs out
     */
    private static void sendPricesToCustomers(Long productId,
            Map<Long, BigDecimal> productIdToPriceMap,
            Map<Long, BigDecimal> customerIdToMarginMap) {

        //This loop is blocking last customer from receiving price until all other customers wil have theri prices calculated. I could create threads, 10000+ customers will be logged in, I cant create so much threads... can I?
        for (Long customerId: customerIdToMarginMap.keySet()){
            BigDecimal customerMargin = customerIdToMarginMap.get(customerId);
            BigDecimal priceResult = productIdToPriceMap.get(productId).add(customerMargin);
            //send priceResult to websocket
        }

    }

A: Here is a simple example of the Listener pattern, I am not sure if this approach will work for you but just throwing out some ideas ...
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;
import java.util.Vector;

import javax.swing.Timer;

public class Demo {

  public static Product[] PRODUCTS = new Product[]{
      new Product("Computer", 400),
      new Product("Desk", 800),
      new Product("Chair", 70),
      new Product("Printer", 300),
      new Product("Television", 200)
  };

  public static void main(String[] args) throws InterruptedException {
      Customer john = new Customer("John", 3);
      john.addProduct(PRODUCTS[1]);
      john.addProduct(PRODUCTS[2]);
      john.addProduct(PRODUCTS[3]);

      Customer mary = new Customer("Mary", 2);
      mary.addProduct(PRODUCTS[1]);
      mary.addProduct(PRODUCTS[2]);
      mary.addProduct(PRODUCTS[4]);

      Thread.sleep(10000);
      System.exit(0);
  }
}

interface IPriceListener {
  public void priceChanged(Product product, int price);
}

class Customer implements IPriceListener {
  String _name;
  int _margin;
  Vector<Product> _products = new Vector<Product>();

  public Customer(String name, int margin){
    _name = name;
    _margin = margin;
  }

  public void addProduct(Product product){
    _products.add(product);
    product.addListener(this);
  }

  public void priceChanged(Product product, int price) {
    System.out.println("[" + _name + "][" + _products.get(_products.indexOf(product)).getName() + "][" + price + "][" + (price + _margin) + "]");
  }
}

class Product implements ActionListener {
  private int _startingPrice;
  private int _currentPrice;

  private String _name;
  private Timer _timer;
  private Vector<IPriceListener> _listeners = new Vector<IPriceListener>();

  public Product(String name, int price) {
    _name = name;
    _startingPrice = _currentPrice = price;
    _timer = new Timer(300, this);
    _timer.start();
  }

  public void addListener(IPriceListener listener) {
    _listeners.add(listener);
  }

  public void removeListener(IPriceListener listener){
    _listeners.remove(listener);
  }

  private void notifyListeners() {
    for(IPriceListener listener : _listeners){
      listener.priceChanged(this, getCurrentPrice());
    }
  }

  public void actionPerformed(ActionEvent e) {
    _currentPrice = _startingPrice + (int)(Math.random() * (5 - (-5))) + (-5);
    notifyListeners();
  }

  public final String getName() {
    return _name;
  }

  private synchronized final int getCurrentPrice() {
    return _currentPrice;
  }
}

A: One way to handle this is to create a single thread whose job is to consume priceResults off a queue and send them down the websocket (I'm assuming you have only one websocket). Your loop would then push priceResults onto the queue every 300ms without blocking the websocket thread. See ConcurrentLinkedQueue javadoc.
Edit:
To avoid the delay between finishing the current loop through customerIdToMarginMap and beginning to loop through the next update, here are some options:

*

*Keep the queue concept and create a fixed thread pool where each thread pulls the next customerId/productIdToPriceMap/customerIdToMarginMapoff the queue. If you have four threads and 10,000 records, each thread will only have to process 2,500 records, thereby starting on the next 300ms data push 4 times earlier than your current implementation. Increase thread count as you find necessary for performance.

*Keep my original queue concept but change the way you receive the price updates. The reason you need to loop is because you're getting pricing updates for every customer at the same time. If you can instead e.g. create a threaded listener for a group of customers that receives a customerIdToMarginMap containing only customerIds it's meant to handle, the iteration time will be significantly decreased.

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Home » World News » Hilaria Baldwin pleads with paparazzi to 'leave my family in peace' following news that husband Alec will be charged in 'Rust' shooting
Hilaria Baldwin pleads with paparazzi to 'leave my family in peace' following news that husband Alec will be charged in 'Rust' shooting
A day after prosecutors said they plan to file charges against Alec Baldwin in the death of Rust cinematographer Halyna Hutchins, the actor's wife, Hilaria, was photographed in New York City.
She wore a sweatshirt that read "empathy," as she stopped to talk to the many photographers who surrounded her. They yelled questions at her, as the paparazzo often do, but she dismissed them. "I'm gonna tell you what I'm going to say," she said in footage the Associated Press posted on TikTok. "You're not gonna ask me questions. I'm gonna tell you, OK?"
Baldwin then said that she wanted them to realize the couple have seven children and the media presence as the kids go between school and home is "not good."
"So on a human level, you guys know I'm not gonna say anything to you. You know that," she continued. "So, please, leave my family in peace and let this all play out. OK? So let my kids come home and you stay away from them. Because they ask me, 'Mommy… what are these people doing?' And it's a very hard thing as a mom to try to explain. So please, go home, 'cause I'm not gonna say anything and Alec is not going to say anything."
At that moment, someone interrupted her to ask about the charges.
On Thursday, New Mexico First Judicial District Attorney Mary Carmack-Altwies announced that she would formally charge the Beetlejuice actor with two counts of involuntary manslaughter in Hutchins's October 2021 death on the set of the film. Baldwin, a producer and star of the Western, was holding a gun that discharged during rehearsals, killing Hutchins and wounding director Joel Souza.
While Baldwin has maintained that he hadn't pulled the trigger on the gun before it fired, an FBI report released in August determined that the gun could not have fired without someone pulling the trigger. He also noted that he was told the gun was "cold," meaning it did not contain live ammunition, and that he was pointing it where he had been directed.
Hannah Gutierrez-Reed, who was tasked with handling guns on the set, will also be charged with two counts of involuntary manslaughter. Assistant director David Halls has agreed to enter a guilty plea on a charge of negligent use of a deadly weapon.
The New York Timesreported Friday that, despite the tragedy, the actor planned to complete the unfinished film.
The Baldwins married in 2012 and they share seven children — Carmen, Rafael, Leonardo, Romeo, Eduardo, Maria and Ilaria — all under 10.
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Check out part one of this series, You Do You, my attempt at a brief introduction to expressive individualism (EI).
In part one, we examined the "You do you" mindset, called expressive individualism. EI is the idea that finding and being true to one's deepest self and then articulating that self out in the world is the key to life. In its strictest forms this mentality doesn't invite input from anyone else, including family, friends, or church for fear of not being true to one's self. Thus, it is an unapologetically self-focused method of identity formation that is pervasive in Western culture today.
Today, in part two, I'm going to summarize in my own words a section from Making Sense of God, written by Tim Keller, who ministers in New York City. In chapter six of Making Sense of God Keller looks at the notion of identity formation and some of the issues we face today.
In the segment we are considering here, Keller writes about how the foundation of our sense of self has changed recently in Western society.
He says that in ancient cultures and many non-Western cultures today the hero's story is of self-sacrifice, and his self-worth was grounded in the honor bestowed upon him by his community. By contrast, today in the modern West the hero's story is about self-assertion, and the hero's sense of self-worth is rooted in the honor he bestows upon himself.
Keller pinpoints a few problems with the ways that Western people today define and create their identity. He describes these problems as incoherent desires, the illusion of self-identification, crushing pressure and how he sees modern identity fracturing today. I will seek only seek to summarize and discuss the first three here.
Here's the problem: when I look into my heart, I see all kinds of desires in there. There are some that I wouldn't mind sharing with you and others I will keep to myself lest you never again take me seriously. When I look closely, I find that many of my desires are contradictory. For instance, I want to spend more time preparing sermons (at the office) and I want to spend more time investing in my kids playing ball. I want to be svelte and ripped, but I also want to eat a box a cereal with whole milk every night right before I go to bed. If I'm going to ground my identity in my desire, how do I decide which desires to pursue?
What if I locate my deepest desire and follow it? Keller writes that this solution has its flaws because it assumes that our desires are ordered and that they always remain the same. I can tell you that, at least for me, desires aren't static. When I was twelve years old I wanted to be Navy Fighter Pilot. From time to time, I catch myself daydreaming about going to medical school and becoming a doctor, but tomorrow I might be dreaming of earning a Ph.D in historical theology. When I was in high school I wanted to be 6'4", bench press 350 lbs., run 4.3 40 yard dash, and play Quarterback for the Dallas Cowboys. Oh wait, I still want that and still think it could happen, but I am afraid that is another category altogether, sorry. I have had quite a number of competing occupational desires in my lifetime and I can't begin to number the competing desires I have had in other significant areas of my life.
Our desires are apt to change, and they are also elusive. Has anyone ever asked you what you want and you find that you really don't know? Identity formation grounded on desire assumes that I have a deep understanding of myself, and if it what I want isn't readily apparent, it can be found with a little effort. Given the chaotic nature of our souls, this seems like unstable ground stand on. Identity rooted in elusive desire is inarguably incoherent.
Sigmund Freud believed that our innermost being is filled with a chaos of desire for power, love, comfort, and control and that these desires compete and would trample each other to win out if they could. He wrote, "We are not happy because we are frustrated….we are frustrated because we are, first of all, unhappy combinations of conflicting desires." Indeed.
Reflecting on more of Freud's body of work, Keller writes that "Freud would have shaken his head at many of his descendants in modern psychotherapy who have lost his realism about the inner darkness, incoherence, and destructiveness of the inmost desires." While Freud would give another name to our inner darkness, the bible calls it sin and says that it invades all aspects of our being, which makes basing our identity on our deepest desires problematic and to Keller's point, incoherent.
One of the components of current identity formation is that you alone set the standard and norm for yourself. You declare your significance to yourself. The problem is, this is impossible to do. And we know it instinctively.
We all want to be admired by those we admire, and we want to be respected by those we respect. It is when we find the approval of someone that we love or care about that we find a sense of worth. Keller writes that even though people act like, and even think that, they are "validating themselves," they (we) are adapting our words and behaviors to fit into a new community of people – the kind of people from whom we crave approval. For me, when certain people complement or criticize a sermon I preached it means more to me than when I hear it from others.
Have you ever seen a social media post where someone states an opinion and then says "that is just who I am and I don't care what other people think," and yet it has 57 "likes"? The poster may have rejected what some people think, but he apparently cares what some crowd of people think, otherwise why would he have posted in a public forum at all?
It is an illusion that we have the capacity to form an identity without being influenced by outside factors and other people.
Everywhere I look I see articles on how much anxiety people are experiencing today in comparison with days gone by. What is going on? Why is this the case?
There are multiple factors, I think, but the pressure of "being ones' self" and "forging one's own identity" must be near the top of the list. Imagine the pressure of trying to be myself and to forge my identity without others' help while I struggle to know what my deep desires are only to discover that they are in conflict.
Keller points out that when self-definition of identity is the norm, it leads to a society that idolizes winners and looks down on losers, with tangible contempt for weakness. He writes, "success or failure is now seen as the individual's responsibility alone," in part because "our culture tells us that we have the power to create ourselves." So if what we create fails, we alone are to blame and "failure" becomes a defining characteristic of our identity.
So many of us today feel the pressure to decide on our own where we go to college, our career, and our mate along with our ethos, our stances and our style, and these decisions are the expression of who we "really" are. And we drink this ideology in the air we breathe without recognizing the pressure it brings.
Do you see the irony here? Expressive individualism leaves us more dependent on what others think of us than ever, because, as Keller writes, we feel the pressure to be brilliant, beautiful, hip, and accomplished and we need others to agree to make it true. This overdependence on others enslaves us, e.g. posting "the perfect" picture at just the right time in the evening on social media to maximize our "likes".
To the extent that this is true in our culture and in our own lives, what are we to do? Anxiety and despair are not aspects of the abundant life that we are promised, so what is the answer?
The third and final part of this series, The Christian Identity and EI, will be posted next week.
 Timothy Keller, Making Sense of God, An Invitation to the Skeptical, (Viking, 2016), 118-132.
 It should be noted that Keller acknowledges that there are problems associated with the ancient and non-Western methods of identity formation as well as important positive aspects of identity formation in the West today by comparison with the ancient and non-Western methods. We will consider a biblical way to look at identity formation in part 3.
 Phillip Reiff, Freud: The Mind of the Moralist, 35.
 Keller, Making Sense of God, 124.
"You do you." That's the one I hear most frequently these days.
I'm sure you've heard one of these slogans, or one like them, because they are ubiquitous in American culture today.
There are certain situations where I hear these sentiments voiced and I think they are quite fitting, and I am glad somebody said them. These mantras have certainly been on my lips.
Feel free to judge whether or not I am judgmental much of the time, (of course, I am), but is it possible I am on the right track in certain instances? Is "you do you" really an all-encompassing lens through which we should be viewing all of life?
Trevin Wax, a professor at Wheaton College, recently wrote a compelling series of articles on the "You do you" mentality, otherwise known as expressive individualism, which is a powerful movement sweeping Western culture.
Wax defines expressive individualism as making the purpose of one's life "to find one's deepest self and then express that to the world" while disregarding the input of "family, friends, political affiliations, previous generations, or religious authorities . . ." It means that you alone are the judge of who you are, what you should do, what is best for you, and how you should interact in society.
This isn't about being genuine, though. "Authenticity," is one of the highest values in a culture of expressive individualism, according Charles Taylor, a Roman Catholic philosopher. Yet, authenticity isn't being used as a synonym for honesty, integrity, credibility, and purity. Instead, as Taylor writes, "it pits authenticity against conformity" – including conformity to parental, political, and religious authority.
Western countries are especially vulnerable to fall into expressive individualism due to their historical roots in individual freedom, personal autonomy, self-definition, and self-expression. In the West, we have a great deal of individual freedom and personal autonomy with respect to others, and this is a great thing. In fact, I think it's worth fighting to keep. However, it is problematic to make individual freedom and autonomy the ultimate value. As an ultimate value, it runs counter to the call of Jesus to seek him and his righteousness first. He says the greatest of all is the one who serves rather than the one who expresses whatever one feels the loudest and longest. I do hope you can see the contrast.
I am a Christian minister, but I still live and breathe the cultural air. If I am to follow Jesus, I must consider how our particular cultural moment may pull me away, even unwittingly, from Jesus. If following Jesus matters to you (and even if it doesn't), I invite you to consider the possible impact of expressive individualism in your own life and in the lives of those you love through this series of posts. Let me leave you with one more nugget that I hope will help you better understand expressive individualism and set the stage for a more detailed critique in my next post.
Expressive individualism is all about forging and expressing a personal identity.
American journalist Yuval Levin describes expressive individualism as more than picking your own path in life. It is also "a yearning for fulfillment through the definition and articulation of one's own identity… and to live in society by fully asserting who you are." It isn't just about self-direction in life, but the drive for public self-pronouncement to advance your self-ness.
Is this a problem? What could possibly go wrong?
In part two of this series, Expressive Individualism and Identity, we explore the problems that occur when expressive individualism influences identity formation.


text:
Dating and wedding: Tradition meets tension in Indian-American domiciles
da Tiziano Togati | Gen 12, 2021 | login | 0 commenti
Dating and wedding, an universal source of parent-child friction, could be especially shaky when you look at the houses of Indian-Americans, as U.S.-raised kids of immigrant parents very carefully tread between assimilating into American tradition, and staying real to their moms and dads' old-country opinions and traditions.
Whenever moms and dads have actually invested their critical teenage years in a various nation, generational and cultural chasms can combine to produce delicate situations and force life-changing choices.
"there clearly was a space into the culture . You lose stuff," said Rajni Venkateswar, 55, who was born in New Delhi but now lives in a southwest Chicago suburb when you filter. She along with her spouse had been involved seven days after their really first conference, in the U.S.
Generational distinctions pose challenges that may result in privacy, unknown conversations, compromises and sometimes tough choices. The most difficult: just How, and for the length of time, will adults play the industry? just How, as soon as, will parents manage to get thier daughters hitched off?
"a great deal of mothers i understand keep nagging me personally, 'When will you be getting the child married?' " stated 59-year-old Darshana Brahmbhatt of Milpitas, Calif., whose only child, Flora, 34, is unmarried. Brahmbhatt ended up being hitched in Asia whenever she was 19.
Although Brahmbhatt is employed to questions that are frequent implied judgment, interrogations from Indian relatives and buddies, whether well-meaning or just nosy, can lead to stress for moms and dads of unwed grownups.
" South Asian parents really have actually lots of peer stress," stated Ranu Boppana, a psychiatrist in ny who may have addressed a huge selection of Indian customers. "It's nearly considered neglect on the component as we see it," she added if they don't get sort of over-involved.
Indeed, numerous immigrant moms and dads are fast to direct, lest kids lose all feeling of their heritage.
"the children, if you don't precisely chatiw us directed, are planning to melt when you look at the big melting pot," said Syed Sultan Mohiuddin, a 62-year-old retired electric engineer into the Detroit suburbs, whom married with an arrangement in Asia in 1972. Searching right back, he regrets the eight-year age huge difference they wed between him and his wife, who was 16 when. Finding provided passions happens to be a struggle that is 38-year he stated.
The divergences between South immigrants that are asian their American-raised kiddies seem to be more info on individual experiences than other things. Moms and dads start to see the globe through a new lens, colored by growing up in India, severely restricted or no relationship, and a drastically various background that is educational.
"a rather big percentage of second-generation Indian-Americans in this county have actually moms and dads who got hitched in an marriage that is arranged" stated Jasbina Ahluwalia, a California-based matchmaker who may have counseled a huge selection of solitary Indian-Americans, and often their hopeful moms and dads.
In pre-arranged matrimonies, there clearly was maybe not really lot of dating or courtship included, Ahluwalia said. And when moms and dads restrict dating, kids will conceal information about their love everyday lives.
"the youngsters had been utterly confused" about dating and navigating two countries, Detroit retiree Mohiuddin said, "so they really would do things behind our straight back."
"they would like to manage to do their thing that is own without their parents, so they really have a tendency to keep it personal," explained David Popenoe, manager associated with nationwide Marriage venture at Rutgers, hawaii University of the latest Jersey.
Also, the Pew Values Survey unearthed that more youthful Americans are more accepting of interracial relationship than their older counterparts. "Many parents want the youngster to marry a person who is very much indeed like on their own when it comes to race, ethnicity, course," Popenoe stated.
Still, some South Asian parents have actually used more-American views on coupling up.
Flora "wants a guy that is indian if at all possible, exactly what's in our fate no body understands," stated Brahmbhatt, who is associated with Hindu faith. "In this point in time, it doesn't happen," she added if it doesn't happen.
Hindus will be the minimum expected to marry or live by having a partner outside their faith that is own to a study carried out because of the Pew Forum on Religion & Public lifetime.
Buddies whom call to setup Brahmbhatt's daughter up with men are grilled for a things that are few mom will consent to a romantic date. Is he well-educated? Is he at the least 5 legs 10 inches or 5 legs 11 inches?
Like Brahmbhatt, Mohiuddin, in Detroit, relates to the stigma of experiencing a daughter that is single the chronilogical age of 30; two, in fact. Mohiuddin's unmarried daughters are 35 and 34.
Which is "an anathema inside our tradition," he said. "a lot of people are bewildered whenever a lady can be so old and not married," he included.
I want to inform about Andrea Urbina-Franco


text:
Tyler Perry's 'Madea's Farewell Play Tour' coming to Shreveport
Madea's Farewell Play Tour will have a performance in Shreveport on March 14, 2019
By KSLA Staff | December 3, 2018 at 10:59 AM CST - Updated December 3 at 12:04 PM
SHREVEPORT, LA (KSLA) - America's favorite tough-talking, gun-toting grandmother is coming to Shreveport.
Tyler Perry, the actor and playwright who created the Madea franchise, announced Monday that "Madea's Farewll Play Tour" will perform at the Municipal Auditorium on Thursday, March 14, 2019. The production is Perry's 21st stage play.
In addition to Perry, the play will star Tamela Mann, David Mann, and Cassi Davis Patton.
Perry made headlines last month when he announced that he is retiring the character. In addition to the final play, a movie titled Madea's Family Funeral will be released in March.
The nationwide tour will kick off in Oakland, CA on January 18. Tickets for the Shreveport show will go on sale Friday, December 7 at 10:00 a.m.
Copyright 2018 KSLA. All rights reserved.
The petition alleges that more than $3.3 million is owed on a defaulted loan
Shreveport youth football team's equipment, trailer stolen
Kenley Hargett
JOBS: Who's hiring across the ArkLaTex
Jennifer Duckworth
Louisiana fishing and swimming advisories now available as an app
WAFB Staff
BPCC hosting national cyber conference later in July


A few weeks ago I announced that two generous GF donors had stepped forward to match the contributions of all new monthly donors for one year. We are thrilled to be saying THANK YOU to them and to all of our 42 new monthly donors. Wow! In just these few weeks, we have come together to build and multiply our capacity to support and serve peace.
The true meaning of this generosity was apparent at the Fallen Leaf Lake retreat. In a very special moment, two scholarship recipients shared their heartfelt gratitude. One recipient had met Gangaji through the Prison Program. It was his first time to meet her face-to-face. His gratitude is the confirmation that the gifts given are received in a deep and true way.
Those serving within and through Gangaji's Foundation play an essential role that Gangaji could never play. We can share with others the same opportunity we were given: to know the ever-present peace that is alive in the heart of all being.
If your life has somehow been touched by Gangaji and her programs, I invite you to share that gift with the world.


15 days exploring the impressive sights of Egypt & Jordan.
Welcome to Egypt or welcome home as the locals like to say as we begin an amazing 15 days with Travel Talk Tours exploring the sights of Egypt & Jordan. Today the group assembles in Cairo and we spend the first full day exploring the Great Pyramids, Sphinx and the Step Pyramid in Sakkara.
This morning we wake up in Luxor after a late night we are up early again so we can visit the impressive West Bank of Luxor. Our first major site for the day is the Temple of Queen Hatshepsut which in itself is an amazing site, then it is off to the Valley of the Kings followed by our journey to Aswan for the night.
This morning is an exceptionally early morning as we depart at 4:00am to hit the road and visit the impressive Abu Simbel Temples in the deep south of Egypt. Upon our return we visit the Philea Temple before returning to Aswan to begin our cruise down the famous River Nile.
We wake up this morning having sailed during the night to the port of Edfu. After breakfast we are taken by horse and cart to the Temple of Edfu where we have plenty of time to explore. Our chariot then awaits to take us back to the ship where we spend the rest of the day relaxing while the Nile passes us by.
Early in the morning travellers are given the opportunity to participate in an optional extra activity of a Hot Air Balloon ride over Luxor. This was an amazing experience to begin the day as we floated high above the Nile. Later in the morning we explore the Temple of Karnak before making our way back to the capital Cairo.
Today we have a full day to explore Cairo as we made our return last night. This morning we travel to the Egyptian Museum, a highlight for many people on the tour as we get to see real life mummy's and ancient treasures that have been recovered from various tombs. Later in the day we visit the Kahn El Khalili Bazaar for a spot of shopping and later in the evening the optional activity is the pyramids Sound & Light Show.
Time to hit the sea! We make our way from Cairo leaving really early in the morning to arrive in Dahab mid afternoon. Here we have two nights to relax by the Red Sea. We go and visit the famous Blue Hole, one of the best dive sites in the world and do a bit of snorkelling. Later we head off for some fun in the sand as we go Quad Bike riding racing each other through the desert sands.
Our time is up in Egypt as we depart Dahab and head for the border to make our way to Jordan via Israel. We eventually get through all the border crossings and make it to the port city of Aqaba and head for Wadi Rum. On arrival we jump into the back of 'jeeps' to be taken around the desert to be shown the impressive rock formations and to see a traditional Bedouin camp.
Today is one of the real highlights of the tour as we leave our desert camp behind and make our way towards the lost city of Petra. Our hotel is close to the entrance and we make our way down to the city where our guide takes us through the history of the area. We see the famous Treasury and come back at night to also witness it lit us in all its glory.
The tour makes its way closer to our final destination of Amman, before that we make a stop first at Kerak Castle followed by a visit to the Dead Sea, the lowest point on Earth and famous for the high concentrate of salt in the water. Once we were all done trying to defy physics in the Dead Sea we continue onto Amman where awe spend our final two nights together.


text:
\section{Introduction}

The topic of our paper has its origins in the following
mathematical puzzle of K\"orner and Malvenuto \cite{KM}. Call two permutations
of $[n]:=\{1,\dots,n\}$ {\em colliding} if, represented by linear orderings of
$[n]$, they
put two consecutive elements of $[n]$ somewhere in the same
position. For the maximum cardinality $\rho(n)$ of a set of
pairwise colliding permutations of $[n]$ the following conjecture was
formulated.

\begin{conj}\label{conj:clak} {\rm (\cite{KM})}
For every $n\in {\mathbb N}$
$$\rho(n)={n \choose {\left \lfloor \frac{n}{2} \right \rfloor}}.$$
\end{conj}

This conjecture remains open; for the best bounds the interested reader may
consult  \cite{KMS} and \cite{BF}.
In this paper we initiate a systematic study of similar problems for all the
graphs
on the countable vertex set ${\mathbb N}$. Not only do we believe that these problems
are interesting on their own, but beyond this we hope that studying them
within a unified framework may shed more light also on the initial problem on
colliding permutations.

Let  $G$ be an infinite graph whose vertex set is the set ${\mathbb N}$ of the natural
numbers.
We call two permutations of the elements of $[n]$, the first $n$ natural
numbers,
$G$--different if they map some $i \in [n]$ to adjacent vertices of $G.$
(We will often think and write about permutations of $[n]$ as $n$-length
sequences that contain each element of $[n]$ exactly once. In this language,
two permutations are $G$--different if there exists a position where the
corresponding two sequences contain the two endpoints of an edge of $G$.)
We denote by
$T_G(n)$ the maximum cardinality of a set of pairwise $G$--different
permutations
of $G.$ The question about pairwise colliding permutations is that special
case of our
present problem where the graph is the infinite path $L$ defined by
$$V(L)={\mathbb N} \qquad E(L)=\{\{i, i+1\} \; | \; i\in {\mathbb N}\}.$$
We will concentrate our attention on the special class of {\it distance
  graphs}.
Given an arbitrary (finite or infinite) set $D\subseteq {\mathbb N}$ we define the
graph
$G=G(D)$ by setting
$$V(G(D))={\mathbb N} \qquad E(G(D))=\{\{i, i+d\} \; | \; i\in {\mathbb N}, d\in D\}.$$
Clearly, $L=G(\{1\})$.
We  will write
$$T(n, D)=T_{G(D)}(n).$$

In the papers \cite{KM} and \cite{KMS} attention was restricted to those cases
where
the growth of $T(n, D)$ in $n$ is only exponential. Here we consider various
speeds of growth and determine
the exact value of $T(n, D)$ for every $n$ in a non--trivial case.
We are especially interested in the relationship between the values of $T(n,
D)$ and $T(n, \overline{D})$, where $\overline{D}={\mathbb N}\setminus D$.

\section{Superexponential growth}

The determination of $T(n, \overline{\{1\}})$ leads to a surprisingly simple formula.
\begin{thm}\label{thm:com}
$$T(n, \overline{\{1\}})={n! \over 2^{\lfloor{n \over 2}\rfloor}} \quad \hbox{for every} \quad n\in {\mathbb N}.$$

\end{thm}

{\noindent \bf Proof. }
First we prove the upper bound
$$T(n, \overline{\{1\}})\leq {n! \over 2^{\lfloor{n \over 2}\rfloor}}.$$
To this end fix $n$ and define $\sigma_{i,j}$ to be the permutation that
exchanges the entries $i\in [n]$ and $j\in [n]$, that is, for any permutation
$\pi$, $\sigma_{i,j}\pi$ differs from $\pi$ only in the places where the
entries $i$ and $j$ stand, which are exchanged.
For any fixed $\pi$ consider the set of permutations
$$C(\pi):=\{\sigma_{1,2}^{\varepsilon_{1,2}}\sigma_{3,4}^{\varepsilon_{3,4}}\dots\sigma_{k,k+1}^{\varepsilon_{k,k+1}}\pi:
\forall i\ \varepsilon_{i,i+1}\in\{0,1\}\},$$
where $k$ equals  $2\lfloor n/2\rfloor-1$,
$\sigma_{i,j}^0$ is meant to be the identity permutation, while
$\sigma_{i,j}^1:=\sigma_{i,j}$.
Let $B$ be a set of permutations of $[n]$ satisfying our condition that for
any pair of them there is an $i \in [n] $ they map to numbers at distance at
least two and observe that the conditions imply  $|C(\pi)\cap B|\leq 1$,
while $C(\pi)\cap C(\pi')=\emptyset$ if $\pi,\pi'\in B, \pi\neq\pi'$.
Since $|C(\pi)|=2^{\lfloor n/2\rfloor}$ for any $\pi ,$ the foregoing implies
$$|B|\leq {n! \over 2^{\lfloor{n \over 2}\rfloor}},$$
which is the claimed upper bound.

In order to prove the inequality in the opposite direction, for every $n$ we
shall explicitely construct a set of permutations satisfying the requirement.
We start by the odd values of $n$ and build our construction in a recursive manner.
It will be important for the recursion that for every odd $n$ the construction be invariant with respect to cyclic shifts.
For $n=1$ the construction consists of the identical permutation.
Suppose next to have constructed
$$t_{n-2}:={(n-2)! \over 2^{\lfloor{(n-2) \over 2}\rfloor}}$$
permutations yielding a set $B_{n-2}$ that satisfies the pairwise relation we need
and has the additional property of being closed with respect to cyclic shifts.
We will construct a set $A_n$ of ${n-1 \over 2}t_{n-2}$ permutations of $[n]$
satisfying the same pairwise condition and define $B_n$ to be the set
consisting of all the
cyclically shifted versions of the elements of $A_n.$ For an arbitrary
permutation $\pi$ of $[n-2]$ and $1<j\leq n$ we define the transformations
 $\Psi^j$ in the following manner. The permutation $\Psi^j\pi$ is acting on the set $[n]$,
\begin{eqnarray*}
\Psi^j\pi(1)&:=&n\\
 \Psi^j\pi(i)&:=&\pi(i-1)\quad  \hbox{for every}\quad 1<i<j\\
 \Psi^j\pi(j)&:=&n-1\\
\Psi^j\pi(i)&:=&\pi(i-2)\quad  \hbox{for every}\quad j<i\leq n.
\end{eqnarray*}
In other words,  the permutation $\Psi^j\pi$ is obtained from $\pi$ by prefixing $n$ in the position preceding the first number in $\pi$ and inserting $n-1$   in the $j$-th position of the resulting permutation.
For a set $A$ of permutations
we denote by $\Psi^j(A)$ the set of the images by $\Psi^j$ of all the
permutations of $[n-2]$ belonging to $A.$ As a last element of notation, let
us denote by $S^j$ the set of those permutations $\tau$ of $[n]$ for which
$\tau^{-1}(n-2)<j.$
Consider
$$A^j:= \Psi^j(B_{n-2})\cap S^j$$
and set
$$A_n:=\cup_{j=2}^{n}A^j.$$
(The attentive reader may note that $A^2=\emptyset$ but we felt it more natural
not to exclude this set from the above union.)
As every permutation in $A_n$ has $n$ at its first position no two of them
can be cyclic shifts of each other, whence $|B_n|=n|A_n|$.
Therefore in order to check that we have constructed the right number of
permutations it is sufficient to verify that
\begin{equation}\label{eq: rn}
|A_n|={{n-1} \over 2}t_{n-2}
\end{equation}
To this effect, recall that by our hypothesis the set $B_{n-2}$ is  invariant
with respect to cyclic shifts.
This implies that the number of those of its sequences in which a fixed
element, in our case $(n-2),$ is
confined to any particular subset of the coordinates is proportionate to the
cardinality of the coordinate set in question,
and thus
$$|A^j|={j-2 \over n-2}|B_{n-2}|,$$
whence
$$|A_n|=\sum_{j=2}^n |A^j |=\sum_{j=2}^n  {j-2 \over n-2} |B_{n-2}|=
{n-1 \over  2}t_{n-2},$$
which, substituting the value of $t_{n-2}$, yields
$$|A_n|={(n-1)! \over 2^{\lfloor{n \over 2}\rfloor}}.$$
This settles our claim (\ref{eq: rn}) and proves that $B_n$ has the requested
number of permutations.

To conclude the proof it remains to show that every pair of sequences from $B_n$ represents a
$G(\overline{\{1\}})$--different pair of permutations.
We will first prove that such is the case if both sequences are from $A_n$.
If they belong to the same
$A^j$ then this is obvious since the two permutations in such a pair must differ somewhere in those coordinates where they feature
an element of $B_{n-2}$ and thus the corresponding elements of $B_{n-2}$
must be different sequences. This implies, by our hypothesis, that they differ in some
coordinate by strictly more than 1.
If the two sequences, $\pi$ and $\tau,$ do not belong to
the same $A^j$, then we must have, say $\pi\in A^j$ and $\tau \in A^k$ with $j<k.$ But then in the
$k$-th position $\tau(k)=n-1$, while by definition, $\pi(k)<n-2,$ settling
this case as well.

If $\pi$ and $\tau$ are two permutations that do not belong to $A_n$ but have
the value $n$ in the same position, then they are clearly in a similar
relation as their respective cyclic shifts in $A_n$, thus the above argument
still applies.

Finally, we must prove that
any two of our sequences having the symbol $n$ in different positions also
represent a $G(\overline{\{1\}})$--different pair of permutations. Now, unless
the symbol $n$ of both of the two sequences meets the symbol $(n-1)$ of the
other one, we are done. Otherwise they have their respective subsequences
belonging to $B_{n-2}$
positioned in the very same coordinates and it suffices to see that these
subsequences are different. For this purpose
suppose that the two sequences have their symbol $(n-1)$ in the $j$-th and the
$k$-th position, respectively.
 But then, supposing $j<k$ we can say that they must have their respective
 symbols
$(n-2)$ in different positions since the one having its $(n-1)$ in the $k$-th
position has its $(n-2)$ in a position belonging to the open interval $(j, k)$
while the other one has it in the complement of the closed interval $[j, k]$
by construction.
This proves our theorem for every odd $n.$

In order to prove our claim also for even values of $n,$ it is enough to
consider the set $A_{n+1}$ (now $n+1$ is odd) and delete the first entry,
which is $(n+1)$, from each of the permutations in this set. This way we get
the right number of permutations of $[n]$ and their pairwise relations satisfy
the requirement by the previous part of the proof.
\hfill$\Box$

\medskip
\begin{rem}\label{rem:chi}
{\rm Consider the graph whose vertex set
is  the set of permutations of $[n]$  and such that  two permutations form an edge if and only if they satisfy the requirement we dealt with in Theorem~\ref{thm:com}. Denote this graph by
  $H_{\overline{\{1\}}}(n)$. Observe that its clique number
  $\omega(H_{\overline{\{1\}}}(n))=T(n,\overline{\{1\}})$ by definition and
  notice
  that by the proof above its chromatic number
  $\chi(H_{\overline{\{1\}}}(n))$ has the same value. (The sets $C(\pi)$
  defined in the proof can serve as color classes of an optimal coloring.)
  This observation will be used in the proof of the subsequent corollary.}
\end{rem}

\medskip

With some additional argument the above theorem gives the exact
value of $T(n,\overline{\{q\}})$ also for $q\neq 1$.
We will need the following well-known lemma, the roots of which go back to
Shannon \cite{Sh}. We give a short proof for the sake of completeness.

\begin{lem} \label{lem:perf}
Let $G_1,\dots,G_k$ be graphs and let $G_1\cdot\dots\cdot G_k$ denote their
co-normal product, i.e., the graph with vertex set $V(G_1)\times\dots\times
V(G_k)$ in which  two vertices $\mbf x, \mbf y$ are adjacent if there is an $i$
such that the respective  $i$-th entries  $x_i, y_i$ of these sequences satisfy
$\{x_i,y_i\}\in E(G_i)$. If $\chi(G_i)=\omega(G_i)$ holds for every $i$, then
$\omega(G_1\cdot\dots\cdot G_k)=\prod_{i=1}^k \omega(G_i).$
\end{lem}

{\noindent \bf Proof. }
It is easy to verify that $\omega(G_1\cdot\dots\cdot G_k)\ge\prod_{i=1}^k
\omega(G_i)$ always holds. To prove the reverse inequality observe that
$\chi(G_1\cdot\dots\cdot G_k)\leq \prod_{i=1}^k \chi(G_i)$. By
$\omega(G_1\cdot\dots\cdot G_k)\leq\chi(G_1\cdot\dots\cdot G_k)$ the
conditions $\chi(G_i)=\omega(G_i)$ imply the statement.
\hfill$\Box$

\begin{cor}\label{cor:alt}
Let $q$ be an arbitrary fixed natural number and let $n$ have the form $aq+m$,
where $m\in \{0,\dots,q-1\}$. Then
$$T(n,\overline{\{q\}})=
\frac{n!}{\left(2^{\lfloor\frac{a}{2}\rfloor}\right)^{q-m}
  \left(2^{\lfloor\frac{a+1}{2}\rfloor}\right)^m}.$$
\end{cor}

{\noindent \bf Proof. }
Let $S_n$ be the set of all permutations of $[n]$ represented as sequences and
consider a largest possible set $B_n$ of sequences from $S_n$ which satisfies
the requirements for $D=\overline{\{q\}}$.
Let $h:{\mathbb N}\to \{0,\dots, q-1\}$ be the residue map modulo $q$, or, in fact, any
map for which
$h(k)=h(\ell)$ if and only if $q$ divides $|k-\ell|$. For sequences $\mbf
x=x_1\dots x_n$ extend $h$ as $h(\mbf x):=h(x_1)\dots h(x_n)$. Partition $S_n$
according to the image of $h$, i.e., put $\mbf x$ and $\mbf y$ into the same
partition class iff $h(\mbf x)=h(\mbf y)$. The number of partition classes so
obtained is
$$t:=\frac{n!}{(a!)^{q-m}((a+1)!)^m}={n\choose{a,\dots,a,a+1,\dots,a+1}}$$
We call the classes $W_1,\dots, W_t$. If two sequences $\mbf x, \mbf y$ belong
to different
$W_j$'s then there must be a position $i$ for which $|x_i-y_i|$ is not
divisible by $q$, in particular, it is not equal to $q$. Thus
$T(n,\overline{\{q\}})$ is just the sum of the maximum possible cardinalities
of sets of sequences one can find within each $W_j$ such that each pair of
these sequences satisfies the condition.

Fix any class $W_j$. For each $\mbf x\in W_j$ and each position $i$ the
value $h(x_i)$ is the same by definition. Let $h^j_i$ denote this common
value. For $k\in\{0,\dots,q-1\}$ set $E_k=\{i\;|\; h^j_i=k\}$. Consider the
subsequence of each $\mbf x\in W_j$ given by the entries at the positions
belonging to $E_k$. Note that the size of $|E_k|$ is either $a$ or $a+1$.
Let $H_k$ be the following graph. Its vertex set consists of $|E_k|$-length
sequences of different
numbers from $[n]\cap \{\ell\;|\; h(\ell)=k\}$. Two such sequences
$\mbf x$ and $\mbf y$ are adjacent in $H_k$ iff at some coordinate $i$ we have
$|x_i-y_i|\neq q$. It is straightforward that $H_k$ is isomorphic to the graph
$H_{\overline{\{1\}}}(|E_k|)$ defined in Remark~\ref{rem:chi}. Whence its
clique number is
$T(|E_k|,\overline{\{1\}})$, while, by Remark~\ref{rem:chi}, its
chromatic number
has this same value. Let $\hat H_j$ be the graph with vertex set $W_j$ where
two vertices are adjacent if they satisfy the requirement that at some
position their difference is neither $0$ nor $q$. One easily verifies that
$\hat H_j$ is
isomorphic to the co-normal product (for the definition see
Lemma~\ref{lem:perf}) of the graphs $H_0,\dots, H_{q-1}$, which
is, by the foregoing, isomorphic to $\prod_{k=0}^{q-1}
H_{\overline{\{1\}}}(|E_k|)$. We are interested in the clique number of this
graph. By Lemma~\ref{lem:perf} and Remark~\ref{rem:chi} this value is
equal to $\prod_{k=0}^{q-1}\omega(H_{\overline{\{1\}}}(|E_k|))$. Noticing that $q-m$
of the sets $E_k$ have size $a$ and $m$ of them have size $a+1$,
this is further equal to
$\left({{a!}/{2^{\lfloor\frac{a}{2}\rfloor}}}\right)^{q-m}
\left({{(a+1)!}/{2^{\lfloor\frac{a+1}{2}\rfloor}}}\right)^{m}$
by Theorem~\ref{thm:com}.

The latter value is the same for all sets $W_j$ and the number of these sets
is ${n\choose{a,\dots,a,a+1,\dots,a+1}}$ (with $a$ and $a+1$ appearing $q-m$
and $m$ times, respectively). Thus we have obtained
$$T(n,\overline{\{q\}})={n\choose{a,\dots,a,a+1,\dots,a+1}}
\left({{a!}\over{2^{\lfloor\frac{a}{2}\rfloor}}}\right)^{q-m}
\left({{(a+1)!}\over{2^{\lfloor\frac{a+1}{2}\rfloor}}}\right)^{m}$$
$$=
\frac{n!}{\left(2^{\lfloor\frac{a}{2}\rfloor}\right)^{q-m}
  \left(2^{\lfloor\frac{a+1}{2}\rfloor}\right)^m}.$$
\hfill$\Box$

\section{Graph pairs}
It seems interesting to study the relationship of the values of $T(n, D)$
for pairs of disjoint sets (graphs) and their union, especially in case of pairs of
complementary sets.

Let us define
$$\phi(D, \overline{D}):=
\limsup_{n\rightarrow \infty}{1 \over n}\log{T(n,D)T(n, \overline{D}) \over n!}$$
and call it the {\it split strength} of the partition $\{D, \overline{D}\}$ of the natural numbers. Consider the case $D:=\{1\}.$ We know from \cite{KMS} that
$${10}^{n-4\over 4}\leq T(n,\{1\})\leq 2^n.$$
(We do not need the sharper form of the upper bound here. For an exponential improvement in the above lower bound the reader is invited to consult Brightwell and
Fairthorne \cite{BF}.)
Using this in combination with Theorem \ref{thm:com} yields

\begin{prop}\label{prop:sp1}
$$0.33 <\phi(\{1\}, \overline{\{1\}})\leq {1 \over 2}$$
\end{prop}

We continue with other examples. Denoting by $2{\mathbb N}$ the set of the even
numbers, we would like to determine $\phi(2{\mathbb N}, \overline{2{\mathbb N}}).$
To this end, notice first that
$$T(n, \overline{2{\mathbb N}})={n \choose \lfloor {n \over 2}\rfloor}.$$
In fact, this easily follows, as in \cite{KM}, by observing that two
permutations differ in every position by an even number if and only if the
even numbers occupy the same set of positions in both.

Somewhat surprisingly, $T(n, 2{\mathbb N})$ seems hard to determine and we only have
some easy  bounds.

\begin{prop}\label{prop:sp2}
$$\frac {n!\left(\left\lceil\frac{n}{2}\right\rceil +1\right)} {2{n \choose
    \lfloor\frac{n}{2}\rfloor}}
\leq T(n, 2{\mathbb N}) \leq {n! \over 2^{\lfloor{n \over 2}\rfloor}}$$
\end{prop}

{\noindent \bf Proof. }
The upper bound is a trivial consequence of (the upper bound part of) Theorem 1.
Although the lower bound follows from the lower bound on $\kappa(K_n)$ in
\cite{KMS}, yet for the reader's
convenience
we give the details without explicit reference to said paper.
(Those needing more details may however consult \cite{KMS}).
We consider the set $[n]$ as the disjoint union of its respective subsets of odd and even numbers. Correspondingly, we divide the coordinate set in two (with a little twist).
In the first $\lceil \frac{n}{2}\rceil+1$ coordinates we write the even
permutations of
the set $A$ consisting of all the odd numbers from $[n]$ with the addition of the
extra symbol $\star.$ (More precisely, first we represent these $\lceil \frac{n}{2}\rceil+1$
many symbols bijectively by the first natural numbers up to their cardinality, then extend this bijection to the permutations of both sets and consider only those permutations of the elements of $A$ that correspond to the even permutations of the first $|A|$ natural numbers).  We represent an arbitrary permutation of $A$ in form of a sequence
${\mbf x}$ and similarly let $\mbf y$ be an arbitrary permutation of the set $B$ of the even elements
of $[n].$ We will say that $\mbf y$ is {\it hooked up to} $\mbf x$ if we replace the $\star$ in
${\mbf x}$ by the first coordinate of $\mbf y$ and concatenate the rest of $\mbf y$ as a suffix
to the resulting sequence. Let us denote by $\mbf x \leftharpoonup \mbf y$ the
permutation of $[n]$ so obtained. Define
$A\leftharpoonup B$ to be the set of all these permutations as $\mbf x$ and
${\mbf y}$ take all of their possible values. Clearly,
$$|A\leftharpoonup B|=\frac{1}{2}
\left (\left \lceil \frac {n}{2}\right \rceil+1\right )! \left \lfloor
  \frac{n}{2}\right \rfloor!$$
which in turn equals the claimed lower bound in the statement of the Proposition.
It is very easy to see on the other hand that all the pairs of permutations from
$A\leftharpoonup B$ differ by an even number in some coordinate.

\hfill$\Box$
\begin{cor}\label{cor:odd}
$$0\leq\phi(2{\mathbb N}, \overline{2{\mathbb N}})\leq \frac{1}{2}$$
\end{cor}

\hfill$\Box$

Next we quickly review the following immediate consequence of our
hitherto results on split strength.

\begin{prop}\label{prop:d}
Let $q$ be an arbitrary but fixed natural number. Then $\phi(\{q\},
\overline{\{q\}})$
is independent of the actual value of $q$.
\end{prop}

{\noindent \bf Proof. }
We prove more, namely that the asymptotics of $T(n, \{q\})$
is independent of the value of $q$ we fix and the same is true for $T(n,
\overline{\{q\}}).$ For the latter it follows immediately from the
formula given in Corollary \ref{cor:alt}.

Now we turn to $T(n, \{q\}).$ Consider the distance graph $G(\{q\})$ of the
set $\{q\}$ and
look at the graph it induces on $[n].$ Since the latter is isomorphic to a
subgraph of $P_n,$ the path on $n$ vertices that the analogous distance graph
$G(\{1\})$ induces on the same set, we immediately see that

\begin{equation}\label{eq:iso}
T(n, \{q\})\leq T(n, \{1\}).
\end{equation}
In the reverse direction, we just have to observe that, for every $m\in
\{0,1,\dots, q-1\}$,
the graph $G(\{q\})$ induces an infinite path on the residue class
$q{\mathbb N}+m$ of the numbers congruent to $m$ modulo $q.$ This implies
\begin{equation}\label{eq:mp}
T(n, \{q\})\geq \prod_{m=0}^{q-1}T\left (\left \lfloor\frac{n-m}{q}\right
  \rfloor,  \{1\}\right)
\end{equation}
by concatenating the respective constructions of permutations for each fixed
$m$.
Whence it is immediate that $T(n, \{q\})$ and $T(n, \{1\})$ have the same
exponential growth rate.
\hfill$\Box$

We know very little about split strength and thus there are many questions to
ask.
Is it always true that $\phi(D, \overline{D})$ is finite and non--negative as
it seems by these examples?
In order to see the greater picture, we have to look at different kind of
growth rates as well.

\section{Intermediate growth}

So far we have only seen growth rates at an exponential factor away from
either 1 or $n!\ .$ We intend to show here, however, that
in between growth rates are also possible. In particular, we will see that
$T(n, D)$ and $T(n, \overline{D})$ can have essentially the same growth rate,
while their product is still about $n!\ $.

Let $ex(n)$ denote the largest  exponent $s$ for which $2^s$ is a divisor of
$n.$
We define

\begin{equation}\label{eq:exp}
E:=\{n\;|\; n \in {\mathbb N}, \; ex(n)\equiv 0 (\hbox{mod 2})\}.
\end{equation}

\begin{thm}\label{thm:self}
If $n$ is a power of $4$, then we have
\
\begin{description}
\item[(a)]
$$(\sqrt{n})!^{\sqrt{n}}\leq T(n, E)\leq \frac{n!}{(\sqrt{n})!^{\sqrt{n}}},$$
\item[(b)]
$$(\sqrt{n})!^{\sqrt{n}}\leq T(n, \overline{E})\leq
\frac{n!}{(\sqrt{n})!^{\sqrt{n}}}.$$
\end{description}
\end{thm}

{\noindent \bf Proof. }
We prove the lower bound part of (a) first.
It will be convenient to consider the elements of $[n]$ as binary sequences of
length
$t:=\lceil \log n \rceil$, with each natural number from $[n]$ represented by
its {\it binary expansion}. (Integer parts could be deleted by our assumption
on $n$, moreover, we also know that $t$ is an even number.) In fact, instead of
permuting the $n$ integers in $\{1,\dots,n\}$, now we will permute the $n$
numbers in $\{0,\dots, n-1\}$. With a shift by $1$, the two are obviously
equivalent for our purposes.
For simplicity, we will index the coordinates of the binary expansions
from right to left. Hence in particular $m$ is odd if in its binary expansion
$\mbf x=x_t x_{t-1}\dots x_1$ the rightmost coordinate $x_1$ is 1 and even
else. Let further $\mbf x^{odd}$ and $\mbf x^{even}$ denote the subsequence of
the odd and the even indexed coordinates of $\mbf x$, respectively. Finally,
let $\nu(\mbf x)$ be the smallest (i. e., rightmost)
index $i$ for which $x_i=1.$ By a slight abuse of notation we will consider
the various subsets of $\{0,\dots, n-1\}$ as subsets of $\{0,1\}^t.$  Quite
clearly, for every $\mbf x\in \{0,1\}^t$ we have $$\nu(\mbf x)=ex(\mbf x)+1$$
and, in particular, $\mbf x \in E$ if and only if
$\nu(\mbf x)\equiv 1$ modulo 2.
In order to prove the lower bound, let us consider the partition induced  on
$\{0,1\}^t$ (i.e., on $\{0,\dots,n-1\}$) by
the mapping $f:\{0,1\}^t \rightarrow \{0,1\}^{\frac{t}{2}}$
where
$$f(\mbf x):=\mbf x^{even} \;\hbox{for every}\; \mbf x\in \{0,1\}^t .$$
(The classes of the partition are the full inverse images corresponding to the
various values of $f.$)
It follows by construction that

\begin{equation}
f(\mbf x)=f(\mbf y)\quad \hbox{implies} \quad |\mbf x- \mbf y| \in E
\end{equation}
where by the difference of the vectors $\mbf x$ and $\mbf y$ we mean the
difference in ordinary arithmetics of
the natural numbers they represent. Indeed, executing the subtraction in the
binary number system we
are using here one sees that both $\mbf x-\mbf y$ and $\mbf y-\mbf x$ have
their rightmost $1$ in the position where, scanning
the binary expansions of $\mbf x$ and $\mbf y$ from right to left, we find the
first position in which they differ.
Now, since $\mbf x^{even}=\mbf y^{even}$ by assumption, the position in
question must have an odd index.
In other words,  $\nu(|\mbf x-\mbf y|)\equiv 1$ modulo 2.
For every $\mbf z \in  \{0,1\}^\frac{t}{2}$ we denote by
$S(\mbf z)$ the set of all the permutations of the elements of the full
inverse image $f^{-1}(\mbf z)$ of $\mbf z.$ Thus, by our previous argument,
all these permutations are pairwise $G(E)$--different.
Consider the Cartesian product
\begin{equation}\label{eq:bic}
C:=\prod_{\msbf z \in \{0,1\}^\frac{t}{2}} S(\mbf z).
\end{equation}
Note that the elements of $C$ are permutations of the numbers in
$\{0,\dots,n-1\}$.
The above consideration implies that $C$ is a set of pairwise
$G(E)$--different permutations.
Further, observing that for every $\mbf z \in  \{0,1\}^{\frac{t}{2}}$
$$|f^{-1}(\mbf z)|=2^{\frac{t}{2}}$$
we have
\begin{equation}\label{eq:crd}
|C|=(2^{\frac{t}{2}})!^{2^{\frac{t}{2}}}
\end{equation}
proving the lower bound in (a).
(One might get a somewhat larger set by using the hookup operation instead of
straightforward direct product
but we do not intend to increase the complexity of the presentation for
this relatively small gain here.)

\medskip

Next we prove the upper bound part of (b).
Notice that the set $C$ we have constructed above has a stronger property than
needed so far. In fact, in every coordinate, the absolute difference of our
permutations is either 0 or else it belongs to $E.$ This observation
will be the basis for our upper bound.

Consider the auxiliary graph $H_{\overline E}$ the vertices of which are the
permutations of $[n]$ and two are adjacent if they satisfy the
requirement that at some position they have two numbers such that their
difference is in $\overline E$. Clearly, $T(n,\overline E)=\omega(H_{\overline
  E})$, the clique number of this graph by definition. The above observation
about $C$ implies that its independence number $\alpha(H_{\overline E})$ is at
least $|C|$. Note that $H_{\overline E}$ is vertex transitive (any permuting
of the coordinates in the vertices gives an automorphism), thus by a well-known
fact (cf., e.g., in \cite{SchU}) its fractional chromatic number
$\chi_f(H_{\overline E})$ is equal to the ratio of the number of vertices and
the independence number. Using also that the clique number cannot acceed the
fractional chromatic number (cf. \cite{SchU}) we obtained

$$T(n,{\overline E})=\omega(H_{\overline E})\leq \chi_f(H_{\overline E})=
{|V(H_{\overline E})|\over {\alpha(H_{\overline E})}}\leq {n!\over |C|}=
{n!\over (2^{\frac{t}{2}})!^{2^{\frac{t}{2}}}}$$
proving the upper bound in part (b).
\medskip

Exchanging the role of even and odd above we obtain the lower bound in (b) and
the upper bound in (a) in a similar way.
\hfill$\Box$
\medskip

Theorem~\ref{thm:self} shows that the investigated values of $T(n, E)$ and
$T(n, \overline{E})$ have the same growth rate at around $\sqrt{n!}\ .$
The following statement is a straightforward generalization of the above.

\begin{thm}
For every rational number $\alpha \in (0, 1)$, there is a set
$E_{\alpha}\subseteq {\mathbb N}$ such that for infinitely many values of $n$ we have
$${(n^{\alpha})!^{n^{1-\alpha}}}\leq T(n,E_{\alpha})\leq {n!\over
    {(n^{1-\alpha})!^{n^{\alpha}}}}$$ and
$${(n^{1-\alpha})!^{n^{\alpha}}}\leq T(n,\overline{E_{\alpha}})\leq {n!\over
    {(n^{\alpha})!^{n^{1-\alpha}}}}.$$
\end{thm}

\begin{rem}
{\rm Notice that taking logarithm and using the estimate $\log (k!)\approx
  k\log k$ the above inequalities give that $\log T(n,E_{\alpha})$ is about
  $\alpha\log (n!)$, while $\log T(n,{\overline E_{\alpha}})$ is about
  $(1-\alpha)\log (n!)$.}
\end{rem}

{\noindent \bf Proof. }
Let
$$E_{\alpha}:=\{m\;|\; m \in {\mathbb N}, \; ex(m)\equiv 0,1,\dots,p-1 (\hbox{mod}\; q)\}.$$

Set $\alpha=p/q$ and suppose $n$ is a power of $2^q$. The reasoning is essentially the
same as in Theorem~\ref{thm:self}, which is the case $\alpha=1/2, q=2$.
Instead of $[n]$ we again permute the
elements of $\{0,\dots,n-1\}$ and represent each of these numbers by their
binary expansion. We collect into one group those numbers of $\{0,\dots,n-1\}$
whose binary expansion has the very same subsequence in those
positions which are indexed by numbers congruent to $p,\dots,q-1$
modulo $q$. There
are $n^{{q-p}\over q}=n^{1-\alpha}$ different such groups each containing
$n^{\alpha}$ different numbers. Permuting the numbers within a group we
get $(n^{\alpha})!$ permutations of those numbers and these are bound to differ
at some position by the difference of two different numbers in the
group. Such a value belongs to $E_{\alpha}$ by construction. Putting all
permutations of all our groups together we obtain
$(n^{\alpha})!^{n^{1-\alpha}}$ different permutations altogether that not only
satisfy the requirements given by the set $E_{\alpha}$ but no two of which
satisfy the requirements prescribed by $\overline E_{\alpha}$. This gives the
upper bound for $T(n,{\overline E_{\alpha}})$ in a similar way as the upper
bound on $T(n,{\overline E})$ is proven in Theorem~\ref{thm:self}. The rest is
also similar to what we have seen there.
\hfill$\Box$

\section{Exponential growth and Shannon capacity}
In this section we return to the more familiar territory of distance graphs
with finite chromatic number. The relevance of this parameter is shown by the
following simple observation.
\begin{prop}\label{prop:chr}
Let $G$ be an infinite graph with finite chromatic number $\chi(G).$ Then
$$T_G(n)\leq (\chi(G))^n$$
\end{prop}

{\noindent \bf Proof. }
Let $c:V(G) \rightarrow [\chi(G)]$ be an optimal coloring of the vertices of $G$ and let
$c_n:V(G)^n \rightarrow [\chi(G)]^n$ be its usual extension to sequences. Notice
that none  of the full inverse images $c_n^{-1}$ of the elements of $
[\chi(G)]^n$ can contain two pairwise $G$--different permutations of
$[n].$

\hfill$\Box$

In particular, distance graphs of "rare" sets of distances have finite
chromatic number.
More precisely, by a result of Ruzsa, Tuza and Voigt \cite{RTV},
if the set $D:=\{d_1,d_2,\dots d_n,\dots \}$ has the density of a geometric
progression
in the sense that $\liminf_{n \rightarrow \infty} \frac{d_{n+1}}{d_n} >1,$
then the distance graph
$G(D)$ has finite chromatic number. Clearly, this density condition is
sufficient but not necessary for the chromatic number to be finite (cf., e.g.,
the set of odd numbers as differences that result in a bipartite graph.)

However, for some graphs $G$ with finite chromatic number one can give a
better upper bound on $T_G(n)$.
This bound is easily obtained once we realize the
tight connection of our present problem with the classical concept of Shannon
capacity of a graph \cite{Sh}.

Given a sequence $\mbf x\in V^n$ we shall denote by $P_{\msbf x}$ the
probability
distribution on the elements of $V$ defined by
	$$P_{\msbf x}(a)={1\over n}|\{i\;|\; x_i=a,\ i=1,2,\ldots,n\}|$$
for every $a \in V$; here $\mbf x=x_1\dots x_n$. The probability distribution
$P_{\msbf x}$ is called the {\it type} of $\mbf x$.
Let $V^n(P,\varepsilon)$ denote the set of those $\mbf x\in V^n$ for which
	$$|P_{\msbf x}-P|=\max_{a\in V}|P_{\msbf x}(a)-P(a)|\leq\varepsilon.$$
Let $G$ be a finite graph with vertex set $V=V(G)$ and edge set $E(G).$
As always, we will say that the sequences $\mbf x=x_1x_2\dots x_n \in V^n$ and
$\mbf y=y_1y_2\dots y_n  \in V^n$ are $G$--different if there is at least one index $i \in [n]$
with
$\{x_i, y_i\}\in E(G).$
Let $\omega(G,n)$ and
$\omega(G,P,\varepsilon,n)$ be the largest cardinality of any set $C\subseteq
V^n$ and $C'\subset V^n(P,\varepsilon)$, respectively, of pairwise $G$-different
sequences.
The Shannon capacity of $G$ (or of $\overline G$ in the more usual notational
convention,
cf. \cite{Sh}) can be defined as $$C(G)=\limsup_{n\to\infty}\frac{1}{n}{\log
       \omega(G,n)},$$
while the {\it capacity $C(G,P)$ of $G$ within the type $P$} is given (cf.
\cite{CK}) as
$$C(G,P):=
     \lim_{\varepsilon\to 0}\limsup_{n\to\infty}\frac{1}{n}{\log
       \omega(G,P,\varepsilon,n)}.$$
It is immediate from the definitions that $C(G,P)\leq C(G)$ always holds and
using the methods of \cite{CsK} it is easy to prove that in fact $C(G)=\max_P
C(G,P)$.

In what follows we will restrict attention to graphs we call {\it residue
  graph}s.
We say that an infinite graph $G$ with vertex set ${\mathbb N}$ is a residue graph if
there exists a natural number $r$ and a finite graph $M=M(G)$ with vertex set
$\{0,1,\dots, r-1\}$ such that
$$\{a,b\}\in E(G)\quad \hbox{if and only if}
\quad \{(a)_{{\rm mod}\; r}, (b)_{{\rm mod}\; r} \} \in E(M)$$

Let $Q$ be the uniform distribution on $\{0,1,\dots, r-1\}.$ We have

\begin{thm}\label{thm:ty}
$$\lim_{n \rightarrow \infty}\frac{1}{n}\log T_G(n)=C(M(G), Q).$$
\end{thm}

{\noindent \bf Proof. }
To prove
$$\limsup_{n \rightarrow \infty}\frac{1}{n}\log T_G(n)\ge C(M(G), Q)$$
consider, for every $n$ those
sequences  $\mbf x \in \{0,1,\dots, r-1\}^n$ whose type $Q_n$ satisfies
$$Q_n(a)=\frac{1}{n}|\{m \;| \;m\leq n,  (m)_{{\rm mod}\; r}=a\}|$$
for every $a\in \{0,1,\dots, r-1\}.$
Let $M^n$ be the graph whose vertices are the $n$--length sequences
of vertices of $M$ and whose vertices are adjacent if the corresponding
sequences are $M$--different.
Let $C_n$ be a complete subgraph of maximum cardinality $M^n$ induces on
the set of sequences of type $Q_n.$
To any sequence $\mbf x \in V(C_n)$ we associate a permutation of $[n]$ by
replacing the occurrences of $a$ in the sequence by the different numbers
congruent to $a$ modulo $r$, in a stricly increasing order. The result is a set
of permutations which is $G$-different.

For the reverse inequality let $n$ be a multiple of $r$ and consider any
construction achieving $T_G(n)$, i.e., a set of
permutations of the elements of $[n]$ that are pairwise $G$-different, while
the cardinality of the set is $T_G(n)$. Substitute the occurence of the
number $i$ in each of these permutations by the unique $j\in\{0,\dots,r-1\}$
which is congruent to it modulo $r$. Doing this for all $i\in [n]$ we get
$T_G(n)$ different sequences of vertices of $M$ each having type $Q$ that actually
form a clique in $M^n$.
\hfill$\Box$
\medskip

The following corollary is immediate.

\begin{cor}
$$\lim_{n \rightarrow \infty}\frac{1}{n}\log T_G(n)\leq C(M(G)).$$
\hfill$\Box$
\end{cor}

\medskip\noindent
Let us consider the following

\par\noindent{\bf Example}
Let $G$ have vertex set ${\mathbb N}$ and set
$$\{a, b\}\in E(G) \quad \hbox{if}\quad |a-b|\equiv 1\;{\rm or}\;
4\;(\rm{mod}\; 5).$$
As an easy consequence of Lov\'asz' celebrated formula \cite{LL} for the
Shannon capacity of the pentagon graph we obtain, using the last theorem (and
also that the Shannon capacity of $C_5$ is obtained by sequences the type of
which is the uniform distribution), that
$$\lim_{n \rightarrow \infty}\frac{1}{n}\log T_G(n)=\frac{1}{2}\log 5.$$

\bigskip

It is an easy observation that for any graph $M$ and any rational probability
distribution $P$ on its vertex set one can construct (by simply substituting
each vertex by an independent set of appropriate size) a graph $\hat M$ for
which $C(M,P)=C(\hat M,Q)$, where $Q$ is again the uniform distribution. It is
then easy to construct an infinite graph $G$, which is a residue graph with
respect to $\hat M$ and thus the asymptotics of $T_G(n)$ is in an analogous
relationship with the capacity $C(M, P)$ as the one expressed in
Theorem~\ref{thm:ty}. Taking into account the remark that $C(M)$ can be
expressed as the maximum of the values $C(M,P)$ over $P$, we can conclude that
the class of problems asking for the asymptotics of $T_G(n)$ for various
infinite graphs $G$ contains the Shannon capacity problem of all such graphs
for which Shannon capacity is attained as the capacity within a type for some
rational distribution.

redpajama_set_name: RedPajamaArXiv


text:
In the Mek Mi Rich case, BTL to pay cost of court of almost $59,000
Uncategorized	— 27 July 2016	— by Rowland A. Parks
BELIZE CITY, Mon. July 25, 2016–In February, the Supreme Court found that Belize Telemedia Limited (BTL) had pirated the concept of the SMS lottery texting game Mek Mi Rich from a proposal made to the company by Corozal businessman Curtis Swasey and the court awarded Swasey $25,000 in damages, but what was left was for the court to determine the cost of court that BTL must pay Swasey's attorneys, Kareem Musa and Wayne Piper.
BTL eventually paid Swasey last month after a Supreme Court marshall went to the company's St. Thomas Street headquarters to crowfoot its assets to secure payment of the court judgment.
Last Wednesday, the court arrived at a figure for the cost of the litigation, and Swasey's attorneys were awarded cost of court of $58,790.10.
Kareem Musa, the lead attorney in the case, told Amandala today, Monday, that he is appealing the amount that Supreme Court Justice Courtenay Abel has awarded to his client.
BTL, too, has filed appeal papers against the court's ruling against the company. While that appeal works its way through the court, BTL is on the hook for the almost $59,000 of legal fees that the judge has ordered it to pay.
In the meantime, the Mek Mi Rich game show, the airing of which was discontinued after a relatively brief period, has been put on indefinite hold.
Mek Mi Rich (MMR) production is a company that was set up by Andre Vega, a son of the former Deputy Prime Minister Gasper Vega, and his business partner SanJay Hotchandani.
The court found, however, that BTL had violated the confidentiality agreement it had with Swasey.
So far, Mek Mi Rich has cost BTL $83,780.10, and this figure does not include what the company had to pay to its attorney, Magali Perdomo.


text:
/**
 * @fileoverview Utility methods for string manipulation.
 * These methods are not specific to Blockly, and could be factored out into
 * a JavaScript framework such as Closure.
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'use strict';

/**
 * Utility methods for string manipulation.
 * These methods are not specific to Blockly, and could be factored out into
 * a JavaScript framework such as Closure.
 * @namespace Blockly.utils.string
 */
goog.module('Blockly.utils.string');

/**
 * Fast prefix-checker.
 * Copied from Closure's goog.string.startsWith.
 * @param {string} str The string to check.
 * @param {string} prefix A string to look for at the start of `str`.
 * @return {boolean} True if `str` begins with `prefix`.
 * @alias Blockly.utils.string.startsWith
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const startsWith = function(str, prefix) {
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};
exports.startsWith = startsWith;

/**
 * Given an array of strings, return the length of the shortest one.
 * @param {!Array<string>} array Array of strings.
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 * @alias Blockly.utils.string.shortestStringLength
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const shortestStringLength = function(array) {
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exports.shortestStringLength = shortestStringLength;

/**
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 * @param {!Array<string>} array Array of strings.
 * @param {number=} opt_shortest Length of shortest string.
 * @return {number} Length of common prefix.
 * @alias Blockly.utils.string.commonWordPrefix
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const commonWordPrefix = function(array, opt_shortest) {
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};
exports.commonWordPrefix = commonWordPrefix;

/**
 * Given an array of strings, return the length of the common suffix.
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 * @param {!Array<string>} array Array of strings.
 * @param {number=} opt_shortest Length of shortest string.
 * @return {number} Length of common suffix.
 * @alias Blockly.utils.string.commonWordSuffix
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const commonWordSuffix = function(array, opt_shortest) {
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  }
  for (let i = 1; i < array.length; i++) {
    const letter = array[i].charAt(array[i].length - len - 1);
    if (letter && letter !== ' ') {
      return wordPrefix;
    }
  }
  return max;
};
exports.commonWordSuffix = commonWordSuffix;

/**
 * Wrap text to the specified width.
 * @param {string} text Text to wrap.
 * @param {number} limit Width to wrap each line.
 * @return {string} Wrapped text.
 * @alias Blockly.utils.string.wrap
 */
const wrap = function(text, limit) {
  const lines = text.split('\n');
  for (let i = 0; i < lines.length; i++) {
    lines[i] = wrapLine(lines[i], limit);
  }
  return lines.join('\n');
};
exports.wrap = wrap;

/**
 * Wrap single line of text to the specified width.
 * @param {string} text Text to wrap.
 * @param {number} limit Width to wrap each line.
 * @return {string} Wrapped text.
 */
const wrapLine = function(text, limit) {
  if (text.length <= limit) {
    // Short text, no need to wrap.
    return text;
  }
  // Split the text into words.
  const words = text.trim().split(/\s+/);
  // Set limit to be the length of the largest word.
  for (let i = 0; i < words.length; i++) {
    if (words[i].length > limit) {
      limit = words[i].length;
    }
  }

  let lastScore;
  let score = -Infinity;
  let lastText;
  let lineCount = 1;
  do {
    lastScore = score;
    lastText = text;
    // Create a list of booleans representing if a space (false) or
    // a break (true) appears after each word.
    let wordBreaks = [];
    // Seed the list with evenly spaced linebreaks.
    const steps = words.length / lineCount;
    let insertedBreaks = 1;
    for (let i = 0; i < words.length - 1; i++) {
      if (insertedBreaks < (i + 1.5) / steps) {
        insertedBreaks++;
        wordBreaks[i] = true;
      } else {
        wordBreaks[i] = false;
      }
    }
    wordBreaks = wrapMutate(words, wordBreaks, limit);
    score = wrapScore(words, wordBreaks, limit);
    text = wrapToText(words, wordBreaks);
    lineCount++;
  } while (score > lastScore);
  return lastText;
};

/**
 * Compute a score for how good the wrapping is.
 * @param {!Array<string>} words Array of each word.
 * @param {!Array<boolean>} wordBreaks Array of line breaks.
 * @param {number} limit Width to wrap each line.
 * @return {number} Larger the better.
 */
const wrapScore = function(words, wordBreaks, limit) {
  // If this function becomes a performance liability, add caching.
  // Compute the length of each line.
  const lineLengths = [0];
  const linePunctuation = [];
  for (let i = 0; i < words.length; i++) {
    lineLengths[lineLengths.length - 1] += words[i].length;
    if (wordBreaks[i] === true) {
      lineLengths.push(0);
      linePunctuation.push(words[i].charAt(words[i].length - 1));
    } else if (wordBreaks[i] === false) {
      lineLengths[lineLengths.length - 1]++;
    }
  }
  const maxLength = Math.max.apply(Math, lineLengths);

  let score = 0;
  for (let i = 0; i < lineLengths.length; i++) {
    // Optimize for width.
    // -2 points per char over limit (scaled to the power of 1.5).
    score -= Math.pow(Math.abs(limit - lineLengths[i]), 1.5) * 2;
    // Optimize for even lines.
    // -1 point per char smaller than max (scaled to the power of 1.5).
    score -= Math.pow(maxLength - lineLengths[i], 1.5);
    // Optimize for structure.
    // Add score to line endings after punctuation.
    if ('.?!'.indexOf(linePunctuation[i]) !== -1) {
      score += limit / 3;
    } else if (',;)]}'.indexOf(linePunctuation[i]) !== -1) {
      score += limit / 4;
    }
  }
  // All else being equal, the last line should not be longer than the
  // previous line.  For example, this looks wrong:
  // aaa bbb
  // ccc ddd eee
  if (lineLengths.length > 1 &&
      lineLengths[lineLengths.length - 1] <=
          lineLengths[lineLengths.length - 2]) {
    score += 0.5;
  }
  return score;
};

/**
 * Mutate the array of line break locations until an optimal solution is found.
 * No line breaks are added or deleted, they are simply moved around.
 * @param {!Array<string>} words Array of each word.
 * @param {!Array<boolean>} wordBreaks Array of line breaks.
 * @param {number} limit Width to wrap each line.
 * @return {!Array<boolean>} New array of optimal line breaks.
 */
const wrapMutate = function(words, wordBreaks, limit) {
  let bestScore = wrapScore(words, wordBreaks, limit);
  let bestBreaks;
  // Try shifting every line break forward or backward.
  for (let i = 0; i < wordBreaks.length - 1; i++) {
    if (wordBreaks[i] === wordBreaks[i + 1]) {
      continue;
    }
    const mutatedWordBreaks = [].concat(wordBreaks);
    mutatedWordBreaks[i] = !mutatedWordBreaks[i];
    mutatedWordBreaks[i + 1] = !mutatedWordBreaks[i + 1];
    const mutatedScore = wrapScore(words, mutatedWordBreaks, limit);
    if (mutatedScore > bestScore) {
      bestScore = mutatedScore;
      bestBreaks = mutatedWordBreaks;
    }
  }
  if (bestBreaks) {
    // Found an improvement.  See if it may be improved further.
    return wrapMutate(words, bestBreaks, limit);
  }
  // No improvements found.  Done.
  return wordBreaks;
};

/**
 * Reassemble the array of words into text, with the specified line breaks.
 * @param {!Array<string>} words Array of each word.
 * @param {!Array<boolean>} wordBreaks Array of line breaks.
 * @return {string} Plain text.
 */
const wrapToText = function(words, wordBreaks) {
  const text = [];
  for (let i = 0; i < words.length; i++) {
    text.push(words[i]);
    if (wordBreaks[i] !== undefined) {
      text.push(wordBreaks[i] ? '\n' : ' ');
    }
  }
  return text.join('');
};

/**
 * Is the given string a number (includes negative and decimals).
 * @param {string} str Input string.
 * @return {boolean} True if number, false otherwise.
 * @alias Blockly.utils.string.isNumber
 */
const isNumber = function(str) {
  return /^\s*-?\d+(\.\d+)?\s*$/.test(str);
};
exports.isNumber = isNumber;

redpajama_set_name: RedPajamaGithub


A centuries-old poetic form, the interlocking rubaiyat is a fairly straightforward form used by Robert Frost and others.
A number of poetic forms exist. Probably the most popular forms - at least seen frequently in the work of poets and in songs (including hymns) are pairs of rhyming lines, presented as ABAB or AABB.
Was blind but now I see. (B).
I subscribe to Writer's Digest magazine. A recurring column is "Poetic Asides," written by Robert Lee Brewer. The poetic form featured in the May/June 2016 issue was the interlocking rubaiyat.
Each successive stanza uses the unrhymed (e.g., B) for the rhyme in that stanza. The flow of a three stanza poem would be AABA, BBCB, CCDC or CCCC. In the last stanza, the third line can remain unrhymed or rhyme with the other lines.
The meter is usually tetrameter or pentameter.
Robert Frost's "Stopping by Woods on a Snowy Evening" follows the interlocking rubaiyat structure.
I encourage you to take a few minutes and write an interlocking rubaiyat. If you do, copy it into the comments below.


text:
Neycho Hristov Neev () (born 9 February 1948) is a Bulgarian politician who served as vice-Prime Minister and Minister of Transport in the Berov government from December 1992 to June 1993.

Between 1980 and 1989, he was employed as the main energy specialist in Kremikovtzi. In the early 1990s, Neev was a member of the UDF before turning to business in 1995.

References

1948 births
Living people
People from Haskovo Province
Union of Democratic Forces (Bulgaria) politicians
Government ministers of Bulgaria

redpajama_set_name: RedPajamaWikipedia