Hello, I hope I haven't kept you waiting. Yes, this is the crime club. I'm the librarian. Murder rents a room. Yes, we have that crime club book for you. Come right over. Take the easy chair by the window. Comfortable? The book is on this shelf. Here it is. Murder rents a room. By Sarah Elizabeth Mason. The very intriguing story of an old homestead that sold hospitality until death became a border. Let's look at it under the reading lamp. The Great Plantation of Cliff's Edge had once been the pride and the show place of the community. But that had been a long time ago and nobody remembered. Now it was just thousands of acres of dried out, desolate land. And on the house that had once been magnificent, there was a sign. Hunting, fishing, rooms for rent. Yes, times had certainly changed on the Great Plantation of Cliff's Edge. It was late morning of a fine spring day. Just three days after the body of Uncle Brock Curtis had been found in the swamp. A billet in his head. Now Uncle Brock was lying peacefully among his ancestors on the hill. And his survivors, having paid their last respects, were in the huge living room of the house, waiting. Yes, they were waiting for Lee Randall to read the will. And so was Sheriff Bill Davies as he stood unnoticed by the door with Tom, his deputy. Well, how about it, Lee? All right, Kitty. Give me a moment to break the seal. You want me to get you an axe? What's the matter with you, Kate? You might be a little more respectful. Uncle Brock hasn't been in his grave more than half an hour. I've shed all the tears I'm going to. I'm tired, honey. Very, very tired. I'd like to get some rest for a change. In the name of God, amen. I, Brock Curtis, being of sound mind and body, but aware of the uncertainty of this life... Can't you skip the forest and get right down to the mushrooms? All right, Kitty. I'll take this house and the grounds. So this is what happens when my back is turned. I stayed here with Uncle Brock and helped him run the place. Yes. How I wish you'd stayed in New York. Oh, there. Go ahead, Lee. Thanks. The rest of the property goes as follows. One-third to Kate, one-third to me, and one-third to you, Kitty. Well, Lee, what about Jeff? Isn't there anything...? I'd better read that part, Kate. I make no provision for my worthless nephew Jeffrey Gaines for reasons that my said nephew should clearly understand. Poor Jeff. And he's the one who really needs help. How much do we get, Lee? A little over 2,000 a piece, Kitty. 2,000? Are you kidding? Uncle Brock was a rich man. My father once said he was the richest man in the township. That was a long time ago. When he died, all he had to his name was this property and a little over $6,000. That's true, Kitty. Kate and I checked his papers yesterday. I'm afraid Uncle Brock made some pretty bad investments none of us knew about. You don't say. We found a pile of no-good stock certificates in the safe. I don't believe it. I've got them in my office at the bank. Anytime you want to see them, I'll be glad to show them to you. I still wouldn't believe it. Look here, Kitty. Are you suggesting that Lee and I... No, dear. Not while the sheriff and his deputy are listening. Don't stop now, Miss Kitty. You're doing fine.  My Sheriff Davies, I haven't said a thing. About them stock certificates being phony? I knew Uncle Brock. We all knew that dear Uncle Brock loved Cash and that he trusted nobody to count it for him. That's no reason to be making statements, young lady. Did I do such a terrible thing, Sheriff? I'm sorry. I guess I'm just tired. I'm going to my room. Will I see you later, Lee? I don't think so. Then you might be wrong. That nasty little troublemaker. She hasn't changed a bit. Hello, dear Miss Kitty. I told you everything I knew three days ago, Sheriff Bill. Uncle Brock disappeared right after breakfast. I went to look for him and I found his body in the swamp. But no gun. Where was Miss Kitty? Didn't she tell you that she'd gone riding? Yeah. Just checking. Stories got a way of changing with time. Now, Mr. Randall, when can I see about them stocks that belong to Brock? Any time after lunch. I'll be in my office all afternoon. Okay. Good one, Tom. Have you gone, Chief? When I say come on, that means we're going. Sure. Well, goodbye for now, Miss Kitty. Oh, by the way, what's Jeff Gaines using for eating and drinking money these days? I help him whenever I can. Was he counting on some of Brock's criticism, Cash? Sheriff Davis. Just checking. You never know what's what when there's been a murder. Hey, Chief, we're passing Jeff's farm. I know. Okay, but I figured you wanted to talk to him. In the meantime, we're going back to headquarters. Uh-huh. You got an angle? Maybe. Gonna relax and let nature take its course. Oh, I see. There are four parts to this puzzle, and Jeff's only one of them. Sure. We do nothing until all four parts get moving. If them stock certificates are phony, then somebody's got a lot of cash that belong to Brock. Yeah, but what's that got to do with nature? Human nature, Tom. Nobody likes to be left out in the cold. Well, goodbye, Kate. I've got to rush back to town. I'll phone you later. But, Lee, I was expecting you to have lunch. Some other time. This is a business day. I may be only second vice president, but the bank still needs me. All right. Do me a favor, please. Forget what Kitty said. I can't. But I told you she didn't mean it. She was disappointed about the inheritance and got excited. She knew Sheriff Bill Davis was in the living room with us, listening to every word. But she didn't realize, and she apologized. Oh, yes. It's about time you recognized her favorite pastime. Getting people into trouble and then apologizing. Well, I'm not going to worry about it. If those stocks are fraudulent, then Kitty will have to prove how she knew. She's just guessing. Goodbye, Kate. And please don't tangle with her. All right, Lee. Remember that. I'll speak to you later. What were you doing in Uncle Brock's room? Looking around. What for? The little things that always point to a murderer. You know, like a safe that can be opened by anybody. You know there was never any secret about that safe, Kitty. No. And nobody ever went to it, except Uncle Brock. Nobody, dear. Kitty, I'm warning you. If you're bent on making trouble. What will happen? Will my body be found in the swamp, too, with a bullet in my head? What do you want? My share of Uncle Brock's money. You're getting it. Yes, darling, but not where I like it. Think of Jeff. He gets nothing. You're changing the subject, dear. Kitty, why don't you try to be reasonable? It isn't my fault or Lee's fault that Uncle Brock made those investments. You know how he kept things to himself. Yes, I know. Lee checked those stocks through the bank. And when he told me they were worthless. You sat right down and quiet.  How do you think I feel about losing my share of $150,000? I could never imagine. Honing a lot of paper and companies that don't even exist anymore. Do you think I like it? Then why don't we make a deal? What do you mean?  Did I? Don't misunderstand. I'm not complaining. Just give me half the money and we'll kiss and make up. You're very shrewd, Kitty, but not smart. Business before compliments, cousin dear. Do you think asking me for money is going to make you look innocent? Oh, a twist to the table and I'm under it, huh? Why did you come back from New York? I couldn't resist the call of the wild. You were gone for two years and not one letter from you. But four days ago, just one day before Uncle Brock was murdered, you came home. Why? Don't get wound up, Cade. You needed money, didn't you? You're liable to break a spring. Didn't you? And you knew exactly where to find it. Then you admit those stocks are phony. You were the first one to bring that up.  Was it part of your plan to get the sheriff to suspect everyone but you? No, dear. It was my plan to get what I was entitled to. But I'll take what's left. You've done such a good job of convincing me. What have you done with the money? You don't seem to understand, Cade. I'm agreeing with you. There is no money except $6,000 and a lot of worthless paper. Hello, Jeff. Kitty. Kitty, sweet. What do you want here? Are you still mad at me? You've got no right to come to my house. Leave me alone. Jeff. Get away from me. You still love me, don't you?  What would I say? Let's go in. Kitty, are you asking me what I would say? Look at this place, Jeff. How can you live in such filth? You get used to a lot of things when you don't care. Oh, I'm sorry. I wasn't nice two years ago when I broke our engagement. Went to New York. I can forget that now.   Yeah, I sure hated you, Kitty. I swore if I ever saw you again, I'd kill you. That's funny, isn't it?   I guess I am, but I can see you. That's enough for me. Jeff, please. Plus the matter. Every reunion starts off with a kiss. No, not now. We've got so much to talk about. Sure. You don't like the smell of my breath. And I need a shave. Okay, let's talk. Jeff, why didn't you come to Uncle Brock's house these last four days? Because he told me you were there. He told you when? The night before he was killed. I met him on the road. That old miser didn't even own a car. Cost too much. Horses are cheaper. Carriages don't use gasoline. I know, he gave me plenty when I breathed in with my car. I got taxes doing this property and a payment on the mortgage. I asked him to give me a loan to help me out. What do you think he said? And him with all that cash and it's safe. Jeff, did you ever see it? Nobody ever saw it. You know that. Or smelled it either. But it was there. Everybody knew it was there. And it's still there waiting to be spent. No, it isn't. What did you say? Somebody stole it. Now, see here, kitty, I've got a great sense of humor, but don't take any bets on it. Don't get excited. You didn't lose anything. I didn't expect anything. He told me he was going to cut me out of his will. But Kate promised to pay off my debt so I can keep my home. Kate! She was here the day before yesterday after Sheriff Bill and Tom got through asking me a lot of silly questions. She told me not to worry. Jeff, don't believe everything you hear. What are you talking about? You're not getting any help. Sheriff Davies is going to arrest you for the murder of Uncle Brock. Well, me? That's right. He's out building up his case right now. But I didn't kill him. How do you think that'll sound in court? Coming from you. Oh, yeah. From a no-good bum. Hope I put him up to it, kitty. Well, I don't know for sure, Jeff. But after the funeral, after Lee went back to town, the sheriff and Kate had a long conference. That's a lie, kitty. What? Kate's the one person in this world who wouldn't try to hang me. Would I? Thank you for being so frank. Goodbye, Jeff. I'll see you in the death house. Kitty, wait. Look, I didn't mean to give you any wrong impression. I'm just not too clear upstairs yet. You've only got one real friend now, darling. Keep that in mind. Oh, but it doesn't seem possible that Kate would...  She loves me. $150,000 worth? The little thing she's done that proved it over and over again. She'd bring me money. She knew I was going to buy whiskey with it.   Once when she thought I was out cold, she kissed me and said how much he loved me. I wanted to cry because I was thinking of you. Jeff, are you going to listen to me? What do you want me to do, Kitty? Believe that Kate is framing me. Believe that I love you. Sure. All right. Now, we've got to get away from here as soon as possible. I'm ready. But I want Uncle Brock's money. What? My share of it. He left it to me and we'll need it, Jeff. Do you want me to find it? Yes, speak to Kate. Ask her to marry you. Yes. Yes, that's good. Tell her her husband can't testify against his wife. She'll go for that, Jeff. Maybe I... Maybe I ought to shave first. Oh, you think of everything, don't you? No. You do.  Miss Kitty, what brings you to town? I've got to talk to you about something very important. Mm-hmm. Want to come into my office? No, no. I can tell you right here. I've just been to see Jeff Gaines. Was he sober enough to see you? Take a look at this. Okay. What's so special about a $50 bill? I found it on the floor in Jeff's house. You did what? Jeff was drunk. He didn't realize I picked it up. That should solve your mystery, Sheriff. Yeah. Where would he get a $50 bill, except from... You talk to anyone else about this, Miss Kitty? Not a soul. Good for you. How do all the talking that has to be done? Oh, to think that I came so close to marrying him two years ago. Are you going to arrest him now? No. But that $50 bill... Don't make a case, young lady. But maybe after I've seen Mr. Lee Randall and them stock certificates at the bank, it will make a case. Go on, Tom. Okay, Chief. Let's go. A lot of gold mining stocks, huh, Mr. Randall? Uncle Brock seems to have had a preference for them, Sheriff. Yeah, and here's some shares in an oil company in Peru. Are these all the stocks you got? All of them, representing 18 different companies. And all of them out of business? For years. Before Brock put money into? Before? Didn't you check that when you checked the company? My note never occurred to me. Hey, Tom. Yeah, Chief? Get some paper and make a bundle. We're taking this stuff to headquarters. Okay, Chief. But, Sheriff Davies, I need those certificates. So do I. I'm filing Uncle Brock's will for probate. These prove what's become of his estate. They might prove a lot more, Mr. Randall, when I get through with him. Who else? The Queen of Sheba? I brought you some lunch. I'm sorry I'm so late, but... What, Jeff? What are you dressed all up in your Sunday suit for? Still looks good, doesn't it? It's the first time I've worn it in two years. Why? I thought you might like to see me as a human being for a change. You've never looked like anything else to me, Jeff. Thanks. I was going to call on your cliff's edge. It's clean there, and... Kate, there's something I wanted... I haven't wanted you to talk to you about for a long time. Yes. Wait a minute. Well, what's the matter? There's a car coming. Oh, for pity's sake. What are you so nervous about? I've been watching every car for the past hour. Looking out of the window like a hunted criminal?  Why? It's turning into my driveway. It's the sheriff and Tom.  No! You sent for them, didn't you? You told them to meet you here. You were bringing me lunch. A cooked lunch. But that takes time to eat. What are you taking down that shotgun for? You're not going to frame me, Kate. I'm going to be just outside this back window. And I'm going to be listening.  What? Sheriff Bill. Hello, Miss Kate. Tell Jeff we'd like to see him. He isn't here. Since when? Well, I don't know. I came here a few minutes ago with some lunch I'd fixed for him. And there was nobody home, huh? Okay, we'll just look around.  Nobody ever does, Miss Kate. Go ahead, Tom. You know what to do. Okay, Chief. And don't forget to look in the closets, too. Okay, Chief. Sheriff Bill. Would you mind telling me what this is all about? We've got a tip. There was some money in this house. Here? Who told you? Miss Kitty. She found a $50 bill on the floor. When? A couple of hours ago. I see. It ain't funny, Miss Kate. Oh, Sheriff, it's a riot. Kitty walks in and there's $50. I've never had such luck. Are you saying she didn't find it? I'm saying she did. But not here. Mm-hmm. Where? In Uncle Brock's safe. Can you prove it? Why did she come back from New York? She hates Cliff's edge and everything that goes with it. That ain't proof, Miss Kate. All right. Then account for the fact that Uncle Brock was murdered just one day after she came back. That still ain't proof. And until you got it, I'm going to keep on thinking... what I'm thinking. Hey, Chief. Yeah, what is it, Tom? He ain't here. You searched the whole house? Four rooms and two closets. That's all there is. Okay. Now we'll start taking this place apart. You won't find any money. You sure of that? I'm sure Jeff isn't a murderer. Mm-hmm. You got a pretty good crush on that fella, ain't you? Now see here, Sheriff Bill. Well, that's all right. Being in love is a mighty fine thing. But don't let it go to your head. Kate, are you sure? Yes, of course, Lee. There's still this house looking for the money. Good grief. But what about the stocks? They must be fraudulent. Lee, I'm sure now Uncle Brock had all that money and is safe. Where's Jeff? I don't know. Kate, don't start lying to me now. You've told me everything else. I don't know where he is. I thought he was outside the window. When I left his house, I walked around to the side to see if he was still there. Of course he wasn't. And he's not here at Cliff's there, either. Oh, Lee, what am I going to do? He might be wandering around in the swamp. I can't think of a place that suits him better. He didn't kill Uncle Brock. Stop talking nonsense. I'm not. Katey did it. She went to Jeff's house and then she went right to the sheriff with a $50 bill Jeff never saw. She's framing him. Is that what you want to believe? It's the truth, Lee. It's just what Katey would do. Well, I'm sorry, Kate, but I can't agree with you. Not in this case. All right. Be like the rest of them. Just because Jeff is down and out. What's all the excitement about Lee? You ought to know, Kitty, you found the evidence. Oh. Yes, sir. But I'm not so sure they'll be able to convict Jeff. What's that? Well, he may be innocent, you know. Now look here, Kitty.  It could have been planted. Why should it give me peace of mind?  What? Don't look so wild. I can keep a secret.  No, dear. And how can you make such a fantastic accusation? Because I'm a little adding machine. And when I get to $150,000, the bill rings. Yeah. Well, thanks for the laugh. Kitty, you're trying too hard. You always were a big bluffer. If you had had any evidence against me. What were you doing in the swamp this morning? Huh? Where Uncle Brock's body was found. What's this, a new angle?  I followed you when you left Kate's house. Then you must have seen me drive right here to the bank. I saw you drive to the swamp. Were you looking for this little cuff link with your initials on it? You found it in Kate's house, didn't you? With all this mud on it, Lee? Don't say that to Kate. She'd be insulted. I lost it there sometime this morning. And I didn't realize it until I was halfway to town. You're getting weaker, darling. I meant to phone her about it. When I got to the office, people were waiting. I completely forgot. Keep punching. You put that mud on there. Swamp mud. It'll match if the sheriff makes a test. But he doesn't have to, Lee.  And that you were framing Jeff. That $50 bill just like that cuff link. Eh, kitty? What made you turn on me? Am I easier to frame? I can do business with you. Give me $150,000. Why, you get out of here. All right. But it's either you or Jeff. I can make it, Jeff. It's up to you. Jeff, you almost scared me into an early grave. Don't ever come from behind a tree without blowing your horn. I've been waiting for you for a long time, kitty. I was in town. Yeah, I found that out. Where's Kate? I don't know. The last time I saw her, she was at my house talking to the sheriff. Why the shotgun, Jeff? Are you worried? No. I was standing up against this tree. Oh, you've got nothing to worry about, kitty.  Jeff, listen to me. Sure. But let's see how convincing you can be. No. Now go ahead, kitty. I'm listening. Tell me how much you love me. You want to marry me, don't you? You're the only real friend I've got, eh?  Where's the sheriff, Tom? He's in there, Mr. Randall, in the inquiry room. Thanks. Hey, you can't go in. He's got someone... It's all right. It's all right. He sent for me. I think you've got here so fast. What's this about Kitty having been murdered? We found her body in the trunk compartment of her own car. It was parked in front of Kate's house. Good Lord. Kate. Yeah, we figured somebody was going to take it for a ride tonight and maybe drop it into a river someplace. Jeff, maybe. What do you think, Mr. Randall? Naturally. She'd have been too good a witness for his prosecution. But why is Kate here? She claims she done it. Kate. I've got nothing more to say, Lee. That's how it's been going for a couple of hours. I thought you'd be a member of the family. Yes, Sheriff, would you mind? I'll be in the other room. There's no one crazier than a woman in love. Oh, eh, there's one thing you ought to know, Mr. Randall. Yes? I got a report on those stocks. The companies they were issued for never went out of business. What? I got told they never even existed. Lee, what's the matter? What's the matter?  I'm sick and tired of having to make excuses for this family. Lee! You would make you murdered Kitty very well. Then you must have killed Uncle Brock too.  There you go, Mr. Randall.  I've got orders to stick with you. The sheriff ain't through yet. You're not going anywhere, Randall. Get back inside. You'd better have a perfect explanation for this sheriff, Davis. Close that door, Tom. Okay, Chief. Now, Randall, let's have the whole story from the beginning.  I can, too. Yeah. All right. Guess I'll have to do the talking. Did you ever see this paper before? I've seen hundreds of government bonds. This one in particular for $10,000 and with that serial number. It's the negotiable kind anybody can sell it. Well, Mr. Randall, how'd you get it? I had an idea your Uncle Brock was too smart a man to keep $150,000 in cash in a safe that wasn't safe. How did you get that money? So I checked with Mr. Brewster, who used to be the president of your bank. He's a pretty old man, and he's remembering what it used to be. But he remembered them bonds Brock Curtis bought 18 years ago. 15 of them at $10,000 apiece. That's when I got a notion to check the brokerage houses out of town. Yeah, and we found one in boredom all the time. That'll do, Tom. I think Mr. Randall wants to talk now. Well, well, a few weeks ago, Uncle Brock asked me to draw up a will for him. He told me about the bonds. Where were they? In the wall of the cellar. He took me down there and showed me the exact place. I was the only one who knew, so I killed him. Why did you kill Miss Kitty? How did you know I killed her? When I drove up to Kate's house, I saw Jeff running away. Kitty was unconscious, nearly dead from strangulation. Thanks for telling me. Now you've made it solid. You can come in now, Jeff. Sure. I don't mind joining the party. What? Has this been a trap? Didn't you know? Maybe not as much as I know now. Jeff gave himself up to me for choking Kitty to death. When Tom and me went up to Kate's house to have a look, we found a trail of blood. Kitty had been stabbed. Yeah. She was making too much trouble. She always was a troublemaker. Wasn't she, Kate? Okay, Tom. Let's go, Mr. Randall. I'm pretty tired. I'll go into my office and put my feet on the desk.  Remember your own county property. And so closes tonight's crime club book, Murder Rents a Room, based on the story by Sarah Elizabeth Mason. Stedman Coles did the radio adaptation. Roger Bauer produced and directed. Elspeth Eric played Kitty Bowling. Sherling Oliver was Lee Randall. Helen Shields was Kate Frazier. Bill Smith was heard as Jeff Gaines. And Cameron Poudin played Sheriff Bill Davies. Oh, I beg your pardon. Hello, I hope I haven't kept you waiting. Yes, this is the crime club. I'm the librarian. Yes, come over a week from tonight. Good. We have the very intriguing story of a romance that had murder in its heart. It's called Death is a Knockout. In the meantime? Well, in the meantime, there is a new crime club book available this week. And every week, it bookstores everywhere. Yes, it's available now. Fine. And we'll look for you next week. This program came from New York. This is the Mutual Broadcasting System.


Senate presidents have made law on today, submitted its expression of interest in nomination forms as the national secretariat of the APC in Abuja. The forms were submitted on its behalf by Governor Mai Malaboni of the U.S. State, Governor Hope Uzodema of the U.S. State, and some members of the Senate. Political associates of the Senate presidents have paid 100 million dollars to purchase the forms on May 9, saying the country needs a Nigerian president, and not one given to ethnic proclivities.


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We have several accommodation options available to course participants including in-house accommodation or rooms at local guest houses and hotels. We have ongoing arrangements with several accommodation providers in Kokong that offer our trainees special monthly rates. All our recommended options have been inspected by our staff to ensure that they provide everything you need for a comfortable stay during your course. All accommodation is conveniently located within walking distance of our training center. Once your place on the course has been confirmed we will send you further information on the accommodation options available. Please note that you are not obliged to use a recommended accommodation and you are free to make your own arrangements. If you have any queries regarding the accommodation available please let us know and we will do all we can to assist you.


Thank you So I am Sylvain. I am part of the Savanna Hacker's group and I'm going to talk to you about Savanna So what is it? If you already saw this website, then you have already been to the one of the project that is hosted on your platform Savanna is essentially like other hosting platforms such as Thoughtforge or Davian Aliott or Launchpad It is sponsored by the GNU project. We have 3,200 projects around more than 40,000 users What do we provide? We provide tools for you to work efficiently on your project in your team So we provide VCS-E such as CVS, SVN, Git, Marker, Al-Bazar We provide back trackers so you can report back to your project We have an upload area for you to put files, which is mirrored worldwide We also have mailing lists and webpages, static webpages, which are Those are on another infrastructure. So I'm not going to present them in details. We have jobs. That means you want to look for a developer for your project and You can say I'm looking for a C developer or for a technical writer You have an interface to post the jobs and people can browse the jobs and check what projects they want to contribute to And all of this is free advertising. On our hosting platform, we use advertising We can see that when you look at the project, you can see advertising for proprietary competitors So we don't run ads on the server A brief history. So it all started in 2001 Well, we decided to take the existing serious infrastructure and put it more structured And we used the source-forge source code. Source-forge is a website. It was also a source code A piece of source-forge that you could install on your own computer to make your own source-forge So we use that and install it at Savannah's because it was a good methodology to be efficient when you work in team Sadly, it quickly became proprietary So we decided to run our own fork of this code We've been maintaining it ever ever since in 2002 we opened the platform for non-new project Previously it was only for new projects. I know everybody can contribute to the project Host a project on the platform If they meet a few requirements In 2004 we decided to make things more clearer The code was called Savannah and the website was called Savannah also So we decided to call the code Savannah to make things clearer And then at the top of the dates, but we since then we've added more support for more VCSs I mentioned Geed Bazaar We have cleaned up the system level with a lot of virtualization first with the server on with the exam We've introduced mirrors and our latest Work is rewriting the code. I will mention it later So what is different in all platform first? We are we're showing that we can make a free hosting platform using only 100% free software. We don't use any non-free software. You don't have any need to We use in particular old package from from devian on all the code that we code including seven Is under the AGPL license the AGPL license is a GPL for web applications If you make a GPL web application and somebody takes it make changes on put it on a website He did not have to give you back the changes with the AGPL He asked to provide you to the changes so you have to contribute back to the community I recommend AGPL for your web-based applications And also we want to promote base licensing practices When a project is submitted to our platform we review it for legal issues Ever since the SEO attack on the Linux kernel We know that we have to be perfectly clean on licensing and so we have to warn people we want to educate them about What's dangerous as I might Encounter on what they should do a bit of what this is recent apparently So so sort of Google code are no blocking countries Based on the origin of the visitors So our IP to countries databases and they decided to block completely users from using or contributing to the platform So this is a problem or some people are moving towards our team platform because of this So currently I don't have an official position, but Because this is a very recent that we do not intend to apply such such blocks If there is a problem if the US government because the service located at Boston if there was governments For us to make such Such blocks we have a backup hosting plan in France that you might migrate to hopefully not So now let's talk a bit about the technical details So first it is not exactly this hard one. I found the perfect picture later So we recently moved to another hardware It is donated by the free software foundation it is quite a good computer we hadn't changed it for five or six years so it is relieving The other identity but fsf on it is hosted at the global maps, which is donated Which is donating the colocation that is a space to physically place the server on the bandwagon So we have good hosting condition Now we can implement more services easily Yes, we recently switched to exam because fsf infrastructure is also using them, which means that if this particular box is crashing then we might easily migrate to other Hardware easily enough I have to mention now also. This is another installation of the sovereign software, which is located in France So again everything is torn on a single box and it is easily hosting 1300 projects, so you do not if you don't have to have a lot of expensive hardware to run your own forge It is very efficient So let's have a look at the architecture So So seven is the central component. It is essentially the glue that makes everything together People use the web interface This changes the database at this level and then so is a lot of system level scripts That replicates the database to all the services that we provide Vcs is the main list the websites etc So we have two distinct parts Above the front and which is the web interface on the system back and which is the system All the system stuff Now we're going to say what? We are going to make changes. We are making changes to both of those. I'm going to introduce what we are working on right now So first the front end I Said we forked the search for original code which is Like ten years old. It wasn't really so great. It didn't use a model view controller models So we try to clean it up, but we face the fact that this is not enough We also saw that there are a lot of new web frameworks that allow to program efficiently and we consider that we might more easily we write All the the web front end using far less code was doesn't try to clean up the existing code So that's what we are working right now we also wanted to Use something else on PHP because PHP wasn't so satisfying as a programming language and are a few legal issues because since the PHP license is not compatible with the GPL license. You might have problems with PHP extensions. Don't use a PHP license So we try to find a good web framework and a good programming language and essentially we choose Python and Django Python might not be the greatest language which it is a good compromise and Django has a Lot of advantages in particular. It is very well documented so that newcomers might Quickly learn how it works and it is also not so complicated. So again Somebody who wants to contribute does not need months of experience. You can learn the basic of Django in a couple weeks There are some limitations because It is not really meant for reusable applications. So we had to choose a few naming conventions Do some tests, but I think we did it right now at the same time. We try to clean up the backhand So the system level Previously We are the huge cron job that was supposed to take the database and Merge it with the system trying to preserve the system changes and it was not very maintainable on We decided to use NSS. NSS is a name service switch Which is a standard UNIX way to access your users. You can put your users out there in a slash etc slash password or such this group and with NSS you can put it in an LDAP directory And you can also put it in our SQL database and that's exactly what we did so now each time you we make a change using the web front end to the Database it is immediately replicated to the system. So there is no replication. It is directly available It greatly simplifies the code because all of the complexity is delegated to NSS Or cron jobs, we still have cron jobs. For example, we have a cron job That is used to replicate all the usage keys to the system and source cron jobs are really simpler to write This is a time-training slide. I'm going to talk about it later So what I wanted to say is that we are greatly looking for contributors in particular, we are looking for Moderators, that is we want people to review the projects and verify the licensing good practices so that we can So that we can approve projects more quickly We are also looking for web designers because we think that our design is a bit old-ish now and we would really like to make it more attractive We are also looking for coders. This I explained for programming the new web front-end adding new features and We're also looking for system administrators because each time we add a new feature. It's like we add a new VCS it requires experiences, skills and also requires time to deal with users, request, upgrade the system, make sure it is good that it's a greater level and make it evolve So currently we would like to introduce more services and we would like more system administrators to Implement these changes So what are the challenges if you want to contribute? First you have 80% of history. So you have to make sure that all the history is kept clean It is quite easy to make a new service from scratch without any data. Here you have to respect all the data that is existing or maybe migrate it to something more clean Also, this is a web service, which means there can only be one version at once. You cannot fork the service and put it on on your computer because you want it to be available for everybody. You can Just copy parts hardware. You can copy parts the bandwagon So we have to you only have one website. Only one can work at once So we have to first please everybody And then we when we make a change we have to make sure that it is perfectly clean before putting it into the public. We cannot really make release candidates So we have to have a good testing environment Also, I said that we are 80% of data We have to think about the future in 80 years. We will also have to maintain this data. So we we cannot just go to the fashion service implemented if we know that we might have trouble to maintain it in the long run. We have to think about it At the security level we decided to Restrict shell access. If there is a vulnerability in the kernel People won't be able to exploit it because they are using a restricted shell. We will of course try to patch the security issues as soon as possible But at least the window the vulnerable window won't be as critical So this has a certain number of impacts in particular people can do things on their owner This needs to be implemented. We need to have current jobs for everything that people need to do And it needs to be scalable. It is easy to make a service for 10 projects But when it comes to 300 projects 3000 projects Then we can't manually fix things. So we have to make sure that everything works Usually what we do is we implement it as beta for a few selected projects and when it is fine We implement it for all the platforms and thank you Do you have any questions?


INSAR technology uses the geometry between antennas and targets to obtain DM and deformation. Orbit information from satellites is necessary for accurate geometry, but some satellite orbit information may be inaccurate. This inaccuracy will be reflected in the interferogram and make it difficult to remove, leading to incorrect results. To address this issue, a new interferogram reflatting method was developed. It combines local processing and residual fringe removal to ensure the removal of unevenly varying residual fringes. Additionally, an adaptive-adjusted local flattening window was designed to ensure that the residual fringes within the local window can be removed cleanly. Experiments using three sets of Galfin3 data and one pair of Sentinel-1A data showed that the proposed method outperformed traditional refinement and reflatting methods in terms of accuracy. This article was authored by Dijuang, Lamei Zhang, and Binzou.


All right, thanks everybody for making it out today. Today we're excited to have Chris Cox from Iowa State University who will be talking to us about the maximum number of paths and cycles and planar graphs. Go ahead and take us away, Chris. Yeah, well, thanks Drew. Yeah, so I'm gonna talk about some recent work with Ryan Martin. That's exactly about what the title says. Trying to maximize the number of paths or cycles in a planar graph. So in general, for graphs of G and H we can define N of G, H to just be the number of copies of H and G. So here we're talking about unlabeled copies. And by a copy, I just mean legitimately a subgraph. I don't care if it's induced or not. So a clique has every graph on that number of vertices as a copy. The actual question that we're interested in is a parameter called N sub curly P of N comma H, which is the maximum number of copies of the graph H that can appear in an N vertex planar graph. So the curly P here actually talks about the number, sorry, the curly P is just my notation for the set of all planar graphs. This is the same as the maximum of all the N of G H's where G is an N vertex planar graph. So this is just a parameter that we can study. And of course, the main question that I care about is the asymptotic of this function. So for various planar graphs H, what is the asymptotic of N P of N H as N goes to infinity? So what is the map? Yes. I have one question, N G H, by the copies you mean induced or just subgraph? Just subgraph. So the clique has every graph on at most that many vertices as a copy. Also people can feel free to turn on their cameras. I don't know why everyone just went dark. Now I'm just talking to myself, but okay. I think we're a little bit worried that it's distracting or sometimes we don't wanna be staring at our cameras all the time. Well, whatever you wanna do. All right, yeah, so here I just wanna know how many copies of H can possibly appear in the planar graph. One thing, if I were to replace P with the set of all graphs, so now I'm just maximizing the number of copies of H in A graph, that's a silly question because the correct answer is however many copies of H appears in a clique. So we should restrict the family of graphs that we're actually looking at. All right, so let me tell you a bit about some previous results. Well, really the first graph, the first results on this, although it wasn't phrased in this way, really go back to Euler with the invention of planar graphs. Well, NP of NK2, that's just asking what is the maximum number of edges that can be in a planar graph? And this is well known to be about three N. So one thing, I'm only caring about the asymptotics of this function. Yes, really, it's like three N minus six or whatever, but I'm just going to care about the leading term. Also, because K5 and K33 aren't planar, well, NP of those guys is just always zero. So those all just go back to Euler and anyone who's known anything about planar graphs knows these facts. But the actual study of this extremal function really goes back to Hakimi and Schmeichel in 1979, who are really concerned with how many copies of various cycle lengths can appear in planar graphs. So they said that the maximum number of triangles is asymptotic to three N and the maximum number of four cycles is asymptotic to N squared over two. A bit later, Alon and Caro actually generalize their four cycle result and were able to get the maximum number of K2Ks for K at least two. Also when K equals one, they were able to get it in which case K21 is just the path on three vertices. I didn't write that result down, mostly because I forgot. More recently, Wood was able to figure out the maximum number of K4s that can appear in a planar graph. And just a couple of years ago, Yuri and friends were able to figure out the maximum number of paths on four vertices and the maximum number of five cycles. So these were two different papers that were put out about two years ago. One thing that I should mention, I'm only caring about asymptotics here, but all of the results up to this point, there are actually exact answers, but I'm only recalling the asymptotics, but these numbers are known exactly. And just last year, Goshen friends were able to figure out the maximum number of P5s in a planar graph. And this is the first result that actually it is asymptotic. They did not figure this out exactly. So this is actually just an asymptotic result that they got. And they had a conjecture of what should happen for all odd paths. So paths on an odd number of vertices. And they conjecture that there should be four m times n over m to the n plus one, whatever that means. I'll explain briefly where this comes from in a bit. And this is really the starting point of Ryan and my work trying to figure out this conjecture. So what happens for odd pads on odd numbers of vertices? So let's talk about that a bit. So our results, talking about pads on an odd number of vertices. First off, we verify the conjecture of Goshen friends for the seven path. Namely, if you plug things into their formula, you would get four over 27 times n to the fourth. And that's indeed what we were able to prove. For all other odd-legged paths, we were able to significantly improve the upper bound. So one thing that I should mention is that the order of magnitude here, so n to the n plus one, is actually an easy result that I'm sure that any of you could sit down and prove in a couple of minutes. It's very simple to get the actual order of magnitude of this function. So we're really fighting for the leading constants. And in fact, in your proof where you would actually get, say, the order of magnitude, you would probably prove a leading constant of something like six to the m, which is pretty bad. So we're able to actually bring that down to n to the n plus one over about m-factory, which is pretty big. So as I said, Ryan and my motivation was just this conjecture about odd paths, but as we were working, we figured out that actually these same ideas apply to even cycles. In fact, the ideas apply to a much more broad class of graphs if we want to actually compute this parameter, but I'll just leave it at even cycles and odd paths for the sake of this talk. So when it comes to even cycles, if you remember Hakimi and Schmeichel already figured out the four cycle. So we figured out the next two cases, namely six cycles and eight cycles. And again, significantly improved the upper bound for even cycles in general. Again, the order of magnitude of n to the m is actually very simple and any of you could prove it quickly.  Again, it would be something like six to the m probably. And in fact, we have a conjecture all attributed to us, not because I think that we're the only ones who've thought of it, but because I haven't seen it in the literature, that this pattern that we see for six cycles and eight cycles should continue for all two n cycles. So the answer for the maximum number of cycles on two m vertices should be n over m to the m. Now, where is this actually coming from? Is the following construction. Let's start with an m cycle. So vertices one, two, three, et cetera up to m. And now take each edge and blow it up into an independent set of size n over m and connect all the vertices that you just blew up to both of the end points of the original edge. So now if we want to actually count two m cycles, well, we first lay down every other vertex at the vertices of the original cycle. And now for all of the intermediate vertices, we have n over m choices for each of those. So we have n about n over m to the m, two m cycles in this construction. And you can quickly verify that this is in fact a planar graph. I've drawn it. One interesting thing, oh, in fact, this is the exact construction that gives the lower bound of the conjecture of Gaussian friends. So that silly number of like four m, whatever, they think that this is also the asymptotic extremal example for paths on an odd number of vertices as well. One thing I should notice about this construction is it's not a maximal planar graph. It only has about two n edges instead of three n edges. So it's a bit weird that we're missing out on n edges that we could possibly still add to this construction, but asymptotically, it doesn't matter at all. This is really the base structure. One more thing that I'll mention about all of our results before I tell you a bit about the ideas behind how to prove them, is that instead of dealing with planar graphs, it doesn't matter, none of our results care that the host graph is actually planar. All we care about is that it has linearly many edges, which planar graphs do, they have at most three n edges, but in fact, any linear number of edges is okay. And there's no legitimate copy of K33. So planar graphs have no subdivided copy of K33 or a topological K33, but we're not even taking advantage of that fact. We're just going to take advantage of the fact that legitimately there is no actual copy of K33 in all of our results. So these are very broad. Should ask if there are any questions before I go on. So as I said, we were really motivated by the question of odd paths, but I actually wanna first tell you about how to get, say the sixth cycle. I wanna walk you through a sketch of this proof, mostly because, well, turns out that even cycles are easier than odd paths in many regards. So what I'd like to do is I'd like to sketch for you a proof of the maximum number of C6s in a planar graph and really harp on the main ideas behind us. All right, so we'll start by fixing a planar graph G on N vertices, and I wanna get a bound on the number of copies of C6 that can appear in this graph. Well, let's look at a C6. I'll draw one right here. So we have our lovely six vertices, great. And let's pick three of these vertices to be, I'll call them anchors. So we have some X, some Y, and some Z. Now imagine that we were to lay down these six, these three vertices inside of G and ask how many ways are there to complete those three vertices, those anchor vertices to a sixth cycle? Well, all I have to do is figure out how many choices there are for each of these vertices right here. Well, this vertex has to be adjacent to both X and Y. So the number of choices is the co-degree of X and Y in G. So I'll write just degree of X comma Y to denote the co-degree of two vertices. So the number of common neighbors that they have. Same thing here, this has to be adjacent to both Z and X. So they're at most degree of Z and X choices. And this one has to be adjacent to both Y and Z. In other words, we can write down a quick and dirty bound for this guy, namely, we'll sum over all triples X, Y, Z in the vertices of G, choose three, of the number of ways to complete those three vertices to a sixth cycle like this. So it's the co-degree of X and Y times the co-degree of Y and Z, times the co-degree of Z and X. And really, if we wanted to, we could divide this by two because there are two choices for the three anchor vertices. They could either be these three vertices or they could be these three vertices. We've counted each sixth cycle twice. So we could divide by two if we really want to. All right, now this alone is enough to get the order of magnitude. If you just play with this expression and use the fact that planar graphs, the sum of the degrees is at most six N, you'll easily get the right order of magnitude, but getting the leading constant is a bit more tricky. So in order to actually get the leading constant, we need to use two big ideas. So idea number one, idea number one is to partition the vertices of G into two sets. So the first set I'll call big and the second set I'll call small. Well, what is big? Big is the set of all vertices that have very large degrees. So let's say that their degree is at least epsilon N where you can think about epsilon as a very, very tiny but fixed constant. So that's why big is much smaller than small, at least in this picture, simply because since there are only linearly many edges, the size of big is about one over epsilon, order one over epsilon. So the number of big degree vertices is about constant whereas small is about everything else. All right, okay, I've partitioned my vertices like this and now the idea is to show that almost every six cycle actually alternates between big and small. So in other words, almost every single six cycle looks like this. It has three vertices in big and three vertices in small. Any six cycle which, you know, does anything else, there are very, very few of them. So essentially every six cycle is controlled by the big degree vertices. Well, if we go back to this picture, that means that I can count six cycles at least asymptotically by only considering these anchor vertices inside of the big set. So in other words, we can bound NGC six asymptotically by the sum over triples XYZ but now that are just big vertices. So that's the same thing. So degree of X comma Y, co-degree of Y and Z, co-degree of Z and X. One thing I should notice is that we do not get this factor of a half anymore because there's only one, this only counts, each six cycle is only counted once because it's only counted by putting the anchors on the big degree vertices. Okay, well I'm very happy at least with this reduction right now because before we were summing over N choose three different things. So N cubed different things but now the size of big is constant. So we're summing over only a constant number of things. Already I think that this is a great improvement. So we only care about actually anchoring our C sixes on the big degree vertices. And now idea number two comes in. And idea number two says that the co-neighborhoods of vertices in big, so if I look at all of the co-neighborhoods between big degree vertices, they are asymptotically disjoint. So there are very, very few vertices that are in more than one co-neighborhood of big degrees. In other words, what our graph really looks like, at least up to approximation, is we have a constant number of big degree vertices and in between them are living a bunch of small degree vertices in these kind of, well Ryan and I like to call them tumors.  Okay. Well, we're making some pretty good progress. Let's do the following. Let's define a function mu of x, y to be the co-degree of x and y divided by n. So mu is a function big choose to we're only going to consider co-degrees between big vertices and mu is asymptotically because the co-neighborhoods of the vertices in big are asymptotically disjoint. That means that if I add up all the co-degrees it's at most n plus a little bit. So dividing by n, that says that if I add up all of the mu of x, y's I'm getting about one. So mu is asymptotically a sub probability but seven quotes. It's a probability mass on big choose to and this is all because the co-neighborhoods of big degree vertices are asymptotically disjoint. So let's rewrite this bound in terms of mu. If we rewrite this bound, the n of g, c6 is asymptotically at most the sum over x, y, z in big choose three of mu of x, y, mu of y, z, mu of z, x all multiplied by n, q to take care of the dividing by n for each of the mu's. Well, this is fantastic. This tells me that now if I just want to bound this guy that this is at most the supremum and I'll tell you over what in a second over x, y, z in capital X choose three where capital X is just sunset mu of x, y, mu of y, z mu of z, x times n cubed where this supremum is taken over capital X a finite set and mu a probability mass on x choose two. So in other words, if we want to get an upper bound, at least asymptotic upper bound on the maximum number of copies of c6, well the n cubed is already there, that's the right order of magnitude and this will be the constant out in front of it the supremum of this optimization problem on probability masses. Now let's talk a little bit about this function. Well, mu is a probability masses on, you can think about it as x is just some vertex set and mu is a probability mass on the clique that has those vertices, I mean that's x choose two and what is this expression? Well, this expression right here if we were to write it down it's a constant, which is pretty easy to compute I think it's like one six in this case of the probability that if we sample independently three edges according to the distribution mu that they form a triangle. So in other words what we're really asking is I'm going to allow you to put some probability distribution on the edges of a clique and I want to maximize the probability that if I sample independently three different edges that they actually form a triangle and I want to know which is the best probability mass for this. Are there any questions so far? No. Okay, so again let's bear this in mind I want to maximize the probability that three edges sampled independently so it may be that I pick the same edge three times they're completely independent samples that they form a triangle. So let me just give you let me just call this a name I'll just call it beta mu and I'll call it C3 because again we're looking for triangles this will just be this definition x, y, z in capital X choose three mu of x, y, mu, y, z, mu, z, x. All right and now our goal is to show that the supremum of the beta mu C3s is at most one third cubed because again that is the constant that we're gunning for to show that it's n cubed divided by three cubed. Okay well one thing that you can notice with this if you think about it for a moment is that this is the exact answer that you would get if you took the uniform distribution on just the edges of a triangle to start with. So in other words we're trying to show that if I want to maximize the probability of actually getting a triangle then I should just take the uniform distribution on the edges of a triangle to start with. So that's what we're going to try to argue. So let me show you why this is true. All right. So let's fix an optimal mu and let's suppose that it has minimum support. So it gives very few edges have positive mass. Now you should go away Chris what do you mean optimal mu? This is a supremum how do you know the supremum exists? There is a way to argue that the supremum exists but that's actually not necessary to do the rest of this proof. You can use a silly little trick to get around this fact but let's ignore that for a moment and pretend that the supremum does indeed exist and therefore we can fix a mu that achieves the supremum just to make our lives easier. All right. So we have at our disposal an optimal mu and its support is as small as possible. So let's define what I call the support graph of mu which I'll just denote by g sub mu and the support graph is exactly kind of how it sounds. Mu is a probability mass on the edges of a clique. So g mu its edge set is precisely the support of mu and its vertex set is you know whatever whatever vertices those edges span. In other words g mu is just telling us which edges actually have a positive probability to be sampled. Okay. So I have the following claim about g mu and the claim is that every pair of edges in g mu are contained in a common triangle in g mu. In other words if I give you any pair of edges that have positive mass under mu then there's a triangle which contains them and also has positive mass. So there's always some way to pick a triangle that uses this pair of edges no matter which pair I give you. The astute among you will notice that if we can prove this we're basically done but I'll leave that for a moment. All right. So why is this true? Well let's define something. So for an edge let's say x, y let's define the function t of x, y to be the sum over all vertices z which are not x or y of mu of x, z times mu of y, z. So if we think about this in probabilistic language this is essentially a conditional probability. It's the probability of picking a triangle conditioned on already having picked the edge x, y. This is the probability of completing the edge x, y into a triangle essentially up to a constant. All right. So for the actual problem let's suppose not, suppose that e and s aren't in a common triangle. So e and s both have positive mass they're in the support but they're not in a common triangle of g mu. If that's the case then we can rewrite beta of mu c3 in the following way. Let's first look at all of the triangles which actually use e. Well those would give us mu of e times t of e. Okay. Plus mu of s, t of s. Now notice that we haven't over counted because by assumption any triangle which contains both e and s has zero mass. So there's no chance of picking it. So we only have to worry about triangles that are actually triangles in g mu. So these are the ones that use e. These are the ones that use s. And then there's I'll just call it the rest. Triangles that use neither e nor s. Sure. All right. Well we'll log we can assume that t of e is at least t of s. Fine. No big deal. Now let's look at a new probability mass which we'll call mu prime. And mu prime simply says move the mass on the edge s to e. So literally e the mass of e under mu prime is the mass of e and plus the mass of s under mu. And s now has no mass on it. Well let's see what happens if we compute beta of mu prime c3. Well first we look at all the triangles that contain e. Well the new mass of e is mu of e plus mu of s. And because s and e were never in a common triangle, this t of e function stays the same. It doesn't know whether it's under mu or mu prime. All right, plus the triangles with choose s are now zeroed out because s now has no mass. And the rest, and this is the same rest as above because again we've only changed the mass on e and on s. And the rest were triangles that use neither of those. But now we use the fact that t of e is at least t of s. So we get at least mu of e times t of e plus mu of s t of s. Plus the rest. And that's exactly the beta of mu c3. But that's a contradiction. Because we started by saying that mu is an optimal mass with minimum support. But we just constructed a new mass mu prime, which does at least as well as mu, but has strictly smaller support. Because, well, we moved all of the mass from s to e. So mu prime has one fewer edge in its support. So this shows us that every pair of edges in the support graph must be contained in a common triangle in the support. Are there any questions? Okay. Well, now we're basically done. Because there's only one graph that has this property. If every pair of edges are contained in a common triangle, then the graph must be a triangle. And possibly there are isolated vertices, but because we only care about edges. This just says that g mu is a triangle. Let's say that the triangle has vertices x, y and z. But now, if we compute it beta mu c3, which is the best one. This is equal to, well, there's only one triangle. So when we sum over the triangle, it's just mu of x, y, mu of y, z, x, apply the A and G M inequality. And now it's a probability mass at the top adds up to one. And there's our proof. So if we go all the way back, we've shown, where did it go? Now I'm losing all my pages. Okay, I don't know where it went. But this says the maximum number of copies of c6 is asymptotically at most we said it was the supremum over mu of beta mu c3 times n cubed. And this is n cubed over three cubed, just as we wanted. So this idea works for any even cycle. Namely, NP of n c2 m. This corresponds to the function beta of mu cm where this is exactly what you'd think this is some constant which can be computed times the probability that if you sample m edges, independently from the probability mass mu, that they form a copy of an m cycle. So in other words, if I want to figure out this extremal problem for the two m cycle, I actually just need to understand this optimization question for the m cycle, namely which mass maximizes the probability that if I sample m independent edges from it, they form a cycle. So our conjecture is that beta of mu cm is at most one over m to the m. In other words, it's maximized by the uniform distribution on the edges of CM. So far we only have a proof for m equals three. That's the one I just showed you. I showed it for m equals four, but it's more involved. So in other words, that's why we get the six cycle and the eight cycle. I think that it should be easy to get this beta of mu C five and C six, because in order to get the four cycle we had to prove a number of structural results on this mass. And essentially with those things you should be able to get it but I don't know how to get those two, but it gets more and more difficult as m gets larger and larger. Right. So one thing that I'll quickly mention is that there was nothing really special about cycles. What was really important is that C2M is the one subdivision of Cm. I subdivide each edge once of Cm, I get a C2M. And this general idea about relating this, this extremal problem to an optimization problem over probability masses really works for any graph that's formed by a one subdivision. There are some restrictions. But generally, you can phrase this correspondence for about any graph. There's like a subdivision. Are there questions? No. Yeah. Oh, okay. So I now want to actually tell you a little bit about pads. So if I bring back up our results on pads. We actually got the right answer for the seven path. So let's use the very similar idea. I want to give you an idea of, you know, the slight difference between pads and cycles. And I don't want to prove our P7 result because it's somewhat complicated. But I do want to re prove the result of Goshen friends who proved the asymptotic answer for the number of pads on five vertices. So this is very different than their proof. And it uses the same ideas for our cycles. So let's prove least sketch a proof that the maximum number of P5s in a planar graph is at most and cute asymptotically. This is Goshen. All right. So we're going to start by doing a very similar idea with the anchor vertices. So if we were to draw a P5, here's a P5 right here. We're going to anchor these two vertices X and Y and count how many ways there are to extend this to a five path. Well, there are co-degree of X and Y many choices for this guy. Degree of X many choices here and degree of Y many choices here. So we can bound N of GP five by we sum over all pairs X and Y in features to of degree of X, the co-degree of X and Y times the degree of Y. And we get to divide by two because there are two orderings on this path. Oh, sorry. I want to sum over just X not equal to Y let's say that way I can divide by two. So now we apply our two ideas. So again, we partition into big and small. And we show that almost every single P5 has these two anchor vertices in the big degree in the big set. So almost every single P5 looks like this. Automatically, we can improve the bound to say asymptotically at most the sum over X not equal to Y but now in big get to divide by two apologize. Degree of X degree of X comma Y times degree of Y note that here I actually get to maintain this factor of two, because there are still two different paths I get I get the same path. I pick this vertex first and then second, as I would if I pick them in the opposite order. And now we apply our second big idea, which is that these co neighborhoods are almost disjoint. And actually, we need a little bit more that says that, well these guys don't live in co neighborhoods these are just literally the degrees of the vertices X and Y. So we need to do a little bit more effort to say that actually the graph really does look like this blow up structure.  So we have an argument, but it's doable. So what we'll wind up getting is that the maximum number of p fives can be asymptotically bounded above by the following. Again, because they're disjoint we're going to change degrees into some probability mass. So we'll sum over X not equal to why in some set X of, and I'll define these in a moment mu bar of X times mu of XY times mu bar of Y times, in this case, n cubed over two, because we divided by n three times. So we get an n cubed, where again the supremum is just over X is a finite set, and mu is a probability mass on x choose two. And now the only term I haven't told you what it means is this guy right here. This is the weighted degree of a vertex if I think about the probability masses being weights on the edges of a clique mu bar is the weighted degree so this is the sum over all why not equal to X of mu of XY. In other words, if I think about it, if I sample an edge from the probability distribution mu mu bar of X is the probability that that edges incident to the vertex X. So if we actually want are there questions about I know I didn't tell you too many details, but the details are roughly the same slightly more complicated than for cycles. But we again get down to this optimization question over probability masses. The goal is to show that for any probability mass mu, the sum over X not equal to why of mu bar of X, mu of XY, mu bar of Y, that this is at most two. Because then that to cancels that to, and we get the correct asymptotic result of just one times n cubed. Well, how can we prove this. Well, let's define a matrix. So capital X. All right so mu, mu is some probability mass on you know the edges of a clique some x choose to let's define a matrix. Those in columns are indexed by X, where the XY coordinate is just the mass between X and Y. And we'll say where we'll just say that mu of X X is zero. So the diagonal entries of M are zeros. All right. So we're going to make a few observations one. M is non negative. I mean its entries are either zero or some number between zero and one. Two, it's symmetric, simply because the probability mass doesn't have an ordering on it. It's just the probability mass of the edge. And three, it's row sums are all at most one. Simply because the row sums of the M matrix are precisely the mu of the mu bars which are just a probability. A row sum is just the probability that an edge is incident to that vertex. Okay. Well number one and number three together tell me that the maximum eigenvalue of M is at most one. It's a non negative matrix, all of whose row sums are at most one so it's max eigenvalue is as well. And the symmetric component. This tells me that I know something about the Rayleigh quotient, namely, this guy is at most Lambda max times the inner product of X with itself. Well, and Lambda max is at most one. So if I take the inner product of over M so X inner product MX, that's at most the inner product of X with itself. So now we rewrite this new bar of X new X Y new bar of Y. And by rewriting it and thinking about mu bar is a vector. I mean it's a function from capital X to R so it's a vector index by X. And this is precisely the inner product of the vector mu bar with M times new bar, which is now at most the inner product of new bar with itself. And if we just write what this means this is the sum overall X new bar of X squared. But now mu bar is a probability. So it's at most one. So I can just delete the squared and only get bigger. The question is what is the sum of the mu bars. Well, mu bar is a degree. And the handshaking lemma tells me that the sum of the degrees is twice the sum of the edges. I mean in this weighted sense the sum of the weights of the degrees of the weighted degrees is twice the sum of the weighted edges. But the edges are a probability mass, which is one. The sum of the mu bars by the handshaking lemma is two. So we have in fact proved exactly what we needed to. All right. So, generally, so I want to reiterate generally, and P and C to M corresponds to the question of maximize the probability. So if you sample edges e one through E M from some probability distribution that they form an M cycle so a copy of C M. We can actually read you everything that we did with P five to show that NP of N and then here P two M plus one. This corresponds to maximize the following function. So the sum overall X one not equal to X M of mu bar of X one times mu of X one X two mu of X M minus one of X M bar of X M overall. Expression is slightly more difficult to actually write down as a clean probability like this. But it can be written as a probabilistic event. It's just these mu bars kind of make it a bit difficult to describe or more just not very enlightening to describe. And these are wide open pretty much. So, again, all we know is this one is good for well technically we can do it when M is two. So we'll say two, three and four. And this one is only good when M is either two or three, but they're wide open otherwise. What I mentioned is that this problem is significantly easier than this problem. For instance I say maximize here really it's a supremum overall probability masses in this question I can prove that the supremum is actually achieved so it's literally a maximum. In this problem, I have no idea. I have absolutely no idea how to even prove that the supremum is even achieved. And essentially the entire difficulty comes from the dangly leaves, the fact that these new bars show up make things very very difficult. And I have no good way around it. So at least when M is say five or six for the cycle question. You, I mean if you had a computer program that could do nonlinear optimization really well, you could solve this problem. Even if you had a computer that could do nonlinear optimization well, I don't know how you'd even start to solve this problem simply because I have no idea how many edges should even be in play. It could be that infinitely many edges, you know that there is no supremum and hence you keep getting better and better and better as I have as you have more and more edges. I don't believe that to be the case, but it technically could be. All right, so all in there. Please let me know if you have any questions. Thanks Chris. If we could all thank our speaker in some way and then we'll go ahead and open it up for for some questions. We have any questions for our speaker. Yes, I do. Yeah. Steered completely around it but I'm guessing you're completely aware that you just did a bunch of like graph on maximization problems right. The number of triangles was like your maximize the number of triangles and a graph on. I mean, well, you say graph on but we're in a very sparse setting. Yeah. So aren't the graph on trivial when you're in this because all of our results apply to only when you're grabbed so again planard never mattered throughout all of this it was only linearly many edges and no legitimate copy of K33. But don't graphons require you to have n squared edges to even exist and not be trivial. Well, I mean so there are these sparse graph models with like linearly many edges. And so maybe it fits into that paradigm but also like that. I mean this question about the way you have written there with the CMs. So that's over any new word any so all I'm saying so the new here is just a probability mass on x choose to where x is just some finite second x is some finite set. So this new it this new is a graph on that's true. That's yeah. Yeah, so the new itself is a graph on absolutely. I mean I know essentially nothing about graphons but the new is a graph on yes. So how is this not just maximizing and I mean of course there's all this like run up and hitting down the structure before you can use the optimization but one thing is that well I don't know if this again. I don't know anything about graphons. So let me preface it with this one. The one thing that might make it not a graph on is the fact that this x is literally finite. So I say that the mu is a graph on but it's in some sense not really because it's important that the vertex set is actually finite here. But all of the examples that we believe actually optimize this do have finitely many vertices. So you're not going to get better as you get more and more vertices which may actually hinder you representing this is a graph on. I mean in some sense it still is because I can take step functions right, but I don't know the right way to think about that. Another thing that's important here that I think actually makes this. I don't know again you tell me because I don't know about graphons is that one thing that could happen here when I select these m edges is that I may have picked the same edge twice. In which case I will never form a copy of CM. Well, a CM on every edges only picked once right every edges multiplicity one. And I have a feeling that well I don't really know with the graph on model if you could represent this clean possibly you could. But I don't actually know in the limit these, these kinds of, you know, the event that you pick something twice ends up. The main issue though is that in some sense, this new is not a limit object because we're actually gunning for a new that is very finite. Because this probability I can check sure that it's optimized by the uniform distribution on the edges of CM itself. Right so I don't know what the what limit object there even is. Another way to avoid the graph is single body line graph, and they consider the maximum large in the maximum what a large in the larger of the way it's just like a switch weighted. Some equal to one so if you consider the learning of the graph and maybe you can apply that. But is it this because if I'm in a line graph triangle in a line graph, well, doesn't correspond to an actual m cycle does it. No, it doesn't because I may have revisited the same vertices when I pull it back to the original graph. Yeah, I think so you probably you have to like condition I forbid this. Yeah, probably I don't actually know I haven't thought about them as line graphs. Again, this is a very general problem as I mentioned for this guy right here. I can really replace C2M by any subdivision of a graph, provided that it's like minimum degree is at least three or two, two or three. Basically, so I have a one subdivision of some graph that doesn't have dangly leads like the path does. Then again we're trying to maximize picking a bunch of edges that they form a copy of the graph that we actually subdivided. This is an interesting question because well you might conjecture that the right answer for if I want to maximize the probability of getting a copy of H that the right thing to do is to take the uniform distribution on the edges of the graph is not actually correct for for infinitely many grass age. I have lots of counter examples. But for cycles I think it is true. And again we can prove it for three and four. Technically this problem is very slightly different when MS to because what what do I mean by a two cycle here. It's really I want to maximize the probability that when I pick two edges I actually pick the same edge. And that's what I mean by two cycle, which is clearly optimized by literally only having one edge in your support. So yeah, Josh. Yeah, sorry, I, I know essentially nothing about graph on the main thing that I would be saying for why. This mu is a graph on because I could think about there being infinitely many vertices, but at least for every, every one of the problems that fit into this situation I can prove that the supremum is achieved by something on not too many vertices. And there are not very many vertices in all of these maybe like the number of vertices is, I don't know, like twice the number of edges is like a strict upper bound, say, remember what exactly I can prove. And for the paths for odd paths. I can't prove the supremum exists but again the conjectured answer for this guy what is the maximum here. The conjecture that we have is that this is actually maximized again by uniform on the edges of an M cycle that this that this thing is optimized by uniform on the edges of an M cycle. Again, we can prove this for two and three. But if these are indeed the optimizers then I don't know what graphons would have to say about it, because there's no real limit object here, but perhaps I just don't know enough to say anything. My mental picture was having X set. Grow without bound, and then you've got a. Yeah, it point here is that theoretically X could grow grow without bound right, because in our reduction step. The epsilon needed to go to zero. So your number of big vertices may tend toward infinity. So theoretically the X could grow without bound. The kind of point is that in what we're saying is that if the X grows without bound actually what happens is that most of those vertices are just isolated. They're not incident to any edge that has mass. Your entire mass is concentrated on some finite sub graph. That should be what's going on in these problems it is going on. I don't have this problem, I have no idea but I suspect that it's also going on but I just don't have any idea on how to, but I'm not going to discourage you from trying to think of it find out a different way to think about this. I just don't have any idea how graphons held. Yeah, the sparseness is making it so that the limit as a size of the X goes to infinity is irrelevant. Another thing is that maybe I should point out is that this mu has nothing to do with planarity right it's literally an arbitrary probability mass on the edges of a clique, where we started with, you know, linearly many edges this mu in its support graph may have all of the edges having positive mass, who knows. What I thought is that taking, taking the ground set to be larger and larger and larger doesn't help you. You should really, and again actually this makes sense. I mean if I had a new that was actually spreading out all of its mass, but I want to maximize the probability that if I pick a bunch of edges it forms a particular sub graph. If a new spreads out all of its mass, then I'm going to pick these edges to be really far away from one another. I have no chance of forming this finite, this like very compressed object. So that kind of gives you an intuition on why you shouldn't spread mass out everywhere that it really should concentrate on some small parts. I'm curious more about like what's actually going on in these functions. The paper is on archive where we do a lot more than than what I talked about and study these types of functions more in depth. If you want to take a look. And maybe there's something there if you see not my hand waving nonsense when talking about this. Maybe you can get a better idea, but I'm great at hand waving nonsense. Do we have any other questions for Chris. Okay, if not, thanks again Chris and thanks everybody for for making it out and have a good weekend everybody. Thank you.


Hey everybody, welcome to another exciting episode, another edition of A Week in Geekdom here on YouTube and just like I did with previous movies, this is sort of a series of topics on the Marvel, Captain Marvel movie I should say, from Disney slash Marvel. This part we're going to be talking about the entertainment weekly pictures that were released and then my thoughts on everything and then we'll do a second video with like a trailer impression I guess and a final one with a movie review when that movie hits theaters next year 2019. But yeah, entertainment weekly got the exclusive, they released the pictures and let's let's let's talk about them. Let's dissect a little bit about the Captain Marvel movie. Now this movie is a very important link for two reasons. One, you're tying up Marvel cinematic universe history by setting it up in the 90s. You're tying up the older stuff, you know, from the Soviet Cold War era and the World War II stuff with Captain America and now with the modern heroes after 2008 with Iron Man. You're missing a huge chunk of information in the 90s and Carol fills that perfectly with her story. I'd like that aspect very much. I am a sucker for continuity and all, you know, bridging the gap between movies and stories and all that stuff. I love it. And the second aspect of course is, well actually, let's say second out of three aspects. Now the second one is the fact that this is a heavy cosmic movie and it fills that cosmic puzzle. It fits that puzzle with the character, you know, after the success of Guardians. It has been proven that people really do like the cosmic side of things. I am a huge cosmic fanboy. I think Marvel cosmic is the best side of the brand, in my honest opinion, when it comes to the comic books and now with the movies as well. You can tell so many rich, awesome stories with cosmic characters, especially with the Guardians, with now that you've got Fantastic Four, with the alien races, and of course with Captain Marvel, etc, etc. And the third aspect that I just completely made up now, she fills in that gap for the Avengers follow-up after the events of Infinity War. She's obviously going to be a powerhouse when it comes to cinematic power levels, redundant, I know. But after Infinity War, I was a little bit disappointed that the geek fanboy in me was disappointed at the power levels. Look, I get it. I'm probably in the minority here, but I kind of did not appreciate the fact that freaking Thanos with a fully-powered Infinity Gauntlet couldn't stop a Stormbreaker, whatever. I knew it was going to happen.  So with that, all of that said, I am very interested in this movie because of that and the fact that we're getting sort of an adaptation, if you will, not only of her origin, but of the Kree vs. Skrull war. I love the Skrulls. They're one of my favorite alien races in comic books. I love the idea that, you know, you've got some really heavy hitters with the Super Skrulls and the Queens and the different levels in their government kingdom structure, if you will, Empire structure and they are shapeshifters and they can be really dubious and infiltrate like they did in the Secret Invasion storyline where they invaded the whole planet Earth and took up residence in different areas from governments to superheroes to secret agencies, all that stuff. So I love that espionage, secrecy level that the Skrulls are involved with. I love that stuff and it looks like the movie, you know, it won't be an origin story. It will be set in the 90s. Carol comes back from space to deal with the Skrulls, but somehow the Kree are involved as well. I'm not too sure. We did get to see Star Force and other aspects that are pretty friggin interesting, but whatever. Let's talk about the images and let's get the negative out of the way. The cover photo. I'm not a fan of it at all and to be perfectly honest with you, I love Rilaris in this Captain Marvel. Don't get me wrong, but I think that image could have been 10,000 times better. It was actually quite dull, boring and obviously touched on Photoshop or whatever, you know, photo manipulation and whatnot because it looked really corny. You know, you could have had her in sort of this hovering flight pose. I don't know. Her expression is very boring. I could have, you know, you could have done sort of like this confident grin. You know, just the composition of the image is a little off to me and I just found it bland, but most of the superhero entertainment weekly covers are like that, unfortunately. And what we do get are behind the scenes photographs. They're not actual spills of the movie per se because I think the movie's still in post-production or something like that. I'm not quite sure as of this video, but it's just behind the scenes photographs. One of the first photos which I really liked and we're seeing here is of Carol in her Captain Marvel costume and it looks pretty friggin fantastic. I wish she had her Mohawk, but I think she has to earn that Mohawk, but still she looks great and looks really awesome and just it really hits the geek spot to see a new character that you love from the comic books get transported into live action. That is always exciting to me. Second photo we get to see the Star Force, which is basically this elite Kree space force, if you will. And we see Korath, the pursuer, as well as Ronan. They're going to be in there because it's a flashback, the Kree Empire, et cetera, et cetera. And you sort of see the origin of Carol's costume except it's a green and black with gold details on it. We do see Jude Law and Brie Larson. Obviously Jude Law is supposed to be playing Marvel, the original Captain Marvel. So I'm very interested in seeing how their dynamic works in the movie and of course how she gets her powers and all that stuff.  Next up one of my favorite photographs and the one I geeked out the most. Obviously the scrolls. We see four scrolls on the beach. A lot of people were making fuss about the makeup. Like I said it's a behind the scenes photograph of production stills or whatever you want to call it. It's not like the actual finished product. I think they look great. If you compare it to the original scroll drawings, it looks like a modernized contemporary look. Obviously some people have pointed out that it looks very Star Trek-y. I can sort of see that but at the same time it retains the essentials of the scrolls. You know you've got that green skin with the wrinkled chin and you've got pointy ears. I think it looks great. Obviously the leader of this ragtag group is Talos. He was very infamous in the comic book so I'm very curious to see if he survives the movie. I want to see more scrolls in the future. I have made comments on this youtube channel that I think the face for a marvel thing is going to be scroll heavy. You know you're getting this movie with the scrolls and just the fact that they are here on earth since the 90s. Who's to say that they haven't been planning you know of course in the plot of the movie they're going to do an invasion but who's to say they didn't they don't stay you know after the bad guys are beaten or whatever. I mean now you're throwing a scroll into the equation now I have to ask myself who do I trust who who's to say that a hero is actually a scroll in disguise. You don't know that I think would be an amazing twist similar to Brian Michael Bendis' secret invasion storyline where you find out that electro-spoilers was actually a scroll in disguise. It could happen man it not necessarily been electro but you know what I mean it could happen some sort of variation some sort of version of that story and that is really exciting plus now you've got the fantastic war with the fox perches they can interact with the scrolls you can do so many cool things with the super scrolls I'm just floored with excitement it's a little bit of a spoiler but you do get to see Talos in his human form infiltrating shield plus some familiar faces you get to see Ronin the accuser which I love Ronin he's such a badass in the comic books I was a little disappointed that we basically got this crazy-ass Zilok version in the Guardians movie but really poor character development hopefully in this movie we get that backstory I love the idea that we get a revisit of old character nice little easter egg right there with a nine inch nails t-shirt and she looks the part I think this is one of the better photographs that was released I really did it I love the idea of the aging of character we saw that with Robert Downey Jr in Captain America Civil War if I'm not mistaken and it looked a little weird it was cool but a little bit weird here it looks much better I'm really excited to see Sam 90s Sam Jackson that's actually pretty cool that they're going they're gonna show us her origin at least in some shape of corn with her past as an Air Force pilot and of course we're seeing Maria Rambo which I think she's the mother of Monica later becomes a Captain Marvel herself so it's pretty cool it's a nice easter egg to spot in these photographs so yeah that's sort of my ramblings I guess on the Captain Marvel Entertainment Weekly reveal hopefully we do get to see a teaser soon I'm thinking by Christmas or something like that I don't know the movie's coming out in March April something somewhere in spring so I'm very interested in seeing how the movie ends up looking you know because like I mentioned these are just stills photographs not actual film frames and stuff so I'm really excited about that what do you guys think about the Captain Marvel stills all these photographs and information the whole idea the plot of it all let me know down below I'm very curious to know what you guys think thank you so much for liking commenting subscribing sharing this content thank you it really means the world to me you are the best find me on your favorite social media platform so we can keep the conversation going and I will catch all of you on our next episode what do you guys think about the cart cap


What's going on infinite fan welcome back to another video real quick guys if you guys are new to the channel Make sure you guys hit that subscribe button today's video is about to be a banger your boys energetic. I'm feeling good today I'm not gonna lie yesterday. I really really hurt my back. I don't know how I think I fell asleep wrong But yeah, your boy woke up really really in pain So that's why there hasn't been a video for the past couple days, but today that's gonna change team Isaiah We're coming with a banger so guys right now Janice went to the bank. She's pulling out some money She's doing something. I don't know she's doing sugar mama stuff As you guys can tell by title and thumbnail the video your boy is gonna be gaming with other girls today, right?  You know, we're just chilling or whatever and Janice is gonna blow her top This is gonna be crazy.   It gets crazy She acts wild so today's video is definitely gonna be a banger So if you guys are ready make sure you guys hit that thumbs up and comment down below team Isaiah all day every day I'm gonna go set up upstairs and we're gonna wait for her to get home Damn Why was that door closed? I'm trying to keep it cold in here Yeah, but he was downstairs No, that's not a roommate That's gone What did you just say?  That's fuuuuck Come on Baby, that is crazy, right? Yeah Keeps going foul, I think I hit the pole, hit the pole Which one is it? I hit the pole just like my roommate What?  What are you talking about? Baby, I'm not talking to you, I'm talking to my friend Yeah, I know, but what roommate are you talking about like in the game? Do you have like roommates or something like that? Hey, what baseball players have roommates? What are you talking about? Cause you just said roommates so I'm like confused No, right now I gotta love them Make that 12 cause I just hit them with a 452 I'm using Bryce Harper Yeah, Bryce Harper and Peter Lanza Bro, the Mets had a chance last year I'm madly glued in the playoffs Against the Padres at that How are you using the wild carpet?  They've been talking to my friend, I said The person I met online So who are you calling your roommate? That's another one Why are you wanting my question? Who are you calling your roommate? Give me a second, give me a second cause my roommate is tweaking What? What happened? You're calling me your roommate? Who's calling you the roommate? You literally just said my roommate is tweaking Yeah, I'm saying roommate, I don't wanna see my girlfriend Why not? To me Don't just me Stop Who are you talking to? Babe, wait I'm telling you, I'm a friend that I met online Babe, listen, can you stop? You're starting to get crazy, relax I'm trying to not let them know that I have a girlfriend Because they're gonna find out that we do YouTube and all that And you know how people gather and we find out that they do YouTube They're like, oh yeah, okay, I'm gonna come over Nobody cares that much Okay, babe, I don't know this person Why are you laughing? No, give me Babe, who are you talking to?  How did that go out?  Babe, does that really matter? Yes, it does matter!  What am I waiting for?  Roommate who? Yeah, that's her in the background Yeah, that's me His girlfriend Not a roommate Don't listen to anything she says We're just roommates She's trying to make things more than friends And I told her I just want to keep a guy out of it Oh, hell no! Who are you talking to?  I'm his girlfriend No, she's... I'm his girl She's not my girlfriend We've been together for seven years We've been roommates We've been together for seven years And you know they're roommates No, we've been roommates for seven years Because I know you're talking to a female You think you're slick Why does it have to be a female? What? What?  You want to play? I'm not playing with you I don't have another controller I don't have another controller You can't play Move Stop, look, I just got a strike I just missed it Give me your headphones Stop No, stop, nothing Babe Why are you scared?  Because you won't get rocked What's up?   No Yo, I'm such a beast When it comes to prayer Man, just once Here we go Now I gotta play hide and seek Janice, where are you? I'm about to show you the outside world You don't start acting, you know, civil What are you talking about?    Just one time You hear that Janice? One time I just want to lose once, please I'm tired of winning It's not fun anymore So with all that being said, guys If you guys enjoyed today's video Make sure you guys smash that thumbs up button down below And then comment down below again one more time Team Isaiah all day, every day And I'll see you guys in the next video