Hi everyone and welcome back to another episode of our Road to Runt roulette series. We kicked off this brand new series at the start of the week and it has been so much fun so far. This is a team that we've ended up with, we've used all of our bonus buttons of the last couple of episodes. We had Houndoom in there, we had Shuckle in there and we had Chandelure in there and we've replaced them, we've used our Legend Maker button bringing Palkia, we've used our Switcher button which brought in the Arcanine and we've used our Randomizer which has brought in Malamar. So we had really good episode yesterday, we had a really good victory to kick us off with. If you'd like to go back, check those out. I'll put a card up there for you before coming into the episode. But this is going to be the penultimate one with this team before we choose our new roulette team Monday next week. So as I mentioned in previous episodes, we have over on the community section of the channel. If you go over there, there will be a post over there asking you to nominate Pokemon for next week's roulette wheel. So I'll go over there and nominate a Pokemon. It can be any Pokemon as well. It can be anything from Caterpie, anything right through to Mewtwo X or just any Pokemon. Your favorite Pokemon, it can be your favorite Pokemon, a Pokemon that you want to see, compare every paid and just nominate and we will throw it into the wheel and hopefully it does come out and we can have that randomised team for next week. So we are seeing on a record of three wins, four losses at the minute, which isn't great, but I feel confident, like I said in the last episode, that we will finish positive. And that is the goal for this. So without further ado, let's get into this first one today. I'm going to just change screen so we can see our bottom screen as well. See what our rating is currently and finishing over 1500 is a good goal. But we want to see if we can improve on this. This is the first week doing it. It's a 14-19, man. So bad, so bad. Let's start with Elite Four. Let's go back to some old favorite music of ours. And team as always is down in the description below. As always, if you do enjoy this content, please remember to drop a like on the video to subscribe to the channel for more Pokemon content and do leave your comments down below as well, because I love hearing your thoughts on the series in general, on the team, on the roulette, on anything. So make sure to get the nominations on that post in the community section, though, but if you want to leave them down below, I will grab them and throw them in. So hopefully it doesn't take too long to find our first opponent today. As always, if it does take a little bit longer than usual, I will just cut this and we'll come straight back to when we find that first opponent of the episode. And we've got our first opponent of the episode. It's not Geo. So is it Geo? Is it Giovanni Castor? Is it not? Who knows? But we'll hop straight over into Team Preview and we'll find out what they're going to be running today. OK, so our first opponent today is running a team of Gengar, which could be potentially the mega of the team. It's likely to be the mega of the team. Then you've got Alolan Marowak, not something you see so commonly on these in this format in general. So then we've got Kyogre, Primal Kyogre, probably. Then we've got Bruxish, which has that dazzling ability to prevent any priority attacks. But fortunately for us, we don't have any priority on that team. Then you've got the Duskman, the Krosma and the Alolan Persian. The Alolan Persian will provide fake out support to help set things up like the speed control, which is probably coming from the Duskman. The Krosma, if it's not alternate Krosma, it could be quite a slow team here, kind of indicated from the Marowak. And then the Gengar as well, potentially could have icy wind or something like Perish songs. So right, what are we going to do? Could Malmo be decent here? I feel like Malmo could be decent here. Let's bring Malmo up front. The knockoff could be very good. Superpile are going to be very good. What are we going to lead with it? Do we bring Superior? We're not brought Superior for ages, although it's not really doing too much here. Arcanine. I'm going to bring Palkia. I'm going to bring Arcanine and I'm going to bring Mantine as well. There's an argument for Sougalail here, but I feel like it's kind of threatened pretty hard from the Marowak, especially if the tricking goes up. The Persian puts a lot of pressure on with foul play, especially if we get the Z Splash off and we're kind of lacking a little bit of speed control. So let's see if we've got. And it is our good mate, our good friend, Giovanni Costa. He has just just tweet me now saying, and I thought my team was crazy. So I'm just going to send him a quick message just to say. Yeah, nice team. Good luck. Have fun, my friends. So we're going to see the Marowak, the Bruxish come out. Right. Malmo not in a bad spot here. And now we could get rid of what's the Bruxish going to do? Does it? I'm just going to go knock off into that slot, for sure. And I'm going to go. Do we go earth power into the Marowak? Let's just do it. Let's just be straight up with attacks. We've not really got too many options. What is this Marowak doing? It's tightening. Let's focus after it's like focus punch. What is this? Wow, we actually take it. So that works out all right. Man, after you, that is crazy. The memes are for real there. I've got to take your hat off to Geo for pulling out this sort of stuff. This is legit, legit memes. Can we get the Bruxish? Oh, we miss out. But we do, we do get rid of that choice scarf, which is good. OK, we've got to protect Malmo because we'll go down to another one of those. We'll just go for another earth power into the Marowak. It might protect here or switch out, maybe. But we still have to be careful with this after you, after you Bruxish. But we do get rid of the choice scarf, which makes it a little bit easier to deal with going forward and focus punch. What a great attack. It's so much damage as well. It's kind of neutral to Malmo, but Marowak. Marowak is a cool Pokemon, isn't it? There's a protect from the Marowak for sure. Yeah, makes sense. Let's watch this Bruxish going to do Psychic Fangs into a Parker. Doing decent damage, you know, like nothing to be scoffed at. Do we need Malmo for the rest of the game? Probably. It's probably going to be quite useful to keep around and have. I'm going to bring in Arcanine. And I'm just going to go for that earth power again. How do we go trick room at this point? Do we go trick room? Trick room could be kind of good, but we might be able to get away with a trick room. Maybe a few turns down the line, maybe, I don't know. Well, we're bringing Arcanine. We'll get the Intimidate off into this Marowak, which will help us out hugely. The nice thing about Arcanine as well is, OK, we're going to see another after you. Yeah, well, is it going to be a palky this time? Maybe, maybe, yeah, makes a lot more sense. But the Intimidate helping us out there a bunch. So we'll just get out very activated. We'll get the Marowak now with the earth power. And I think the thing that we're missing on the roulette is the memes. But hopefully next week with our new roulette selection, we'll be able to get the memes going. With a new set of Pokemon, random Pokemon. Right, the Bruxish, I think. I kind of want to say that Arcanine will outspeed it. What's this Kyogre going to do, though? It's not really threatening. I want to say that the Arcanine will probably outspeed the Kyogre as well. But off the top of my head, I don't know, Bruxish is just base speed.  I'm just going to go for a earth power into the Bruxish here. It definitely outspeeds Palkia. So we'll just protect Arcanine for now. Yeah, there's you see the after you and imagine it at speeds. Yeah, there's a water spot. So Palkia taking this like a champ, the water driving, typing. It's good. So get rid of this Bruxish, get rid of this pesky after you strategy. She's so good, though. OK, and now, now they're down to two Pokemon. Likely both these restricteds, though. I wonder what a life orb, wild charge will do. And do we outspeed the Kyogre? It's like we should outspeed the Kyogre with Arcanine. I'm just going to double up into the Kyogre. And the problem will be here is if we say protect Kyogre and trick room does mean which we're probably going to see right now. Ah, no. OK, let's see. Double the Kyogre fell straight into that one. There it is. Oh, OK. Hmm. Right, let's get Arcanine out of here. Let's bring in Mantine and let's just protect Palkia for a turn. So I imagine we're probably seeing the Cosmogor for that. Photon Geyser, if it's not Ultron or Cosmogor, it doesn't seem like it's going to be. Yeah, Photon Geyser into the Palkia. I'm going to see what it's about. Yeah, I'd imagine the Kyogre probably has Kyogre probably has Thunder. Let's see. Let's see the team make it. Maybe not, because then it does conflict with the Marowak a little bit. But I don't know. I don't know. Let's Skald and Earth Power. Then the Cosmogor. Let's see if we can get a couple of attacks off. I mean, the Earth Power into the Cosmogor would be pretty big here. Skald's not going to really do very much. Mantine's pretty safe here. I think he'd concentrate down more on the Palkia with the Photon Geyser here. Unless the Kyogre does have Thunder. No, we don't take it. Wow. That's brutal. Skald, at least we underspeed the Kyogre. It's good. I mean, if you can get a burn here, that's huge. No. And it does have Thunder. Changes things quite a bit. All right. But we do have Malmo in the back. We do have Malmo. So we can bring Malmo in pretty safely here. You can knock off that Nekrosma. We can, we can wide guard. Because how many turns of trick room you got left? Two. So this, this kind of will work out pretty perfectly. So a wide guard will go for that knockoff. I don't know if the knockoff will get the Nekrosma, though. That's the only problem. But I think Kyogre probably feels safe going for a water spout here. Send Steel Strike. Has this into the Malomar? Yeah, it will be. Hopefully this doesn't take us down. This is where the Intimidate would be very useful. Oh, God. Damn, that would be done for now, I think. Yeah, because we can't, we can't deal with the Nekrosma. And then the trick room just gets set up once again. Yeah. That's a good game, my friend. Love the team. I'm just, the Nekrosma's hitting us too hard. I think like the thing that we needed to do was get an Intimidate onto it, which gives us a little bit more leeway. But like, I just forget how hard Duskman and Nekrosma can actually hit. I mean, we can protect here. There's not really much we can do. We can't flare blitz the Nekrosma, which is the sad thing with the Rainer. So we can't really get rid of it that way. Photon Geyser, Water Spout will come out. And then it's so easy for Gia just to protect the Kyogre this next turn. And then just trick room once again.  I mean, we can go for a wild charge into the Kyogre if he doesn't protect it. Then we probably get it. But yeah, I mean, it's just this, the super safest play here to make. Oh, it's so sad it takes a losing streak down to five. So three wins, five losses. I mean, yeah, trick room.  Good game Geo, great team, nice strategy, good memes, all good. But yeah, just not got the, not what we need in the tank. Unfortunately, which is pretty sad. Duskman Nekrosma hitting like a truck. I don't have expected Marmar to take a Sunsteel Strike from that range, but I mean, we are more specially defensive. Nevermind, nevermind. All we can do is move on. We're gonna have a brand new roulette team Monday anyway. So it's fine. I'm not gonna get too sad about it.  I'm sad about our rating. I'm sad about our win record as well at the minute. Like at least though we finish positive. We've got Jack from the United States up next. So let's hop straight into team preview. See what Jack's brought. He's brought the Rayquaza, the Kyoga, Crobat, Bravery, the Ferrothorn and Tabacoco. So kind of standard looking Rayoga team apart from the Bravery, which is a nice option to kind of stop the intimidate support from our side to neuter things like that Rayquaza, which really does get hurt by it quite a lot. But I think Arcanine is still good to bring here because we will outspeed it and we will be able to nuke it with a wild charge as well. I think Mantine's generally quite good here, but the problem with Mantine is it is thrown by the Tabacoco if it has got the Z-Move, but other than that we're not too bad. We can match the Crobat to Tailwind as well. Do we bring Sogaleo here? Yeah, I think we do. And I think we bring Palkia as well and Bench the other boys. The superior to be honest, it's definitely one of those Pokemon we could have probably sacked off. We've not really used it since like the first couple of episodes this week. So it's definitely a Pokemon that we could have said we could switch out. They bring something else in. But never mind, it's all all right. It's all all right. So see if we can core a win to finish up this week. I feel like the team deserves it. So Jack's sending out the Bravery and the Rayquaza. We are gonna get the intimidate onto both targets, which is nice, so that's always good. And we will activate that defiant ability on the Bravery. Bird up, my friends. But we've got one bird of our own in man time. We'll go for the Wild Charge into that slot and we'll go for a Tailwind, Icy Wind. I'll go for an Icy Wind just because the Rayquaza could be sashed and breaking the sash is gonna be quite useful for us. If it does make a revolver, it's gonna be hitting very hard, but just breaking the sash and lowering that speed a good thing for us. And definitely after an intimidate, we'll take a Dragon Ascent from the Rayquaza. So whether or not, I don't know, I don't know. Like I worry about Earth Power into Arcanine for sure, but we should take at least an Earth Power from Rayquaza. Problem is though, the Bravery is now protected by the strong winds, which I do forget about and we do like nothing and take massive recoil damage and Arcanine not in a great place. Then we are gonna see, as we move, what if it didn't be just Super Sonic Sky Strike, brave bird, yeah, it's probably gonna be into man time. 100%, I think we lost this already. Man time, no, yeah, no way we take a plus one from, yeah, that's not good. That's not good at all. Sougalale isn't a bad option, but we're like, I mean, we're just earthquake, yeah, we're just earthquake fodder. I mean, could a close combat, would a close combat do more to Bravery? It probably would, and then an Earth Power, we're probably better off switching into Parker and then banking on the Ray going for an Earth Power into the Sougalale and the Bravery going for maybe a tail end or just attacking into the Arcanine. Crunch, okay, yeah, that's what we wanna see. Close combat, we've gotta hope this takes it down. It does, okay, we get rid of the bird. Still not over though by a long way because we've got to deal with that crunch. Although that is minus one, the Rayquaza, so it's not too bad. Now if we can get a trick room up, things are looking a little bit brighter for us, but we're so far behind right now, it's not ideal. But I've been in worse positions before with Pokemon, Bals, and we've still come back, so we can't give up. Depends what my opponent brings out. It's gonna be the Kyogre, okay.  It's just, I don't think we're gonna be able to get. I don't think we'll get, as they splash off. That's the only thing. So we could wide guard, but does the Kyogre water spell arching pose? Probably, we could spatial rend into the Rayquaza, or we could just trick room. Depends on what my opponent's got really in the back. Do they have pherothone? Probably, or tapacoco? Imagine maybe tapacoco. Our bankon have been tapacoco, although you're kind of more likely to lead with it, but I'm gonna trick room with Parkia, and I'm gonna wide guard. I'm gonna hope that the Kyogre's faster than Parkia as well. There's the Earth Par, okay. Wow, the Crit Earth Par. I don't think we take that combination anyway, but the Crit just kind of rubs salt into the wound a little bit. Right, okay. So spatial rend, I mean, the Kyogre's not really putting on too much pressure at all, so it's gonna be a solvest Rayquaza, though. It hasn't revealed protect. That the Ray is such a hard matchup, I think for like this team generally. Like when I first did the wheel, the very start of the week, I think like highlighting that Rayquaza was gonna be like one of those Pokemon that's really difficult to deal with, just because of its versatility. Is that a Crit? No, okay, Ice Beam, and yeah, Dragon Ascent, minus one. Be nice if this proc'd our berry. Not even a chance. Minus one, we would have taken that 100%, put ourselves back into this game, but the RNG said no, good game, Jack, and that ends it up with this team this week, so I feel like it's not been the best in results wise, six losses, three wins. It could have been better, it could have been way better, but it could have been a lot worse as well. At least we had three wins. I mean, we're taking a real silver lining from this, but I think it's been good for one, just a learning experience about how the roulette's gonna work and what we're gonna do going into next week. So with a brand new team, I feel like it's gonna get a lot better performance wise anyway. I think one of the things that we need to really do is drill down and look for these real gimmicks. If we're using Pokemon like we're using, it's having the gimmicks and making sure that we've got stuff that worked. I think it was a bit harder this week because we didn't really have a lot of cohesiveness between the Pokemon themselves. I think we obviously got the Z Splash Togalaya working really nicely and a few times the man team did so much work as well, but I think for the kind of memes that we're looking for wasn't really that redirection. There wasn't really that after you use it. There wasn't really the gravity user that we needed to take advantage of something else. So hopefully when we go forward with this, we'll have selections where we have picks and things really kind of like just jump out at us and we can make these memes work and have a little bit more success with it. So this has been a really fun week. I hope you guys have enjoyed the episodes this week. I'm looking forward to starting with the brand new team next week and get your suggestions in right now. Remember it is over on the community post. They're on the channel. Just click on the community icon. It will take you over to the post and nominate Pokemon for that roulette wheel next week and we'll be getting under where the brand new team's starting on Monday. So thank you so much for tuning in guys. Have a great weekend. What are you up to? We'll be back with more Rotorank roulette next week. So until then, take care and bye bye.


Hello ladies and gentlemen and thank you for flying with us on Fort Worth TV Flight 190. The captain has informed me that we are about to arrive at our final destination. The Hops and Props Warbird and Craft Beer Festival at the Fort Worth Aviation Museum. If you look out the window on our approach you can see helicopter rides, live music, a free museum, open house, craft beer tasting and many more family friendly events. No please fasten your seat belts and return all trays to the upright positions as we learn more about this unique Fort Worth Festival. Hops and Props is an aircraft flying and a craft beer festival. We've decided to combine an event that has the best of a lot of things that people like. Most of our visitors come out to see our airplanes so they'll have an opportunity to see our airplanes. We're going to have an open house for free all day long and then we also have some special airplanes that are going to be flying in for the day so people get a chance to see a bunch of other airplanes and then in the afternoon we've got ten of local brewers from the area here that are going to come in and they're going to show off their craft beers. In addition they will have aircraft flights that have beer flights. Beer flights and breweries are the little testers. You get testers of four or five different types of their beer and they refer to that as a beer flight. So we're going to be offering beer flights on vintage airplanes. The flagship Detroit is going to be here and also off of the Southern Cross which is a World War II vintage transport airplane and so we're going to be selling beer flights on those and Epic Helicopter is going to be here giving helicopter rides for anybody who would like to buy a helicopter ride. We've got family friendly vendors here, they're going to have about thirty of them or so, bounce houses for the kids, Buffalo West is going to be here and they're going to be serving food all day. It's a fundraiser for Fort Worth Aviation Museum so they can continue to do educational programs with local schools and to preserve the history of aviation here in North Texas. The history of aviation here in North Texas goes back a hundred and five years and starting in nineteen forty when we started building airplanes here aviation really changed the whole economy culture and the lives of tens of thousands now hundreds of thousands millions of people in the area. For instance from nineteen fifty to nineteen seventy when the B fifty eight hustler was being built here the population of the city increased almost thirty percent and that was pretty much all attributable to the B fifty eight that was the first supersonic bomber ever built in the world and so Fort Worth was the center of aviation in the entire world at that point in time. There are very few B fifty eight still around but there is one available right now in Illinois. It's going to be very expensive to move it here from Illinois so we are working very hard to find supporters to help us bring that airplane here so that we can use it as a tribute to not only the people who designed it built it fluid and maintained it but also as a point of inspiration for young people here and for the city to take some pride in the fact of the things that the city has accomplished here. To learn more about Fort Worth Aviation Museum and to purchase the hops and props tickets visit their website. Here's a look at other events happening in and around Fort Worth in the next few weeks. For a listing of more events and city meetings you always can check the calendar at the city's website.


So now we will discuss the first Grigori group. I have sketched the definition here on this blackboard, and now we discuss it more in detail. So this is a group acting on a root of 3, and the action of the generator's ABC is defined recursively. That is, A just permits the branches, and the action of BCD is defined recursively in the way presented on these pictures. Maybe I mentioned also some equivalent language. So these are groups acting on root of 3. And initial language of the original Grigori-Chuk paper was using intervaling changes. So you can see the interval from 0 to 1. Let's say 0, not included 1 included. Then A permits 2 halves of this interval. B acts in the following way. So we have 0, 1 half, 3, 4, 7, 8, and so on. We can see the points of the form 1 minus 2 to some integer power. And we say B acts in the following way. On the first interval that we have drawn, B acts like A, meaning permits the halves of the interval. On the second interval from 1 half to 3 fourth, B permits x like A, that is, permits the halves of this interval. Then on the third one from 3 fourth to 7 eighth, it acts 3 value, identity on this one. And then again, with period 3, we have again like A from 7 eighth until 15, 16th, like A, like A, and so on. So we have a sequence. In this notation, one can say a B acts like A, A, identity, A, A, identity, and so on with period 3. Now C acts like A, meaning my drug, you can picture, so. On the first interval from 0 to 1 half, it acts like A permitting the halves from 1 half to 3 fourth, it acts 3 value. Then from 3 fourth to 7 eighth, again permitting the halves of the intervals and so on. So C acts like A, identity A, identity A, and so on. Again, with period 3. And D, the action of D is a negligence, or only we start with identity, that is meaning from 0 to 1 half, A, D acts like identity. And then from 1 half to 3 fourth, permitting the halves of the interval, from 3 fourth to 7 eighth permitting the halves of the intervals, and so on. Identity A, A, and with period 3, identity A, A, and so on. This is just the same definition I explain drawing trees. The halves of intervals correspond to the branches of the tree. And though what is nice with BCDs, it's convenient to use both notation, so it's easy on these pictures to draw explicitly the action of BCDs. What was good that BCDs didn't permute the first level, and the only interesting action is occurring along this ray. So something really interesting in the language of the interval occurs only close to 1, and that's made possible to make these drawings. In general, for some more complicated groups, usually it's more convenient to use nice and relevant language of trees. But I wanted to mention the languages of intervals just to mention that some orientation not related to what we discussed today, just pointing out these groups defined as groups exchanging subintervals, but it is important in this definition there is an infinite number of intervals involved in the definition. But just to mention that even if we consider very different or priori class of groups of interval exchanges, when we allow only finite number of intervals, many group theoretical questions seem to be open. So here there are people who know, who are really specialists in this subject, who can say much more than I can do. But apparently, if you can see, for example, a group with finite number of intervals, real group of interval exchanges, it's not known whether the growth is always exponential or at least, you can confirm, right? Exponential or polynomial, but this group somehow fits in a more general context where infinite number of intervals is considered. So this is our group. So by definition, it has four generated by four elements, a, b, c, d. But actually the first observation that three of them suffice. So the generating system will be considered, and which is really an important generating system, a, b, c, d. But in principle, this group is generated by three elements, not only by four. Just because there are some straightforward relations in this group, I firmly that there is a remark. First of all, a square is identity, and the same for b square is identity, c square is identity, d square is identity. And d is equal to the product of b and c. And actually, so actually, b, c, d generate a finite abelian group for elements. So c is equal to product of b and d. And b is product of c and d. How do we see that? First of all, a square is just in volution. So we see from the definition a square is identity, right? Now what about b? On each interval, b acts either like a or like identity. So a square is identity, and identity square is obviously identity. So just on each interval, we see that the restriction of b square to these intervals, we consider that the definition is identity. So b square is identity. And the same for b, c, d. Just on each interval, we had only a, or identity. The square is identity. Now, y, d is a product of b and c. What happens if you take the product of b and c? So on the first interval, so let's write the action of b and c, right? On the first interval from 0 to 1 half, b acted like a and c acted as a as well. So we have a square, meaning we have identity on this first interval, right? On the second interval from 1 half to 3 fourth, a acted like a, I mean b acted like a, and c acted like identity, like identity. So we had a times e, we had a. On the third one from 3 fourth to 7 eighth, a acted trivially, and c acted non-trivially as a. So we have identity times a. We see, and then we know with period three we have the same action, right? So on the first interval in our definition, we have identity. On the second, we have a. On the third, we have a, and so on. So we see the product of b and c is the d, and analogously, we have these relations, which will be important for us. And so we can reformulate this statement saying that if we have an element, take some element in our group. This group is called first Grigorych group. This is the first Grigorych group. Take some element in our first Grigorych group. Suppose it has some length. So with respect to s of g is equal to some n. Then there is a nice standard form to write, to express g in terms of a, b, c, d. We can say then, then g can be written in the following way. We write star for b, c, d's. The first star could be present or not. Then we have a, when we have some among b, c, d's. Then we have a, and so on, the last a, and maybe at the end some star present or not. So each star denotes b, c, d's, but not necessarily the same one. So why do we have that? Just since a squared is equal to identity, no need to write a squared, so shortest word for g cannot have a squared. And since the product of two generators among b, c, d's is the third one, so there is no sense to write a product of b, c, or c, d. So if we have a shortest, actually not only that there exists an expression like this, actually any shortest word for g will have this form in the generating set s. And this will be important for all what we will prove about this group. And now we want to show that the growth of this group is sub-exponential. So this was a crucial discovery of Grigorychuk. We want to show some main lemma, so it's a lemma, but it's actually a theorem rather than a lemma. We will show that there exists alpha smaller than 1, such that the growth function of the first Grigorychuk group is smaller than x, some constant, and to the alpha. We'll write a constant if you don't specify which alpha we take, we can forget. To prove this, we will use a certain self-similarity property of this group. This group is self-similar, and actually it's self-similar in several possible meanings. So for us, we will need the following. Actually, what we really will show, which is stronger than the previous claim, we will show the following. That there exists h, a finite index subgroup in G, and a map phi, there exists phi, from G to the direct product of several copies of G, such that the following holds. So s is the standard generative set. s will be as always a, b, c, d. From h, we will need an application from this finite index subgroup, h to Gk. Such that the following holds, we can see the, so we can see that for any h in h, we can see that its length, ls of h. And we can see that the length of, so in Gk, we can see that the length function in Gk will be just the sum of ls of h1 plus ls of g2 plus ls of Gk. And so we can see that the length of the image of h, meaning, so we can see the l of, let's say, h1 plus l of h. We want that this is much smaller than the length of the element we started with. We want to show that this is smaller than some constant gamma. So we want to say that there exists gamma smaller than 1, such that for any h, this is smaller gamma times lh of h plus some constant. Just for, yes, these are where phi of h is equal to h1 hk. So this is some very strong actually contracting property. For example, even if you would require that gamma is strictly smaller than 1, it's already a quite strong property. But for sub-exponential growth, we needed that it's strictly smaller than 1. Just before we explain why this, we have this for the first Grigatu group, maybe I make some general comments on important groups. The existence of such map is a kind of multidimensional analog of a dilation in important groups. If you have an important group and if you have some map of an important group, let's say n, important or virtually important group. A priori, we can start with any group, but only virtually important groups would admit the dilation that we define right now. So we can see the map, we can see the endomorphous psi from n to n. And it's called to be dilation. If for any n, the length is multiplied by some constant greater than 1 such that for any n in n, the length, we fix some generating set. So we can see the sum group n and s is generating set. And so ls, we want that ls of any element is greater than 1 plus something, some constant, times of the image psi of n, of psi of n, is greater or equal 1 plus c, the length of our n. Some very easy example. If n is equal to zd, if n is equal to zd, if you can see the psi, just multiplication by some number, for example, by 2, if you can see the psi, that sends z1, zd to, for example, z1 to zd. It's clearly a group of homomorphism, right? And yes, I didn't finish the definition. And so I defined one property here. So this property to multiply by some constant greater than 1 is clearly satisfied for this endomorphism of zd. And I forgot to say, we also ask that the image psi of n is a finite index of nn. So here, for example, it's clear for zd, we have this map that multiplies all coordinates with some number. So clearly, the length is multiplied by something. And it's clear the index is finite. So the index here is 2 to the power of d, for example. Not all important groups admit dilation, but many do admit. So another example, if you consider, for example, a group like this, of upper triangular matrices. So in one of the exercises, you had the group of upper triangular matrices, right? So with one on diagonal and some coefficients, so some a, b, c, n, z. And if you multiply now, if we send it to something, we multiply by some large number. Actually, 2 is already not bad. But if we multiply by, let's say, if we send it p, a, p, b, we can't multiply c by p. It wouldn't be a group homomorphism. But if you multiply it by p squared, then this is very good. This is a group homomorphism. From our matrix to some finite index subgroup. So here the index will be p times p times p squared. It will be the subgroup in this p squared in our group, right? And one can check that this is a dilation. So I guess it's already dilation with the definition I gave. But it's even easier if we take some iteration of psi. Actually, there are several versions of possible definition of dilation one uses. Sometimes one requires the property I have explained not for psi, but rather for some power of psi. But anyway, so if one understands just how we prove it, so the length of this element is approximately a plus b plus square root of c up to some constant. But having this in mind, so yes, in a quality way. Here it's smaller, so the image is smaller. And here the image is bigger. We wanted the image, yes, here the image, I have not yet explained. Yes, we want it to be, no, the lemma stated is correct. So h is a finite index subgroup that I stated. I explained why we have inequality in different directions for the Grigory group and here. So this map for the first Grigory group that we will discuss now, and that's what we need for the proof of the upper bound, is not an analog of dilation. It's an analog of the inverse map to the dilation. So just for important groups, for a billion groups, in a natural way, usually one looks dilation as a map from g to a finite index subgroup, but one can reformulate. It looks artificial in the case of a billion groups, but becomes not artificial, but really essential for the first Grigory group. If we have a dilation, just reformulation. We can see that just the image of our dilation, right? Let's phi be the inverse to our upside. Phi defined not everywhere, but phi is so. Phi is defined on a finite index, right? Phi is defined on psi of n, and this is a finite, by definition, it's a finite index subgroup, n. Finite index subgroup. It should be the identity map, sending everything to the identity and you would have a map that's satisfied. I just injected. Yes, sorry, I didn't say this is injective map. You will do injective. Thank you. Yes, phi, injective. Actually, that is, it's not so important. In our example for the first Grigory group, it will be a morphism. But the dilemma is true even if you would not assume that. We will construct what we will construct will be a group homomorphism. But just for this general conclusion about the growth, it's important we have just some map. It's enough. So just this analogy, to finish this analogy with a billion on an important case, so if you consider the inverse to the dilation, then the inverse divides the length by some constant, right? So the longer than for this inverse, we have phi of anything, of any n, is smaller than the length of its length, will be smaller than some constant gamma, smaller than 1. We have some constant gamma smaller than 1. And the length is smaller than gamma length of our element. This is our analogy. So now, I have mentioned so many groups admit dilations. For example, groups of upper triangle matrices, no matter the size, m times m's upper triangle matrices admits a dilation. It's not difficult to see. Just looking on homomorphism of this form. For this group, actually, I don't know if it was explained good in the formulation exercise, but there is a natural generating set which consists of you put of elementary matrices. You put one here. This is the first generator that corresponds to our coefficient A. And you put here one. This is another generator. Actually, these two generators are enough. We could consider them a generating set, or we can add also the matrices. It's not important. We could add the matrices like this. But this matrix is a commutator of these two. The statement is true for both of them, because the length is like this. Actually, it doesn't help at all. So you had some indication for this in one of the exercises. So actually, it's not in any way up to constant. Let's say up to constant that it's clearly up to some constant. But to understand that it's really dilation, if you don't know the constant a priori, we don't see it. Why it's really dilation? If it would be really equal like this, it would be straightforward. We have dilation. But that's why it's a bit more convenient, though not really necessary, to take a power of our phi. Then we see no matter whatever. Even if you don't control good a priori, this constant, it doesn't matter. We see its power certainly multiplies by some number greater than 1. So this is our analogy. As I said, for example, all Unitreaga-Marx group do have a dilation. And this is one of the reasons why nilpotent groups have polynomial growth. There are many ways to see why nilpotent groups have polynomial growth. Even those that don't admit a dilation do have polynomial growth. But for our analogy, it's helpful to know just something which is called Frank's lemma, that if you have a group with dilation, a priori, we don't assume nilpotent just any finitely generated group with dilation, admitting a dilation, dilation, then V of n is polynomial is smaller than n to some power. Why? Because let's estimate the growth function. We have some element, G. It's with a finite distance of the image of our dilation. So there is some constant related to the index. So G exists. We know that it's constant C. And for any element G, we can choose some G prime in the image of our dilation, which was phi psi, I think, right? Psi, it was psi, right? For any element, we can find an element which is close to it and which is in the image of our dilation psi. And since it's in the image of the dilation, so it's equal psi of something, psi of some other element H. And we know that the length of H is much shorter than the length of G. The length of H is smaller than some constant gamma times L of G. What does it show us? We see that V of n is smaller than some constant times V of n divided by gamma. Let's write it several times. V of n is smaller than, or not divided by gamma, but smaller than one, multiplied by gamma. We can write this is smaller than this. Then applying it again, this is smaller than C square V of n gamma square, smaller than C cubed V n gamma cubed, and so on. So we want to take log n times. If we log n times, divide n by some, so multiply by some number smaller than one, but divide by some number greater than one, we will obtain some constant. So we can deduce that V of n is smaller than C to the power some constant log n. Is it OK a lot? So if we take x, so it's exponent of constant log n, so yes, it's smaller than n to some power. So this front lemma gives you a very easy explanation why if you have a dilation, the growth is at least polynomial. Now our main lemma formulated above will be this multidimensional analog. We don't have a dilation, because we have a map not to the group itself, but to a product of several copies of the group. So we certainly cannot guarantee it's not true at all for aggregature groups that the growth is polynomial. But there is an analogous, also not very difficult, analytical lemma of the same client that will show if we have this situation with the contraction, then the growth is at most x n to some power, smaller than one. And the main idea behind this construction is really to check why we have this contraction, why we have this map that contracts from a finite index subgroup to a product of several copies of our group. So maybe the main thing is to explain why we have the contraction. What is our map? What is the letter used for aggregature group phi? What is our map phi? And what is our finite index subgroup? And how many copies of the group we'll have, which we did not by K? Actually what is possible is our H. We could consider a stabilizer of the first level of our tree or in the language of the intervals, meaning elements that don't, that all points of the first half sent to the first half and all points of the second half sent to the second half. But it's a little bit easier if you don't care too much about our coefficient alpha, only if you want to claim that it's smaller than one. The easiest way is to go to the third level. So we consider the stabilizer of the third level in our tree or on the interval, meaning we divide by 1, 2 to the power 3. We consider the subintervals of the length 1, 8, 2, 8, so on. And we consider only the elements that don't interchange points of some intervals with the points of another interval. This is clearly a finite index subgroup in our group, right? So what is left? So any element making some permutation on the corresponding level, we can multiply it. And so we get some element in the stabilizer, meaning the index is not larger than the permutation group on 8 elements, right? Actually, it is smaller, but for us, we just need finite index. So the index is not more than, index is not more than cardinality of symmetric group on 8 elements. So this group is good for our setting. And we want to consider a map. So we have this group h will be this stabilizer. We want to consider phi from h to 8 to g to the 8. And what is this map? This map, simply on each branch of the tree, or clearly on each small interval of the size 1 over 8, writes the corresponding element. So h, we can write h1, h8. Just in our definition for our generators, the restrictions to the branches of the first level, that we can repeat and take whatever level we like, right? So restriction of the generators of the group to some level also expressed in terms of the generators if we reinterpret smaller branches as if we really identify them with the whole tree, right? So this is the map we are considering. Since it's true for generators, it's true also for any element in the group. So actually, any element in the group, we can write this way. So we have this third level, third level, I think. And then just we look what we have on the first branch, what we have on the second branch, and so on. We have 8 elements like this. And our main goal is to check that this map is contrasting in the sense explained on the other blackboard so that the sum of all the lengths of h i will be much smaller than the length of our element. Why is it so? Here we use this normal form that is explained here. So any element we write in this form, we take some element h in our subgroup h. We write it as stated. So something, a, then sum of bcds, a, and so on. And we want to see what happens if we consider its restriction to some level. But first of all, before we go into the third level, let's consider just restriction to the first level. So we write suppose that h was in the stabilizer on the third level, in particular the stabilizer of the first level. Meaning we can write h as, it's not the same h, maybe because h was on the, let's write it as g1, g2. I don't use h1, h2 because they, in my notation, corresponded to the third level. So we write h like this, right? So on the three meaning h is like this, g1, g2. And big branches are not interchanged. Oh, in the language of interval, they have g1, g2. This is our h. And we want to show two things. First of all, we want to show that even for distinction on the first level, the length of g1 plus the length of g2 is not greater than the length of h. I don't say yet that we multiply by constant smaller than 1. But it's already helpful to know at least it's not greater than the length of h plus constant. And then we want to say another thing. We look on this normal form here. Our stars are BCDs, right? And we look how many among the stars let n be the number of B's among our stars in our normal form, numbers of B's. And analogously, we consider nc, nd, which is the number of C's and the numbers of D's. Then the number, just because we have this form, the number of A's. I don't make a notation for the number of A's. Because the number of A's, we know that it's nb plus nc plus nd plus minus 1 or plus 0, right? So I say up to constant 1, which is not important for our argument, the number of A's is the same as the total number of BCDs just because they're alternated, right? And now, a second claim that we want to say that if there were many D's among BCDs, if we assume that nd, the proportion of nd among nb plus nc plus among BCDs, BCDs was at least 1 third, then we want to say that then L of G1 plus L of G2 would be greater or equal to 5 over 6 L of H plus constant. So if you would have to work only with words that have important proportion of D's, it wouldn't be necessary for us to go to third level. We would see already for this restriction on the stabilizer of the first level, we have gained this constant smaller than 1. So why so? Just let us see how we can estimate the length of G1 and how we can estimate the length of G2. Again, so we have written our word with A's and BCDs. And to write, we'll find now some words, not necessarily minimal ones, but some words representing G1 and G2. And to do this, we do the following. Just each time we multiply by something, we look what happens on the branches. So before we multiply it by A, here we had some BC or D. We just need to see what B gives us on each branch. So just by definition of the action, we knew that B acted like AC, C acted like AD, D acted like identity and B. We see that B gives a contribution A on one branch and contribution C on the other branch. C similarly gives contribution A on one branch, D on the other branch, and D gives trivial contribution on one branch and contribution B on the other branch. And when I say one branch, it depends on how many multiplications by A we had until some position we are considering right now. If we had an even number of multiplications, for example, if we had at the very beginning of our word BC or D, we really looked like written on the first branch, it's A, on the second branch it's C. But if we had some not even number of A's, for example, if we have some odd number of A's, for example, 1A, if you look on this star in our normal form, then meaning we had that the halves of our interval or the big branches of our tree are interchanged, meaning that if we're controlling the contribution of this element, if it was B on the first branch, it acts like C and on the second like A. But what is important for us is just the total number of contribution our element BCD can give for G1 and G2. So we see the total contribution of each B is 2 because 2 generates A and C is 2, of C is 2, of D is just 1. This shows that the length of G1 plus length of G2 is not greater than 2 times NB plus 2 times NC plus ND. Plus, yes, if it was, we suppose this element in the stabilizer of the first level, that's it. Because we just observed that we don't see contribution of A's at all. A only tells us whether intervals corresponding to the I interchanged or not when we look on their contribution. So it's really BCD who defined these contributions. And A's only shows if the big branches I interchanged at the moment we control it or if they're on the initial places. Then in the length and the length of H, we knew that it's more or less the length of H. It was 2 times NB plus NC plus ND plus minus 1. So we see if the proportion of ND is at least 1 third, we obtained our claim, right? So our second claim on the middle blackboard. So we have proved this claim. Now what if not? First of all, it's not so important, but we won't use it, that the proportion was 1 third. Actually, any positive proportion would give us some contraction, the worst one, but some. But we don't use it. We just observe that at least for one element BC and D's, the proportion at least 1 third, right? So obviously, for some either as we used in this claim, ND divided by NB plus NC plus ND by the total number of BCD's is at least 1 third. Or this happens for maybe for C and or NC divided by NC plus ND is at least 1 third, or maybe the maximum proportion is the number of BCD's is NB plus ND is at least 1 third. If the proportion of D is good, we already proven our statement because on the first level already, we have this contraction with coefficient of 5 over 6. And then we go to second and third, and by the first part of the claim, the total sum doesn't increase. That is up to a constant only. And so if there are many D's in our normal form passing to the first level, we have the contraction and we can pass to the third level as we claim, then the contraction is not worse than on the first level. Now suppose this doesn't happen. Consider the case, then there is a big number of C's. Use this rewriting procedure we have used. So instead of our H, we have this element G1 and G2. Then each C becomes ND, right? So for total contribution of C to G1 and G2, all C's become D. So we keep this G1 and G2 as we have obtained it. It's important we don't rewrite it as a shortest word. We just use this rewriting system and look on this G1 and G2. And we see that we can apply the second part of the claim just for these elements, right? So if the proportion of C's was large, then we can think then passing to the first level the proportion of D's now is large and then passing to one more level, we have a contraction. And similarly, so actually if at least one of the first cases holds, we wouldn't need the third level. It would be enough to go to the second level. But sometimes we need the third level because it may happen so that the proportion of B is large but maybe of C's is not large. Then we have to go to the third level. Then again, for the first passing, B's become C's. For there's one more passing, C's becomes D's. And we can always claim that the total sum, so finally considering in all these cases, right? In any case, so in any of these three cases, in any case, we obtain that the length of H1 plus the length of H2 plus the length of H8 is not greater than 5 over 6 L of H plus constant 3 or something. Maybe if constant is not important. Now this is the main observation behind the observation of Grigorychuk that the growth is sub-exponential. So the only thing which is left to prove, I have four or five minutes, right? So the only thing which is left to show that is this lemma, did I give the number? I did not give the number. The second lemma for the proof of sub-exponential bound for the growth function of the first Grigorychuk group, it's enough to prove the lemma written on the highest blackboard saying that if we have this contraction then the growth is at most xn to some alpha. And this lemma is really analogous to the Frank's lemma and we can formulate it just a simple lemma about function. So consider an increasing function. So let's f of n be increasing function, positive edit increasing function, such that the following holds suppose that for any sufficient large n, f of n is smaller than the sum, some constant, and then the sum of vn1 times vn2 times vnk. Yes, the sum of the products like this. Where the sum is taken over n1 nk, such that n1 plus n2 plus nk is smaller than gamma times n. Gamma times n. If we have a function satisfying this inequality, then we say there exists alpha, which depends in principle on our coefficient, contraction coefficient here, gamma. It's f, sorry, it's f. We will apply later for the growth function v, but I wanted to formulate just an abstract lemma about functions. Nothing to do with group theory. It's just a fundamental lemma about functions analogous to Frank's lemma. So suppose we have this. Then we say there exists some alpha smaller than 1, fn2 times fnk. So we have some k, so we have some k. A product. No, it's a product, just product. So we have some function, right? Defined on integer numbers. And we take a product, fn1, fn2, and so on, where we take all possible products. We take all possible n1 and so on, nk, with the total sum smaller than gamma times n. And gamma was a constant, so gamma was some constant smaller than 1. And k was some integer number. We are interested, it's true even for k equal to 1, but the claim is not interesting for k equal to 1, because much better claim of Frank's lemma. We have already proven it's valid. We are interested in k, at least, or 2. We say that there exists alpha, depending on gamma and k, such that f of n is smaller than x, some constant n to the alpha. And this would be enough to finish our proof. Because we do the same thing as we have done in the Frank's lemma, so far we have this analytic lemma. If we have proven lemma free, just we bound the growth. So we have something we use again. So for the first Greek integer group, we have our subgroup h, which is a finite index. So any element is at some finite distance from some element of h. So it will give some possible sum constant. It's enough to control the intersection of the ball of some radius. As we discussed with our subgroup h. And in the subgroup h, just we see how many choices we have for element of h. If the length of h is some l, then h, the image, this phi of h, is h1 hk. And the total sum, just by definition of contraction, l of h1 plus l of hk is smaller than gamma times l. It's not important here. We can also allow this constant. We can start with some constant and write the sum is greater. Like this. So just considering this intersection of the balls with our subgroup h and applying this analytical lemma, we did use the growth. Indeed, it will be xn to the alpha. So the only thing which is left to prove is this analytical lemma. And again, so it's done directly like front lemma here. So we just use the fact that proof of lemma 3, we want to prove that log f of n is smaller than some constant a, let's say, n to some alpha, with some alpha smaller than 1. We assume that this holds for any n smaller equal to some, strictly smaller, let's say, than some m. And we prove this claim for, we will show that if we choose a large enough and if we choose alpha close enough to 1, that we will be able to prove the claim of lemma 3 just by induction on our n. So we assume that for smaller numbers, we have this conclusion. Now, we look on what we know about f of n. For f of n, we have this inequality that we will use. Observe that if n is large enough, each n i is smaller than, because the total sum is smaller than gamma n plus c. So in particular, each n i is smaller than gamma n plus c. And this is strictly smaller than n if n is large enough. I mean, if we wouldn't have a constant here, it's true for all n. But if you want to formulate here with a constant here for any n large enough, we can see that the elements that figure in this inequality are smaller. And we can apply induction claim. So we can see each term here, so each term. We have some sum, but each term satisfies the following. So we take log, let's say log of each term. We take log, we have log of n1 plus log of n2 plus log of nk. And we just use the fact that n to the alpha is a concave function, right? So just here we use induction hypothesis. So it's a smaller by induction claim by n1 to the alpha plus nk to the alpha. And we will see that this is smaller that, well, it's certainly smaller than n to the alpha. But it's not enough for induction because we don't have one term. We have many terms. So we want to claim a little bit better. But still, using the same concavity, we can claim that this is smaller than 1 minus epsilon, the same constant a, and n to the alpha. Time to give it a state, but this is really elementary. You have something like gamma n divided the k to the alpha times k. And it is sufficient to choose alpha such that gamma n divided by k to the alpha times k is smaller than 1. It will give us some epsilon, 3k smaller than 1. Why we can do it? But by our assumption, gamma is strictly smaller than, sorry, this is gamma n, yes, I want gamma k. Yes, we want to choose some alpha such that gamma divided by k to the alpha times k is strictly smaller than 1 to have this conclusion. But we know if alpha would take alpha equal to 1, we would get just alpha divided by k times k gamma, which is strictly smaller than 1. And since it's strictly smaller one, if we can take some alpha a little bit smaller, and we still have a stricter in the quality, so we can always choose appropriate alpha. And so we have this claim. And now just we count how many terms we have here. The number of terms is at most, so each term is at most gamma times n. And the number of terms is at most the number of terms is not greater than gamma n times k. So this is just some polynomial. And here for each term, we have exponential So since we have an extra epsilon here, this epsilon will kill the polynomial term. So even taking the sums of all terms, we can conclude that the value of n is smaller than xn to the alpha. And this finishes the proof. And I have to stop here.


I did it. I did it. I did it. I did I did I did I did I did I did it. Yeah, yeah Ducky busting the beat dub me on the flow about the yo ho chillin with the bro playing Valorant Smokin weed bro kind of hot though pushing mid bro. It's over dead.  You're juicy.  You should Was he just there the whole time? I guess so Roseman's the imposter with the Someone else friend. Oh Ash was here. There's one here on the close up. There's two gen two gen two gen. I'm gonna check him in We're in the gen. He's not seeing anyone. They made I'm gonna Let's go boys I got the other one as well Me remaining at their knees


Thank you for coming. We are here today to announce and celebrate Data Privacy Day, an international effort to raise awareness and promote privacy and data protection best practices. Also I am joined by legislative leaders from the House Commerce Committee to spotlight H-121, the Comprehensive Data Privacy Bill. After my remarks, we will hear from the Chair of that committee, Representative Michael Marcotte, and the Vice-Chair Representative Stephanie Jerome. And then we will turn it over to questions. I greatly appreciate this committee's interest and promotion of critical data privacy protections for Vermonters, a passion I share. Privacy is a Vermont value. It is embodied in our very motto freedom and unity. Part of being free means being free from unwanted intrusion. We have a long history and tradition of protecting us from unwanted intrusion, laws to protect decisions about our own bodies, trespass, peeping toms, unwarranted search and seizures, and so on. But today, 24-7 corporate surveillance can happen. Companies can know where you go, what you eat, how you're feeling. Technology is changing rapidly and our laws need to change with them. In 2018, the legislature here in Vermont was the very first to pass a law regulating data brokers. I am very proud of the work that this legislature did along with my office in getting up to speed and passing that law. And other states followed suit. But since that time, 13 states have gone several steps beyond to pass their own comprehensive data privacy bills. And we're playing catch-up at this point. So it's wonderful that we have had the space and time after the pandemic and the worker shortage and the flood. I'm so grateful that this committee is taking the time to focus on data privacy. Artificial intelligence, use of biometric information and the increasing use purchase and sale of consumer data have the potential to compromise consumer privacy for these reasons. I strongly support the work of the legislature that they're doing on the bill H-121. And that bill has highlights that I care deeply about. One is ensuring data minimization to limit the consumer information businesses collect and store.  Protections of biometric data, that is data about us that we cannot change, like our faces, our fingerprints or DNA, that require express consumer consent to collect or use. Our personal data is being collected, bought, stored and sold every second.  Vermonters have a right to control their own data and to take action when their data privacy has been violated. I want to list a few tips that I think can be useful when you are trying to protect your own data and then I'll turn things over to Chair Marca. One, set your social media settings to private and review data sharing features of the social media sites you use. Be aware that when you don't have to pay for an app, you're paying for it with your data. Employ your own data minimization strategy. Do not give out your personally identifiable information if you don't have to. Don't share photos of yourself or your children if you don't have to. Delete your online history. If you don't want to see ads based on your previous online activity, delete cookies and clear your browsing and search history. On your phone, delete or reset identifiers used to track you. You can disable Siri on your phone and resist the urge to install a smart speaker in your home. Deactivate or permanently delete any social media or online accounts that you no longer use. You can change your password frequently. Use dual factor authentication for all online accounts. Just a few tips. We probably have lots more. As always, I like to remind everyone of the Consumer Assistance Program who can help you through consumer issues including identity theft and data privacy and they can be reached at 1-800-649-2424. And now I will turn things over to Representative Marquardt who's chair of the House Commerce Committee. Thank you Attorney General Clark. We've worked together a long time on consumer protection, which this is a consumer protection, data privacy. I think we've started, I started working in this realm as vice chair of the committee under and working with Chair Bill Botzo. And we continue to continue to focus on the data privacy throughout my time as chair. And I really appreciate the work that the working relationship that we have with the Attorney General's office. I appreciate all the work that my committee is doing on this subject as well. We know that this subject is really something that the federal government should be focusing on. And unfortunately, they have not been able to get together and figure this out. So it's been left up to the states to deal with this. Right now there's 15 states that have data privacy legislation passed, but it's not all the same. And what we've strived to do here is to work with other states to better understand what they're doing. We've had conversations with Senator Moroney in Connecticut, also with Representative Farley-Bouvier in Massachusetts, where we're trying to make sure that what we do isn't confusing to our businesses or confusing to our consumers. And so as we continue to work on this, I think that we have the ability now to look at those 15 states that have passed legislation and how can we better align together on how these laws work so that we're not confusing consumers or confusing business world. We want to make sure that they have the ability and the understanding on how to protect consumers' privacy, especially our Vermont consumers, making sure that our Vermont consumers have the ability to opt out from data collection whenever they choose to. So we're looking forward to the draft of our bill, which should be out next week, and starting that conversation with everyone that has an interest in this. And my understanding is there's a long list of people that want to chat with us about this. So again, we look forward to that. I do want to especially call out my vice chair, Representative Jerome and Representative Priestley. They have taken, I've asked them to take this on, and they have been the ones that have been collecting a lot of the information and putting this bill together. So I certainly appreciate them and I appreciate my committee for their willingness to take this up as well, because this is a really difficult subject to walk through and to come up with the right parameters around how we deal with this. But I know they can do it. I know that our colleagues in the Senate will be able to also work through this, and we look forward to finished product. Thank you. Thank you. And now we'll hear from the vice chair, Stephanie Jerome. Thank you. Thank you, Chair Marca. Thank you, Charity. I appreciate being here today. So we'll be continuing our discussions on H-121 next week. And reflecting on data privacy day, I was thinking about some of the conversations we have with people when talking about data privacy, and we say things like, well, who really owns my face and who owns my fingertips? And these sort of questions go on when we start diving into data privacy. I truly believe that data privacy cuts across all party lines. It is not a political issue. Protecting our personal information, our face, our fingerprints, our gate, our social security number, our address, our shopping habits, our travel, our biometric and our personal data. This all belongs to us. And is it right that businesses can sell information that belongs to me or how much of this data can be sold and used? What kind of permissions do I give? How can I opt out of providing information who benefits from my information, and what are the safeguards? As we stated, I think it's 15 states have already passed data privacy bills to protect consumers, to protect our information, and probably most importantly, to protect our children. I believe there are about 10 other states that are currently working on data privacy bills. It's a nationally important issue, and as Chair Marcot said, it really should be a law that has passed federally but has been left to the state's responsibilities. The House Commerce Committee is going to be building a data privacy bill that is based on work that has been done across the United States and has done, and we're looking at states such as Connecticut and California and others who have done in-depth research and have really solid bills to create a bill that works for Vermonters and works for Vermont businesses and one that works well with all the other states in our region. And I'm confident that we can get it done right. I'd like to turn it over now to Representative Monique Priestley, who I am indebted to for all the work that she has done on this topic, and is an expert in the field as we currently talk about it. And I'd like to turn it over to Monique, please. I'm thankful for all that Attorney General Clark, Chair Marcot and Vice Chair Jerome have done to protect Vermont consumers. Where federal action lags, we alongside visionary legislators across the United States are collaborating to bridge critical gaps in data privacy, crafting laws that not only pave the way for future tech policy, but also stand as beacons of safety and security in our citizens' digital lives. Thank you so much. I think all of the committee is here, almost all the committee is here. So it says a lot about the commitment that they have to this really important issue. And data privacy day is officially on Sunday, which seems like a sleepy day for a day, but there it is. So we're celebrating today. But happy to answer any questions you have. And I'm sure the committee can answer questions you have about the bill if you have questions about that. Or we can go get lunch. Alright, well, thanks so much for coming.


What is up ladies and gents hydro here with another best list talents and runes video this time we're looking at mage first of all we're going to look at the best gear we're going to go for it really quickly keep in mind that the mindset again here is we're going to look for maximum stamina and intellect so we're not going to be looking to get too much spell power as we're going to sacrifice way too many stats for that so we're going to start out with the lucky fishing hat next we've got spectral necklace of the eagle sage's mantle of the eagle cutthroat's cape of the eagle and then the black velvet robes you also have the option of the clam weave tunic here then we've got durable braces of the eagle we've got laurel keeper's staff there are again a few staves you can go for they're all pretty good for mage rainclaw emits of the eagle gamura's cinch watches leggings of the eagle extra plane of spider silk boots seal of rin planes ring or just an eagle ring arena grandmaster insignia of the alliance and then dancing flame now keep in mind all these eagle items may be subject to change depending on what bfd loop becomes available


We are already halfway through season three of The Mandalorian.  There's only eight episodes. We just got done watching the fourth. I'm gonna talk about it. Let's begin. If this is your first time here, I've been doing these recap videos every week, so make sure to subscribe so you don't miss a video. I also do a lot of other non-mando related content. It's called Adam Does Movies. So naturally the bulk of the material is movie based. Anyway, this is gonna be filled with spoilers. I am gonna be breaking down everything that happened. I'm not some hardcore Star Wars guy, so names allude me, locations I might trip up on. Basically, I'm just breaking down this show. It's entertainment value, the purpose of it at this point, and how I feel about it. To me, the first two seasons of Mandalorian have been fantastic. It's taken a major nosedive since season three. Really, since season two and a half, which was the book of Boba Fett, complete disaster. And since then they have not fully recovered. Here we are on episode four. Are things better? Well, there's certainly more entertaining than the last one, which was a side story about Moth Gideon's crew and what they're up to. How that ties into the overall narrative is anyone's guess, including the writers of this show. Because there's no consistency. Initially I thought, okay, they're gonna focus on Din Djarin becoming a Mandalorian again.  Three episodes in, he's already a fully-fledged member. And so is Bo-Katan, and she didn't even wanna be one.  Why does he wanna be part of this group? It's embarrassing at this point, we'll break that down. I just have to start out with the fact that, okay, the rules say that if he bathes, if an unwashed, un-peer, unclean Mando from the past has lost his way, but he then cleans himself off by bathing in the living waters on Planet Mando, or whatever the hell it's called, then he can be brought back into the fold, or she. In this case, it's both. Din Djarin trips over his own stupidity, falls into the living waters, and has to be saved by Bo-Katan, who is essentially tails to Sonic the Hedgehog at this point. He's constantly in harm's way.  And man, did the tribe accept her quickly. She's already leading them in her fur.   This episode opens with these bounty hunters on the beach, that same beach where there was a giant crocodile creature that almost killed half of them. Yeah, they're out there again training, like nothing happened the day before. Anyway, this scene is embarrassing. It looks like a bunch of bros doing crossfit in the gym. There's no semblance of structure. A couple guys are like blocking. Others are just randomly shooting into the water. It's bedlam. It's chaos. It's pandemonium.  So they play a game of darts. Each contestant has a little shooter on the side of their wrist where they point and hit him with like a paintball gun, basically. Three shots, you're done. And it would seem based on the rules, you take a shot, you reset, you do it again, kind of like karate. You get a good kick in, reset to the next one. Except for what happens here is the kid hits Grogu two times, like boom, point one, boom, point two. And then instead of the kid finishing him, Grogu does a double flip, somersault kiss, twist, and then boom, boom, boom. Hits him with three of them at once. I assumed he could only hit him one time and then get that point and then we would go for the next two. But no, he just straight up circumvents the rules and he wins because of showmanship, I guess. Kind of bizarre. Before any real celebration can ensue, a new challenger appears. A dragon flies down, swoops in, kidnaps a child and then just takes off on the horizon.  I don't know his name, we're gonna call him Gatlin Joe for the remainder of this episode. He's like, we can never catch this guy. It's always getting away. Their jet packs aren't strong enough or something. Even though in the previous episode we said didn't jar and free fall from space and use this jet pack as like a little, just a little booster stop. He's going hundreds of miles an hour and just a little boost from the jets keeps him from splatting on the ground. There's no consistency with these packs. Bo Katan, who is now in charge of all the Mandalorians for some reason, assembles a crew and she also knows this planet better than they do. Nothing makes sense, I don't understand. As that's happening, little Grogu or Grogu, I'm gonna switch it up depending on how I feel in the moment.  We've seen her make like 80,000 pieces of Mandalorian armor now. That seems to be the only thing she does. As she's making this, sparks fly and sparks also erupt inside of our little Grogu mind. We're treated to a prequel flashback. Gotta love these. But this time, baby Grogu's getting saved not just by some basic bitches. He's got a Jedi with him, this cool new Jedi who looks somewhat familiar and has a voice that's unmistakable to the ear. Probably because it's Jar Jar Binks, now in human form. That's right, Disney's putting their Ahmed best foot forward giving this guy a little bit of redemption. He gets to play a cool Jedi.  And before this flashback really goes anywhere or builds to anything, we're back in present time hunting the dragon again.  We have a group of Mandalorians at night, setting up shop, they have their fire going, they're cooking some dinner and that's when Boca-Tan goes. So how do we eat if we can't take our helmets off?  It's like, oh yeah, we all just kind of sneak away into different corners where we can't be seen by each other.   And even if I was to take this from the mind of like an assassin or a hunter, I don't think it's wise to at night disband your entire crew into different corners of the earth just so that you can have a meal in peace and not break your code. The next morning after all of them presumably had a great night's sleep in their armor with their masks on unless they also slept in different areas where they couldn't see each other. Hey, hey, hey, Dan, Dan, I see your feet. If I see that helmet off, you're done. You're done, so buddy. And you just got back into the crew. So if you get kicked out again, then what are you gonna do? Take your super fast spaceship back to the planet, go back into the living water again. Good luck, that took you one episode to do last time. After a rocky sequence of wall climbing, we are then treated to an actual good section of the show where they are pursuing this dragon in the air. It feels like you're up there with them. It's pretty intense. I kind of dug this. It only lasts for a few minutes, but it's solid. It ends like all things do now with the dragon falling into the water and then another Jurassic world-esque monster comes out of nowhere and eats it. They're just down there waiting. They're always waiting. I can't wait for another dragon to fall into my mouth. That's how they sound. That's how they talk. This whole section led to so many questions. One, why were there other helmets up there? I think that's obvious. Other kids have been taken and killed, so great job, guys. Two, why was this kid even still alive? The dragon kept him overnight. Wanted to wait for breakfast to feed his younglings. Whatever. Three, why are the Mandalorians on this planet? The waters are filled with massive monster creatures that can wipe out half of them without even batting an eye. Last time they probably would have died, had dinjara nut conveniently been flying in the sky that same day. There's also dragons here that are kidnapping the kids and eating them. So you have threats in the water. You have threats in the sky. You probably have threats on land, too.  It's just insane. It's asinine. After the beast has slayed, the orphaned baby dragons are now gonna come back with the Mandalorians. They may have said what they're gonna do with them. They were in shackles. I would assume they're gonna eat them, but I don't know if it was clearly ever stated. Maybe they'll just be pets.  I don't know. I wasn't disrespecting how to train your dragon. That trilogy's fantastic. We should be watching that instead of the Mandalorian. That's really what should be happening right now. Before we finished watching the episode, there was a nice forceful intermission thanks to the Disney Plus app freezing.  It's always a fun time. It's consistent. I'll say that. It is consistent.  She has a new armor piece, replaced the shoulder with the one that she lost during the dragon flight. Now she has a mythosaur symbol on there because yeah, of course, fully invested, drank the Kool-Aid, she's all in. This is really unfortunate because Bo-Katan was the only character I actually thought was cool anymore because she didn't have the helmet on. She was a badass.  She's a Mandalorian in her own right. And I genuinely have no idea where this show is going. My investment is a zero at this point. I don't care about anyone. Even little cute Grogu who I used to go, oh, every time you would show up, I don't care at all. Now he really is nothing more than a little prop that sells merchandise. I mean, I'm so cynical on this whole thing because they lost me. They lost the plot. It has no purpose anymore at all. And I don't know how it's gonna tie in with the Andorish kind of garbage they did in the last episode. I don't care. I really don't. We're gonna finish because I'm dedicated to bringing these to you. Maybe it'll somehow turn the corner and all these pieces will fit together. I just don't see it. I don't have the faith that I once had in this production. Let me know your thoughts.  What a wonderful experience this was. I can't wait for more misadventures with this Mandalorian crew. Or are you like me and you appreciate it when it was more mysterious? And it was just Din Djarin and Little Grogu going on their last of us style adventure to get the thing to safety. It was simpler. It was more of a Wild West show. And now it really just feels like another Disney production where they're just throwing anything at the wall. Like the video if you had a good time. Again, please subscribe if you haven't. I post these every week. I post a lot of movie reviews in between. So I'd love to have you stick around and try to keep this channel growing. All right, thank you very much and hopefully you'll see you next time. Boom, boom, boom, boom, boom, boom. F-F-F-F-F-F-F-F-F-F-F. Boom, boom, boom, boom, boom, boom, boom, boom. Scrum-dum-dum-dum, dum-dum. This show is a cash grab. Boom, boom, boom, boom, boom, boom, boom. Thanks for sticking around since you're still here. I should also point out I have a Patreon. Patreon.com slash adam does movies with that membership which starts at $1 a month. You're showing the channel support. One man operation could use the help. YouTube payout is embarrassingly low. You could also become a member here on YouTube via the YouTube join button. It works the same way. Different tiers. If you jump up to myth roll, which is 30 bucks, you get a movie review on the house. You say, Adam, I want you to watch this movie and review it and I will and give you a shout out. It's pretty awesome. Those are options.