Today we find out if new gold is better than old gold. What's up guys we're back once again to answer the age-old question about gold series. Is the older gold premium gold and today we have King of Bling as well as Return of the Bling. Is the old gold better than the new maximum gold? We have done a video like this before but this time I've added King of Bling so we have three different gold series I want you to let me know in the comments what your favorite gold series is we do have a quick giveaway. I'll be giving away these three iconic magic cards all you have to do is like the video be subscribed turn on notifications of course let me know your favorite gold series from here and if it's not from one of these three let me know the other gold that you like better. All right let's crack into these boxes we're gonna open them one at a time. All right I have removed the maximum gold from their special edition or whatever it's called because it's obnoxiously big it's like twice as big as these other ones and I can barely like it's just very obnoxious to open. Premium gold return of the Bling is what we're gonna start off with. I think these have like three packs inside or something like that let's find out. What is all this? Premium gold return of the Bling thanks brand new cards plus updates and upgrades to dozens of your favorite golden age up to the present day. 22 gold rares provides with our new and improved advanced gold technology. It's like the ultra rare technology I love when they use technology to describe their rarity. More shine than ever upgraded 48 top cards very nice brand new 21 new cards all right so they got a lot of explanation on the back of that thing. All right we'll at least look at like the mini boxes for at least one of them so let's start off with premium gold return of the Bling let's see if we can pull anything cool. Golds oh it's right off the bat yeah it's the secret rare gold we have I can't read these what in the world blackwing tamer obsidian hawk joe very difficult to read these names I'm not a fan of that blackwing chris the crack of dawn he's at it again call of the hunted that's a beautiful card rekindling and we have traptrix a tracks not bad all right let's do a we'll go back and forth a little bit with the maximum gold as well I probably should have just looked at one of each that would have made more sense oh well let's do a maximum gold El Dorado fossil dino patch cephalo not bad urgent schedule also not bad scrap dragon also decent rose striker dragon and the hot red dragon archfiend bane all right let's take a look at the premium gold real quick and then we'll get into it okay premium gold they say the king of bling but it's actually called premium gold so I don't remember exactly what is in premium gold but in here they had include some of the greatest hits from old gold series sets okay pretty cool we have a premium gold via the other that will be available everywhere that you normally buy Yu-Gi-Oh TCG not anymore the new gold secrets take all of that and add secret rare in the texture foil pattern okay these are going to be a little bit insane aren't they all right let's check this out premium gold first edition inside we have how many packs is it three packs again all right we have three packs inside just like the return of the bling so premium gold I assume this isn't was this the first premium gold after all the regular golds gold one through three all right right off the bat we have an honest gold rare that's pretty beautiful we'll take that wing dragon of raw secret gold or I think that's like 20 or 30 bucks that's a great hit right off the bat that's our best hit for so far we got the tri edge levia then we have a morphing jar that's pretty good too and then a mind control wow that was a good pack after that pack I think our premium gold return of the bling or premium gold king of the gold aka just premium gold is definitely in first place of our battle here so let's see what we can you know do here with the uh the maximum gold El Dorado I guess you got a pull access code I don't know that's still like 30 bucks uh fire formation 10 key not bad pot of extravagance that'll do it that's a good one I still don't think this is over 10 bucks but still not bad definitely gonna be pretty solid for the value though I really like these gold ulcers I think they're like good I mean I guess they have the weird like lettering but I think the outside looks pretty nice but once they get old they kind of like get brown and that's when they get ugly okay let's go back to a little premium gold there's more maximum gold I think we get like what five five or no four per mini box you have an extra few packs number 14 greedy sarah yam sarah mea okay stardust re-spark we're at it again ooh doom cow that's a pretty sick card I feel like that's like not that expensive though but still pretty nice all right the our our our mateys our mateys our mateys I think this was or mateys maybe won one of the first synchros in duolinks so pretty cool to see that and then oh you can get exodia left leg of the forbidden one this would be pretty cool if we got exodia I would be down for that I'm gonna do like one of the older gold and then one of the newer gold because there's more maximum gold so we'll do it like that we've got alter guys multi faker we've got scrap dragon sigh averse gadget the weather painter snow uh-oh ruckson special between both of them okay we got a ray not bad rays pretty decent hot red dragon archfiend bane and numeron wall not gonna lie I just really enjoy opening the gold series and it seems like you guys like the gold series too because the last two gold series videos one was with was with semo and then the other one was the is old gold better than new gold video you guys I mean that did those videos did really well so I think you guys like this stuff tanky we definitely need to do one already open all gold at some point that'd be pretty sick gimmick puppet Humpty Dumpty number c40 gimmick puppet of dark strings so if you do have oh dark arm dragon that's a if you do have like all the golds maybe hit me up we've got a titanial princess of camellias camellias so many good Edison cards because these are printed like around Edison time not bad we'll take that dad and the princess both pretty good we got the weather thundery canvas phantom knights sideframe learn lambda cursed eldland is it the next card evils or doleka not bad for the dinos a caro and a prank kids fansies all right pretty solid pretty solid let's go back to premium gold turn of the bling I don't remember which one we opened last time knocking alive I know we opened uh eldorado but the one before that I don't remember black wing penaki the waxing moon look at that little bow and arrow that's pretty cute we got a parallel twister all right veil of darkness I think this is like a secret wearing glass like a very strange secret rare for some reason the grand spellbook tower there's those spell books and sinister serpent on another that's a great goat card so very nice in terms of the pulls from gold right now we're looking pretty good I should probably actually pull out the other ones just so I don't open all the maximum gold eldorado by accident all right we are finally uh we have everything open at least uh it takes it took a lot all right that's probably been eight minutes of this video is me opening the actual boxes all right let's keep going number four trap trick hodra dragon archfiend king calamity cursed eldlands took the same package last time and Rygeki there we go all right that's pretty good in that pranks is pranks prank kids roxies that I'm trying to say premium gold return of the bling let's do this let's see what we can pull with the bling we have our rose ballad pretty cool parallel twister number two oh we got a herald of perfection upside down pretty cool then we have a grand spellbook and sinister soup premium gold time let's go premium gold we have a light pulsar dragon we have gimmick puppet dreary doll number 31 embodiment of punishment mizuki pretty cool gold rare and a black wing armor master there's so many good Edison cards just for a bunch of different decks in here it's pretty cool all right looking pretty solid I think uh all the golds are doing pretty well so far I think the worst one is definitely uh maximum gold it's not doing so good so the new gold is not looking so good trishelon nice card we've got a prank it's lamp sees prank it's bow a bark exodia the forbidden one so now we have the left leg I think and then the exodia it's just different sets you know different gold rears and then a sign at mining not bad so pretty cool actually because what if we got exodia from all different golds that actually be kind of sick that would be a pretty cool uh you know little thing to happen we got a light pulsar dragon very cool number c40 gimmick puppet of dark strings chronomaly guardian knot or gordian oh spirit reaper gold rare I just think they look good man necro gardener the old is the old gold better than new gold let me know in the comments do you like the way the old gold looks or the way the new gold looks and we're talking premium gold we haven't gotten all the way back to gold one through three here but we have premium gold which is pretty old at this point 2014 2015 stuff like that rose lover for all your plant lovers out there junk jiaming compare these what do you think 20 things better that's the secret gold and then there's just premium gold return the black maximum I keep wanting to call it premium gold oh gores why are these upside down gore is the emissary of darkness nice another upside down our mateys and another left leg of the forbidden one okay another left leg so we got two lefties not bad let's go back to maximum gold pretty cool pretty cool let's see we can get scrap wyvern for the dino deck with danger nessie pretty good card we got baby serisores for the dinos we got a ruckson special right into another dino card soul eating over raptor we've got weather painter rain and then number 41 baguska okay premium gold return of the bling let's return to the bling I don't know what the best card out of these sets is I feel like the wing dragon of raw is one of the best that we could have pulled we have a black wing tamer the I can't read him because he's too shiny true the tristan knight I'm not sure gore is the emissary of darkness beautiful exchange of the spirit formerly a retro pack one secret rare and then a sinister server we have at least two of those pretty nice I keep putting these in the wrong pile back to maximum gold eldorado some amazing pulls so far I really have no idea how long this video is going to be because I spent half the time opening the boxes rainbow neos verte anaconda band we've got a ruckson special right into the side frame of your gamma not bad pretty nice zeta pretty good card and weather washer return of the bling I feel like I did I need to open a uh not return of the bling the other one we got a cat shark that card's pretty decent we've got it wait where's my pile I lost it oh no oh here it is here it is okay we found it rose lover don't lose the piles naturia beast a good card madolce queen tier misu actually I did see this one this set list this is the best card in the set madolce queen tier misu I think it's only like $8 which is pretty insane and then double the king so there's like not a lot of really expensive cards but there's a lot of decently expensive cards this is the one I was looking up king of bling I didn't realize it's just premium gold so I couldn't find it and then I just gave up and just open the packs I was like who cares oh forbidden lance that card's pretty nice I've sold a few of these on tcg player recently my store is ruxin 34 if you guys don't know and if you guys want to support me check on my tcg player and ebay affiliate links down below very nice and then we've got oh an upside down card junk puppet the chronomaly tula guardian icarus attack there we go pretty strong card and a chaos sword that's a beautiful version of that card one of the coolest I would say all right max gold max gold eldorado not premium gold is this the best gold out there I don't know it's the newest gold what do you guys think gold sarcophagus beautiful oracle of zephyra elvis mad golden lord prank its place and the prank kids fanzies all right so far not doing so hot over there with the maximum gold eldorado but we do have premium gold return of the bling stardust respark is it going to be upside down no santa claus will the next heart be upside down hey merry christmas we've got a hero prisma trap tricks a tracks and a right leg let's go so we now have the legs and the head of exodia we're looking strong we just need the arms we could do it we could pull exodia premium gold will it happen oh a dark magician girl i forgot this was in here oh my goodness that's a big one dark magician girl okay we then we got a huge belly knight this looks like something from one piece if i had to pick something we got a dragon necro nether soul dragon another good card kai is the shadow monarch this pack is insane this is such a good pack and then a sword's a revealing light which is pretty cool as well wow what a pole talk about some great poles dark magician girl gold rare we will take that to go okay premium gold is really good i feel like premium gold is aged really well compared to like return of the bling that is super strong all right we'll take it we got a rainbow neos very nice rocket tracer deep sea diva that's a great card i'm gonna pull that out phantom nights break sword we got ourselves a sky striker ace ray shooting shooting code talker shooting code talker and then pandemonium all right pretty cool wow some amazing poles today we have a lot of packs to go i love opening gold series it's super fun we have the ghost chair in the underworld boat man beautiful mast hero angie for the mast hero decks i might keep this this one i want to make a mast hero an elemental hero deck at some point i really like those for uh in dual lengths so i would probably like in real life even though they're probably awful and then i'm messy on the time lord pretty cool all right premium gold you've been good to us so far we pulled ourselves a wing dragon a raw and we pulled ourselves a dark magician girl now we got a solar recharge which is nice for those lights weren't edison decks we've got a gaga got academy emergency network number c9 chaos dyson sphere black luster soldier envoy of the beginning let's go and then a bottomless another amazing card wow we are pulling some really nice stuff out of here that is a set right there premium gold that's probably why it costs so much all right okay pretty cool let's see what you can do out of max gold i think premium gold is running away with this one right now i don't actually know the value but it feels like they're probably running away verite would have been great you know a few months ago number two solding over raptor i don't we haven't pulled an access code we haven't pulled much at all let's see premium gold you've pulled some decent stuff i think it's probably uh premium gold then return to the bling and then maximum gold eldorado in terms of value right now but let me know your favorite i want to know your favorite rarity a sold a sold a whatever you want to call it rose paladin the star light road actually need a minty one of this for my edison deck and pull that out preparation rights also good trap chicks de anaya pretty cool pretty nice now it's premium gold time can we pull a big card a car that we're looking for another solar solar recharge i should say i was gonna say soul exchange that's not the same card uh gimmick puppet magnet doll dragon necro nether soul mazuki for those zombie decks and a rhoda that's a pretty nice one as well wow just like every single card is really nice maximum gold eldorado it's interesting that the other ones are like sideways to look at them you have to like look at the pack like this for it to to read it you know straight up and this one you know they change that rainbow neos all ruckson special right into exodia all right exodia the forbidden one we now have two of his head two of his left arm we got a book of moon sign that mining and an altar guys not bad premium gold's up next we need the left arm and the right arm that's what we need we got a santa claws ho ho ho merry christmas absolute king blackjack that's worth a few bucks you got a ruckson special or not a ruckson special upside down card naturey bark and like a reflective of this card is you can almost see me in there and then you see me back there okay upside down again tumble the kings animus will be the jackal these go perfectly together that's pretty cool premium gold time at probably the front runner right now in terms of winning this battle let's go we have colossal fighter beautiful silent wambi or wabi the gimmick puppet dreary doll these are creepy cards mind control and gold or lord of dark world that's probably a good thing because or wood lord i should say because you know the dark world structure deck just came up maximum gold eldorado time let's do it let's dorado this stuff we have a ilti man's yoken we got the weather thundered canvas trisheloth dragon of the ice barrier alter guy silquitos the heat of the fire charmer number c1000 all right and then a haquero of the golden land i think our best card is the pot of extravagance and it's not close i think we have one extra pack here so we're going to open another maximum gold they have maximum a little as extra packs and they're still kind of getting waxed uh white dragon wyvern buster invocation there's a good one it's probably like three bucks but it's still pretty decent broken line alter guys mellow seek ride gecky that's pretty good as well we'll take it prank its fancies back to premium gold the best one so far i would say we have the eclipse wyvern oh what is this it's like a blackwing dragon black feather dark rage dragon okay that's pretty cool i can't read it though gimmick puppet shadow feeler stardust dragon there we go that's a pretty one that's one of the ones i actually do like the name and then a gravekeeper spy also another strong card gravekeeper spy so many good cards at a premium gold it's insane all right we are still a couple ahead on maximum gold versus the return of the blanks we're going to try and catch up make them even here we have catch up and mustard here we go eldorado dark magician let's go we got a dark magician and an apple sourdark uh evils evils are dolka there we go i am struggling right now summon limit that's probably a pretty good guard all right back to premium because we have five of these and five of the eldorado and only four the return of the bling we're going to try to get those even we have ourselves a light pulsar dragon gimmick puppet at it again junk puppet a gold lord of dark world and a black armed or an armed wing dragon black wing armed wing there we go now there's no dragon involved it's just armed away all right now this will make them even we'll now have four of each left eldorado is up prank its pranks with a fiend weather painted rainbow a siphon lord lambda and looks are the black awakening zeta and weather washer okay nothing too exciting there back to return of the bling it's been a while but now we got a return to the bling let's go we have black wing something cat shark five headed dragoon lavalval chain and ring of the d the ring of destruction pretty nice we will take that premium gold let's go big potential in this set here's a fire fist guy so probably not a great pull there fire fist tiger king number c-88 utopia buster a cross attack and a prime material dragon very nice prime material junior journey maximum gold eldorado let's go let's see what we can get weather painter thunder do new the dududududu whether better snowy canvas oh gamma Again and IP Mascarena. There we go. We're getting some good cards now and a buguska. Not bad We will take that we're getting some good cards Finally nine packs to go in this super battle that we have here We have another Anki if we want to run two of those that's pretty sick number 95 Naturia Barkeon Armadies and a down nerd magician. Okay, not our best pack out of the return of the bling That's not too bad. I'll take the Anki pretty cool next up premium gold We get the dark flare in here as well. So you've had dark flare. We've had solar flare, whatever it's called We got the power tool mecha dragon. I don't even remember that guy city Babylon Necro Gardner and a Exploder dragon pretty cool. That was good enough speed duel for a little bit. All right maximum gold Eldorado What can we pull will it be access code? Will it be access epic? Let's go lost world ultimate conductor tyranto number two Evels or log yet those are rucks and special by the way, and then a firewall and a giant wreck Dude, this is the dino set literally all the guards are in here All right, I'm gonna do another Eldorado because I want to end with Premium gold. So we're gonna do it like this. We're gonna have weather painter cloudy neos. There he is broken line Scrap golem and a ghost mourner and moonlet shield not a bad card either pretty cool Let's go to premium gold or turn of the bling They call premium gold Eldorado so I'm gonna go I think I did I don't know we got the Tristan is sold We've got ourselves an upside down Cephalon. Oh, there we go Left arm of the forbidden one and rank up magic barrier in force. We are one arm away. Is this not familiar? We're one piece away and it's an arm. We just need instead this time We need the right arm so we've got one more pack to potentially do that. Let's go premium gold I don't think they have exotic pieces in this one Colossal fighter pretty nice guard there. We got a Chronomaly Gordian knot number C88 Traps done also good and then breaker the magical warrior That's beautiful as well with a little reprint action three packs to go Let me know in the comments. What is the best gold out there? Is it one of these three and what is your favorite of these three in my opinion? Premium gold is just an awesome set and I think they look the best so that's what I'm picking for side frame driver We got a mystic mind. Sorry, buddy. You're banned straight out of Nim Nim town. There we go We got the ossa numeron storm. All right Very cool two packs to go our last chance Will it be the right arm of the forbidden one? Come on. That would be so cool All right, not so far. It is a number 21 frozen lady justice black wing Chris the crack of dawn Gladdy or beast guys are it's I need this for my glad to be stuck. That's actually hype I'm so sad. We didn't pull two actually advance a ritual art. I think it's usually the last card. Come on, baby. Let's go Many on the time Lord, okay Well, but I still I'm really happy about this guy's errors because I only had one for my my Gladby's deck I need one more so I finished out the main deck, but I didn't get that card I might do a video showing off my Edison decks if you guys want to see him Let me know in the comments. All right premium gold time. Let's do this First edition can we pull another big card? Let's go We have an eclipse wyvern very nice Chronomaly Tula guardian Nether soul dragon also pretty good Gold sarcophagus and our final card of this gold series video. Make sure to let me know in the comments your favorite gold series of all time Here we go Kias the shadow mind. That's pretty cool for Edison. We'll take that. All right. That was a pretty awesome opening We've got some really amazing polls Hope you guys enjoyed it because it was pretty expensive if you did make sure to like the video and subscribe to the Channel and let me know if you want to see every gold series opened up in a single video Shout out to tone foe show Dax or tomato juice JT show TCG trust of cards puffins of the doom Ernesto D'Anda Dizzy flexi boy hoppers choice three through three miss cycle James Jance Frankie Martina as an unattied show Christopher Ward Ian Musa junior barding Mike Nance mimic echo shadow fall and Thomas McClane. Thank you guys for supporting the channel I'll see you guys in the next one. Peace


We'll give you the second lecture. Third, sorry, sorry, third lecture. I can do the second lecture again if you want. Back by popular demand, my second lecture. OK. All right. So at the very end of the second lecture, we were talking about recurrences for polynomial powers. We're going to dive into that a bit more. So let me remind you briefly what we were doing. So I had my polynomial f. OK. d is 2g plus 2. And we were looking at f to the m. We called that h. And here m is p minus 1 over 2.  Del is the degree-preserving differential operator. This is m times del f times h. So this is the differential equation. And we wrote down a recurrence for the coefficients of h. So we had hk equals 1 over k times f0 times some stuff. I'll write it down. I think I should probably just write it down. OK. We had some. What was it? Half j minus k fj hk minus j. OK. And this expresses hk as linear combination of the previous d values, provided that our k is not 0 modulo p. So I want to take a quick look at what happens when we try to use this recurrence to compute the matrix a sub f. So remember, we have our matrix a sub f. And its entries are given by hvp minus u. This is a g by g matrix that we're trying to compute. It's the Hassavit matrix. So what do we do? Well, we start with h0. Well, h0 is very easy to compute, right? Because it's just the constant term. So I just compute the mth power of the constant term of f0. And I can do that very quickly using repeated squaring again. It's also sometimes called binary powering. OK, so I compute h0. That's great. And now I compute h1 using the recurrence. And there was actually a question in the problem session yesterday. What about the h with negative indices? What's h of minus 1? h of minus 1 is 0. All the negative indices is 0. I guess that's just by convention. But you can also check when you prove this that's what it has to be. They all have to be 0. OK, so h1 is a linear combination of h0, h minus 1, h minus 2, the previous d values. So you get h1. That's great. And then you get h2. And you keep going all the way up to hp minus 1. And at this point, we have the first row of a sub f. The first row is going to be these last g coefficients that we computed. This is the first row of af. That's v equals 1. But now we have a problem. So the problem is that to get to hp, we need to divide by p, because k equals p. And we can't do that. Cannot get hp, need to divide by p. Oh, I shouldn't write that. That's confusing. Actually, it's kind of true. We have to divide by 1 and by 2 and by 3 and by 4 I'll put it back. OK, there we go. So this is a bit of a problem. And we actually need to do hp because we need to get to the second row. The second row, we need to get all the way, we need to keep going. We want to get all the way to h2p minus 1. That's for the second row. Now you might say, OK, well, I can't get hp, but I could try k is p plus 1, for example. That's going to give me some relation involving hp. Does that tell me what I need? Well, you can play around with it, and it doesn't. The problem is there's always going to be one degree of freedom that you haven't dealt with. So it doesn't work. OK, so that doesn't work. So we're going to have to try something completely different, which is that we are going to start again from the beginning, but we are going to lift the whole problem to working modulo power of p instead of modulo p. So we are going to lift the whole computation. So the first place I learned this idea from was this paper of Boston, Gaudrey, and Schost, which I mentioned many times in these notes, very important paper in this area. But I'm sure it's a much older idea than that. I don't know where it goes to, but obviously, a much older thing. So we are going to lift to z mod p to the mu, z for some mu, let's say at least 2. And one of our jobs is going to be to figure out what mu we need, what is the smallest mu we can get away with, because larger mu is more expensive computation, for several reasons. OK, so here's what I'm going to do. I'm going to choose a lift, capital F, which is F0 plus F1x, Fd, x to the d. This is in z mod p to the mu, z, x. So polynomial over these integers mod p to the mu. And by lift, I mean that each Fj is congruent to our original Fj, modulo p. OK, so any lift is fine. You just choose it once and for all. The most obvious thing to do is, if your little f's are in the range, if the little f's are in the range 0 to p, it's a pretty reasonable way to represent them, then you just take the capital F to be literally the same integers, but now thinking of the modulo high power of p. So we're going to lift like that, and then we are going to put capital H to be capital F to the m. And this is also polynomial modulo p to the mu.  You can't just pick another lift of h, a little h. That doesn't make any, I mean you can do that, but nothing will work. It has to be consistent with this choice of f. And now, what we want to do is it's enough to compute coefficients of h modulo p, enough to compute the target coefficients, I'll say, Hj modulo p. And by the target coefficients, I mean the ones coming up in this matrix. And we only need the modulo p. So we might compute them incorrectly modulo p squared or p cubed or whatever other powers. As long as we get them correct, modulo p, we're happy. OK, so since we have h is f to the m, we can actually go through some of the same setup as before. There's a differential equation and a recurrence satisfied by coefficients of capital H. So let's write that down. So it satisfies the same differential equation, which is capital F times del h is m del f times h. So remember, this is now a congruence modulo p to the mu. Everything here is happening modulo p to the mu. And we can write down a recurrence using exactly the same method. So here's what you get. I'm going to be slightly careful. I'm not going to put the k in the denominator. So I'm going to write it like this. k hk is equal to 1 over f0. Now, I should just say about this capital F0. I think I forgot to mention it at the beginning. I think I mentioned this last time. Remember, we are assuming that little f0 is not 0. And that tells us that f0 is invertible. Capital F0 is invertible, so we can divide by f0, no problem. OK, so here is the recurrence. It looks slightly different than the other one, because I need to remember that p is no longer 0. Remember over here, that half is actually p plus 1 over 2. But now I'm working modulo high power of p. So I have to keep track of the p, and it's hidden in here. This is really, so I've left it as m, but m plus 1 is p plus 1 over 2. That's p plus 1 over 2. This is going to cause much pain when we get to section 7. But let's just write it like that for the moment. And then we have capital F, j, and h, k minus j. OK, so again, remember, this is all modulo p to the mu. Now, what we're going to do is we're going to try to do the same things before. We're going to start with h0, and we're going to try and compute h1, h2, and so on using the recurrence. And occasionally, we're going to have to divide by something nasty, something when k is divisible by p. We have to divide by p. And that is going to introduce an error into our computed values of h. So here's how I'm going to do it. So we've got these coefficients h0, h1, h2, and so on. But I'm going to define a sequence of approximations, which I'll call h0, twiddle, h1, twiddle, h2, twiddle. These are going to be the values that we actually compute. And they may be different to the h's. As long as we hit our target values modulo p, I'm happy. So let me show you how this works. So we'll start with h0, twiddle. Well, that's just going to be h0. And we can calculate that because that's just f0 to the m. So that's actually equal to h0. There's no error there. And we can do that very quickly. That's just repeated squaring. But now we're working modulo p to the mu. And then we compute, inductively we find, so we compute h1, twiddle, h2, twiddle, et cetera, by solving that recurrence over there but with the h's replaced by h, twiddle. So by solving k, h, twiddle, k, it's 1 over f0, m plus 1, j, yes, j minus k, fj. And now this is the only difference, really, is this one is hk minus j, twiddle. And this is all, of course, modulo p to the mu. So we use the previously computed approximations in this recurrence to compute subsequent values. I hope that's clear. I want to show an example of what happens when we do this because there's some pretty funky stuff happens. Let's try an example. I'm not actually going to compute any numbers, of course. That would be too hard. You can use your computer for that. Let's take a genus 3 example. So my degree is going to be, I'll use 2g plus 2, which is 8. And I'm going to take mu equal to 3. So I'm working mod p cubed. And you'll see why in a moment. So let's see what happens. So we start with h0, twiddle. That's great. So that's correct modulo p cubed. And then we apply the recurrence. And you can see that I'm not dividing by p anywhere. So h1 twiddle is still going to be correct modulo p cubed. What I mean is it's congruent to h1 modulo p cubed.  You divide by f0. There's no division by p. You have to divide by k, but k is just 1. So there's no problem. And you keep going h2 twiddle all the way up to h p minus 1 twiddle. And all these guys are correct modulo p cubed. Correct modulo p cubed. And at this point, again, we have the first row of a sub f. So these last few coefficients, this gives me the first row. Now I get to h twiddle p. This is the interesting point. So we get to h twiddle p. Now, when we solve for h twiddle p, h p twiddle, we have to divide by p. So what's going to happen is our value is only going to be correct modulo p squared. And there's no hope of computing it modulo p cubed. There is not enough information there. We've really lost that information. It was never there in the first place. So this is only correct modulo p squared. And we keep going. Now we look at h p plus 1. Now for h p plus 1, we don't have to divide by p again. We're dividing by p plus 1. But the problem is that the h p is one of the values in our linear combination. And so it's going to muck up our value of h p plus 1. So it's also only going to be correct modulo p squared. And we keep going, and the same thing happens here. So all these guys are correct. I mean, that you might get lucky. They might be correct modulo p cubed, but probably not. They're probably correct modulo p squared. And now we have the second row. Because these guys are correct modulo p squared. So they're also correct modulo p, which means we have our second row. And I'll quickly write out the third row. And the reason I'm doing this in such detail is it's something weird actually happens. So third row is we get to h 3p. So we have to, sorry, 2p, 2p. So we had to divide by p again. So now we're only correct modulo p, and we keep going. And we get to h 3p minus 1. So these are correct modulo p. And here we get the last row, the third row. And my genus was 3, so that's why I chose p cubed, so that we would have all three rows correct modulo p. And then, of course, we can stop. We don't need to go any further. We don't need any more coefficients. If we went up to h 2 or 3p, then it wouldn't be correct at all. It would look like random garbage. So if you generalize this argument, in general, it looks like we need to take mu. Well, what do we need? We basically divided by all the integers from 1 up to 3p minus 1. So in general, what it should be is the valuation. So vp is the number of powers of p of gp minus 1 factorial. These are all the denominators that show up. And then you need to add 1, because we need the result to be correct modulo p. OK? Now, the incredible thing, which, of course, those of you who've read ahead in the notes, will have spoiled the surprise. But I'm sure no one read ahead, right? You're not that kind of student. It really warmed my heart the other day when I said, does anyone want me to keep going for five minutes? Everyone's like, whew. Normally, my students, they wouldn't be answering, because they would already be asleep. So here's the amazing thing, is you can actually do better than this. And it's very, very surprising, because this argument looks tight, right? There was really no room. There was no room to do anything. But you can actually do better than this. So corollary 5.3.1 says that actually it can take mu to be smaller than this. So this is a logarithm in base p, gp minus 1 floor plus 1. So this mu, in our example, is 3. But you can check that in this formula, it's 2. This is in our example. In the example. So it's completely crazy. So look at this. Here's what happens. I'm saying you can start mod p squared. So you start mod p squared. OK. Here, these are only going to be correct modulo p. And then you keep going. And then you divide by p again. I'm saying it doesn't matter. These will still be correct modulo p. Like, how can this happen? Impossible. Well, there's a proof in the notes. I'm not going to go through it in the lectures now. It has to do with the fact that at every point, these coefficients satisfy this recurrence by design, which means they satisfy the differential equation, which I seem to have erased. There's a corresponding. Basically, these h twiddles satisfy the same differential equation as the h's. And magically, that is enough to sort of resurrect this mod p information. I wish I had more time. I could go into that. But this is actually one of the most annoying things in all these sorts of algorithms, because you actually have to work a bit to prove these things. And in one variable, it's OK. In two variables, it starts getting very difficult. And I don't completely understand what's going on in two variables. But it's a very interesting phenomenon. So it helps us in the sense that it means we need less precision than we expect in these sorts of algorithms. But it also hurts us because it hurts our brains, basically. And I think there's a lot that's not understood about this sort of phenomenon yet, especially in a larger number of variables. And I really encourage you to try this on the machine. Just try it. It's kind of incredible. It's like, where does it come from? OK. So let me just quickly say now what complexity bound we get. So the complexity of, I'll call it the recurrence strategy, of the recurrence strategy for computing a sub f. And the recurrence strategy, I mean, do what I've just done here. So take your f, lift it to this capital F, start computing the sequence of h twiddles until you have all the rows of a sub f. So it's not too hard to work out because you just need to count the number of operations you needed to do to compute all these coefficients. And it turns out to be o of g squared times p operations in, oh, hang on, have I got this right? Yeah, OK. So here's what's happening. There's, yeah. So the degree of the, the number of terms you're trying to compute is g times p. The extra factor of g is coming from evaluating all the terms of the sum is o of g terms. And I'm using the fact that we're actually working modulo p squared, not p to the g, but p squared everywhere. So I think I need to assume something like p is at least g. So this, this is always just 2, OK? So we're just working modulo p squared everywhere, OK? And then the cost of an operation modulo p squared, as before, is log p to the 1 plus epsilon, OK? Yeah, so I was confused briefly about why it was g squared. I think that's OK. So if we compare this to the expansion method, where you just compute the whole polynomial, then I believe what happens. I think we had, previously we had something like o of g times p, and there might have been a log p squared, I think. I think it was a little more, I think it simplifies to this if you assume p is greater than or equal to g. So if you compare these two, you see that our dependence on p is still pretty much linear, which is not surprising. You have to compute. You still compute all the coefficients, all right? The dependence on g has gone up. You can actually get rid of that if you work a little bit harder. I'll leave that. I think I had that as a problem at some point, and I got rid of it. But you can think about that if you like, how to improve that power of g. And the dependence on p has actually gotten slightly better. It's actually a bit misleading. This algorithm in practice is much, much faster than the expansion polynomial one. And the main reason is because it uses almost no memory. With this algorithm, you only have to keep track of the previous decoefficients. So you just have decoefficients. You compute the new one. You throw the other one away. So you're operating on very little memory, whereas taking a large power of a polynomial uses a huge amount of memory. And that makes an enormous difference. Memory is very slow. So Drew Sutherland has some lots of data on this. I'm not sure about for this algorithm. It's true here? Maybe not. But we have a paper on a similar algorithm, which is not for hyperliptic curves, but for plane quartics, which is going to be at ants next week. And he's got some data showing that this kind of algorithm is much, much faster than. Actually, no, he doesn't even compute the powers, because it's so hopelessly slow. OK. Yeah. Yeah. Sorry, I'm not sure I understand the question. Well, you have to work modulo p squared everywhere. So you can't represent elements. You can't represent things as elements of fp anymore. You have to represent them as integers modulo p squared. So they take twice as much space. Is that what you're asking? Yeah, you do. OK, I should have said more about this. You're right. So the question was, how do you do the division by p? So the answer is that whenever you need to solve this congruence, you just solve it. You write down any solution. So suppose k is p, for example. What you'll find is you'll find this right-hand side is divisible by p. So you divide the right-hand side by p, and then you're basically done. So whenever there's some indeterminacy, whenever k is divisible by p, there is not going to be a unique solution. And it doesn't matter.  And then you continue. And there'll be some error because you didn't know which was the right solution. And what's proved in the notes is it doesn't matter which solution you pick. Everything will work out. Thanks, good question. Any other questions before I go into section 6? Yes. And in fact, what will happen is, yeah. So what will happen is that this first line here will be correct modulo p squared. And after that, everything will be correct modulo p, even though you've divided by p lots of times. It's very, very strange. Yeah, as I said, it's one of these things that's hard to believe. And you don't really believe it until you try it on the computer, and you're like, OK. That's what he said. I mean, I had to try it on the computer before I came here. But I wanted to make sure. OK. All right, section 6. Oh, there's a question up there. You mean these powers here, the big O constants? Yeah, so I have no idea. I suspect that the big O constants here are smaller for this algorithm. See, the problem is there's this tension between what is the theoretical answer and what is the answer in practice. I just know from a lot of experience that in practice, this is much, much faster. And the other problem is that the way I've written it, this log p to the 1 plus epsilon is not very realistic. I mean, what will typically happen is that log p, you're dealing with operations modulo p squared. And typically, that will fit into one machine word, or maybe two. So the operations mod p or mod p squared are all being done in hardware. So it'll just be some constant. It doesn't really depend on how big p is. So this is a sort of theoretical bound. It's not very meaningful. The p part is very meaningful. It really will scale linearly like p, because that's the number of operations you need to do. Now here, this is much more complicated because here you're doing polynomial arithmetic on enormous polynomials. And roughly what's happening is I think one factor of log p is coming from the size of the coefficients. And the other log p is coming from the fast Fourier transforms. And I think you really will see both of those factors of log p. I bet that you could probably measure it and roughly see them happening in the data for large enough p, maybe. I'm not sure if I've really answered your question, but yeah. OK. Let's move on to section 6. And this is where we get to some really fun stuff. This is the square root algorithm, or square root time algorithm. So everything we've seen so far is linear in p. And now we're going to get this down to square root of p. And I'm going to first of all do a bit of a warm up just to get the ideas. And that's going to be Wilson primes. So this has nothing to do with point counting. But this is another fun problem that I've worked on before and still think about occasionally. So what is a Wilson prime? So we all know Wilson's theorem. It says p minus 1 factorial is minus 1 modulo p. So Wilson prime is when this congruence holds modulo p squared. So here are the Wilson primes we know about. 5, because 4 factorial is 24, which is 1 less than 25. 13. And this was discovered when people were still doing these things by hand. Takes a little while to do it by hand. Good exercise. 563. I believe this one was discovered the first time anyone threw a computer at this problem. OK. And there are no others less than 2 times 10 to the 13. And this was a computation that I did with Edgar, who might be here, and Robert Gerber. It's about 10 years ago now. Oh, where's my eraser? Here it is. So our goal is given a prime p is to compute p minus 1 factorial modulo p squared. Given any prime p, we want to compute p minus 1 factorial modulo p squared. And if it's minus 1, we have a Wilson prime. And otherwise, we don't. I don't know of any way of finding Wilson primes, apart from just testing every prime in some range. And you can sort of work out the heuristics, what you expect to happen. And everyone expects that there are infinitely many and that they're extremely sparsely distributed. And now we've gotten out so far, it's getting pretty hopeless to think that we'll find any more. But you never know. So my theory is if I look for Wilson primes, there's a very small chance of finding one. But if I also look for some other kind of prime, well, there's also a small chance of that. And if I do enough different types of primes, then maybe I'll find one interesting one.   And you keep doing this. And it's a hopeless algorithm. But it's pretty good when you've got GPUs doing lots and lots of primes all at the same time. And some people have tried doing it that way. So that's a linear time algorithm. You've got to do roughly p multiplications. So it still takes a while when p is large. So I want to show you Strassen's idea for getting this down to square root of p. So here's Strassen's idea. So let me write s to be p minus 1. This is the number of terms we need to multiply together. So p minus 1 factorial is now 1 times 2 times 3 all the way up to s. And what we're going to do is we're going to split this product into about square root of s blocks of size square root of s. So let me write t to be the floor square root of s. So that's about square root of s, which means I can write s is t squared plus some leftover thing, t prime. And you can check that t prime is not too big. It's actually o of square root of s. I think I worked out some precise bound in the notes. OK, and then I'm going to split up this product into t blocks of t plus t prime leftover terms. So I get p minus 1 factorial. It's going to be 1 times 2 all the way up to t. That's my first block. And then I'm going to have t plus 1 up to 2t. That's my second block. And I keep going. And then my last block is going to be, check I get this right, t minus 1 times t plus 1. Sorry about all the brackets. All the way up to t squared, right. So that's block number 0, block number 1, block number t minus 1. So up to t squared. And then I have the leftover terms, which is t squared plus 1 up to t squared plus t prime. So this is leftover terms. And this stuff is all t blocks of t. Now, Strasson's idea is to introduce a polynomial, which I'm going to write as q of k. So q of k is going to be k plus 1 times k plus 2, all the way up to k plus t. And Strasson knew a thing or two about polynomials. And he realized there's a way of rewriting this sum, sorry, this product, so that you could compute it much more quickly using some techniques of fast polynomial arithmetic. OK. So what happens is that p minus 1 factorial, you see the first block is actually q of 0. And the next block is q of t. And so on up to the last block, which is q of t minus 1 times t. So it's all the multiples of t. There are t multiples of t here. And then there's the other stuff, the leftovers. OK, so Strasson's idea is we're actually going to compute this polynomial, q of k. And by compute it, I mean work out its coefficients, right, working modulo p squared. Because everything here is going to be p squared. And then we're going to evaluate that polynomial at these t points. And there are fast algorithms for these things. And I'm going to explain briefly how these algorithms work. So here's the algorithm. So the first step is we're going to compute q k. Now, as a polynomial with coefficients modulo p squared, OK, now those of you who are not computing people will say, but I already have the polynomial q of k, right? Here it is. I need to actually expand it out, right? Because I need the actual coefficients as the input to the next part of the algorithm. So let's compute this q of k. And I'm going to use something called a product tree. And product trees are discussed in general in section 2 of the notes. But I'm just going to show you what it looks like here. And just to keep my life simple, I'm going to assume t is a power of 2. Let's assume t is a power of 2, just for this picture, OK? You don't really need to assume t is a power of 2. It works perfectly well. This is actually the difference between Drew Sutherland and me, is in my code, t doesn't have to be a power of 2. And in his code, t is always a power of 2. So my code is more painful, but more general. And we sometimes have fights about this, but anyway. So OK, so how does this work?  And then I'm just going to multiply them together in pairs. So I multiply together k plus 1, k plus 2. So I get, and I actually have to do the multiplication, right? k squared plus 3k plus 2. I'll do one more. k squared plus, oh dear, 7k plus 12, et cetera. OK, there's something here, some quadratic. And I keep going, I multiply these two together. This is k to the fourth plus something, plus 24, and so on. And all these things keep going down to the bottom of the tree. And at the bottom of the tree, I have, well, what do I have? I have this product, right? Which I can't really write it out, but I'll just write it out like the way I wrote before. But you know what I mean. It's the actual expanded product. So this is a very simple example of a product tree, right? These guys up here are called the leaves. This is called the root, the root node. I think in my notes, they're upside down, right? I just realized this. I have the, I mean, the root should be at the bottom, right? I guess this is an Australian tree, right? It's the other way around. Yeah, I think in my notes, the root is always at the top. I mean, I just used some package, right? And it puts the root at the top. So all these American trees are upside down. It's very strange. OK. OK. Now, the idea is that, of course, we're going to use fast polynomial multiplication to compute all these products. Yeah, question. No, it doesn't. You could put the linear factors in any order. That doesn't matter in this particular case, because in fact, the only thing I really care about here is this polynomial at the bottom. But in section 7, we are going to have some product trees coming back with a vengeance. And there, it very much matters which order you put the nodes in. So you'll see, hopefully, we'll get to that in section 7 on Friday. OK, no more questions. OK. Now, I'm just going to briefly go over the complexity analysis. You see, the idea is that, let's say at this top level, we had two big polynomials of degree roughly t over 2 that we had to multiply together. So the top level, we just do multiplying two ginormous polynomials. The next layer up, we multiplied two pairs of polynomials that were half the size. But at each level of the tree, the total amount of polynomial is the same. The total degree, the sum of all the degrees at each level is the same. We started off with t here. Here, there's half as many polynomials of twice the degree and so on. So it turns out that when you estimate the cost at each level, the cost is actually the same at each level, up to a constant maybe. In practice, what you find is it's this bottom level that is the most expensive. And it gets cheaper as you go down to the smaller polynomials. So what happens is the cost at each layer is going to be o of, so I can work it out. It's going to be t times the coefficient size, which is going to be log p. And then we need log of all this stuff, oops, log p. And I think that's it. I think that's going to be OK, which turns out to be o of p to the half log squared p. And the reason is that our t, remember, is squared of p. And then the number of levels of the tree is o of log p. It's log t, which is log p. So the total cost is p to the half log cubed p. And the key point here is that it's basically squared of p. And this is discussed. This analysis is done a little bit more detail in the notes. And since I've only got 10 minutes, I don't want to go into that in any more detail. OK, so now I have to explain how do we evaluate q at these t points. And the idea is you use this technique called multipoint evaluation. So now, so step two is evaluate qk at k equals 0t2t up to t minus 1 times t. So we have a polynomial of degree roughly t, in fact, exactly t, which we're evaluating at exactly t points. And this is exactly what multipoint evaluation does. You see, if you evaluated separately at each point, you would get a quadratic time bound, quadratic in t. Because first, you'd have to evaluate a t that takes t operations. Then you evaluate it 2t that takes another t operations. The multipoint evaluation lets you do them all at the same time much faster. So use multipoint evaluation. And let me explain briefly how that works. So the first thing you do is you compute another product tree. So let's say, suppose we want to evaluate at, let's say, points alpha 1 up to alpha t. These are in whatever ring we're working in. Here we're working in z mod p squared. So if we want to evaluate at t points, in this example would be these t points, what we do is we write down, first of all, this product tree. We take all these linear polynomials which have the alpha i's as the roots. And we keep going down this tree until at the bottom we have the product of all of these terms. So this is our second product tree. And the cost of this is going to be exactly the same as the analysis over there. So the cost is p to the 1 half log cubed p. So first, we compute that tree. And then, this is really the key part, is we're going to use this tree to figure out q of k modulo all those polynomials. So now we do what's called a remainder tree. So I'm going to start at the top. Now my tree will be upside down, because I always want to start at the top. So this tree is like hanging from the ceiling.  And I don't really have to do anything here, because this guy is degree t. And this guy is degree t. I suppose I could do one little reduction to get rid of the degree t term, but it's not a big deal. And then what I do is I split this like this. I figure out q of k modulo, just the first half of the terms, k minus alpha 1 up to k minus alpha t over 2. And then over this branch, I have q of k modulo the rest of the terms. So this is k minus alpha t over 2 plus 1 dot dot dot k minus alpha t. And the way I compute these is I just divide by these polynomials and take the remainders. And I know those polynomials, because they are the values near the bottom of this tree here. They're the two polynomials here and here that went into this root node. And then I just keep splitting like this. These are both half the degree, right? So I keep splitting. Now I have four polynomials that are a quarter of the degree. And I go all the way to the bottom. So at the bottom of the tree, I have q of k modulo k minus alpha 1. Well, if I modulo a linear polynomial, that's just the value of that polynomial at that point. So this is equal to q of alpha 1. And similarly with all the others, I get q of alpha 2 and so on all the way up to q of alpha t. So I know if you haven't seen this before, I realize it's a lot to absorb. And I really think the best way to understand this algorithm is go and implement it. It's not hard to do as long as you have a computer algebra system which can do all this polynomial arithmetic for you. As long as you can multiply and divide polynomials, this is pretty straightforward to implement. Now this step is also of p to the half log qp. And the reason is that at each stage, we're doing a polynomial division. And you can prove that, well, I think I mentioned this before, division is big O of multiplication. OK. So this is how you compute these main parts of that product. Then you have to multiply all these results together. So I guess step three is multiply together all of these q multiples of t, qiit, and leftover terms. And remember there are only something like square root of p leftover terms, so that's also square root of p. So altogether you get square root of p. So conclusion, you can compute p minus 1 factorial modulo p squared in time p to the half log cube p. Now I don't really have time to get back to the point counting, which is really what we want to do with this. So I'll do that on Friday. But I might just make a few more comments about this. And so I've got a minute or two left. So firstly, the most important improvement on this algorithm, if you're interested, is you can get this 3 down to 2. There's an algorithm, I mentioned this before, Boston, Gaudry, and Schost. Very nice algorithm, which is a similar idea but you never actually compute this polynomial q. You don't actually expand this out. What you do instead is you sort of compute values of some of these sub-products at certain points. And it's quite clever and saves a factor of log p. And that is a big deal, right? A factor of log p. I mean, your p might be in the billions or trillions even, so that can be a factor of 30 or 40 or something like that. Huge, huge difference. So we actually used that algorithm that I just described. We did this big computation of Wilson primes up to 2 times 10 to the 13. We didn't use this for the main part of the computation, but we used it as a check, right? So after we've computed all these remainders, we pick some little sample of them and we verify them individually using this sort of algorithm. This is pretty much the fastest way I know how to compute these Wilson remainders. Anything else I want to say? This algorithm uses a lot of memory, which is a problem. You really have to keep track of all these polynomials. So that hurts. Yeah, I think that's all I'm going to say today. So thank you very much. Any questions?


This actually I know won't last long though Maybe until that next sign into destiny 2 but we will see is what it is My hopes are still high though of this game becoming better than what it is right now I mean can it get any worse? I don't think it can so today I want to look to the future and actually not too far into the future Maybe that five days into the future and that is with the night full exclusive rewards now these rewards We know which has been confirmed by bungee have been delayed once before so the design team had more time making them look Super unique this alone is epic in my opinion. It kind of shows that there might not be reskins Which is good news they were originally meant to drop in the February updates But now we are getting them on March 27th with the gold fast update, which is literally five or six days away Now what I want to talk about today is bungee's ability right now to offer us more grind than what we've ever had From this sequel and this can be done via night full exclusive loot alone Now we all miss a grind destiny one offered the fact We had to replay the same thing over and over and over again to get that certain item Whether that be Omnigul for the grasp of Malak Vanguard strikes for rep for the 1,000 yards there Crucible matches for the IS Luna. It was a grind we need indeed to but there is just none of it Now with bungee adding night full exclusive loot It gives them the chance to bring us what we want even if it's only from the single part of the game If this is the case and they bring us what I'm about to mention It builds towards a better game and it's all good and dandy us telling bungee what to do and what to change But without them being able to test it with us They really can't sacrifice other changes made for ones not tested which could indeed seek their game even more So what I suggest they do is test the waters with this night full loot make these night full exclusive rewards Extremely rare and I mean as rare as that perfect role IS Luna a weapon I still don't have by the way in doing this not everyone will get that night for loot within the first run But here is the best part if a said weapon is a part of the week's night full exclusive loot Make that weapon have a random rules Even if the rules ain't randomized make that single weapon have about five or ten different versions each after in different roles Also, do not cap how many times per week we can run the night full So if we run this night full up to 50 times per week There's a chance this exclusive weapon will drop on that 50th run.  Let's go with the poor medium vex strike Let's say the exclusive weapon for this strike is the vex myth of class or something of such I know this weapon won't ever make a return, but hey is what it is Imagine this weapon had 10 different available roles each as rare as each other two roles from the 10 were considered the best There was no limit on how many times you could run a night full within that week is trying to obtain this weapon You could just keep on grinding this I know for a fact will give people that nostalgia feeling missing from the game Which made them fall in love with destiny one that element of surprise when killing a boss and waiting for that Angram to drop you running over to it Sliding into it and seeing that weapon pop up on the right of your screen slowly going into your inventory to check its role Praying it's the role you want that build up then the motions that excitement that relief This is all missing from destiny two bungee here have as good an opportunity to deliver us this with these night full Exclusive rewards and with them only dropping one night full per week Also, it makes sense from their perspective in having limited content to offers at the moment System as such for night full exclusive rewards is the best possible outcome in my opinion And it's this kind of system bungee need to offer on a 27th of March when this update drops It's all good and dandy night fools having exclusive loot But if everybody gets them on their first run, what is the point seriously? There needs to be a system in place where we have to grind for that weapon or that item There needs to be a system in place where these are super rare to get There needs to be a system in place where we can run these night falls as many times per week as we want and each and every time Having a chance of that exclusive loot dropping. That's the kind of system We need in place for these night full exclusive loot rewards We really do now I said the Vex as an example by the way, although I know it won't ever make a return It was just that one weapon I was so desperate to get from destiny one most of a destiny one weapons I grinded for did look good in my collection, but we weren't really of much use to me But the original Vex myth of class is like nothing else. I ever had those were also the best times for me in Destiny as a destiny player now I know my idea of how bungee should play these night falls out probably won't ever happen But if it did it would in turn make many people return and like I said earlier in the video Be the basis of bungee realizing what we actually want by testing the water first Such a system could be the start of many new Incorporations of better additions to the game and on that note I am out I do hope bungee see the video, but I ain't hopeful maybe you guys could pass the message on that would be awesome But guys, thanks as always for stopping by if you did enjoy the video leaving a like really does help me out And hopefully guys I will see you on that next one