Hi everyone, Joe for Jasper's case breaks.com. Happy Saturday. We're coming at you with 2020 tops chrome baseball Jumbo edition eight box pick your team number four Big thanks all of these folks right here for getting into the action We got on the eight eight doing an eight box crazy eights Ladies and gentlemen Brad with that last spot mojo star next to his name If you got rooftops next to your name that means you won those spots in In the the pack break that we did just a few moments ago Alright, I think we got everything and we got some break credit that we're giving away at the end too So that's pretty nice that I upload that pack break. I did excellent So if you want to watch the video results there, you can All right, good luck Yeah, Cali. We were giving away national baseball card day packs. We've had a few people stop by already I don't know where they are We've been given them away Unfortunately this year's national baseball card day not as exciting because not many people out and about but all right five chrome autographs per box in the jumbo the the What should I call it the hobby has three Two or three, but jumbo has five So it's big. I'm gonna stack some of these up here to give myself a little bit of room Good luck. Ladies and gents I think we'll be closed by then Cali Sorry I'm sure we'll have some left over you can pick some up next time Baseball scores. What do we got here? I've been watching the last some NASCAR the last handful of laps have been a bit of a mess They red flag that they had everyone stop at a certain point Now it's crazy three laps to go Kevin Harvick in the lead Then I'll give you baseball scores. Yeah, Kevin Harvick is like like four car lengths ahead of Number nine is that chase Elliott? Wow, Denny Hamlin. What happened? Do you run out of gas? You've dropped down to like fifth or sixth. Yeah, there's chase Elliott Ryan Blaney's and third battling with keselowski Another wreck. Come on guys What happened? That's like the eighth caution in the last two laps The 14th caution here. Come on Who is this guy? Austin Dillon? Come on Austin Dillon. Who pushed you? Maybe some push him into the wall He's got some turf in his grill And another yellow flag out there. All right Here we go first box of chrome pick your jumbo pick your team number four vet commons don't ship But autos and you know, that's a refractor of Tim Anderson.  Hey, so so Lazzardo made a good start the other day Gavin Luxe don't sleep on Gavin Luxe. I know the shortened season what didn't quite make it on the major league team, but They're definitely planning for him to be there in the future nice negative refractor parallel For the tribe Carlo Santana Mark Bissette with the tribe Gavin Luxe will go to Greg who has my Dodgers Kyle Lewis goes to the M's Mark Bissette has the Mariners as well Maurizio Dubon one of the Giants big prospects for I Coppola and the Higantes my rivals that can Fordo will ship of course and what do we have here? We've got Kansas City. You've got Josh Stalmont blue autograph for the Royals Jarell Taylor with the Royals There you go JT 67 out of 150 on that one Trout Sean Jaspy pulled a nice trout super last night in that 1985 Throwback design that was pretty sharp over shed DJ. La Mayhew Jean Segura Hank Aaron die cut Dustin May and Tim Lopes Mark Bissette with the M's Future stars John means and Jorge Soler for Jarell Taylor green that's 16 out of 99 Matt Tice and Box one in the books Next box we're in overtime in NASCAR ladies and gentlemen now I can give you baseball scores as they're getting ready to restart this race Nationals up on the Orioles in DC to nothing bottom of the fifth and with runners on first and second In Philadelphia the Phillies are leading the braves for nothing in the top of the sixth scoreless in Texas bottom of the second zero zero In Kansas City Royals are up one nothing early on top of the second Another early early on Marlins up one nothing in New York Mets Reds up early three nothing on the Brewers and the bottom of the second in Milwaukee Any other live games going on? Red Sox and Blue Jays and Boston are scoreless top of the first bottom of the fifth raise up 5-1 on the Yankees Rockies merit Rockies at Mariners tonight that hasn't started dining backs at Padres Giants at my Dodgers early game Yankees bit the raise in the first of the doubleheader 8-4 It needs to be the White Sox 7-1 Tigers pounding on the Pirates 11-6 A's etched out the Astros in Oakland 3-1 And of course the Cubs Cardinals series postponed This is just gonna stack double headers on double headers at the end of the season Not sure what is happening with Miguel on Duvarr? Is he playing? Is he not playing? What's the deal? Yeah, Marlin still cruising COVID no problem. We'll just Let's get some guys together and just keep playing some ball We got Andres Muñoz That goes to the Padres that'll be for Icapola we've got speaking of the Marlins Robert Duggar For Miami that'll be for Brad M and the Marlins Nolan Ryan die cut we got Buster Posey negative refractor Aristides Aquino refractor and a Restart here for the NASCAR in Michigan three to go Doubleheader weekend to do another one of these tomorrow Aristides Aquino refractor for EA and the Reds the Buster Posey negative will go to Capola and the Giants and the Luis Robert will go to Matthew Shearer and the White Sox Jose Rodriguez is your angels autograph ASA who won the angels in that team. That's nice autograph Got Bobby Bradley rookie auto for the tribe mark Basette 377 out of 499 on that one news on Corey seager. He injured his lower back Which is a little concerning. There's Kyle Lewis for the Mariners Mark each row die cut And Joe Palumbo Rangers. Oh, is it gonna be Brad keselowski ladies and gentlemen? Or Harvick gets a little push push us forward two to go Rangers Matthew Stubblefield with Texas Now Harvick's like two car lengths ahead now He's got that clean air you guys white flag 41 out of 50 Peter Lonzo. They're really no Michigan drivers one from Michigan I don't think keselowski's catching him. That's checkered flag right there Harvick's got it Impress your friends folks Kevin Harvick the first driver from Michigan to win at Michigan according to the NBC Sportsnet Announcers here What's he driving today? Oh, he's driving a bush car today the old Hunt Brothers Pizza bush on the On the front there. There you go. All right now. I think I have I have some MBA on TNT Now that NASCAR love all this sports happening here my Lakers Playing the Pacers I know what kind of is it on my local network now direct TV search Search for the program on other channels. I Guess TNT for you unless you're in LA And then the Lakers network if you're in Los Angeles I'm gonna have a lot of baseball going on tonight, too. Lots of sports happening anything crazy and hockey today I feel like yesterday people were telling me about about a Lot of big upsets a couple big upsets. I feel like I think the Oilers got knocked out And then the maple leaves on an Austin Matthews overtime goal advanced That's getting crazy, too.  They got a big monitor set up and people can just drive out there and watch Lakers games Maybe for the playoffs, that'll be a lot of fun. All right LeBron LeBron 23 points TJ Warren 27 point TJ Warren on fire dust off your TJ Warren rookie cards. He'll be he'll be in a Sun's uniform Trent Grisham Padres, Ica Polo with the Friars gets the Trent Grisham rookie auto That's 22 out of 50 gold Kyle Lewis in that 1985 design Darth. What's up?  Yeah, you're welcome I think he might be first or second in the NL in home runs Trent Grisham And we got Aaron judge this guy in the AL has been raking Aaron judge Negative refractor Jeremy Port who won the Yankees in the team random That's right. Yeah, talent Horton Tucker, baby Take that three nails it Nails it nothing but the bottom of the net splash From talent Horton Tucker and there's Dom Nunez rookie auto for Travis Neve Yeah, I think NT3 is pretty close to if we don't want to go out of order nice Darth Join you we can rule the no, no, thanks You just bought a box of 13 for it. Did you buy a box from us? Nice. Thanks And we've got Hunter Harvey rookie auto for Jarell Taylor and the O's 459 out of 499 you got Ted Williams die cut Niko Horner for the Cubs 85 design going to Peter and the Cubbies and Willie Castro for the Tigers. That's gonna be for Brad last spot mojo We got Liam Hendricks blue shimmer 11 out of 75 for the A's That's gonna be for Matt Stubblefield. All right next box. Good luck boys and girls. Thank you Getting in We got more in the store. We got another jumbo case that we can do tonight We've got another hobby case we can do tonight. We can do both tonight if you want to Should be a lot of fun. Love this baseball just dropped yesterday Next box. We got Rogelio Armenteros for Justin gutting and Houston Out of 499 on that refractor autograph on Rivera die cut for the Yankees Nice Mike trout right there That was the super fracture version. We pulled yesterday. It was pretty amazing How many Zion one of ones have I pulled? I Don't know three Me personally three or four and we've got another Houston all right, Brian of bray you Brian I Think you need to get this a little Make better use of the canvas here Brian. It's got to work on that Astros Justin gutting maybe the offseason something something to work on the offseason for next year We've got Andrew Heaney angels to 299 Kyle Lewis 1985 throwback That's for the Ems Mark Bissette and the angels will get this ASA with the angels. He won that in a team read And we got for the Tigers Justin Verlander game use memorabilia I Think these are generally one per case. Are they not?  Oh, is that another relic in there? What's that? Oh? That gave it away right Toronto and he guesses on that I Think I think we know who it's gonna be. I do like that. They put these team names on there this This year, that's a thing. Yeah, it's gonna be Boba Shet Blue wave parallel. That's nice blue J's blue wave Nick de Kramer. I Was about to write Toronto on it, but I don't need to do that anymore Let's just force out it. So what's this down here? There's well There's Rogelio Armenteros again for the Astros Justin gutting Come on. All right behind Andres Muñoz is gonna be Nice Cody Belanger Two-color patch It's a much thicker card stock. What does that mean? Lower number perhaps. Oh Four out of five Nice a little Dodger Joe mojo going to Greg be Greg be in the belly nice Out of fives and under as you know get the train whistle all aboard the big head Express Dodgers will be playing the Giants tonight. We'll be watching that game together So two relics in the same box They're standing into the lineup die cut for the Dodgers freshman flash Kyle Lewis Gold wave gold shimmer Hunter Dozier nine out of fifty for the Royals that'll be for Jarrell and a raise autograph Colin posh Going to EA and the Rays No one impressed by that Belanger, huh? Fair enough. All right, we are halfway through this eight box jumbo case break We're about what we have 26 minutes here. All right, so probably another 25 30 minutes to go And then we will be able to move on to the next thing.  Maybe that Donner's optic choice. Someone said was sold out So yeah, we'll get all that on the schedule. Let's go pass it to Talon Horton Tucker LeBron He's open. He's at the top of the key Who's one didn't want that three Talon Horton Tucker wanted the three Give him the three We got Danny Mendic kind of spells out his name there instead of signing it. It's all good So that's his style Matthew Shearer with the White Sox and we got Logan Allen rookie autograph for the tribe That's gonna be Mark Bassett with the Cleveland Indians Mark McGuire Die cut got Walker Bueller blue to 150 for the Dodgers Louise Robert 1985 throwback Get there Dwight. There you go There's Louise Robert 1985 designed for Matthew Shearer and the White Sox Walker Bueller going to Greg And we got Jake Rogers autograph for Brad We've got another Indian coming up It's another Logan Allen this time a different parallel for Mark and the tribe That's to 250 on that one mark freshman slash Louise Robert Randy Johnson die cut Diamondbacks Gavin Lux Dodgers 1985 got a diamond back coming up. It's gonna be Zach Gallin Logan Allen to Zach Gallin Robert with the snakes There's Walker Bueller 88 out of 99 green shimmer for the Dodgers Once again, Greg with my Dodgers Got that nice Cody Belendr out of five a little bit earlier Three boxes to go. Good luck everyone. Is everyone alive out there? I haven't seen anyone chat. Is my chat broken if my chat might be broken. I need to refresh the chat Maybe people are lagging behind Okay, okay Just doing a chat check Eric Rose out there. There you go. So there People just kicking back and relaxing. It's all good Saturday a lot of team sports on There you go Kevin Crouch.  You want to see what happened in this break. All right Oh some red red shimmer And autograph Kyle Lewis What a break Is that out of five or out of ten? Mark Bissette with the Mariners two out of five Kyle Lewis has been raking so far this season. Wow, that's a big hit right there All aboard the big hit Express Nice nice Kyle Lewis are the Mariners playing today. Yeah Rockies are in Seattle Yeah, Kyle Lewis batting third in center field tonight It's 25 years old. He was a first-round pick in 2016 pick number 11 his 2020 stats In 56 at bats. He's hitting 375 an OPS of 980 Three home runs 10 RBI's in a stolen base Not too bad, right? Not too shabby Not too shabby at all. All of these Kyle loses will go to you mark and the M's There you go. Yeah, Eric. I heard I guess we we have we could build a rainbow from all of the From all these Kyle Lewis if we could if we collectively put all the Kyle Lewis side together This guy's been playing well too Couple youngsters delivering on the on the hype. I like that. That's all that always helps There's Alex Rodriguez for the Rangers die cut and John's and Hernandez Rookie Otto for the Rangers Matthew Stubblefield No, a Cinder guard to 299 Mets Gavin Luxe 1985 And we've been pulling a lot of heat out of these cases ladies and gentlemen if you want to get more Jasper's case breaks calm We got more jumbo and more hobby Whatever your heart desires And we got Zach Collins rookie Otto for Matthew shear and the White Sox Oh Seth Brown rookie Otto for the A's Matthew Stubblefield Johnny Bench die cut Yeah, probably Eric Everybody wants to fill that quickly Tommy Edmond 55 at a 99 green shimmer. I think we got a basketball mixer. Maybe a couple other things That should only take maybe couple hours or so and Then we can dive right back into another tops chrome case There's a bird alzalee cubs Peter alliance with the Cubby's Yeah, we do we Are trying to get a around-the-clock breaking operation happening. Yeah, Sean was here pretty early in the morning We've got to work on his efficiency a little bit I think the amount of time that he was on I probably would have been able to do double the amount of breaks But he's doing a great job. I mean he's only a couple weeks in yeah, he'll get he'll get faster He'll get faster Thankfully that's really the only thing that we got a that he has to work on that's just with repetition Just getting used to products like this and doing it at a different like he started on Instagram But uh, so like he's going box at a time So so I think that's just a different vibe when you have to do a group break Thanks, great. Oh, I've been doing this for five or six years Five years five and a half years. Maybe I've been doing this for a while Next box watch out Robert There you go and Jose Rodriguez blue shimmer autograph rookie angels ASA won that spot in a team random 22 out of 150 Racing Instagram breaks run. Oh, but the window is like literally the the viewing window are you saying? It's a little hard on the mobile device the kids love it though Another Luis Robert.  We've got it Babe Ruth.  There's Alex young Robert with the Diamondbacks.  What's up? Text. Oh, yeah, right that very important text. I sent you I remember that Freshman flash Gavin Luxe Catching them. Excellent. Good. I haven't had time to catch them all All right, Jake Rogers gold shimmer Yeah, male is weird. I mean ever since the pandemic hit and things were shut down and people were stuck at home There's such a huge spike in volume. This goes to Brad. By the way, nice gold shimmer. There was a Such a high volume of people just ordering stuff on Amazon and online and stuff like that that shipping times have just gotten all all wacky That's Jesus, Tanoko For the rocks that will be for Travis me Yeah, forget unpacking Jason. Wait, Jason. Do you do you have a do you have a location? Within reach from your apartment. There's alex rodriguez die cut for the Rangers that goes to Texas Aquino Tim Anderson prison refractor Tim Anderson regular regular Ronald Acuna junior Right there negative Which is awesome. I like Brandon McKay. What's he been doing this season? Is he even playing? I don't know, but I want I want it to be like the next show. You're tiny EA Nice one for the Bravo's Brian Duncan Gotcha. Gotcha. Gotcha. Yeah, check. You might be able to you might be able to spin a stop Right from right from your bed. It's convenient. All right, Tony Gonsolin Greg and the Dodgers he loves cats too. He can he can spin it pretty well and and he loves cats You're you're Tony Gonsolin trivia of the day. I Wish Jeremy port we all wish We all wish there was a super factor per case, but no, it's not it's far more rare than that I've been working on it Jason. I've been trying to get to level 40 last box ladies and gentlemen last five autographs. Good luck And then we'll go through some more orders See what else we can what other kinds of trouble we can get into Who raised asking if we all watched the Warren ball today Who's Warren ball Alonzo? TJ. Oh, I see. I see what you're saying.  So I was like, who's Warren ball? TJ Warren Fallen out.  He had dust off your TJ Warren cards folks He's been one of the darlings of the restart. He's been playing excellent basketball He's on the Pacers now playing my Lakers and But he'll you'll find his rookies in in his son's uniform Late bloomers. That's why you got to hang out to all those rookies. All right last box last five autos We got Domingo Leyba Robert with the Diamondbacks There's autograph one Ricky Henderson the man of steel die cut for the Oakland A's Louise Robert and for the Rays we got Michael brusso rookie autograph for EA and the Rays Second auto Three more to go. We got my miles amicolas to 299 Another Kyle Lewis right here. Nice for the M's Wait for the recap to see that Kyle Lewis autograph again. That was real nice freshman flash Gavin Lux Luxie the Luxter Oh, Talon Horton Tucker with ten points dust off your Talon Horton Tucker Rookie cards ladies and gentlemen if you've got stuck with the Lakers and you're hoping for LeBron's and ended up with Talon Horton Tucker's well now guess what? There you go Logan Webb rookie auto for the Giants I cupola with the Higantists Now you got a Brandon McKay refractor, right? I think we've seen a lot of bass, but there's a refractor for you yeah, and the fourth autograph of five is gonna be a cardinal junior Fernandez and That will be for Ryan s and the Cardinals he won that spot in that pack break in the team random And there you go. Oh Tyler hero has 12 points in 10 minutes Dust off your Tyler heroes.  I think I should have one more, right? Yeah. Oh Is that an autograph No, it's not Scott Kingery though orange Shimmer 11 out of 25 those that looks really sharp Phillies.  That was like a blue parallel made We'll see in the in the recap in just a mo and there you go Aristides Aquino at the end for the Reds EA will get all those Aquino rookies Including the ones I forgot to sleep. You'll get all of those man. There you have it and there you go Recap after we give away money Can't forget about that All right, so if you bought at least two team Let's flip back to the the product page right here if we if you bought at least two teams You'll get you'll get entries into you know two teams one entry for break credit four teams two entries so on and so forth We'll randomize just that group top five. We'll get 50 bucks a break credit each. So let's alphabetize your names Asa got to so that's an entry Brad got to that's an entry EA got to that's an entry Coppola you got to that's an entry Jarell you got to that's an entry John Wilkins with two that's an entry Garza you got to that's an entry Mark got to that's an entry stubble feel you got to that's an entry Ryan you you want to that's still it's still an entry still counts. All right So top five out so half this group Half the spots are gonna get 50 bucks each. That's a good deal. Let's roll it. Let's randomize it three times top five after three one two and Three easy There's the top five right there Garza Coppola Jarell stubble feel mark five of you 50 bucks of break credit each Congrats. Thanks everybody else for giving that a shot. We got more in the store jumbo and hobby Check it out. Jasper's case great calm. What a break. Look at this. What a break Bang some orange shimmer some Luis Robert some cat mojo Right there Some color in the autograph some Zach Collins a lot of rookie autos more Roberts that red Kyle Lewis two out of five Whoo, I got saved that for picture time. That was nice There was another train whistle you're too belly. Yeah Two color patch four out of five that was a beaut as well. Hey, what about a blue wave refractor parallel? Boba shed. That's not too bad a second rally like Justin Verlander. Man Good stuff. I feel like a decent spread of teams. That was a gold Grisham back there to Trent Grisham for the Padres another Padres auto Dubon Roberts Zooks There you go and a Domingo Leba right there and there you have it That was eight box jumbo case pick your team number four brand new release 2020 tops chrome baseball Just dropped yesterday more in the store jazby's case breaks calm. Thanks everyone. I'm Joe. I'll see you next time. Bye. Bye


Okay. So the next talk is random oracle reducibility and it's by Paul Becher and Mark Fishland. I hope I said your name correctly, but you can correct it. And the tech person, I thought you were on your way to me. I'm stuck here. Okay. Okay. They put a person who has no technical skills at the first chair. Okay. Paul? Paul? Yeah. Well, you're getting ready. I want to draw your attention to the fact that there is a monitor there with the time. It seemed like Evgeny was missing it. So you see it on the floor? Yeah. Okay.  Okay. Okay. So there you go. Okay. Thanks for the introduction. This work is joint work with Mark Fishland and is in a nutshell on relating the random oracles of two cryptographic schemes. But let's start at the beginning. So consider this situation. We have two cryptographic schemes, A and B. And these schemes come with a proof, as they should be, under certain assumptions. So for scheme A, we have some set of assumptions A. And for scheme B, we have possibly a different set of assumptions B. So if we assume furthermore that these schemes are for the same purpose, we may ask ourselves which one of this scheme is better. So for example, one may be more efficient than the other. Or the assumptions of the one scheme might be milder or even subset of the assumptions of the other scheme. And there are purpose-specific properties such as the ciphertext size, for example, if we're talking about an encryption scheme. So this is rather easy to assess in the standard model if we want to see which one of the schemes is better. Now, if we consider the very same situation, but this time these proofs come in the random-oracle model. So scheme A is proven secure under A in the random-oracle model, and scheme B is proven secure under B in the random-oracle model. It is now much less clear which one of the schemes is better. So the comparison is kind of biased by this random-oracle dependency. What do we mean by that? For example, we could have the case that scheme A, the assumptions of scheme A are a strict subset of the assumption of scheme B. But scheme A might use more of the power that the random-oracle gives it. So, for example, it might need that the hash function is perfectly one-way or collision-resistant or has some other properties which we don't know or cannot assess. On the other hand, scheme B could use this random-oracle to, could use much less power than we're given by this random-oracle. Or in the extreme case, the random-oracle of scheme A might even be uninstangible. So what's the solution to this? We could look at how the random-oracle is being used by isolating all the properties that are being used. But this is sometimes not clear how to do this, or it can be very tedious. Instead, what we propose here in this work is the standard cryptographers approach. That is, we want to use a reduction. And we want to be able to make statements such as if scheme A, given access to random-oracle H, is secure, then scheme B is secure using the very same random-oracle. So intuitively speaking, we want to say that any hash function that makes scheme A secure also makes scheme B secure. And from this it follows that if the random-oracle of scheme B is uninstangible then the random-oracle of scheme A is uninstangible as well. So you might have noticed this function H, this article T here, which we would like to view as a transformation function because the random-oracles of two schemes are not necessarily compatible in the sense that in the one scheme we may want to have bit strings of a certain length and in the other scheme we might work with some group elements. So we slot in here a transformation function that guarantees this structural compatibility. So what we get with this notion is a relative security notion. That is, we're not saying scheme A is secure but if scheme A is secure with this hash function then scheme B will be secure with this hash function as well. So let's look at this in a more formal way. We say that scheme A reduces to scheme B. If for every hash function there exists a transformation such that, and here's the first kind of reducibility we have, strictly if A is secure under some set of assumptions then B is secure using this very same hash function under a possibly different set B. This is very strong as we will see and therefore, so our G defines a security game. So for the sake of this talk we can think of this as say IND, CCA, security game for example. So this is the very strict notion. We also propose a WECA notion which says if scheme A is secure under A then scheme B is secure under A and B. So here we allow that, we allow that proving this we can assume that both sets of assumptions hold at the same time. And we also have a notion that is in between these two worlds that is if A is secure under A then B is secure under A union B. But B could also be secure for some other function assuming only B. So but maybe assuming some other assumptions for the hash function. So how does this work? Our main example is an encryption scheme. And we are looking here at the twin hashed algorithmal encryption scheme due to cache kilts and chub which is an extension of some other original hashed algorithmal encryption scheme. But it comes with Milo assumptions specifically to prove the twin hashed algorithmal encryption scheme secure is sufficient to assume the computational Diffie-Hellman assumption as opposed to the strong computational Diffie-Hellman assumption. And the security then follows if we furthermore assume that we have a symmetric encryption scheme that is CND, CCA, and secure. So this is an example where we have such a situation where one might be tempted to say that this other scheme is clearly better than the original scheme. And it turns out our example here is we look at a variant of this twin hashed algorithmal scheme and it turns out that this variant is in fact strongly reducible to the original scheme. Some details. The proof is done in two steps. So first to show that we have a weak reducibility from the twin scheme to the original scheme and then we prove that the twin scheme itself on its own is secure in a random oracle model by which we conclude that we also have a strong reducibility. So I won't go into the full details of these two schemes but I would like to draw your attention to the use of the random oracle which is being used to derive a symmetric key here from two group elements randomness and the public key to the y to randomness. And in the twin hashed algorithmal scheme we have three group elements. We then use only half of this key to encrypt the actual message. So here it is clear that we cannot use directly this hash function because we have here three group elements as opposed to two group elements. So this is where we need some kind of transformation function. And it turns out that this transformation function which takes triples of group elements and applies the hash function to the oracle here twice and concatenates is indeed a good transformation function for this. So let's have a look at the proof. So we assume we have an adversary against the twin the twin hashed algorithmal scheme and this adversary is going to ask hash queries random oracle queries and this will be triples. So we are going to construct an adversary A that attacks the original scheme and this adversary has to answer these queries and it does this by simply implementing the transformation function. So that is for a given triple ABC it will ask two queries to its own hash oracle AB and AC and simply concatenate the results just like this transformation function. A will also have to handle decryption queries. We have one public key from the original scheme. The other public key which is used in the twin hashed algorithmal scheme. This one we simulate ourselves. So we have the private key X1. We give the two public key parts to adversary B and B will ask some kind of decryption queries which are these. Since we simulate the one half of the key by ourselves which is this part, we can check if this is indeed, this is indeed a properly, proper safer text and we will do this by reusing the randomness that we obtain here and taking this to our private key. So if this is equal we will go on and ask our own decryption oracle about this randomness and the safer text and simply return the message. We return but if this is not equal here because then we can be sure that the safer text is not properly calculated. The challenge query is answered analogously. So we are given two messages M1 and M0. We forward these to our own encryption oracle, get back a safer text Y which is the randomness and C. We use again this randomness to calculate our part of the key and simply feed this back to the adversary B. Finally the attacker A outputs whatever B outputs and it turns out that this simulation is perfect. So adversary A is successful whenever adversary B is successful and by that we get that we have a weak reducibility from the twin hashed algorithm scheme to our variant of the scheme to the original scheme. The next step of the proof would be to show that this variant is also secure on its own which is a bit more technical so we will have to look at the paper for that and by these two results we then conclude that we also have a strong reducibility. We have some more applications of this notion. Specifically we have that variant of probabilistic RSA FDH signatures is reducible to GQ signatures. We have also that probabilistic RSA FDH signatures are reducible to PSS signatures and finally that Schnorr signatures are reduced to BLS signatures. These reductions are all strict and since this notion allows us to argue about un-instantiability, yeah this is very interesting for these schemes. Okay so I saved some time. I would like to thank the sponsors for the stipends and I would like to thank you for your attention. We do have time for a question. Okay so thank you again.


So, we are in the process of establishing classification of surfaces and last time we began with the canonical polygons and carried out two of the steps in the reduction process. One was to get pairs of type 1 which are adjacent eliminated. The second one was to see that there is only one vertex class. All the vertices on the canonical polygon should be identified at a single point. So, this we have achieved. Now, the next step, next there are three more steps I will say. So, step 3 here now to make any pair of edges of type 2 adjacent. So, that means that we are assuming there is a pair of adjacent type 2. There is a pair of type 2. It may not be adjacent. It is adjacent, there is nothing to do. If it is not adjacent, what is the meaning of that? There will be an edge A here and then there will be some sequence and then the edge A occurs again A followed by A and capital A followed by A and again another gap here. If B or A is empty, then A and this A will be adjacent. So, essentially you must assume that B and A are non-empty. So, this may be any length, this may be any length. So, what I do? I take the tip of first A, the tip of the second A. So, it is a convex polygon. So, you can join them by a straight line to take that one. So, name it as C and I am going to cut it along that. So, this polygon will be cut along this line. The top portion here is beta, alpha and beta. So, two parts. Beta is kept as it is. The top portion is turned around. If you have turned it around through 5.5, 4. So, this C which is directed like this is now turning around to come directed like this. That is not the object. The object is to bring this A which is downward arrow to match this A which is upward arrow. So, match them and identify. Identify these two areas. So, these parts of the polygon are identified now. To another thing, which may not be a convex polygon. That is not a problem plus. All that we need is the sequence and then we rewrite the convex polygon. That is what we do. So, what has happened? A has disappeared. A needs to have disappeared but a new edge has come which is of type C under C. So, this is the step 3. So, we have brought this together. Now, again we look at the rest of the sequence whether there is any type 2 which is not adjacent. Again we do the same thing. When we do that, this whole thing will be in one part either in A part or in B part of this one. So, whatever is I have, we have brought it that will not get disturbed. So, that is our point here. So, this way repeating this process again and again all the edges of type 2 will come adjacent. They may not be together. All of them may not, each of them will be adjacent. C C, then some gap, B B, some gap, A A, like that. So, that is the idea of this step 3 and I have explained how to do that. So, this is what we do. What is the corresponding relation code? The relation code is look at this one A capital A, A capital B. That is the sequence gets converted into, you can start from wherever you like C, C, B, A inverse, the other way around. This B is A, that is precisely what we have done. See A capital A, A, B is C, C, B, A inverse.  So, that is the inverse sequence. Exactly the opposite direction you have taken. Note that such an operation does not disturb some of the other pairs B, B, C, C, etc., which is already present. Therefore, repeated application of this will make all the pairs of type 2 adjacent. Of course, it may be necessary to perform step 1 again, namely a new A A inverse, the type 1 may might have appeared which are adjacent. You just do the same thing, process of folding it up and eliminating them. On the other hand, what happens is the number of vertex class will never be more. This is what you have to observe. Number of vertex class will never be more. Look at here, for example. So, all these vertices were in the same class, one single class. What I have done? I will cut it here. So, this will take end of this one and beginning of A, wherein maybe they may have gone into different class. It will not happen because this beginning of A is beginning of C. Beginning of C was the point here. So, that is not in a different class. Now, the other one is end of A here. They have come together already. So, only one vertex class will be there in the new thing also. Check that there is only one vertex class. So, that is what they have to satisfy. At this stage, so, stage step 3 is also done. So, let us take a look what is happening. At this stage, there may be no pair of adjacent of type 1 left. Then what is the meaning of this? Everything is A, A, B, B, C, C and so on. That means that we have the list 3, A1, A1, A2, A2, A3. This is the list 3. So, this is a canonical, this is a normal form. So, they are done. So, what is that, what is the other case left out? The other case is precisely there is a, there is at least one pair of type 1 and they are not adjacent. So, what to do with that one? That is the point. So, this is the step 4 handling a pair of edges of type 1. Suppose there is at least one pair of edges of type 1, say labeled A and then other one somewhere A inverse. Definitely, if it is adjacent, we have already eliminated it. So, it is not adjacent. But A, some capital A, then A inverse, some capital B and then that. So, A and B, A capital B may not be, should not be, this is what the meaning of this. So, P has the form as shown in figure 36. A something non-empty, A inverse and something non-empty. All right. If P or A is empty, then I can immediately eliminate it by step 1. So, that is not a point. So, we can, if no edge in A is identified any edge in P, now let me examine what happens. So, there must be some identification. Each edge here is identified some other edge. But if suppose edges inside A are identified within themselves, namely inside A itself, B inside B itself, then what happens? Look at this endpoint here, the endpoint here. That goes to one single class which is coming from A. This endpoint, the beginning and beginning of A, they will be in a different class because A, when you identify A, they are not getting identified. They will be in a coming out of B. There is no identification from A to B. That means that there will be two vertex classes. That is not allowed now because we have already made one vertex class. Therefore, some edge here must have identified some edge here which will make finally all the vertex class, single vertex class, all the vertex is being identified. So, one of the edges here must be identified one of the edges here. Therefore, what we get is a better idea namely this figure one here, A, A find inside this upper part there will be some B which will be identified with a B here. So, A, B little B, capital B, capital A inverse this one is C and this is again B and B inverse. B is this way, B inverse this way, D and A. It cannot be B because B and B are all adjacent. So, A, A, B, B both of them are pairs of type one. Such a thing is called interlocked pair of type one. Interlocked means what they are not together. They are A to A that is something in between B and B something in between A something in between and so on. If these are empty then there is nothing to prove. So, you better assume they are non-empty otherwise whatever you are doing is already somewhat useless thing whatever it is already achieved what you want to do. So, let us see what we want to do now. So, this time so cut top of A which is top of two line from top of A to top of A. Earlier we did top of A to top here all time right top of A, top of A cut off, cut off this part all right. Whatever remains is drawn here is alpha this beta will be now brought back so that B and B are getting identified. You just slide it down and put it here identify the edges B and B. The new sequence will be what now we just read from wherever you take. So, it starts B and then C is the new edge then A. So, this B is inside now. So, this B you pick up from here D, A this is a new edge C, A and then C that is a new sequence. You redraw it as a convex polygon here in diagram 3. Same thing I have C I have now read A to D this is a this is a black box right we do not know what sequence it is some a sequence nonempty sequence. So, you write it as A to D A D. Okay this is A C A this will be what this is C B okay now we have this C and C which is again type 1 C C inverse right.  So, new phenomena here what it is C okay C D inverse C inverse A D C B D okay. So, you can redraw it and then you see that D C D inverse C inverse. So, this whole thing is now together and then it is really something D C D inverse C inverse okay. So, there is no in between things here now. So, this is the best thing you can do of course you cannot bring them together D D inverse will not come together there will be C D C D inverse C inverse. So, this was precisely the sequence for the torres. So, this portion of the surface will give you a torres and then this will be connected some with something more. So, we will transform the polygon so that the two interlocker pairs of edges become consecutive okay. So, the corresponding thing is starting with A A B B A inverse C B inverse D D finally got turned into this one A D B A D C B something D C D inverse C inverse. You can call it as X Y X inverse Y inverse if you want what label you do is immaterial okay. Now, look at A D C B do the same operation on this part by calling this something nice already you will never this never be disturbed. If there is a pair here which is not adjacent you perform the same operation here okay. So, keep doing this all the interlocker systems of type one they will come together okay. So, this is what we have by repeated application of this get the canonical polygon of the form A 1 B 1 A 1 inverse B 1 inverse A 2 B 2 A 2 inverse B 2 inverse and so on. And then all the other ones are type 2 C 1 C 1 C 2 C 2 C 3 C 3 C M C N if either M is 0 or N is 0 then M is 0 means it is type 3 list 3 and N is 0 means list 2 okay. We had in the classification we had three of them first list was just the A A inverse which corresponded to the sphere 2 S okay. Then list 2 was only this kind of things up to A A inverse B A inverse the list 3 was this one. This is combination this is neither in it is not in the list. So, this is not normal so we have to do one more hat trick here to make it a normal sequence okay. So, that will be the last step to be handling the case of mixed type unmixed type means what either only A A inverses are there or C 1 C 1 C 2 C 3 these are type one these are type two. If both of them are there what to do okay. So, now what we do is we concentrate on the middle portion is here A M B M A inverse B A inverse C 1 C 2 one of the type one pair here and the other one is type two of them together okay you have take you cannot take just one of them at the other one is type 2 pair. So, how what to do with this okay. You will convert this one into entirely type 2 okay that is the whole idea. Combining results of the two cut and paste done in figure we obtain the following relation okay. So, what I am calling is A here up to here A M minus 1 B M minus 1 A M inverse A minus 1 inverse B M minus 1 inverse you call it capital A then you take this part call it as A B A inverse B inverse then take the C 1 C 1 C 1 C 1 call it as just C C. So, this portion is getting converted into A into x y x z y z okay. So, let us see how this gets. So, this is what is A B A inverse B inverse okay then C C and then a big gap which I have not written down anything this capital A you can write as capital A here A B A inverse B inverse C okay look at the tip of A okay to tip of C. So, you have to make this guess where to cut and so on this is not at all obvious okay. So, there are two of them you have to break this C at this point. So, you must be really worried because you have brought this one with so much of effort now you are asking cut it okay. So, you have to do that cut it here okay and bring this part beta part is kept as it is bring this one on this side so that this A and this A are together okay this part of A and this part of these two things together identify them. So, this A disappears what is the new thing that appears this D will appear now okay D is this way and you have turned it around. So, this will become downward here okay D inverse B D C then this capital A here which was which was blank thing here okay after that what you have this part here now B inverse and C ending with D here okay rewrite this one okay just like this way now one more cut you have to do look at the starting part of B to the end of C okay B D C are there already cut it along this this is E now cutting along E okay this portion is brought back here so that the C here will match C here. So, you are actually destroying the old C the old A but you are getting some new of them okay. So, what you get is now E D inverse E again some gap okay then this is what this is B inverse D inverse B inverse okay you better start from here and keep reading and then capital A at the last okay D inverse D inverse B inverse okay then E D inverse E all right. So, we can now apply three successively to make this into A X X Y Y Z Z finally what we have what is this process we have got A into X Y X Z Y Z.  So, this is this is not what we wanted okay it should be X X Y Y and Z but that you can always do by step three right whenever you have X Y X so this X and X can be brought together okay and anyway there are X X Y Y Z Z three of them all the three of them can be brought together X X Y Y Z Z Y step three okay so you have to repeat step three here again to make it what we want. So, what has happened finally is the following you started with one one A B A B inverse of this step and one of this one okay rest of this part and this part you could have taken all this part in single here and assumed that this is the at the end by cyclically permitting okay then that all that A remains as it is and this part becomes X X Y Y Z Z so it becomes three pairs of type two okay so one plus one here one type of this one one point of this one we do three pairs of type two whatever is remaining you repeat the same process again on that all right each time you see there are plenty of type two things to mix with that one till all the these AIs and BIs have disappeared one by one if there is only one by now it would have become one here one here it would have become three C C three C one seven C two C two three three there there are two of them again again three more okay each time with mixing which is one this will convert into two C two of the sequence okay by repeating this process so in the middle you keep changing that is what I have told in middle this will be replaced by X Y X Y Y Z Z and rewrite the sequence this part again take the next one and you will make it X X Y Y Z Z and rewrite okay finally you will be like with X 1 X 1 X 2 X K Y X K X K repeated okay this will be the sequence okay actually this K will be precisely equal to whatever n was there each of this m will be get into two of them so two n plus n okay so this completes the proof of part A of the theorem 6.48 starting with a polygon the first case is A A inverse which is S 2 we do not want to do anything that is it is declared as normal otherwise what should happen it will be of the of the future two let me just go back the list two is is like a repeated torus A1 B1 A1 inverse B1 A1 A1 A1 A1 A2 A2 etc projective space repeated projective space okay the second part is A2 distinct members of this list Q surfaces which are non-homomorphic so that is the part that will okay all the surfaces you can see here is one part which is already a big thing now the list is not a redundant list there is no unnecessary elements you know duplication it is not there this is what you have to assure so that part we will prove using fundamental group it will give a very easy proof then just to justify all these homologies which will use a simple homology and give a proof we shall also give a proof which will say cellular homology okay so today we will stop here thank you