1 00:00:24,640 --> 00:00:27,029 let's get back to space complexity um so what uh just to review what we've been doing um from last lecture which feels like a long time ago but it was only two days ago um we were looking at um those three theorems which all had this basically the same proof one was the latter about the latter problem of uh where you're trying to see if you can get from one string to another string changing one symbol at a time um and all of the strings were in the language of some finite automaton or you had some reasonable rule for saying which of the valid strings which are not and then uh we built on that idea we showed savage's theorem that going from any non-deterministic machine to a deterministic machine only squares the amount of space and then um we finally uh proved that tqbf uh this problem about quantified boolean formulas testing whether they are true uh that that problem is complete for p space um so we're going to build off of that last last theorem today and talk about uh one other complete problem um and also show a connection between p space and testing uh which side uh can win in certain kinds of games um so um and then the at the end of the lecture we will the second half of the lecture we'll talk about diving deeper into space complexity we're going to talk about a different part of the parameter space where instead of looking at polynomial space we're going to talk about what happens if you have logarithmic space which is another natural point to look at for from for space complexity but we'll get to that after the after the break okay um uh talk about game complexity and games um so when i was little uh we used to play me and my sisters would play a game called geography i don't know if how many of you have heard it or maybe it go goes under a different name um but uh it was a game about basically about words and places and it has a very simple um set of rules um let's say there were two people you take turns picking names of places and then each person has to pick a place which starts with the letter that the previous person's place ended with for example you know you have two people playing um perhaps you have you know you agree on some starting place like a major city nearby like boston um so one you know one player the first player takes boston and then after that the next player has to pick a place that starts with n because boston ends with n and so maybe nebraska would be a possible response to boston and then the first player would have to respond to that with a place that starts with an a because nebraska ends with an a it's like maybe alaska uh which also starts with a ends with a so then arkansas maybe would be a nat reasonable response to that and then that ends with s um so uh san francisco and then so on and so on um and uh of course you want to forbid people from reusing names because then you'll just get into a loop you know saying alaska alaska alaska won't make the game very interesting so you have to forbid that possibility names cannot be reused they can be used at most once and then um so eventually one side or the other is going to run out of names uh and not have a either because of the lack of their geographical knowledge or maybe you've just exhausted all the possibilities and i will will not have a response and that person will be stuck and that person is the loser um the other person has won the game so the objective is to try to avoid getting stuck um so let's just take a look um i'm going to think about that games in a somewhat slightly more abstracted formalized way um so you write down all of the legitimate places um they become nodes of a graph such as i've shown here and then you draw an edge from one node to another if um that if that corresponds to a legal move in the game a legal step of the game so you know boston connects to places that start within like new york and nebraska um so the first person might pick boston the second person might pick either of these two um and then back to the first person if they pick nebraska they could pick arkansas they can pick alaska and so on just as i described believe it or not we actually played this game i can remember playing this game when i was a kid that was of course before we had um uh you know league of legends and other fun fun stuff but this is what kids used to do uh so anyway um um so just to get it down on the slide the rules are we'll assume we have two players we'll call them player one and player two player one goes first um and they take turns picking places that start with the letter which ended the previous place not no repeats allowed and the first person players the first player stuck loses you know the person who doesn't have a move to make so what we're going to look at um today is a generalized version of that where we're kind of um just going to take away the names but just leave the underlying graph and we're going to call that generalized geography so here um that can be played on any directed graph you take away the names of the edges the names of the nodes now you just have a some arbitrary graph um you're going to designate a particular node as the starting node and then the players take turns following those edges um and uh what uh you're because you're gonna forbid having any loops reusing any um any nodes uh what you end up doing is working together you're going to be constructing a simple path in that graph simple path just follows those edges and never intersects itself and the first person the first player to be stuck is the loser so the other one is the winner okay so that's we're going to call that generalized geography and we're going to look at the computational complexity of deciding for a given graph and starting point which side would win if both uh players played optimally as best uh the best possible okay and we'll name that problem gg or look at its associated language so here is a graph and the starting node and you want to say um that pair is in the language gigi if the first player player one who must play at that node a um to start off with has a forced win um uh in that general geography game on that graph starting at a okay where again what i mean by a force when i'm not going to define this formally that we could do that it's not hard to do but a forced wind also it's also called a winning strategy um that just means that if both sides play optimal they pay the best possible not up their capability i mean the absolutely the best possible play then that player one would still win there's nothing that player two can do to prevent player one from winning um if player one has a winning strategy or a forced win um you may have seen for example uh there are examples of this that i mean i don't even know if they still exist but it used to be in the newspaper they used to be examples of chess boards and they would say you know if they would start from a certain position and they would say white to win in three and so that means white has a forced win no matter what black does white is going to end up winning but you have to think of what the strategy is it may not be so obvious to think through what are the right moves that what makes but the point is that no matter what black would do white would end up winning um and what you can show we're not going to do that but for a class of games that includes this generalized geography either one side or the other is going to be guaranteed to have a forced win that's not necessarily true for all possible games obviously but for a large class of games um in one side or the other is guaranteed to have a winning strategy or forced win okay so let's just review that because this is going to be essential for you know the first half of the lecture to understand what we mean by a game and what we mean by one side or the other to have a force to win okay so let's let's just uh oh yeah and what we're going to show is that gg is piece based complete that the problem of giving one of these graphs and a starting point figuring out does player 1 have a force win or is it player 2 that has a force win um which side is guaranteed to win under optimal play that's a piece based complete problem um and so let's do a little checking on that and so i'm going to give you a very small example of the generalized geography game you have to figure out whether it's player one or player two that has the forced win the winning strategy or maybe neither or both those are the four options okay and you understand player one has to play here because that's the starting place of this general geography game so then it's up to player two to continue on all right so let's um launch that poll uh you'll have to think about it a little bit um and uh let me know what you think which side has a forced win which player makes the first move it's player one makes the first move and player one plays as i showed you here this is player one playing at a so player one picks a picks and then player two has to pick one of these two okay all right i think we're gonna end the poll everybody's in pick something um you you guys did not do well on this bowl uh at least not too many of you picked d that's that's true that's a that's a little reassuring because you can't have both players having a forced win i did say that in games of this kind one side or the other is going to have a force win so c is not a very good choice either um so you're maybe you should spend less time reading your email and more time listening to the lecture uh um the actual correct answer is b player two um which is obviously a is a minority position here player two has the forced win let's understand why um so player one plays here as i said that's the b that's the first move player two can either take the upper node or the lower note okay player two takes the upper note then player one has no choice but to take the rightmost node and now there's no move for player two uh so player two will lose um if player two takes the upper note the upper choice there because player two will go here player one will go there and it'll be game over and player one will have one because player two was stuck however player two had another choice um uh player two had another choice um player two could have also gone down here um so player two uh if it played down here um things are looking a lot better for player two because player two goes here player one's move is forced goes here but now there's another move left uh that player two could make so play player so two goes here one goes there player two goes up there and now player one is stuck so at player two's choice um player two can make a move which will end up causing player one to get stuck and so because it you know that's the nature of what we mean by uh having a winning strategy there is some way to guarantee if you're playing optimally and you're certainly playing optimally mean you'll take that lower move uh then you will take that so under uh optimal play player two will win this game and so player two has a winning strategy it has a forced win okay um so i i see so i'm not going to define playing optimally i'm going to you know there's some questions here about that about that that's um you know you can think about i'm going to i'm going to leave it um you know somewhat informal um but we could make that all precise i i don't think it would be clarifying to to go through a precise definition of that um uh you know you you can just think you we don't even have to think about it in terms of optimal play you you can say that there is um a uh forever uh a um you say a player has a force win if uh they have moves that they could make in response to every possible move the opponent could make uh that'll guarantee that they will end up winning so it doesn't say we don't even have to talk about optimality here just talk about every possible opponent every possible opponent is going to lose to somebody who um uh you know has a has a winning strategy and pl and follows that strategy okay um okay i can see there was some confusion about who was going first and so on i'm glad we that was part of the reason i gave this check and just so hopefully to clear up some of those points um so player one plays here then player two uh goes and i got some comments that some people were confused about what does it mean for the first uh where does the first person move okay hopefully that's that's uh clear all right so let's move on because we're gonna make we're gonna be spending a while um proving something about generalized geography and these graphs um okay so let's let's continue so in order to make a connection between um games and complexity we're going to need to talk about a problem one that we've actually related to one that we've already seen where we can reformulate that problem as a game and that's going to be a problem on quantified boolean formulas called the formula game okay so this is a different game now um not one that you're likely to end up playing but in in in in a sense in which you'll see this is a still a reasonable game where two players can play against one another and one of them is going to end up winning and one of them is going to end up having uh one or the other will have a winning strategy so let's understand what the game is uh the game is played the game is played on a formula so you so you write down a quantified boolean formula remember we talked about that you know for several lectures now uh you know it's a bunch of quantum variables with quantifiers and um then there's the part that has no quantifiers um which you know uh you can always put into conjunctive normal form if you want but for the purposes of this discussion so far it doesn't matter later on we'll actually want it to be in quant in in cnf but okay but but for now we just have some formula [Music] and associated to that formula there is a game in a very natural sense so first of all let me they're going to be two players but now the names of those players are going to be player exist and player for all okay and the way the way this is how the game goes the uh the player exists but both players are going to be picking values for variables they're going to say you know variable x1 is true variable x2 is false variable x3 is false variable x4 is true and and they're going to be that that's how the moves of the game go they're just assigning values to variables so that in the end all of the variables have been assigned a value a boolean value and so we end up with an assignment but before we get to that stage which players get to pick which variables and the way it works is the exist player is going to assign values to the exist variables so exist we'll pick x1 and x3 and assign them val values the for all player is going to pick the very ver assigned values to the variables that have referral quantifier attached so in this case would be x2 would be um the the the for all player will pick the value for x the x2 variable um fading out here okay now um so and the order of play is according to the order of the appearance of the quantifiers in the formula so the way i have it written out in this particular formula it's exist x1 so exist will pick the value for x1 then it's for all's turn for all pick the value of x2 then it's going to be exist picking the value for x3 and and so on until they get to the end and all of the variables have been assigned a value by one side or the other and now the game is over okay so what's left is to understand who has won at that point um and the way we say who has won is exist wins if in the end this part psi which is the part without quantifiers ended up being satisfied by those um by that assignment that the two players together ended up picking and then for all will win if that part of the formula is not satisfied so let me just write that down here after the variables have been assigned values we determine the winner exist wins if the assignment that they co-op together built they built cooperatively together uh satisfied psy and otherwise for all ones if it doesn't satisfy psi okay so you kind of think about it this way you know exist is picking the values of those variables he's trying to make this uh part of the formula satisfied trying to satisfy this part of the formula for all is picking values but she's trying to make this part of the formula unsatisfied so they're you know they're uh in opposition to one another and um in the end one of them is going to have succeeded and the other one will have failed uh and now thinking about it as a game one side depending upon the formula one side or the other is going to have the ability to dominate the other one and always win um and so the question is which side has that um ability to always win which side has the winning strategy of course it'll depend upon the formula some formulas it'll be the exist player who has the winning strategy and other formulas it'll be the for all player who has the winning strategy and computationally we'd like to make that determination which side is going to win um you know for a given formula and there's a very very nice and actually very easy um and simple way to understand that because we've already run into that problem before the problem of determining which side has a force force win is exactly the same thing as determining whether this quantified boolean formula is true because the exist player has a force win when that formula game on that um on that formula um let me say it again the exist player has a force win when that formula is true um so if this quantified boolean formula were a true quantified bullying formula then exist is guaranteed to be able to win that game if it if it plays its hand right if this is a false formula for all the for all player will always be able to win um if it plays its hand right okay so um why is that and it really follows kind of almost you know without doing any work at all this is kind of a you know this proof even though it looks superficially like it might be not so easy to prove it follows for a very simple reason because the meaning of what it means for exist to have a winning strategy is exactly captured by the semantics of the quantifiers let's see what that means what does it mean that exists has a force to win let's say up here that means that well exist has a move that it could make that exists can pick some value for x1 so there was some move for x for there was some way to assign x1 so that no matter what the opponent does no matter what for all does there's going to be a response that it exists can make um some some assignment to it x3 um uh so that no matter what for all does there's going to be an assignment to the next variable um and so on and so on so that this thing ends up being true so let me just say that again somehow i didn't feel it came out very clear uh so exist has a winning strategy exactly when there was some assignment to x1 that exists can make such that no matter what assignment that for all makes to x2 there is some assignment to x3 um such that no matter what assignment that um for all makes to x4 and so on makes this thing true that's what it means for exist to have a winning strategy that's exactly what the quantifiers are saying in the first place so even really without doing anything you can see that exists has a winning strategy means exactly the same thing as the formula the quantified being formula being true so making the test of whether uh which side has a winning strategy is the same as testing whether the formula is true okay um [Music] so let's let me try to turn to the chat now um before before we move on to anything else um all right uh whoops uh here we go [Music] okay so i'm getting some questions about the order of play and about the alternation of the quantifiers here so the way i'm showing it is that the quantifiers always alternate exist for all exists for all that doesn't really matter you know you could have several exists in a row and that would correspond to exist making several moves in a row before turning it over to parole but alternatively we could just add extra variables which don't play any role in psi and um they're just kind of dummy variables that serve as spacers so if you have two exists in a row and you don't like that you can just add add a for all of some variable that you never see again in between so you can always get it to be alternating if you want so um um also question the difference between pha fee uh and psi so psi is the part here that does not have any quantifiers so the way we always write down our qbfs is their leading quantifiers with the variables which uh you know and all variables have to be within the scope of one of the quantifiers of quantified variables um so you know there are no free variables that are unquantified in our qbfs um and so there's a you know all of these qbfs have a run of quantifiers and then a part without quantifiers where there's this sort of boolean logic part i'm psi here is the boolean logic part and phi is the whole thing together with the quantifiers um so i'm not sure i understand some of these ques why not have the for all player have a force to win the for all player might have a full force win no see one side or the other is guaranteed to have a forced win so if exist does not have a force to win then then parole will have a forced win that you know i'm not going to prove that but that's a fairly simple proof by induction which i'm not going to go through just take it as a fact somebody's asking why does for all want to not satisfy the uh this expression the the psy part well i mean that's the way we set up the game um you know in order to make this correspond to tqbf um just the truth of the of this of this expression we want to make exist try to satisfy the the this part and for all try to unsatisfy make it not satisfied um and so i'm not sure because that's what works i'm not sure what else whatever whether why i can answer there okay somebody i'm going to answer some of the you know most basic questions here somebody's asking how do we know how many variables to use so the the variables are the variables of the formula so somebody's going to hand you a formula that's going to have x1 to x10 whatever that formula has some number of variables and so that game is going to have 10 moves you know if the quantifiers alternate then the moves will alternate but whatever the pattern of the quantifiers is going to be the pattern of the moves so exist is going to follow the um you know the places where the exist quantifier occurs and fraud is going to follow the places where the for all quantifier occurs so if you have you know exist for all exist for all exist will move first then for all move then exist will move then for all will move together they're picking values to their respective variables with with their opposing goals existed trying to make the um the psi part satisfied for all is trying to avoid making it satisfied to make to to make make the assignment not satisfy that part and if you think about what that means just in terms of the meaning of that that it's going to mean exactly the same thing that the formula is true um so anyway if you don't let's see i don't know if there's any more questions that i can try to answer here um well so it's still about the numbers of variables the um each formula is going to define a different game you see the formula so the formula however big it may have 50 variables that may have two variables but we'll do an example maybe that'll help um the next check-in which is coming up for pretty soon might actually be now uh we'll give you an example um so then we'll see who's who's understanding and who's not okay so let's continue here um so so therefore the problem of determining does exist have a force win is exactly the tqbf problem because exist has a force when exactly when the formula is true and what we're going to show is that t qbf is a polynomial time reducible to generalized geography but it's conceptually we're going to think about tqbf now as a game because that's how geography is generally geography is set up so we're going to given a formula like this we're going to construct an instance of generalized geography we're going to construct a graph where play on that graph is going to mimic play in the formula game so making the moves in that graph are going to correspond to setting variables true and false because the graph is going to be specially designed to to force that behavior okay so i think we have a check-in coming in yeah so let's just see how um i i suggest you pay attention to this check-in and really try to think about it and solve it because i think that'll help you make the connection between the truth of the formula and exist having a winning strategy okay so so if you take this formula here here's you know this is some particular formula exist x for all y blah blah blah um this is going to be associated to that formula a formula game where first exist moves assigns a value for x and depending upon that value for all is going to move and they're going to assign a value for y and after that's done this part here is either going to be true or false it's either going to be satisfied or unsatisfied and i want to know can exist always find a way can it does to to guarantee that this part is going to be satisfied or can for all always get find a way to guarantee that this part is not satisfied okay so it's you have to stare you have to look at this uh formula to understand which side is going to end up succeeding um so you understand you have to if you're not clear on the rules look back here exist is trying to make this part satisfied for all is trying to make this part unsatisfied but they neither of them has the no the totally upper hand because existence depicting one of the variables for all is picking the other variable but they're doing it in a certain order okay first exist is going to pick then a for all is going to pick okay so that's the way the game works exist picks a value then for all picks a value and then you see who wins so who wins who's guaranteed to win um you know can can we make sure exist always wins or can we make sure that it for all always wins for this particular formula so there's just two variables here and you can think about this in either of two ways strictly speaking purely as a game or you can look at understand whether that formula is true or not um they're equivalent ways of looking at at the problem in fact in a certain sense it's the same that's what i'm trying to trying to get across thinking about this as a game or thinking about in terms of quantifiers doesn't make any difference um okay almost done here so all right last call closing the poll okay i think this one you got you you you did pretty well on so yeah the correct answer is the for all player has a uh winning strategy has a forced win uh and um also similarly the expression um is fee is is false because you know if you try to find some x such that no matter what y you pick um this is going to be true this is going to be satisfied you're out of luck because you know if you make x true well um then if you make x true now i'm confused uh what is going on if you make x true then uh y oh if you make x true then uh x bar is false um so y bar has to be true uh um now i'm completely confused oh no if make x true it has to work for all y so this thing is clearly false if you make x true um uh then then then this x bar is going to be false and it's not the case that for every y of the setting for y because y is forced here y is going to have to be true well y bar is going to have to be true so y is going to have to be false somebody if you make it very clearly but if you make if you and now if you if you try again to make x false um maybe that's a one to think about first if you make x false so you're going to be stuck on the first clause because you have it has to work for both settings of y but maybe thinking about it as a game is better you know if you no matter what x does a y is going to find have a way of making one of those two clauses unsatisfied so if x is true why can uh uh can set itself like why we can set y to make the second clause uh unsatisfied and if x is false we can find y can be assigned to make the first clause unsatisfied um so anyway um the right answer is uh the for all player uh has a winning strategy here in the end of the day it's not critical that you understand this correspondence but but you have to then if you don't quite understand that you're going to have to take on faith that the formula game is the same as tqbf because that's we're going to use and by the way i would like to also mention here that um that this correspondence between uh games and quantifiers is some things that mathematicians use all the time um because if you have a if you have some expression which comes up often in mathematics where there are you know a whole run of quantifiers you know in front of some uh some statement it's really kind of hard for anybody uh to really uh get a feeling for what that means if you have like six alternating quantifiers which happens often in mathematics you're going to have statements that have a lot of alternations with quantifiers very hard to get a feeling for what that means but if you think about it as a game it's much more intuitive and it's completely equivalent to think about quantifiers go back and forth between quantifiers and games anyway um so let's look at the at the reduction the the construction that shows uh generalized geography is uh complete took a little longer than i thought maybe this is i just want to make sure everybody's with me on this so why don't we call the break now and then we'll come back um and we'll look at the construction um after that because the construction itself is going to take about 10 minutes to to work through to figure out how to reduce tqbf and build a graph that simulates the formula okay so i'm going to move right on to uh the um and we'll come back okay so feel free to ask some questions um and uh get ready for us diving in to a construction uh for the um uh for the reduction of tqbf to generalize geography so there's kind of an interesting question about you know what's a relationship you know between a formula and when you swap the order of the quantifiers um uh you know the the the the question is asking does that somehow relate to the negating the um the you know the part the quantifier free part the part without the quantifiers and i um i don't think that that's the right correspondence there it actually is uh some relationship when you say um you know there exists for all is actually a stronger statement than than saying that for all there exists in general the um that actually implies you know exist x for a why is this implies uh for a why there exists an x um because what you're saying is that there is the the choice of y if there exists an x uh for all y there was one x that works for every one of the y's if you're saying for y there exists an x the choice of that x can depend on the y um so there there is a connection but um you have to think through what it means in order to understand that connection not i think you won't need to know that in order to process what we're going to be doing but maybe it just helps you think about it um other questions here okay i think we're um we're out of time here so let's return to our our lecture um so i'm gonna go back hopefully that won't crash everything [Music] here we go okay so now we're going to be reducing tqbf to generalize geography are we all together now um uh okay so i'm gonna illustrate this construction which is very nice construction by the way i'm gonna illustrate this construction just by doing an example or a partial example okay but i think it'll give you the idea so you understand so what i'm trying to do here is i'm starting off with a quantified boolean formula so here it is um i'm going to assume it's in conjunctive normal form which i can always convert it into that form uh maybe by adding some additional variables but without doing anything too drastic to it and so now what i'm gonna uh starting from that formula i'm gonna now build a graph we're playing the geography game on that graph which is remember taking turns picking nodes which may form a path is going to correspond to um playing the formula game which is picking the variables of that formula and then you want uh the if the exist player wins in the formula then player one is supposed to win in the geography game okay um so here is how the graph is going to look okay so good to follow try to follow this um so it's going to be kind of there's going to be kind of two parts one that's going to correspond to the variables and the other part that's going to correspond to the clauses um and so for each variable i'm going to have like a diamond structure here so there's a little starting thing that's going to be unique to the to the first variable but then every variable is going to have like a little diamond structure here um and they're going to be attached one to the next and we'll to understand how this what this means when we play geography on this piece of the graph so but let's just understand the structure first okay so this is the start node and now um here i'm player one is going to play the role of exist player two is going to play the role of parole and in fact i'm going to identify that so i'm going to call because it's going to be helpful just to have uh helpful just to think about player 1 is being the exist player so the exist player is going to be playing on this graph the exist and the for all players are going to be playing on this graph it's really just players one and two so exist the exist player player one has to b by the rules has to pick the start note and now with player two's move these the four players move now if you're with me on this graph the for all for the for all players move um is now not very interesting because it's forced it has to go to here because again they're just picking the um the nodes of a simple path in this graph so for the fraud player to the for all player has to go here is if exist started over there now um now it's now it's the exist player's turn and now now something interesting happens the exist player has a choice you can either go left or right there are two possibilities and then after that the four player's turn who's again just forced so so far for all has not had much interesting stuff to do in this hasn't has not had to have has not had to think hard um so for all players no matter whether exists went left or right the for all players move is forced and ends up over here and now let's see if this player's turn again this is kind of an easy one oh so before i do that let's just look at how that play could go so the first two moves that's kind of i was suggesting exist goes here and then for all goes there i'm gonna i'm gonna illustrate kind of possible plays through this graph by tracing them out in green um so now exists goes here for all goes there those those were forced but now the exist player goes could either go this way or could go that way so those are two possible different ways of playing this the game that we said we're building um and now uh now whose turn is it here this was the for all player pick this one so the the the next turn is the exist player and that's forced but now notice now for the first time the for all player has to think because the for all player can either go left or right so could either go left or it could go right um and so then there's going to be a sequence of these diamond structures where kind of they're constructed so that alternately either the exist has a choice or the for all player has a choice until you get all the way down to the bottom okay that is the graph we've built so far now if we stopped here um then whoever ended up at this point whether it was existing for all would be the winner because there's nowhere for the other for the opponent to go of course that wouldn't be very interesting so there's going to be more stuff we're going to add on which are going to correspond to the clauses now how to think about what's happened so far now so maybe as some of you might be guessing is that um when the exist player over here has a choice could have either gone left or right that's going to correspond because this is supposed to be mimicking the formula game that's going to correspond to the uh exist player in the formula picking the variable either true or false so that the left or right is going to correspond to true or false and so um let's just arbitrarily say left is going to correspond to picking it true and right is going to correspond to picking false okay so that so the way i've said it so far is exist ended up picking the first variable false then for all pick the second variable false and the empty variable also got set folds everything got set for so far who knows what happened over here of course um uh okay so just to understand what we've done so far now i'm going to show you how to build the rest of the graph so let's take away the green part um and now we're just back to the construction the green part is actually how you use that construction so back to the construction here now what do we want to achieve now by the time the the players on the graph have got down to here they've effectively made an assignment to the variables by going either left or right at each one of these diamonds so the assignment is done from the game from the perspective of the formula game the game is over and now one side or the other has one here we want to build some extra structure here as a kind of an end game to make sure that the for all player gets stuck if the exist player has made an assignment which satisfies the formula satisfies this part of the formula and the exist player should get stuck if the for all player if the assignment that they made does not satisfy the form okay so let's see how we're going to achieve that with some additional structure so there's going to be some extra node over here let me just tell you what the structure is and then we'll argue why it works so there's going to be going from this bottom node we're going to start a kind of you know the second part of the graph there's going to be a node which fans out to a node for each one of the clauses and each one of those clauses in turn fans out to a node for each one of its literals so you see we have clause c1 and it has the literals x1 x2 bar x3 x1 x2 bar x3 so there's no note for each of the literals and clause c1 same for clause c2 and so on now we're almost done now i'm going to tell you how to connect up these literal nodes so x1 each node is going to get correspond back to its own diamond so now we're going to tie it back to the the first part where we made the assignment the assignment part of the graph x1 is going to connect back to the true part of the x1 diamond and x2 because it's negated is going to tie back to the false part of the x2 diamond because it's an x2 variable similarly x1 bar now in you know in because x1 bar you know here's x1 bar and clause 2. x1 bar is going to connect up now to the full side of the x1 diamond and x2 bar is going to connect up to the full side of the x2 diamond and so on so if we had i don't have the other diamonds here but the x4 would connect up to the true side of the x4 diamond for example that's the whole construction let's understand why it works okay so the end game here is we want exist to win um if the assignment satisfied all the clauses and for all should win if there was some unsatisfied clause okay so why why does this happen so let's put back an assignment here that was made so here's here for the in the one i'm i'm putting back is you know the assignment that they cooperatively built had x1 being true x2 being false and some other stuff now why does this work so what we want to have happen now now so now the move proceeds over to here and you need to arrange it so that exists is the one who picks that note if you have a um if you know if you you know if this would have been for all just add an extra node to switch who's whose turn it is but you want exist to be up here and what you want is for all to be picking the clause the clause note now here is here's the part to understand what's going on um we want for all to win if this is not satisfied so the for all is going to make a claim um if this is not satisfied there is some clause which is not satisfied there's some clause which ended up being false in the assignment so the for all player gonna say i think i won and i won because clause number one is unsatisfied so it's going to pick clause number one so here's the four players more going his exist player and so the four player picks the clause claimed unsatisfied so over here full player says claude c1 is unsatisfied i'm going to move here if this player says i don't agree with you the assignment you know your your line um uh the assignment actually makes one of the literals true in clause c1 let's say it made x1 was true as it as in fact it is here x1 in fact is true so exist is now its turn it's going to pick the the literal that was true in the assignment in that clause that the fraud player is claiming is unsatisfied one of them is lying at this point this is either now true in here which means a for all player did not pick an unsatisfied clause or the exist player picked a false literal which is the the the exist player picked the false letter over to the exist players lie so now we're gonna now the moment of truth because let's see if exist gold picked here now it's um the for all's players turn it's connected back to the true side of the diamond so that node was already used that's the only place that for all could possibly go for all is stuck and an exist has won the game which is what you want because um for all claims are false and the and the exist was was was was correct in saying that x1 satisfied clause c1 so x1 is true so um there's nowhere for for all to go but compare that with the situation if the assignment on this part of the graph had gone through the false had assigned x1 to be false so the path had gone through the full side of the diamond now this node would still be unoccupied would still be um available so now the for role play so if if exist was claiming that x1 was true and it really was false the fraud player would be able to move on to that node and now it's the exist player's turn and the exist would be stuck because this node down here is guaranteed to have been used okay so that is the idea it's really very beautiful um and actually not that complicated you have to you have to stare at it a bit so let's just see what happens for example if um i had one other case maybe it's not necessary at this point um but you know if the for all player claimed well it's the second clause which is unsatisfied in the mutually selected assignment um then uh so this is sort of the the case where the for all players is correct um you know in in that if the exist player now says well i think x1 bar satisfied that uh second clause but you know this assignment made x1 true so x1 bar is false and so the exist player is lying this time and so now the for all player can take this last edge and go here and has one one last move to make and then the exist player stuck so the for all will win in that case so let's turn now shifting gears entirely to a different um uh part of the spate you know of space complexity instead of looking at polynomial space which is like very powerful we're going to look at log space which is comparatively speaking much much weaker so log space are the things that you can do when you only have enough space to write down essentially a pointer or some fixed number of pointers into the input that's what you get when you have logarithmic space and you know because log space is enough to write down an index of something um so you can just this is what you can do with a bunch of pointers and in order to make sense of this we have to change the model of computation because if we use the ordinary one tape turing machine just scanning across the input will just reading the input with the way we've defined it would cost space n and so you can't even read the input if you have less than n space available like log space so we're going to introduce a different model just to allow us to define this and that's a model where it's going to be have two tapes where the part that contains the input doesn't count and that's in order to make sure it doesn't you know it's not being cheated the the input tape is going to be read only um but it doesn't count toward the space used only the work tape which is now has can read can be read or or written or red is going to count toward the space bound okay um so now what we're going to define is you know using this definition for our space complexity um we'll define space log n the things that you can do if you have only a log amount of space here on the work tape so the length of the input is n the length of the um of the work tape is lo order log n so we have uh the deterministic and non-deterministic associated classes uh space log n and n space log n we're calling that l and n l log space and not deterministic log space okay um and as i said that's just enough space to write down some pointers into the input and let's do some quick examples um so if you take the language of palindromes essentially or ww reverse um that's in log space uh so here is a string i don't know i was here's a string that's in www reverse and i mean the ordinary way you might use to test whether a string is of this form in ww reverse might be to cross off corresponding locations here but you can't do that it's a read-only input tape so you have to somehow avoid writing on the input tape but still testing whether you're in this language it's not hard to do you can use the work tape just to keep track of where your pointers would be um and in so doing you can just make sure you're matching off corresponding locations with each other in the input so log space because of its ability to store some fixed number of pointers it gives you enough to test membership in this language so this is in this is solvable in log space uh the path problem which we've seen before you know you're given a graph and a start node and an ending node or a start and a target you're given a graph of an s and t it's a directed graph and you know can i know can i get from s to t um so that problem is solvable in non-deterministic log space uh because as um the way we would do that i'm not not writing this down in in any detail but uh what you would do in your um not you know non-deterministic log space machine you have your input graph written here in the input and you're just going to guess the sequence of nodes one by one which takes you from the start node to the target node you're not going to write down that whole sequence because that in advance because that would be cost you way too much space to write down the only space that you're going to use is to remember the current location we are sitting so you start out you know you write down on the work tape the start node then you're going to non-deterministically pick one of the outgoing edges from the start node and look at its associated node and replace that on your work tape and keep keep repeating that if you ever get to the node t then you can accept and you also have to be a little careful i don't have this on on the slide you also have to make sure that if the graph has a loop in it or because that's not allowed in space complexity uh so you're going to need also a counter to make sure that if you count up to a um you've you you you you've gone through a number of nodes which exceeds the total number of nodes in the graph uh then you can shut down that branch of the non-determinism uh because you know it's you know just it's just going in a loop um if there's any path that connects s to t there's certainly going to be a path that has at most the number of nodes of the graph in it okay um so path is in nl this language here is an in l um what's the relationship between l and n l well certainly the deterministic class is contained in the non-deterministic class that's all that's known whether these two collapse down to be the same is unsolved and so we're going to spend a little time next lecture exploring this um but but let's let's first look at some of the basic facts about log space um sort of setting ourselves up for next lecture so first of all anything that you can do in log space you can do in polynomial time in a sense kind of trivially and in a sense we almost kind of really proved this already because if you remember we said that anything that you can do in a certain amount of space you can do in that much uh you can do in time in that's exponential in the amount of space in the corresponding amount of space so going from space to time blows you up exponentially and order log the exponential of order log n is polynomial so the way we're going to prove this and again we kind of proved this already but let's just go through it again specific to log space uh we'll say and this is going to be a useful definition for us anyway um a configuration for m on w so we would also we already talked about a configuration of turing machine m which is just a snapshot which is the tape the state and the head position when we have an input which is sort of an uh read-only input which is not being counted toward the space we don't include that in the configuration um we just say it's a configuration of m on that particular input but we're going to be counting configurations and i don't want to count all the different inputs as well i'm going to fix the input and count all of the configurations relative to that input and so the number of such configurations is just simply going to be the number of states times the number of head positions for the two heads now times the number of possible tape contents and which as i mentioned here is the number of different possible tape contents it's d which is the tape alphabet to the order log n and that's just going to be n to the k for some k so it's going to be polynomial so that tells us that because the total number of configurations um for m on w is polynomial this machine can only be running for a polynomial number of steps otherwise it will be looping people are repeating a configuration and so therefore that machine has to run itself in polynomial time so you don't really in a sense have to do any work if a machine is deterministically designing a language in log space it's also deterministically deciding the language in polynomial time because that's all the time you can use when you have log space unless you're looping which is not allowed okay so therefore the same machine shows that you're also in pain oh i'll get to that in a second one thing i forgot to mention on the previous slide when i'm talking about the model is by the way this model is not so unreasonable um where you have kind of a imagine having a very large read-only input and and your local storage is much smaller it's much too small to pull in the entire input the way i used to explain this years ago was like you have a your input is a cd-rom which you guys probably barely even know what it is anymore but used to distribute software that way so the rom was like on a dvd uh or cd which contain your whatever you're trying to like yourself whether you're trying to distribute you know it was some large thing uh relatively and so you would imagine having a smaller amount of memory relative to that so you didn't want to necessarily copy that whole thing into your storage maybe even a better example now is like you think of the input as being the entire internet um obviously you can't download unless you're google you can't download the whole internet into your own uh local memory but you're going to have references pointers into the input into different places that's perhaps more analogous to this sort of read-only input touring machine model that i'm describing um okay uh so um and it's another fact i want to mention is that i'm not going to savage just the anything that you can do non-deterministic log space you can do in deterministic uh space but now with a squaring and that's using the same proof that's using savage's theorem which you have to check also works down to log space same proof um so anything that you do not deterministically in log n space you can use deterministically log squared n space okay so let me just see if there's a couple of questions i want to answer here [Music] the relationship between l and nl is not known to be strict nobody knows of an example no one knows whether they're equal and and have people looked at um sublinear time classes yes generally for when you have non-deterministic or probabilistic which we're not haven't defined yet but we will um people have looked at um those sub-linear time classes as well deterministically doesn't make so much sense because then you can't even talk about either the whole input um okay uh let me move on um okay so this this is my last slide let me see if we can do this before we break um not only is l contained within p but this much stronger statement is that nl is contained within p and that we you'll that for that you'll have to do some work because converting your non-deterministic log space machine to a deterministic machine um obviously you're going to have to change the machine um and so we'll introduce a new method here maybe we'll quickly go over this at the beginning of the next lecture i don't like rushing through things at the end but for this if i if i'm given um a non-deterministic machine that runs in log space i want to make a new machine that runs in polynomial time deterministically for the same language and i'm going to define something called the configuration graph for m on an input w and that's just you take m and it's and w its input and you look at all of the configurations um for m w actually configuration graph actually that's should be called the computation graph that's what it's called in the book but okay this is a typo yeah i'll fix that next time uh it doesn't matter computation graph configuration will graph uh all the same basically you're going to take all of the configurations of m on w of which we already observed only number or only polynomial in number and may they become the nodes of a graph and the edges connect two configurations if one follows another according to the rules of the machine okay so here's a picture these are ins this is some graph the nodes of this graph are the configurations of m on w right so each node here corresponds to the machine a snapshot of the machine a tape contents head position and state so writing down all those different configurations i connect one to the other if i can get to cj cj from ci in one step legally on the machine and then uh the non-deterministic machine m accepts w exactly when there's a path from the start configuration to the to to to an accept configuration let's assume we have just a single accept configuration as we argue we can do one or two lectures back um if we clean up the tapes so testing whether you can get from the start to the accept um uh is the same as testing whether the machine accepts its input and so because we can test whether there's a path in a graph connecting two nodes in polynomial time we can solve this problem on this computation configuration graph in polynomial time and so we can figure out whether the non-deterministic machine accepts its input sorry that came a little faster than i like to do so we'll we'll we'll see it again um so here's the polynomial time algorithm you construct the graph you accept if there's a path from the star to the accept and you reject it does not um and so that tells us that not only is l contained with an nl but nl itself is also contained within p so here's that here's a kind of a nice hierarchy of languages we don't not only do we not know whether l equals n l we don't know whether l equals p it's possible that anything you can do in polynomial time you can do deterministically in log space shocking as that might be because it's a pretty weak class but we don't know how to prove that uh there's anything different anything in p that's not in here um last check in uh so we showed that path is in nl what's the best thing we can do about the deterministic space complexity of path so deterministic so this is non-deterministic log space what can we say deterministically about path um hint this should not be this should not be hard uh if you think back to what we've shown very recently okay get your check in points closing up closing stop here all set one two three yes so the correct answer is log squared space because this is just savages there we can do it in log space non-deterministically so you can do it in log squared space deterministically okay so that's um this is what we did today um and i as i mentioned i will uh do this again on at tuesday's lecture just to recap that all right so i'll stick around a little bit for questions and um okay someone says to be about the nomenclature why is it l in l space um i you know because there's not really people don't usually talk about l time so l is sort of everybody knows it's the only reasonable option is space so people just say l and then l um i mean some of these things have had you know you know these things these names have a little bit evolved over time and even now some people say talk about uh um you know i call time classes some call some people call it d time classes you know it's just you know it uh you can make different choices there um okay uh let's see good um why does savage's theorem work for log n you have to you have to look back and see what you needed um and all you needed to be able to do was to um write down uh the configurations um and uh if you look back at how savage's theorem works you're just needing to write down the configuration so the deterministic machine can write down the configurations for the non-deterministic machine they take exactly the same size and then you look at the recursion and the the depth of the recursion is going to be exponential in um the size of the configurations and so you're going to end up with a squaring again you have to go back and just rethink that proof and you'll see nowhere did it need a linear amount of space it works down to a lot it actually does not work for less than log n log n is sort of the lower threshold there and the reason for that is because you also need to store the input location and that already takes log space uh the you know not the the the the head tape heads the tape heads already kind of um have kind of a log space aspect to them and so if you're going to you use less than log space then funky things happen with the tape head storing the tape heads and so less than log space usually turns out not to be interesting very specific to turing machines and not sort of general general models yeah so somebody's asking suppose in the reduction um to generalize geography from tqbf if the formula had two exists in a row um then you would do kind of the natural thing in the graph instead of having that spacer edge between the two diamonds you could just have one diamond connecting directly to the other diamond without a spacer edge and that would be give you the effect of not switching whose turn it is uh somebody's asking me just a general question are people thinking about these open problems um i don't know people don't don't don't say um there was a lot of work on problems related that seemed to be related to those those many open questions like p versus np l versus nl or l versus p and so on p versus p space we'll say we'll talk a little bit more about some of that there's some very interesting things that have been developed i think there's a sense within the community that people are stuck and it's not something you're going to need some sort of major new uh idea in order to push the thing forward so i don't know how many people are still thinking about them i hope people are because i would like to see the answer at some point but or get the answer i think about them sometimes myself but one has to acknowledge chances of success are not high uh okay um so we're a little after past the end of the hour here um unless there's any other questions if i didn't answer your question i may have missed it so you can ask again um but otherwise i'm going to close this close the session here okay bye bye all have a good weekend see you next week take care you 2 00:00:27,029 --> 00:00:28,230 let's get back to space complexity um so what uh just to review what we've been doing um from last lecture which feels like a long time ago but it was only two days ago um we were looking at um those three theorems which all had this basically the same proof one was the latter about the latter problem of uh where you're trying to see if you can get from one string to another string changing one symbol at a time um and all of the strings were in the language of some finite automaton or you had some reasonable rule for saying which of the valid strings which are not and then uh we built on that idea we showed savage's theorem that going from any non-deterministic machine to a deterministic machine only squares the amount of space and then um we finally uh proved that tqbf uh this problem about quantified boolean formulas testing whether they are true uh that that problem is complete for p space um so we're going to build off of that last last theorem today and talk about uh one other complete problem um and also show a connection between p space and testing uh which side uh can win in certain kinds of games um so um and then the at the end of the lecture we will the second half of the lecture we'll talk about diving deeper into space complexity we're going to talk about a different part of the parameter space where instead of looking at polynomial space we're going to talk about what happens if you have logarithmic space which is another natural point to look at for from for space complexity but we'll get to that after the after the break okay um uh talk about game complexity and games um so when i was little uh we used to play me and my sisters would play a game called geography i don't know if how many of you have heard it or maybe it go goes under a different name um but uh it was a game about basically about words and places and it has a very simple um set of rules um let's say there were two people you take turns picking names of places and then each person has to pick a place which starts with the letter that the previous person's place ended with for example you know you have two people playing um perhaps you have you know you agree on some starting place like a major city nearby like boston um so one you know one player the first player takes boston and then after that the next player has to pick a place that starts with n because boston ends with n and so maybe nebraska would be a possible response to boston and then the first player would have to respond to that with a place that starts with an a because nebraska ends with an a it's like maybe alaska uh which also starts with a ends with a so then arkansas maybe would be a nat reasonable response to that and then that ends with s um so uh san francisco and then so on and so on um and uh of course you want to forbid people from reusing names because then you'll just get into a loop you know saying alaska alaska alaska won't make the game very interesting so you have to forbid that possibility names cannot be reused they can be used at most once and then um so eventually one side or the other is going to run out of names uh and not have a either because of the lack of their geographical knowledge or maybe you've just exhausted all the possibilities and i will will not have a response and that person will be stuck and that person is the loser um the other person has won the game so the objective is to try to avoid getting stuck um so let's just take a look um i'm going to think about that games in a somewhat slightly more abstracted formalized way um so you write down all of the legitimate places um they become nodes of a graph such as i've shown here and then you draw an edge from one node to another if um that if that corresponds to a legal move in the game a legal step of the game so you know boston connects to places that start within like new york and nebraska um so the first person might pick boston the second person might pick either of these two um and then back to the first person if they pick nebraska they could pick arkansas they can pick alaska and so on just as i described believe it or not we actually played this game i can remember playing this game when i was a kid that was of course before we had um uh you know league of legends and other fun fun stuff but this is what kids used to do uh so anyway um um so just to get it down on the slide the rules are we'll assume we have two players we'll call them player one and player two player one goes first um and they take turns picking places that start with the letter which ended the previous place not no repeats allowed and the first person players the first player stuck loses you know the person who doesn't have a move to make so what we're going to look at um today is a generalized version of that where we're kind of um just going to take away the names but just leave the underlying graph and we're going to call that generalized geography so here um that can be played on any directed graph you take away the names of the edges the names of the nodes now you just have a some arbitrary graph um you're going to designate a particular node as the starting node and then the players take turns following those edges um and uh what uh you're because you're gonna forbid having any loops reusing any um any nodes uh what you end up doing is working together you're going to be constructing a simple path in that graph simple path just follows those edges and never intersects itself and the first person the first player to be stuck is the loser so the other one is the winner okay so that's we're going to call that generalized geography and we're going to look at the computational complexity of deciding for a given graph and starting point which side would win if both uh players played optimally as best uh the best possible okay and we'll name that problem gg or look at its associated language so here is a graph and the starting node and you want to say um that pair is in the language gigi if the first player player one who must play at that node a um to start off with has a forced win um uh in that general geography game on that graph starting at a okay where again what i mean by a force when i'm not going to define this formally that we could do that it's not hard to do but a forced wind also it's also called a winning strategy um that just means that if both sides play optimal they pay the best possible not up their capability i mean the absolutely the best possible play then that player one would still win there's nothing that player two can do to prevent player one from winning um if player one has a winning strategy or a forced win um you may have seen for example uh there are examples of this that i mean i don't even know if they still exist but it used to be in the newspaper they used to be examples of chess boards and they would say you know if they would start from a certain position and they would say white to win in three and so that means white has a forced win no matter what black does white is going to end up winning but you have to think of what the strategy is it may not be so obvious to think through what are the right moves that what makes but the point is that no matter what black would do white would end up winning um and what you can show we're not going to do that but for a class of games that includes this generalized geography either one side or the other is going to be guaranteed to have a forced win that's not necessarily true for all possible games obviously but for a large class of games um in one side or the other is guaranteed to have a winning strategy or forced win okay so let's just review that because this is going to be essential for you know the first half of the lecture to understand what we mean by a game and what we mean by one side or the other to have a force to win okay so let's let's just uh oh yeah and what we're going to show is that gg is piece based complete that the problem of giving one of these graphs and a starting point figuring out does player 1 have a force win or is it player 2 that has a force win um which side is guaranteed to win under optimal play that's a piece based complete problem um and so let's do a little checking on that and so i'm going to give you a very small example of the generalized geography game you have to figure out whether it's player one or player two that has the forced win the winning strategy or maybe neither or both those are the four options okay and you understand player one has to play here because that's the starting place of this general geography game so then it's up to player two to continue on all right so let's um launch that poll uh you'll have to think about it a little bit um and uh let me know what you think which side has a forced win which player makes the first move it's player one makes the first move and player one plays as i showed you here this is player one playing at a so player one picks a picks and then player two has to pick one of these two okay all right i think we're gonna end the poll everybody's in pick something um you you guys did not do well on this bowl uh at least not too many of you picked d that's that's true that's a that's a little reassuring because you can't have both players having a forced win i did say that in games of this kind one side or the other is going to have a force win so c is not a very good choice either um so you're maybe you should spend less time reading your email and more time listening to the lecture uh um the actual correct answer is b player two um which is obviously a is a minority position here player two has the forced win let's understand why um so player one plays here as i said that's the b that's the first move player two can either take the upper node or the lower note okay player two takes the upper note then player one has no choice but to take the rightmost node and now there's no move for player two uh so player two will lose um if player two takes the upper note the upper choice there because player two will go here player one will go there and it'll be game over and player one will have one because player two was stuck however player two had another choice um uh player two had another choice um player two could have also gone down here um so player two uh if it played down here um things are looking a lot better for player two because player two goes here player one's move is forced goes here but now there's another move left uh that player two could make so play player so two goes here one goes there player two goes up there and now player one is stuck so at player two's choice um player two can make a move which will end up causing player one to get stuck and so because it you know that's the nature of what we mean by uh having a winning strategy there is some way to guarantee if you're playing optimally and you're certainly playing optimally mean you'll take that lower move uh then you will take that so under uh optimal play player two will win this game and so player two has a winning strategy it has a forced win okay um so i i see so i'm not going to define playing optimally i'm going to you know there's some questions here about that about that that's um you know you can think about i'm going to i'm going to leave it um you know somewhat informal um but we could make that all precise i i don't think it would be clarifying to to go through a precise definition of that um uh you know you you can just think you we don't even have to think about it in terms of optimal play you you can say that there is um a uh forever uh a um you say a player has a force win if uh they have moves that they could make in response to every possible move the opponent could make uh that'll guarantee that they will end up winning so it doesn't say we don't even have to talk about optimality here just talk about every possible opponent every possible opponent is going to lose to somebody who um uh you know has a has a winning strategy and pl and follows that strategy okay um okay i can see there was some confusion about who was going first and so on i'm glad we that was part of the reason i gave this check and just so hopefully to clear up some of those points um so player one plays here then player two uh goes and i got some comments that some people were confused about what does it mean for the first uh where does the first person move okay hopefully that's that's uh clear all right so let's move on because we're gonna make we're gonna be spending a while um proving something about generalized geography and these graphs um okay so let's let's continue so in order to make a connection between um games and complexity we're going to need to talk about a problem one that we've actually related to one that we've already seen where we can reformulate that problem as a game and that's going to be a problem on quantified boolean formulas called the formula game okay so this is a different game now um not one that you're likely to end up playing but in in in in a sense in which you'll see this is a still a reasonable game where two players can play against one another and one of them is going to end up winning and one of them is going to end up having uh one or the other will have a winning strategy so let's understand what the game is uh the game is played the game is played on a formula so you so you write down a quantified boolean formula remember we talked about that you know for several lectures now uh you know it's a bunch of quantum variables with quantifiers and um then there's the part that has no quantifiers um which you know uh you can always put into conjunctive normal form if you want but for the purposes of this discussion so far it doesn't matter later on we'll actually want it to be in quant in in cnf but okay but but for now we just have some formula [Music] and associated to that formula there is a game in a very natural sense so first of all let me they're going to be two players but now the names of those players are going to be player exist and player for all okay and the way the way this is how the game goes the uh the player exists but both players are going to be picking values for variables they're going to say you know variable x1 is true variable x2 is false variable x3 is false variable x4 is true and and they're going to be that that's how the moves of the game go they're just assigning values to variables so that in the end all of the variables have been assigned a value a boolean value and so we end up with an assignment but before we get to that stage which players get to pick which variables and the way it works is the exist player is going to assign values to the exist variables so exist we'll pick x1 and x3 and assign them val values the for all player is going to pick the very ver assigned values to the variables that have referral quantifier attached so in this case would be x2 would be um the the the for all player will pick the value for x the x2 variable um fading out here okay now um so and the order of play is according to the order of the appearance of the quantifiers in the formula so the way i have it written out in this particular formula it's exist x1 so exist will pick the value for x1 then it's for all's turn for all pick the value of x2 then it's going to be exist picking the value for x3 and and so on until they get to the end and all of the variables have been assigned a value by one side or the other and now the game is over okay so what's left is to understand who has won at that point um and the way we say who has won is exist wins if in the end this part psi which is the part without quantifiers ended up being satisfied by those um by that assignment that the two players together ended up picking and then for all will win if that part of the formula is not satisfied so let me just write that down here after the variables have been assigned values we determine the winner exist wins if the assignment that they co-op together built they built cooperatively together uh satisfied psy and otherwise for all ones if it doesn't satisfy psi okay so you kind of think about it this way you know exist is picking the values of those variables he's trying to make this uh part of the formula satisfied trying to satisfy this part of the formula for all is picking values but she's trying to make this part of the formula unsatisfied so they're you know they're uh in opposition to one another and um in the end one of them is going to have succeeded and the other one will have failed uh and now thinking about it as a game one side depending upon the formula one side or the other is going to have the ability to dominate the other one and always win um and so the question is which side has that um ability to always win which side has the winning strategy of course it'll depend upon the formula some formulas it'll be the exist player who has the winning strategy and other formulas it'll be the for all player who has the winning strategy and computationally we'd like to make that determination which side is going to win um you know for a given formula and there's a very very nice and actually very easy um and simple way to understand that because we've already run into that problem before the problem of determining which side has a force force win is exactly the same thing as determining whether this quantified boolean formula is true because the exist player has a force win when that formula game on that um on that formula um let me say it again the exist player has a force win when that formula is true um so if this quantified boolean formula were a true quantified bullying formula then exist is guaranteed to be able to win that game if it if it plays its hand right if this is a false formula for all the for all player will always be able to win um if it plays its hand right okay so um why is that and it really follows kind of almost you know without doing any work at all this is kind of a you know this proof even though it looks superficially like it might be not so easy to prove it follows for a very simple reason because the meaning of what it means for exist to have a winning strategy is exactly captured by the semantics of the quantifiers let's see what that means what does it mean that exists has a force to win let's say up here that means that well exist has a move that it could make that exists can pick some value for x1 so there was some move for x for there was some way to assign x1 so that no matter what the opponent does no matter what for all does there's going to be a response that it exists can make um some some assignment to it x3 um uh so that no matter what for all does there's going to be an assignment to the next variable um and so on and so on so that this thing ends up being true so let me just say that again somehow i didn't feel it came out very clear uh so exist has a winning strategy exactly when there was some assignment to x1 that exists can make such that no matter what assignment that for all makes to x2 there is some assignment to x3 um such that no matter what assignment that um for all makes to x4 and so on makes this thing true that's what it means for exist to have a winning strategy that's exactly what the quantifiers are saying in the first place so even really without doing anything you can see that exists has a winning strategy means exactly the same thing as the formula the quantified being formula being true so making the test of whether uh which side has a winning strategy is the same as testing whether the formula is true okay um [Music] so let's let me try to turn to the chat now um before before we move on to anything else um all right uh whoops uh here we go [Music] okay so i'm getting some questions about the order of play and about the alternation of the quantifiers here so the way i'm showing it is that the quantifiers always alternate exist for all exists for all that doesn't really matter you know you could have several exists in a row and that would correspond to exist making several moves in a row before turning it over to parole but alternatively we could just add extra variables which don't play any role in psi and um they're just kind of dummy variables that serve as spacers so if you have two exists in a row and you don't like that you can just add add a for all of some variable that you never see again in between so you can always get it to be alternating if you want so um um also question the difference between pha fee uh and psi so psi is the part here that does not have any quantifiers so the way we always write down our qbfs is their leading quantifiers with the variables which uh you know and all variables have to be within the scope of one of the quantifiers of quantified variables um so you know there are no free variables that are unquantified in our qbfs um and so there's a you know all of these qbfs have a run of quantifiers and then a part without quantifiers where there's this sort of boolean logic part i'm psi here is the boolean logic part and phi is the whole thing together with the quantifiers um so i'm not sure i understand some of these ques why not have the for all player have a force to win the for all player might have a full force win no see one side or the other is guaranteed to have a forced win so if exist does not have a force to win then then parole will have a forced win that you know i'm not going to prove that but that's a fairly simple proof by induction which i'm not going to go through just take it as a fact somebody's asking why does for all want to not satisfy the uh this expression the the psy part well i mean that's the way we set up the game um you know in order to make this correspond to tqbf um just the truth of the of this of this expression we want to make exist try to satisfy the the this part and for all try to unsatisfy make it not satisfied um and so i'm not sure because that's what works i'm not sure what else whatever whether why i can answer there okay somebody i'm going to answer some of the you know most basic questions here somebody's asking how do we know how many variables to use so the the variables are the variables of the formula so somebody's going to hand you a formula that's going to have x1 to x10 whatever that formula has some number of variables and so that game is going to have 10 moves you know if the quantifiers alternate then the moves will alternate but whatever the pattern of the quantifiers is going to be the pattern of the moves so exist is going to follow the um you know the places where the exist quantifier occurs and fraud is going to follow the places where the for all quantifier occurs so if you have you know exist for all exist for all exist will move first then for all move then exist will move then for all will move together they're picking values to their respective variables with with their opposing goals existed trying to make the um the psi part satisfied for all is trying to avoid making it satisfied to make to to make make the assignment not satisfy that part and if you think about what that means just in terms of the meaning of that that it's going to mean exactly the same thing that the formula is true um so anyway if you don't let's see i don't know if there's any more questions that i can try to answer here um well so it's still about the numbers of variables the um each formula is going to define a different game you see the formula so the formula however big it may have 50 variables that may have two variables but we'll do an example maybe that'll help um the next check-in which is coming up for pretty soon might actually be now uh we'll give you an example um so then we'll see who's who's understanding and who's not okay so let's continue here um so so therefore the problem of determining does exist have a force win is exactly the tqbf problem because exist has a force when exactly when the formula is true and what we're going to show is that t qbf is a polynomial time reducible to generalized geography but it's conceptually we're going to think about tqbf now as a game because that's how geography is generally geography is set up so we're going to given a formula like this we're going to construct an instance of generalized geography we're going to construct a graph where play on that graph is going to mimic play in the formula game so making the moves in that graph are going to correspond to setting variables true and false because the graph is going to be specially designed to to force that behavior okay so i think we have a check-in coming in yeah so let's just see how um i i suggest you pay attention to this check-in and really try to think about it and solve it because i think that'll help you make the connection between the truth of the formula and exist having a winning strategy okay so so if you take this formula here here's you know this is some particular formula exist x for all y blah blah blah um this is going to be associated to that formula a formula game where first exist moves assigns a value for x and depending upon that value for all is going to move and they're going to assign a value for y and after that's done this part here is either going to be true or false it's either going to be satisfied or unsatisfied and i want to know can exist always find a way can it does to to guarantee that this part is going to be satisfied or can for all always get find a way to guarantee that this part is not satisfied okay so it's you have to stare you have to look at this uh formula to understand which side is going to end up succeeding um so you understand you have to if you're not clear on the rules look back here exist is trying to make this part satisfied for all is trying to make this part unsatisfied but they neither of them has the no the totally upper hand because existence depicting one of the variables for all is picking the other variable but they're doing it in a certain order okay first exist is going to pick then a for all is going to pick okay so that's the way the game works exist picks a value then for all picks a value and then you see who wins so who wins who's guaranteed to win um you know can can we make sure exist always wins or can we make sure that it for all always wins for this particular formula so there's just two variables here and you can think about this in either of two ways strictly speaking purely as a game or you can look at understand whether that formula is true or not um they're equivalent ways of looking at at the problem in fact in a certain sense it's the same that's what i'm trying to trying to get across thinking about this as a game or thinking about in terms of quantifiers doesn't make any difference um okay almost done here so all right last call closing the poll okay i think this one you got you you you did pretty well on so yeah the correct answer is the for all player has a uh winning strategy has a forced win uh and um also similarly the expression um is fee is is false because you know if you try to find some x such that no matter what y you pick um this is going to be true this is going to be satisfied you're out of luck because you know if you make x true well um then if you make x true now i'm confused uh what is going on if you make x true then uh y oh if you make x true then uh x bar is false um so y bar has to be true uh um now i'm completely confused oh no if make x true it has to work for all y so this thing is clearly false if you make x true um uh then then then this x bar is going to be false and it's not the case that for every y of the setting for y because y is forced here y is going to have to be true well y bar is going to have to be true so y is going to have to be false somebody if you make it very clearly but if you make if you and now if you if you try again to make x false um maybe that's a one to think about first if you make x false so you're going to be stuck on the first clause because you have it has to work for both settings of y but maybe thinking about it as a game is better you know if you no matter what x does a y is going to find have a way of making one of those two clauses unsatisfied so if x is true why can uh uh can set itself like why we can set y to make the second clause uh unsatisfied and if x is false we can find y can be assigned to make the first clause unsatisfied um so anyway um the right answer is uh the for all player uh has a winning strategy here in the end of the day it's not critical that you understand this correspondence but but you have to then if you don't quite understand that you're going to have to take on faith that the formula game is the same as tqbf because that's we're going to use and by the way i would like to also mention here that um that this correspondence between uh games and quantifiers is some things that mathematicians use all the time um because if you have a if you have some expression which comes up often in mathematics where there are you know a whole run of quantifiers you know in front of some uh some statement it's really kind of hard for anybody uh to really uh get a feeling for what that means if you have like six alternating quantifiers which happens often in mathematics you're going to have statements that have a lot of alternations with quantifiers very hard to get a feeling for what that means but if you think about it as a game it's much more intuitive and it's completely equivalent to think about quantifiers go back and forth between quantifiers and games anyway um so let's look at the at the reduction the the construction that shows uh generalized geography is uh complete took a little longer than i thought maybe this is i just want to make sure everybody's with me on this so why don't we call the break now and then we'll come back um and we'll look at the construction um after that because the construction itself is going to take about 10 minutes to to work through to figure out how to reduce tqbf and build a graph that simulates the formula okay so i'm going to move right on to uh the um and we'll come back okay so feel free to ask some questions um and uh get ready for us diving in to a construction uh for the um uh for the reduction of tqbf to generalize geography so there's kind of an interesting question about you know what's a relationship you know between a formula and when you swap the order of the quantifiers um uh you know the the the the question is asking does that somehow relate to the negating the um the you know the part the quantifier free part the part without the quantifiers and i um i don't think that that's the right correspondence there it actually is uh some relationship when you say um you know there exists for all is actually a stronger statement than than saying that for all there exists in general the um that actually implies you know exist x for a why is this implies uh for a why there exists an x um because what you're saying is that there is the the choice of y if there exists an x uh for all y there was one x that works for every one of the y's if you're saying for y there exists an x the choice of that x can depend on the y um so there there is a connection but um you have to think through what it means in order to understand that connection not i think you won't need to know that in order to process what we're going to be doing but maybe it just helps you think about it um other questions here okay i think we're um we're out of time here so let's return to our our lecture um so i'm gonna go back hopefully that won't crash everything [Music] here we go okay so now we're going to be reducing tqbf to generalize geography are we all together now um uh okay so i'm gonna illustrate this construction which is very nice construction by the way i'm gonna illustrate this construction just by doing an example or a partial example okay but i think it'll give you the idea so you understand so what i'm trying to do here is i'm starting off with a quantified boolean formula so here it is um i'm going to assume it's in conjunctive normal form which i can always convert it into that form uh maybe by adding some additional variables but without doing anything too drastic to it and so now what i'm gonna uh starting from that formula i'm gonna now build a graph we're playing the geography game on that graph which is remember taking turns picking nodes which may form a path is going to correspond to um playing the formula game which is picking the variables of that formula and then you want uh the if the exist player wins in the formula then player one is supposed to win in the geography game okay um so here is how the graph is going to look okay so good to follow try to follow this um so it's going to be kind of there's going to be kind of two parts one that's going to correspond to the variables and the other part that's going to correspond to the clauses um and so for each variable i'm going to have like a diamond structure here so there's a little starting thing that's going to be unique to the to the first variable but then every variable is going to have like a little diamond structure here um and they're going to be attached one to the next and we'll to understand how this what this means when we play geography on this piece of the graph so but let's just understand the structure first okay so this is the start node and now um here i'm player one is going to play the role of exist player two is going to play the role of parole and in fact i'm going to identify that so i'm going to call because it's going to be helpful just to have uh helpful just to think about player 1 is being the exist player so the exist player is going to be playing on this graph the exist and the for all players are going to be playing on this graph it's really just players one and two so exist the exist player player one has to b by the rules has to pick the start note and now with player two's move these the four players move now if you're with me on this graph the for all for the for all players move um is now not very interesting because it's forced it has to go to here because again they're just picking the um the nodes of a simple path in this graph so for the fraud player to the for all player has to go here is if exist started over there now um now it's now it's the exist player's turn and now now something interesting happens the exist player has a choice you can either go left or right there are two possibilities and then after that the four player's turn who's again just forced so so far for all has not had much interesting stuff to do in this hasn't has not had to have has not had to think hard um so for all players no matter whether exists went left or right the for all players move is forced and ends up over here and now let's see if this player's turn again this is kind of an easy one oh so before i do that let's just look at how that play could go so the first two moves that's kind of i was suggesting exist goes here and then for all goes there i'm gonna i'm gonna illustrate kind of possible plays through this graph by tracing them out in green um so now exists goes here for all goes there those those were forced but now the exist player goes could either go this way or could go that way so those are two possible different ways of playing this the game that we said we're building um and now uh now whose turn is it here this was the for all player pick this one so the the the next turn is the exist player and that's forced but now notice now for the first time the for all player has to think because the for all player can either go left or right so could either go left or it could go right um and so then there's going to be a sequence of these diamond structures where kind of they're constructed so that alternately either the exist has a choice or the for all player has a choice until you get all the way down to the bottom okay that is the graph we've built so far now if we stopped here um then whoever ended up at this point whether it was existing for all would be the winner because there's nowhere for the other for the opponent to go of course that wouldn't be very interesting so there's going to be more stuff we're going to add on which are going to correspond to the clauses now how to think about what's happened so far now so maybe as some of you might be guessing is that um when the exist player over here has a choice could have either gone left or right that's going to correspond because this is supposed to be mimicking the formula game that's going to correspond to the uh exist player in the formula picking the variable either true or false so that the left or right is going to correspond to true or false and so um let's just arbitrarily say left is going to correspond to picking it true and right is going to correspond to picking false okay so that so the way i've said it so far is exist ended up picking the first variable false then for all pick the second variable false and the empty variable also got set folds everything got set for so far who knows what happened over here of course um uh okay so just to understand what we've done so far now i'm going to show you how to build the rest of the graph so let's take away the green part um and now we're just back to the construction the green part is actually how you use that construction so back to the construction here now what do we want to achieve now by the time the the players on the graph have got down to here they've effectively made an assignment to the variables by going either left or right at each one of these diamonds so the assignment is done from the game from the perspective of the formula game the game is over and now one side or the other has one here we want to build some extra structure here as a kind of an end game to make sure that the for all player gets stuck if the exist player has made an assignment which satisfies the formula satisfies this part of the formula and the exist player should get stuck if the for all player if the assignment that they made does not satisfy the form okay so let's see how we're going to achieve that with some additional structure so there's going to be some extra node over here let me just tell you what the structure is and then we'll argue why it works so there's going to be going from this bottom node we're going to start a kind of you know the second part of the graph there's going to be a node which fans out to a node for each one of the clauses and each one of those clauses in turn fans out to a node for each one of its literals so you see we have clause c1 and it has the literals x1 x2 bar x3 x1 x2 bar x3 so there's no note for each of the literals and clause c1 same for clause c2 and so on now we're almost done now i'm going to tell you how to connect up these literal nodes so x1 each node is going to get correspond back to its own diamond so now we're going to tie it back to the the first part where we made the assignment the assignment part of the graph x1 is going to connect back to the true part of the x1 diamond and x2 because it's negated is going to tie back to the false part of the x2 diamond because it's an x2 variable similarly x1 bar now in you know in because x1 bar you know here's x1 bar and clause 2. x1 bar is going to connect up now to the full side of the x1 diamond and x2 bar is going to connect up to the full side of the x2 diamond and so on so if we had i don't have the other diamonds here but the x4 would connect up to the true side of the x4 diamond for example that's the whole construction let's understand why it works okay so the end game here is we want exist to win um if the assignment satisfied all the clauses and for all should win if there was some unsatisfied clause okay so why why does this happen so let's put back an assignment here that was made so here's here for the in the one i'm i'm putting back is you know the assignment that they cooperatively built had x1 being true x2 being false and some other stuff now why does this work so what we want to have happen now now so now the move proceeds over to here and you need to arrange it so that exists is the one who picks that note if you have a um if you know if you you know if this would have been for all just add an extra node to switch who's whose turn it is but you want exist to be up here and what you want is for all to be picking the clause the clause note now here is here's the part to understand what's going on um we want for all to win if this is not satisfied so the for all is going to make a claim um if this is not satisfied there is some clause which is not satisfied there's some clause which ended up being false in the assignment so the for all player gonna say i think i won and i won because clause number one is unsatisfied so it's going to pick clause number one so here's the four players more going his exist player and so the four player picks the clause claimed unsatisfied so over here full player says claude c1 is unsatisfied i'm going to move here if this player says i don't agree with you the assignment you know your your line um uh the assignment actually makes one of the literals true in clause c1 let's say it made x1 was true as it as in fact it is here x1 in fact is true so exist is now its turn it's going to pick the the literal that was true in the assignment in that clause that the fraud player is claiming is unsatisfied one of them is lying at this point this is either now true in here which means a for all player did not pick an unsatisfied clause or the exist player picked a false literal which is the the the exist player picked the false letter over to the exist players lie so now we're gonna now the moment of truth because let's see if exist gold picked here now it's um the for all's players turn it's connected back to the true side of the diamond so that node was already used that's the only place that for all could possibly go for all is stuck and an exist has won the game which is what you want because um for all claims are false and the and the exist was was was was correct in saying that x1 satisfied clause c1 so x1 is true so um there's nowhere for for all to go but compare that with the situation if the assignment on this part of the graph had gone through the false had assigned x1 to be false so the path had gone through the full side of the diamond now this node would still be unoccupied would still be um available so now the for role play so if if exist was claiming that x1 was true and it really was false the fraud player would be able to move on to that node and now it's the exist player's turn and the exist would be stuck because this node down here is guaranteed to have been used okay so that is the idea it's really very beautiful um and actually not that complicated you have to you have to stare at it a bit so let's just see what happens for example if um i had one other case maybe it's not necessary at this point um but you know if the for all player claimed well it's the second clause which is unsatisfied in the mutually selected assignment um then uh so this is sort of the the case where the for all players is correct um you know in in that if the exist player now says well i think x1 bar satisfied that uh second clause but you know this assignment made x1 true so x1 bar is false and so the exist player is lying this time and so now the for all player can take this last edge and go here and has one one last move to make and then the exist player stuck so the for all will win in that case so let's turn now shifting gears entirely to a different um uh part of the spate you know of space complexity instead of looking at polynomial space which is like very powerful we're going to look at log space which is comparatively speaking much much weaker so log space are the things that you can do when you only have enough space to write down essentially a pointer or some fixed number of pointers into the input that's what you get when you have logarithmic space and you know because log space is enough to write down an index of something um so you can just this is what you can do with a bunch of pointers and in order to make sense of this we have to change the model of computation because if we use the ordinary one tape turing machine just scanning across the input will just reading the input with the way we've defined it would cost space n and so you can't even read the input if you have less than n space available like log space so we're going to introduce a different model just to allow us to define this and that's a model where it's going to be have two tapes where the part that contains the input doesn't count and that's in order to make sure it doesn't you know it's not being cheated the the input tape is going to be read only um but it doesn't count toward the space used only the work tape which is now has can read can be read or or written or red is going to count toward the space bound okay um so now what we're going to define is you know using this definition for our space complexity um we'll define space log n the things that you can do if you have only a log amount of space here on the work tape so the length of the input is n the length of the um of the work tape is lo order log n so we have uh the deterministic and non-deterministic associated classes uh space log n and n space log n we're calling that l and n l log space and not deterministic log space okay um and as i said that's just enough space to write down some pointers into the input and let's do some quick examples um so if you take the language of palindromes essentially or ww reverse um that's in log space uh so here is a string i don't know i was here's a string that's in www reverse and i mean the ordinary way you might use to test whether a string is of this form in ww reverse might be to cross off corresponding locations here but you can't do that it's a read-only input tape so you have to somehow avoid writing on the input tape but still testing whether you're in this language it's not hard to do you can use the work tape just to keep track of where your pointers would be um and in so doing you can just make sure you're matching off corresponding locations with each other in the input so log space because of its ability to store some fixed number of pointers it gives you enough to test membership in this language so this is in this is solvable in log space uh the path problem which we've seen before you know you're given a graph and a start node and an ending node or a start and a target you're given a graph of an s and t it's a directed graph and you know can i know can i get from s to t um so that problem is solvable in non-deterministic log space uh because as um the way we would do that i'm not not writing this down in in any detail but uh what you would do in your um not you know non-deterministic log space machine you have your input graph written here in the input and you're just going to guess the sequence of nodes one by one which takes you from the start node to the target node you're not going to write down that whole sequence because that in advance because that would be cost you way too much space to write down the only space that you're going to use is to remember the current location we are sitting so you start out you know you write down on the work tape the start node then you're going to non-deterministically pick one of the outgoing edges from the start node and look at its associated node and replace that on your work tape and keep keep repeating that if you ever get to the node t then you can accept and you also have to be a little careful i don't have this on on the slide you also have to make sure that if the graph has a loop in it or because that's not allowed in space complexity uh so you're going to need also a counter to make sure that if you count up to a um you've you you you you've gone through a number of nodes which exceeds the total number of nodes in the graph uh then you can shut down that branch of the non-determinism uh because you know it's you know just it's just going in a loop um if there's any path that connects s to t there's certainly going to be a path that has at most the number of nodes of the graph in it okay um so path is in nl this language here is an in l um what's the relationship between l and n l well certainly the deterministic class is contained in the non-deterministic class that's all that's known whether these two collapse down to be the same is unsolved and so we're going to spend a little time next lecture exploring this um but but let's let's first look at some of the basic facts about log space um sort of setting ourselves up for next lecture so first of all anything that you can do in log space you can do in polynomial time in a sense kind of trivially and in a sense we almost kind of really proved this already because if you remember we said that anything that you can do in a certain amount of space you can do in that much uh you can do in time in that's exponential in the amount of space in the corresponding amount of space so going from space to time blows you up exponentially and order log the exponential of order log n is polynomial so the way we're going to prove this and again we kind of proved this already but let's just go through it again specific to log space uh we'll say and this is going to be a useful definition for us anyway um a configuration for m on w so we would also we already talked about a configuration of turing machine m which is just a snapshot which is the tape the state and the head position when we have an input which is sort of an uh read-only input which is not being counted toward the space we don't include that in the configuration um we just say it's a configuration of m on that particular input but we're going to be counting configurations and i don't want to count all the different inputs as well i'm going to fix the input and count all of the configurations relative to that input and so the number of such configurations is just simply going to be the number of states times the number of head positions for the two heads now times the number of possible tape contents and which as i mentioned here is the number of different possible tape contents it's d which is the tape alphabet to the order log n and that's just going to be n to the k for some k so it's going to be polynomial so that tells us that because the total number of configurations um for m on w is polynomial this machine can only be running for a polynomial number of steps otherwise it will be looping people are repeating a configuration and so therefore that machine has to run itself in polynomial time so you don't really in a sense have to do any work if a machine is deterministically designing a language in log space it's also deterministically deciding the language in polynomial time because that's all the time you can use when you have log space unless you're looping which is not allowed okay so therefore the same machine shows that you're also in pain oh i'll get to that in a second one thing i forgot to mention on the previous slide when i'm talking about the model is by the way this model is not so unreasonable um where you have kind of a imagine having a very large read-only input and and your local storage is much smaller it's much too small to pull in the entire input the way i used to explain this years ago was like you have a your input is a cd-rom which you guys probably barely even know what it is anymore but used to distribute software that way so the rom was like on a dvd uh or cd which contain your whatever you're trying to like yourself whether you're trying to distribute you know it was some large thing uh relatively and so you would imagine having a smaller amount of memory relative to that so you didn't want to necessarily copy that whole thing into your storage maybe even a better example now is like you think of the input as being the entire internet um obviously you can't download unless you're google you can't download the whole internet into your own uh local memory but you're going to have references pointers into the input into different places that's perhaps more analogous to this sort of read-only input touring machine model that i'm describing um okay uh so um and it's another fact i want to mention is that i'm not going to savage just the anything that you can do non-deterministic log space you can do in deterministic uh space but now with a squaring and that's using the same proof that's using savage's theorem which you have to check also works down to log space same proof um so anything that you do not deterministically in log n space you can use deterministically log squared n space okay so let me just see if there's a couple of questions i want to answer here [Music] the relationship between l and nl is not known to be strict nobody knows of an example no one knows whether they're equal and and have people looked at um sublinear time classes yes generally for when you have non-deterministic or probabilistic which we're not haven't defined yet but we will um people have looked at um those sub-linear time classes as well deterministically doesn't make so much sense because then you can't even talk about either the whole input um okay uh let me move on um okay so this this is my last slide let me see if we can do this before we break um not only is l contained within p but this much stronger statement is that nl is contained within p and that we you'll that for that you'll have to do some work because converting your non-deterministic log space machine to a deterministic machine um obviously you're going to have to change the machine um and so we'll introduce a new method here maybe we'll quickly go over this at the beginning of the next lecture i don't like rushing through things at the end but for this if i if i'm given um a non-deterministic machine that runs in log space i want to make a new machine that runs in polynomial time deterministically for the same language and i'm going to define something called the configuration graph for m on an input w and that's just you take m and it's and w its input and you look at all of the configurations um for m w actually configuration graph actually that's should be called the computation graph that's what it's called in the book but okay this is a typo yeah i'll fix that next time uh it doesn't matter computation graph configuration will graph uh all the same basically you're going to take all of the configurations of m on w of which we already observed only number or only polynomial in number and may they become the nodes of a graph and the edges connect two configurations if one follows another according to the rules of the machine okay so here's a picture these are ins this is some graph the nodes of this graph are the configurations of m on w right so each node here corresponds to the machine a snapshot of the machine a tape contents head position and state so writing down all those different configurations i connect one to the other if i can get to cj cj from ci in one step legally on the machine and then uh the non-deterministic machine m accepts w exactly when there's a path from the start configuration to the to to to an accept configuration let's assume we have just a single accept configuration as we argue we can do one or two lectures back um if we clean up the tapes so testing whether you can get from the start to the accept um uh is the same as testing whether the machine accepts its input and so because we can test whether there's a path in a graph connecting two nodes in polynomial time we can solve this problem on this computation configuration graph in polynomial time and so we can figure out whether the non-deterministic machine accepts its input sorry that came a little faster than i like to do so we'll we'll we'll see it again um so here's the polynomial time algorithm you construct the graph you accept if there's a path from the star to the accept and you reject it does not um and so that tells us that not only is l contained with an nl but nl itself is also contained within p so here's that here's a kind of a nice hierarchy of languages we don't not only do we not know whether l equals n l we don't know whether l equals p it's possible that anything you can do in polynomial time you can do deterministically in log space shocking as that might be because it's a pretty weak class but we don't know how to prove that uh there's anything different anything in p that's not in here um last check in uh so we showed that path is in nl what's the best thing we can do about the deterministic space complexity of path so deterministic so this is non-deterministic log space what can we say deterministically about path um hint this should not be this should not be hard uh if you think back to what we've shown very recently okay get your check in points closing up closing stop here all set one two three yes so the correct answer is log squared space because this is just savages there we can do it in log space non-deterministically so you can do it in log squared space deterministically okay so that's um this is what we did today um and i as i mentioned i will uh do this again on at tuesday's lecture just to recap that all right so i'll stick around a little bit for questions and um okay someone says to be about the nomenclature why is it l in l space um i you know because there's not really people don't usually talk about l time so l is sort of everybody knows it's the only reasonable option is space so people just say l and then l um i mean some of these things have had you know you know these things these names have a little bit evolved over time and even now some people say talk about uh um you know i call time classes some call some people call it d time classes you know it's just you know it uh you can make different choices there um okay uh let's see good um why does savage's theorem work for log n you have to you have to look back and see what you needed um and all you needed to be able to do was to um write down uh the configurations um and uh if you look back at how savage's theorem works you're just needing to write down the configuration so the deterministic machine can write down the configurations for the non-deterministic machine they take exactly the same size and then you look at the recursion and the the depth of the recursion is going to be exponential in um the size of the configurations and so you're going to end up with a squaring again you have to go back and just rethink that proof and you'll see nowhere did it need a linear amount of space it works down to a lot it actually does not work for less than log n log n is sort of the lower threshold there and the reason for that is because you also need to store the input location and that already takes log space uh the you know not the the the the head tape heads the tape heads already kind of um have kind of a log space aspect to them and so if you're going to you use less than log space then funky things happen with the tape head storing the tape heads and so less than log space usually turns out not to be interesting very specific to turing machines and not sort of general general models yeah so somebody's asking suppose in the reduction um to generalize geography from tqbf if the formula had two exists in a row um then you would do kind of the natural thing in the graph instead of having that spacer edge between the two diamonds you could just have one diamond connecting directly to the other diamond without a spacer edge and that would be give you the effect of not switching whose turn it is uh somebody's asking me just a general question are people thinking about these open problems um i don't know people don't don't don't say um there was a lot of work on problems related that seemed to be related to those those many open questions like p versus np l versus nl or l versus p and so on p versus p space we'll say we'll talk a little bit more about some of that there's some very interesting things that have been developed i think there's a sense within the community that people are stuck and it's not something you're going to need some sort of major new uh idea in order to push the thing forward so i don't know how many people are still thinking about them i hope people are because i would like to see the answer at some point but or get the answer i think about them sometimes myself but one has to acknowledge chances of success are not high uh okay um so we're a little after past the end of the hour here um unless there's any other questions if i didn't answer your question i may have missed it so you can ask again um but otherwise i'm going to close this close the session here okay bye bye all have a good weekend see you next week take care you 3 00:00:28,230 --> 00:00:31,429 4 00:00:31,429 --> 00:00:33,430 5 00:00:33,430 --> 00:00:33,440 6 00:00:33,440 --> 00:00:34,470 7 00:00:34,470 --> 00:00:36,470 8 00:00:36,470 --> 00:00:37,990 9 00:00:37,990 --> 00:00:39,430 10 00:00:39,430 --> 00:00:42,549 11 00:00:42,549 --> 00:00:45,350 12 00:00:45,350 --> 00:00:47,990 13 00:00:47,990 --> 00:00:50,310 14 00:00:50,310 --> 00:00:50,320 15 00:00:50,320 --> 00:00:51,670 16 00:00:51,670 --> 00:00:53,750 17 00:00:53,750 --> 00:00:56,709 18 00:00:56,709 --> 00:00:58,389 19 00:00:58,389 --> 00:01:00,389 20 00:01:00,389 --> 00:01:02,389 21 00:01:02,389 --> 00:01:05,509 22 00:01:05,509 --> 00:01:07,510 23 00:01:07,510 --> 00:01:10,390 24 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--> 00:21:35,669 641 00:21:35,669 --> 00:21:37,750 642 00:21:37,750 --> 00:21:40,230 643 00:21:40,230 --> 00:21:40,240 644 00:21:40,240 --> 00:21:41,990 645 00:21:41,990 --> 00:21:42,000 646 00:21:42,000 --> 00:21:42,830 647 00:21:42,830 --> 00:21:45,750 648 00:21:45,750 --> 00:21:47,190 649 00:21:47,190 --> 00:21:49,270 650 00:21:49,270 --> 00:21:51,590 651 00:21:51,590 --> 00:21:54,230 652 00:21:54,230 --> 00:21:56,070 653 00:21:56,070 --> 00:21:58,230 654 00:21:58,230 --> 00:22:00,789 655 00:22:00,789 --> 00:22:00,799 656 00:22:00,799 --> 00:22:01,909 657 00:22:01,909 --> 00:22:03,590 658 00:22:03,590 --> 00:22:05,350 659 00:22:05,350 --> 00:22:09,029 660 00:22:09,029 --> 00:22:11,909 661 00:22:11,909 --> 00:22:13,510 662 00:22:13,510 --> 00:22:15,270 663 00:22:15,270 --> 00:22:16,950 664 00:22:16,950 --> 00:22:18,549 665 00:22:18,549 --> 00:22:20,149 666 00:22:20,149 --> 00:22:22,070 667 00:22:22,070 --> 00:22:24,630 668 00:22:24,630 --> 00:22:24,640 669 00:22:24,640 --> 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699 00:23:38,950 --> 00:23:40,710 700 00:23:40,710 --> 00:23:43,269 701 00:23:43,269 --> 00:23:44,789 702 00:23:44,789 --> 00:23:47,029 703 00:23:47,029 --> 00:23:48,630 704 00:23:48,630 --> 00:23:50,710 705 00:23:50,710 --> 00:23:50,720 706 00:23:50,720 --> 00:23:52,230 707 00:23:52,230 --> 00:23:55,669 708 00:23:55,669 --> 00:23:58,470 709 00:23:58,470 --> 00:24:00,549 710 00:24:00,549 --> 00:24:00,559 711 00:24:00,559 --> 00:24:02,070 712 00:24:02,070 --> 00:24:03,909 713 00:24:03,909 --> 00:24:06,870 714 00:24:06,870 --> 00:24:10,230 715 00:24:10,230 --> 00:24:12,470 716 00:24:12,470 --> 00:24:14,149 717 00:24:14,149 --> 00:24:14,159 718 00:24:14,159 --> 00:24:14,870 719 00:24:14,870 --> 00:24:17,190 720 00:24:17,190 --> 00:24:19,350 721 00:24:19,350 --> 00:24:21,909 722 00:24:21,909 --> 00:24:23,430 723 00:24:23,430 --> 00:24:26,470 724 00:24:26,470 --> 00:24:29,029 725 00:24:29,029 --> 00:24:30,710 726 00:24:30,710 --> 00:24:33,669 727 00:24:33,669 --> 00:24:35,590 728 00:24:35,590 --> 00:24:35,600 729 00:24:35,600 --> 00:24:37,990 730 00:24:37,990 --> 00:24:40,470 731 00:24:40,470 --> 00:24:42,310 732 00:24:42,310 --> 00:24:43,990 733 00:24:43,990 --> 00:24:46,789 734 00:24:46,789 --> 00:24:48,630 735 00:24:48,630 --> 00:24:50,070 736 00:24:50,070 --> 00:24:51,990 737 00:24:51,990 --> 00:24:53,430 738 00:24:53,430 --> 00:24:55,909 739 00:24:55,909 --> 00:24:58,310 740 00:24:58,310 --> 00:25:00,549 741 00:25:00,549 --> 00:25:03,510 742 00:25:03,510 --> 00:25:06,470 743 00:25:06,470 --> 00:25:08,149 744 00:25:08,149 --> 00:25:11,029 745 00:25:11,029 --> 00:25:12,549 746 00:25:12,549 --> 00:25:14,630 747 00:25:14,630 --> 00:25:17,510 748 00:25:17,510 --> 00:25:18,870 749 00:25:18,870 --> 00:25:21,029 750 00:25:21,029 --> 00:25:22,630 751 00:25:22,630 --> 00:25:25,110 752 00:25:25,110 --> 00:25:27,750 753 00:25:27,750 --> 00:25:27,760 754 00:25:27,760 --> 00:25:28,870 755 00:25:28,870 --> 00:25:30,789 756 00:25:30,789 --> 00:25:32,710 757 00:25:32,710 --> 00:25:33,830 758 00:25:33,830 --> 00:25:36,390 759 00:25:36,390 --> 00:25:39,029 760 00:25:39,029 --> 00:25:40,789 761 00:25:40,789 --> 00:25:43,110 762 00:25:43,110 --> 00:25:43,960 763 00:25:43,960 --> 00:25:43,970 764 00:25:43,970 --> 00:25:45,350 765 00:25:45,350 --> 00:25:47,510 766 00:25:47,510 --> 00:25:49,590 767 00:25:49,590 --> 00:25:50,630 768 00:25:50,630 --> 00:25:50,640 769 00:25:50,640 --> 00:25:52,470 770 00:25:52,470 --> 00:25:57,269 771 00:25:57,269 --> 00:26:00,880 772 00:26:00,880 --> 00:26:00,890 773 00:26:00,890 --> 00:26:08,830 774 00:26:08,830 --> 00:26:11,590 775 00:26:11,590 --> 00:26:13,269 776 00:26:13,269 --> 00:26:15,510 777 00:26:15,510 --> 00:26:17,350 778 00:26:17,350 --> 00:26:19,750 779 00:26:19,750 --> 00:26:21,909 780 00:26:21,909 --> 00:26:24,710 781 00:26:24,710 --> 00:26:26,390 782 00:26:26,390 --> 00:26:28,149 783 00:26:28,149 --> 00:26:30,710 784 00:26:30,710 --> 00:26:33,269 785 00:26:33,269 --> 00:26:33,279 786 00:26:33,279 --> 00:26:34,950 787 00:26:34,950 --> 00:26:34,960 788 00:26:34,960 --> 00:26:36,789 789 00:26:36,789 --> 00:26:39,430 790 00:26:39,430 --> 00:26:42,870 791 00:26:42,870 --> 00:26:44,950 792 00:26:44,950 --> 00:26:47,110 793 00:26:47,110 --> 00:26:48,950 794 00:26:48,950 --> 00:26:50,950 795 00:26:50,950 --> 00:26:52,870 796 00:26:52,870 --> 00:26:55,269 797 00:26:55,269 --> 00:26:56,470 798 00:26:56,470 --> 00:27:00,149 799 00:27:00,149 --> 00:27:00,159 800 00:27:00,159 --> 00:27:01,350 801 00:27:01,350 --> 00:27:01,360 802 00:27:01,360 --> 00:27:06,470 803 00:27:06,470 --> 00:27:06,480 804 00:27:06,480 --> 00:27:07,510 805 00:27:07,510 --> 00:27:10,230 806 00:27:10,230 --> 00:27:10,240 807 00:27:10,240 --> 00:27:11,190 808 00:27:11,190 --> 00:27:14,149 809 00:27:14,149 --> 00:27:16,310 810 00:27:16,310 --> 00:27:19,350 811 00:27:19,350 --> 00:27:21,269 812 00:27:21,269 --> 00:27:23,029 813 00:27:23,029 --> 00:27:24,950 814 00:27:24,950 --> 00:27:26,310 815 00:27:26,310 --> 00:27:28,870 816 00:27:28,870 --> 00:27:31,269 817 00:27:31,269 --> 00:27:35,830 818 00:27:35,830 --> 00:27:38,470 819 00:27:38,470 --> 00:27:40,789 820 00:27:40,789 --> 00:27:42,310 821 00:27:42,310 --> 00:27:44,710 822 00:27:44,710 --> 00:27:47,669 823 00:27:47,669 --> 00:27:49,590 824 00:27:49,590 --> 00:27:51,269 825 00:27:51,269 --> 00:27:51,279 826 00:27:51,279 --> 00:27:53,590 827 00:27:53,590 --> 00:27:53,600 828 00:27:53,600 --> 00:27:59,110 829 00:27:59,110 --> 00:28:00,470 830 00:28:00,470 --> 00:28:02,230 831 00:28:02,230 --> 00:28:03,990 832 00:28:03,990 --> 00:28:07,029 833 00:28:07,029 --> 00:28:08,870 834 00:28:08,870 --> 00:28:11,029 835 00:28:11,029 --> 00:28:13,029 836 00:28:13,029 --> 00:28:14,630 837 00:28:14,630 --> 00:28:17,350 838 00:28:17,350 --> 00:28:19,510 839 00:28:19,510 --> 00:28:20,710 840 00:28:20,710 --> 00:28:23,110 841 00:28:23,110 --> 00:28:26,389 842 00:28:26,389 --> 00:28:30,710 843 00:28:30,710 --> 00:28:32,870 844 00:28:32,870 --> 00:28:35,029 845 00:28:35,029 --> 00:28:36,470 846 00:28:36,470 --> 00:28:38,710 847 00:28:38,710 --> 00:28:38,720 848 00:28:38,720 --> 00:28:39,909 849 00:28:39,909 --> 00:28:41,350 850 00:28:41,350 --> 00:28:42,630 851 00:28:42,630 --> 00:28:45,590 852 00:28:45,590 --> 00:28:48,950 853 00:28:48,950 --> 00:28:50,230 854 00:28:50,230 --> 00:28:53,750 855 00:28:53,750 --> 00:28:55,909 856 00:28:55,909 --> 00:28:57,430 857 00:28:57,430 --> 00:28:59,430 858 00:28:59,430 --> 00:29:02,149 859 00:29:02,149 --> 00:29:03,750 860 00:29:03,750 --> 00:29:05,350 861 00:29:05,350 --> 00:29:07,350 862 00:29:07,350 --> 00:29:08,789 863 00:29:08,789 --> 00:29:09,830 864 00:29:09,830 --> 00:29:12,310 865 00:29:12,310 --> 00:29:14,149 866 00:29:14,149 --> 00:29:15,350 867 00:29:15,350 --> 00:29:17,990 868 00:29:17,990 --> 00:29:19,350 869 00:29:19,350 --> 00:29:21,990 870 00:29:21,990 --> 00:29:23,830 871 00:29:23,830 --> 00:29:26,630 872 00:29:26,630 --> 00:29:28,549 873 00:29:28,549 --> 00:29:30,389 874 00:29:30,389 --> 00:29:32,230 875 00:29:32,230 --> 00:29:34,789 876 00:29:34,789 --> 00:29:34,799 877 00:29:34,799 --> 00:29:35,830 878 00:29:35,830 --> 00:29:37,830 879 00:29:37,830 --> 00:29:40,310 880 00:29:40,310 --> 00:29:41,430 881 00:29:41,430 --> 00:29:43,830 882 00:29:43,830 --> 00:29:45,909 883 00:29:45,909 --> 00:29:47,590 884 00:29:47,590 --> 00:29:49,350 885 00:29:49,350 --> 00:29:51,029 886 00:29:51,029 --> 00:29:53,190 887 00:29:53,190 --> 00:29:54,549 888 00:29:54,549 --> 00:29:56,310 889 00:29:56,310 --> 00:30:00,310 890 00:30:00,310 --> 00:30:02,149 891 00:30:02,149 --> 00:30:04,630 892 00:30:04,630 --> 00:30:07,190 893 00:30:07,190 --> 00:30:10,230 894 00:30:10,230 --> 00:30:12,870 895 00:30:12,870 --> 00:30:14,870 896 00:30:14,870 --> 00:30:18,710 897 00:30:18,710 --> 00:30:18,720 898 00:30:18,720 --> 00:30:20,389 899 00:30:20,389 --> 00:30:20,399 900 00:30:20,399 --> 00:30:22,230 901 00:30:22,230 --> 00:30:24,710 902 00:30:24,710 --> 00:30:26,230 903 00:30:26,230 --> 00:30:28,230 904 00:30:28,230 --> 00:30:28,240 905 00:30:28,240 --> 00:30:29,269 906 00:30:29,269 --> 00:30:29,279 907 00:30:29,279 --> 00:30:35,269 908 00:30:35,269 --> 00:30:35,279 909 00:30:35,279 --> 00:30:37,430 910 00:30:37,430 --> 00:30:38,870 911 00:30:38,870 --> 00:30:40,710 912 00:30:40,710 --> 00:30:40,720 913 00:30:40,720 --> 00:30:43,110 914 00:30:43,110 --> 00:30:45,750 915 00:30:45,750 --> 00:30:49,190 916 00:30:49,190 --> 00:30:51,430 917 00:30:51,430 --> 00:30:53,510 918 00:30:53,510 --> 00:30:55,110 919 00:30:55,110 --> 00:30:55,120 920 00:30:55,120 --> 00:30:55,909 921 00:30:55,909 --> 00:30:55,919 922 00:30:55,919 --> 00:30:57,509 923 00:30:57,509 --> 00:30:59,269 924 00:30:59,269 --> 00:31:01,269 925 00:31:01,269 --> 00:31:02,789 926 00:31:02,789 --> 00:31:04,710 927 00:31:04,710 --> 00:31:06,950 928 00:31:06,950 --> 00:31:09,509 929 00:31:09,509 --> 00:31:09,519 930 00:31:09,519 --> 00:31:11,909 931 00:31:11,909 --> 00:31:11,919 932 00:31:11,919 --> 00:31:13,830 933 00:31:13,830 --> 00:31:15,509 934 00:31:15,509 --> 00:31:17,750 935 00:31:17,750 --> 00:31:20,149 936 00:31:20,149 --> 00:31:22,149 937 00:31:22,149 --> 00:31:22,159 938 00:31:22,159 --> 00:31:24,630 939 00:31:24,630 --> 00:31:26,549 940 00:31:26,549 --> 00:31:26,559 941 00:31:26,559 --> 00:31:27,830 942 00:31:27,830 --> 00:31:27,840 943 00:31:27,840 --> 00:31:28,789 944 00:31:28,789 --> 00:31:31,190 945 00:31:31,190 --> 00:31:33,029 946 00:31:33,029 --> 00:31:34,630 947 00:31:34,630 --> 00:31:37,269 948 00:31:37,269 --> 00:31:39,029 949 00:31:39,029 --> 00:31:40,549 950 00:31:40,549 --> 00:31:42,549 951 00:31:42,549 --> 00:31:44,789 952 00:31:44,789 --> 00:31:46,310 953 00:31:46,310 --> 00:31:48,710 954 00:31:48,710 --> 00:31:51,110 955 00:31:51,110 --> 00:31:51,120 956 00:31:51,120 --> 00:31:52,710 957 00:31:52,710 --> 00:31:54,950 958 00:31:54,950 --> 00:31:56,710 959 00:31:56,710 --> 00:31:59,350 960 00:31:59,350 --> 00:32:01,190 961 00:32:01,190 --> 00:32:03,430 962 00:32:03,430 --> 00:32:06,230 963 00:32:06,230 --> 00:32:08,230 964 00:32:08,230 --> 00:32:10,310 965 00:32:10,310 --> 00:32:13,029 966 00:32:13,029 --> 00:32:14,389 967 00:32:14,389 --> 00:32:16,070 968 00:32:16,070 --> 00:32:18,310 969 00:32:18,310 --> 00:32:18,320 970 00:32:18,320 --> 00:32:20,230 971 00:32:20,230 --> 00:32:22,230 972 00:32:22,230 --> 00:32:22,240 973 00:32:22,240 --> 00:32:23,190 974 00:32:23,190 --> 00:32:25,350 975 00:32:25,350 --> 00:32:28,630 976 00:32:28,630 --> 00:32:31,190 977 00:32:31,190 --> 00:32:32,870 978 00:32:32,870 --> 00:32:35,590 979 00:32:35,590 --> 00:32:37,669 980 00:32:37,669 --> 00:32:39,669 981 00:32:39,669 --> 00:32:41,509 982 00:32:41,509 --> 00:32:44,070 983 00:32:44,070 --> 00:32:45,909 984 00:32:45,909 --> 00:32:47,990 985 00:32:47,990 --> 00:32:50,470 986 00:32:50,470 --> 00:32:51,750 987 00:32:51,750 --> 00:32:52,870 988 00:32:52,870 --> 00:32:52,880 989 00:32:52,880 --> 00:32:54,710 990 00:32:54,710 --> 00:32:57,350 991 00:32:57,350 --> 00:32:58,870 992 00:32:58,870 --> 00:33:00,710 993 00:33:00,710 --> 00:33:02,950 994 00:33:02,950 --> 00:33:04,950 995 00:33:04,950 --> 00:33:07,190 996 00:33:07,190 --> 00:33:09,190 997 00:33:09,190 --> 00:33:11,750 998 00:33:11,750 --> 00:33:14,149 999 00:33:14,149 --> 00:33:16,549 1000 00:33:16,549 --> 00:33:16,559 1001 00:33:16,559 --> 00:33:18,389 1002 00:33:18,389 --> 00:33:19,750 1003 00:33:19,750 --> 00:33:21,830 1004 00:33:21,830 --> 00:33:23,509 1005 00:33:23,509 --> 00:33:25,350 1006 00:33:25,350 --> 00:33:27,750 1007 00:33:27,750 --> 00:33:29,669 1008 00:33:29,669 --> 00:33:31,830 1009 00:33:31,830 --> 00:33:33,269 1010 00:33:33,269 --> 00:33:33,279 1011 00:33:33,279 --> 00:33:34,230 1012 00:33:34,230 --> 00:33:36,830 1013 00:33:36,830 --> 00:33:39,590 1014 00:33:39,590 --> 00:33:43,669 1015 00:33:43,669 --> 00:33:45,990 1016 00:33:45,990 --> 00:33:48,070 1017 00:33:48,070 --> 00:33:51,190 1018 00:33:51,190 --> 00:33:52,789 1019 00:33:52,789 --> 00:33:54,630 1020 00:33:54,630 --> 00:33:57,509 1021 00:33:57,509 --> 00:33:59,350 1022 00:33:59,350 --> 00:34:01,430 1023 00:34:01,430 --> 00:34:03,430 1024 00:34:03,430 --> 00:34:03,440 1025 00:34:03,440 --> 00:34:05,190 1026 00:34:05,190 --> 00:34:06,389 1027 00:34:06,389 --> 00:34:08,310 1028 00:34:08,310 --> 00:34:09,349 1029 00:34:09,349 --> 00:34:11,829 1030 00:34:11,829 --> 00:34:14,550 1031 00:34:14,550 --> 00:34:14,560 1032 00:34:14,560 --> 00:34:19,990 1033 00:34:19,990 --> 00:34:21,909 1034 00:34:21,909 --> 00:34:23,589 1035 00:34:23,589 --> 00:34:25,990 1036 00:34:25,990 --> 00:34:27,349 1037 00:34:27,349 --> 00:34:29,190 1038 00:34:29,190 --> 00:34:30,310 1039 00:34:30,310 --> 00:34:31,990 1040 00:34:31,990 --> 00:34:35,030 1041 00:34:35,030 --> 00:34:35,040 1042 00:34:35,040 --> 00:34:37,589 1043 00:34:37,589 --> 00:34:39,349 1044 00:34:39,349 --> 00:34:39,359 1045 00:34:39,359 --> 00:34:40,869 1046 00:34:40,869 --> 00:34:41,750 1047 00:34:41,750 --> 00:34:44,310 1048 00:34:44,310 --> 00:34:48,550 1049 00:34:48,550 --> 00:34:50,869 1050 00:34:50,869 --> 00:34:52,869 1051 00:34:52,869 --> 00:34:55,589 1052 00:34:55,589 --> 00:34:58,150 1053 00:34:58,150 --> 00:34:58,160 1054 00:34:58,160 --> 00:34:59,109 1055 00:34:59,109 --> 00:34:59,119 1056 00:34:59,119 --> 00:35:00,950 1057 00:35:00,950 --> 00:35:04,390 1058 00:35:04,390 --> 00:35:06,630 1059 00:35:06,630 --> 00:35:09,030 1060 00:35:09,030 --> 00:35:11,349 1061 00:35:11,349 --> 00:35:11,359 1062 00:35:11,359 --> 00:35:12,310 1063 00:35:12,310 --> 00:35:15,030 1064 00:35:15,030 --> 00:35:17,109 1065 00:35:17,109 --> 00:35:18,550 1066 00:35:18,550 --> 00:35:20,470 1067 00:35:20,470 --> 00:35:20,480 1068 00:35:20,480 --> 00:35:21,510 1069 00:35:21,510 --> 00:35:23,109 1070 00:35:23,109 --> 00:35:23,119 1071 00:35:23,119 --> 00:35:28,790 1072 00:35:28,790 --> 00:35:30,470 1073 00:35:30,470 --> 00:35:30,480 1074 00:35:30,480 --> 00:35:32,230 1075 00:35:32,230 --> 00:35:37,829 1076 00:35:37,829 --> 00:35:40,710 1077 00:35:40,710 --> 00:35:40,720 1078 00:35:40,720 --> 00:35:41,670 1079 00:35:41,670 --> 00:35:41,680 1080 00:35:41,680 --> 00:35:42,550 1081 00:35:42,550 --> 00:35:44,550 1082 00:35:44,550 --> 00:35:48,150 1083 00:35:48,150 --> 00:35:50,790 1084 00:35:50,790 --> 00:35:50,800 1085 00:35:50,800 --> 00:35:53,270 1086 00:35:53,270 --> 00:35:53,280 1087 00:35:53,280 --> 00:35:55,589 1088 00:35:55,589 --> 00:35:59,510 1089 00:35:59,510 --> 00:36:01,829 1090 00:36:01,829 --> 00:36:03,270 1091 00:36:03,270 --> 00:36:04,790 1092 00:36:04,790 --> 00:36:04,800 1093 00:36:04,800 --> 00:36:05,829 1094 00:36:05,829 --> 00:36:08,069 1095 00:36:08,069 --> 00:36:10,390 1096 00:36:10,390 --> 00:36:12,870 1097 00:36:12,870 --> 00:36:15,829 1098 00:36:15,829 --> 00:36:17,589 1099 00:36:17,589 --> 00:36:18,950 1100 00:36:18,950 --> 00:36:20,310 1101 00:36:20,310 --> 00:36:22,550 1102 00:36:22,550 --> 00:36:24,470 1103 00:36:24,470 --> 00:36:26,390 1104 00:36:26,390 --> 00:36:28,069 1105 00:36:28,069 --> 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