1 00:00:24,560 --> 00:00:27,589 so welcome everybody welcome back uh let's get started with today's lecture we um where were we on tuesday we covered uh the main topic was the recursion theorem which allows programs to self-reference and we saw some applications of that too so we gave a new proof that atm is undecidable we looked at this language of minimal turing machine descriptions and we had a short digression into mathematical logic where we looked at how one shows that there are true but unprovable statements in any uh reasonable formal system okay so today we're i'm gonna shift gears entirely um and we're moving into the second half of the course uh where we are beginning a study of uh computational complexity theory and uh we'll say a little bit about that during the course of the lecture um of course but um you know the main things that we're going to cover in terms of content that you will need um is defining the complexity classes and the class p um so and we'll prove a few theorems along the way but that's the main objective of today's lecture computability theory which was the subject of the first half of the course and which is what going what the midterm exam is going to cover um was a subject that was an active area of of mathematical uh study in the first part of the 20th century i really got it really dates back um into the late 19th century in fact where people were trying to figure out how to formalize mathematical reasoning um [Music] but that really really got going um in the 1930s with the work of girdle and church and turing um who um really formalized for the first time what we mean by algorithm and that that allowed the study of algorithms to really get started um with and it had its impact as i mentioned on the actual uh design building and thinking about real computers um [Music] the main question if you kind of boil the subject down to a a single question is is some language decidable or not um in complexity theory which got started kind of when uh computability theory more or less wrapped up as a subject largely because they answered many of the questions that they they answered pretty much all of the questions that they were asking so there really aren't uh interesting unsolved questions left in that field and you really need mathematical questions to keep a subject alive unsolvable questions so complexity theory got its start in the 1960s and it continues on as an activator area of research to the present day um and i guess if you could boil it down it would be is is a language decidable with some restriction on the resources such as the amount of time or memory or some other or some other kinds of resources that you might provide you know randomness and so on although those are followed within the area of computational complexity theory so let's get ourselves started with an example here is a language that we've looked at in the past a to the k b to the k and um let's look at it from now from the perspective of complexity all of the languages that we're going to be studying in complexity are all going to be decidable languages um so the question of undecidability undecidability in complexity theory is not really of interest it's all decidable languages but the question is how decidable you know what sort of resources do you need to do the deciding um so um for this language a how many steps are needed well we're going to spend a little time just kind of setting up the definitions of the subject um and and motivating them so in the in this for this language a when i ask how many steps are needed well it's going to depend upon which input you have in mind you know some inputs might require more steps than others so the way we're going to set the subject up which is the the the standard way that uh people in this field look at it um and i think you know that applies to you know lots of examples outside as well um is that we're going to kind of group all of the inputs of the same length together and look at the maximum uh uh cost the maximum number of steps you need to solve any one of those inputs of a given length and we'll do that for each length and the way we're going to frame it is in terms of giving a maximum or what's called an upper bound on the amount of time that you need uh to solve all of those inputs of length n that's what's uh sometimes called worst case complexity i'm sure many of you seen this already but just to make sure we're all together on this um uh you might contrast that for example with what's called average case complexity where instead of looking at the most difficult case among all inputs of length n you take the average of all inputs of length n and then you have to then it's a little bit more complicated because then you need to have a probability distribution on those inputs and so on we're not going to look at it in this course from that perspective we're only going to be looking at what's called worst case complexity um so let's um begin then uh uh by taking uh looking at this in more detail and uh taking as our starting point the theorem that says that on a one tape turing machine which is deciding this language a you know a to the k b to the k um you can do this on a one tape turing machine um m we're calling it in at most some constant times n squared steps for any input of length n where the constant is going to be fixed independent of it okay so this is going to be you know having a constant in in in in factor in the complexity is going to come all come up often and so instead of saying this over and over again we're going to use the notation um that m uses order in squared steps um i'm sure many of you seen that terminology as well but just just for the purposes of making sure we're all together on that there is this big o and little o notation i'm expecting i'm expecting you to be familiar with that big o is when you apply applied to functions as it's done you say uh f is big o of g as for two functions f and g it's basically if f is less than or equal to g if you're ignoring constant factors and you say f is little o of g if f is strictly less than g if you're ignoring constant factors that's kind of one sort of informal way of looking at it the precise definition is given up there on the slide and if you haven't seen it before make sure you look at it in the book where it's all carefully described and defined so that you're comfortable with these notions because it's really we're going to be using this without any further discussion uh from here on okay so let's get to the proof then of this theorem that you can do uh the language a in order n squared steps not super hard if i think if i ask you to come up with an algorithm to solve a this would be the algorithm that you would find basically first you would start off by scanning the input w to make sure it's of the right form so a run of a's followed by a run of b's of some lengths um and if it's not of that form uh then you're going to reject right away the next thing you'll do is then go through a repeat loop um in the turing machine and you know if you imagine here is your machine here is the input w you're going to go through that repeating repeatedly of course you can do this in a number of different ways but here's the way i i'm i have in mind for you uh on this slide anyway we're going to scan the entire tape crossing off one a single a and a single b um on a on a scan and then you're going to keep doing that until you've crossed off everything um unless you run out of a's or you run out of b's in that case you're going to reject um if you run out of a's or b's before you run out of uh the other type then you know that you started out with an unequal number and so the machine is going to reject if you've managed to cross them all off without running out of one before the other then you'll accept i know this is kind of obvious but i think it's important to get us all together on this at the beginning so here's a little um animation of the turing machine doing that i'm not showing the motion of the head but imagine the head scanning back and forth crossing off these a's and b's one a and one b on each pass until they're all crossed off and then it accepts unless of course it runs out of a's or b's before the other type then it rejects okay now let's do a very quick informal analysis of how much time this has taken so the very first stage i'm calling each of these things stages of the turing machine to distinct distinguish them from the steps of the turing machine which the individual transition function moves so this is the entire stage of the machine the very first stage takes order n steps because you have to make a scan across the input and then you know i'm not giving all the full detail of course you're going to scan and then you're going to return the head back to its starting position i'm not that's just going to be an extra factor of n an extra an extra an extra n um so that's what we're talking about being order n steps for the very first stage um and then as you do through goes with the repeat loop each time you go through the repeat loop you're gonna cross off one a and one b so um that's gonna mean you're gonna have to do this roughly order you're gonna have to do this order n times in order to cross off all the a's and b's so there's gonna be edit order and iterations of this repeat loop each one of them is going to again require a state a scan so that's order n steps uh for each one of the iterations um so adding that all up um the very top row gives us the order n and then we have the order and iterations times order n steps is order n squared steps and so the sum of these two is order n square order n squared steps due to the nature of arithmetic when you have the big o notation the dominant term overrides all of the others so that's in a nutshell how we well this is our very first example of analyzing the turing machine algorithm for a language okay so let me ask you now whether there is some other turing machine some other one tape turing machine which can do better than this turing machine um does in terms of how much time it takes so one idea that you might have instead of crossing off a single a or in the single b maybe you can cross off um two a's and two b's or ten a's and ten b's well that'll cut down the running time by a factor of ten but from the standpoint of this theorem it's still going to be an order n squared algorithm um and so that's not gonna that really doesn't change um the amount of time used from from the perspective of um from our perspective where we're going to be ignoring the constant factors on the running time um so i ask you here can you do better than just improving things by a constant factor there we go okay so here's a check-in on this on this uh on this problem on the problem a solving deciding a on a one tape turing machine can we do better than uh order n squared as i just described in that uh in that theorem and that algorithm for that theorem or um you know can you get it down to n log n um or maybe you can get it down to order n um what do you think uh obviously you know we're we're just getting started but just um just make your best guess and let me just post that for you as a poll what's most important to me is that you understand the terminology that we're using and the way we're discussing with the way we're talking about it because that's going to be setting setting us up for the definitions that we're going to come a little bit later in the lecture okay so um i'm going to close the poll um uh we're kind of all over the place on it but that's that's that that's good um uh since we haven't really covered this material yet material yet in fact b is the correct answer we can improve this algorithm down to order n log n but not all the way down to order n um so let me get let me show you how you do it in order n log n on the next slide and so here is a one tape turing machine that can decide a by using only order and log n steps instead of order and squared steps so this is a significant improvement um so i'm just i'll describe the turing machine here again is the picture of the machine on an input um and the very first thing i need to say is that we're going to scan to make sure the input is of the right form and now we're going to again it's going to be making repeated passes over the input but we're going to do it a little differently now instead of crossing off a single a and a single b or some fixed number of a's and a fixed number of b's we're going to cross off every other a and every other b in in that way we're going to essentially cut the number of a's and b's in half and that's why the number of iterations is only going to be a log instead of um linear okay so we're going to cross off every other a and every other b um and at the same time we're going to keep track of the parity the even-odd parity of the number of a's um that we've seen and the s and the parity of the number of b's that we've seen um that have not yet been crossed off um and we're going to compare those parodies to make sure that they agree if they ever disagree we know we started off with different numbers of a's and b's um and i'll i'll illustrate this in a second just to make sure you where um we understand this algorithm okay so i'm going to write down as a little table of the parities that we've seen and i'm going to illustrate the algorithm with a little animation okay so again we're now going to scan across crossing off every other a and then every other b but now but before we get to the b's we observe as we cross these off that we had six a's now i'm not saying we count them we just keep track of the even odd parity that can be done in the finite control of the machine counting them would be more complicated but karen but just keeping track of the parity is not is is something that the finite automaton could do so the parity in this case because they were six is going to be even now we cross off the b's same even parity now we're going to return the head back to the beginning i'm obviously not showing the head moving here we return the head back to the big beginning and now we scan across again crossing off every other remaining a and counting the parity of the remaining a's so here now it's going to be this one and this one are going to get crossed off and there were three a's so that was odd parity and the same for the b's three b's odd parity and now we return our head back to the beginning cross again off uh every other a and every other b every a so that was odd parry there was just one crossing off the b odd parity because it's just one they all agree um so the machine is going to accept um because everything is now crossed off and and the parity is agreed along the way let me just say for a second you know obviously if you ever get disagreement on the parities then you know that the two uh the number of a's and number b's had to disagree um but how do we know that if the parities always agree that we actually did start out with the same number of a's as b's and that you could see that in a number of different ways but perhaps you know uh a cute way to see it is there is actually if you look at these parodies here the sequence of parodies actually in reverse so if you let's say odd odd even um and you look at the binary representation of the number of a's which is six so the binary representation would be one one zero the fact that you get odd odd even and one one zero is not a coincidence in fact the sequence of parodies in reverse that you get is um exactly um the same as the binary representation you'd have to you know confirm that with a little proof it's not hard uh but um that once you've confirmed that you can see that um if the sequence of parities agree then the numbers have to be the same because the binary uh representations agreed okay so now getting to the analysis here again order n steps to do the check login iterations each scan takes order n steps so the total uh running time here is going to be order n log n because it's going to be the log n times n that's where the log n log n comes from now questions you might ask could i do even better than that can i beat n log n by more than any constant factor um so can i do little o of n log n um and the answer is no this is the best possible a one tape turing machine for this language um uh so a one day turing machine cannot decide a by using little o of n log n steps we're not going to prove that i'm not going to ask you to be responsible for the proof but in fact um what you can show is that any language that you can do on a one tape turing machine in little o of n log n steps turns out to be regular okay so um uh you can prove a rather strong theorem here not super difficult to to prove but i don't want to spend a lot of time on um uh proving bounds for turing machines because you know really the whole purpose of setting this up using turing machines is to talk about algorithms in general algorithms in general i'm not going to be focusing on the nitty gritty of turing machines um okay uh so um what is my um okay yeah so i wanted to just stop and make sure that we were we are together here um so a brief pause and feel free to send me any questions um okay this is a good question if we can keep track of parodies why can't we just keep track of the number of a's or b's um well you could keep track of the parity in the finite memory and so you can do that with effectively no work but the finite memory cannot is not enough for doing an account up to you know some arbitrarily large number now you could store the count on the tape but that's going to cost you time to maintain that counter and so um that's not going to be so simple as keeping track of the parity in the finite memory um okay somebody's asking if a star b star is regular yes a star b star is regular so what um but a to the k b to the k which is our language is not regular uh right so that's the language we're looking at um okay so getting questions about what happens with multiple tapes we'll talk about that in a second um yes we could do an order end steps on a regular computer sure but uh you know right for this slide anyway we're looking at one tape turing machines um okay getting questions about a big o and little o um for something to be little little o means it's less than any constant factor times the function um so you have to look at the definition in the book i think almost enough people in the class probably know this and have seen this already i don't want to spend everybody's time on it but uh please review it in the book somebody's asking if you need to store the parody on the tape no you can just throw it in the finite memory um uh i mean storing the finite memory seems to be the simplest thing okay when we move on please feel free to ask questions to the tas as well we have two tas at least here attending uh with me so um good um right um all right so we cannot do better than a one-tape turing machine on a one-tape turing machine and order n log n um and that's something we can prove that we're not going to do it here however if you if you have if you change the model for example you use a two-tape turing machine then yes as a lot of you are suggesting in the chat you can do better than that so if we now have a two-tape turing machine or a multi-tape turing machine you can do it in order n steps and that's actually the point i i'm really trying to make here um so if you have here your um two tape turing machine um then two tips same language now what we're going to do is copy the a's to the second tape that we can do on a single pass um and then once the a's have been copied to the second tape we can continue on reading the b's and match them off with the a's that appear on the second tape so in order n steps you can do the comparison instead of order n log n steps and of course if they match you're going to accept otherwise reject so let's just see here's a little animation uh uh uh demonstrating that of course it's very simple um so here is that here that if you could see that it came maybe a little too fast here let's just show it again here is the heads moving across and the a is coming down uh to the bottom and now the head the upper head is going to continue on reading the bees the lower head is going to go back to the beginning on the a's and matching off the b's with the a's and that's how we can verify or check that they are the same number okay so now in this case they were the same number so the machine would accept if they were the different number the machine would not accept um and the analysis is very simple um uh each stage here is going to take a linear number of steps order n steps um because it just consists of a single uh scan there are no uh there are no loops in this machine no no no repeat loops um okay so um question on this um all right why don't we move on then um now observe and the point i'm really trying to make is that uh on a one type turing machine you can do it in n log n but not any better but on a two tape turning machine you can do it in order n um so there's a difference in how much time you need to spend how many steps you need to spend depending upon the model and that's significant for us um so the number of steps depends on the model one tape turing machine was order and login multitape was order n we call them model dependence if you contrast that with the situation in the complexities in the computability section of the course we had model independence the choice of the model didn't matter and that was nice for us because the theory of decidability didn't depend upon whether you had a one tape turing machine or a multi-tape turing machine it was all the same set of decidable and recognizable languages uh so we didn't have to worry about which model we're actually going to work with we could work with any model even just an informal model of algorithm um was going to would be good enough because we're going to end up with the same notion in the end now that's goes away in complexity theory now we have a difference depending upon the model and from a mathematical standpoint that's a little less nice because um you know which model do you work with if you want to understand the complexity of some problem you know that you have at hand um now you have to make a choice you're going to work with the turing machine around how many tapes or you're going to look at some other model and you're going to get different results so it's somewhat less natural from a mathematical standpoint just to talk about the complexity of some problem but we're going to kind of um bring back something close enough to model independence by observing that even though we don't have uh model independence as we did in computability theory we can limit how much dependence there is so the amount of dependence is going to be low as we will see provided you stick with a reasonable class of deterministic models so the dependence though it does exist is not going to be um that much it's going to be polynomial dependence and we'll say exactly what that means in a second um and from our standpoint that's going to be uh um a small difference uh a negligible difference that we're going to ignore um so we're going to focus on questions that do not depend on the model choice among these reasonable deterministic models now you may say well that's not interesting from a practical standpoint because polynomial differences say the difference between n squared and n cubed certainly make a difference in practice but it really uh depends on the um the uh um you know what what kinds of questions you're focusing in on so if you want to look at something that's very uh a very precise distinction say between you know n squared and n cubed um then you might want to focus in on which model you want to be working with and that's going to be more the domain of an algorithms class but from our standpoint we're going to be looking at other still important questions but there are questions that um don't uh depend upon exactly which polynomial you're going to have we're going to be looking more at distinctions between polynomial and exponential and still there are important practical questions that arise in in that somewhat different um setting okay so with that in mind we're going to continue to use the one tape turing machine as our basic model of complexity um since the the model um among the reasonable deterministic models in the end is not going to matter from the from the perspective of the kinds of questions we're going to be asking um okay um so with that so we are going to continue then it's important to remember that from going forward we're going to stick with the one tape turing machine model maybe that's something you would have expected us to do anyway but i'm trying to justify that you know through this little discussion that we had so far thus far so now we're going to start defining things with the one tape model in mind um so so first of all if you have a turing machine we're going to say it runs in a certain amount of time so if t is some sort of time bound function like n squared or n log n we'll say the machine runs in that amount of time like n squared or n log n if that machine m always halts within that number of steps on all inputs of length n so it always holds within t of n steps on inputs of length n then we'll say that the machine runs in t of n time so in other words if the machine runs in n squared time then the machine when you give it an input of length 10 it's gotta it's got to be guaranteed to hold within a hundred steps uh 10 10 squared 100 steps on every input of length 10. um that's what it means um for the machine to be running in that much time and it has to do that for every every n for every input length and with that we're going to come to the following definition which is highlighted in color because it's going to be um we're going to be using this definition throughout the semester so it's important to understand it this is the definition of what's called the time complexity classes and what i'm going to do is take some bound t of n and again think of t of n like a bound like n squared so if you have time t of n or like time n squared that's going to be the collection of all languages that you can decide within time and square or within time t of n so in other words it's a collection of all languages b such that there's some one tape turing machine here we're focusing again on the one tape turning machine there is some deterministic one tape turing machine that decides b and that machine runs in that amount of time so this is a collection of languages the time collect complexity class is a set of languages i'm going to draw it now as a diagram um so if you take the language again that we've been using as our standard example a to the kb to the k um that's in time and log n as we observed for uh uh two slides two slides back or three slides back so on a one tape turning machine you can do this language a in time n log n so it's in the time complexity class n log n this this region here captures all of the languages that you can do in order and log in time for example that also includes all of the regular languages why is that well any regular language can be done on a one tape turing machine in time order n because the turing machine only just needs to scan across doesn't even need to write just need to scan across from left to right on the on the tape and in n steps it has the answer so all of the regular languages are actually in time n certainly a subset of time n log n and these all form a kind of a hierarchy so if you increase the bound you can imagine that the collect class of languages grows as you allow the machine to have more and more steps to do its computing um so these are all the languages that you can do in n squared uh order n square time on a one tape turing machine and cube time on a one tape turing machine and so on to exponential time two to the end time on a one tape turning machine these are all collections of languages um getting larger and larger um as we increase the bound um so someone is asking kind of is okay uh let's see let me let me get to some of these questions here i'll try to get them in order um uh okay um so somebody says if you have a good question if you have a regular computer so an ordinary sort of random access computer so we'll talk about that in a second can do it in order and uh can you do it on a multi-tape during machine also in order and time actually i don't know the answer teller offhand i suspect the answer is no that ordinary computers have a random access capability that turing machines do not and so that there are going to be some examples of problems that you can do with a random with a regular computer that you cannot do with a multi-tape turing machine in order end time uh i'd have to double check that though so we can and so some question what what we believe is true and what we can prove to be true um as we'll see there are a lot of things that we believe to be true in this subject that we don't know how to prove um somebody's asking kind of an interesting sort of more advanced question is there some function f uh some some some some some function t where um it's so big that it so that time t of n gives you all of the decidable problems um it would be a very big t but the the answer actually to that question is yes um but that's a little bit exotic so let let's not spend a lot of time on that right here but i'm happy to talk about that offline um it's a good question here somebody's asking me does it mean that there are no languages between order n and order n log n because i pointed out that anything below n log n is going to be regular and so as soon as you get below n log n you can do it in order n and yes there is what's called a gap between order n and order n log n on a one take turing machine this you don't get anything new from order n until you jump up so from order n to order n log log n nothing new shows up um so we'll talk about those kinds of things a little a little bit down the road when we look at actually the relationship among these various uh classes and what we call a hierarchy theorem which shows you know what how much bigger do you have to make the bound in order in order to be sure you'll get something new um all right [Music] okay somebody's asking is there a model which has the same time complexity as a normal computer uh well i mean there's the random access model which is supposed to capture a normal computer um okay so let me let me um these are all great questions kind of more uh riffing off of this into in more into more advanced directions uh let's move on um here's another check-in um suppose we take this is a little bit of a check to see how well your how comfortable you are with the notions we've just presented and whether you can think about some of the arguments that we've made and apply them in into a new language so take the language ww reverse um you know strings followed by their um uh this followed by themselves backwards this language via the even length palindromes if you will um what's the smallest uh bound that you need uh to be able to solve uh that language b and i'll pose it as a pose that is a question for you so which which time complexity class is um that language b in is it time order n or n log n n squared so on what do you think so we're about to come to the coffee break so uh why don't we um uh i'll answer any questions that come up i think we're got everybody uh answered so i'm gonna end the polling okay make sure you're in if you want to be in um so the correct answer is in fact uh order n squared um you it would be hard for you know you you know every reasonable guess here would be order n log n i mean you can come up with the same procedure that the as the one we showed at the beginning uh the order n squared procedure for eight to the kb to the k works for ww reverse as well um you can just cross off sweep back and forth crossing off um the symbol from w and going across to the other side crossing off the symbol from w reverse and that procedure will give you an n squared an order n squared algorithm um you might imagine you can improve it to order n log n but you cannot you can prove that order n squared is the best possible um i i'm a little unhappy that a lot of you came up with order n frankly because i already told you that order n is uh these are just regular languages anything that you can do in less than little o of n log n is going to be regular and we know this language is not regular so this is a this was uh not a good answer you know so please pay attention and um okay so let us stop sharing uh i will turn now to the uh our our our break uh for five minutes and i'm happy to you know try to take questions along the way as we're um uh we're waiting for the time to to end so let's see let me put this up here okay let me try to take some of your questions um so someone has asked me about quantum computers as reasonable models of you know you may say a quantum computer is a reasonable model of computation and that's fine i would not say it's a reasonable model deterministic computation um uh at least from our standpoint though let's not quibble about the words um i'm not including quantum computers in um the collection of um machines that i have in mind right now when i'm talking about the the the reasonable models of deterministic computation that we're going to be we're going to be discussing um uh let's see okay a bunch of people apparently are asking the tas why all regular languages can be done in order n um [Music] so if you think about a dfa which you know processes a an input of length n within steps and the dfa is um going to be a type of turing machine that never writes on its tape uh so if the dfa can do it in in in n steps the turing machine can do it n steps and so therefore every regular language can be done in order and steps on a turing machine um not sure where the confusion is so please message me if you're not still not getting it um okay somebody's saying why are we using one-take turing machines instead of random access wouldn't it be better to use the random access machines um if you were using if you're trying to do algorithms yes that's a more reasonable model we're trying to prove things about the computation and from that standpoint we want to use as simple as model as possible uh trying to prove things um using random access computers it's possible be very messy so that's why we don't use uh random access machines to prove the kinds of things we're going to be proving um about computation you know that that are really the meat and potatoes of this course so uh i mean there's compelling reasons why you would want to use a simple model like a turing machine but not a powerful model like a random access computer so somebody's asking me does um the class the time uh order and log log n have any elements yes it has all the regular languages but nothing else order n log log n is only read only the regular languages you have to go all the way up to or and log in before you get something non-regular uh somebody's asking me are we going to talk about how the random access model works no you know there's uh beyond the scope of this course outside of what we're going to be doing we're going to talk about turing machines not because we care so much about turing machines but i'm trying to prove things about computation and the turing machines are a convenient vehicle for doing that um okay our candle has burned out uh why don't we return then to um the next uh slide so everybody come back so this answers one of the questions i got on the chat is what's what actually is the dependency between multi-tape turing machines and one tape turing machines can we bound that in general yes we can we're going to show that converting a multi-tape turing machine to a one-tape turing machine can at most blow up the amount of time that's necessary by a squaring um no i acknowledge it's a lot but it still allows you um uh but you know it's still small compared with an exponential increase and we're going to be focusing in this course on things like the difference between polynomial and exponential not between the different not not between the difference uh of n not between not the difference between n squared and n cubed that's going to be less of a fact less of an issue from us uh okay um [Music] so the way i'm framing uh showing this theorem is that if you have a multi-tape turing machine that can do a language in a certain amount of time then it's in the time complexity class of that time bound squared and the way i'm just saying that is because this is the bound that's utilizing the one tape model so another way of saying that is converting multi-tape to one tape squares the amount of time you need at most so the way we're going to prove that is simply by going back and remembering the conversion that we already presented from multi-tape to one tape and observe that if we analyze that conversion it just uh ends up squaring the amount of time that the multitape used okay so why is that so if you remember let's just make sure we're all altogether on this the way the single tape machine s simulates the multi-tape turing machine m is that it takes the contents of each of m's tapes you know up to the place where there's infinitely many blanks obviously you don't store the infinite part but the the the the active portion of each of them's tapes you're going to store them consecutively in separate blocks on s's tape on estes only tape um and now every time m makes one move s has to scan its entire tape to uh to uh to see what's under each of the heads and to do all the updating so to simulate one take step of m's computation s is going to use order t of n steps where t of n is the total running time that m is going to use so why is t of n steps coming up here well that's because you have to measure how s is going to make a scan of course its tape how big can its tape be well m if it's trying to use as much tape as possible can use at most t of n um tape on each of the t event cells on each of its tapes so all together there's just going to be some constant number of times t of n cells on s's tape do you see that so each one of these is going to be at most t of n long so this all together is going to be order t of n log because you know what can m do it could send its head out say the head on this tape here moving as fast as possible to the right using as much tape as it can but you can only use t event cells in t of n time so this is going to be ordered t of n so one step of m's computation is going to be t of n steps on s's computation but m itself has t of n steps so it's going to be t of n times t of n for the total number of steps that s is going to end up using and that's where the squaring comes from okay similar results i'm not going to do lots of simulations of one model by another i think that you'll get the idea and you can if you're interested you can uh study those on your own but you can convert multi-dimensional turing machines to one one tape turing machines um you know one tape ordinary uh linear one tape one dimensional uh machines and the bottom line is that among all of the reasonable models they're all what are called polynomially related uh if each can simulate the other with at most the polynomial overhead so if one of the machines can use this t of n time the other machine that's simulating it would use t to the k of n time for some k um that's what it means for the two machines to be polynomially related and all reasonable deterministic models are polynomially related so as we've already seen one tape in multi-tape turing machines are polynomial related because converting multi-tape to one tape blows you up by most squaring so k equals two in this case um multi-dimensional turing machines again polynomial related the random access machine which i'm not going to define but it's the machine that you might imagine you know you would i'm sure they must define in some form in the algorithms classes polynomial related cellular automata which are just arrays of a finite automata that can communicate with each other similar similarly um you know all the reasonable deterministic models um you know again classical models i'm not talking about quantum computing are polynomial polynomially related so we are um uh that kind of justifies our uh choice in picking one of them as long as we're going to ask questions which don't depend upon the polynomial so let's then talk about the class p so the class p this is an important definition for us this is um the collection of all languages that you can do in time into the k for some k on a one take turing machine or as i've written it over here i don't know if this notation is um unfamiliar to you but this is like just a big sum um but here it's a big union symbol it's union over all values of k of the time class n to the k so this is time n union to time n squared union time n cubed union time into the fourth and so on um and we call these the polynomial time decidable languages okay so we're going to be spending um you know uh a certain amount of effort uh exploring this class p and other similar classes somebody's asking why is it a union i'm not sure how else you would write it so if somebody hasn't if you have a proposal for a different way to write it that's fine but this is you know k this is for all k i don't know you know there's if you only had a limit finite number of k's you could just take the biggest one but since it's for all k you need to write it as a union um now i want to argue that the class p um is an important class and you know why is it had so much impact on the subject um and in terms of applications as well um so one thing is that the class p is invariant for all reasonable deterministic models and what do i mean by that so we have defined the class p in terms of these time classes here which in turn are defined in terms of the one tape model so we have defined p by using one tape turing machines now if we had to find p in terms of multi-tape turing machines we get exactly the same class because one tape and multi-tape turing machines are polynomially related to one another and since we're taking the union over all polynomials that polynomial difference is going to wash out um similarly we could define p using any of the other reasonable deterministic models we get exactly the same class so in a sense we get back what we um the situation that we had in computability theory when the class of decidable languages didn't depend on the choice of model here the class p does not matter depending upon the choice of reasonable deterministic model um and we also kind of even in the case of computability theory we have to stick with kind of reasonable models that cannot do an infinite amount of work in one step um so uh you know when i'm not going to define what it means to be reasonable and that's in a sense an informal notion but among all of those reasonable models you're going to get the same class p the other thing that makes p important is that p roughly corresponds to the problems that you can solve in some reasonable practical sense not exactly you know problems that require into the hundredth time um you could argue cannot be uh solved in any reasonable sense um but uh you know if you think about it for example from the perspective of cryptography um uh cryptographic codes that people come up with are typically designed to require or or the hope is that they require an exponential amount of effort to crack if someone found even an into the hundredth algorithm to crack uh that would crack a code people would feel that the code is not secure even though into the hundredth is still a large um so you know it's a it's a rough kind of test but it's still used as a kind of litmus litmus test for practical solvability if you can solve it in polynomial time you basically figured out how to avoid um large searches if you can solve problems in in polynomial time we'll say more about that later but you know what i want to bring out here is that we have combined here in the class p something that's mathematically nice mathematically element elegant with something that's practically relevant and when you have a combination of the two then you know you have a winner then you know how you have a concept that's going to make a difference and that's been true for the class b this has been very influential within and without the theory of computation um so let's look at an example um let's let's define a new language we haven't seen before though it's similar to procedures that we've looked at before the path language which is where i'm going to give you a graph g two nodes in the graph s and t um where i'm thinking of g is a directed graph so directed means that the um the connections between the nodes in g are going to be redirected and that they have arrows on them they're not just lines but they have an orientation with an arrow um so g is a directed graph that has a path from s to t that respects the directions um so such a i think i might even have a picture here yeah so imagine here here is your graph if you can see it there are little arrows connecting the uh the nodes and i want to know is there a path from the node s to the node t so that is you know a picture of a problem in you know an instance of a graph of a path problem and i want to find uh an algorithm for that and i can show that there is an algorithm that operates in polynomial time for this uh path problem okay and the algorithm uh you know any of the standard searching algorithms uh would work here but let's just for completeness sake um include the breadth first search algorithm that we have explored previously when we talked about finite automata um so we'll mark s and they'll keep repeating until nothing new is new is marked and we'll mark all of the nodes that were reachable by a single arrow from a previously marked node um and then cft is marked after you could mark everything you can get to so you're going to mark let's see pictorially i think i have this in indicated yeah you're going to mark all of the things that are reachable from the start not from the from the node s and then see after you can't mark anything new whether the node is marked and if it is you'll accept if it is not you reject okay um now we can analyze this too um and i'm not going to be spending a lot of time analyzing algorithms here um but let's just uh do the do it kind of this one time uh again um we're gonna we're get we're doing a bunch of iterations here so gonna be repeating until nothing new is marked so each time we mark something new we can only do that at most end times at which point we've marked everything so the number of iterations here is going to be at most n and now for each each time we've marked something we have to um look at all of the previously marked nodes um and see which things they point at to mark them too so this is going to be an inner loop which again has at most n iterations because it's going to all of the previously marked nodes and then once we have that we can scan g to see to mark all of the new uh all of the nodes which we have not yet marked uh whether they're connected with a previously marked node by an edge and i'm being generous here because i don't really care um this can be done in the most n squared steps on a one-take turing machine i'm not going to describe the implementation but you know i'll leave it to you as an exercise but this is straightforward so the total number of steps here would be n iterations times n iterations times and squared so you're going to be at most into the fourth steps needed so this is a polynomial algorithm and whether i ended up with n to the fourth or n to the fifth or n cubed i don't really care um because i'm just trying to illustrate um that the total is polynomial and that's all i'm going to be typically asking you to do um so to show polynomial time what i'd be asking you to do is to show that each stage each stage of this algorithm should be clearly polynomial and that the total number of stages i'm sorry this should say stages here should be polynomial so each stage is polynomial and after you're doing all the iterations the total number of stages that are executed is polynomial and so therefore all together the total running time the total number of steps is going to be polynomial okay so that's um the uh the way we would write up um polynomial algorithms in this class um so let's see if there's any questions here i don't want to get too far ahead of people um let's see um yes in this theorem i'm talking about one tape turing machines um because we're defining everything in terms of one type machines but now at this point you know we're talking about polynomial time we can you know my analysis is based on one tape turning machines but in general you know you could use any reasonable deterministic model on which to carry out your analysis uh because they're all polynomially equivalent so from the perspective of coming up with showing that a problem is in polynomial time isn't p um you can use any of the models um you know that you that you wish that for convenient okay oh that's a good question what is n thank you for asking that question n is always going to be reserved to indicate the length of the input so here n is going to be when we encode g s and t and here also i haven't said this but i'm assuming that the encoding that you're using is also somehow reasonable you know i think we'll talk a little bit about more about that in the next lecture um which is going to be after the midterm um but you know you can cause problems if you intentionally try to come up with nasty encodings which will you know represent things uh with unnecessarily many characters but um you know if if you try to be reasonable then um you just use any one of those encodings and you'll be fine um so yeah um so n is the length of the representation of the input um let's see someone's trying to you're trying to dig into the my the actual how this is running here scan g to mark all y where x y is an edge i'm saying that you can do an n squared steps um uh um you know if you have x you can mark it in a certain place on the tape and then as you're going through every other node every other edge um you can you can go back and compare x with the x of the edge and then c and then find the y i mean it's just going to be too messy to talk about here i mean i i'll leave it this is an exercise to you i'm not going to try to fumble my way through explaining why you can do this in n squared steps but it's it's not hard um uh you guys are all like hardcore algorithms folks you want to know the algorithms for this i'm not going to do that sorry uh high level picture here um you know so so you know if you want you know if you want to look at detailed analyses this is not the right course for you um can k equal n what is k um um [Music] no this if you're talking about this k here k cannot equal n we're not looking at n to the n um this is these are all fixed k um it's like n squared and q but not into the end is going to be an exponential bound and so that's not going to be included within uh this union okay so we're near the end of the hour um i'm gonna introduce one last language here um called handpath and the handpath problem is um i'm going to ask now again for a path from s to t but now a different kind of a path one that goes through every node of g along the way um so i'm looking for a path that goes that hits every note of g not just the shortest most direct path but in a sense the most indirect pair the longest path that goes through from from s to t that visits everything else along the way um a path that of that kind that hits every node of the graph is called the hamiltonian path um because the mathematician hamilton studied um those made some definitions about that i'm not gonna i don't actually know the history there but i just know they're called hamiltonian pets um so here's a picture i want to get from snt but i want to sort of pick up everything else along the way so as you remember the path problem itself can be solved in p and what i'd like to know can the simple modification where i'm asking you to visit everything else along the way is that problem also in p okay and i'm going to put it pose this as a checking for you um but actually before i get to that let me give you an algorithm for handpath um that doesn't work to show it's in p um because it's exponential so here's an algorithm for handpath um uh where let's m be the number of nodes in g um and what i'm going to do is i'm going to try every can every possible path in g um and see if it's actually works as a hamiltonian path and except if it is um and then if all paths fail uh then i'll reject so i'm going to try every possible routing through g if you want to think about it think of m as every possible permutation of the nodes of g and then you're going to see whether that's actually constitutes a path in g um that takes you from s to t and goes through all of the nodes um so this algorithm would work this would give you a correct algorithm for the hand path problem the problem is the difficulty is that there are so many possible paths that it's going to take you an exponential number of steps to execute this algorithm it's not a polynomial time algorithm because there are many possible pairs that you could go through um if you're looking at you know with a very crude bound but you really can't improve that significantly there would be um m-factorial which is going to be you know much greater than 2 to the m paths of length m um so the algorithm is going to run for exponential time and not polynomial time so my question for you is i'm going to pose it as a check-in problem is whether you could actually do this problem in polynomial time so why don't you think about that as i'm setting up the question so take the hand path problem just like the path problem which i described with that marking algorithm but now you want to hit every node along the way can you show that problem is solvable in p and there's a whole range of possibilities here where either the answer is yes you can see what the polynomial time algorithm is uh to uh to definitely know where you can prove there is no such polynomial time algorithm and i'll put this as a checking for you and see i'm curious to see what you come up with most people are getting it wrong well i mean wrong i don't you know i'm not clear what wrong is here um okay are we done please uh check something you can see a few of you have not answered but the poll is running out okay time is up um so uh in fact you know as as i think many of you know but not all of you uh this is an unsolved problem this is a very famous unsolved problem which is equivalent to the p versus np problem that we're going to be talking about very soon which among other things would be worth a million dollars if you solve it so for those of you who have answered a or e please talk to me after lecture um and maybe we can work on it together um [Music] uh no so um yeah i i think most people would believe that the answer is no um uh but no one knows how to prove it at this time uh so i'm interested in the uh folks who have uh come up with uh what they think are solutions um and i should say that there are some folks who um believe that there might be other outcomes besides uh just a simple note um which might be proven eventually um so we're going to talk more about this um but this is the answer to the question just for your uh just to make sure you understand is that it's an unsolved problem right now so we don't we don't know uh definitely yes and definitely no at least according to the state of knowledge of which i'm aware is not um are not correct answers um but any of the others well who knows um so i think that's the end of what i had to say for today we covered complexity theory as an introduction looked at different possible models focused on the one tape model introduced based on the one tape model these uh complexity classes the class p and we showed an example of this path problem being in p talked also about this hand path problem which we'll talk about more um after the midterm okay so um i'll stick around for a few minutes uh if you have any further questions otherwise um uh okay so let me just take questions here um somebody's asking me about my personal opinion on p versus np um my personal opinion on people versus np is that p is not equal to np and that we will prove it someday when i was a graduate student back in the mid 70s i thought it would be solved by now and in fact i made a bet with len adelman who i subsequently ended up becoming the a of the rsa code um that we would solve it by the year 2000 and i bet what was then a lot of money for me which was an ounce of gold which i didn't end up then which i did end up paying off to lend in the year 2000 so i'm not making any more bets but i still believe that it will be solved uh hopefully i'll get a chance to see the solution um i've spent a lot of time thinking about it myself you know obviously i haven't solved it otherwise we would know but um uh you know uh hopefully somebody will i'm getting asked a question here that's kind of an interesting question but i don't really know um i'm sure i understand it what's the largest possible run time of a decidal decidable problem what is the larger largest decidable runtime so you anything that i can describe can be um there are going to be algorithms that run for longer uh you can define an algorithm you can define a runtime which would in a sense beats all other runtimes which so that anything any runtime is going to be um dominated by that extremely slow runtime but it's not something that one can describe um i can describe it to you by perceive mathematical procedure but there's it's not going to be something like you know uh 2 to the 2 to the n somebody's here proposing they prepare a solution to the hand path problem i presume in polynomial time why is the following flawed if s goes through all nodes and ends up at t well s will not you mean you mean i presume you mean starting at s if we end up going through all nodes and and end of t um the proposal is a little complicated here basically you you want to if if i can try to rephrase it you want to try to from every possible node you want to try to calculate a path to t and also a path from s to that node um and you can do that for all possible nodes but there's no way to really combine them into a single path that visits all you know solving for each node separately is not going to do the trick because you have to somehow combine all that information into a single path just one path that goes from s to t and visits all the other nodes along the way and that that is not uh i i don't see how what your proposal um how that's how that's going to actually work professor spitzer yes um a question on the um like we were kind of talking about earlier but like what we talked about today was defined for the one tape turn machines correct yep so um and you said like we could have applied for the multi multitape ones but um are like is well i don't know if it's we talked about earlier but if something is accepted by the one tape turn machine like can be applied to the multi-tape turn machine and vice versa like are they interchangeable like that well they're interchangeable only in the sense that the amount of time that you would need it's a different machine if you have a multi-tape turing machine for some language um you can convert it to a one-tape turing machine using a procedure that we described it earlier in the term and she'll get a different machine it's going to run for a different amount of time by using their procedure the point is is that the amount of time that the one-take turing machine is going to run for is not that much worse than the multi-tape turing machine's time so if the multi-tape turning machine's time was you know uh end cubed the one tape turning machines uh is going to be the square of that so it's going to be into the sixth okay but it's not going to be it's not going from multi-tape to one tape is not going to convert you from polynomial to exponential that's that's the only point i'm trying to make it's going to convert from one polynomial to a somewhat bigger polynomial but it's still going to leave you polynomial [Music] i don't know you don't seem to i i can't see your face no no yeah i guess i guess my opinion is that like especially when we were talking about earlier um when you're bringing it up it just seemed like you could just turn anything to a multi-tape turn machine and like completely like cut the timeout like if i had like something to n log n like if i do it in a multi-tape turn machine i have it in like um like big o of n yeah i mean like it just seemed like the multi-tape was so much more powerful but then i guess i guess not with the like with the explanations and like the the models we were talking about today yeah i would not say the multi multi-tape turing machines are still pretty limited in their capabilities and they're um you know don't forget when you have a multi-tape turing machine you have only a fixed number of tapes i mean you could also define variations of multi-tape turing machines that have increasing number of tapes as the input uh you know under you know either under program control it can launch new tapes or it could um um you know just have more tapes depending upon the size of the input that would also be a possibility that's not the model that we have defined but you could define a model like that um but as long as the total amount of work being done by the machine at any step it's going to be a polynomial amount of work um then you can convert it to a one tape turing machine uh with only a polynomial increase in the bound all right so you know you want to be careful of machines you know you know one one thing i meant to say but didn't say you know so he would be an unreasonable model um which you might think of as a plausible model but it's not going to be a reasonable model from our standpoint and that would be a model um for example that can do um full precision say integer arithmetic um uh with a unit cost per operation so each operation counts as a costs one um but i'm going to allow you to do for example you know addition and multiplication uh the the thing that's bad about that um in terms of being uh unreasonable is that uh after k step each time you do a step you could double the size of the integer by by squaring it yeah so after k steps you could have an integer which is 2 to the k long and now doing operations there is going to involve an exponential amount of work even in any reasonable sense um you know in a theoretical sense and also in a practical sense so um you know a model that operates like that is not going to be able to convert to a one-tape turing machine with only a polynomial increase because it's doing an exponential amount of work potentially you know uh within a polynomial number of steps um so that's within a linear number of steps within n steps so that's um an example of an uh you know of an unreasonable deterministic model yeah thank you sure so i'm just curious some idea just occurred to me um i guess like if you had an oracle turing machine uh basically just so that you could look at a uh a turing machine description and decide whether it's a decider or not yeah um then it'd be a little bit interesting to think about uh what happens if you look at all uh you know if you have a description of a pair of a turing machine and an input string and you can look at for all size n descriptions of a pair uh like what's the most steps that it takes uh for a such a machine to converge so then so then you'd have combined turing machine and input stream descriptions uh where you can look at what's the longest it takes to converge it's just i don't know just a random thought that occurred yeah so we we actually we're gonna we we will talk about oracle turing machines uh later on in the term um you know these are machines that have access to sort of free information um and that actually turns out to be uh i mean there's some interesting interesting things you can say about what what happens when you're providing a machine with in a sense information for free that you know you might otherwise want to charge it for actually computing that information but let's just say we're going to allow it to uh get that information without being charged and then how does that affect the complexity of other problems for example um and so uh we will talk about that later but too much if i think of a digression at this moment to try to define all that um but happy to chat with you about it on piazza if you if you want to raise a question there i think sure oh no problem um so he's asking me about strategies for solving the p versus np problem you know we will talk also talk about that a little later in the term as well but you know asking you know clearly you know it seems beyond the reach of our present techniques to be able to prove that some problems you know really take a long time like the hamiltonian path problem seems like you know there's nothing really you can do that's significantly better than trying all possibilities as i described in that exponential algorithm you know on the last slide uh but um how do you prove that nobody knows um so um lots of people have tried including yours truly i mean i spent a lot of time thinking about it having succeeded with it but um there is uh um you know somebody i i think somebody someday someday somebody will come up with the the new idea that's needed to solve it but it's going to clearly take some sort of a breakthrough some sort of a new idea it's not just going to be a combination of existing ideas the existing methods um i think we're about 10 minutes past a few of you are still here i'm going to say goodbye to you folks and um um shortly i'm going to join my tas for our t weekly ta meeting so see you guys thanks for being here 2 00:00:27,589 --> 00:00:29,189 so welcome everybody welcome back uh let's get started with today's lecture we um where were we on tuesday we covered uh the main topic was the recursion theorem which allows programs to self-reference and we saw some applications of that too so we gave a new proof that atm is undecidable we looked at this language of minimal turing machine descriptions and we had a short digression into mathematical logic where we looked at how one shows that there are true but unprovable statements in any uh reasonable formal system okay so today we're i'm gonna shift gears entirely um and we're moving into the second half of the course uh where we are beginning a study of uh computational complexity theory and uh we'll say a little bit about that during the course of the lecture um of course but um you know the main things that we're going to cover in terms of content that you will need um is defining the complexity classes and the class p um so and we'll prove a few theorems along the way but that's the main objective of today's lecture computability theory which was the subject of the first half of the course and which is what going what the midterm exam is going to cover um was a subject that was an active area of of mathematical uh study in the first part of the 20th century i really got it really dates back um into the late 19th century in fact where people were trying to figure out how to formalize mathematical reasoning um [Music] but that really really got going um in the 1930s with the work of girdle and church and turing um who um really formalized for the first time what we mean by algorithm and that that allowed the study of algorithms to really get started um with and it had its impact as i mentioned on the actual uh design building and thinking about real computers um [Music] the main question if you kind of boil the subject down to a a single question is is some language decidable or not um in complexity theory which got started kind of when uh computability theory more or less wrapped up as a subject largely because they answered many of the questions that they they answered pretty much all of the questions that they were asking so there really aren't uh interesting unsolved questions left in that field and you really need mathematical questions to keep a subject alive unsolvable questions so complexity theory got its start in the 1960s and it continues on as an activator area of research to the present day um and i guess if you could boil it down it would be is is a language decidable with some restriction on the resources such as the amount of time or memory or some other or some other kinds of resources that you might provide you know randomness and so on although those are followed within the area of computational complexity theory so let's get ourselves started with an example here is a language that we've looked at in the past a to the k b to the k and um let's look at it from now from the perspective of complexity all of the languages that we're going to be studying in complexity are all going to be decidable languages um so the question of undecidability undecidability in complexity theory is not really of interest it's all decidable languages but the question is how decidable you know what sort of resources do you need to do the deciding um so um for this language a how many steps are needed well we're going to spend a little time just kind of setting up the definitions of the subject um and and motivating them so in the in this for this language a when i ask how many steps are needed well it's going to depend upon which input you have in mind you know some inputs might require more steps than others so the way we're going to set the subject up which is the the the standard way that uh people in this field look at it um and i think you know that applies to you know lots of examples outside as well um is that we're going to kind of group all of the inputs of the same length together and look at the maximum uh uh cost the maximum number of steps you need to solve any one of those inputs of a given length and we'll do that for each length and the way we're going to frame it is in terms of giving a maximum or what's called an upper bound on the amount of time that you need uh to solve all of those inputs of length n that's what's uh sometimes called worst case complexity i'm sure many of you seen this already but just to make sure we're all together on this um uh you might contrast that for example with what's called average case complexity where instead of looking at the most difficult case among all inputs of length n you take the average of all inputs of length n and then you have to then it's a little bit more complicated because then you need to have a probability distribution on those inputs and so on we're not going to look at it in this course from that perspective we're only going to be looking at what's called worst case complexity um so let's um begin then uh uh by taking uh looking at this in more detail and uh taking as our starting point the theorem that says that on a one tape turing machine which is deciding this language a you know a to the k b to the k um you can do this on a one tape turing machine um m we're calling it in at most some constant times n squared steps for any input of length n where the constant is going to be fixed independent of it okay so this is going to be you know having a constant in in in in factor in the complexity is going to come all come up often and so instead of saying this over and over again we're going to use the notation um that m uses order in squared steps um i'm sure many of you seen that terminology as well but just just for the purposes of making sure we're all together on that there is this big o and little o notation i'm expecting i'm expecting you to be familiar with that big o is when you apply applied to functions as it's done you say uh f is big o of g as for two functions f and g it's basically if f is less than or equal to g if you're ignoring constant factors and you say f is little o of g if f is strictly less than g if you're ignoring constant factors that's kind of one sort of informal way of looking at it the precise definition is given up there on the slide and if you haven't seen it before make sure you look at it in the book where it's all carefully described and defined so that you're comfortable with these notions because it's really we're going to be using this without any further discussion uh from here on okay so let's get to the proof then of this theorem that you can do uh the language a in order n squared steps not super hard if i think if i ask you to come up with an algorithm to solve a this would be the algorithm that you would find basically first you would start off by scanning the input w to make sure it's of the right form so a run of a's followed by a run of b's of some lengths um and if it's not of that form uh then you're going to reject right away the next thing you'll do is then go through a repeat loop um in the turing machine and you know if you imagine here is your machine here is the input w you're going to go through that repeating repeatedly of course you can do this in a number of different ways but here's the way i i'm i have in mind for you uh on this slide anyway we're going to scan the entire tape crossing off one a single a and a single b um on a on a scan and then you're going to keep doing that until you've crossed off everything um unless you run out of a's or you run out of b's in that case you're going to reject um if you run out of a's or b's before you run out of uh the other type then you know that you started out with an unequal number and so the machine is going to reject if you've managed to cross them all off without running out of one before the other then you'll accept i know this is kind of obvious but i think it's important to get us all together on this at the beginning so here's a little um animation of the turing machine doing that i'm not showing the motion of the head but imagine the head scanning back and forth crossing off these a's and b's one a and one b on each pass until they're all crossed off and then it accepts unless of course it runs out of a's or b's before the other type then it rejects okay now let's do a very quick informal analysis of how much time this has taken so the very first stage i'm calling each of these things stages of the turing machine to distinct distinguish them from the steps of the turing machine which the individual transition function moves so this is the entire stage of the machine the very first stage takes order n steps because you have to make a scan across the input and then you know i'm not giving all the full detail of course you're going to scan and then you're going to return the head back to its starting position i'm not that's just going to be an extra factor of n an extra an extra an extra n um so that's what we're talking about being order n steps for the very first stage um and then as you do through goes with the repeat loop each time you go through the repeat loop you're gonna cross off one a and one b so um that's gonna mean you're gonna have to do this roughly order you're gonna have to do this order n times in order to cross off all the a's and b's so there's gonna be edit order and iterations of this repeat loop each one of them is going to again require a state a scan so that's order n steps uh for each one of the iterations um so adding that all up um the very top row gives us the order n and then we have the order and iterations times order n steps is order n squared steps and so the sum of these two is order n square order n squared steps due to the nature of arithmetic when you have the big o notation the dominant term overrides all of the others so that's in a nutshell how we well this is our very first example of analyzing the turing machine algorithm for a language okay so let me ask you now whether there is some other turing machine some other one tape turing machine which can do better than this turing machine um does in terms of how much time it takes so one idea that you might have instead of crossing off a single a or in the single b maybe you can cross off um two a's and two b's or ten a's and ten b's well that'll cut down the running time by a factor of ten but from the standpoint of this theorem it's still going to be an order n squared algorithm um and so that's not gonna that really doesn't change um the amount of time used from from the perspective of um from our perspective where we're going to be ignoring the constant factors on the running time um so i ask you here can you do better than just improving things by a constant factor there we go okay so here's a check-in on this on this uh on this problem on the problem a solving deciding a on a one tape turing machine can we do better than uh order n squared as i just described in that uh in that theorem and that algorithm for that theorem or um you know can you get it down to n log n um or maybe you can get it down to order n um what do you think uh obviously you know we're we're just getting started but just um just make your best guess and let me just post that for you as a poll what's most important to me is that you understand the terminology that we're using and the way we're discussing with the way we're talking about it because that's going to be setting setting us up for the definitions that we're going to come a little bit later in the lecture okay so um i'm going to close the poll um uh we're kind of all over the place on it but that's that's that that's good um uh since we haven't really covered this material yet material yet in fact b is the correct answer we can improve this algorithm down to order n log n but not all the way down to order n um so let me get let me show you how you do it in order n log n on the next slide and so here is a one tape turing machine that can decide a by using only order and log n steps instead of order and squared steps so this is a significant improvement um so i'm just i'll describe the turing machine here again is the picture of the machine on an input um and the very first thing i need to say is that we're going to scan to make sure the input is of the right form and now we're going to again it's going to be making repeated passes over the input but we're going to do it a little differently now instead of crossing off a single a and a single b or some fixed number of a's and a fixed number of b's we're going to cross off every other a and every other b in in that way we're going to essentially cut the number of a's and b's in half and that's why the number of iterations is only going to be a log instead of um linear okay so we're going to cross off every other a and every other b um and at the same time we're going to keep track of the parity the even-odd parity of the number of a's um that we've seen and the s and the parity of the number of b's that we've seen um that have not yet been crossed off um and we're going to compare those parodies to make sure that they agree if they ever disagree we know we started off with different numbers of a's and b's um and i'll i'll illustrate this in a second just to make sure you where um we understand this algorithm okay so i'm going to write down as a little table of the parities that we've seen and i'm going to illustrate the algorithm with a little animation okay so again we're now going to scan across crossing off every other a and then every other b but now but before we get to the b's we observe as we cross these off that we had six a's now i'm not saying we count them we just keep track of the even odd parity that can be done in the finite control of the machine counting them would be more complicated but karen but just keeping track of the parity is not is is something that the finite automaton could do so the parity in this case because they were six is going to be even now we cross off the b's same even parity now we're going to return the head back to the beginning i'm obviously not showing the head moving here we return the head back to the big beginning and now we scan across again crossing off every other remaining a and counting the parity of the remaining a's so here now it's going to be this one and this one are going to get crossed off and there were three a's so that was odd parity and the same for the b's three b's odd parity and now we return our head back to the beginning cross again off uh every other a and every other b every a so that was odd parry there was just one crossing off the b odd parity because it's just one they all agree um so the machine is going to accept um because everything is now crossed off and and the parity is agreed along the way let me just say for a second you know obviously if you ever get disagreement on the parities then you know that the two uh the number of a's and number b's had to disagree um but how do we know that if the parities always agree that we actually did start out with the same number of a's as b's and that you could see that in a number of different ways but perhaps you know uh a cute way to see it is there is actually if you look at these parodies here the sequence of parodies actually in reverse so if you let's say odd odd even um and you look at the binary representation of the number of a's which is six so the binary representation would be one one zero the fact that you get odd odd even and one one zero is not a coincidence in fact the sequence of parodies in reverse that you get is um exactly um the same as the binary representation you'd have to you know confirm that with a little proof it's not hard uh but um that once you've confirmed that you can see that um if the sequence of parities agree then the numbers have to be the same because the binary uh representations agreed okay so now getting to the analysis here again order n steps to do the check login iterations each scan takes order n steps so the total uh running time here is going to be order n log n because it's going to be the log n times n that's where the log n log n comes from now questions you might ask could i do even better than that can i beat n log n by more than any constant factor um so can i do little o of n log n um and the answer is no this is the best possible a one tape turing machine for this language um uh so a one day turing machine cannot decide a by using little o of n log n steps we're not going to prove that i'm not going to ask you to be responsible for the proof but in fact um what you can show is that any language that you can do on a one tape turing machine in little o of n log n steps turns out to be regular okay so um uh you can prove a rather strong theorem here not super difficult to to prove but i don't want to spend a lot of time on um uh proving bounds for turing machines because you know really the whole purpose of setting this up using turing machines is to talk about algorithms in general algorithms in general i'm not going to be focusing on the nitty gritty of turing machines um okay uh so um what is my um okay yeah so i wanted to just stop and make sure that we were we are together here um so a brief pause and feel free to send me any questions um okay this is a good question if we can keep track of parodies why can't we just keep track of the number of a's or b's um well you could keep track of the parity in the finite memory and so you can do that with effectively no work but the finite memory cannot is not enough for doing an account up to you know some arbitrarily large number now you could store the count on the tape but that's going to cost you time to maintain that counter and so um that's not going to be so simple as keeping track of the parity in the finite memory um okay somebody's asking if a star b star is regular yes a star b star is regular so what um but a to the k b to the k which is our language is not regular uh right so that's the language we're looking at um okay so getting questions about what happens with multiple tapes we'll talk about that in a second um yes we could do an order end steps on a regular computer sure but uh you know right for this slide anyway we're looking at one tape turing machines um okay getting questions about a big o and little o um for something to be little little o means it's less than any constant factor times the function um so you have to look at the definition in the book i think almost enough people in the class probably know this and have seen this already i don't want to spend everybody's time on it but uh please review it in the book somebody's asking if you need to store the parody on the tape no you can just throw it in the finite memory um uh i mean storing the finite memory seems to be the simplest thing okay when we move on please feel free to ask questions to the tas as well we have two tas at least here attending uh with me so um good um right um all right so we cannot do better than a one-tape turing machine on a one-tape turing machine and order n log n um and that's something we can prove that we're not going to do it here however if you if you have if you change the model for example you use a two-tape turing machine then yes as a lot of you are suggesting in the chat you can do better than that so if we now have a two-tape turing machine or a multi-tape turing machine you can do it in order n steps and that's actually the point i i'm really trying to make here um so if you have here your um two tape turing machine um then two tips same language now what we're going to do is copy the a's to the second tape that we can do on a single pass um and then once the a's have been copied to the second tape we can continue on reading the b's and match them off with the a's that appear on the second tape so in order n steps you can do the comparison instead of order n log n steps and of course if they match you're going to accept otherwise reject so let's just see here's a little animation uh uh uh demonstrating that of course it's very simple um so here is that here that if you could see that it came maybe a little too fast here let's just show it again here is the heads moving across and the a is coming down uh to the bottom and now the head the upper head is going to continue on reading the bees the lower head is going to go back to the beginning on the a's and matching off the b's with the a's and that's how we can verify or check that they are the same number okay so now in this case they were the same number so the machine would accept if they were the different number the machine would not accept um and the analysis is very simple um uh each stage here is going to take a linear number of steps order n steps um because it just consists of a single uh scan there are no uh there are no loops in this machine no no no repeat loops um okay so um question on this um all right why don't we move on then um now observe and the point i'm really trying to make is that uh on a one type turing machine you can do it in n log n but not any better but on a two tape turning machine you can do it in order n um so there's a difference in how much time you need to spend how many steps you need to spend depending upon the model and that's significant for us um so the number of steps depends on the model one tape turing machine was order and login multitape was order n we call them model dependence if you contrast that with the situation in the complexities in the computability section of the course we had model independence the choice of the model didn't matter and that was nice for us because the theory of decidability didn't depend upon whether you had a one tape turing machine or a multi-tape turing machine it was all the same set of decidable and recognizable languages uh so we didn't have to worry about which model we're actually going to work with we could work with any model even just an informal model of algorithm um was going to would be good enough because we're going to end up with the same notion in the end now that's goes away in complexity theory now we have a difference depending upon the model and from a mathematical standpoint that's a little less nice because um you know which model do you work with if you want to understand the complexity of some problem you know that you have at hand um now you have to make a choice you're going to work with the turing machine around how many tapes or you're going to look at some other model and you're going to get different results so it's somewhat less natural from a mathematical standpoint just to talk about the complexity of some problem but we're going to kind of um bring back something close enough to model independence by observing that even though we don't have uh model independence as we did in computability theory we can limit how much dependence there is so the amount of dependence is going to be low as we will see provided you stick with a reasonable class of deterministic models so the dependence though it does exist is not going to be um that much it's going to be polynomial dependence and we'll say exactly what that means in a second um and from our standpoint that's going to be uh um a small difference uh a negligible difference that we're going to ignore um so we're going to focus on questions that do not depend on the model choice among these reasonable deterministic models now you may say well that's not interesting from a practical standpoint because polynomial differences say the difference between n squared and n cubed certainly make a difference in practice but it really uh depends on the um the uh um you know what what kinds of questions you're focusing in on so if you want to look at something that's very uh a very precise distinction say between you know n squared and n cubed um then you might want to focus in on which model you want to be working with and that's going to be more the domain of an algorithms class but from our standpoint we're going to be looking at other still important questions but there are questions that um don't uh depend upon exactly which polynomial you're going to have we're going to be looking more at distinctions between polynomial and exponential and still there are important practical questions that arise in in that somewhat different um setting okay so with that in mind we're going to continue to use the one tape turing machine as our basic model of complexity um since the the model um among the reasonable deterministic models in the end is not going to matter from the from the perspective of the kinds of questions we're going to be asking um okay um so with that so we are going to continue then it's important to remember that from going forward we're going to stick with the one tape turing machine model maybe that's something you would have expected us to do anyway but i'm trying to justify that you know through this little discussion that we had so far thus far so now we're going to start defining things with the one tape model in mind um so so first of all if you have a turing machine we're going to say it runs in a certain amount of time so if t is some sort of time bound function like n squared or n log n we'll say the machine runs in that amount of time like n squared or n log n if that machine m always halts within that number of steps on all inputs of length n so it always holds within t of n steps on inputs of length n then we'll say that the machine runs in t of n time so in other words if the machine runs in n squared time then the machine when you give it an input of length 10 it's gotta it's got to be guaranteed to hold within a hundred steps uh 10 10 squared 100 steps on every input of length 10. um that's what it means um for the machine to be running in that much time and it has to do that for every every n for every input length and with that we're going to come to the following definition which is highlighted in color because it's going to be um we're going to be using this definition throughout the semester so it's important to understand it this is the definition of what's called the time complexity classes and what i'm going to do is take some bound t of n and again think of t of n like a bound like n squared so if you have time t of n or like time n squared that's going to be the collection of all languages that you can decide within time and square or within time t of n so in other words it's a collection of all languages b such that there's some one tape turing machine here we're focusing again on the one tape turning machine there is some deterministic one tape turing machine that decides b and that machine runs in that amount of time so this is a collection of languages the time collect complexity class is a set of languages i'm going to draw it now as a diagram um so if you take the language again that we've been using as our standard example a to the kb to the k um that's in time and log n as we observed for uh uh two slides two slides back or three slides back so on a one tape turning machine you can do this language a in time n log n so it's in the time complexity class n log n this this region here captures all of the languages that you can do in order and log in time for example that also includes all of the regular languages why is that well any regular language can be done on a one tape turing machine in time order n because the turing machine only just needs to scan across doesn't even need to write just need to scan across from left to right on the on the tape and in n steps it has the answer so all of the regular languages are actually in time n certainly a subset of time n log n and these all form a kind of a hierarchy so if you increase the bound you can imagine that the collect class of languages grows as you allow the machine to have more and more steps to do its computing um so these are all the languages that you can do in n squared uh order n square time on a one tape turing machine and cube time on a one tape turing machine and so on to exponential time two to the end time on a one tape turning machine these are all collections of languages um getting larger and larger um as we increase the bound um so someone is asking kind of is okay uh let's see let me let me get to some of these questions here i'll try to get them in order um uh okay um so somebody says if you have a good question if you have a regular computer so an ordinary sort of random access computer so we'll talk about that in a second can do it in order and uh can you do it on a multi-tape during machine also in order and time actually i don't know the answer teller offhand i suspect the answer is no that ordinary computers have a random access capability that turing machines do not and so that there are going to be some examples of problems that you can do with a random with a regular computer that you cannot do with a multi-tape turing machine in order end time uh i'd have to double check that though so we can and so some question what what we believe is true and what we can prove to be true um as we'll see there are a lot of things that we believe to be true in this subject that we don't know how to prove um somebody's asking kind of an interesting sort of more advanced question is there some function f uh some some some some some function t where um it's so big that it so that time t of n gives you all of the decidable problems um it would be a very big t but the the answer actually to that question is yes um but that's a little bit exotic so let let's not spend a lot of time on that right here but i'm happy to talk about that offline um it's a good question here somebody's asking me does it mean that there are no languages between order n and order n log n because i pointed out that anything below n log n is going to be regular and so as soon as you get below n log n you can do it in order n and yes there is what's called a gap between order n and order n log n on a one take turing machine this you don't get anything new from order n until you jump up so from order n to order n log log n nothing new shows up um so we'll talk about those kinds of things a little a little bit down the road when we look at actually the relationship among these various uh classes and what we call a hierarchy theorem which shows you know what how much bigger do you have to make the bound in order in order to be sure you'll get something new um all right [Music] okay somebody's asking is there a model which has the same time complexity as a normal computer uh well i mean there's the random access model which is supposed to capture a normal computer um okay so let me let me um these are all great questions kind of more uh riffing off of this into in more into more advanced directions uh let's move on um here's another check-in um suppose we take this is a little bit of a check to see how well your how comfortable you are with the notions we've just presented and whether you can think about some of the arguments that we've made and apply them in into a new language so take the language ww reverse um you know strings followed by their um uh this followed by themselves backwards this language via the even length palindromes if you will um what's the smallest uh bound that you need uh to be able to solve uh that language b and i'll pose it as a pose that is a question for you so which which time complexity class is um that language b in is it time order n or n log n n squared so on what do you think so we're about to come to the coffee break so uh why don't we um uh i'll answer any questions that come up i think we're got everybody uh answered so i'm gonna end the polling okay make sure you're in if you want to be in um so the correct answer is in fact uh order n squared um you it would be hard for you know you you know every reasonable guess here would be order n log n i mean you can come up with the same procedure that the as the one we showed at the beginning uh the order n squared procedure for eight to the kb to the k works for ww reverse as well um you can just cross off sweep back and forth crossing off um the symbol from w and going across to the other side crossing off the symbol from w reverse and that procedure will give you an n squared an order n squared algorithm um you might imagine you can improve it to order n log n but you cannot you can prove that order n squared is the best possible um i i'm a little unhappy that a lot of you came up with order n frankly because i already told you that order n is uh these are just regular languages anything that you can do in less than little o of n log n is going to be regular and we know this language is not regular so this is a this was uh not a good answer you know so please pay attention and um okay so let us stop sharing uh i will turn now to the uh our our our break uh for five minutes and i'm happy to you know try to take questions along the way as we're um uh we're waiting for the time to to end so let's see let me put this up here okay let me try to take some of your questions um so someone has asked me about quantum computers as reasonable models of you know you may say a quantum computer is a reasonable model of computation and that's fine i would not say it's a reasonable model deterministic computation um uh at least from our standpoint though let's not quibble about the words um i'm not including quantum computers in um the collection of um machines that i have in mind right now when i'm talking about the the the reasonable models of deterministic computation that we're going to be we're going to be discussing um uh let's see okay a bunch of people apparently are asking the tas why all regular languages can be done in order n um [Music] so if you think about a dfa which you know processes a an input of length n within steps and the dfa is um going to be a type of turing machine that never writes on its tape uh so if the dfa can do it in in in n steps the turing machine can do it n steps and so therefore every regular language can be done in order and steps on a turing machine um not sure where the confusion is so please message me if you're not still not getting it um okay somebody's saying why are we using one-take turing machines instead of random access wouldn't it be better to use the random access machines um if you were using if you're trying to do algorithms yes that's a more reasonable model we're trying to prove things about the computation and from that standpoint we want to use as simple as model as possible uh trying to prove things um using random access computers it's possible be very messy so that's why we don't use uh random access machines to prove the kinds of things we're going to be proving um about computation you know that that are really the meat and potatoes of this course so uh i mean there's compelling reasons why you would want to use a simple model like a turing machine but not a powerful model like a random access computer so somebody's asking me does um the class the time uh order and log log n have any elements yes it has all the regular languages but nothing else order n log log n is only read only the regular languages you have to go all the way up to or and log in before you get something non-regular uh somebody's asking me are we going to talk about how the random access model works no you know there's uh beyond the scope of this course outside of what we're going to be doing we're going to talk about turing machines not because we care so much about turing machines but i'm trying to prove things about computation and the turing machines are a convenient vehicle for doing that um okay our candle has burned out uh why don't we return then to um the next uh slide so everybody come back so this answers one of the questions i got on the chat is what's what actually is the dependency between multi-tape turing machines and one tape turing machines can we bound that in general yes we can we're going to show that converting a multi-tape turing machine to a one-tape turing machine can at most blow up the amount of time that's necessary by a squaring um no i acknowledge it's a lot but it still allows you um uh but you know it's still small compared with an exponential increase and we're going to be focusing in this course on things like the difference between polynomial and exponential not between the different not not between the difference uh of n not between not the difference between n squared and n cubed that's going to be less of a fact less of an issue from us uh okay um [Music] so the way i'm framing uh showing this theorem is that if you have a multi-tape turing machine that can do a language in a certain amount of time then it's in the time complexity class of that time bound squared and the way i'm just saying that is because this is the bound that's utilizing the one tape model so another way of saying that is converting multi-tape to one tape squares the amount of time you need at most so the way we're going to prove that is simply by going back and remembering the conversion that we already presented from multi-tape to one tape and observe that if we analyze that conversion it just uh ends up squaring the amount of time that the multitape used okay so why is that so if you remember let's just make sure we're all altogether on this the way the single tape machine s simulates the multi-tape turing machine m is that it takes the contents of each of m's tapes you know up to the place where there's infinitely many blanks obviously you don't store the infinite part but the the the the active portion of each of them's tapes you're going to store them consecutively in separate blocks on s's tape on estes only tape um and now every time m makes one move s has to scan its entire tape to uh to uh to see what's under each of the heads and to do all the updating so to simulate one take step of m's computation s is going to use order t of n steps where t of n is the total running time that m is going to use so why is t of n steps coming up here well that's because you have to measure how s is going to make a scan of course its tape how big can its tape be well m if it's trying to use as much tape as possible can use at most t of n um tape on each of the t event cells on each of its tapes so all together there's just going to be some constant number of times t of n cells on s's tape do you see that so each one of these is going to be at most t of n long so this all together is going to be order t of n log because you know what can m do it could send its head out say the head on this tape here moving as fast as possible to the right using as much tape as it can but you can only use t event cells in t of n time so this is going to be ordered t of n so one step of m's computation is going to be t of n steps on s's computation but m itself has t of n steps so it's going to be t of n times t of n for the total number of steps that s is going to end up using and that's where the squaring comes from okay similar results i'm not going to do lots of simulations of one model by another i think that you'll get the idea and you can if you're interested you can uh study those on your own but you can convert multi-dimensional turing machines to one one tape turing machines um you know one tape ordinary uh linear one tape one dimensional uh machines and the bottom line is that among all of the reasonable models they're all what are called polynomially related uh if each can simulate the other with at most the polynomial overhead so if one of the machines can use this t of n time the other machine that's simulating it would use t to the k of n time for some k um that's what it means for the two machines to be polynomially related and all reasonable deterministic models are polynomially related so as we've already seen one tape in multi-tape turing machines are polynomial related because converting multi-tape to one tape blows you up by most squaring so k equals two in this case um multi-dimensional turing machines again polynomial related the random access machine which i'm not going to define but it's the machine that you might imagine you know you would i'm sure they must define in some form in the algorithms classes polynomial related cellular automata which are just arrays of a finite automata that can communicate with each other similar similarly um you know all the reasonable deterministic models um you know again classical models i'm not talking about quantum computing are polynomial polynomially related so we are um uh that kind of justifies our uh choice in picking one of them as long as we're going to ask questions which don't depend upon the polynomial so let's then talk about the class p so the class p this is an important definition for us this is um the collection of all languages that you can do in time into the k for some k on a one take turing machine or as i've written it over here i don't know if this notation is um unfamiliar to you but this is like just a big sum um but here it's a big union symbol it's union over all values of k of the time class n to the k so this is time n union to time n squared union time n cubed union time into the fourth and so on um and we call these the polynomial time decidable languages okay so we're going to be spending um you know uh a certain amount of effort uh exploring this class p and other similar classes somebody's asking why is it a union i'm not sure how else you would write it so if somebody hasn't if you have a proposal for a different way to write it that's fine but this is you know k this is for all k i don't know you know there's if you only had a limit finite number of k's you could just take the biggest one but since it's for all k you need to write it as a union um now i want to argue that the class p um is an important class and you know why is it had so much impact on the subject um and in terms of applications as well um so one thing is that the class p is invariant for all reasonable deterministic models and what do i mean by that so we have defined the class p in terms of these time classes here which in turn are defined in terms of the one tape model so we have defined p by using one tape turing machines now if we had to find p in terms of multi-tape turing machines we get exactly the same class because one tape and multi-tape turing machines are polynomially related to one another and since we're taking the union over all polynomials that polynomial difference is going to wash out um similarly we could define p using any of the other reasonable deterministic models we get exactly the same class so in a sense we get back what we um the situation that we had in computability theory when the class of decidable languages didn't depend on the choice of model here the class p does not matter depending upon the choice of reasonable deterministic model um and we also kind of even in the case of computability theory we have to stick with kind of reasonable models that cannot do an infinite amount of work in one step um so uh you know when i'm not going to define what it means to be reasonable and that's in a sense an informal notion but among all of those reasonable models you're going to get the same class p the other thing that makes p important is that p roughly corresponds to the problems that you can solve in some reasonable practical sense not exactly you know problems that require into the hundredth time um you could argue cannot be uh solved in any reasonable sense um but uh you know if you think about it for example from the perspective of cryptography um uh cryptographic codes that people come up with are typically designed to require or or the hope is that they require an exponential amount of effort to crack if someone found even an into the hundredth algorithm to crack uh that would crack a code people would feel that the code is not secure even though into the hundredth is still a large um so you know it's a it's a rough kind of test but it's still used as a kind of litmus litmus test for practical solvability if you can solve it in polynomial time you basically figured out how to avoid um large searches if you can solve problems in in polynomial time we'll say more about that later but you know what i want to bring out here is that we have combined here in the class p something that's mathematically nice mathematically element elegant with something that's practically relevant and when you have a combination of the two then you know you have a winner then you know how you have a concept that's going to make a difference and that's been true for the class b this has been very influential within and without the theory of computation um so let's look at an example um let's let's define a new language we haven't seen before though it's similar to procedures that we've looked at before the path language which is where i'm going to give you a graph g two nodes in the graph s and t um where i'm thinking of g is a directed graph so directed means that the um the connections between the nodes in g are going to be redirected and that they have arrows on them they're not just lines but they have an orientation with an arrow um so g is a directed graph that has a path from s to t that respects the directions um so such a i think i might even have a picture here yeah so imagine here here is your graph if you can see it there are little arrows connecting the uh the nodes and i want to know is there a path from the node s to the node t so that is you know a picture of a problem in you know an instance of a graph of a path problem and i want to find uh an algorithm for that and i can show that there is an algorithm that operates in polynomial time for this uh path problem okay and the algorithm uh you know any of the standard searching algorithms uh would work here but let's just for completeness sake um include the breadth first search algorithm that we have explored previously when we talked about finite automata um so we'll mark s and they'll keep repeating until nothing new is new is marked and we'll mark all of the nodes that were reachable by a single arrow from a previously marked node um and then cft is marked after you could mark everything you can get to so you're going to mark let's see pictorially i think i have this in indicated yeah you're going to mark all of the things that are reachable from the start not from the from the node s and then see after you can't mark anything new whether the node is marked and if it is you'll accept if it is not you reject okay um now we can analyze this too um and i'm not going to be spending a lot of time analyzing algorithms here um but let's just uh do the do it kind of this one time uh again um we're gonna we're get we're doing a bunch of iterations here so gonna be repeating until nothing new is marked so each time we mark something new we can only do that at most end times at which point we've marked everything so the number of iterations here is going to be at most n and now for each each time we've marked something we have to um look at all of the previously marked nodes um and see which things they point at to mark them too so this is going to be an inner loop which again has at most n iterations because it's going to all of the previously marked nodes and then once we have that we can scan g to see to mark all of the new uh all of the nodes which we have not yet marked uh whether they're connected with a previously marked node by an edge and i'm being generous here because i don't really care um this can be done in the most n squared steps on a one-take turing machine i'm not going to describe the implementation but you know i'll leave it to you as an exercise but this is straightforward so the total number of steps here would be n iterations times n iterations times and squared so you're going to be at most into the fourth steps needed so this is a polynomial algorithm and whether i ended up with n to the fourth or n to the fifth or n cubed i don't really care um because i'm just trying to illustrate um that the total is polynomial and that's all i'm going to be typically asking you to do um so to show polynomial time what i'd be asking you to do is to show that each stage each stage of this algorithm should be clearly polynomial and that the total number of stages i'm sorry this should say stages here should be polynomial so each stage is polynomial and after you're doing all the iterations the total number of stages that are executed is polynomial and so therefore all together the total running time the total number of steps is going to be polynomial okay so that's um the uh the way we would write up um polynomial algorithms in this class um so let's see if there's any questions here i don't want to get too far ahead of people um let's see um yes in this theorem i'm talking about one tape turing machines um because we're defining everything in terms of one type machines but now at this point you know we're talking about polynomial time we can you know my analysis is based on one tape turning machines but in general you know you could use any reasonable deterministic model on which to carry out your analysis uh because they're all polynomially equivalent so from the perspective of coming up with showing that a problem is in polynomial time isn't p um you can use any of the models um you know that you that you wish that for convenient okay oh that's a good question what is n thank you for asking that question n is always going to be reserved to indicate the length of the input so here n is going to be when we encode g s and t and here also i haven't said this but i'm assuming that the encoding that you're using is also somehow reasonable you know i think we'll talk a little bit about more about that in the next lecture um which is going to be after the midterm um but you know you can cause problems if you intentionally try to come up with nasty encodings which will you know represent things uh with unnecessarily many characters but um you know if if you try to be reasonable then um you just use any one of those encodings and you'll be fine um so yeah um so n is the length of the representation of the input um let's see someone's trying to you're trying to dig into the my the actual how this is running here scan g to mark all y where x y is an edge i'm saying that you can do an n squared steps um uh um you know if you have x you can mark it in a certain place on the tape and then as you're going through every other node every other edge um you can you can go back and compare x with the x of the edge and then c and then find the y i mean it's just going to be too messy to talk about here i mean i i'll leave it this is an exercise to you i'm not going to try to fumble my way through explaining why you can do this in n squared steps but it's it's not hard um uh you guys are all like hardcore algorithms folks you want to know the algorithms for this i'm not going to do that sorry uh high level picture here um you know so so you know if you want you know if you want to look at detailed analyses this is not the right course for you um can k equal n what is k um um [Music] no this if you're talking about this k here k cannot equal n we're not looking at n to the n um this is these are all fixed k um it's like n squared and q but not into the end is going to be an exponential bound and so that's not going to be included within uh this union okay so we're near the end of the hour um i'm gonna introduce one last language here um called handpath and the handpath problem is um i'm going to ask now again for a path from s to t but now a different kind of a path one that goes through every node of g along the way um so i'm looking for a path that goes that hits every note of g not just the shortest most direct path but in a sense the most indirect pair the longest path that goes through from from s to t that visits everything else along the way um a path that of that kind that hits every node of the graph is called the hamiltonian path um because the mathematician hamilton studied um those made some definitions about that i'm not gonna i don't actually know the history there but i just know they're called hamiltonian pets um so here's a picture i want to get from snt but i want to sort of pick up everything else along the way so as you remember the path problem itself can be solved in p and what i'd like to know can the simple modification where i'm asking you to visit everything else along the way is that problem also in p okay and i'm going to put it pose this as a checking for you um but actually before i get to that let me give you an algorithm for handpath um that doesn't work to show it's in p um because it's exponential so here's an algorithm for handpath um uh where let's m be the number of nodes in g um and what i'm going to do is i'm going to try every can every possible path in g um and see if it's actually works as a hamiltonian path and except if it is um and then if all paths fail uh then i'll reject so i'm going to try every possible routing through g if you want to think about it think of m as every possible permutation of the nodes of g and then you're going to see whether that's actually constitutes a path in g um that takes you from s to t and goes through all of the nodes um so this algorithm would work this would give you a correct algorithm for the hand path problem the problem is the difficulty is that there are so many possible paths that it's going to take you an exponential number of steps to execute this algorithm it's not a polynomial time algorithm because there are many possible pairs that you could go through um if you're looking at you know with a very crude bound but you really can't improve that significantly there would be um m-factorial which is going to be you know much greater than 2 to the m paths of length m um so the algorithm is going to run for exponential time and not polynomial time so my question for you is i'm going to pose it as a check-in problem is whether you could actually do this problem in polynomial time so why don't you think about that as i'm setting up the question so take the hand path problem just like the path problem which i described with that marking algorithm but now you want to hit every node along the way can you show that problem is solvable in p and there's a whole range of possibilities here where either the answer is yes you can see what the polynomial time algorithm is uh to uh to definitely know where you can prove there is no such polynomial time algorithm and i'll put this as a checking for you and see i'm curious to see what you come up with most people are getting it wrong well i mean wrong i don't you know i'm not clear what wrong is here um okay are we done please uh check something you can see a few of you have not answered but the poll is running out okay time is up um so uh in fact you know as as i think many of you know but not all of you uh this is an unsolved problem this is a very famous unsolved problem which is equivalent to the p versus np problem that we're going to be talking about very soon which among other things would be worth a million dollars if you solve it so for those of you who have answered a or e please talk to me after lecture um and maybe we can work on it together um [Music] uh no so um yeah i i think most people would believe that the answer is no um uh but no one knows how to prove it at this time uh so i'm interested in the uh folks who have uh come up with uh what they think are solutions um and i should say that there are some folks who um believe that there might be other outcomes besides uh just a simple note um which might be proven eventually um so we're going to talk more about this um but this is the answer to the question just for your uh just to make sure you understand is that it's an unsolved problem right now so we don't we don't know uh definitely yes and definitely no at least according to the state of knowledge of which i'm aware is not um are not correct answers um but any of the others well who knows um so i think that's the end of what i had to say for today we covered complexity theory as an introduction looked at different possible models focused on the one tape model introduced based on the one tape model these uh complexity classes the class p and we showed an example of this path problem being in p talked also about this hand path problem which we'll talk about more um after the midterm okay so um i'll stick around for a few minutes uh if you have any further questions otherwise um uh okay so let me just take questions here um somebody's asking me about my personal opinion on p versus np um my personal opinion on people versus np is that p is not equal to np and that we will prove it someday when i was a graduate student back in the mid 70s i thought it would be solved by now and in fact i made a bet with len adelman who i subsequently ended up becoming the a of the rsa code um that we would solve it by the year 2000 and i bet what was then a lot of money for me which was an ounce of gold which i didn't end up then which i did end up paying off to lend in the year 2000 so i'm not making any more bets but i still believe that it will be solved uh hopefully i'll get a chance to see the solution um i've spent a lot of time thinking about it myself you know obviously i haven't solved it otherwise we would know but um uh you know uh hopefully somebody will i'm getting asked a question here that's kind of an interesting question but i don't really know um i'm sure i understand it what's the largest possible run time of a decidal decidable problem what is the larger largest decidable runtime so you anything that i can describe can be um there are going to be algorithms that run for longer uh you can define an algorithm you can define a runtime which would in a sense beats all other runtimes which so that anything any runtime is going to be um dominated by that extremely slow runtime but it's not something that one can describe um i can describe it to you by perceive mathematical procedure but there's it's not going to be something like you know uh 2 to the 2 to the n somebody's here proposing they prepare a solution to the hand path problem i presume in polynomial time why is the following flawed if s goes through all nodes and ends up at t well s will not you mean you mean i presume you mean starting at s if we end up going through all nodes and and end of t um the proposal is a little complicated here basically you you want to if if i can try to rephrase it you want to try to from every possible node you want to try to calculate a path to t and also a path from s to that node um and you can do that for all possible nodes but there's no way to really combine them into a single path that visits all you know solving for each node separately is not going to do the trick because you have to somehow combine all that information into a single path just one path that goes from s to t and visits all the other nodes along the way and that that is not uh i i don't see how what your proposal um how that's how that's going to actually work professor spitzer yes um a question on the um like we were kind of talking about earlier but like what we talked about today was defined for the one tape turn machines correct yep so um and you said like we could have applied for the multi multitape ones but um are like is well i don't know if it's we talked about earlier but if something is accepted by the one tape turn machine like can be applied to the multi-tape turn machine and vice versa like are they interchangeable like that well they're interchangeable only in the sense that the amount of time that you would need it's a different machine if you have a multi-tape turing machine for some language um you can convert it to a one-tape turing machine using a procedure that we described it earlier in the term and she'll get a different machine it's going to run for a different amount of time by using their procedure the point is is that the amount of time that the one-take turing machine is going to run for is not that much worse than the multi-tape turing machine's time so if the multi-tape turning machine's time was you know uh end cubed the one tape turning machines uh is going to be the square of that so it's going to be into the sixth okay but it's not going to be it's not going from multi-tape to one tape is not going to convert you from polynomial to exponential that's that's the only point i'm trying to make it's going to convert from one polynomial to a somewhat bigger polynomial but it's still going to leave you polynomial [Music] i don't know you don't seem to i i can't see your face no no yeah i guess i guess my opinion is that like especially when we were talking about earlier um when you're bringing it up it just seemed like you could just turn anything to a multi-tape turn machine and like completely like cut the timeout like if i had like something to n log n like if i do it in a multi-tape turn machine i have it in like um like big o of n yeah i mean like it just seemed like the multi-tape was so much more powerful but then i guess i guess not with the like with the explanations and like the the models we were talking about today yeah i would not say the multi multi-tape turing machines are still pretty limited in their capabilities and they're um you know don't forget when you have a multi-tape turing machine you have only a fixed number of tapes i mean you could also define variations of multi-tape turing machines that have increasing number of tapes as the input uh you know under you know either under program control it can launch new tapes or it could um um you know just have more tapes depending upon the size of the input that would also be a possibility that's not the model that we have defined but you could define a model like that um but as long as the total amount of work being done by the machine at any step it's going to be a polynomial amount of work um then you can convert it to a one tape turing machine uh with only a polynomial increase in the bound all right so you know you want to be careful of machines you know you know one one thing i meant to say but didn't say you know so he would be an unreasonable model um which you might think of as a plausible model but it's not going to be a reasonable model from our standpoint and that would be a model um for example that can do um full precision say integer arithmetic um uh with a unit cost per operation so each operation counts as a costs one um but i'm going to allow you to do for example you know addition and multiplication uh the the thing that's bad about that um in terms of being uh unreasonable is that uh after k step each time you do a step you could double the size of the integer by by squaring it yeah so after k steps you could have an integer which is 2 to the k long and now doing operations there is going to involve an exponential amount of work even in any reasonable sense um you know in a theoretical sense and also in a practical sense so um you know a model that operates like that is not going to be able to convert to a one-tape turing machine with only a polynomial increase because it's doing an exponential amount of work potentially you know uh within a polynomial number of steps um so that's within a linear number of steps within n steps so that's um an example of an uh you know of an unreasonable deterministic model yeah thank you sure so i'm just curious some idea just occurred to me um i guess like if you had an oracle turing machine uh basically just so that you could look at a uh a turing machine description and decide whether it's a decider or not yeah um then it'd be a little bit interesting to think about uh what happens if you look at all uh you know if you have a description of a pair of a turing machine and an input string and you can look at for all size n descriptions of a pair uh like what's the most steps that it takes uh for a such a machine to converge so then so then you'd have combined turing machine and input stream descriptions uh where you can look at what's the longest it takes to converge it's just i don't know just a random thought that occurred yeah so we we actually we're gonna we we will talk about oracle turing machines uh later on in the term um you know these are machines that have access to sort of free information um and that actually turns out to be uh i mean there's some interesting interesting things you can say about what what happens when you're providing a machine with in a sense information for free that you know you might otherwise want to charge it for actually computing that information but let's just say we're going to allow it to uh get that information without being charged and then how does that affect the complexity of other problems for example um and so uh we will talk about that later but too much if i think of a digression at this moment to try to define all that um but happy to chat with you about it on piazza if you if you want to raise a question there i think sure oh no problem um so he's asking me about strategies for solving the p versus np problem you know we will talk also talk about that a little later in the term as well but you know asking you know clearly you know it seems beyond the reach of our present techniques to be able to prove that some problems you know really take a long time like the hamiltonian path problem seems like you know there's nothing really you can do that's significantly better than trying all possibilities as i described in that exponential algorithm you know on the last slide uh but um how do you prove that nobody knows um so um lots of people have tried including yours truly i mean i spent a lot of time thinking about it having succeeded with it but um there is uh um you know somebody i i think somebody someday someday somebody will come up with the the new idea that's needed to solve it but it's going to clearly take some sort of a breakthrough some sort of a new idea it's not just going to be a combination of existing ideas the existing methods um i think we're about 10 minutes past a few of you are still here i'm going to say goodbye to you folks and um um shortly i'm going to join my tas for our t weekly ta meeting so see you guys thanks for being here 3 00:00:29,189 --> 00:00:29,199 4 00:00:29,199 --> 00:00:31,189 5 00:00:31,189 --> 00:00:31,199 6 00:00:31,199 --> 00:00:32,150 7 00:00:32,150 --> 00:00:32,160 8 00:00:32,160 --> 00:00:33,510 9 00:00:33,510 --> 00:00:36,630 10 00:00:36,630 --> 00:00:38,229 11 00:00:38,229 --> 00:00:38,239 12 00:00:38,239 --> 00:00:39,430 13 00:00:39,430 --> 00:00:39,440 14 00:00:39,440 --> 00:00:40,549 15 00:00:40,549 --> 00:00:42,389 16 00:00:42,389 --> 00:00:42,399 17 00:00:42,399 --> 00:00:44,150 18 00:00:44,150 --> 00:00:46,869 19 00:00:46,869 --> 00:00:49,750 20 00:00:49,750 --> 00:00:52,150 21 00:00:52,150 --> 00:00:54,150 22 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--> 00:16:45,910 522 00:16:45,910 --> 00:16:47,189 523 00:16:47,189 --> 00:16:49,430 524 00:16:49,430 --> 00:16:52,470 525 00:16:52,470 --> 00:16:56,790 526 00:16:56,790 --> 00:16:58,470 527 00:16:58,470 --> 00:17:01,189 528 00:17:01,189 --> 00:17:03,590 529 00:17:03,590 --> 00:17:05,510 530 00:17:05,510 --> 00:17:07,510 531 00:17:07,510 --> 00:17:09,510 532 00:17:09,510 --> 00:17:11,029 533 00:17:11,029 --> 00:17:12,870 534 00:17:12,870 --> 00:17:14,390 535 00:17:14,390 --> 00:17:16,789 536 00:17:16,789 --> 00:17:18,870 537 00:17:18,870 --> 00:17:21,029 538 00:17:21,029 --> 00:17:23,750 539 00:17:23,750 --> 00:17:28,069 540 00:17:28,069 --> 00:17:30,150 541 00:17:30,150 --> 00:17:31,590 542 00:17:31,590 --> 00:17:33,350 543 00:17:33,350 --> 00:17:35,190 544 00:17:35,190 --> 00:17:37,430 545 00:17:37,430 --> 00:17:39,430 546 00:17:39,430 --> 00:17:41,270 547 00:17:41,270 --> 00:17:42,549 548 00:17:42,549 --> 00:17:44,070 549 00:17:44,070 --> 00:17:45,990 550 00:17:45,990 --> 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580 00:18:33,990 --> 00:18:36,150 581 00:18:36,150 --> 00:18:38,150 582 00:18:38,150 --> 00:18:40,549 583 00:18:40,549 --> 00:18:42,310 584 00:18:42,310 --> 00:18:44,470 585 00:18:44,470 --> 00:18:46,630 586 00:18:46,630 --> 00:18:48,230 587 00:18:48,230 --> 00:18:49,990 588 00:18:49,990 --> 00:18:52,230 589 00:18:52,230 --> 00:18:56,310 590 00:18:56,310 --> 00:18:58,870 591 00:18:58,870 --> 00:19:01,590 592 00:19:01,590 --> 00:19:03,990 593 00:19:03,990 --> 00:19:04,000 594 00:19:04,000 --> 00:19:05,029 595 00:19:05,029 --> 00:19:07,110 596 00:19:07,110 --> 00:19:08,549 597 00:19:08,549 --> 00:19:10,789 598 00:19:10,789 --> 00:19:12,710 599 00:19:12,710 --> 00:19:12,720 600 00:19:12,720 --> 00:19:14,310 601 00:19:14,310 --> 00:19:15,909 602 00:19:15,909 --> 00:19:15,919 603 00:19:15,919 --> 00:19:17,029 604 00:19:17,029 --> 00:19:19,590 605 00:19:19,590 --> 00:19:21,270 606 00:19:21,270 --> 00:19:22,549 607 00:19:22,549 --> 00:19:26,070 608 00:19:26,070 --> 00:19:28,390 609 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--> 00:20:36,870 639 00:20:36,870 --> 00:20:38,710 640 00:20:38,710 --> 00:20:41,750 641 00:20:41,750 --> 00:20:45,909 642 00:20:45,909 --> 00:20:49,029 643 00:20:49,029 --> 00:20:51,029 644 00:20:51,029 --> 00:20:53,430 645 00:20:53,430 --> 00:20:55,029 646 00:20:55,029 --> 00:20:55,039 647 00:20:55,039 --> 00:20:56,310 648 00:20:56,310 --> 00:20:58,149 649 00:20:58,149 --> 00:20:59,590 650 00:20:59,590 --> 00:21:01,430 651 00:21:01,430 --> 00:21:03,110 652 00:21:03,110 --> 00:21:04,950 653 00:21:04,950 --> 00:21:07,110 654 00:21:07,110 --> 00:21:08,630 655 00:21:08,630 --> 00:21:08,640 656 00:21:08,640 --> 00:21:11,029 657 00:21:11,029 --> 00:21:13,669 658 00:21:13,669 --> 00:21:14,950 659 00:21:14,950 --> 00:21:21,510 660 00:21:21,510 --> 00:21:23,830 661 00:21:23,830 --> 00:21:25,590 662 00:21:25,590 --> 00:21:29,029 663 00:21:29,029 --> 00:21:31,350 664 00:21:31,350 --> 00:21:34,230 665 00:21:34,230 --> 00:21:34,240 666 00:21:34,240 --> 00:21:38,070 667 00:21:38,070 --> 00:21:39,909 668 00:21:39,909 --> 00:21:41,990 669 00:21:41,990 --> 00:21:46,230 670 00:21:46,230 --> 00:21:46,240 671 00:21:46,240 --> 00:21:47,110 672 00:21:47,110 --> 00:21:49,029 673 00:21:49,029 --> 00:21:51,190 674 00:21:51,190 --> 00:21:53,750 675 00:21:53,750 --> 00:21:55,510 676 00:21:55,510 --> 00:21:57,190 677 00:21:57,190 --> 00:21:59,830 678 00:21:59,830 --> 00:22:01,590 679 00:22:01,590 --> 00:22:01,600 680 00:22:01,600 --> 00:22:02,789 681 00:22:02,789 --> 00:22:04,789 682 00:22:04,789 --> 00:22:06,630 683 00:22:06,630 --> 00:22:09,510 684 00:22:09,510 --> 00:22:13,430 685 00:22:13,430 --> 00:22:13,440 686 00:22:13,440 --> 00:22:14,470 687 00:22:14,470 --> 00:22:14,480 688 00:22:14,480 --> 00:22:15,909 689 00:22:15,909 --> 00:22:18,630 690 00:22:18,630 --> 00:22:21,430 691 00:22:21,430 --> 00:22:21,440 692 00:22:21,440 --> 00:22:22,950 693 00:22:22,950 --> 00:22:22,960 694 00:22:22,960 --> 00:22:25,110 695 00:22:25,110 --> 00:22:26,710 696 00:22:26,710 --> 00:22:29,110 697 00:22:29,110 --> 00:22:29,120 698 00:22:29,120 --> 00:22:30,390 699 00:22:30,390 --> 00:22:32,549 700 00:22:32,549 --> 00:22:35,190 701 00:22:35,190 --> 00:22:37,190 702 00:22:37,190 --> 00:22:37,200 703 00:22:37,200 --> 00:22:39,590 704 00:22:39,590 --> 00:22:39,600 705 00:22:39,600 --> 00:22:40,390 706 00:22:40,390 --> 00:22:40,400 707 00:22:40,400 --> 00:22:42,070 708 00:22:42,070 --> 00:22:43,750 709 00:22:43,750 --> 00:22:45,669 710 00:22:45,669 --> 00:22:48,789 711 00:22:48,789 --> 00:22:50,549 712 00:22:50,549 --> 00:22:53,669 713 00:22:53,669 --> 00:22:55,830 714 00:22:55,830 --> 00:22:57,590 715 00:22:57,590 --> 00:23:04,230 716 00:23:04,230 --> 00:23:06,950 717 00:23:06,950 --> 00:23:10,390 718 00:23:10,390 --> 00:23:12,149 719 00:23:12,149 --> 00:23:14,070 720 00:23:14,070 --> 00:23:18,070 721 00:23:18,070 --> 00:23:19,909 722 00:23:19,909 --> 00:23:22,390 723 00:23:22,390 --> 00:23:24,070 724 00:23:24,070 --> 00:23:25,830 725 00:23:25,830 --> 00:23:27,990 726 00:23:27,990 --> 00:23:29,990 727 00:23:29,990 --> 00:23:31,350 728 00:23:31,350 --> 00:23:33,110 729 00:23:33,110 --> 00:23:37,110 730 00:23:37,110 --> 00:23:37,120 731 00:23:37,120 --> 00:23:38,710 732 00:23:38,710 --> 00:23:40,230 733 00:23:40,230 --> 00:23:42,390 734 00:23:42,390 --> 00:23:45,350 735 00:23:45,350 --> 00:23:47,750 736 00:23:47,750 --> 00:23:50,310 737 00:23:50,310 --> 00:23:51,430 738 00:23:51,430 --> 00:23:51,440 739 00:23:51,440 --> 00:23:55,110 740 00:23:55,110 --> 00:23:55,120 741 00:23:55,120 --> 00:23:56,710 742 00:23:56,710 --> 00:23:56,720 743 00:23:56,720 --> 00:24:00,549 744 00:24:00,549 --> 00:24:02,950 745 00:24:02,950 --> 00:24:04,710 746 00:24:04,710 --> 00:24:06,950 747 00:24:06,950 --> 00:24:08,390 748 00:24:08,390 --> 00:24:12,149 749 00:24:12,149 --> 00:24:13,669 750 00:24:13,669 --> 00:24:14,870 751 00:24:14,870 --> 00:24:16,630 752 00:24:16,630 --> 00:24:17,669 753 00:24:17,669 --> 00:24:19,190 754 00:24:19,190 --> 00:24:21,190 755 00:24:21,190 --> 00:24:23,430 756 00:24:23,430 --> 00:24:24,630 757 00:24:24,630 --> 00:24:24,640 758 00:24:24,640 --> 00:24:26,549 759 00:24:26,549 --> 00:24:28,310 760 00:24:28,310 --> 00:24:30,149 761 00:24:30,149 --> 00:24:32,549 762 00:24:32,549 --> 00:24:34,789 763 00:24:34,789 --> 00:24:36,870 764 00:24:36,870 --> 00:24:39,510 765 00:24:39,510 --> 00:24:41,350 766 00:24:41,350 --> 00:24:41,360 767 00:24:41,360 --> 00:24:42,710 768 00:24:42,710 --> 00:24:42,720 769 00:24:42,720 --> 00:24:44,789 770 00:24:44,789 --> 00:24:46,390 771 00:24:46,390 --> 00:24:48,230 772 00:24:48,230 --> 00:24:49,830 773 00:24:49,830 --> 00:24:51,510 774 00:24:51,510 --> 00:24:54,390 775 00:24:54,390 --> 00:24:56,310 776 00:24:56,310 --> 00:24:58,390 777 00:24:58,390 --> 00:25:01,350 778 00:25:01,350 --> 00:25:03,430 779 00:25:03,430 --> 00:25:05,110 780 00:25:05,110 --> 00:25:07,510 781 00:25:07,510 --> 00:25:09,590 782 00:25:09,590 --> 00:25:09,600 783 00:25:09,600 --> 00:25:12,470 784 00:25:12,470 --> 00:25:13,750 785 00:25:13,750 --> 00:25:16,230 786 00:25:16,230 --> 00:25:17,350 787 00:25:17,350 --> 00:25:18,950 788 00:25:18,950 --> 00:25:20,549 789 00:25:20,549 --> 00:25:22,230 790 00:25:22,230 --> 00:25:22,240 791 00:25:22,240 --> 00:25:23,190 792 00:25:23,190 --> 00:25:23,200 793 00:25:23,200 --> 00:25:24,950 794 00:25:24,950 --> 00:25:27,269 795 00:25:27,269 --> 00:25:29,510 796 00:25:29,510 --> 00:25:30,950 797 00:25:30,950 --> 00:25:33,190 798 00:25:33,190 --> 00:25:37,029 799 00:25:37,029 --> 00:25:39,350 800 00:25:39,350 --> 00:25:41,190 801 00:25:41,190 --> 00:25:42,789 802 00:25:42,789 --> 00:25:45,350 803 00:25:45,350 --> 00:25:47,269 804 00:25:47,269 --> 00:25:49,269 805 00:25:49,269 --> 00:25:55,029 806 00:25:55,029 --> 00:25:56,950 807 00:25:56,950 --> 00:25:58,870 808 00:25:58,870 --> 00:26:00,149 809 00:26:00,149 --> 00:26:01,830 810 00:26:01,830 --> 00:26:01,840 811 00:26:01,840 --> 00:26:02,710 812 00:26:02,710 --> 00:26:04,789 813 00:26:04,789 --> 00:26:04,799 814 00:26:04,799 --> 00:26:07,350 815 00:26:07,350 --> 00:26:10,149 816 00:26:10,149 --> 00:26:12,470 817 00:26:12,470 --> 00:26:14,390 818 00:26:14,390 --> 00:26:17,430 819 00:26:17,430 --> 00:26:19,110 820 00:26:19,110 --> 00:26:21,510 821 00:26:21,510 --> 00:26:21,520 822 00:26:21,520 --> 00:26:22,789 823 00:26:22,789 --> 00:26:22,799 824 00:26:22,799 --> 00:26:23,669 825 00:26:23,669 --> 00:26:23,679 826 00:26:23,679 --> 00:26:24,789 827 00:26:24,789 --> 00:26:24,799 828 00:26:24,799 --> 00:26:27,510 829 00:26:27,510 --> 00:26:31,190 830 00:26:31,190 --> 00:26:31,200 831 00:26:31,200 --> 00:26:33,830 832 00:26:33,830 --> 00:26:37,029 833 00:26:37,029 --> 00:26:41,269 834 00:26:41,269 --> 00:26:43,190 835 00:26:43,190 --> 00:26:45,669 836 00:26:45,669 --> 00:26:47,110 837 00:26:47,110 --> 00:26:47,120 838 00:26:47,120 --> 00:26:48,710 839 00:26:48,710 --> 00:26:50,870 840 00:26:50,870 --> 00:26:53,430 841 00:26:53,430 --> 00:26:55,029 842 00:26:55,029 --> 00:26:56,470 843 00:26:56,470 --> 00:26:56,480 844 00:26:56,480 --> 00:26:57,669 845 00:26:57,669 --> 00:27:00,230 846 00:27:00,230 --> 00:27:01,750 847 00:27:01,750 --> 00:27:01,760 848 00:27:01,760 --> 00:27:02,630 849 00:27:02,630 --> 00:27:04,149 850 00:27:04,149 --> 00:27:05,990 851 00:27:05,990 --> 00:27:06,000 852 00:27:06,000 --> 00:27:07,110 853 00:27:07,110 --> 00:27:10,230 854 00:27:10,230 --> 00:27:12,149 855 00:27:12,149 --> 00:27:14,070 856 00:27:14,070 --> 00:27:16,710 857 00:27:16,710 --> 00:27:20,149 858 00:27:20,149 --> 00:27:22,470 859 00:27:22,470 --> 00:27:24,870 860 00:27:24,870 --> 00:27:27,909 861 00:27:27,909 --> 00:27:30,630 862 00:27:30,630 --> 00:27:33,830 863 00:27:33,830 --> 00:27:36,870 864 00:27:36,870 --> 00:27:40,310 865 00:27:40,310 --> 00:27:41,990 866 00:27:41,990 --> 00:27:44,149 867 00:27:44,149 --> 00:27:46,549 868 00:27:46,549 --> 00:27:49,029 869 00:27:49,029 --> 00:27:49,039 870 00:27:49,039 --> 00:27:50,549 871 00:27:50,549 --> 00:27:50,559 872 00:27:50,559 --> 00:27:51,350 873 00:27:51,350 --> 00:27:52,870 874 00:27:52,870 --> 00:27:54,310 875 00:27:54,310 --> 00:27:55,990 876 00:27:55,990 --> 00:27:58,310 877 00:27:58,310 --> 00:27:59,909 878 00:27:59,909 --> 00:28:01,750 879 00:28:01,750 --> 00:28:03,909 880 00:28:03,909 --> 00:28:03,919 881 00:28:03,919 --> 00:28:05,029 882 00:28:05,029 --> 00:28:07,269 883 00:28:07,269 --> 00:28:09,669 884 00:28:09,669 --> 00:28:11,269 885 00:28:11,269 --> 00:28:13,510 886 00:28:13,510 --> 00:28:14,710 887 00:28:14,710 --> 00:28:17,669 888 00:28:17,669 --> 00:28:20,310 889 00:28:20,310 --> 00:28:23,430 890 00:28:23,430 --> 00:28:23,440 891 00:28:23,440 --> 00:28:24,950 892 00:28:24,950 --> 00:28:26,870 893 00:28:26,870 --> 00:28:28,470 894 00:28:28,470 --> 00:28:30,870 895 00:28:30,870 --> 00:28:31,990 896 00:28:31,990 --> 00:28:33,909 897 00:28:33,909 --> 00:28:35,269 898 00:28:35,269 --> 00:28:37,430 899 00:28:37,430 --> 00:28:38,950 900 00:28:38,950 --> 00:28:41,269 901 00:28:41,269 --> 00:28:43,350 902 00:28:43,350 --> 00:28:44,789 903 00:28:44,789 --> 00:28:47,510 904 00:28:47,510 --> 00:28:50,630 905 00:28:50,630 --> 00:28:53,269 906 00:28:53,269 --> 00:28:56,230 907 00:28:56,230 --> 00:28:58,310 908 00:28:58,310 --> 00:29:00,549 909 00:29:00,549 --> 00:29:02,630 910 00:29:02,630 --> 00:29:05,110 911 00:29:05,110 --> 00:29:06,950 912 00:29:06,950 --> 00:29:08,149 913 00:29:08,149 --> 00:29:09,430 914 00:29:09,430 --> 00:29:11,350 915 00:29:11,350 --> 00:29:16,870 916 00:29:16,870 --> 00:29:19,190 917 00:29:19,190 --> 00:29:21,350 918 00:29:21,350 --> 00:29:23,269 919 00:29:23,269 --> 00:29:23,279 920 00:29:23,279 --> 00:29:24,389 921 00:29:24,389 --> 00:29:25,830 922 00:29:25,830 --> 00:29:26,950 923 00:29:26,950 --> 00:29:29,590 924 00:29:29,590 --> 00:29:32,789 925 00:29:32,789 --> 00:29:34,149 926 00:29:34,149 --> 00:29:35,430 927 00:29:35,430 --> 00:29:35,440 928 00:29:35,440 --> 00:29:37,510 929 00:29:37,510 --> 00:29:39,430 930 00:29:39,430 --> 00:29:42,070 931 00:29:42,070 --> 00:29:43,909 932 00:29:43,909 --> 00:29:43,919 933 00:29:43,919 --> 00:29:44,870 934 00:29:44,870 --> 00:29:46,549 935 00:29:46,549 --> 00:29:48,470 936 00:29:48,470 --> 00:29:50,470 937 00:29:50,470 --> 00:29:52,470 938 00:29:52,470 --> 00:29:56,789 939 00:29:56,789 --> 00:29:56,799 940 00:29:56,799 --> 00:29:57,669 941 00:29:57,669 --> 00:29:58,870 942 00:29:58,870 --> 00:30:01,750 943 00:30:01,750 --> 00:30:03,190 944 00:30:03,190 --> 00:30:03,200 945 00:30:03,200 --> 00:30:04,950 946 00:30:04,950 --> 00:30:11,350 947 00:30:11,350 --> 00:30:11,360 948 00:30:11,360 --> 00:30:12,310 949 00:30:12,310 --> 00:30:14,549 950 00:30:14,549 --> 00:30:17,590 951 00:30:17,590 --> 00:30:20,710 952 00:30:20,710 --> 00:30:23,350 953 00:30:23,350 --> 00:30:26,549 954 00:30:26,549 --> 00:30:29,350 955 00:30:29,350 --> 00:30:30,950 956 00:30:30,950 --> 00:30:32,549 957 00:30:32,549 --> 00:30:34,230 958 00:30:34,230 --> 00:30:36,310 959 00:30:36,310 --> 00:30:38,950 960 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