1 00:00:25,519 --> 00:00:26,710 all right why don't we get why don't we get started it's uh 2 35 time to begin so everyone welcome i'm back good to see you all and um well this is a few of you glad that you're here um uh so um let's see why don't i uh get a layout as i usually do what where we have been where we're going today i'll talk a little tiny bit about the homework because i've gotten a couple questions on that and then we'll jump in uh so we have been um talking about turing machines which is going to be our preferred model for the rest of the semester since turing machines are the model which we use to capture general purpose general purpose computing and we looked at turing machines and various um variations on turing machines multi-tapes non-deterministic and so on um it's a bit recapping the history of the subject when people looked at a variety of different ways of formalizing the notion of algorithm and they showed that all of those different formalizations were equivalent to one another which led to the church turing thesis that all of these models each of these models really capture our intuitive idea of what we mean by a procedure or an algorithm um first for at least for addressing um you know things like mathematical problems and and precise uh precise problems of that kind so we talked about the church turing thesis we also talked about a notation for encodings and turing machines we'll review that briefly so today we're going to give a bunch of examples of turing machine algorithms for solving a variety of different problems i should just say algorithms are really nothing special about turing machines during machines are just going to be our formal counterpart but from now on we're going to use our um church touring thesis in a way to just to talk about algorithms in general because that's really our interest turing machines is just you know our way of reasoning about them mathematically but we're really ultimately interested in understanding algorithms um i'm gonna in terms of the format of today's class i'm gonna try a little experiment um we're gonna have little breaks along the way and uh in as well as the big coffee break in the middle uh we're gonna have a little mini breaks um because i sometimes feel that there really isn't enough time for people to be writing chat uh you know questions in the chat um because things just are racing on and so this way we'll have a little break after the they have pretty much after each slide some of them will be a little longer than others for in case you have questions you can pose them to me or to the tas and we'll try to get back to you on those so um moving on then so today just as a quick review uh turing machines as we set them up they have on any input w they have three possible outcomes the turing machine can hold and accept w can hold and reject w or it can go forever on w which is rejecting by looping in in our language a turing machine can recognize a language the collection of all strings that it accepts and if the turing machine always halts we say it decides the language and it's a decider and so therefore that language is a decidable language or turing decidable often we just say decidable because we don't really have a notion of deciding in other models so we often talk about touring recognizable or just decidable well we'll sometimes just say recognizable when we understand that we're talking about turing machines we briefly talked about encodings um when you write it inside brackets some objects whatever they are it could be strings could be machines could be graphs it could be polynomials we're representing them as strings and maybe a collection of objects together represented as a string in order so that we can present that information as an input to a machine and we talk about our languages or collections of strings and a notation for writing a turing machine is going to be simply that english description put inside quotation marks to represent our informal way of talking about the formal object that we could produce if we wanted to but we're never going to ask you to do that okay um so now um let me see uh let's just take a quick break as i promised um um if there's any quick questions here you can ask [Music] um uh let's see we get a lot of questions here um all right why don't we move on and uh some of these things are good questions if there was something um you know somebody asked can you how do you tell if the machine is looping or if it's really just taking a very long time um you can't that's what we're gonna prove not in today's lecture but we're gonna prove that on thursday um and so that's uh you just can't tell so you're gonna if if that's in reference to problem five you're gonna have to find some other way of doing solving that problem without knowing whether the machine is actually ever going to hold um and somebody asked how can we tell if an english description is possible for a turing machine well i mean um you can always write down an english description uh for any machine um it's uh might be very uh technical looking but if you can write it down in a formal way as states and transitions then you can certainly write down an english description which would just be follow those follow those states um okay let's let's move on um so uh okay this is going to be our first example of a um algorithm that's going to answer a question about automata um and it's a very simple problem it's a very simple um problem to solve because we want to start out easy um and it's the problem we'll get the name of the problem is the acceptance problem for deterministic finance acceptance problem for dfas and i'm going to express it as i always do as a language so this language here i'm calling it a dfa which stands for the acceptance problem for dfas is the collection of pairs b and w b is it is a dfa w is uh considered to be some other string which will be an input to b we'll say well me thinking of it as an input to be i put the two of them in brackets to represent the pair of them as a single string and it's the you know we're not going to make explicit what the form of the encoding is the only important thing is is that the encoding should be something simple um but that the turing machine can decode back into the dfa and this input string to that dfa so anything reasonable is going to be a satisfactory encoding from us for us so this is an encoded of the two of them into a string and where b is a dfa and b accepts w so now if you want to test if something's a member of adfa a then first of all you want to make sure that the string itself that you're getting really encodes a dfa and a string so it has to be of the right form and uh once you know that then you have the dfa you have the string w and you're then gonna do the obvious thing which would be to simulate b on w and see if it's actually accepting w okay so that's that's what the content of this slide is i'm just going to write it down for you um so i'm going to give a turing machine which i'm going to call i'm going to give the name of that turing machine it's going to be d adfa to help you remember the function of this machine this is a decider for the language below the adfa language this is just a name but so nothing fancy going on here but just to help us remember because i'm going to refer to some of these during machines later on so this is the decider for adfa and i'm going to describe that machine for you and that machine decides the adfa language um so what does that mean so that machine i'm describing and now you in english as i promised i'm going to take an input string s um and first it's going to check that s has the right form uh as i mentioned has the form which is the encoding of a dfa and a w um if it's not of that form the turing machine should reject that input right away now i'm not going to go through the details of how that turing machine is going to work though i'll say a little bit more this time this time only just to give you a sense of how it actually might carry that out but you should believe that if you don't believe it you can do it with a turing machine believe you could do it with your favorite programming language that's good enough because that's going to be that's equivalent to a turing machine so first you check that the inputs are the right form then you're going to simulate the computation of b on w and then if b ends up in an except state then we know b is accepting the input and we should accept and we do if b does not end up in an accept state at the end of the end of w um then we should reject because b is not accepting w okay that's my description of this machine let's just turn to a little bit of details just to make sure we're comfortable with this um so here is our turing machine d the decider for adfa the input to the turing machine as i mentioned is going to be uh b and w provided it's of the right form so that's what the string s is supposed to be of that form um and what does that mean it's just an encoding of the of the machine w and the string the machine b and the string w so here is let me just write it down here is b written down in some just completely explicit way where you're just listing the states of b the alphabet of b the transition function as a table the start state and the set of um accepting states just writing it down explicitly as you might do it if you just want to describe that machine in a completely formal way and then writing down what the string is um whatever it is once d8fa has that as its input it can then go ahead and do that computation and you know just to try to make it a little bit more explicit i'm going to capture that here by saying let's give uh that turing machine an extra tape because we already know that the multi-tape model is equivalent to the single tape model make it our life perhaps a little easier and in the course of doing that simulation what do we want to keep track of well what is the current state of uh b the dfa as we're reading the the the symbols of w and where in w are we at right now so i'll call that k which is the uh input head location on w how many symbols of w have we read okay and that's what all i'm going to say about uh what this turing machine d looks like but i'm happy to um oh there's one more thing i do want to say further that's coming up um uh because pretty much all of the turing machines that we're going to talk about today and and going forward are going to often want to check that their input is of the correct form i don't want to repeat this every time because that's going to be kind of assumed so my shorthand for that is to say my input is of the form i'm looking for and that has built into it the check that the string the input string is of that form and we're going to reject if it is if it is not so all of our turing machines are going to start out on input the string is of a certain form and then go on and do something with it okay um and uh i thought this is a good way to um uh okay so let me try to answer a few of the questions um that that i've gotten here uh um because i think this is important as a way of getting us all started um now somebody asked me can we use arguments of this form um ken is somebody asking can we use can we uh give our description of a procedure if i'm understanding this correctly as uh using some other programming language well typically you just want to make sure you're understood if you want to do that on a homework i wouldn't advocate writing your algorithm in in java because it's just going to be hard to read but to write it down in some pseudo programming language if you want just to make sure it's clear that what what you're doing um probably english is going to be the easiest for you um um even though this person says he feels it would be easier to do it in terms of a formal language well whatever's easier as long as we can understand it um now what another this is a good question here that i got asked what if b uh here's our b gets into a loop on w well that's not going to happen b is a dfa dfas on an um they move from from state to state every time they read a symbol of w when they get to the end of w it's the end of the story there's no more computation to be done so we know in exactly how many steps b is ever going to take on it's going to take the same number of steps as the length of the input that's how many moves it gets to make so there's never b as the dfa never loops so that would be a problem if it did loop but it doesn't and there that input never does a loop so have we verified that d as a decider well i think i just did d is never going to be this d from my standpoint we've said enough to be sure that d is a decider um uh there's never any reason for d to be for that dadfa turing machine to be getting into a loop um the input head location is referring to where we are on the string w yes [Music] and somebody's asking me is this the level of detail for the homework yes that's all i want it's all i'm looking for okay um [Music] let us move on uh i'm gonna have to uh otherwise never get anywhere there's a lot of questions here they're good questions um so why don't we uh go on some of these may get resolved as we're going to look at additional examples because that's all of today is pretty much examples uh let's talk about the similar problem the acceptance problem but now for anaphase um so now actually anaphase can loop um so we have to think about what that possibly could look at but let's before we get ahead of ourselves so now the acceptance problems for anaphase looks very similar except b is going to be an nfa that's going to be a decidable language too um and now we have turing machine a d a n f a the decider for a nfa that decides a nfa okay so now uh as promised here is my new form for writing our turing machine on input bw we're we're assuming that b you know based on on the context sometimes you may want to say at this point what b and w are but from the context we know what they are they're going to be an nfa and an input w for that nfa oh so what um i don't want to jump into the solution uh first before we actually look at the solution of this we could you know d uh the turing machine could simulate the nfa on input w and you have to be careful on that simulation that you don't end up looping um because don't forget uh an nfa could have um epsilon moves and could be looping on those epsilon moves and so that would be a problem if you're not careful about how you do that simulation um now i think if you were going to simulate an nfa um you would be you wouldn't follow around loops forever um uh and i think you can uh without getting because this is not the way i'm going to solve the problem you could find a way to avoid getting caught in in getting stuck in loops for an nfa um so uh even though um that looks like it could be a problem in terms of looping forever it turns out that it won't be a problem um it wouldn't be a problem if you're careful but i'm not going to solve it that way anyway because i'm going to illustrate a different method for solving this problem and that is we have exhibited before a way of converting anaphase to dfas so my turing machine is going to solve the a n f a problem by converting its input which is an n of a to an equivalent dfa i'm calling the nfab and the dfa that i got converting it into b prime and what's nice about that is that first of all we already know how to do that conversion because we um we essentially went over that in lecture a few lectures back and it's spelled out in full detail in the textbook so that is a conversion we know how to do we'll assume we know how to do and we can implement that during that conversion on a turing machine then once we have the equivalent dfa what do we do with that well we've in the previous slide we showed how to solve the problem for dfas so if we can convert the anapha to a dfa and then we already know how to solve the problem for dfas then we're then we're done so that's how i'm going to say we're going to convert the nfa to a dfa and then i'm going to run that dadfa problem on the new machine that i produced so remember that the um this machine here decides the adfa prop and now i'm going to accept if that new machine uh if that if that previous turing machine accepts the a dfa the the adfa problem um machine accepts and um i'm gonna i'm gonna basically do whatever it does if it accepts i'll accept if it rejects then i'll reject um now um this is um [Music] uh so i guess the thing that this illustrates is this idea of using a conversion construction inside a turing machine and then a previously constructed turing machine basically as a subroutine um all this is perfectly legal and it's the kind of thing we're going to be doing a lot of and that you should be used to doing that get ready to be doing that on your homework too um and in fact i'll give you a another little extra hint that problem six can be solved in this way okay um all right so here let's pause briefly um uh okay uh got some interesting questions here coming up um uh you know do somebody ask me do we meet do we need to be explicit about how we're going to simulate that uh nfa or the dfa because we don't know how many states it has not knowing how many states you you do know how many states it has once it's presented to you on the input you can see oh this is a machine that has 12 states because you're given the machine and then you can do the simulation um let's see um and someone is asking me about limits on the simulation power of a turing machine uh that's a question that i'm going to postpone to later because you're asking if a turing machine can simulate itself and we'll talk about things like that but that's you know in in a few weeks from now so i'll hold off on that one um decidable languages the collection of decided somebody's asking me a good question about closure properties of the class of decidable languages are they closed under intersection um union and so on yes the design and the decidable languages are also closed under complement that should be something you should think about but yes that is true um the recognizable languages however are closed under union and intersection but they are not closed under complement um so that would that we will prove on thursday's lecture so another question is could we have just run b on w in this in solving this problem instead of using the reduced converting it to a new turing machine a new mission the b prime well you know yes we could have just done that um i'm okay i think we better move on uh i don't want to get too bogged down um here got a lot of questions there so sorry if i'm not getting to all of the questions okay um or you can write to the tas too obviously i'm sure you are okay so let's talk about a different problem emptiness problem for dfas okay um so now we have here i'm going to give you now a just a dfa b and no input and i'm going to ask is the language of that dfa the the empty set the empty language you understand the problem first of all i'm just handing you an um a dfa and i want to know does this dfa accept any strings at all or is it just some dumb dfa it's just always very negative dfa it just rejects everything um and has the empty language okay how do you tell not super hard um if you think about how you would write a program to do that or how you do it yourself so that's a decidable language again um so gonna give it now a touring machine decider for that language um that decider says uh well i'm given that dfa i want to know if its language is empty and the idea is just kind of what you would think i'm going to see is there a path from the start state of that dfa to the an except state of the dfa if there is such a path then that dfa is going to have an input which goes along that path and will be accepted and so the language won't be empty there's no such path then clearly this dfa can never accept anything and so the language will be empty okay now there are many different uh you know there are many different path algorithms um i think would be a little bit uh sparse of me just you know some of you know path algorithms some of you don't know path checking algorithms i would like you to just to if you were giving this kind of an answer on homework to give me some sense of how that algorithm is going to work um don't be too high level about it so the one i have in mind is kind of breadth first search er if you've heard of that but uh it's very simple algorithm what i'm going to use is a kind of a marking procedure so i'm going to start off by coloring the here is this i should have indicated this is my dfa b this is b over here um should i try taking a chance of writing on this thing oops this is b oh great that didn't help uh so this is b here and the way i'm going to test if it has a path if it accepts any input is by seeing if there's a path from the start state to any one of the except states and i'm going to start it by marking the start state and then marking all states that i can get to from previously marked states and keep doing that i can't tell i can't get to anything new okay so here there's going to be like a series of iterations here marking more and more states until there's nothing new to mark and then i say did i mark any except states okay so let's just see how i write that down in english um so i started marking the start state uh i repeat until no new state is marked and that uh i'm going to say mark every state that has an incoming arrow from our previously marked state um accept and then once i'm done with that repeat loop except if no accept state is marked because that means don't forget it's a little bit inverted from what you might think i'm going to accept if there's no marked accept state because that means there's no path to an accept state from the start state which means the language is empty and edfa is the dfas that have empty language so i should be accepting those so except if there's no um way to get to an accept state no except status is marked and rejected if some accept state is marked because then the language is not empty okay so that's all i wanted to say about that um can you somebody's asking can you just say it's something like breadth first search as long as you know the sketchier you are the more chances that you're going to get caught by the greater um who's knocking who's going to be um you know we have an army of people grading these problems and just to stay on this be on the safe side of sketchiness and say don't cut too many corners because you might miss something um chances are would be okay just to say breadth research but you know i would prefer if you you'd be safer if you said a little bit more than that okay oh um this is a good question here uh somebody asks uh can we just run the dfa on all shorts on all strings well first of all you know one thing um to say something bad would be we'll just feed in all possible strings into the dfa and if any one of them except if it accepts any one of them then we know its language is not empty well that's not a good algorithm because that's going to potentially run forever um because you know there's lots of strings there's infinitely many strings to feed in to that um to that dfa and so a turing machine you know if it's trying to be a decider had better not do any infinite operations that are potentially going to go forever um now this is to be fair to the uh proposer here the questioner here didn't ask that says well can we feed an old strings up to some bound um which would be the number of states of the machine in this case and yes that would work but then you would have to argue that that's enough and yes it is enough um but it wouldn't would not be enough in answering the problem just to say feed it in that number of them and you know we're done you would have to say what um um okay um uh um a lot of questions here i'm gonna i'm gonna move on uh uh let's see equivalence problem for dfas now we're going to take take things to the next level ask are there two um uh i'm going to give you two dfas and i want to know do they describe the same language do they do they recognize the same language okay um [Music] so how are we going to do that so that's a decidable language um here's the decider um [Music] my input now is going to be two dfas again represented as a string because that's what turing machines deal with as their inputs but they can unpack that string into two dfas and there are several different ways to do this problem and i'm sure i'm going to get uh suggestions with other other ways to go one thing you could do is uh just to feed in strings up to a certain length well because you can't just like before you can't feed in all possible strings and see if the machines ever behave differently on any of them um because that's infinite an infinite operation and we're a decider we can't do that now if you're if you want to talk about this being a recognizer instead of a decider then you might be able to do something like that just to make sure you're end up you have to be just careful um let me not say more on that right now but certainly for a decider you can't go forever you can't have infinite operations um so uh you would have to have a cutoff so you can feed in all possible strings up to some length say into a and b and see if there's any difference now we actually had a problem on that in the whole in problem set one um which said if two two dfas have um unequal languages then they're going to see a pl a difference that there's going to be a string which acts differently on them which is of length at most the product of the number of states of those two machines so it you know you can either reference that uh problem that would be an out that would be an adequate solution or reprove it or whatever that would be fine in fact there's you can do even better than that as the optional problem showed you only have to go up to the sum of the number of states not up to the product but that's actually very difficult to show but i don't i'm not going to prove it that way i'm going to prove it an entirely different way which doesn't require any analysis at all no proofing proving something about bounds um i'm going to take advantage of something we've already shown and which is i'm going to make a new fine art automaton a new dfac built out of a and b which accepts all the strings on which a and b disagree i'll show you how to that's easy to do so first of all let's in terms of a picture let's understand what this is so here we have this is the language of a this is the language of b written as a venn diagram and where are those places where they disagree um well they're you know either in a but not in b or in b but not in a i'm showing it to here in terms of the blue region um that actually has a name called the symmetric difference of these two of these two sets these are this this all of the members which are in exactly one out of the two but not both if you can make a machine c that would accept uh all of the strings in the blue region then what do we do with that machine we test if its language is empty which is what we've already shown how to do because if the blue region is empty that means that l of a equals l of b okay so i'm going to make a machine a dfac where the language of c is exactly that symmetric difference it's all the strings in a intersect with the strings that are not in b so in a and not in b or in b and not it then not an a and take the union of those two parts and how do we know we can make c well we have those closure constructions which we showed several lectures back those closure constructions can be implemented on a turing machine so a turn machine can build the dfac and then use that test from a few slides back the emptiness though last slide the emptiness tester for dfas on c to see whether its language is empty and now if c's language is empty then we know we can accept because that means the two the l of a equals l of b otherwise we can reject okay so here's a now i'm going to ask you a check um you can also use that time to send me a few more questions if you want okay here's my check-in okay now instead of testing equivalents of finite of dfas i want to test the equivalence of regular expressions um to r1 r2 regular expressions are called the eq regular expressions problem equivalence of regular expressions can we now conclude that this problem is also decidable either immediately from stuff we've already shown or yes but would take some significant extra work or no because intersection is not a regular operation so let's see if i can pull that out here launch the polling um here we go okay i think we're just about done here um five oh five seconds more okay ready set end all right yes the ant's correct answer is a we have basically already shown this fact um because what why is that because we have shown that we can convert if you're given two regular expressions and we want to test whether they're equivalent or whether they generate the same language we can convert them both to to finite automata we can convert them to n phase and then the n phase to dfas and um then we can apply what we just showed about testing equivalence of dfas so yes it really follows immediately from stuff we've already shown the conversion number one and then the testing equivalents for dfas um so there's not really any work to do here and in fact what i'm trying to illustrate with this uh check-in that once you've shown um how to do some kind of a test for one model it have going to apply for all of the equivalent models that we've that we've shown to be equivalent because the turing machine can can do the constructions which show the equivalence um okay uh so let's move on um somebody didn't get the pole did most people not get the pole well i think most of you have gotten it did you i'm not seeing a lot of complaints so if you didn't get the poll something is wrong with your setup and you're going to have to take the recorded uh uh check-ins um that are going to launch right after this lecture is over sorry but we just you should figure out what's wrong with the setup there because i think most people are getting these okay and with that we're going to take a little break and then we'll continue on afterwards now let me know how this is is this slowing us too you know should i be speed a little speedier about the um the about the little mini breaks that we're taking or um is this good for you you know feedback is always i'm trying to make this as good as i can for for all of you okay i i think there's a mixture between people saying that these breaks are good some says they're a little a little over long so i'll try to tighten them up a little um some some some folks are saying what more people should be uh asking the tas i don't know how the tas are doing in terms of their i'll check with them afterward how well this is going for them in terms of uh questions but i think some amount of break is good um so that we don't uh so there's time to be asking asking questions um otherwise what's the point of having a live lecture so we will start promptly when this timer runs down um so uh be ready to go in 55 seconds from now just to confirm to show the decidability of the equivalence of two regular expressions do we need to show that we can use a turing machine to convert them to dfas first well i mean any procedure that you can give if you want to test whether two regular expressions are equivalent you can give any procedure for doing that deciding you want um i'm just offering one one simple way to do it that takes advantage of stuff we've already shown but if you want to dig down and do some analysis of those regular expressions to show that they describe the same language they generate the same language be my guest i think that's actually would be complicated to do that that way okay so we're just about um ready to go here uh so let's continue um uh and let me take this timer off here um all right now we are going to talk a little about context-free grammars so we talked about decision problems for finite automata let's talk about some for grammars um okay now i'm going to give us an analogous problem i'm going to give you a the except i'm calling it the acceptance problem but just just for consistency with everything else but it's really the generating problem so i'm giving you a grammar g a c a context free grammar g and a string that's in a string and i want to know is it in the language of g so does g generate w i'm calling that the a cfg problem so that's going to be a decidable again so now let's just think about how that these are getting slightly harder as we're moving along so i'm giving you a grammar and a string and i want to know does the grammar generate that string um well you know it's not totally trivial to solve that you know one thing you might try doing is looking at all possible things all possible derivations from that grammar and see if any one of them leads to w leads you to generate w well you have to be careful because uh as i as we've shown in some of our examples um you actually could have um because you're allowed to have variables that can get converted to empty string you might have very long intermediate strings being generated from a grammar which then ultimately give you a small string of terminals that get generated a small string in the language that gets produced and so it's not clear how you know you certainly don't want to try every possible derivation because there's infinitely many different derivations most of them generating of course things that are irrelevant to the string w but you have to know how to cut things off and it's not immediately obvious how you do that unless you have done the homework problems that i asked you to do which i'm sure very many of you have not done the zerobot point x problems because there that's going to come in handy right now and so i'll i'll help you through that um remember i'm hoping you should remember but you may not have looked at it uh something called chomsky normal form which is for context-free grammars but only allows the rules to be of a special kind and they have to look like this they can be a variable that goes to two variables on the right-hand side or a variable that goes to a terminal those are the only kinds of rules that you're allowed there's a special case for the empty string but let's ignore that you know for if you have you want to put the the start variable can also go to the empty string if you want to have a the empty string in the language but that's that's a special case let's ignore that these are the only two kinds of rules that you can have in the chomsky normal form gram once you have a chomsky normal form grammar um well first of all there's two things first of all you can always convert any any context free grammar into chomsky normal form so that's given in the textbook it's very it's sort of the you do the obvious thing i'm not going to spend time on it and you don't have to know it it's it's kind of just a little bit um tedious but it's it's straightforward and it's there if you if you're curious um but the second line is the thing that's important to us uh which is that if you have a grammar which in is in chomsky normal form and a string that's generated every derivation of that string has exactly two times the length of the string minus one steps it's actually if you think about it i mean this dilemma is very easy to prove um i think the homework problem asks you to prove that in the 0.2 or whatever it was um and problem set one or pub set two when i remember um because you can in order to do you know rules of this kind allow you to make the string one longer and rules of this kind allow you to produce an uh an x a new terminal symbol and so you're really going to end up having um you know if if the length of w is n you're going to have n minus one of these and n of those that's why you get two n minus one steps but the point is i don't really care exactly how many it's just that we have a bound and once you have that bound life is good from the point of view of um giving a turing machine which is going to decide this language because here's the turing machine what's going to do with the first thing is it's going to convert g into chomsky normal form that we assume we know how to do now we're going to try all derivations but only those of length 2 twice the length of the string minus 1. um because if any derivation is going to generate w it has to be this this many steps um now that we know the grammar is in chomsky normal form okay so we have converted this problem of one that might be a very unboundedly lengthy problem into one way you know it's going to terminate after some fixed number of steps and so therefore we can accept if any of those uh generate w and reject not okay and so oh so this is uh before moving on so this uh answers the problem shows that this language is decidable the the acfg language okay so that's something that you know make sure you understand um uh we're basically trying all derivations of up to of a certain length because that's all that's needed when the grammar is in that form now we're going to use that to prove a corollary which is um which is important for understanding how everything fits together and that corollary states that every context-free language is a decidable language every content-free language is decidable now why does that follow from this well um suppose you have a context-free language a we know that language is generated by some context free grammar g okay that's what it means so then you can take that grammar g and you can build it into a turing machine so there's going to be a turing machine which ha which is uh um constructed with the knowledge of g um and uh that turing machine is going to take its w and run the acfg algorithm um with w combined with the g that's already built into it so it's just going to stick that grammar g in front of w and now run the ac of g decide it's going to say does g generate w well that's going to be yes every time w is an a and so this machine here now uh is going to end up accepting every string that's in a because it's every string that g generates okay so um uh that is i think what i wanted to say about this now i feel that this corollary here throws a little bit of a curveball curveball um and um uh we can just pause for a moment here just to uh make sure you're understanding this so let me the tricky thing about this corollary is that i often get uh a question i often get where is you know when we start off with a context-free language who gives a w how do we get g because we need g to build the turing machine um m sub g so we know a is a context free language but how do we know what g is well the point is that uh we may not know what g is but we know g exists because that's a definition of a being a context-free language it must have a context-free grammar by definition and so because g exists i now know my turn machine m sub g exists and that's enough to know that a is decidable because it has a decider that exists now if you're not going to tell me the grammar for g i'm not going to tell you the turing machine which decides a but i know that but both of them exist so if you tell me g then i can tell you the turing machine okay so it's a subtle tricky issue there a certain little element maybe of sometimes people call it non-constructive because we're just in a sense showing just that something is existing but um you know it does the job for us because it shows that a is a decidable language okay so not so many questions here maybe the tas are getting them so let's move on um here's another check-in uh so now can we conclude that a instead of a cfg a pda is decidable hey people are getting this one pretty fast another 10 seconds three seconds okay i'm gonna shut it down um end polling share results yeah you know i uh so this uh problem here is apda is decidable it's really in a sense i was almost wondering whether or not to give this poll because it has it's true for the same reason as the as poll number one um because we know how to convert pdas or we stated we can convert pdas to cfgs and you know that conversion is given in the book we didn't do in lecture but i just stated you have to know that fact but not necessarily know how to do it that's okay but the fact is true um the conversion is in the book and it could be implemented on a turing machine so if you want to know whether a pda's language is empty you convert it to a context-free grammar and then use this procedure here to test whether it's a language not empty i'm sorry the acceptance problem i just want to not accept i just want to know does the pda accept some input i'm saying it wrong so i want to know does the pda accept some input i convert that pda to a grammar and then i see if the grammar generates that input because it's an equivalent grammar um so again this is using the fact that grammars and pdas are equivalent and interconvertible from one to another just like regular expressions in dfas from the previous poll so you need to be comfortable with that uh because we're going to not even talk about it anymore we're just going to be treating those things going back and forth between them without sometimes without even any comment okay so let's move on uh emptiness problem for cfgs hopefully you're getting comfortable with the know you know the terminology i'm using um so now the emptiness problem for context-free grammars i'm just going to give you a grammar and i want to know if its language is empty okay so remember we did this refined automata by testing a path can't you know we don't really have paths here you might think testing pairs and push down automata that's not going to really work because the stack is there so how are we going to do that test well there's something sort of like testing a path um just working with the grammar itself kind of working backwards from the terminals now um i'll kind of illustrate that here here's imagine i'm giving you here's a very simple grammar and i want to know does it generate any strings any strings of term obviously only care about strings of terminals because those are the things that are in the in the language so does this grammar generate any strings of terminals so what we're going to answer that is by a marking procedure but in a sense we're going to start from the um from the terminals and work backwards to see if we can get to the start variable so first we're going to mark all the terminal symbols and then we're going to mark anything that goes to a string of completely marked symbols either terminals or variables because anything that's marked we know can derive a string of terminals that's what it means anything that's that's blue can derive a string of just terminals and so now if you have a collection of those that are all marked they together can generate some you know string of terminals um together and so then you can mark the associated variable so like t goes to a so that may be over simple but we're going to mark t here everywhere in the grammar so all these t's are going to get marked because we know that t can generate a string of terminals now let's take a look at r r is going to a string of symbols that are all marked and that means those symbols can in turn generate the strings of terminals so we can mark r2 we can't get mark s because we don't know yet whether s has some unmarked thing that it goes to so we don't know yet whether s can generate a string of terminals but r we know right now so we're going to mark r but we then now that gets us to mark this r and so then we can go backwards and we can mark s and we keep doing that until there's nothing new to mark and here we've marked everything so clearly there's nothing more you can mark but you might stop before you've marked everything and then you see whether you've marked the stark variable or not and if you have you know the language is not empty okay so let's just go through this in in in in text for mark all occurrences of terminals in g then repeat until no new variables are marked we mark all occurrences of variables a if a goes to a string of symbols and all of those symbols were already marked because those are the things that already have been shown to generate a string of terminals and so now we're going to reject if the start status start variable is marked because that means that the language is not empty and except if it's not okay so um let me just just i'll take a couple of quick quick questions here okay somebody um somebody asked whether uh if i understand other regular languages also decidable well remember the regular languages are context-free languages and the context-free languages are decidable so yes the regular languages are decidable um okay i don't know some of those are going to be too long to answer and they're trying to come up with alternative solutions so i think we're going to move on um all right we talked about um just like we did for the final automata we talked about acceptance we talked about emptiness we talked about equivalence so how about the equivalence problem for context-free grammars i'm going to give you now two context-free grammars and i'd like to know are those two contexts free grammars generating the same language so how might you think about that well you know one thing uh you know following some of the ideas that we've already had um you could try feeding strings in to those grammars we know how to test whether those individual grammars can generate those strings so you can just try feeding strings in to g and to h and seeing whether those grammars ever disagree or whether they generate some string find some string that say g generates but h does not generate and we can test that string by string unfortunately we've got a lot of strings to test right that's gonna you know we we would need to give some bound if we were going to use that for a procedure not clear what the bound is um another idea um might be to use the uh closure construction uh that we had from before um so let's let's mull that over um whether that might work um but in fact um [Music] uh let me give away the answer here um this language is not decidable there's no algorithm out there no turing machine but therefore no algorithm which can take two grammars and test whether they generate the same language or not seems at first glance kind of surprising such simple things as context-free grammars can nonetheless be so complicated that there's no procedure out there which can tell whether the two gram those two grammars generate the same language we will show that um next week a related problem which is related to your homework this uh that's due on uh thursday uh is testing whether a grammar is ambiguous so given a grammar i'd like to know is is that grammar and ambiguous grammar or not i mean does it generate some string you know in the language of that grammar in two possibly different ways two or more different ways so is there some string um that can be generated you know with two different parsers um so um the problem of testing whether grammar is ambiguous not decidable so i'm asking you to do something hard something hard when you have to produce that grammar which is unambiguous for that language um in general testing whether grammar is ambiguous or not is not a decidable problem now hopefully that the grammar that you'll produce uh to show that um you know to get that unambiguous grammar for that language that i'm asking you to produce is not going to require our graders to solve some undecidable problem that it'll be clear based on the construction of that grammar why it's uh not ambiguous so you'll have to hopefully say some some explanation of what you're thinking there uh okay um so that's uh uh and we will prove that uh the problem of testing ambiguity is not decidable that's gonna be a homework problem in problem set three um okay uh let's check in here kind of something that i alluded to but didn't quite didn't want to give away uh why not use the same technique that we use to show equivalence of dfas is decidable to show that equivalence of context-free grammars is decidable obviously something goes wrong because eqcfg is not decidable why doesn't that same technique just work um well um what's the answer oh this is got a real race here another 15 seconds this one it gives you something to mull over um all right let's end it okay you're good to go at least click something um okay ending polling sharing results okay a bunch of you have thought well cfgs are generators and dfas are recognizers and that's the issue well not really because you know you can we could test um equivalence of regular expressions those are generators it's nothing to do with being a generator or not um um because you know we can convert regular expressions to dfas and then and then test equivalence for the dfa so that it's not really a matter of a matter of being generated that's not the issue the issue is that we can't follow the same construction that we did to show eqdfa as decidable because the context-free languages are not closed under those operations we needed to make that symmetric difference machine if you remember how that worked so then i closed under complementation and not closer intersection we needed both in order to build that machine c um which accepted all the strings that are sort of in the difference okay so let's continue on um let's move now to turing machines uh this is where stuff is really going to start to get interesting i'm not hoping it's been interesting all along but maybe even more interesting um so let's talk about the acceptance problem for turing machines atm this is going to be because this language is going to become an old friend we're just getting just getting to know it uh so this is the problem you're given a turing machine now in an input and i want to know does m accept that input that term does that turning machine accept its input um okay so that's going to not be a decidable problem either so it kind of shifted gears from a bunch of decidable things to a bunch of undecidable things so this is not a decidable language we will prove that um on thursday that's going to be the whole point of thursday's lecture is proving that atm is undecidable and that's going to be really a jumping off point for showing other problems are undecidable um so that's going to be our first uh proof of undecidability um but we do know that atm is recognizable and that's worth um understanding um for multiple reasons but for one thing it's going to give us an example of a problem which we know is recognizable but not decidable as we will as will prove undecidability but it's also um i think of of of historical importance um uh uh this this algorithm for showing it recognizable so let's go through that algorithm for it's very simple sort of doing the obvious thing um the following turing machine i'm calling you going to call it u for a reason that i'll make clear in a second that's going to be a recognizer for atm so it takes as input an m and a w and it's just going to simulate m w pretty much the way the um algorithm the the decider for adfa worked but now the machine instead of being a dfa it's a turing machine and the turing machine might go forever and so the simulation may not stop and that's the key difference which makes it from a decider into a recognizer so that simulation um so you're going to be simulating just keeping track of the tape of m the current state of m um where you know and proceeding to modify the tape as m modifies it and then if m enters an accept state then you know m is accepted as input so you also will enter and accept state so the machine u enters the accept state if m observes during the simulation that m enters an accept state uh furthermore if n m enters a reject state then u also enters a reject state okay now i want to say something beyond that which i i want you to pay attention to um which is the kind of thing i do see sometimes people saying uh we want you to reject if m never holds i mean that's also seems like what we want because if it never holds then m is rejecting by looping so you should also reject but i don't like that i don't like that line here step four of the turing machine because there's no way for a turing machine to determine or at least as we no obvious way uh and in fact it will not be any way but certainly at this point no obvious way for m to even tell for you to tell whether m is holding or not uh well certainly tell that is to say it never holds how can you make that determination so this is not if i saw this on a um on a solution either on an exam or a homework i would mark you off this is no good it's not a legal touring machine action if m does not hold on w um then you should reject that is correct and you can make that reasoning external to the algorithm of you but you is going to end up rejecting because it never holds either it never actually knows that m is rejecting w in that case if m is rejecting by looping it's just blindly going along and doing the simulation of m on w um and we'll end up holding rejecting by looping if m is rejecting by looping but that's something that you can argue if you need to make a proof or make some sort of reasoning about the machine but it's not part of the algorithm of the machine um okay now what what's i think from a historical standpoint it's interesting about this machine you and why i'm calling it you is because this appeared this al this machine you appeared in turing's original paper where he laid out turing machines he didn't call them touring machines by the way he called them computing machines people afterward called them turing machines uh but uh turing called this the universal computing machine that's his language actually i just looked at the paper yesterday um just to refresh my memory of it um and he gives the description of the operation of view in gory detail with all of the transitions and the states he nails the whole thing down takes pages and pages and pages so he gives it gives it there um so this is the original universal uh computing mission universal turing machine and this it's more than just an idle curiosity that disappeared in touring's paper because this actually turned out to be profoundly influential in the in computer science because it really was the first example of a machine that operated based on a stored program it really was a revolutionary idea in the in those days if you wanted to make a machine that did something different you had to wire the machine rewire the machine but here's a machine that operated based on instructions um and the instructions in a sense are no different than the data uh so this is what's been come to be known as the von neumann architecture but von neumann himself gave credit to turing machine for having inspired inspired him to think of this and some people argue that it's really calling it the binomial a bunch of people came up with this concept uh around the same time maybe other people too you know there's babbage and so on and others who um you know out of love lies also who came up with notions of programming but i think is a little different than um uh than this in concept but anyway uh this nevertheless played an important role in the history of the subject so with that i think we're at a time i'm going to um uh quickly review where we've been um so we should just show the decidability of various problems um these are all languages that we uh showed are decidable we showed that atm is touring recognizable and i think that was all we had for today um so um i will uh we will stop right here i will stick around take a few questions and our tas can take a few questions if you want by chat um and then i also have my office hours which will start in like five minutes or so once i get everything set up on my head um okay somebody wanted me to review this point here which i'm happy to do um uh you know this code here that i'm describing in english um needs to be something that you can implement on a turing machine um you know we're never going to go through the effort of building the transition functions and the states and so on but we need to be sure that we could if if we had to and how could you make a turing machine do the test that m doesn't hold that's something we don't know how to do i mean i can see if m holds enter i mean i can during the simulation i can see that m has entered the q reject state or the q accept state so i can tell while i'm simulating that it has halted but you know how would the machine or how would you know that m is now how would you you know someone says well do x if m never holds well how do you know him how how can you do the test that m is not holding um not there's no obvious way to do that in fact there is no way to do that but you know as it stands right now you know the proof would be on you to um to show how to implement that on a machine and there's just no obvious way to do that so i think that's why you should you should only put down things here which you're sure you could implement um you know even if it might take a long time so you don't have to worry about how long it would take but you have to put things down that you are sure you could at least implement in principle so someone has asked me about um the equivalence problem for context-free grammars uh being unsolvable uh why couldn't the machine be a human-level system of logic so that the computer could logically deduce whether or not it was decidable um well sort of that you know we're taking a turn into the subject of mathematical logic which is supposed to kind of uh formalize reasoning in a way um and uh in the end what it really is kind of come down to that there are certain uh grammars uh which are equivalent to one another but there's not going to be any way to prove that they're equivalent in any reasonable system um in equivalence you can always prove right you can exhibit a string that one you can show that this grammar is generating it this grammar is not generating this other one is not so uh in equivalence you can always prove but equivalence there are going to be certain pairs of grammars which are going to be beyond the capability of any reasonable formal system to be able to prove that they're equivalent because you can even convert them and make those grammars into something which talk about the system itself ultimately you're going to end up you know with a russell paradox kind of thing that's maybe going beyond more than you want to know and i'm happy to talk about it offline but you know there's just no way to make a turing machine which is going to implement sort of human reasoning and then get the right answer on all pairs of grammars just cannot be done bye bye i'm going to shut down the meeting though you 2 00:00:26,710 --> 00:00:29,269 all right why don't we get why don't we get started it's uh 2 35 time to begin so everyone welcome i'm back good to see you all and um well this is a few of you glad that you're here um uh so um let's see why don't i uh get a layout as i usually do what where we have been where we're going today i'll talk a little tiny bit about the homework because i've gotten a couple questions on that and then we'll jump in uh so we have been um talking about turing machines which is going to be our preferred model for the rest of the semester since turing machines are the model which we use to capture general purpose general purpose computing and we looked at turing machines and various um variations on turing machines multi-tapes non-deterministic and so on um it's a bit recapping the history of the subject when people looked at a variety of different ways of formalizing the notion of algorithm and they showed that all of those different formalizations were equivalent to one another which led to the church turing thesis that all of these models each of these models really capture our intuitive idea of what we mean by a procedure or an algorithm um first for at least for addressing um you know things like mathematical problems and and precise uh precise problems of that kind so we talked about the church turing thesis we also talked about a notation for encodings and turing machines we'll review that briefly so today we're going to give a bunch of examples of turing machine algorithms for solving a variety of different problems i should just say algorithms are really nothing special about turing machines during machines are just going to be our formal counterpart but from now on we're going to use our um church touring thesis in a way to just to talk about algorithms in general because that's really our interest turing machines is just you know our way of reasoning about them mathematically but we're really ultimately interested in understanding algorithms um i'm gonna in terms of the format of today's class i'm gonna try a little experiment um we're gonna have little breaks along the way and uh in as well as the big coffee break in the middle uh we're gonna have a little mini breaks um because i sometimes feel that there really isn't enough time for people to be writing chat uh you know questions in the chat um because things just are racing on and so this way we'll have a little break after the they have pretty much after each slide some of them will be a little longer than others for in case you have questions you can pose them to me or to the tas and we'll try to get back to you on those so um moving on then so today just as a quick review uh turing machines as we set them up they have on any input w they have three possible outcomes the turing machine can hold and accept w can hold and reject w or it can go forever on w which is rejecting by looping in in our language a turing machine can recognize a language the collection of all strings that it accepts and if the turing machine always halts we say it decides the language and it's a decider and so therefore that language is a decidable language or turing decidable often we just say decidable because we don't really have a notion of deciding in other models so we often talk about touring recognizable or just decidable well we'll sometimes just say recognizable when we understand that we're talking about turing machines we briefly talked about encodings um when you write it inside brackets some objects whatever they are it could be strings could be machines could be graphs it could be polynomials we're representing them as strings and maybe a collection of objects together represented as a string in order so that we can present that information as an input to a machine and we talk about our languages or collections of strings and a notation for writing a turing machine is going to be simply that english description put inside quotation marks to represent our informal way of talking about the formal object that we could produce if we wanted to but we're never going to ask you to do that okay um so now um let me see uh let's just take a quick break as i promised um um if there's any quick questions here you can ask [Music] um uh let's see we get a lot of questions here um all right why don't we move on and uh some of these things are good questions if there was something um you know somebody asked can you how do you tell if the machine is looping or if it's really just taking a very long time um you can't that's what we're gonna prove not in today's lecture but we're gonna prove that on thursday um and so that's uh you just can't tell so you're gonna if if that's in reference to problem five you're gonna have to find some other way of doing solving that problem without knowing whether the machine is actually ever going to hold um and somebody asked how can we tell if an english description is possible for a turing machine well i mean um you can always write down an english description uh for any machine um it's uh might be very uh technical looking but if you can write it down in a formal way as states and transitions then you can certainly write down an english description which would just be follow those follow those states um okay let's let's move on um so uh okay this is going to be our first example of a um algorithm that's going to answer a question about automata um and it's a very simple problem it's a very simple um problem to solve because we want to start out easy um and it's the problem we'll get the name of the problem is the acceptance problem for deterministic finance acceptance problem for dfas and i'm going to express it as i always do as a language so this language here i'm calling it a dfa which stands for the acceptance problem for dfas is the collection of pairs b and w b is it is a dfa w is uh considered to be some other string which will be an input to b we'll say well me thinking of it as an input to be i put the two of them in brackets to represent the pair of them as a single string and it's the you know we're not going to make explicit what the form of the encoding is the only important thing is is that the encoding should be something simple um but that the turing machine can decode back into the dfa and this input string to that dfa so anything reasonable is going to be a satisfactory encoding from us for us so this is an encoded of the two of them into a string and where b is a dfa and b accepts w so now if you want to test if something's a member of adfa a then first of all you want to make sure that the string itself that you're getting really encodes a dfa and a string so it has to be of the right form and uh once you know that then you have the dfa you have the string w and you're then gonna do the obvious thing which would be to simulate b on w and see if it's actually accepting w okay so that's that's what the content of this slide is i'm just going to write it down for you um so i'm going to give a turing machine which i'm going to call i'm going to give the name of that turing machine it's going to be d adfa to help you remember the function of this machine this is a decider for the language below the adfa language this is just a name but so nothing fancy going on here but just to help us remember because i'm going to refer to some of these during machines later on so this is the decider for adfa and i'm going to describe that machine for you and that machine decides the adfa language um so what does that mean so that machine i'm describing and now you in english as i promised i'm going to take an input string s um and first it's going to check that s has the right form uh as i mentioned has the form which is the encoding of a dfa and a w um if it's not of that form the turing machine should reject that input right away now i'm not going to go through the details of how that turing machine is going to work though i'll say a little bit more this time this time only just to give you a sense of how it actually might carry that out but you should believe that if you don't believe it you can do it with a turing machine believe you could do it with your favorite programming language that's good enough because that's going to be that's equivalent to a turing machine so first you check that the inputs are the right form then you're going to simulate the computation of b on w and then if b ends up in an except state then we know b is accepting the input and we should accept and we do if b does not end up in an accept state at the end of the end of w um then we should reject because b is not accepting w okay that's my description of this machine let's just turn to a little bit of details just to make sure we're comfortable with this um so here is our turing machine d the decider for adfa the input to the turing machine as i mentioned is going to be uh b and w provided it's of the right form so that's what the string s is supposed to be of that form um and what does that mean it's just an encoding of the of the machine w and the string the machine b and the string w so here is let me just write it down here is b written down in some just completely explicit way where you're just listing the states of b the alphabet of b the transition function as a table the start state and the set of um accepting states just writing it down explicitly as you might do it if you just want to describe that machine in a completely formal way and then writing down what the string is um whatever it is once d8fa has that as its input it can then go ahead and do that computation and you know just to try to make it a little bit more explicit i'm going to capture that here by saying let's give uh that turing machine an extra tape because we already know that the multi-tape model is equivalent to the single tape model make it our life perhaps a little easier and in the course of doing that simulation what do we want to keep track of well what is the current state of uh b the dfa as we're reading the the the symbols of w and where in w are we at right now so i'll call that k which is the uh input head location on w how many symbols of w have we read okay and that's what all i'm going to say about uh what this turing machine d looks like but i'm happy to um oh there's one more thing i do want to say further that's coming up um uh because pretty much all of the turing machines that we're going to talk about today and and going forward are going to often want to check that their input is of the correct form i don't want to repeat this every time because that's going to be kind of assumed so my shorthand for that is to say my input is of the form i'm looking for and that has built into it the check that the string the input string is of that form and we're going to reject if it is if it is not so all of our turing machines are going to start out on input the string is of a certain form and then go on and do something with it okay um and uh i thought this is a good way to um uh okay so let me try to answer a few of the questions um that that i've gotten here uh um because i think this is important as a way of getting us all started um now somebody asked me can we use arguments of this form um ken is somebody asking can we use can we uh give our description of a procedure if i'm understanding this correctly as uh using some other programming language well typically you just want to make sure you're understood if you want to do that on a homework i wouldn't advocate writing your algorithm in in java because it's just going to be hard to read but to write it down in some pseudo programming language if you want just to make sure it's clear that what what you're doing um probably english is going to be the easiest for you um um even though this person says he feels it would be easier to do it in terms of a formal language well whatever's easier as long as we can understand it um now what another this is a good question here that i got asked what if b uh here's our b gets into a loop on w well that's not going to happen b is a dfa dfas on an um they move from from state to state every time they read a symbol of w when they get to the end of w it's the end of the story there's no more computation to be done so we know in exactly how many steps b is ever going to take on it's going to take the same number of steps as the length of the input that's how many moves it gets to make so there's never b as the dfa never loops so that would be a problem if it did loop but it doesn't and there that input never does a loop so have we verified that d as a decider well i think i just did d is never going to be this d from my standpoint we've said enough to be sure that d is a decider um uh there's never any reason for d to be for that dadfa turing machine to be getting into a loop um the input head location is referring to where we are on the string w yes [Music] and somebody's asking me is this the level of detail for the homework yes that's all i want it's all i'm looking for okay um [Music] let us move on uh i'm gonna have to uh otherwise never get anywhere there's a lot of questions here they're good questions um so why don't we uh go on some of these may get resolved as we're going to look at additional examples because that's all of today is pretty much examples uh let's talk about the similar problem the acceptance problem but now for anaphase um so now actually anaphase can loop um so we have to think about what that possibly could look at but let's before we get ahead of ourselves so now the acceptance problems for anaphase looks very similar except b is going to be an nfa that's going to be a decidable language too um and now we have turing machine a d a n f a the decider for a nfa that decides a nfa okay so now uh as promised here is my new form for writing our turing machine on input bw we're we're assuming that b you know based on on the context sometimes you may want to say at this point what b and w are but from the context we know what they are they're going to be an nfa and an input w for that nfa oh so what um i don't want to jump into the solution uh first before we actually look at the solution of this we could you know d uh the turing machine could simulate the nfa on input w and you have to be careful on that simulation that you don't end up looping um because don't forget uh an nfa could have um epsilon moves and could be looping on those epsilon moves and so that would be a problem if you're not careful about how you do that simulation um now i think if you were going to simulate an nfa um you would be you wouldn't follow around loops forever um uh and i think you can uh without getting because this is not the way i'm going to solve the problem you could find a way to avoid getting caught in in getting stuck in loops for an nfa um so uh even though um that looks like it could be a problem in terms of looping forever it turns out that it won't be a problem um it wouldn't be a problem if you're careful but i'm not going to solve it that way anyway because i'm going to illustrate a different method for solving this problem and that is we have exhibited before a way of converting anaphase to dfas so my turing machine is going to solve the a n f a problem by converting its input which is an n of a to an equivalent dfa i'm calling the nfab and the dfa that i got converting it into b prime and what's nice about that is that first of all we already know how to do that conversion because we um we essentially went over that in lecture a few lectures back and it's spelled out in full detail in the textbook so that is a conversion we know how to do we'll assume we know how to do and we can implement that during that conversion on a turing machine then once we have the equivalent dfa what do we do with that well we've in the previous slide we showed how to solve the problem for dfas so if we can convert the anapha to a dfa and then we already know how to solve the problem for dfas then we're then we're done so that's how i'm going to say we're going to convert the nfa to a dfa and then i'm going to run that dadfa problem on the new machine that i produced so remember that the um this machine here decides the adfa prop and now i'm going to accept if that new machine uh if that if that previous turing machine accepts the a dfa the the adfa problem um machine accepts and um i'm gonna i'm gonna basically do whatever it does if it accepts i'll accept if it rejects then i'll reject um now um this is um [Music] uh so i guess the thing that this illustrates is this idea of using a conversion construction inside a turing machine and then a previously constructed turing machine basically as a subroutine um all this is perfectly legal and it's the kind of thing we're going to be doing a lot of and that you should be used to doing that get ready to be doing that on your homework too um and in fact i'll give you a another little extra hint that problem six can be solved in this way okay um all right so here let's pause briefly um uh okay uh got some interesting questions here coming up um uh you know do somebody ask me do we meet do we need to be explicit about how we're going to simulate that uh nfa or the dfa because we don't know how many states it has not knowing how many states you you do know how many states it has once it's presented to you on the input you can see oh this is a machine that has 12 states because you're given the machine and then you can do the simulation um let's see um and someone is asking me about limits on the simulation power of a turing machine uh that's a question that i'm going to postpone to later because you're asking if a turing machine can simulate itself and we'll talk about things like that but that's you know in in a few weeks from now so i'll hold off on that one um decidable languages the collection of decided somebody's asking me a good question about closure properties of the class of decidable languages are they closed under intersection um union and so on yes the design and the decidable languages are also closed under complement that should be something you should think about but yes that is true um the recognizable languages however are closed under union and intersection but they are not closed under complement um so that would that we will prove on thursday's lecture so another question is could we have just run b on w in this in solving this problem instead of using the reduced converting it to a new turing machine a new mission the b prime well you know yes we could have just done that um i'm okay i think we better move on uh i don't want to get too bogged down um here got a lot of questions there so sorry if i'm not getting to all of the questions okay um or you can write to the tas too obviously i'm sure you are okay so let's talk about a different problem emptiness problem for dfas okay um so now we have here i'm going to give you now a just a dfa b and no input and i'm going to ask is the language of that dfa the the empty set the empty language you understand the problem first of all i'm just handing you an um a dfa and i want to know does this dfa accept any strings at all or is it just some dumb dfa it's just always very negative dfa it just rejects everything um and has the empty language okay how do you tell not super hard um if you think about how you would write a program to do that or how you do it yourself so that's a decidable language again um so gonna give it now a touring machine decider for that language um that decider says uh well i'm given that dfa i want to know if its language is empty and the idea is just kind of what you would think i'm going to see is there a path from the start state of that dfa to the an except state of the dfa if there is such a path then that dfa is going to have an input which goes along that path and will be accepted and so the language won't be empty there's no such path then clearly this dfa can never accept anything and so the language will be empty okay now there are many different uh you know there are many different path algorithms um i think would be a little bit uh sparse of me just you know some of you know path algorithms some of you don't know path checking algorithms i would like you to just to if you were giving this kind of an answer on homework to give me some sense of how that algorithm is going to work um don't be too high level about it so the one i have in mind is kind of breadth first search er if you've heard of that but uh it's very simple algorithm what i'm going to use is a kind of a marking procedure so i'm going to start off by coloring the here is this i should have indicated this is my dfa b this is b over here um should i try taking a chance of writing on this thing oops this is b oh great that didn't help uh so this is b here and the way i'm going to test if it has a path if it accepts any input is by seeing if there's a path from the start state to any one of the except states and i'm going to start it by marking the start state and then marking all states that i can get to from previously marked states and keep doing that i can't tell i can't get to anything new okay so here there's going to be like a series of iterations here marking more and more states until there's nothing new to mark and then i say did i mark any except states okay so let's just see how i write that down in english um so i started marking the start state uh i repeat until no new state is marked and that uh i'm going to say mark every state that has an incoming arrow from our previously marked state um accept and then once i'm done with that repeat loop except if no accept state is marked because that means don't forget it's a little bit inverted from what you might think i'm going to accept if there's no marked accept state because that means there's no path to an accept state from the start state which means the language is empty and edfa is the dfas that have empty language so i should be accepting those so except if there's no um way to get to an accept state no except status is marked and rejected if some accept state is marked because then the language is not empty okay so that's all i wanted to say about that um can you somebody's asking can you just say it's something like breadth first search as long as you know the sketchier you are the more chances that you're going to get caught by the greater um who's knocking who's going to be um you know we have an army of people grading these problems and just to stay on this be on the safe side of sketchiness and say don't cut too many corners because you might miss something um chances are would be okay just to say breadth research but you know i would prefer if you you'd be safer if you said a little bit more than that okay oh um this is a good question here uh somebody asks uh can we just run the dfa on all shorts on all strings well first of all you know one thing um to say something bad would be we'll just feed in all possible strings into the dfa and if any one of them except if it accepts any one of them then we know its language is not empty well that's not a good algorithm because that's going to potentially run forever um because you know there's lots of strings there's infinitely many strings to feed in to that um to that dfa and so a turing machine you know if it's trying to be a decider had better not do any infinite operations that are potentially going to go forever um now this is to be fair to the uh proposer here the questioner here didn't ask that says well can we feed an old strings up to some bound um which would be the number of states of the machine in this case and yes that would work but then you would have to argue that that's enough and yes it is enough um but it wouldn't would not be enough in answering the problem just to say feed it in that number of them and you know we're done you would have to say what um um okay um uh um a lot of questions here i'm gonna i'm gonna move on uh uh let's see equivalence problem for dfas now we're going to take take things to the next level ask are there two um uh i'm going to give you two dfas and i want to know do they describe the same language do they do they recognize the same language okay um [Music] so how are we going to do that so that's a decidable language um here's the decider um [Music] my input now is going to be two dfas again represented as a string because that's what turing machines deal with as their inputs but they can unpack that string into two dfas and there are several different ways to do this problem and i'm sure i'm going to get uh suggestions with other other ways to go one thing you could do is uh just to feed in strings up to a certain length well because you can't just like before you can't feed in all possible strings and see if the machines ever behave differently on any of them um because that's infinite an infinite operation and we're a decider we can't do that now if you're if you want to talk about this being a recognizer instead of a decider then you might be able to do something like that just to make sure you're end up you have to be just careful um let me not say more on that right now but certainly for a decider you can't go forever you can't have infinite operations um so uh you would have to have a cutoff so you can feed in all possible strings up to some length say into a and b and see if there's any difference now we actually had a problem on that in the whole in problem set one um which said if two two dfas have um unequal languages then they're going to see a pl a difference that there's going to be a string which acts differently on them which is of length at most the product of the number of states of those two machines so it you know you can either reference that uh problem that would be an out that would be an adequate solution or reprove it or whatever that would be fine in fact there's you can do even better than that as the optional problem showed you only have to go up to the sum of the number of states not up to the product but that's actually very difficult to show but i don't i'm not going to prove it that way i'm going to prove it an entirely different way which doesn't require any analysis at all no proofing proving something about bounds um i'm going to take advantage of something we've already shown and which is i'm going to make a new fine art automaton a new dfac built out of a and b which accepts all the strings on which a and b disagree i'll show you how to that's easy to do so first of all let's in terms of a picture let's understand what this is so here we have this is the language of a this is the language of b written as a venn diagram and where are those places where they disagree um well they're you know either in a but not in b or in b but not in a i'm showing it to here in terms of the blue region um that actually has a name called the symmetric difference of these two of these two sets these are this this all of the members which are in exactly one out of the two but not both if you can make a machine c that would accept uh all of the strings in the blue region then what do we do with that machine we test if its language is empty which is what we've already shown how to do because if the blue region is empty that means that l of a equals l of b okay so i'm going to make a machine a dfac where the language of c is exactly that symmetric difference it's all the strings in a intersect with the strings that are not in b so in a and not in b or in b and not it then not an a and take the union of those two parts and how do we know we can make c well we have those closure constructions which we showed several lectures back those closure constructions can be implemented on a turing machine so a turn machine can build the dfac and then use that test from a few slides back the emptiness though last slide the emptiness tester for dfas on c to see whether its language is empty and now if c's language is empty then we know we can accept because that means the two the l of a equals l of b otherwise we can reject okay so here's a now i'm going to ask you a check um you can also use that time to send me a few more questions if you want okay here's my check-in okay now instead of testing equivalents of finite of dfas i want to test the equivalence of regular expressions um to r1 r2 regular expressions are called the eq regular expressions problem equivalence of regular expressions can we now conclude that this problem is also decidable either immediately from stuff we've already shown or yes but would take some significant extra work or no because intersection is not a regular operation so let's see if i can pull that out here launch the polling um here we go okay i think we're just about done here um five oh five seconds more okay ready set end all right yes the ant's correct answer is a we have basically already shown this fact um because what why is that because we have shown that we can convert if you're given two regular expressions and we want to test whether they're equivalent or whether they generate the same language we can convert them both to to finite automata we can convert them to n phase and then the n phase to dfas and um then we can apply what we just showed about testing equivalence of dfas so yes it really follows immediately from stuff we've already shown the conversion number one and then the testing equivalents for dfas um so there's not really any work to do here and in fact what i'm trying to illustrate with this uh check-in that once you've shown um how to do some kind of a test for one model it have going to apply for all of the equivalent models that we've that we've shown to be equivalent because the turing machine can can do the constructions which show the equivalence um okay uh so let's move on um somebody didn't get the pole did most people not get the pole well i think most of you have gotten it did you i'm not seeing a lot of complaints so if you didn't get the poll something is wrong with your setup and you're going to have to take the recorded uh uh check-ins um that are going to launch right after this lecture is over sorry but we just you should figure out what's wrong with the setup there because i think most people are getting these okay and with that we're going to take a little break and then we'll continue on afterwards now let me know how this is is this slowing us too you know should i be speed a little speedier about the um the about the little mini breaks that we're taking or um is this good for you you know feedback is always i'm trying to make this as good as i can for for all of you okay i i think there's a mixture between people saying that these breaks are good some says they're a little a little over long so i'll try to tighten them up a little um some some some folks are saying what more people should be uh asking the tas i don't know how the tas are doing in terms of their i'll check with them afterward how well this is going for them in terms of uh questions but i think some amount of break is good um so that we don't uh so there's time to be asking asking questions um otherwise what's the point of having a live lecture so we will start promptly when this timer runs down um so uh be ready to go in 55 seconds from now just to confirm to show the decidability of the equivalence of two regular expressions do we need to show that we can use a turing machine to convert them to dfas first well i mean any procedure that you can give if you want to test whether two regular expressions are equivalent you can give any procedure for doing that deciding you want um i'm just offering one one simple way to do it that takes advantage of stuff we've already shown but if you want to dig down and do some analysis of those regular expressions to show that they describe the same language they generate the same language be my guest i think that's actually would be complicated to do that that way okay so we're just about um ready to go here uh so let's continue um uh and let me take this timer off here um all right now we are going to talk a little about context-free grammars so we talked about decision problems for finite automata let's talk about some for grammars um okay now i'm going to give us an analogous problem i'm going to give you a the except i'm calling it the acceptance problem but just just for consistency with everything else but it's really the generating problem so i'm giving you a grammar g a c a context free grammar g and a string that's in a string and i want to know is it in the language of g so does g generate w i'm calling that the a cfg problem so that's going to be a decidable again so now let's just think about how that these are getting slightly harder as we're moving along so i'm giving you a grammar and a string and i want to know does the grammar generate that string um well you know it's not totally trivial to solve that you know one thing you might try doing is looking at all possible things all possible derivations from that grammar and see if any one of them leads to w leads you to generate w well you have to be careful because uh as i as we've shown in some of our examples um you actually could have um because you're allowed to have variables that can get converted to empty string you might have very long intermediate strings being generated from a grammar which then ultimately give you a small string of terminals that get generated a small string in the language that gets produced and so it's not clear how you know you certainly don't want to try every possible derivation because there's infinitely many different derivations most of them generating of course things that are irrelevant to the string w but you have to know how to cut things off and it's not immediately obvious how you do that unless you have done the homework problems that i asked you to do which i'm sure very many of you have not done the zerobot point x problems because there that's going to come in handy right now and so i'll i'll help you through that um remember i'm hoping you should remember but you may not have looked at it uh something called chomsky normal form which is for context-free grammars but only allows the rules to be of a special kind and they have to look like this they can be a variable that goes to two variables on the right-hand side or a variable that goes to a terminal those are the only kinds of rules that you're allowed there's a special case for the empty string but let's ignore that you know for if you have you want to put the the start variable can also go to the empty string if you want to have a the empty string in the language but that's that's a special case let's ignore that these are the only two kinds of rules that you can have in the chomsky normal form gram once you have a chomsky normal form grammar um well first of all there's two things first of all you can always convert any any context free grammar into chomsky normal form so that's given in the textbook it's very it's sort of the you do the obvious thing i'm not going to spend time on it and you don't have to know it it's it's kind of just a little bit um tedious but it's it's straightforward and it's there if you if you're curious um but the second line is the thing that's important to us uh which is that if you have a grammar which in is in chomsky normal form and a string that's generated every derivation of that string has exactly two times the length of the string minus one steps it's actually if you think about it i mean this dilemma is very easy to prove um i think the homework problem asks you to prove that in the 0.2 or whatever it was um and problem set one or pub set two when i remember um because you can in order to do you know rules of this kind allow you to make the string one longer and rules of this kind allow you to produce an uh an x a new terminal symbol and so you're really going to end up having um you know if if the length of w is n you're going to have n minus one of these and n of those that's why you get two n minus one steps but the point is i don't really care exactly how many it's just that we have a bound and once you have that bound life is good from the point of view of um giving a turing machine which is going to decide this language because here's the turing machine what's going to do with the first thing is it's going to convert g into chomsky normal form that we assume we know how to do now we're going to try all derivations but only those of length 2 twice the length of the string minus 1. um because if any derivation is going to generate w it has to be this this many steps um now that we know the grammar is in chomsky normal form okay so we have converted this problem of one that might be a very unboundedly lengthy problem into one way you know it's going to terminate after some fixed number of steps and so therefore we can accept if any of those uh generate w and reject not okay and so oh so this is uh before moving on so this uh answers the problem shows that this language is decidable the the acfg language okay so that's something that you know make sure you understand um uh we're basically trying all derivations of up to of a certain length because that's all that's needed when the grammar is in that form now we're going to use that to prove a corollary which is um which is important for understanding how everything fits together and that corollary states that every context-free language is a decidable language every content-free language is decidable now why does that follow from this well um suppose you have a context-free language a we know that language is generated by some context free grammar g okay that's what it means so then you can take that grammar g and you can build it into a turing machine so there's going to be a turing machine which ha which is uh um constructed with the knowledge of g um and uh that turing machine is going to take its w and run the acfg algorithm um with w combined with the g that's already built into it so it's just going to stick that grammar g in front of w and now run the ac of g decide it's going to say does g generate w well that's going to be yes every time w is an a and so this machine here now uh is going to end up accepting every string that's in a because it's every string that g generates okay so um uh that is i think what i wanted to say about this now i feel that this corollary here throws a little bit of a curveball curveball um and um uh we can just pause for a moment here just to uh make sure you're understanding this so let me the tricky thing about this corollary is that i often get uh a question i often get where is you know when we start off with a context-free language who gives a w how do we get g because we need g to build the turing machine um m sub g so we know a is a context free language but how do we know what g is well the point is that uh we may not know what g is but we know g exists because that's a definition of a being a context-free language it must have a context-free grammar by definition and so because g exists i now know my turn machine m sub g exists and that's enough to know that a is decidable because it has a decider that exists now if you're not going to tell me the grammar for g i'm not going to tell you the turing machine which decides a but i know that but both of them exist so if you tell me g then i can tell you the turing machine okay so it's a subtle tricky issue there a certain little element maybe of sometimes people call it non-constructive because we're just in a sense showing just that something is existing but um you know it does the job for us because it shows that a is a decidable language okay so not so many questions here maybe the tas are getting them so let's move on um here's another check-in uh so now can we conclude that a instead of a cfg a pda is decidable hey people are getting this one pretty fast another 10 seconds three seconds okay i'm gonna shut it down um end polling share results yeah you know i uh so this uh problem here is apda is decidable it's really in a sense i was almost wondering whether or not to give this poll because it has it's true for the same reason as the as poll number one um because we know how to convert pdas or we stated we can convert pdas to cfgs and you know that conversion is given in the book we didn't do in lecture but i just stated you have to know that fact but not necessarily know how to do it that's okay but the fact is true um the conversion is in the book and it could be implemented on a turing machine so if you want to know whether a pda's language is empty you convert it to a context-free grammar and then use this procedure here to test whether it's a language not empty i'm sorry the acceptance problem i just want to not accept i just want to know does the pda accept some input i'm saying it wrong so i want to know does the pda accept some input i convert that pda to a grammar and then i see if the grammar generates that input because it's an equivalent grammar um so again this is using the fact that grammars and pdas are equivalent and interconvertible from one to another just like regular expressions in dfas from the previous poll so you need to be comfortable with that uh because we're going to not even talk about it anymore we're just going to be treating those things going back and forth between them without sometimes without even any comment okay so let's move on uh emptiness problem for cfgs hopefully you're getting comfortable with the know you know the terminology i'm using um so now the emptiness problem for context-free grammars i'm just going to give you a grammar and i want to know if its language is empty okay so remember we did this refined automata by testing a path can't you know we don't really have paths here you might think testing pairs and push down automata that's not going to really work because the stack is there so how are we going to do that test well there's something sort of like testing a path um just working with the grammar itself kind of working backwards from the terminals now um i'll kind of illustrate that here here's imagine i'm giving you here's a very simple grammar and i want to know does it generate any strings any strings of term obviously only care about strings of terminals because those are the things that are in the in the language so does this grammar generate any strings of terminals so what we're going to answer that is by a marking procedure but in a sense we're going to start from the um from the terminals and work backwards to see if we can get to the start variable so first we're going to mark all the terminal symbols and then we're going to mark anything that goes to a string of completely marked symbols either terminals or variables because anything that's marked we know can derive a string of terminals that's what it means anything that's that's blue can derive a string of just terminals and so now if you have a collection of those that are all marked they together can generate some you know string of terminals um together and so then you can mark the associated variable so like t goes to a so that may be over simple but we're going to mark t here everywhere in the grammar so all these t's are going to get marked because we know that t can generate a string of terminals now let's take a look at r r is going to a string of symbols that are all marked and that means those symbols can in turn generate the strings of terminals so we can mark r2 we can't get mark s because we don't know yet whether s has some unmarked thing that it goes to so we don't know yet whether s can generate a string of terminals but r we know right now so we're going to mark r but we then now that gets us to mark this r and so then we can go backwards and we can mark s and we keep doing that until there's nothing new to mark and here we've marked everything so clearly there's nothing more you can mark but you might stop before you've marked everything and then you see whether you've marked the stark variable or not and if you have you know the language is not empty okay so let's just go through this in in in in text for mark all occurrences of terminals in g then repeat until no new variables are marked we mark all occurrences of variables a if a goes to a string of symbols and all of those symbols were already marked because those are the things that already have been shown to generate a string of terminals and so now we're going to reject if the start status start variable is marked because that means that the language is not empty and except if it's not okay so um let me just just i'll take a couple of quick quick questions here okay somebody um somebody asked whether uh if i understand other regular languages also decidable well remember the regular languages are context-free languages and the context-free languages are decidable so yes the regular languages are decidable um okay i don't know some of those are going to be too long to answer and they're trying to come up with alternative solutions so i think we're going to move on um all right we talked about um just like we did for the final automata we talked about acceptance we talked about emptiness we talked about equivalence so how about the equivalence problem for context-free grammars i'm going to give you now two context-free grammars and i'd like to know are those two contexts free grammars generating the same language so how might you think about that well you know one thing uh you know following some of the ideas that we've already had um you could try feeding strings in to those grammars we know how to test whether those individual grammars can generate those strings so you can just try feeding strings in to g and to h and seeing whether those grammars ever disagree or whether they generate some string find some string that say g generates but h does not generate and we can test that string by string unfortunately we've got a lot of strings to test right that's gonna you know we we would need to give some bound if we were going to use that for a procedure not clear what the bound is um another idea um might be to use the uh closure construction uh that we had from before um so let's let's mull that over um whether that might work um but in fact um [Music] uh let me give away the answer here um this language is not decidable there's no algorithm out there no turing machine but therefore no algorithm which can take two grammars and test whether they generate the same language or not seems at first glance kind of surprising such simple things as context-free grammars can nonetheless be so complicated that there's no procedure out there which can tell whether the two gram those two grammars generate the same language we will show that um next week a related problem which is related to your homework this uh that's due on uh thursday uh is testing whether a grammar is ambiguous so given a grammar i'd like to know is is that grammar and ambiguous grammar or not i mean does it generate some string you know in the language of that grammar in two possibly different ways two or more different ways so is there some string um that can be generated you know with two different parsers um so um the problem of testing whether grammar is ambiguous not decidable so i'm asking you to do something hard something hard when you have to produce that grammar which is unambiguous for that language um in general testing whether grammar is ambiguous or not is not a decidable problem now hopefully that the grammar that you'll produce uh to show that um you know to get that unambiguous grammar for that language that i'm asking you to produce is not going to require our graders to solve some undecidable problem that it'll be clear based on the construction of that grammar why it's uh not ambiguous so you'll have to hopefully say some some explanation of what you're thinking there uh okay um so that's uh uh and we will prove that uh the problem of testing ambiguity is not decidable that's gonna be a homework problem in problem set three um okay uh let's check in here kind of something that i alluded to but didn't quite didn't want to give away uh why not use the same technique that we use to show equivalence of dfas is decidable to show that equivalence of context-free grammars is decidable obviously something goes wrong because eqcfg is not decidable why doesn't that same technique just work um well um what's the answer oh this is got a real race here another 15 seconds this one it gives you something to mull over um all right let's end it okay you're good to go at least click something um okay ending polling sharing results okay a bunch of you have thought well cfgs are generators and dfas are recognizers and that's the issue well not really because you know you can we could test um equivalence of regular expressions those are generators it's nothing to do with being a generator or not um um because you know we can convert regular expressions to dfas and then and then test equivalence for the dfa so that it's not really a matter of a matter of being generated that's not the issue the issue is that we can't follow the same construction that we did to show eqdfa as decidable because the context-free languages are not closed under those operations we needed to make that symmetric difference machine if you remember how that worked so then i closed under complementation and not closer intersection we needed both in order to build that machine c um which accepted all the strings that are sort of in the difference okay so let's continue on um let's move now to turing machines uh this is where stuff is really going to start to get interesting i'm not hoping it's been interesting all along but maybe even more interesting um so let's talk about the acceptance problem for turing machines atm this is going to be because this language is going to become an old friend we're just getting just getting to know it uh so this is the problem you're given a turing machine now in an input and i want to know does m accept that input that term does that turning machine accept its input um okay so that's going to not be a decidable problem either so it kind of shifted gears from a bunch of decidable things to a bunch of undecidable things so this is not a decidable language we will prove that um on thursday that's going to be the whole point of thursday's lecture is proving that atm is undecidable and that's going to be really a jumping off point for showing other problems are undecidable um so that's going to be our first uh proof of undecidability um but we do know that atm is recognizable and that's worth um understanding um for multiple reasons but for one thing it's going to give us an example of a problem which we know is recognizable but not decidable as we will as will prove undecidability but it's also um i think of of of historical importance um uh uh this this algorithm for showing it recognizable so let's go through that algorithm for it's very simple sort of doing the obvious thing um the following turing machine i'm calling you going to call it u for a reason that i'll make clear in a second that's going to be a recognizer for atm so it takes as input an m and a w and it's just going to simulate m w pretty much the way the um algorithm the the decider for adfa worked but now the machine instead of being a dfa it's a turing machine and the turing machine might go forever and so the simulation may not stop and that's the key difference which makes it from a decider into a recognizer so that simulation um so you're going to be simulating just keeping track of the tape of m the current state of m um where you know and proceeding to modify the tape as m modifies it and then if m enters an accept state then you know m is accepted as input so you also will enter and accept state so the machine u enters the accept state if m observes during the simulation that m enters an accept state uh furthermore if n m enters a reject state then u also enters a reject state okay now i want to say something beyond that which i i want you to pay attention to um which is the kind of thing i do see sometimes people saying uh we want you to reject if m never holds i mean that's also seems like what we want because if it never holds then m is rejecting by looping so you should also reject but i don't like that i don't like that line here step four of the turing machine because there's no way for a turing machine to determine or at least as we no obvious way uh and in fact it will not be any way but certainly at this point no obvious way for m to even tell for you to tell whether m is holding or not uh well certainly tell that is to say it never holds how can you make that determination so this is not if i saw this on a um on a solution either on an exam or a homework i would mark you off this is no good it's not a legal touring machine action if m does not hold on w um then you should reject that is correct and you can make that reasoning external to the algorithm of you but you is going to end up rejecting because it never holds either it never actually knows that m is rejecting w in that case if m is rejecting by looping it's just blindly going along and doing the simulation of m on w um and we'll end up holding rejecting by looping if m is rejecting by looping but that's something that you can argue if you need to make a proof or make some sort of reasoning about the machine but it's not part of the algorithm of the machine um okay now what what's i think from a historical standpoint it's interesting about this machine you and why i'm calling it you is because this appeared this al this machine you appeared in turing's original paper where he laid out turing machines he didn't call them touring machines by the way he called them computing machines people afterward called them turing machines uh but uh turing called this the universal computing machine that's his language actually i just looked at the paper yesterday um just to refresh my memory of it um and he gives the description of the operation of view in gory detail with all of the transitions and the states he nails the whole thing down takes pages and pages and pages so he gives it gives it there um so this is the original universal uh computing mission universal turing machine and this it's more than just an idle curiosity that disappeared in touring's paper because this actually turned out to be profoundly influential in the in computer science because it really was the first example of a machine that operated based on a stored program it really was a revolutionary idea in the in those days if you wanted to make a machine that did something different you had to wire the machine rewire the machine but here's a machine that operated based on instructions um and the instructions in a sense are no different than the data uh so this is what's been come to be known as the von neumann architecture but von neumann himself gave credit to turing machine for having inspired inspired him to think of this and some people argue that it's really calling it the binomial a bunch of people came up with this concept uh around the same time maybe other people too you know there's babbage and so on and others who um you know out of love lies also who came up with notions of programming but i think is a little different than um uh than this in concept but anyway uh this nevertheless played an important role in the history of the subject so with that i think we're at a time i'm going to um uh quickly review where we've been um so we should just show the decidability of various problems um these are all languages that we uh showed are decidable we showed that atm is touring recognizable and i think that was all we had for today um so um i will uh we will stop right here i will stick around take a few questions and our tas can take a few questions if you want by chat um and then i also have my office hours which will start in like five minutes or so once i get everything set up on my head um okay somebody wanted me to review this point here which i'm happy to do um uh you know this code here that i'm describing in english um needs to be something that you can implement on a turing machine um you know we're never going to go through the effort of building the transition functions and the states and so on but we need to be sure that we could if if we had to and how could you make a turing machine do the test that m doesn't hold that's something we don't know how to do i mean i can see if m holds enter i mean i can during the simulation i can see that m has entered the q reject state or the q accept state so i can tell while i'm simulating that it has halted but you know how would the machine or how would you know that m is now how would you you know someone says well do x if m never holds well how do you know him how how can you do the test that m is not holding um not there's no obvious way to do that in fact there is no way to do that but you know as it stands right now you know the proof would be on you to um to show how to implement that on a machine and there's just no obvious way to do that so i think that's why you should you should only put down things here which you're sure you could implement um you know even if it might take a long time so you don't have to worry about how long it would take but you have to put things down that you are sure you could at least implement in principle so someone has asked me about um the equivalence problem for context-free grammars uh being unsolvable uh why couldn't the machine be a human-level system of logic so that the computer could logically deduce whether or not it was decidable um well sort of that you know we're taking a turn into the subject of mathematical logic which is supposed to kind of uh formalize reasoning in a way um and uh in the end what it really is kind of come down to that there are certain uh grammars uh which are equivalent to one another but there's not going to be any way to prove that they're equivalent in any reasonable system um in equivalence you can always prove right you can exhibit a string that one you can show that this grammar is generating it this grammar is not generating this other one is not so uh in equivalence you can always prove but equivalence there are going to be certain pairs of grammars which are going to be beyond the capability of any reasonable formal system to be able to prove that they're equivalent because you can even convert them and make those grammars into something which talk about the system itself ultimately you're going to end up you know with a russell paradox kind of thing that's maybe going beyond more than you want to know and i'm happy to talk about it offline but you know there's just no way to make a turing machine which is going to implement sort of human reasoning and then get the right answer on all pairs of grammars just cannot be done bye bye i'm going to shut down the meeting though you 3 00:00:29,269 --> 00:00:30,550 4 00:00:30,550 --> 00:00:32,630 5 00:00:32,630 --> 00:00:35,510 6 00:00:35,510 --> 00:00:36,549 7 00:00:36,549 --> 00:00:38,389 8 00:00:38,389 --> 00:00:40,630 9 00:00:40,630 --> 00:00:42,830 10 00:00:42,830 --> 00:00:44,150 11 00:00:44,150 --> 00:00:46,790 12 00:00:46,790 --> 00:00:48,069 13 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--> 00:12:41,110 367 00:12:41,110 --> 00:12:43,110 368 00:12:43,110 --> 00:12:44,310 369 00:12:44,310 --> 00:12:45,990 370 00:12:45,990 --> 00:12:47,590 371 00:12:47,590 --> 00:12:49,190 372 00:12:49,190 --> 00:12:50,470 373 00:12:50,470 --> 00:12:52,150 374 00:12:52,150 --> 00:12:55,430 375 00:12:55,430 --> 00:12:56,710 376 00:12:56,710 --> 00:12:59,590 377 00:12:59,590 --> 00:13:00,949 378 00:13:00,949 --> 00:13:03,269 379 00:13:03,269 --> 00:13:06,069 380 00:13:06,069 --> 00:13:09,509 381 00:13:09,509 --> 00:13:12,150 382 00:13:12,150 --> 00:13:14,550 383 00:13:14,550 --> 00:13:18,230 384 00:13:18,230 --> 00:13:19,430 385 00:13:19,430 --> 00:13:21,430 386 00:13:21,430 --> 00:13:21,440 387 00:13:21,440 --> 00:13:22,389 388 00:13:22,389 --> 00:13:23,910 389 00:13:23,910 --> 00:13:26,949 390 00:13:26,949 --> 00:13:29,110 391 00:13:29,110 --> 00:13:30,949 392 00:13:30,949 --> 00:13:32,629 393 00:13:32,629 --> 00:13:34,949 394 00:13:34,949 --> 00:13:36,870 395 00:13:36,870 --> 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425 00:14:28,639 --> 00:14:31,189 426 00:14:31,189 --> 00:14:34,790 427 00:14:34,790 --> 00:14:34,800 428 00:14:34,800 --> 00:14:36,389 429 00:14:36,389 --> 00:14:38,470 430 00:14:38,470 --> 00:14:40,629 431 00:14:40,629 --> 00:14:43,030 432 00:14:43,030 --> 00:14:44,310 433 00:14:44,310 --> 00:14:45,750 434 00:14:45,750 --> 00:14:47,590 435 00:14:47,590 --> 00:14:50,470 436 00:14:50,470 --> 00:14:52,629 437 00:14:52,629 --> 00:14:54,150 438 00:14:54,150 --> 00:14:55,350 439 00:14:55,350 --> 00:14:57,189 440 00:14:57,189 --> 00:14:58,949 441 00:14:58,949 --> 00:15:00,790 442 00:15:00,790 --> 00:15:02,470 443 00:15:02,470 --> 00:15:02,480 444 00:15:02,480 --> 00:15:04,629 445 00:15:04,629 --> 00:15:04,639 446 00:15:04,639 --> 00:15:06,150 447 00:15:06,150 --> 00:15:08,150 448 00:15:08,150 --> 00:15:10,389 449 00:15:10,389 --> 00:15:12,790 450 00:15:12,790 --> 00:15:14,790 451 00:15:14,790 --> 00:15:17,030 452 00:15:17,030 --> 00:15:20,870 453 00:15:20,870 --> 00:15:23,829 454 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--> 00:16:26,069 484 00:16:26,069 --> 00:16:29,060 485 00:16:29,060 --> 00:16:29,070 486 00:16:29,070 --> 00:16:34,389 487 00:16:34,389 --> 00:16:35,670 488 00:16:35,670 --> 00:16:37,829 489 00:16:37,829 --> 00:16:39,350 490 00:16:39,350 --> 00:16:39,360 491 00:16:39,360 --> 00:16:41,350 492 00:16:41,350 --> 00:16:42,630 493 00:16:42,630 --> 00:16:42,640 494 00:16:42,640 --> 00:16:44,310 495 00:16:44,310 --> 00:16:47,030 496 00:16:47,030 --> 00:16:48,470 497 00:16:48,470 --> 00:16:50,230 498 00:16:50,230 --> 00:16:51,509 499 00:16:51,509 --> 00:16:51,519 500 00:16:51,519 --> 00:16:52,550 501 00:16:52,550 --> 00:16:55,269 502 00:16:55,269 --> 00:16:57,030 503 00:16:57,030 --> 00:16:58,389 504 00:16:58,389 --> 00:16:59,829 505 00:16:59,829 --> 00:16:59,839 506 00:16:59,839 --> 00:17:01,189 507 00:17:01,189 --> 00:17:03,910 508 00:17:03,910 --> 00:17:05,350 509 00:17:05,350 --> 00:17:05,360 510 00:17:05,360 --> 00:17:07,669 511 00:17:07,669 --> 00:17:07,679 512 00:17:07,679 --> 00:17:09,110 513 00:17:09,110 --> 00:17:11,909 514 00:17:11,909 --> 00:17:13,669 515 00:17:13,669 --> 00:17:15,590 516 00:17:15,590 --> 00:17:17,429 517 00:17:17,429 --> 00:17:19,110 518 00:17:19,110 --> 00:17:20,870 519 00:17:20,870 --> 00:17:23,909 520 00:17:23,909 --> 00:17:25,350 521 00:17:25,350 --> 00:17:25,360 522 00:17:25,360 --> 00:17:26,230 523 00:17:26,230 --> 00:17:26,240 524 00:17:26,240 --> 00:17:27,270 525 00:17:27,270 --> 00:17:29,669 526 00:17:29,669 --> 00:17:29,679 527 00:17:29,679 --> 00:17:30,630 528 00:17:30,630 --> 00:17:32,870 529 00:17:32,870 --> 00:17:34,950 530 00:17:34,950 --> 00:17:36,150 531 00:17:36,150 --> 00:17:38,470 532 00:17:38,470 --> 00:17:41,110 533 00:17:41,110 --> 00:17:44,950 534 00:17:44,950 --> 00:17:46,789 535 00:17:46,789 --> 00:17:48,789 536 00:17:48,789 --> 00:17:50,549 537 00:17:50,549 --> 00:17:52,390 538 00:17:52,390 --> 00:17:55,350 539 00:17:55,350 --> 00:17:56,789 540 00:17:56,789 --> 00:17:59,110 541 00:17:59,110 --> 00:18:01,430 542 00:18:01,430 --> 00:18:03,510 543 00:18:03,510 --> 00:18:06,789 544 00:18:06,789 --> 00:18:09,029 545 00:18:09,029 --> 00:18:11,430 546 00:18:11,430 --> 00:18:13,190 547 00:18:13,190 --> 00:18:15,590 548 00:18:15,590 --> 00:18:15,600 549 00:18:15,600 --> 00:18:16,470 550 00:18:16,470 --> 00:18:16,480 551 00:18:16,480 --> 00:18:17,430 552 00:18:17,430 --> 00:18:21,510 553 00:18:21,510 --> 00:18:23,430 554 00:18:23,430 --> 00:18:25,350 555 00:18:25,350 --> 00:18:27,590 556 00:18:27,590 --> 00:18:31,510 557 00:18:31,510 --> 00:18:31,520 558 00:18:31,520 --> 00:18:33,430 559 00:18:33,430 --> 00:18:34,950 560 00:18:34,950 --> 00:18:37,029 561 00:18:37,029 --> 00:18:37,039 562 00:18:37,039 --> 00:18:37,830 563 00:18:37,830 --> 00:18:40,150 564 00:18:40,150 --> 00:18:41,830 565 00:18:41,830 --> 00:18:44,549 566 00:18:44,549 --> 00:18:47,430 567 00:18:47,430 --> 00:18:48,549 568 00:18:48,549 --> 00:18:50,390 569 00:18:50,390 --> 00:18:54,070 570 00:18:54,070 --> 00:18:58,150 571 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--> 00:19:53,190 601 00:19:53,190 --> 00:19:55,270 602 00:19:55,270 --> 00:19:56,630 603 00:19:56,630 --> 00:19:58,470 604 00:19:58,470 --> 00:19:59,750 605 00:19:59,750 --> 00:20:01,830 606 00:20:01,830 --> 00:20:04,149 607 00:20:04,149 --> 00:20:07,590 608 00:20:07,590 --> 00:20:09,350 609 00:20:09,350 --> 00:20:11,510 610 00:20:11,510 --> 00:20:15,110 611 00:20:15,110 --> 00:20:17,430 612 00:20:17,430 --> 00:20:18,710 613 00:20:18,710 --> 00:20:20,710 614 00:20:20,710 --> 00:20:22,630 615 00:20:22,630 --> 00:20:24,549 616 00:20:24,549 --> 00:20:27,190 617 00:20:27,190 --> 00:20:27,200 618 00:20:27,200 --> 00:20:29,990 619 00:20:29,990 --> 00:20:32,230 620 00:20:32,230 --> 00:20:32,240 621 00:20:32,240 --> 00:20:33,350 622 00:20:33,350 --> 00:20:35,669 623 00:20:35,669 --> 00:20:39,669 624 00:20:39,669 --> 00:20:41,750 625 00:20:41,750 --> 00:20:45,590 626 00:20:45,590 --> 00:20:49,669 627 00:20:49,669 --> 00:20:51,830 628 00:20:51,830 --> 00:20:53,270 629 00:20:53,270 --> 00:20:55,430 630 00:20:55,430 --> 00:20:57,750 631 00:20:57,750 --> 00:21:00,830 632 00:21:00,830 --> 00:21:00,840 633 00:21:00,840 --> 00:21:02,549 634 00:21:02,549 --> 00:21:04,149 635 00:21:04,149 --> 00:21:05,669 636 00:21:05,669 --> 00:21:05,679 637 00:21:05,679 --> 00:21:06,200 638 00:21:06,200 --> 00:21:06,210 639 00:21:06,210 --> 00:21:07,750 640 00:21:07,750 --> 00:21:07,760 641 00:21:07,760 --> 00:21:10,950 642 00:21:10,950 --> 00:21:10,960 643 00:21:10,960 --> 00:21:11,750 644 00:21:11,750 --> 00:21:14,549 645 00:21:14,549 --> 00:21:16,549 646 00:21:16,549 --> 00:21:18,549 647 00:21:18,549 --> 00:21:19,750 648 00:21:19,750 --> 00:21:21,750 649 00:21:21,750 --> 00:21:24,950 650 00:21:24,950 --> 00:21:24,960 651 00:21:24,960 --> 00:21:25,990 652 00:21:25,990 --> 00:21:28,230 653 00:21:28,230 --> 00:21:29,750 654 00:21:29,750 --> 00:21:31,430 655 00:21:31,430 --> 00:21:33,110 656 00:21:33,110 --> 00:21:37,669 657 00:21:37,669 --> 00:21:37,679 658 00:21:37,679 --> 00:21:38,549 659 00:21:38,549 --> 00:21:40,710 660 00:21:40,710 --> 00:21:44,149 661 00:21:44,149 --> 00:21:48,710 662 00:21:48,710 --> 00:21:51,029 663 00:21:51,029 --> 00:21:58,870 664 00:21:58,870 --> 00:21:58,880 665 00:21:58,880 --> 00:22:01,350 666 00:22:01,350 --> 00:22:03,190 667 00:22:03,190 --> 00:22:04,789 668 00:22:04,789 --> 00:22:07,350 669 00:22:07,350 --> 00:22:07,360 670 00:22:07,360 --> 00:22:09,909 671 00:22:09,909 --> 00:22:11,750 672 00:22:11,750 --> 00:22:13,350 673 00:22:13,350 --> 00:22:16,070 674 00:22:16,070 --> 00:22:17,990 675 00:22:17,990 --> 00:22:19,830 676 00:22:19,830 --> 00:22:21,750 677 00:22:21,750 --> 00:22:23,750 678 00:22:23,750 --> 00:22:26,390 679 00:22:26,390 --> 00:22:28,630 680 00:22:28,630 --> 00:22:28,640 681 00:22:28,640 --> 00:22:29,750 682 00:22:29,750 --> 00:22:31,909 683 00:22:31,909 --> 00:22:38,830 684 00:22:38,830 --> 00:22:44,230 685 00:22:44,230 --> 00:22:46,630 686 00:22:46,630 --> 00:22:48,789 687 00:22:48,789 --> 00:22:48,799 688 00:22:48,799 --> 00:22:49,590 689 00:22:49,590 --> 00:22:50,789 690 00:22:50,789 --> 00:22:53,350 691 00:22:53,350 --> 00:22:55,270 692 00:22:55,270 --> 00:22:56,630 693 00:22:56,630 --> 00:22:59,190 694 00:22:59,190 --> 00:23:00,950 695 00:23:00,950 --> 00:23:05,350 696 00:23:05,350 --> 00:23:07,110 697 00:23:07,110 --> 00:23:08,630 698 00:23:08,630 --> 00:23:11,110 699 00:23:11,110 --> 00:23:13,590 700 00:23:13,590 --> 00:23:16,230 701 00:23:16,230 --> 00:23:17,909 702 00:23:17,909 --> 00:23:19,430 703 00:23:19,430 --> 00:23:19,440 704 00:23:19,440 --> 00:23:20,710 705 00:23:20,710 --> 00:23:22,470 706 00:23:22,470 --> 00:23:24,070 707 00:23:24,070 --> 00:23:25,750 708 00:23:25,750 --> 00:23:27,510 709 00:23:27,510 --> 00:23:30,310 710 00:23:30,310 --> 00:23:32,870 711 00:23:32,870 --> 00:23:35,669 712 00:23:35,669 --> 00:23:35,679 713 00:23:35,679 --> 00:23:36,390 714 00:23:36,390 --> 00:23:38,390 715 00:23:38,390 --> 00:23:45,669 716 00:23:45,669 --> 00:23:47,350 717 00:23:47,350 --> 00:23:49,830 718 00:23:49,830 --> 00:23:51,669 719 00:23:51,669 --> 00:23:53,430 720 00:23:53,430 --> 00:23:55,430 721 00:23:55,430 --> 00:23:58,390 722 00:23:58,390 --> 00:24:01,750 723 00:24:01,750 --> 00:24:01,760 724 00:24:01,760 --> 00:24:02,789 725 00:24:02,789 --> 00:24:04,390 726 00:24:04,390 --> 00:24:07,029 727 00:24:07,029 --> 00:24:08,950 728 00:24:08,950 --> 00:24:12,789 729 00:24:12,789 --> 00:24:15,029 730 00:24:15,029 --> 00:24:16,549 731 00:24:16,549 --> 00:24:18,549 732 00:24:18,549 --> 00:24:19,990 733 00:24:19,990 --> 00:24:20,000 734 00:24:20,000 --> 00:24:21,190 735 00:24:21,190 --> 00:24:24,390 736 00:24:24,390 --> 00:24:27,110 737 00:24:27,110 --> 00:24:27,120 738 00:24:27,120 --> 00:24:27,909 739 00:24:27,909 --> 00:24:30,310 740 00:24:30,310 --> 00:24:33,029 741 00:24:33,029 --> 00:24:35,110 742 00:24:35,110 --> 00:24:35,120 743 00:24:35,120 --> 00:24:36,630 744 00:24:36,630 --> 00:24:38,470 745 00:24:38,470 --> 00:24:39,909 746 00:24:39,909 --> 00:24:42,310 747 00:24:42,310 --> 00:24:45,990 748 00:24:45,990 --> 00:24:47,590 749 00:24:47,590 --> 00:24:50,310 750 00:24:50,310 --> 00:24:52,310 751 00:24:52,310 --> 00:24:55,269 752 00:24:55,269 --> 00:24:57,830 753 00:24:57,830 --> 00:24:59,590 754 00:24:59,590 --> 00:24:59,600 755 00:24:59,600 --> 00:25:01,110 756 00:25:01,110 --> 00:25:03,909 757 00:25:03,909 --> 00:25:06,630 758 00:25:06,630 --> 00:25:08,870 759 00:25:08,870 --> 00:25:11,430 760 00:25:11,430 --> 00:25:14,310 761 00:25:14,310 --> 00:25:16,950 762 00:25:16,950 --> 00:25:19,190 763 00:25:19,190 --> 00:25:20,950 764 00:25:20,950 --> 00:25:20,960 765 00:25:20,960 --> 00:25:22,230 766 00:25:22,230 --> 00:25:23,990 767 00:25:23,990 --> 00:25:25,350 768 00:25:25,350 --> 00:25:26,710 769 00:25:26,710 --> 00:25:28,549 770 00:25:28,549 --> 00:25:28,559 771 00:25:28,559 --> 00:25:29,750 772 00:25:29,750 --> 00:25:31,669 773 00:25:31,669 --> 00:25:33,430 774 00:25:33,430 --> 00:25:36,310 775 00:25:36,310 --> 00:25:38,230 776 00:25:38,230 --> 00:25:39,510 777 00:25:39,510 --> 00:25:41,350 778 00:25:41,350 --> 00:25:43,750 779 00:25:43,750 --> 00:25:43,760 780 00:25:43,760 --> 00:25:44,789 781 00:25:44,789 --> 00:25:47,110 782 00:25:47,110 --> 00:25:49,350 783 00:25:49,350 --> 00:25:51,510 784 00:25:51,510 --> 00:25:53,990 785 00:25:53,990 --> 00:25:56,789 786 00:25:56,789 --> 00:25:58,230 787 00:25:58,230 --> 00:26:00,549 788 00:26:00,549 --> 00:26:03,269 789 00:26:03,269 --> 00:26:05,990 790 00:26:05,990 --> 00:26:07,110 791 00:26:07,110 --> 00:26:08,710 792 00:26:08,710 --> 00:26:11,269 793 00:26:11,269 --> 00:26:12,710 794 00:26:12,710 --> 00:26:13,990 795 00:26:13,990 --> 00:26:16,149 796 00:26:16,149 --> 00:26:17,830 797 00:26:17,830 --> 00:26:17,840 798 00:26:17,840 --> 00:26:18,950 799 00:26:18,950 --> 00:26:22,470 800 00:26:22,470 --> 00:26:25,430 801 00:26:25,430 --> 00:26:27,350 802 00:26:27,350 --> 00:26:29,350 803 00:26:29,350 --> 00:26:31,269 804 00:26:31,269 --> 00:26:33,590 805 00:26:33,590 --> 00:26:35,430 806 00:26:35,430 --> 00:26:35,440 807 00:26:35,440 --> 00:26:36,310 808 00:26:36,310 --> 00:26:37,830 809 00:26:37,830 --> 00:26:39,110 810 00:26:39,110 --> 00:26:41,669 811 00:26:41,669 --> 00:26:45,269 812 00:26:45,269 --> 00:26:46,789 813 00:26:46,789 --> 00:26:49,590 814 00:26:49,590 --> 00:26:49,600 815 00:26:49,600 --> 00:26:51,510 816 00:26:51,510 --> 00:26:54,310 817 00:26:54,310 --> 00:26:56,230 818 00:26:56,230 --> 00:26:57,430 819 00:26:57,430 --> 00:26:57,440 820 00:26:57,440 --> 00:26:59,110 821 00:26:59,110 --> 00:27:01,350 822 00:27:01,350 --> 00:27:01,360 823 00:27:01,360 --> 00:27:02,070 824 00:27:02,070 --> 00:27:04,870 825 00:27:04,870 --> 00:27:07,269 826 00:27:07,269 --> 00:27:08,710 827 00:27:08,710 --> 00:27:11,430 828 00:27:11,430 --> 00:27:13,510 829 00:27:13,510 --> 00:27:15,510 830 00:27:15,510 --> 00:27:17,510 831 00:27:17,510 --> 00:27:19,110 832 00:27:19,110 --> 00:27:21,750 833 00:27:21,750 --> 00:27:23,110 834 00:27:23,110 --> 00:27:24,470 835 00:27:24,470 --> 00:27:26,710 836 00:27:26,710 --> 00:27:29,510 837 00:27:29,510 --> 00:27:31,269 838 00:27:31,269 --> 00:27:31,279 839 00:27:31,279 --> 00:27:34,630 840 00:27:34,630 --> 00:27:36,549 841 00:27:36,549 --> 00:27:37,830 842 00:27:37,830 --> 00:27:37,840 843 00:27:37,840 --> 00:27:38,950 844 00:27:38,950 --> 00:27:41,590 845 00:27:41,590 --> 00:27:44,710 846 00:27:44,710 --> 00:27:45,909 847 00:27:45,909 --> 00:27:45,919 848 00:27:45,919 --> 00:27:46,950 849 00:27:46,950 --> 00:27:48,310 850 00:27:48,310 --> 00:27:49,669 851 00:27:49,669 --> 00:27:52,549 852 00:27:52,549 --> 00:27:54,549 853 00:27:54,549 --> 00:27:57,669 854 00:27:57,669 --> 00:28:00,310 855 00:28:00,310 --> 00:28:01,830 856 00:28:01,830 --> 00:28:03,269 857 00:28:03,269 --> 00:28:05,269 858 00:28:05,269 --> 00:28:07,430 859 00:28:07,430 --> 00:28:09,029 860 00:28:09,029 --> 00:28:10,870 861 00:28:10,870 --> 00:28:13,190 862 00:28:13,190 --> 00:28:15,750 863 00:28:15,750 --> 00:28:17,669 864 00:28:17,669 --> 00:28:19,669 865 00:28:19,669 --> 00:28:22,710 866 00:28:22,710 --> 00:28:24,310 867 00:28:24,310 --> 00:28:26,549 868 00:28:26,549 --> 00:28:27,909 869 00:28:27,909 --> 00:28:30,149 870 00:28:30,149 --> 00:28:32,149 871 00:28:32,149 --> 00:28:32,159 872 00:28:32,159 --> 00:28:33,430 873 00:28:33,430 --> 00:28:33,440 874 00:28:33,440 --> 00:28:34,870 875 00:28:34,870 --> 00:28:36,230 876 00:28:36,230 --> 00:28:37,350 877 00:28:37,350 --> 00:28:37,360 878 00:28:37,360 --> 00:28:38,470 879 00:28:38,470 --> 00:28:40,389 880 00:28:40,389 --> 00:28:42,870 881 00:28:42,870 --> 00:28:44,070 882 00:28:44,070 --> 00:28:46,549 883 00:28:46,549 --> 00:28:46,559 884 00:28:46,559 --> 00:28:47,430 885 00:28:47,430 --> 00:28:49,750 886 00:28:49,750 --> 00:28:51,350 887 00:28:51,350 --> 00:28:53,990 888 00:28:53,990 --> 00:28:55,990 889 00:28:55,990 --> 00:28:57,909 890 00:28:57,909 --> 00:29:00,549 891 00:29:00,549 --> 00:29:01,990 892 00:29:01,990 --> 00:29:03,430 893 00:29:03,430 --> 00:29:05,590 894 00:29:05,590 --> 00:29:07,110 895 00:29:07,110 --> 00:29:07,120 896 00:29:07,120 --> 00:29:08,789 897 00:29:08,789 --> 00:29:08,799 898 00:29:08,799 --> 00:29:11,190 899 00:29:11,190 --> 00:29:13,830 900 00:29:13,830 --> 00:29:17,350 901 00:29:17,350 --> 00:29:19,909 902 00:29:19,909 --> 00:29:20,789 903 00:29:20,789 --> 00:29:22,549 904 00:29:22,549 --> 00:29:25,909 905 00:29:25,909 --> 00:29:28,549 906 00:29:28,549 --> 00:29:30,070 907 00:29:30,070 --> 00:29:32,070 908 00:29:32,070 --> 00:29:33,909 909 00:29:33,909 --> 00:29:36,389 910 00:29:36,389 --> 00:29:38,710 911 00:29:38,710 --> 00:29:41,669 912 00:29:41,669 --> 00:29:43,669 913 00:29:43,669 --> 00:29:45,190 914 00:29:45,190 --> 00:29:48,230 915 00:29:48,230 --> 00:29:49,510 916 00:29:49,510 --> 00:29:51,990 917 00:29:51,990 --> 00:29:54,070 918 00:29:54,070 --> 00:29:55,510 919 00:29:55,510 --> 00:29:58,310 920 00:29:58,310 --> 00:30:00,389 921 00:30:00,389 --> 00:30:02,470 922 00:30:02,470 --> 00:30:05,269 923 00:30:05,269 --> 00:30:09,110 924 00:30:09,110 --> 00:30:10,789 925 00:30:10,789 --> 00:30:12,870 926 00:30:12,870 --> 00:30:14,710 927 00:30:14,710 --> 00:30:17,110 928 00:30:17,110 --> 00:30:18,149 929 00:30:18,149 --> 00:30:20,310 930 00:30:20,310 --> 00:30:21,590 931 00:30:21,590 --> 00:30:22,870 932 00:30:22,870 --> 00:30:24,789 933 00:30:24,789 --> 00:30:26,310 934 00:30:26,310 --> 00:30:26,320 935 00:30:26,320 --> 00:30:27,510 936 00:30:27,510 --> 00:30:27,520 937 00:30:27,520 --> 00:30:30,149 938 00:30:30,149 --> 00:30:30,159 939 00:30:30,159 --> 00:30:32,389 940 00:30:32,389 --> 00:30:37,190 941 00:30:37,190 --> 00:30:37,200 942 00:30:37,200 --> 00:30:39,029 943 00:30:39,029 --> 00:30:39,039 944 00:30:39,039 --> 00:30:39,990 945 00:30:39,990 --> 00:30:41,590 946 00:30:41,590 --> 00:30:45,110 947 00:30:45,110 --> 00:30:47,190 948 00:30:47,190 --> 00:30:47,200 949 00:30:47,200 --> 00:30:48,789 950 00:30:48,789 --> 00:30:50,470 951 00:30:50,470 --> 00:30:53,909 952 00:30:53,909 --> 00:30:58,950 953 00:30:58,950 --> 00:31:00,789 954 00:31:00,789 --> 00:31:02,549 955 00:31:02,549 --> 00:31:04,950 956 00:31:04,950 --> 00:31:05,860 957 00:31:05,860 --> 00:31:05,870 958 00:31:05,870 --> 00:31:07,750 959 00:31:07,750 --> 00:31:09,350 960 00:31:09,350 --> 00:31:12,310 961 00:31:12,310 --> 00:31:13,669 962 00:31:13,669 --> 00:31:13,679 963 00:31:13,679 --> 00:31:14,090 964 00:31:14,090 --> 00:31:14,100 965 00:31:14,100 --> 00:31:15,990 966 00:31:15,990 --> 00:31:19,350 967 00:31:19,350 --> 00:31:21,269 968 00:31:21,269 --> 00:31:22,870 969 00:31:22,870 --> 00:31:24,230 970 00:31:24,230 --> 00:31:26,950 971 00:31:26,950 --> 00:31:26,960 972 00:31:26,960 --> 00:31:28,310 973 00:31:28,310 --> 00:31:28,320 974 00:31:28,320 --> 00:31:29,350 975 00:31:29,350 --> 00:31:31,029 976 00:31:31,029 --> 00:31:32,789 977 00:31:32,789 --> 00:31:35,110 978 00:31:35,110 --> 00:31:37,269 979 00:31:37,269 --> 00:31:39,830 980 00:31:39,830 --> 00:31:39,840 981 00:31:39,840 --> 00:31:41,110 982 00:31:41,110 --> 00:31:44,149 983 00:31:44,149 --> 00:31:45,990 984 00:31:45,990 --> 00:31:48,070 985 00:31:48,070 --> 00:31:49,669 986 00:31:49,669 --> 00:31:53,029 987 00:31:53,029 --> 00:31:54,950 988 00:31:54,950 --> 00:31:57,110 989 00:31:57,110 --> 00:31:59,269 990 00:31:59,269 --> 00:32:02,470 991 00:32:02,470 --> 00:32:04,149 992 00:32:04,149 --> 00:32:05,990 993 00:32:05,990 --> 00:32:06,000 994 00:32:06,000 --> 00:32:06,870 995 00:32:06,870 --> 00:32:08,630 996 00:32:08,630 --> 00:32:10,789 997 00:32:10,789 --> 00:32:10,799 998 00:32:10,799 --> 00:32:12,630 999 00:32:12,630 --> 00:32:14,549 1000 00:32:14,549 --> 00:32:16,549 1001 00:32:16,549 --> 00:32:16,559 1002 00:32:16,559 --> 00:32:18,070 1003 00:32:18,070 --> 00:32:19,909 1004 00:32:19,909 --> 00:32:23,350 1005 00:32:23,350 --> 00:32:25,509 1006 00:32:25,509 --> 00:32:27,509 1007 00:32:27,509 --> 00:32:29,190 1008 00:32:29,190 --> 00:32:30,789 1009 00:32:30,789 --> 00:32:33,350 1010 00:32:33,350 --> 00:32:37,750 1011 00:32:37,750 --> 00:32:39,750 1012 00:32:39,750 --> 00:32:42,149 1013 00:32:42,149 --> 00:32:43,909 1014 00:32:43,909 --> 00:32:46,549 1015 00:32:46,549 --> 00:32:48,389 1016 00:32:48,389 --> 00:32:49,590 1017 00:32:49,590 --> 00:32:51,110 1018 00:32:51,110 --> 00:32:53,029 1019 00:32:53,029 --> 00:32:54,870 1020 00:32:54,870 --> 00:32:56,630 1021 00:32:56,630 --> 00:32:58,310 1022 00:32:58,310 --> 00:32:59,750 1023 00:32:59,750 --> 00:33:01,669 1024 00:33:01,669 --> 00:33:03,269 1025 00:33:03,269 --> 00:33:05,269 1026 00:33:05,269 --> 00:33:06,710 1027 00:33:06,710 --> 00:33:06,720 1028 00:33:06,720 --> 00:33:08,549 1029 00:33:08,549 --> 00:33:10,310 1030 00:33:10,310 --> 00:33:11,509 1031 00:33:11,509 --> 00:33:12,789 1032 00:33:12,789 --> 00:33:14,389 1033 00:33:14,389 --> 00:33:16,070 1034 00:33:16,070 --> 00:33:17,990 1035 00:33:17,990 --> 00:33:21,029 1036 00:33:21,029 --> 00:33:23,750 1037 00:33:23,750 --> 00:33:26,310 1038 00:33:26,310 --> 00:33:29,990 1039 00:33:29,990 --> 00:33:31,509 1040 00:33:31,509 --> 00:33:34,310 1041 00:33:34,310 --> 00:33:36,310 1042 00:33:36,310 --> 00:33:39,269 1043 00:33:39,269 --> 00:33:40,630 1044 00:33:40,630 --> 00:33:42,310 1045 00:33:42,310 --> 00:33:45,509 1046 00:33:45,509 --> 00:33:48,070 1047 00:33:48,070 --> 00:33:50,070 1048 00:33:50,070 --> 00:33:52,470 1049 00:33:52,470 --> 00:33:55,269 1050 00:33:55,269 --> 00:33:57,509 1051 00:33:57,509 --> 00:33:59,750 1052 00:33:59,750 --> 00:33:59,760 1053 00:33:59,760 --> 00:34:01,269 1054 00:34:01,269 --> 00:34:03,350 1055 00:34:03,350 --> 00:34:04,950 1056 00:34:04,950 --> 00:34:07,669 1057 00:34:07,669 --> 00:34:07,679 1058 00:34:07,679 --> 00:34:08,389 1059 00:34:08,389 --> 00:34:10,470 1060 00:34:10,470 --> 00:34:15,750 1061 00:34:15,750 --> 00:34:18,470 1062 00:34:18,470 --> 00:34:18,480 1063 00:34:18,480 --> 00:34:20,230 1064 00:34:20,230 --> 00:34:22,310 1065 00:34:22,310 --> 00:34:26,470 1066 00:34:26,470 --> 00:34:28,950 1067 00:34:28,950 --> 00:34:30,710 1068 00:34:30,710 --> 00:34:33,349 1069 00:34:33,349 --> 00:34:38,629 1070 00:34:38,629 --> 00:34:39,829 1071 00:34:39,829 --> 00:34:42,790 1072 00:34:42,790 --> 00:34:44,950 1073 00:34:44,950 --> 00:34:46,230 1074 00:34:46,230 --> 00:34:47,589 1075 00:34:47,589 --> 00:34:49,190 1076 00:34:49,190 --> 00:34:51,990 1077 00:34:51,990 --> 00:34:53,990 1078 00:34:53,990 --> 00:34:55,990 1079 00:34:55,990 --> 00:34:58,230 1080 00:34:58,230 --> 00:35:02,470 1081 00:35:02,470 --> 00:35:04,310 1082 00:35:04,310 --> 00:35:05,910 1083 00:35:05,910 --> 00:35:09,270 1084 00:35:09,270 --> 00:35:12,550 1085 00:35:12,550 --> 00:35:13,589 1086 00:35:13,589 --> 00:35:16,430 1087 00:35:16,430 --> 00:35:19,510 1088 00:35:19,510 --> 00:35:23,510 1089 00:35:23,510 --> 00:35:25,030 1090 00:35:25,030 --> 00:35:27,990 1091 00:35:27,990 --> 00:35:29,510 1092 00:35:29,510 --> 00:35:32,630 1093 00:35:32,630 --> 00:35:35,430 1094 00:35:35,430 --> 00:35:35,440 1095 00:35:35,440 --> 00:35:36,230 1096 00:35:36,230 --> 00:35:37,829 1097 00:35:37,829 --> 00:35:37,839 1098 00:35:37,839 --> 00:35:39,430 1099 00:35:39,430 --> 00:35:42,310 1100 00:35:42,310 --> 00:35:44,470 1101 00:35:44,470 --> 00:35:46,390 1102 00:35:46,390 --> 00:35:48,470 1103 00:35:48,470 --> 00:35:51,270 1104 00:35:51,270 --> 00:35:55,109 1105 00:35:55,109 --> 00:35:55,119 1106 00:35:55,119 --> 00:35:56,150 1107 00:35:56,150 --> 00:35:56,160 1108 00:35:56,160 --> 00:35:57,109 1109 00:35:57,109 --> 00:35:59,349 1110 00:35:59,349 --> 00:36:01,589 1111 00:36:01,589 --> 00:36:03,349 1112 00:36:03,349 --> 00:36:05,349 1113 00:36:05,349 --> 00:36:07,670 1114 00:36:07,670 --> 00:36:07,680 1115 00:36:07,680 --> 00:36:08,710 1116 00:36:08,710 --> 00:36:10,069 1117 00:36:10,069 --> 00:36:12,630 1118 00:36:12,630 --> 00:36:15,030 1119 00:36:15,030 --> 00:36:16,790 1120 00:36:16,790 --> 00:36:18,630 1121 00:36:18,630 --> 00:36:20,550 1122 00:36:20,550 --> 00:36:20,560 1123 00:36:20,560 --> 00:36:22,630 1124 00:36:22,630 --> 00:36:26,310 1125 00:36:26,310 --> 00:36:31,190 1126 00:36:31,190 --> 00:36:31,200 1127 00:36:31,200 --> 00:36:32,230 1128 00:36:32,230 --> 00:36:36,710 1129 00:36:36,710 --> 00:36:42,630 1130 00:36:42,630 --> 00:36:47,430 1131 00:36:47,430 --> 00:36:49,750 1132 00:36:49,750 --> 00:36:52,230 1133 00:36:52,230 --> 00:36:52,240 1134 00:36:52,240 --> 00:36:53,270 1135 00:36:53,270 --> 00:36:53,280 1136 00:36:53,280 --> 00:36:54,470 1137 00:36:54,470 --> 00:36:56,950 1138 00:36:56,950 --> 00:36:59,829 1139 00:36:59,829 --> 00:37:01,670 1140 00:37:01,670 --> 00:37:03,510 1141 00:37:03,510 --> 00:37:04,790 1142 00:37:04,790 --> 00:37:09,030 1143 00:37:09,030 --> 00:37:10,710 1144 00:37:10,710 --> 00:37:12,630 1145 00:37:12,630 --> 00:37:15,589 1146 00:37:15,589 --> 00:37:18,630 1147 00:37:18,630 --> 00:37:18,640 1148 00:37:18,640 --> 00:37:19,670 1149 00:37:19,670 --> 00:37:22,630 1150 00:37:22,630 --> 00:37:24,470 1151 00:37:24,470 --> 00:37:25,829 1152 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01:13:54,400 --> 01:13:55,189 2293 01:13:55,189 --> 01:13:56,870 2294 01:13:56,870 --> 01:13:58,950 2295 01:13:58,950 --> 01:14:01,270 2296 01:14:01,270 --> 01:14:03,350 2297 01:14:03,350 --> 01:14:04,709 2298 01:14:04,709 --> 01:14:06,390 2299 01:14:06,390 --> 01:14:06,400 2300 01:14:06,400 --> 01:14:08,390 2301 01:14:08,390 --> 01:14:08,400 2302 01:14:08,400 --> 01:14:09,430 2303 01:14:09,430 --> 01:14:11,110 2304 01:14:11,110 --> 01:14:12,950 2305 01:14:12,950 --> 01:14:14,870 2306 01:14:14,870 --> 01:14:16,630 2307 01:14:16,630 --> 01:14:18,229 2308 01:14:18,229 --> 01:14:20,790 2309 01:14:20,790 --> 01:14:23,030 2310 01:14:23,030 --> 01:14:24,950 2311 01:14:24,950 --> 01:14:26,790 2312 01:14:26,790 --> 01:14:28,229 2313 01:14:28,229 --> 01:14:28,239 2314 01:14:28,239 --> 01:14:29,350 2315 01:14:29,350 --> 01:14:29,360 2316 01:14:29,360 --> 01:14:30,310 2317 01:14:30,310 --> 01:14:31,669 2318 01:14:31,669 --> 01:14:33,990 2319 01:14:33,990 --> 01:14:36,390 2320 01:14:36,390 --> 01:14:37,910 2321 01:14:37,910 --> 01:14:42,390 2322 01:14:42,390 --> 01:14:47,270 2323 01:14:47,270 --> 01:14:48,870 2324 01:14:48,870 --> 01:14:51,990 2325 01:14:51,990 --> 01:14:53,910 2326 01:14:53,910 --> 01:14:56,709 2327 01:14:56,709 --> 01:14:58,390 2328 01:14:58,390 --> 01:15:00,310 2329 01:15:00,310 --> 01:15:03,189 2330 01:15:03,189 --> 01:15:03,199 2331 01:15:03,199 --> 01:15:05,030 2332 01:15:05,030 --> 01:15:05,040 2333 01:15:05,040 --> 01:15:06,390 2334 01:15:06,390 --> 01:15:08,790 2335 01:15:08,790 --> 01:15:11,110 2336 01:15:11,110 --> 01:15:13,110 2337 01:15:13,110 --> 01:15:14,790 2338 01:15:14,790 --> 01:15:20,229 2339 01:15:20,229 --> 01:15:20,239 2340 01:15:20,239 --> 01:15:21,510 2341 01:15:21,510 --> 01:15:22,870 2342 01:15:22,870 --> 01:15:25,590 2343 01:15:25,590 --> 01:15:27,430 2344 01:15:27,430 --> 01:15:28,950 2345 01:15:28,950 --> 01:15:32,149 2346 01:15:32,149 --> 01:15:32,159 2347 01:15:32,159 --> 01:15:33,669 2348 01:15:33,669 --> 01:15:36,070 2349 01:15:36,070 --> 01:15:36,080 2350 01:15:36,080 --> 01:15:37,430 2351 01:15:37,430 --> 01:15:39,750 2352 01:15:39,750 --> 01:15:41,030 2353 01:15:41,030 --> 01:15:44,470 2354 01:15:44,470 --> 01:15:44,480 2355 01:15:44,480 --> 01:15:45,669 2356 01:15:45,669 --> 01:15:47,910 2357 01:15:47,910 --> 01:15:49,750 2358 01:15:49,750 --> 01:15:51,910 2359 01:15:51,910 --> 01:15:54,070 2360 01:15:54,070 --> 01:15:54,080 2361 01:15:54,080 --> 01:15:55,189 2362 01:15:55,189 --> 01:15:57,510 2363 01:15:57,510 --> 01:15:58,950 2364 01:15:58,950 --> 01:15:58,960 2365 01:15:58,960 --> 01:16:00,709 2366 01:16:00,709 --> 01:16:01,910 2367 01:16:01,910 --> 01:16:04,310 2368 01:16:04,310 --> 01:16:06,630 2369 01:16:06,630 --> 01:16:08,310 2370 01:16:08,310 --> 01:16:10,709 2371 01:16:10,709 --> 01:16:12,229 2372 01:16:12,229 --> 01:16:14,310 2373 01:16:14,310 --> 01:16:15,990 2374 01:16:15,990 --> 01:16:17,510 2375 01:16:17,510 --> 01:16:19,110 2376 01:16:19,110 --> 01:16:21,270 2377 01:16:21,270 --> 01:16:22,550 2378 01:16:22,550 --> 01:16:25,830 2379 01:16:25,830 --> 01:16:27,590 2380 01:16:27,590 --> 01:16:29,270 2381 01:16:29,270 --> 01:16:31,189 2382 01:16:31,189 --> 01:16:32,950 2383 01:16:32,950 --> 01:16:34,630 2384 01:16:34,630 --> 01:16:37,350 2385 01:16:37,350 --> 01:16:50,390 2386 01:16:50,390 --> 01:16:50,400 2387 01:16:50,400 --> 01:16:52,480