1 00:00:25,599 --> 00:00:28,230 okay everyone let's get started uh welcome back um to uh theory of computation lecture number six um so we have been um looking at a number of different models of computation and um last lecture actually was an important one for us because we shifted gears um from our restricted models um finite automata and pushdown automata and their associated um generative models regular expressions and context-free grammars to turing machines which are going to be the model that we're going to which is the model that we're going to be sticking with for the rest of the semester because that's going to be our model as we're going to argue today for a general purpose computer so in the sense everything we've been doing up till now has or up until that point has been kind of a warm up but it's been um nevertheless has gotten the chance for us to introduce some important concepts uh that are um used in practice and also will serve as good examples for us moving forward and also just to get our you know get kind of um uh you know in a sense on the same page uh so what we're going to be doing today is looking at the turing machine model in a little bit more depth one can define turing machines in all sorts of different ways but it's going to turn out not to matter because all of those different ways are going to be equivalent to one another and so we are going to stick with the very simplest version the one we've already defined namely this uh the simple one tape turning machine but um we're going to basically justify that by looking at some of these other models and proving equivalent so we can we'll look at multi-tape turing machines we'll look at non-deterministic turing machines and we'll look at another model which is slightly different but it's still based on turing machines called an enumerator and we'll show that those all in the end um give you the same class of languages and so in that sense they're all equivalent to one another and that's going to be served as a kind of a motivator um kind of to in a sense recapitulate some of the history of the subject um and going to lead to our um discussion of what's called the church turing thesis so we will uh i'll get to that in due course and we'll also talk about uh some notation for um languages and notations for turing machines and for encoding objects to feed into turing machines as input but we'll get to that shortly as well uh so let's move on then we will next go into uh a little review of what we did with for touring machines and um just want to make sure we're all together on that the very important concept for us this term so the turing machine model looks like there's going to be this finite control there's a tape with a head that can read and write on the tape can move in both directions the tape is infinite to the to one side and so on as i mentioned last time the output of the turing machine is either going to be a halt accepting or rejecting or loop that the machine may run forever with three possible outcomes for any particular input the machine may accept that input by entering queue accept may halt and reject by entering q reject and it might reject by looping which means it just never gets to the accept state or the uh it never gets to any holding state it just goes forever but we still consider that to be rejecting the input just it's rejecting by looping okay um and as we defined last time a language is touring recognizable as or as we typically write t recognizable if it's the language of some turing machine the collection of accepted strings from that turing machine again just as before machine may accept zero strings one string many strings but it always has one language the collection of all accepted strings um now if you have a decider which is a machine that never loops which always holds on every input then we say its language is a decidable language so we'll say language is turning decidable or simply decidable if it's the language of some decider which is a turing machine that always holds on all inputs so we have those two distinct notions turning recognizable languages and turing decidable languages um now as we're going to argue this lecture turing machines are going to be our model of a general general purpose computer so that's the way we're going to think of computers as touring machines um and why turing machines why didn't we pick something else well the fact is it doesn't matter and that's going to be the point of this lecture all reasonable models of general purpose computation um you know unrestricted um computation in the sense of not limited memory um are all going to be are all have been shown um all the models that we've ever encountered have also been shown to be equivalent to one another um and so you you're free to pick anyone you like and so for this course we're going to pick turing machines because they're very simple to argue about mathematically and also they have a sort of a some element of familiarity in that you know they're they they feel like um you know you well they um they're more familiar than some of the other models that have been proposed that are out there um you know such as you know rewriting systems lambda calculus and so on because turing machines feel like a primitive kind of computer and in that sense they have a certain familiarity okay um so let's start talking about variations on the turing machine model and we're going to argue that it doesn't matter it doesn't make any difference um and then whether this is going to be kind of a precursor to a discussion of a bit of the history of the subject that we'll go and get to in the second half of the lecture so um a multi-tape turing machine is a turing machine as you might imagine that has more than one well it has one or possibly more tapes and so a single tape turning machine would be a special version of a multi-tape turing machine that's okay but you might have more than one tape um as i've shown in this diagram now uh how do we actually use a multi-tape turing machine well you present the in and we're going to see these coming up for convenience sometimes it's it's nice to be working with multiple tapes so we're going to see these later on in the semester a couple of times as well but for now um you know we'll we're setting the model up so that the input is going to be presented on a special input tape um so that's where the input appears and it's going to be followed by blanks just as we had before and now we have these potentially other tapes possibly other tapes which are we call them work tapes where the machine can write other stuff as it wishes and those tapes are going to be initially blank so just all blanks on them all of the tapes are going to be read and write so you can write on the input tape obviously you can read on from the input tape and you can read and write on the other tapes as well um so what we want to first establish that by having these additional tapes you don't get additional power for the machine in the sense that you're not going to have you won't have additional languages that you can recognize by virtue of having these additional tapes i mean you could imagine that having more tapes will allow you to do more things you know for example if you have a push down automaton with two stacks you can do more languages than you can with one stack so it's conceivable that by having more tapes you can do more languages than you could with one with one tape but in fact that's not the case what one tape is as good as having many tapes uh and we're going to prove that you know we're going to quickly sketch through the proof of that fact um so uh the theorem is that turing machine a language is touring recognizable and when we say turing recognizable for now we mean just with one tape um so that's the way we've defined it so a language is turing recognizable if and only if some multi-tape turing machine recognizes that language okay so really that another way of saying that is if you have a language that you can do with a single tape you can do it with a multi-tape and vice versa so one direction that is immediate because a single tape turing machine is already a multi-tape turing machine that just happens to have one tape uh so the forward direction of that is is um there's nothing to say that's just immediately true um but if we want to prove the reverse then we're gonna have to do some work and the work is going to be showing how do you convert a multi-tape turing machine to a single tape turing machine so if we have something that's recognized by a multi-tape turing machine it's still turing recognizable what that means is you can do it with a single tape turning machine so we have to show how to do the conversion and i'm kind of you know i'll i'll i'll show you that i think in a convincing way but we know without getting into too much detail so here is an image of a multi-tape turing machine in during the course of its input so it has already we've already started it off um you know initially it starts off with the um other tapes the work tapes being all blank but now it's processed for a while and you know the head on the input tape is now has moved off from its starting position at the left end it's somewhere in the middle it's written stuff on the other tapes and what we want to do is show how to represent that same information on a single tape turing machine in a way which allows the single tape turing machine to carry out the steps of the multi-tape turing machine but using only its single tape by virtue of some kind of a data structure which allows for the simulation to cut to go forward okay so how is the simula the single tape turning machine going to be simulating going to be carrying out this the the same effect of having these multiple tapes on the multi-tape turing machine so here's a picture of the single tape turning machine it just has one tape by the way i should mention that all of the tapes are infinite to the right um in the multi-tape touring machine just as we had for the single tape turning machine and now uh with this uh single tape turning machine it's going to represent the information that was on that's present on these multiple tapes but using only the single tape and the way i'm choosing to do that is going to be particularly simple i'm going to just divide up so here i'm saying it in words uh it's going to simulate him by storing the contents of the multi-tape on the single tape in separate blocks so i'm basically going to divide up the single tape uh turing machines tape into separate regions where each one of those regions is going to have the information that was on one of the tapes in the um multitape machine so for example here is on the first tape it's got aabba well that's going to appear here in that first block on the single tape turing machine okay sort of seeing it float down to suggest that it's coming from this multi-tape turing machine but that's really been developed by the simulation that the single tape turning machine has obviously the single tape turning machine doesn't have any direct access to the multi-tape machine but it's going to be simulating so this is how we're showing the data is being stored on the single tape machines tape so in the second block it's going to have the contents of the second uh tape of the turing multi-tape machine and the uh the final um uh region of the single tapes this is the final block so-called of the single tapes tape is going to have single tape machines tape it's going to have their the rest of the um the contents of the last tape of of m and then it's going to be followed by the same infinitely many blanks that each of these um tapes are going to have following the portion that they may have written during the course of their computation okay so that's what the single tape turing machine's tape is going to look like during the course of its computation it's going to have the informations represented in these blocks capturing all of the information that the multi-tape turing machine has in a perhaps somewhat less convenient way because the multi-tape turing machine you know we didn't really make this explicit and in part because it doesn't really matter we didn't say exactly how the multi-tape turning machine operates what i have in mind is that the multi-tape turning machine can read and move all of its heads from all of its heads in one step so in a single step of the multi-tape machine it can it can obtain the information that's underneath each of its heads feed that into its transition function and then together with the state of the multi-tape machine decide how to change each of those locations and then how to move each one of those heads so it's kind of operating on all of the tapes in parallel okay so now uh how how does the actual steps of the simulation of the single tape you know by the single tape um machine uh how does it go so first of all besides storing the contents of each of the tapes in in s's single tape there's some additional information that it needs to record namely um m has a head for each one of its tapes but s has just a single head and each of m's heads could be in some different location you know in this case as i've shown it on the first tape its head is in location three on the second tape it's in location two on the third tape it's a location on the last tape is in location one uh so where were we we were simulating the multi-tape turing machine with the single tape turing machine and we had to keep track of where the heads are and so we're going to do that by writing the locations on those blocks okay so we're going to have a special dot as i've shown here to represent the location of the head in that very first block so the heads on the b i'm going to dot the b and i'm going to do the same thing for the locations of the others um heads and how am i getting that effect well we've had seen something like that before where we just expand the tape alphabet of s to allow for these dotted symbols as well as the uh irregular symbols we also have expanded it to include those delimiter markers that separate the blocks from one one another which i'm writing as a pound sign um so we can just get that effect simply by expanding the tape alphabet of s okay a few more details of s that are just worth looking at just to make sure we're kind of all understanding what's happening so for every time m takes one step s has actually a lot of work to do because one step of m it can move all of those read and move all of those heads all at one shot s has to scan the entire tape to see what those what's underneath those in effect virtual heads which are the locations of the dotted symbols it has to see what's underneath each one of those heads to see how to update the transition you know update its state to the next state and then it has to scan again to change the location change the contents of those of those tape locations and also to move the head left or right by moving the effectively now moving the dot left or right so that's pretty straightforward uh uh there's one complication that can occur however um which is that what happens if um on one of the tapes let's say m starts using writing more content then you have room in that block uh to store that information you know so if the m like moves it has one zero one suppose m you know after a few steps moves his head into the originally blank portion of the tape and writes another symbol there we have to put that symbol somewhere because we have to record all that information and the block is like full so what do we do well in that case s goes to like a little interrupt routine and uh moves shifts all of the symbols um to the right one place to make open up um a location here where it can write a new symbol for uh carrying continuing the simulation so with that in mind uh s can maintain the current contents of each of m's tapes and um let me just note that here shift to add room is needed and that's all that happens in this during the course of the simulation okay and if of course if s as it's simulating m observes that m comes to an accept or reject state uh s should do the same okay so that that's our description of how the simulation of the the single tape simulation of the multi-tape machine works right okay let's turn to non-deterministic machines now if you remember for fine art and i hope you do for finite automata we had equivalence between non-deterministic and deterministic um uh finite automata for pushdown automata we did not have equivalence we didn't prove that but we just stated it as a fact there are certain languages that you can do with non-deterministic push-down automata which is the model we typically hold as our standard so the some things you can do with non-deterministic push-down automata that you cannot do with deterministic push-down automatic um you really need the non-determinism they're not not equivalent for turing machines we're going to get the equivalents back so we'll show that anything you can do with a non-deterministic turing machine in terms of language recognition you can also do with a deterministic turing machine so remember we kind of mentioned this briefly last time the non-deterministic turing machine looks exactly like a deterministic turing machine except for the transition function where we have this power set which allows for multiple different outcomes at a given step instead of just a single outcome that you would have in a deterministic machine so we represent that with this power set here um applied to the possibilities that um the machine would have as the next uh next step okay um so we're now going to prove a very similar theorem that a is recognizable by an ordinary one tape turing machine that kind of goes without saying so that's how we defined it a is turning recognizable if and only if some non-deterministic turing machine recognizes that language okay um again remember we're using non-determinism in the way we always do namely that a non-deterministic machine accepts if there is some branch or some thread if it's not a computation ends up at an except other branches might go forever or might uh reject part with part of the way through but acceptance always overrules them if any branch accepts that's how non-determinism works for us okay so now the forward direction of this is again just like before immediate because a deterministic turing machine is a special kind of not a deterministic turing machine which never branches um not deterministically on any of its steps so that forward direction is immediate the backwards definition is where we're going to have to do some work converting our non-deterministic turing machine to a deterministic machine and we'll do so as follows we'll take a non-deterministic turing machine uh here's our picture and we're going to now um imagine its computation in terms of a tree um and uh so here is the non-deterministic computation tree for n on some input w so n for nondeterministic uh here is the tree um somewhere down here it might end up accepting and as i mentioned that is going to be uh defining the overall computation to be accepting if there is some place here that's accepting and the only way it can be non-accepting computation if there is no except that occurs anywhere and somehow if you're going to convert this non-deterministic machine to a deterministic machine the deterministic machine is going to have to get that same effect though it only has determinant you know it doesn't have the non-determinism so how does it manage that well it's really going to be carrying out the simulation you know kind of as you would imagine doing it yourself if you were told you had to simulate a non-deterministic turing machine which is you're going to search that tree of possibilities looking to see if you find and accept if you do then you accept if you if you don't well maybe you'll go forever or maybe you'll manage to search the entire tree and then you will say no i accept uh no i reject um okay so let's just see again what the data structure of that deterministic machine's tape is going to look like um so the way this simulation is going to carry out and and you know you could program this in a host of different ways and in fact the textbook has a somewhat different simulation but i think this one here is uh lends itself a little bit better um and maybe it's a little simpler for description um in this in this setting uh but there's lots of different ways of improving these things so we're going to m is going to simulate n kind of a little bit similar to the previous uh simulation it's going to break the tape up into blocks and it's going to now store in each block a different thread at a particular point in time of n's computation so you imagine where m is going to be simulating and sort of um doing the same parallelism but doing the getting the effect of the parallelism not through the non-determinism but by maintaining copies of each of the threads on its tape and then updating them accordingly so the idea is i think not not too complicated um so let's just see some of the some of the details here so here here populating those blocks with the contents of um ends tape on different threads at a so you know maybe that corresponds to um ends tape you know after the fifth step of n um you know on each of those threads you have these three possibilities let's say written down as three different blocks um now uh remember in the case of the multi-tape simulation we needed to record some extra information extra information in particular we had to record the location of the heads by using those dotted symbols we're going to do that again because each one of these threads might have their head in some different location but we're going to have to do more but actually so let's do one other one thing at a time we're going to store the head location here they are all written down as dotted symbols but now there's something that goes further because in a multi-tape case there was one global state that the multitape machine was in of course at any given time it's just one has one finite control it's in one state um and uh but but but in the case of the non-deterministic machine that we're simulating now each thread can be in a different state um so whereas before we could keep track of the multitapes during the multi-tape machine state inside the single tape machines finite control now there could be many many different states to keep track of so you won't be able to store that in a finite control anymore because there could be some huge number of threads active at any given stage and some of them can be one state others can be in different state and you know how do you keep track of all that so we're going to have to use the tape we're going to actually write down which state a given thread is in on that block and we'll do that i'll show it like this so writing down the state of each cert in the block by writing down symbols which correspond to those uh states so we're actually in a sense writing the states down right on top of block using symbols that correspond to those states so again we're doing getting that effect by increasing the tape alphabet of m to include um symbols that correspond to each of its states i'm going to write those symbols down so here's the symbol for state q8 that says well one of th that very first thread of n's computation is in state q8 it's in this third position of its t um on its tape uh the head is on the third position there and the tape contents is aaba that's how we represent that thread of the computation um including the state okay now um so am is going to carry out that step by step maintaining that information on its tape there again similar to before there might be some special situations arising so for example n is not non-deterministic so a thread might fork into uh two threads what do we do then um well sort of the you know the reasonable thing if there's a forking thread m then copies that block makes two copies of the block to correspond to the two or however many copies you need to correspond to the possibilities that um that thread is branching into so you want to represent them all on different blocks of m's tape all right um and then if m ever finds out that on one of the threads it's entering the accept state it can just shut down the whole computation at that point and say accept okay so perhaps a little elaborate but i think conceptually hopefully um not too bad um now uh uh let's move on then to talk about uh somewhat of a different uh looking well sort of a different model in a way um uh which has some historical significance and sometimes it's a helpful way to look at uh touring recognizability and you and incidentally you have a homework problem uh on this um model called an enumerator or turing enumerator so uh here it's going to operate somewhat differently first of all we're going to enhance the turing machine model it's going to have just a single tape but it's going to also have a printer okay so we're now adding in this new device into the turing machine called the printer and um the turing machine has tape as before except we never provide any input to the machine the tape always starts out fully blank it's only used for reading and writing uh only used for for like the keeping tr for uh for for work uh it's not where you present the input so how do you talk about the language of the machine well what the way you work this machine is you take this enumerator uh it's a deterministic determination with a printer as i mentioned you start it off in blank tape and it you know runs and runs and runs and periodically it can print a string under some sort of program control which i'm not going to spell out for you but you can imagine you might have to define the machine in such a way that when it goes into a certain print state so we're not going to set that all up then whatever string is in a certain region of the tape maybe from this the start point up into the head location that gets printed out on the printer and then it can print out other strings in due course and so periodically you think of this printer as printing out a string um under the control of the uh the turing machine enumerator um and so these print strings these strings that could print out w1 w2 they appear and the machine might possibly go forever and print infinitely many strings or it might go forever and only print finally many strings or it might stop after a while by entering a holding state accepting or rejecting is irrelevant but enters a halting state and then the number the strings that it's print out the finite really many strings it printed out are just that's the output of the machine now the language of the machine is the collection of all strings that it ever prints out so we're defining language in a somewhat different way here it's not the strings that it accepts it's a string that it prints so we think of this machine almost a little bit analogous to like the regular expression or the grammar in the sense that it's a generative model it produces the language rather than accepts the strings in the language this is not really a recognizer it's a generator and the language of the machine as i'm writing over here for an enumerator here we say that it's language is the collection of strings that it prints okay now we will show that a language is touring recognizable in the previous sense uh you know recognized by some ordinary turing machine if and only if it's the language of summer numerator and actually this is kind of a little bit of an interesting proof uh you have to do some work it's an if and only if now neither direction is immediate you're gonna have to do some work uh in both directions one direction is a little harder and the other will start off with the easier direction let's say we have an enumerator i hope you hopefully you got this concept of this turing machine which periodically prints strings when you've started it off on the empty tape um uh so now what i'm going to construct for you is that turing machine recognizer defined from the enumerator so this recognizer is going to be simulating the enumerator um so it's basically the recognizer going to launch the enumerator it's going to start off simulating it um now if you want if it's convenient we we can use several tapes we can use one tape to simulate the recognizer and you have other tapes available for convenience if you want because we already showed multi-tape and single tape are equivalent so if you want to think of m as having multiple types it's not going to really be that important relevant to my conversation um but you're going to simulate e starting on the blank input as you're supposed to and then you're going to see whenever e prints an x you're going to see if x is the same as the input to the recognizer you you get the idea so we have a recognizer it has an input string maybe like 1 1 0 1. and i want to know i want to accept that string if it's one of the strings that the enumerator e prints out because that's what the recognizer do it's supposed to accept all the strings that the enumerator prints out so i got this one one zero one what do i do i launch i fire up that numerator i get it going and i start watching the strings it prints out comparing them with my input string one one zero one if i ever see it print out a one one zero one great i i know i can accept because i know it's in the enumerator's language but if i compare and compare and compare i never see uh the enumerator ever printing out uh my input string 1101 well i just keep simulating if e holds well then i can hold and reject if i've never seen that string coming out as an output if e doesn't hold well i'm just not going to hold either i'm going to go forever but then i'm going to be rejecting my input by looping okay so that's the proof of converting in this uh backward direction which is the easier direction now let's look at the forward direction which has a certain wrinkle that in it that we're going to have to get to so now we're going to be building our enumerator to simulate our recognizer okay and the way we'll do that is the enumerator has to print out all of the strings that the recognizer would ever accept so you kind of do the obvious thing you're going to take the numerator is going to start simulating the recognizer m on all possible strings sort of you know one by one using doing them in parallel taking turns you never know you know we um maybe we didn't make this so explicit but you don't have to you know you can sort of time share among several different possibilities you run you know it's sort of like having different blocks uh for the machine diff you're gonna run um uh the enumerator is gonna run the recognizer on uh it's you know well let's just say it does it sequentially runs it on the empty string it runs it on the string zero it runs it on the string one runs it on the string zero zero these are all the possible strings in sigma star you you run the on all of them and whenever the enumerator uh oh this is wrong whenever um so whenever m accepts then print i can't write very well print w um w i oops okay so let me just say it in words you want to simulate emma on each wi whenever you notice m accepting wi you just print out w i um because you want to print out all of the strings that m accepts now what have the problem there is a problem here so uh hopefully you're not confused by this typo um uh doing it sequentially like i just described doesn't quite work so let me just back that up here doing it sequentially doesn't quite work because emma might get stuck looping on one of the wi's like you know maybe when i feed w zero into w into m m goes forever because m is rejecting zero by looping but maybe it also accepts one this next string in the list the string one i'll never get to one by just feeding the strings into m one by one like this what i really have to do is run all of the strings in m in parallel and the way i'm going to indicate that is i'm going to simulate m on w1 to wi for i steps for each possible i so i'm going to run m on more and more strings for more and more steps okay until and every time i notice an m accepts something i print it out um so i will fix this in the uh version of the uh file that i um publish on the website so if you're still not getting it you can look there um uh but just to make things even worse i have a check in which is going to be about this one little point here so i got a question where do we get the wi strings the wi strings are simply the list of all strings in sigma star so we can under program control we can make m go through every possible string like if you had an odometer you're going to first get to the empty string then you go to the string zero then the string one you can write a program which is going to do that and the turing machine can similarly write write code to do that to get to each possible string one by one and then that's what the enumerator is doing it's obtaining each of those strings and then feeding them into m seeing what m does what i'm trying to get at here is that it's not good to feed them in run m to completion on each one before going to the next one you really kind of have to do them all in parallel uh to avoid the problem of getting stuck on one string that m is looping on um all right so uh this is relevant to your homework in fact if i convert m to an enumerator um in this way does that enumerator always print the strings out in string order string order is this order here um having them uh the having the shorter strings coming up before the longer strings and within strings of a certain length just doing them in lexicographic order um uh so um hopefully you follow that but since you you know these are not uh correctness that doesn't matter uh for us let me launch that poll let's i'll see how confused how random the answer is here uh so a fair amount of confusion if you're confused obviously you're in good company there um okay about to shut this down uh five seconds to go okay ending in now uh all right so the correct answer as the majority of you did yet is in fact that the order is not going to be uh respected here when he prints things out it might print out some things later in the order before earlier things in the order what's going to control how um when he is going to print it out it really is if m accepts some strings more quickly in fewer steps then e might end up printing out that string earlier in the list um then some other string that might be a shorter string um because uh the order in which e is go going to now identify that m is ex is accepting those strings is is the um depends upon the speed with which m is do actually doing the accepting so you have to um this is relevant to one of your homeworks because if uh what you're going to be asked to show is that if you start with a decidable language instead of a recognizable language then you can make it then you can make an enumerator which always prints out things in order in the string order okay and um and vice versa so if you have and if you have a string which uh the numerator which prints the things out in string order then the language is it's decidable okay so need to think about that one direction is a fairly simple modification of what i just showed the other direction is a little bit more work um so uh anyway why don't we uh uh take our little coffee break here uh okay so that's an interesting question so one question is if there's because you know sigma has an infinite sigma star is infinite um uh if i understand this question quite correctly which says you know it's an infinite number of strings here uh how how does the machine even get started because it has to go through and enumerate all of it no it's not first writing down all of those strings um because yeah you can write down and it you can't do this infinite step of writing all the strings what i had in mind is you're going to take each string in turn first you're going to take the first string on the list of sigma star run m uh on that string and then the next string and sigma star run it on that string and so on string after string and so you're never going to have infinitely many strings to deal with it's just going to be more and more strings as you're going along and um the question i was trying to get at is whether you're going to run each one to complete completion before you get to the next one or you're going to kind of try to do them all in parallel which is what you need to do for to prove this uh this there um won't there be infinitely many wi yes so there are a similar question infinitely many wi there are infinitely many wi but you know it's just an infinite loop that we're in one by one you're taking care of more and more uh wi as you're going um uh let's see here yes so e is another question is are we running it all on all the strings in parallel yes we are running things all on running all of the strings in parallel but you know it's uh running them in parallel but um these are um uh you know it's we're not defining a parallel turing machine it's running them in parallel using time sharing you know it runs a little bit in this block runs a little bit and that block and then another block and sort of shifts around and sort of does a few steps of each and that's how it's getting the effect of the parallelism um okay um question is would the enumerator essentially have a print tape and a work tape yeah you could think of it as having a print tape you know however you want to formalize it it's it it doesn't matter i mean i'm being a little whimsical but attaching it to a picture of a a real printer but yeah it's just you think conceptually you can have a print tape however you like to think about it's fine um um how can we directly say that without knowing the program of the printer uh the printer is just there's no we don't have to get into the code the structure of the printer the printer is just uh something where you say print a string and it print and a string comes out that's all we know about the printer and that's all we need to know um so there's no program for the printer uh sorry if i not understand your question your question but the question literally says how can we say that without knowing the program of the printer so if the machine is decidable why does it have to print all strings in order for example if one string is shorter and can it be decided later we well yeah if the uh the reason why we want so the question is if the machine is decidable why does it have to print all strings in order uh because that's what i'm asking you to do in the problem that's what if you read the problem i think it's number five on the pset that's what i'm asking you to do that's why you have to print it in order otherwise you wouldn't have to worry about printing an order just because i'm asking you to okay so i think we're at a time let's uh let us move back into um our our material um okay second half church touring thesis um so church uh this is going back into the bit of the history of the subject back to the 1930s um back then people were uh interested in formulating the notion of what do we mean by algorithm they don't even call it algorithm um and those they they call it some people call it procedures some people call it effective procedure um some people called it effective calculation um but you know people had in their minds mathematicians have been dealing for centuries thousands of years with procedures for doing things that's a very natural thing and mathematical logicians in particular church and turing are touring somebody you surely obviously have heard of church maybe not um a church was turing's thesis advisor in fact um and they both were coming out of the mathematical logic field of mathematics and uh using trying to use mathematical logic to formalize this intuitive notion of what we have had for for centuries about what what a procedure is what is it what is what is an algorithm and um back in those days you know they came up with different ways of formalizing it um uh so you know here we had this notion of algorithm which is kind of intuitive concept touring proposed turing machine as a way of capturing that informal way in mathematically precise way other people came up with other ways of doing it and back then it wasn't obvious that all of those different formulations would end up giving you equivalent concepts equivalent notions um and in fact they proved in fairly elaborate detail that the different methods that people came up with it was a lambda calculus there was uh rewriting systems there were several methods that were proposed for formalizing um this notion and they all turned out to be equivalent to one another um in today that seems kind of obvious even though i went to some effort to prove that just to give you a feeling for how those things uh go you know that you know if you have programs if you have pascal and java say and thinking about what you can do mathematically in those i'm not talking about their ability to interface with you know windows and so on but just the mathematical capabilities the capability of doing mathematical calculations or functions with a pascal program or a java program you know it would be absurd to think there's some program that you can write in java that you can't write in pascal um or python and the reason is we know you can compile python into java and you can compile java back into python that tells you that the two systems of two programming languages are equivalent in power um that wasn't obvious that from the get-go to these folks so they um observed that uh all of the different efforts they came at formalizing algorithm all were equivalent to one another that was that that was a kind of a breakthrough moment when they realized that they're all that not all the ways that they've come up with and once they got the idea they realized all reasonable ways of doing it always going to be equivalent um and so that suggested that they've really captured this notion of algorithm by any one of those methods say a turing machine and that's what they took you can't prove that because algorithms are an intuitive notion but the fact that we're able to capture that in a formal way that's what we call today the church touring thesis okay um so um that is uh um so any of these methods uh capture the notion of algorithm as we described and that has had a big impact on mathematics uh so you know i just want to just give me one second here all right here's a check-in on this one so you know alan turing so here are some facts which may or may not be true about elementary um so uh now you get to pick all the ones that apply um based on your knowledge obviously this is not uh something this is more for fun but um or historical interest um but here is uh let's launch that poll see what how much you know about uh mr turing um check all that apply okay almost done please let's wrap it up i think that's everybody okay two seconds please get credit for doing this and polling in fact it's kind of interesting here uh um [Music] uh he you all do know that he was a code breaker uh he was a worked uh in fact did was part of a team i think led the team which broke the german code um during world war ii you know the turing test is a famous uh thing for how do you characterize when you have an intelligent machine um so he definitely worked in ai he worked in biology as well um he has a paper that less well known to computer to scientists but if you look him up um on wikipedia where i get all my information uh uh he actually and i knew this anyway uh he has a very famous paper influential paper on how um for example spots arise on um on tiger on leopards and stripes on tigers and so on um he gave a kind of mathematical model for that which actually proves actually is uh quite does capture things in an accurate way uh as was shown subsequent to that uh was imprisoned for being gay in fact as far as i know from his history he was not imprisoned for being gay um he was convicted of being gay um and he was given a choice to go to prison or to take chemicals to cure him from being gay and he opted not to go to prison and take the chemicals uh and uh sadly he committed suicide two years after that so um he was um treated very treated very badly by uh british society and british government uh despite having the the great the work that he had done um and um you might think that he has been honored by appearing on a british banknote that's also not true um uh but the good news is he's not uh currently on a british banknote but he's going to be on a british bank note starting next year um so that is uh um so that's uh you know along with um winston churchill and a number of other notable brits he's going to be on the 50 pound banknote okay um so let us continue um so as i mentioned uh the church thesis is important for mathematics um and it has to do with these hilbert problems i don't know how many of you have encountered that david hilbert was widely considered to be the perhaps the greatest mathematician of his day um and um you know every four years mathematicians get together for an international congress called the international congress mathematicians it was you know email that's been going on for over 100 years and he was invited to the 1900 uh meeting to give a talk about anything he wanted and what he decided to do in during that presentation is give it is present 23 uh problems that would be a challenge for mathematicians for the coming century um and um uh you know i we're running a little short on time so i'm not going to go into them some of these will talk about later uh but the tenth problem here is about algorithms the tenth problem on hilbert's list of called called hilbert's tenth problem is a problem about algorithms and it says given algorithm for solving what are called diaphantine equations um he didn't call it an algorithm you could call it some uh you know finite procedure or something like that but you know you what are the diaphantine equations i'm glad you asked uh diaphantine equations uh what are they well they're very simple they're just polynomial equations like you know does this polynomial equal some constant for example um but where you're looking for the solutions you know the polynomials are variables you're looking for the solutions but you're only allowing those solutions to be integers so here's an example um so here i give you this polynomial um here i'm setting it equal to 7 and i want to solve that equation so it has these three variables x y and z so you have to find a solution there but i'm only going to allow you to have plug-ins integers and in fact there is a solution in integers one two and minus two if you plug it in you'll see it works all right uh now the general problem that hilbert was addressing is suppose i give you a polynomial um let's say it said it it's set to zero so we're looking for roots of the polynomial but just some polynomial equation you can always put it in this form and i want to know does it have a solution in integers um and what i want what i'm looking for is a procedure which will tell me yes or no i want to know um give an algorithm to answer that question for a given polynomial or using the language of this course defining this in terms of a language uh decide this language give a give a turing machine which will decide yes or no for any given polynomial yes there is a solution in integers no there is no solution okay and um uh that is a um uh um that was what um hilbert asked in his tenth problem given algorithm now as we know now we took um 70 years to get the answer that um that there is no such algorithm it's an undecidable problem and we'll talk about that a little bit later in the term but um there is d is not a decidable language now um there was no hope of coming up with that answer in 1900 or even in 1910 because we didn't have a formal idea of what an algorithm is that had to wait until the church turing thesis told us that algorithms really are turing machines that there is a formal way of saying what an algorithm is and once you had that notion then you could prove that there is no turing machine but before that you had only this vague notion intuitive notion of what an algorithm is and so there was no hope of ever answering that because the in fact the answer is there is no such algorithm okay now um i'll give this as a little exercise to you we're a little bit running short on time but this language d is in fact a recognizable language so i would suggest you think about that offline um but basically you know you can try plugging in different values for these variables and if you ever find that you know the um that it evaluates to zero then you can accept but otherwise you just have to keep going and looking so showing recognizability is very simple but decidability is false okay now let's talk a little bit about encodings for turing machines um so we're going to be working encodings and turing machines so we're now going to be working with um turing machines going forward we'll go as i you know we're going to the input to those turing machines might be polynomials might be strings they might be other things too we might want to feed automata into the turing machines so the turing machine can answer questions about the automata well don't forget turing machines take as their input just strings so we have to think about how we're going to represent something more complicated like an automaton as a string and i'm not going to get into the details of that you could spell it out in detail um but i think that's not too interesting to get into that we're just going to have develop a notation that says if you have some object o it could be a polynomial it could be an automaton it could be a graph it could be whatever you're working with a table i'm going to write that o in these brackets to mean an encoding of that object into a string and then you can feed the string into the turing machine okay so that's how we're going to be thinking of presenting turing machines with more complicated objects and strings as input because we're just going to represent them as strings using some ways in which i'm sure you're all familiar i mean that's how you deal with representing stuff when you're you know how to write your programs anyway um but just to make it formal and that's why we're going to write it down in these brackets and if you have a list of several objects that you want to present together to a turing machine we'll just write them together within brackets like this now for writing turing machines down we're going to going forward we're going to be using high-level english descriptions um we're not going to be worrying about managing the like the stuff we've been doing up till now i'm not going to ask you to do it we're not going to do that anymore and i'm not going to ask you to do it managing where stuff goes on the tapes and all that stuff that's too low level because really now that we have the church turing thesis we're really interested in talking about algorithms we're not that interested in talking about turing machines per se with ancient algorithms um and what the power of computation is so um we're going to only be talking about touring machines now in a high higher level way and only when we need to prove something about uh capabilities we're going to come back to turing machines i'm going to be proving things you know about their limitations and so on so our notation for running turing machines is going to be basically going to put the turing machine inside these quotation marks and we're going to know that we could if in principle write out the touring machine in a precise way in terms of states and transition function and so on but we'll never actually go ahead and do that lengthy exercise um so quick check in here so one of the features well okay let me not give this away so if x and y are strings you know i want to now have a way of presenting two strings as a single string as input to my machine because i always think of my machine as getting a single input so would you suggest one way of combining two strings into one which we would be a good encoding is just to concatenate those two strings together is would it be that would would that be the way you would do it is that a good way to do it or not such a good way to do it um so let's uh um let's see here i can get to that next and think what you would want to have happen in a good encoding okay ready set sold uh share results yeah i think most of you got the idea that this is not a good way of combining two strings into one um because the problem is it's kind of ambiguous in a sense you know that you could if you combine two strings this way what's what's important in an encoding is that the machine when it gets the encoding it can decode them back into the original objects and if you're just going to be sticking the two strings together you don't know where one string ends and the next string begins and so it's not going to be a good way of combining things you you should find a little bit more clever way of either introducing another symbol or it's doing something a little bit more sophisticated which allow you to be able to do the decoding as well as the encoding in a unique way okay so uh getting back to that notation for a turing machine so here is the machine we've already seen once before for a to the kb to the kc to the k i would write it now more simply than managing all of the tapes that we had the first time around which we did last lecture i would just say i would give it an input w check if w is of the right form with the things in the right order reject if it's not count the number of a's b's and c's and w except if all the counts are equal and reject if not we're not going to be that's going to be good enough to be writing things at that higher level when you're going to be writing your algorithm descriptions you just have to make sure that whatever you're doing you can implement um you know you you can't make uh you know you you don't want to be doing tests which are impossible or doing infinite infinitely much work in one stage that's not good um but as long as it's clear that what you're doing is doing only a finite amount of work in every stage and it's something that you could really implement you can write it at a high in a high level um okay i wanted to spend a few minutes talking about problem set two but we a little bit running out of time here um uh in particular there was this problem number five where you show that a language is turing recognizable if and only if there's a decidable d um where c is turing recognizable where d is now a collection of pairs we've got some questions about the notation which i hopefully answered now so um c uh is a set of x's such that there exists a a string you can pair it with so that the string x y is in d um let me try to give you a picture of that uh um [Music] so i'm gonna think of d as a collection of pairs of strings and it might be helpful to think of the kind of you know on the axes here so if we have a pair of strings x y which i have not yet packaged into a single string i'm thinking of them as a pair of two objects at this moment uh think of d as so you know just the x part is just below here on the x-axis um d you might want to think just conceptually we want to think of it as a collection of pairs so it's like a subset of all of these pairs and c is all of the the x's that correspond to any pair in here um sometimes we call that the projection because it's all the things that are below or the shadow if you had a light sitting up here it's all the things that are kind of underneath a d um so i've written c here kind of a little thicker if you can see that so that's the c language and um what you're going to what there's a two directions here one is a lot easier than the other if i give you a decidable d and i want to test whether something's in c so i give you an x i want to know is it is x in c well you now have to test is there some y that i can pair with x where the pairs in d um so you know you can start looking for that y if you ever find one you know that x is in c there are infinitely many ways to look for but don't forget you know you're only looking for a recognizer for c so you know on the rejecting side you're allowed to go forever so i've kind of given you a way to think about the easy direction the hard direction um you need to think about um how are you going to come up with that decidable d if i give you c you have to find a decidable d and now you're you're bringing something down lower you're starting with a recognizing recognizable language and a decidable language which is going to sort of be counterpart to that in this sense is a simpler language and the way it becomes simpler is that y is going to help you determine whether x is in c um and i guess the thing that you should i'll leave you with is that maybe we can talk about this a little bit more on tuesday i'll try to leave it a little bit more time if you want to test if something's in c x is in c but i don't want to let you go forever anymore because i want to be a decidable language and i'm going to use why to help you what information would help you guarantee you get the right answer for x being in c but that you would have to be sure you're getting the right answer so it's you know why could just be the answer but then you know you don't know that that's you you have to be convinced that you you have the right answer you know why just saying what it is i mean you could well i could say that x is in d even when it's not true x isn't c even when that's not true but so one information would allow you to check that x is in c what would be helpful information to check the x's and c where you would avoid ever having to go forever a little bit too rushed here for that to be helpful anyway let's just uh conclude what we've done um just brief summary here um and i i i don't want to uh um keep you going keep you over time so i'll just leave this up on the board um i will uh um uh see if there's any so the lecture is over uh and you can take off uh as you wish um i will stick around for a couple of minutes i have another meeting soon um but uh i'll answer try to answer some chats uh all right um people are commenting about this movie about touring the imitation game uh you know if you haven't seen that i recommend it um um and uh is it a good idea to make y to be equal to the repeated loop string hmm i'm not sure about that uh thank you for the uh thank you everybody for sending your kind notes um is it church touring thesis the question is is it in the book yes it's in the book kind of similar to what i've already said but yes okay this is a good question here the church turing thesis somebody asked me is it proved in the book there's nothing to prove the church touring thesis is an equivalence between the intuitive and the formal you can't prove something like that you can just make it's really in a sense a hypothesis that the only thing you'll ever be able to compute um is something that you can do with a turing machine um i mean that has something to do with the nature of the physical world and i'm sure i don't really think that's what people had in mind it's that you know the the kinds of things that we normally think of being able to do with a with a rest with a procedure mathematically it's exactly the kinds of things that you can do with the turing machine um so somebody's pointing out well trying to be helpful here maybe these folks are asking about equivalence of various different computation models being proved in the book not that church drink these sits itself well maybe that's the the things that we proved in lecture are also proved in the book um about the equivalence of single tape machines multitape machines non-deterministic machines um those are all proved in the book we don't those i mean there's lots of models out there so we can spend a lot of time proving equivalences um and there are books that do spend a lot of time proving equivalences but i think you know uh we yeah i i think what we've given is probably enough um uh so somebody that's a good question if if you give an algorithm okay if you give an algorithm you don't need to give the actual machine uh going forward unless somebody unless you're explicitly asked to um uh but yes you don't you don't have to give um the states and transitions anymore um okay so it's a little after four i think we're going to i think i'm going to move on to my next meeting uh so i uh thank you for being here and i will see you on tuesday you 2 00:00:28,230 --> 00:00:31,509 okay everyone let's get started uh welcome back um to uh theory of computation lecture number six um so we have been um looking at a number of different models of computation and um last lecture actually was an important one for us because we shifted gears um from our restricted models um finite automata and pushdown automata and their associated um generative models regular expressions and context-free grammars to turing machines which are going to be the model that we're going to which is the model that we're going to be sticking with for the rest of the semester because that's going to be our model as we're going to argue today for a general purpose computer so in the sense everything we've been doing up till now has or up until that point has been kind of a warm up but it's been um nevertheless has gotten the chance for us to introduce some important concepts uh that are um used in practice and also will serve as good examples for us moving forward and also just to get our you know get kind of um uh you know in a sense on the same page uh so what we're going to be doing today is looking at the turing machine model in a little bit more depth one can define turing machines in all sorts of different ways but it's going to turn out not to matter because all of those different ways are going to be equivalent to one another and so we are going to stick with the very simplest version the one we've already defined namely this uh the simple one tape turning machine but um we're going to basically justify that by looking at some of these other models and proving equivalent so we can we'll look at multi-tape turing machines we'll look at non-deterministic turing machines and we'll look at another model which is slightly different but it's still based on turing machines called an enumerator and we'll show that those all in the end um give you the same class of languages and so in that sense they're all equivalent to one another and that's going to be served as a kind of a motivator um kind of to in a sense recapitulate some of the history of the subject um and going to lead to our um discussion of what's called the church turing thesis so we will uh i'll get to that in due course and we'll also talk about uh some notation for um languages and notations for turing machines and for encoding objects to feed into turing machines as input but we'll get to that shortly as well uh so let's move on then we will next go into uh a little review of what we did with for touring machines and um just want to make sure we're all together on that the very important concept for us this term so the turing machine model looks like there's going to be this finite control there's a tape with a head that can read and write on the tape can move in both directions the tape is infinite to the to one side and so on as i mentioned last time the output of the turing machine is either going to be a halt accepting or rejecting or loop that the machine may run forever with three possible outcomes for any particular input the machine may accept that input by entering queue accept may halt and reject by entering q reject and it might reject by looping which means it just never gets to the accept state or the uh it never gets to any holding state it just goes forever but we still consider that to be rejecting the input just it's rejecting by looping okay um and as we defined last time a language is touring recognizable as or as we typically write t recognizable if it's the language of some turing machine the collection of accepted strings from that turing machine again just as before machine may accept zero strings one string many strings but it always has one language the collection of all accepted strings um now if you have a decider which is a machine that never loops which always holds on every input then we say its language is a decidable language so we'll say language is turning decidable or simply decidable if it's the language of some decider which is a turing machine that always holds on all inputs so we have those two distinct notions turning recognizable languages and turing decidable languages um now as we're going to argue this lecture turing machines are going to be our model of a general general purpose computer so that's the way we're going to think of computers as touring machines um and why turing machines why didn't we pick something else well the fact is it doesn't matter and that's going to be the point of this lecture all reasonable models of general purpose computation um you know unrestricted um computation in the sense of not limited memory um are all going to be are all have been shown um all the models that we've ever encountered have also been shown to be equivalent to one another um and so you you're free to pick anyone you like and so for this course we're going to pick turing machines because they're very simple to argue about mathematically and also they have a sort of a some element of familiarity in that you know they're they they feel like um you know you well they um they're more familiar than some of the other models that have been proposed that are out there um you know such as you know rewriting systems lambda calculus and so on because turing machines feel like a primitive kind of computer and in that sense they have a certain familiarity okay um so let's start talking about variations on the turing machine model and we're going to argue that it doesn't matter it doesn't make any difference um and then whether this is going to be kind of a precursor to a discussion of a bit of the history of the subject that we'll go and get to in the second half of the lecture so um a multi-tape turing machine is a turing machine as you might imagine that has more than one well it has one or possibly more tapes and so a single tape turning machine would be a special version of a multi-tape turing machine that's okay but you might have more than one tape um as i've shown in this diagram now uh how do we actually use a multi-tape turing machine well you present the in and we're going to see these coming up for convenience sometimes it's it's nice to be working with multiple tapes so we're going to see these later on in the semester a couple of times as well but for now um you know we'll we're setting the model up so that the input is going to be presented on a special input tape um so that's where the input appears and it's going to be followed by blanks just as we had before and now we have these potentially other tapes possibly other tapes which are we call them work tapes where the machine can write other stuff as it wishes and those tapes are going to be initially blank so just all blanks on them all of the tapes are going to be read and write so you can write on the input tape obviously you can read on from the input tape and you can read and write on the other tapes as well um so what we want to first establish that by having these additional tapes you don't get additional power for the machine in the sense that you're not going to have you won't have additional languages that you can recognize by virtue of having these additional tapes i mean you could imagine that having more tapes will allow you to do more things you know for example if you have a push down automaton with two stacks you can do more languages than you can with one stack so it's conceivable that by having more tapes you can do more languages than you could with one with one tape but in fact that's not the case what one tape is as good as having many tapes uh and we're going to prove that you know we're going to quickly sketch through the proof of that fact um so uh the theorem is that turing machine a language is touring recognizable and when we say turing recognizable for now we mean just with one tape um so that's the way we've defined it so a language is turing recognizable if and only if some multi-tape turing machine recognizes that language okay so really that another way of saying that is if you have a language that you can do with a single tape you can do it with a multi-tape and vice versa so one direction that is immediate because a single tape turing machine is already a multi-tape turing machine that just happens to have one tape uh so the forward direction of that is is um there's nothing to say that's just immediately true um but if we want to prove the reverse then we're gonna have to do some work and the work is going to be showing how do you convert a multi-tape turing machine to a single tape turing machine so if we have something that's recognized by a multi-tape turing machine it's still turing recognizable what that means is you can do it with a single tape turning machine so we have to show how to do the conversion and i'm kind of you know i'll i'll i'll show you that i think in a convincing way but we know without getting into too much detail so here is an image of a multi-tape turing machine in during the course of its input so it has already we've already started it off um you know initially it starts off with the um other tapes the work tapes being all blank but now it's processed for a while and you know the head on the input tape is now has moved off from its starting position at the left end it's somewhere in the middle it's written stuff on the other tapes and what we want to do is show how to represent that same information on a single tape turing machine in a way which allows the single tape turing machine to carry out the steps of the multi-tape turing machine but using only its single tape by virtue of some kind of a data structure which allows for the simulation to cut to go forward okay so how is the simula the single tape turning machine going to be simulating going to be carrying out this the the same effect of having these multiple tapes on the multi-tape turing machine so here's a picture of the single tape turning machine it just has one tape by the way i should mention that all of the tapes are infinite to the right um in the multi-tape touring machine just as we had for the single tape turning machine and now uh with this uh single tape turning machine it's going to represent the information that was on that's present on these multiple tapes but using only the single tape and the way i'm choosing to do that is going to be particularly simple i'm going to just divide up so here i'm saying it in words uh it's going to simulate him by storing the contents of the multi-tape on the single tape in separate blocks so i'm basically going to divide up the single tape uh turing machines tape into separate regions where each one of those regions is going to have the information that was on one of the tapes in the um multitape machine so for example here is on the first tape it's got aabba well that's going to appear here in that first block on the single tape turing machine okay sort of seeing it float down to suggest that it's coming from this multi-tape turing machine but that's really been developed by the simulation that the single tape turning machine has obviously the single tape turning machine doesn't have any direct access to the multi-tape machine but it's going to be simulating so this is how we're showing the data is being stored on the single tape machines tape so in the second block it's going to have the contents of the second uh tape of the turing multi-tape machine and the uh the final um uh region of the single tapes this is the final block so-called of the single tapes tape is going to have single tape machines tape it's going to have their the rest of the um the contents of the last tape of of m and then it's going to be followed by the same infinitely many blanks that each of these um tapes are going to have following the portion that they may have written during the course of their computation okay so that's what the single tape turing machine's tape is going to look like during the course of its computation it's going to have the informations represented in these blocks capturing all of the information that the multi-tape turing machine has in a perhaps somewhat less convenient way because the multi-tape turing machine you know we didn't really make this explicit and in part because it doesn't really matter we didn't say exactly how the multi-tape turning machine operates what i have in mind is that the multi-tape turning machine can read and move all of its heads from all of its heads in one step so in a single step of the multi-tape machine it can it can obtain the information that's underneath each of its heads feed that into its transition function and then together with the state of the multi-tape machine decide how to change each of those locations and then how to move each one of those heads so it's kind of operating on all of the tapes in parallel okay so now uh how how does the actual steps of the simulation of the single tape you know by the single tape um machine uh how does it go so first of all besides storing the contents of each of the tapes in in s's single tape there's some additional information that it needs to record namely um m has a head for each one of its tapes but s has just a single head and each of m's heads could be in some different location you know in this case as i've shown it on the first tape its head is in location three on the second tape it's in location two on the third tape it's a location on the last tape is in location one uh so where were we we were simulating the multi-tape turing machine with the single tape turing machine and we had to keep track of where the heads are and so we're going to do that by writing the locations on those blocks okay so we're going to have a special dot as i've shown here to represent the location of the head in that very first block so the heads on the b i'm going to dot the b and i'm going to do the same thing for the locations of the others um heads and how am i getting that effect well we've had seen something like that before where we just expand the tape alphabet of s to allow for these dotted symbols as well as the uh irregular symbols we also have expanded it to include those delimiter markers that separate the blocks from one one another which i'm writing as a pound sign um so we can just get that effect simply by expanding the tape alphabet of s okay a few more details of s that are just worth looking at just to make sure we're kind of all understanding what's happening so for every time m takes one step s has actually a lot of work to do because one step of m it can move all of those read and move all of those heads all at one shot s has to scan the entire tape to see what those what's underneath those in effect virtual heads which are the locations of the dotted symbols it has to see what's underneath each one of those heads to see how to update the transition you know update its state to the next state and then it has to scan again to change the location change the contents of those of those tape locations and also to move the head left or right by moving the effectively now moving the dot left or right so that's pretty straightforward uh uh there's one complication that can occur however um which is that what happens if um on one of the tapes let's say m starts using writing more content then you have room in that block uh to store that information you know so if the m like moves it has one zero one suppose m you know after a few steps moves his head into the originally blank portion of the tape and writes another symbol there we have to put that symbol somewhere because we have to record all that information and the block is like full so what do we do well in that case s goes to like a little interrupt routine and uh moves shifts all of the symbols um to the right one place to make open up um a location here where it can write a new symbol for uh carrying continuing the simulation so with that in mind uh s can maintain the current contents of each of m's tapes and um let me just note that here shift to add room is needed and that's all that happens in this during the course of the simulation okay and if of course if s as it's simulating m observes that m comes to an accept or reject state uh s should do the same okay so that that's our description of how the simulation of the the single tape simulation of the multi-tape machine works right okay let's turn to non-deterministic machines now if you remember for fine art and i hope you do for finite automata we had equivalence between non-deterministic and deterministic um uh finite automata for pushdown automata we did not have equivalence we didn't prove that but we just stated it as a fact there are certain languages that you can do with non-deterministic push-down automata which is the model we typically hold as our standard so the some things you can do with non-deterministic push-down automata that you cannot do with deterministic push-down automatic um you really need the non-determinism they're not not equivalent for turing machines we're going to get the equivalents back so we'll show that anything you can do with a non-deterministic turing machine in terms of language recognition you can also do with a deterministic turing machine so remember we kind of mentioned this briefly last time the non-deterministic turing machine looks exactly like a deterministic turing machine except for the transition function where we have this power set which allows for multiple different outcomes at a given step instead of just a single outcome that you would have in a deterministic machine so we represent that with this power set here um applied to the possibilities that um the machine would have as the next uh next step okay um so we're now going to prove a very similar theorem that a is recognizable by an ordinary one tape turing machine that kind of goes without saying so that's how we defined it a is turning recognizable if and only if some non-deterministic turing machine recognizes that language okay um again remember we're using non-determinism in the way we always do namely that a non-deterministic machine accepts if there is some branch or some thread if it's not a computation ends up at an except other branches might go forever or might uh reject part with part of the way through but acceptance always overrules them if any branch accepts that's how non-determinism works for us okay so now the forward direction of this is again just like before immediate because a deterministic turing machine is a special kind of not a deterministic turing machine which never branches um not deterministically on any of its steps so that forward direction is immediate the backwards definition is where we're going to have to do some work converting our non-deterministic turing machine to a deterministic machine and we'll do so as follows we'll take a non-deterministic turing machine uh here's our picture and we're going to now um imagine its computation in terms of a tree um and uh so here is the non-deterministic computation tree for n on some input w so n for nondeterministic uh here is the tree um somewhere down here it might end up accepting and as i mentioned that is going to be uh defining the overall computation to be accepting if there is some place here that's accepting and the only way it can be non-accepting computation if there is no except that occurs anywhere and somehow if you're going to convert this non-deterministic machine to a deterministic machine the deterministic machine is going to have to get that same effect though it only has determinant you know it doesn't have the non-determinism so how does it manage that well it's really going to be carrying out the simulation you know kind of as you would imagine doing it yourself if you were told you had to simulate a non-deterministic turing machine which is you're going to search that tree of possibilities looking to see if you find and accept if you do then you accept if you if you don't well maybe you'll go forever or maybe you'll manage to search the entire tree and then you will say no i accept uh no i reject um okay so let's just see again what the data structure of that deterministic machine's tape is going to look like um so the way this simulation is going to carry out and and you know you could program this in a host of different ways and in fact the textbook has a somewhat different simulation but i think this one here is uh lends itself a little bit better um and maybe it's a little simpler for description um in this in this setting uh but there's lots of different ways of improving these things so we're going to m is going to simulate n kind of a little bit similar to the previous uh simulation it's going to break the tape up into blocks and it's going to now store in each block a different thread at a particular point in time of n's computation so you imagine where m is going to be simulating and sort of um doing the same parallelism but doing the getting the effect of the parallelism not through the non-determinism but by maintaining copies of each of the threads on its tape and then updating them accordingly so the idea is i think not not too complicated um so let's just see some of the some of the details here so here here populating those blocks with the contents of um ends tape on different threads at a so you know maybe that corresponds to um ends tape you know after the fifth step of n um you know on each of those threads you have these three possibilities let's say written down as three different blocks um now uh remember in the case of the multi-tape simulation we needed to record some extra information extra information in particular we had to record the location of the heads by using those dotted symbols we're going to do that again because each one of these threads might have their head in some different location but we're going to have to do more but actually so let's do one other one thing at a time we're going to store the head location here they are all written down as dotted symbols but now there's something that goes further because in a multi-tape case there was one global state that the multitape machine was in of course at any given time it's just one has one finite control it's in one state um and uh but but but in the case of the non-deterministic machine that we're simulating now each thread can be in a different state um so whereas before we could keep track of the multitapes during the multi-tape machine state inside the single tape machines finite control now there could be many many different states to keep track of so you won't be able to store that in a finite control anymore because there could be some huge number of threads active at any given stage and some of them can be one state others can be in different state and you know how do you keep track of all that so we're going to have to use the tape we're going to actually write down which state a given thread is in on that block and we'll do that i'll show it like this so writing down the state of each cert in the block by writing down symbols which correspond to those uh states so we're actually in a sense writing the states down right on top of block using symbols that correspond to those states so again we're doing getting that effect by increasing the tape alphabet of m to include um symbols that correspond to each of its states i'm going to write those symbols down so here's the symbol for state q8 that says well one of th that very first thread of n's computation is in state q8 it's in this third position of its t um on its tape uh the head is on the third position there and the tape contents is aaba that's how we represent that thread of the computation um including the state okay now um so am is going to carry out that step by step maintaining that information on its tape there again similar to before there might be some special situations arising so for example n is not non-deterministic so a thread might fork into uh two threads what do we do then um well sort of the you know the reasonable thing if there's a forking thread m then copies that block makes two copies of the block to correspond to the two or however many copies you need to correspond to the possibilities that um that thread is branching into so you want to represent them all on different blocks of m's tape all right um and then if m ever finds out that on one of the threads it's entering the accept state it can just shut down the whole computation at that point and say accept okay so perhaps a little elaborate but i think conceptually hopefully um not too bad um now uh uh let's move on then to talk about uh somewhat of a different uh looking well sort of a different model in a way um uh which has some historical significance and sometimes it's a helpful way to look at uh touring recognizability and you and incidentally you have a homework problem uh on this um model called an enumerator or turing enumerator so uh here it's going to operate somewhat differently first of all we're going to enhance the turing machine model it's going to have just a single tape but it's going to also have a printer okay so we're now adding in this new device into the turing machine called the printer and um the turing machine has tape as before except we never provide any input to the machine the tape always starts out fully blank it's only used for reading and writing uh only used for for like the keeping tr for uh for for work uh it's not where you present the input so how do you talk about the language of the machine well what the way you work this machine is you take this enumerator uh it's a deterministic determination with a printer as i mentioned you start it off in blank tape and it you know runs and runs and runs and periodically it can print a string under some sort of program control which i'm not going to spell out for you but you can imagine you might have to define the machine in such a way that when it goes into a certain print state so we're not going to set that all up then whatever string is in a certain region of the tape maybe from this the start point up into the head location that gets printed out on the printer and then it can print out other strings in due course and so periodically you think of this printer as printing out a string um under the control of the uh the turing machine enumerator um and so these print strings these strings that could print out w1 w2 they appear and the machine might possibly go forever and print infinitely many strings or it might go forever and only print finally many strings or it might stop after a while by entering a holding state accepting or rejecting is irrelevant but enters a halting state and then the number the strings that it's print out the finite really many strings it printed out are just that's the output of the machine now the language of the machine is the collection of all strings that it ever prints out so we're defining language in a somewhat different way here it's not the strings that it accepts it's a string that it prints so we think of this machine almost a little bit analogous to like the regular expression or the grammar in the sense that it's a generative model it produces the language rather than accepts the strings in the language this is not really a recognizer it's a generator and the language of the machine as i'm writing over here for an enumerator here we say that it's language is the collection of strings that it prints okay now we will show that a language is touring recognizable in the previous sense uh you know recognized by some ordinary turing machine if and only if it's the language of summer numerator and actually this is kind of a little bit of an interesting proof uh you have to do some work it's an if and only if now neither direction is immediate you're gonna have to do some work uh in both directions one direction is a little harder and the other will start off with the easier direction let's say we have an enumerator i hope you hopefully you got this concept of this turing machine which periodically prints strings when you've started it off on the empty tape um uh so now what i'm going to construct for you is that turing machine recognizer defined from the enumerator so this recognizer is going to be simulating the enumerator um so it's basically the recognizer going to launch the enumerator it's going to start off simulating it um now if you want if it's convenient we we can use several tapes we can use one tape to simulate the recognizer and you have other tapes available for convenience if you want because we already showed multi-tape and single tape are equivalent so if you want to think of m as having multiple types it's not going to really be that important relevant to my conversation um but you're going to simulate e starting on the blank input as you're supposed to and then you're going to see whenever e prints an x you're going to see if x is the same as the input to the recognizer you you get the idea so we have a recognizer it has an input string maybe like 1 1 0 1. and i want to know i want to accept that string if it's one of the strings that the enumerator e prints out because that's what the recognizer do it's supposed to accept all the strings that the enumerator prints out so i got this one one zero one what do i do i launch i fire up that numerator i get it going and i start watching the strings it prints out comparing them with my input string one one zero one if i ever see it print out a one one zero one great i i know i can accept because i know it's in the enumerator's language but if i compare and compare and compare i never see uh the enumerator ever printing out uh my input string 1101 well i just keep simulating if e holds well then i can hold and reject if i've never seen that string coming out as an output if e doesn't hold well i'm just not going to hold either i'm going to go forever but then i'm going to be rejecting my input by looping okay so that's the proof of converting in this uh backward direction which is the easier direction now let's look at the forward direction which has a certain wrinkle that in it that we're going to have to get to so now we're going to be building our enumerator to simulate our recognizer okay and the way we'll do that is the enumerator has to print out all of the strings that the recognizer would ever accept so you kind of do the obvious thing you're going to take the numerator is going to start simulating the recognizer m on all possible strings sort of you know one by one using doing them in parallel taking turns you never know you know we um maybe we didn't make this so explicit but you don't have to you know you can sort of time share among several different possibilities you run you know it's sort of like having different blocks uh for the machine diff you're gonna run um uh the enumerator is gonna run the recognizer on uh it's you know well let's just say it does it sequentially runs it on the empty string it runs it on the string zero it runs it on the string one runs it on the string zero zero these are all the possible strings in sigma star you you run the on all of them and whenever the enumerator uh oh this is wrong whenever um so whenever m accepts then print i can't write very well print w um w i oops okay so let me just say it in words you want to simulate emma on each wi whenever you notice m accepting wi you just print out w i um because you want to print out all of the strings that m accepts now what have the problem there is a problem here so uh hopefully you're not confused by this typo um uh doing it sequentially like i just described doesn't quite work so let me just back that up here doing it sequentially doesn't quite work because emma might get stuck looping on one of the wi's like you know maybe when i feed w zero into w into m m goes forever because m is rejecting zero by looping but maybe it also accepts one this next string in the list the string one i'll never get to one by just feeding the strings into m one by one like this what i really have to do is run all of the strings in m in parallel and the way i'm going to indicate that is i'm going to simulate m on w1 to wi for i steps for each possible i so i'm going to run m on more and more strings for more and more steps okay until and every time i notice an m accepts something i print it out um so i will fix this in the uh version of the uh file that i um publish on the website so if you're still not getting it you can look there um uh but just to make things even worse i have a check in which is going to be about this one little point here so i got a question where do we get the wi strings the wi strings are simply the list of all strings in sigma star so we can under program control we can make m go through every possible string like if you had an odometer you're going to first get to the empty string then you go to the string zero then the string one you can write a program which is going to do that and the turing machine can similarly write write code to do that to get to each possible string one by one and then that's what the enumerator is doing it's obtaining each of those strings and then feeding them into m seeing what m does what i'm trying to get at here is that it's not good to feed them in run m to completion on each one before going to the next one you really kind of have to do them all in parallel uh to avoid the problem of getting stuck on one string that m is looping on um all right so uh this is relevant to your homework in fact if i convert m to an enumerator um in this way does that enumerator always print the strings out in string order string order is this order here um having them uh the having the shorter strings coming up before the longer strings and within strings of a certain length just doing them in lexicographic order um uh so um hopefully you follow that but since you you know these are not uh correctness that doesn't matter uh for us let me launch that poll let's i'll see how confused how random the answer is here uh so a fair amount of confusion if you're confused obviously you're in good company there um okay about to shut this down uh five seconds to go okay ending in now uh all right so the correct answer as the majority of you did yet is in fact that the order is not going to be uh respected here when he prints things out it might print out some things later in the order before earlier things in the order what's going to control how um when he is going to print it out it really is if m accepts some strings more quickly in fewer steps then e might end up printing out that string earlier in the list um then some other string that might be a shorter string um because uh the order in which e is go going to now identify that m is ex is accepting those strings is is the um depends upon the speed with which m is do actually doing the accepting so you have to um this is relevant to one of your homeworks because if uh what you're going to be asked to show is that if you start with a decidable language instead of a recognizable language then you can make it then you can make an enumerator which always prints out things in order in the string order okay and um and vice versa so if you have and if you have a string which uh the numerator which prints the things out in string order then the language is it's decidable okay so need to think about that one direction is a fairly simple modification of what i just showed the other direction is a little bit more work um so uh anyway why don't we uh uh take our little coffee break here uh okay so that's an interesting question so one question is if there's because you know sigma has an infinite sigma star is infinite um uh if i understand this question quite correctly which says you know it's an infinite number of strings here uh how how does the machine even get started because it has to go through and enumerate all of it no it's not first writing down all of those strings um because yeah you can write down and it you can't do this infinite step of writing all the strings what i had in mind is you're going to take each string in turn first you're going to take the first string on the list of sigma star run m uh on that string and then the next string and sigma star run it on that string and so on string after string and so you're never going to have infinitely many strings to deal with it's just going to be more and more strings as you're going along and um the question i was trying to get at is whether you're going to run each one to complete completion before you get to the next one or you're going to kind of try to do them all in parallel which is what you need to do for to prove this uh this there um won't there be infinitely many wi yes so there are a similar question infinitely many wi there are infinitely many wi but you know it's just an infinite loop that we're in one by one you're taking care of more and more uh wi as you're going um uh let's see here yes so e is another question is are we running it all on all the strings in parallel yes we are running things all on running all of the strings in parallel but you know it's uh running them in parallel but um these are um uh you know it's we're not defining a parallel turing machine it's running them in parallel using time sharing you know it runs a little bit in this block runs a little bit and that block and then another block and sort of shifts around and sort of does a few steps of each and that's how it's getting the effect of the parallelism um okay um question is would the enumerator essentially have a print tape and a work tape yeah you could think of it as having a print tape you know however you want to formalize it it's it it doesn't matter i mean i'm being a little whimsical but attaching it to a picture of a a real printer but yeah it's just you think conceptually you can have a print tape however you like to think about it's fine um um how can we directly say that without knowing the program of the printer uh the printer is just there's no we don't have to get into the code the structure of the printer the printer is just uh something where you say print a string and it print and a string comes out that's all we know about the printer and that's all we need to know um so there's no program for the printer uh sorry if i not understand your question your question but the question literally says how can we say that without knowing the program of the printer so if the machine is decidable why does it have to print all strings in order for example if one string is shorter and can it be decided later we well yeah if the uh the reason why we want so the question is if the machine is decidable why does it have to print all strings in order uh because that's what i'm asking you to do in the problem that's what if you read the problem i think it's number five on the pset that's what i'm asking you to do that's why you have to print it in order otherwise you wouldn't have to worry about printing an order just because i'm asking you to okay so i think we're at a time let's uh let us move back into um our our material um okay second half church touring thesis um so church uh this is going back into the bit of the history of the subject back to the 1930s um back then people were uh interested in formulating the notion of what do we mean by algorithm they don't even call it algorithm um and those they they call it some people call it procedures some people call it effective procedure um some people called it effective calculation um but you know people had in their minds mathematicians have been dealing for centuries thousands of years with procedures for doing things that's a very natural thing and mathematical logicians in particular church and turing are touring somebody you surely obviously have heard of church maybe not um a church was turing's thesis advisor in fact um and they both were coming out of the mathematical logic field of mathematics and uh using trying to use mathematical logic to formalize this intuitive notion of what we have had for for centuries about what what a procedure is what is it what is what is an algorithm and um back in those days you know they came up with different ways of formalizing it um uh so you know here we had this notion of algorithm which is kind of intuitive concept touring proposed turing machine as a way of capturing that informal way in mathematically precise way other people came up with other ways of doing it and back then it wasn't obvious that all of those different formulations would end up giving you equivalent concepts equivalent notions um and in fact they proved in fairly elaborate detail that the different methods that people came up with it was a lambda calculus there was uh rewriting systems there were several methods that were proposed for formalizing um this notion and they all turned out to be equivalent to one another um in today that seems kind of obvious even though i went to some effort to prove that just to give you a feeling for how those things uh go you know that you know if you have programs if you have pascal and java say and thinking about what you can do mathematically in those i'm not talking about their ability to interface with you know windows and so on but just the mathematical capabilities the capability of doing mathematical calculations or functions with a pascal program or a java program you know it would be absurd to think there's some program that you can write in java that you can't write in pascal um or python and the reason is we know you can compile python into java and you can compile java back into python that tells you that the two systems of two programming languages are equivalent in power um that wasn't obvious that from the get-go to these folks so they um observed that uh all of the different efforts they came at formalizing algorithm all were equivalent to one another that was that that was a kind of a breakthrough moment when they realized that they're all that not all the ways that they've come up with and once they got the idea they realized all reasonable ways of doing it always going to be equivalent um and so that suggested that they've really captured this notion of algorithm by any one of those methods say a turing machine and that's what they took you can't prove that because algorithms are an intuitive notion but the fact that we're able to capture that in a formal way that's what we call today the church touring thesis okay um so um that is uh um so any of these methods uh capture the notion of algorithm as we described and that has had a big impact on mathematics uh so you know i just want to just give me one second here all right here's a check-in on this one so you know alan turing so here are some facts which may or may not be true about elementary um so uh now you get to pick all the ones that apply um based on your knowledge obviously this is not uh something this is more for fun but um or historical interest um but here is uh let's launch that poll see what how much you know about uh mr turing um check all that apply okay almost done please let's wrap it up i think that's everybody okay two seconds please get credit for doing this and polling in fact it's kind of interesting here uh um [Music] uh he you all do know that he was a code breaker uh he was a worked uh in fact did was part of a team i think led the team which broke the german code um during world war ii you know the turing test is a famous uh thing for how do you characterize when you have an intelligent machine um so he definitely worked in ai he worked in biology as well um he has a paper that less well known to computer to scientists but if you look him up um on wikipedia where i get all my information uh uh he actually and i knew this anyway uh he has a very famous paper influential paper on how um for example spots arise on um on tiger on leopards and stripes on tigers and so on um he gave a kind of mathematical model for that which actually proves actually is uh quite does capture things in an accurate way uh as was shown subsequent to that uh was imprisoned for being gay in fact as far as i know from his history he was not imprisoned for being gay um he was convicted of being gay um and he was given a choice to go to prison or to take chemicals to cure him from being gay and he opted not to go to prison and take the chemicals uh and uh sadly he committed suicide two years after that so um he was um treated very treated very badly by uh british society and british government uh despite having the the great the work that he had done um and um you might think that he has been honored by appearing on a british banknote that's also not true um uh but the good news is he's not uh currently on a british banknote but he's going to be on a british bank note starting next year um so that is uh um so that's uh you know along with um winston churchill and a number of other notable brits he's going to be on the 50 pound banknote okay um so let us continue um so as i mentioned uh the church thesis is important for mathematics um and it has to do with these hilbert problems i don't know how many of you have encountered that david hilbert was widely considered to be the perhaps the greatest mathematician of his day um and um you know every four years mathematicians get together for an international congress called the international congress mathematicians it was you know email that's been going on for over 100 years and he was invited to the 1900 uh meeting to give a talk about anything he wanted and what he decided to do in during that presentation is give it is present 23 uh problems that would be a challenge for mathematicians for the coming century um and um uh you know i we're running a little short on time so i'm not going to go into them some of these will talk about later uh but the tenth problem here is about algorithms the tenth problem on hilbert's list of called called hilbert's tenth problem is a problem about algorithms and it says given algorithm for solving what are called diaphantine equations um he didn't call it an algorithm you could call it some uh you know finite procedure or something like that but you know you what are the diaphantine equations i'm glad you asked uh diaphantine equations uh what are they well they're very simple they're just polynomial equations like you know does this polynomial equal some constant for example um but where you're looking for the solutions you know the polynomials are variables you're looking for the solutions but you're only allowing those solutions to be integers so here's an example um so here i give you this polynomial um here i'm setting it equal to 7 and i want to solve that equation so it has these three variables x y and z so you have to find a solution there but i'm only going to allow you to have plug-ins integers and in fact there is a solution in integers one two and minus two if you plug it in you'll see it works all right uh now the general problem that hilbert was addressing is suppose i give you a polynomial um let's say it said it it's set to zero so we're looking for roots of the polynomial but just some polynomial equation you can always put it in this form and i want to know does it have a solution in integers um and what i want what i'm looking for is a procedure which will tell me yes or no i want to know um give an algorithm to answer that question for a given polynomial or using the language of this course defining this in terms of a language uh decide this language give a give a turing machine which will decide yes or no for any given polynomial yes there is a solution in integers no there is no solution okay and um uh that is a um uh um that was what um hilbert asked in his tenth problem given algorithm now as we know now we took um 70 years to get the answer that um that there is no such algorithm it's an undecidable problem and we'll talk about that a little bit later in the term but um there is d is not a decidable language now um there was no hope of coming up with that answer in 1900 or even in 1910 because we didn't have a formal idea of what an algorithm is that had to wait until the church turing thesis told us that algorithms really are turing machines that there is a formal way of saying what an algorithm is and once you had that notion then you could prove that there is no turing machine but before that you had only this vague notion intuitive notion of what an algorithm is and so there was no hope of ever answering that because the in fact the answer is there is no such algorithm okay now um i'll give this as a little exercise to you we're a little bit running short on time but this language d is in fact a recognizable language so i would suggest you think about that offline um but basically you know you can try plugging in different values for these variables and if you ever find that you know the um that it evaluates to zero then you can accept but otherwise you just have to keep going and looking so showing recognizability is very simple but decidability is false okay now let's talk a little bit about encodings for turing machines um so we're going to be working encodings and turing machines so we're now going to be working with um turing machines going forward we'll go as i you know we're going to the input to those turing machines might be polynomials might be strings they might be other things too we might want to feed automata into the turing machines so the turing machine can answer questions about the automata well don't forget turing machines take as their input just strings so we have to think about how we're going to represent something more complicated like an automaton as a string and i'm not going to get into the details of that you could spell it out in detail um but i think that's not too interesting to get into that we're just going to have develop a notation that says if you have some object o it could be a polynomial it could be an automaton it could be a graph it could be whatever you're working with a table i'm going to write that o in these brackets to mean an encoding of that object into a string and then you can feed the string into the turing machine okay so that's how we're going to be thinking of presenting turing machines with more complicated objects and strings as input because we're just going to represent them as strings using some ways in which i'm sure you're all familiar i mean that's how you deal with representing stuff when you're you know how to write your programs anyway um but just to make it formal and that's why we're going to write it down in these brackets and if you have a list of several objects that you want to present together to a turing machine we'll just write them together within brackets like this now for writing turing machines down we're going to going forward we're going to be using high-level english descriptions um we're not going to be worrying about managing the like the stuff we've been doing up till now i'm not going to ask you to do it we're not going to do that anymore and i'm not going to ask you to do it managing where stuff goes on the tapes and all that stuff that's too low level because really now that we have the church turing thesis we're really interested in talking about algorithms we're not that interested in talking about turing machines per se with ancient algorithms um and what the power of computation is so um we're going to only be talking about touring machines now in a high higher level way and only when we need to prove something about uh capabilities we're going to come back to turing machines i'm going to be proving things you know about their limitations and so on so our notation for running turing machines is going to be basically going to put the turing machine inside these quotation marks and we're going to know that we could if in principle write out the touring machine in a precise way in terms of states and transition function and so on but we'll never actually go ahead and do that lengthy exercise um so quick check in here so one of the features well okay let me not give this away so if x and y are strings you know i want to now have a way of presenting two strings as a single string as input to my machine because i always think of my machine as getting a single input so would you suggest one way of combining two strings into one which we would be a good encoding is just to concatenate those two strings together is would it be that would would that be the way you would do it is that a good way to do it or not such a good way to do it um so let's uh um let's see here i can get to that next and think what you would want to have happen in a good encoding okay ready set sold uh share results yeah i think most of you got the idea that this is not a good way of combining two strings into one um because the problem is it's kind of ambiguous in a sense you know that you could if you combine two strings this way what's what's important in an encoding is that the machine when it gets the encoding it can decode them back into the original objects and if you're just going to be sticking the two strings together you don't know where one string ends and the next string begins and so it's not going to be a good way of combining things you you should find a little bit more clever way of either introducing another symbol or it's doing something a little bit more sophisticated which allow you to be able to do the decoding as well as the encoding in a unique way okay so uh getting back to that notation for a turing machine so here is the machine we've already seen once before for a to the kb to the kc to the k i would write it now more simply than managing all of the tapes that we had the first time around which we did last lecture i would just say i would give it an input w check if w is of the right form with the things in the right order reject if it's not count the number of a's b's and c's and w except if all the counts are equal and reject if not we're not going to be that's going to be good enough to be writing things at that higher level when you're going to be writing your algorithm descriptions you just have to make sure that whatever you're doing you can implement um you know you you can't make uh you know you you don't want to be doing tests which are impossible or doing infinite infinitely much work in one stage that's not good um but as long as it's clear that what you're doing is doing only a finite amount of work in every stage and it's something that you could really implement you can write it at a high in a high level um okay i wanted to spend a few minutes talking about problem set two but we a little bit running out of time here um uh in particular there was this problem number five where you show that a language is turing recognizable if and only if there's a decidable d um where c is turing recognizable where d is now a collection of pairs we've got some questions about the notation which i hopefully answered now so um c uh is a set of x's such that there exists a a string you can pair it with so that the string x y is in d um let me try to give you a picture of that uh um [Music] so i'm gonna think of d as a collection of pairs of strings and it might be helpful to think of the kind of you know on the axes here so if we have a pair of strings x y which i have not yet packaged into a single string i'm thinking of them as a pair of two objects at this moment uh think of d as so you know just the x part is just below here on the x-axis um d you might want to think just conceptually we want to think of it as a collection of pairs so it's like a subset of all of these pairs and c is all of the the x's that correspond to any pair in here um sometimes we call that the projection because it's all the things that are below or the shadow if you had a light sitting up here it's all the things that are kind of underneath a d um so i've written c here kind of a little thicker if you can see that so that's the c language and um what you're going to what there's a two directions here one is a lot easier than the other if i give you a decidable d and i want to test whether something's in c so i give you an x i want to know is it is x in c well you now have to test is there some y that i can pair with x where the pairs in d um so you know you can start looking for that y if you ever find one you know that x is in c there are infinitely many ways to look for but don't forget you know you're only looking for a recognizer for c so you know on the rejecting side you're allowed to go forever so i've kind of given you a way to think about the easy direction the hard direction um you need to think about um how are you going to come up with that decidable d if i give you c you have to find a decidable d and now you're you're bringing something down lower you're starting with a recognizing recognizable language and a decidable language which is going to sort of be counterpart to that in this sense is a simpler language and the way it becomes simpler is that y is going to help you determine whether x is in c um and i guess the thing that you should i'll leave you with is that maybe we can talk about this a little bit more on tuesday i'll try to leave it a little bit more time if you want to test if something's in c x is in c but i don't want to let you go forever anymore because i want to be a decidable language and i'm going to use why to help you what information would help you guarantee you get the right answer for x being in c but that you would have to be sure you're getting the right answer so it's you know why could just be the answer but then you know you don't know that that's you you have to be convinced that you you have the right answer you know why just saying what it is i mean you could well i could say that x is in d even when it's not true x isn't c even when that's not true but so one information would allow you to check that x is in c what would be helpful information to check the x's and c where you would avoid ever having to go forever a little bit too rushed here for that to be helpful anyway let's just uh conclude what we've done um just brief summary here um and i i i don't want to uh um keep you going keep you over time so i'll just leave this up on the board um i will uh um uh see if there's any so the lecture is over uh and you can take off uh as you wish um i will stick around for a couple of minutes i have another meeting soon um but uh i'll answer try to answer some chats uh all right um people are commenting about this movie about touring the imitation game uh you know if you haven't seen that i recommend it um um and uh is it a good idea to make y to be equal to the repeated loop string hmm i'm not sure about that uh thank you for the uh thank you everybody for sending your kind notes um is it church touring thesis the question is is it in the book yes it's in the book kind of similar to what i've already said but yes okay this is a good question here the church turing thesis somebody asked me is it proved in the book there's nothing to prove the church touring thesis is an equivalence between the intuitive and the formal you can't prove something like that you can just make it's really in a sense a hypothesis that the only thing you'll ever be able to compute um is something that you can do with a turing machine um i mean that has something to do with the nature of the physical world and i'm sure i don't really think that's what people had in mind it's that you know the the kinds of things that we normally think of being able to do with a with a rest with a procedure mathematically it's exactly the kinds of things that you can do with the turing machine um so somebody's pointing out well trying to be helpful here maybe these folks are asking about equivalence of various different computation models being proved in the book not that church drink these sits itself well maybe that's the the things that we proved in lecture are also proved in the book um about the equivalence of single tape machines multitape machines non-deterministic machines um those are all proved in the book we don't those i mean there's lots of models out there so we can spend a lot of time proving equivalences um and there are books that do spend a lot of time proving equivalences but i think you know uh we yeah i i think what we've given is probably enough um uh so somebody that's a good question if if you give an algorithm okay if you give an algorithm you don't need to give the actual machine uh going forward unless somebody unless you're explicitly asked to um uh but yes you don't you don't have to give um the states and transitions anymore um okay so it's a little after four i think we're going to i think i'm going to move on to my next meeting uh so i uh thank you for being here and i will see you on tuesday you 3 00:00:31,509 --> 00:00:33,670 4 00:00:33,670 --> 00:00:33,680 5 00:00:33,680 --> 00:00:35,270 6 00:00:35,270 --> 00:00:35,280 7 00:00:35,280 --> 00:00:36,310 8 00:00:36,310 --> 00:00:36,320 9 00:00:36,320 --> 00:00:38,790 10 00:00:38,790 --> 00:00:41,030 11 00:00:41,030 --> 00:00:41,040 12 00:00:41,040 --> 00:00:43,510 13 00:00:43,510 --> 00:00:45,590 14 00:00:45,590 --> 00:00:46,950 15 00:00:46,950 --> 00:00:46,960 16 00:00:46,960 --> 00:00:49,110 17 00:00:49,110 --> 00:00:50,790 18 00:00:50,790 --> 00:00:53,270 19 00:00:53,270 --> 00:00:55,910 20 00:00:55,910 --> 00:00:59,670 21 00:00:59,670 --> 00:01:01,430 22 00:01:01,430 --> 00:01:03,670 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464 00:15:42,230 --> 00:15:46,150 465 00:15:46,150 --> 00:15:46,160 466 00:15:46,160 --> 00:15:48,230 467 00:15:48,230 --> 00:15:49,670 468 00:15:49,670 --> 00:15:52,870 469 00:15:52,870 --> 00:15:54,949 470 00:15:54,949 --> 00:15:54,959 471 00:15:54,959 --> 00:15:56,069 472 00:15:56,069 --> 00:15:57,189 473 00:15:57,189 --> 00:16:00,550 474 00:16:00,550 --> 00:16:03,670 475 00:16:03,670 --> 00:16:05,590 476 00:16:05,590 --> 00:16:07,749 477 00:16:07,749 --> 00:16:09,350 478 00:16:09,350 --> 00:16:11,509 479 00:16:11,509 --> 00:16:13,269 480 00:16:13,269 --> 00:16:15,110 481 00:16:15,110 --> 00:16:17,189 482 00:16:17,189 --> 00:16:17,199 483 00:16:17,199 --> 00:16:18,150 484 00:16:18,150 --> 00:16:18,160 485 00:16:18,160 --> 00:16:19,110 486 00:16:19,110 --> 00:16:22,470 487 00:16:22,470 --> 00:16:24,230 488 00:16:24,230 --> 00:16:26,389 489 00:16:26,389 --> 00:16:27,829 490 00:16:27,829 --> 00:16:30,310 491 00:16:30,310 --> 00:16:31,829 492 00:16:31,829 --> 00:16:35,509 493 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--> 00:17:30,630 523 00:17:30,630 --> 00:17:30,640 524 00:17:30,640 --> 00:17:32,710 525 00:17:32,710 --> 00:17:37,029 526 00:17:37,029 --> 00:17:38,950 527 00:17:38,950 --> 00:17:41,110 528 00:17:41,110 --> 00:17:41,120 529 00:17:41,120 --> 00:17:41,830 530 00:17:41,830 --> 00:17:43,750 531 00:17:43,750 --> 00:17:44,870 532 00:17:44,870 --> 00:17:47,029 533 00:17:47,029 --> 00:17:49,350 534 00:17:49,350 --> 00:17:51,909 535 00:17:51,909 --> 00:17:55,110 536 00:17:55,110 --> 00:17:57,590 537 00:17:57,590 --> 00:17:59,350 538 00:17:59,350 --> 00:18:02,630 539 00:18:02,630 --> 00:18:04,390 540 00:18:04,390 --> 00:18:07,510 541 00:18:07,510 --> 00:18:09,830 542 00:18:09,830 --> 00:18:09,840 543 00:18:09,840 --> 00:18:12,230 544 00:18:12,230 --> 00:18:14,470 545 00:18:14,470 --> 00:18:16,390 546 00:18:16,390 --> 00:18:18,470 547 00:18:18,470 --> 00:18:20,470 548 00:18:20,470 --> 00:18:22,470 549 00:18:22,470 --> 00:18:25,990 550 00:18:25,990 --> 00:18:26,000 551 00:18:26,000 --> 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581 00:19:26,870 --> 00:19:28,710 582 00:19:28,710 --> 00:19:28,720 583 00:19:28,720 --> 00:19:29,669 584 00:19:29,669 --> 00:19:33,029 585 00:19:33,029 --> 00:19:34,789 586 00:19:34,789 --> 00:19:34,799 587 00:19:34,799 --> 00:19:36,230 588 00:19:36,230 --> 00:19:39,430 589 00:19:39,430 --> 00:19:40,950 590 00:19:40,950 --> 00:19:43,990 591 00:19:43,990 --> 00:19:46,310 592 00:19:46,310 --> 00:19:48,230 593 00:19:48,230 --> 00:19:50,789 594 00:19:50,789 --> 00:19:53,190 595 00:19:53,190 --> 00:19:53,200 596 00:19:53,200 --> 00:19:54,230 597 00:19:54,230 --> 00:19:56,150 598 00:19:56,150 --> 00:19:58,070 599 00:19:58,070 --> 00:19:59,590 600 00:19:59,590 --> 00:20:02,789 601 00:20:02,789 --> 00:20:04,789 602 00:20:04,789 --> 00:20:07,190 603 00:20:07,190 --> 00:20:08,710 604 00:20:08,710 --> 00:20:10,390 605 00:20:10,390 --> 00:20:10,400 606 00:20:10,400 --> 00:20:11,430 607 00:20:11,430 --> 00:20:13,029 608 00:20:13,029 --> 00:20:13,039 609 00:20:13,039 --> 00:20:14,149 610 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--> 00:21:05,669 640 00:21:05,669 --> 00:21:07,510 641 00:21:07,510 --> 00:21:09,430 642 00:21:09,430 --> 00:21:11,110 643 00:21:11,110 --> 00:21:13,270 644 00:21:13,270 --> 00:21:14,950 645 00:21:14,950 --> 00:21:17,830 646 00:21:17,830 --> 00:21:19,590 647 00:21:19,590 --> 00:21:21,669 648 00:21:21,669 --> 00:21:23,590 649 00:21:23,590 --> 00:21:24,870 650 00:21:24,870 --> 00:21:27,110 651 00:21:27,110 --> 00:21:30,070 652 00:21:30,070 --> 00:21:32,070 653 00:21:32,070 --> 00:21:35,029 654 00:21:35,029 --> 00:21:35,039 655 00:21:35,039 --> 00:21:35,830 656 00:21:35,830 --> 00:21:37,510 657 00:21:37,510 --> 00:21:39,990 658 00:21:39,990 --> 00:21:41,510 659 00:21:41,510 --> 00:21:42,950 660 00:21:42,950 --> 00:21:44,549 661 00:21:44,549 --> 00:21:46,870 662 00:21:46,870 --> 00:21:48,549 663 00:21:48,549 --> 00:21:51,190 664 00:21:51,190 --> 00:21:51,200 665 00:21:51,200 --> 00:21:53,350 666 00:21:53,350 --> 00:21:54,870 667 00:21:54,870 --> 00:21:57,270 668 00:21:57,270 --> 00:21:59,029 669 00:21:59,029 --> 00:22:01,590 670 00:22:01,590 --> 00:22:03,510 671 00:22:03,510 --> 00:22:05,270 672 00:22:05,270 --> 00:22:05,280 673 00:22:05,280 --> 00:22:06,230 674 00:22:06,230 --> 00:22:08,789 675 00:22:08,789 --> 00:22:11,110 676 00:22:11,110 --> 00:22:11,120 677 00:22:11,120 --> 00:22:12,710 678 00:22:12,710 --> 00:22:15,270 679 00:22:15,270 --> 00:22:16,710 680 00:22:16,710 --> 00:22:18,950 681 00:22:18,950 --> 00:22:21,590 682 00:22:21,590 --> 00:22:23,590 683 00:22:23,590 --> 00:22:25,669 684 00:22:25,669 --> 00:22:27,350 685 00:22:27,350 --> 00:22:29,669 686 00:22:29,669 --> 00:22:31,990 687 00:22:31,990 --> 00:22:33,590 688 00:22:33,590 --> 00:22:33,600 689 00:22:33,600 --> 00:22:35,430 690 00:22:35,430 --> 00:22:37,990 691 00:22:37,990 --> 00:22:39,510 692 00:22:39,510 --> 00:22:41,190 693 00:22:41,190 --> 00:22:42,870 694 00:22:42,870 --> 00:22:44,870 695 00:22:44,870 --> 00:22:46,630 696 00:22:46,630 --> 00:22:48,789 697 00:22:48,789 --> 00:22:50,789 698 00:22:50,789 --> 00:22:50,799 699 00:22:50,799 --> 00:22:51,750 700 00:22:51,750 --> 00:22:51,760 701 00:22:51,760 --> 00:22:52,870 702 00:22:52,870 --> 00:22:55,110 703 00:22:55,110 --> 00:22:55,120 704 00:22:55,120 --> 00:22:56,070 705 00:22:56,070 --> 00:22:56,080 706 00:22:56,080 --> 00:22:57,110 707 00:22:57,110 --> 00:22:59,350 708 00:22:59,350 --> 00:23:02,870 709 00:23:02,870 --> 00:23:06,710 710 00:23:06,710 --> 00:23:09,830 711 00:23:09,830 --> 00:23:11,909 712 00:23:11,909 --> 00:23:15,029 713 00:23:15,029 --> 00:23:17,510 714 00:23:17,510 --> 00:23:19,590 715 00:23:19,590 --> 00:23:21,110 716 00:23:21,110 --> 00:23:23,590 717 00:23:23,590 --> 00:23:25,990 718 00:23:25,990 --> 00:23:28,470 719 00:23:28,470 --> 00:23:30,149 720 00:23:30,149 --> 00:23:31,669 721 00:23:31,669 --> 00:23:33,510 722 00:23:33,510 --> 00:23:35,350 723 00:23:35,350 --> 00:23:37,669 724 00:23:37,669 --> 00:23:39,669 725 00:23:39,669 --> 00:23:41,350 726 00:23:41,350 --> 00:23:43,029 727 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--> 00:24:34,230 757 00:24:34,230 --> 00:24:36,549 758 00:24:36,549 --> 00:24:38,310 759 00:24:38,310 --> 00:24:41,029 760 00:24:41,029 --> 00:24:43,430 761 00:24:43,430 --> 00:24:44,549 762 00:24:44,549 --> 00:24:47,669 763 00:24:47,669 --> 00:24:49,830 764 00:24:49,830 --> 00:24:51,190 765 00:24:51,190 --> 00:24:54,470 766 00:24:54,470 --> 00:24:57,590 767 00:24:57,590 --> 00:24:59,590 768 00:24:59,590 --> 00:24:59,600 769 00:24:59,600 --> 00:25:02,310 770 00:25:02,310 --> 00:25:04,230 771 00:25:04,230 --> 00:25:06,470 772 00:25:06,470 --> 00:25:09,510 773 00:25:09,510 --> 00:25:11,430 774 00:25:11,430 --> 00:25:14,630 775 00:25:14,630 --> 00:25:16,149 776 00:25:16,149 --> 00:25:16,159 777 00:25:16,159 --> 00:25:17,590 778 00:25:17,590 --> 00:25:20,390 779 00:25:20,390 --> 00:25:20,400 780 00:25:20,400 --> 00:25:21,269 781 00:25:21,269 --> 00:25:22,549 782 00:25:22,549 --> 00:25:24,390 783 00:25:24,390 --> 00:25:27,029 784 00:25:27,029 --> 00:25:29,110 785 00:25:29,110 --> 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815 00:26:33,750 --> 00:26:35,830 816 00:26:35,830 --> 00:26:38,470 817 00:26:38,470 --> 00:26:40,149 818 00:26:40,149 --> 00:26:41,750 819 00:26:41,750 --> 00:26:46,710 820 00:26:46,710 --> 00:26:49,750 821 00:26:49,750 --> 00:26:52,070 822 00:26:52,070 --> 00:26:53,830 823 00:26:53,830 --> 00:26:56,230 824 00:26:56,230 --> 00:26:56,240 825 00:26:56,240 --> 00:26:57,269 826 00:26:57,269 --> 00:26:57,279 827 00:26:57,279 --> 00:26:58,870 828 00:26:58,870 --> 00:27:00,789 829 00:27:00,789 --> 00:27:02,789 830 00:27:02,789 --> 00:27:04,310 831 00:27:04,310 --> 00:27:08,630 832 00:27:08,630 --> 00:27:08,640 833 00:27:08,640 --> 00:27:09,430 834 00:27:09,430 --> 00:27:11,990 835 00:27:11,990 --> 00:27:13,590 836 00:27:13,590 --> 00:27:16,070 837 00:27:16,070 --> 00:27:19,110 838 00:27:19,110 --> 00:27:20,710 839 00:27:20,710 --> 00:27:22,630 840 00:27:22,630 --> 00:27:24,149 841 00:27:24,149 --> 00:27:25,669 842 00:27:25,669 --> 00:27:27,269 843 00:27:27,269 --> 00:27:29,190 844 00:27:29,190 --> 00:27:30,710 845 00:27:30,710 --> 00:27:31,990 846 00:27:31,990 --> 00:27:33,669 847 00:27:33,669 --> 00:27:35,110 848 00:27:35,110 --> 00:27:37,350 849 00:27:37,350 --> 00:27:41,669 850 00:27:41,669 --> 00:27:44,630 851 00:27:44,630 --> 00:27:46,230 852 00:27:46,230 --> 00:27:48,549 853 00:27:48,549 --> 00:27:52,630 854 00:27:52,630 --> 00:27:56,070 855 00:27:56,070 --> 00:27:57,830 856 00:27:57,830 --> 00:27:59,909 857 00:27:59,909 --> 00:28:01,669 858 00:28:01,669 --> 00:28:04,149 859 00:28:04,149 --> 00:28:06,870 860 00:28:06,870 --> 00:28:08,310 861 00:28:08,310 --> 00:28:10,070 862 00:28:10,070 --> 00:28:10,080 863 00:28:10,080 --> 00:28:11,269 864 00:28:11,269 --> 00:28:13,830 865 00:28:13,830 --> 00:28:13,840 866 00:28:13,840 --> 00:28:15,350 867 00:28:15,350 --> 00:28:16,870 868 00:28:16,870 --> 00:28:19,830 869 00:28:19,830 --> 00:28:22,310 870 00:28:22,310 --> 00:28:25,350 871 00:28:25,350 --> 00:28:29,190 872 00:28:29,190 --> 00:28:30,789 873 00:28:30,789 --> 00:28:33,350 874 00:28:33,350 --> 00:28:33,360 875 00:28:33,360 --> 00:28:35,190 876 00:28:35,190 --> 00:28:36,870 877 00:28:36,870 --> 00:28:38,630 878 00:28:38,630 --> 00:28:40,870 879 00:28:40,870 --> 00:28:40,880 880 00:28:40,880 --> 00:28:42,070 881 00:28:42,070 --> 00:28:42,080 882 00:28:42,080 --> 00:28:43,350 883 00:28:43,350 --> 00:28:43,360 884 00:28:43,360 --> 00:28:44,389 885 00:28:44,389 --> 00:28:47,110 886 00:28:47,110 --> 00:28:47,120 887 00:28:47,120 --> 00:28:47,990 888 00:28:47,990 --> 00:28:50,389 889 00:28:50,389 --> 00:28:51,750 890 00:28:51,750 --> 00:28:53,190 891 00:28:53,190 --> 00:28:55,110 892 00:28:55,110 --> 00:28:57,909 893 00:28:57,909 --> 00:29:00,230 894 00:29:00,230 --> 00:29:02,789 895 00:29:02,789 --> 00:29:06,950 896 00:29:06,950 --> 00:29:08,950 897 00:29:08,950 --> 00:29:10,630 898 00:29:10,630 --> 00:29:13,110 899 00:29:13,110 --> 00:29:15,110 900 00:29:15,110 --> 00:29:17,669 901 00:29:17,669 --> 00:29:19,909 902 00:29:19,909 --> 00:29:23,029 903 00:29:23,029 --> 00:29:24,950 904 00:29:24,950 --> 00:29:28,789 905 00:29:28,789 --> 00:29:32,230 906 00:29:32,230 --> 00:29:34,789 907 00:29:34,789 --> 00:29:36,470 908 00:29:36,470 --> 00:29:37,669 909 00:29:37,669 --> 00:29:43,269 910 00:29:43,269 --> 00:29:44,789 911 00:29:44,789 --> 00:29:46,549 912 00:29:46,549 --> 00:29:48,470 913 00:29:48,470 --> 00:29:48,480 914 00:29:48,480 --> 00:29:49,830 915 00:29:49,830 --> 00:29:52,389 916 00:29:52,389 --> 00:29:54,470 917 00:29:54,470 --> 00:29:59,110 918 00:29:59,110 --> 00:30:01,590 919 00:30:01,590 --> 00:30:01,600 920 00:30:01,600 --> 00:30:02,470 921 00:30:02,470 --> 00:30:04,230 922 00:30:04,230 --> 00:30:04,240 923 00:30:04,240 --> 00:30:05,110 924 00:30:05,110 --> 00:30:06,630 925 00:30:06,630 --> 00:30:07,990 926 00:30:07,990 --> 00:30:10,389 927 00:30:10,389 --> 00:30:12,389 928 00:30:12,389 --> 00:30:14,470 929 00:30:14,470 --> 00:30:14,480 930 00:30:14,480 --> 00:30:15,669 931 00:30:15,669 --> 00:30:18,470 932 00:30:18,470 --> 00:30:20,070 933 00:30:20,070 --> 00:30:20,080 934 00:30:20,080 --> 00:30:23,029 935 00:30:23,029 --> 00:30:23,039 936 00:30:23,039 --> 00:30:24,710 937 00:30:24,710 --> 00:30:26,710 938 00:30:26,710 --> 00:30:27,990 939 00:30:27,990 --> 00:30:28,000 940 00:30:28,000 --> 00:30:30,070 941 00:30:30,070 --> 00:30:32,470 942 00:30:32,470 --> 00:30:34,230 943 00:30:34,230 --> 00:30:35,990 944 00:30:35,990 --> 00:30:39,190 945 00:30:39,190 --> 00:30:41,110 946 00:30:41,110 --> 00:30:42,389 947 00:30:42,389 --> 00:30:45,029 948 00:30:45,029 --> 00:30:47,590 949 00:30:47,590 --> 00:30:50,710 950 00:30:50,710 --> 00:30:50,720 951 00:30:50,720 --> 00:30:51,430 952 00:30:51,430 --> 00:30:53,350 953 00:30:53,350 --> 00:30:55,830 954 00:30:55,830 --> 00:30:57,669 955 00:30:57,669 --> 00:30:57,679 956 00:30:57,679 --> 00:30:59,190 957 00:30:59,190 --> 00:31:01,990 958 00:31:01,990 --> 00:31:04,470 959 00:31:04,470 --> 00:31:07,110 960 00:31:07,110 --> 00:31:09,909 961 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--> 00:32:08,559 991 00:32:08,559 --> 00:32:09,509 992 00:32:09,509 --> 00:32:10,549 993 00:32:10,549 --> 00:32:12,470 994 00:32:12,470 --> 00:32:16,710 995 00:32:16,710 --> 00:32:16,720 996 00:32:16,720 --> 00:32:17,669 997 00:32:17,669 --> 00:32:20,310 998 00:32:20,310 --> 00:32:22,389 999 00:32:22,389 --> 00:32:23,750 1000 00:32:23,750 --> 00:32:25,669 1001 00:32:25,669 --> 00:32:28,230 1002 00:32:28,230 --> 00:32:30,549 1003 00:32:30,549 --> 00:32:32,389 1004 00:32:32,389 --> 00:32:34,710 1005 00:32:34,710 --> 00:32:36,230 1006 00:32:36,230 --> 00:32:37,669 1007 00:32:37,669 --> 00:32:39,430 1008 00:32:39,430 --> 00:32:40,710 1009 00:32:40,710 --> 00:32:43,110 1010 00:32:43,110 --> 00:32:45,190 1011 00:32:45,190 --> 00:32:49,750 1012 00:32:49,750 --> 00:32:51,430 1013 00:32:51,430 --> 00:32:53,110 1014 00:32:53,110 --> 00:32:55,350 1015 00:32:55,350 --> 00:32:55,360 1016 00:32:55,360 --> 00:32:57,430 1017 00:32:57,430 --> 00:32:59,430 1018 00:32:59,430 --> 00:33:01,590 1019 00:33:01,590 --> 00:33:03,909 1020 00:33:03,909 --> 00:33:06,630 1021 00:33:06,630 --> 00:33:08,950 1022 00:33:08,950 --> 00:33:13,190 1023 00:33:13,190 --> 00:33:15,190 1024 00:33:15,190 --> 00:33:16,870 1025 00:33:16,870 --> 00:33:19,190 1026 00:33:19,190 --> 00:33:20,789 1027 00:33:20,789 --> 00:33:22,630 1028 00:33:22,630 --> 00:33:25,509 1029 00:33:25,509 --> 00:33:25,519 1030 00:33:25,519 --> 00:33:26,389 1031 00:33:26,389 --> 00:33:28,470 1032 00:33:28,470 --> 00:33:30,310 1033 00:33:30,310 --> 00:33:32,870 1034 00:33:32,870 --> 00:33:34,630 1035 00:33:34,630 --> 00:33:34,640 1036 00:33:34,640 --> 00:33:35,430 1037 00:33:35,430 --> 00:33:37,509 1038 00:33:37,509 --> 00:33:39,590 1039 00:33:39,590 --> 00:33:40,789 1040 00:33:40,789 --> 00:33:42,549 1041 00:33:42,549 --> 00:33:44,549 1042 00:33:44,549 --> 00:33:46,470 1043 00:33:46,470 --> 00:33:48,710 1044 00:33:48,710 --> 00:33:50,070 1045 00:33:50,070 --> 00:33:51,669 1046 00:33:51,669 --> 00:33:53,830 1047 00:33:53,830 --> 00:33:57,830 1048 00:33:57,830 --> 00:33:59,269 1049 00:33:59,269 --> 00:34:01,350 1050 00:34:01,350 --> 00:34:03,830 1051 00:34:03,830 --> 00:34:05,669 1052 00:34:05,669 --> 00:34:05,679 1053 00:34:05,679 --> 00:34:06,630 1054 00:34:06,630 --> 00:34:06,640 1055 00:34:06,640 --> 00:34:09,190 1056 00:34:09,190 --> 00:34:11,270 1057 00:34:11,270 --> 00:34:12,790 1058 00:34:12,790 --> 00:34:14,470 1059 00:34:14,470 --> 00:34:14,480 1060 00:34:14,480 --> 00:34:17,349 1061 00:34:17,349 --> 00:34:20,470 1062 00:34:20,470 --> 00:34:22,470 1063 00:34:22,470 --> 00:34:22,480 1064 00:34:22,480 --> 00:34:23,990 1065 00:34:23,990 --> 00:34:25,990 1066 00:34:25,990 --> 00:34:28,550 1067 00:34:28,550 --> 00:34:30,069 1068 00:34:30,069 --> 00:34:32,829 1069 00:34:32,829 --> 00:34:32,839 1070 00:34:32,839 --> 00:34:34,389 1071 00:34:34,389 --> 00:34:36,629 1072 00:34:36,629 --> 00:34:38,310 1073 00:34:38,310 --> 00:34:40,629 1074 00:34:40,629 --> 00:34:42,389 1075 00:34:42,389 --> 00:34:43,909 1076 00:34:43,909 --> 00:34:45,669 1077 00:34:45,669 --> 00:34:47,829 1078 00:34:47,829 --> 00:34:49,430 1079 00:34:49,430 --> 00:34:50,950 1080 00:34:50,950 --> 00:34:53,750 1081 00:34:53,750 --> 00:34:55,589 1082 00:34:55,589 --> 00:34:58,390 1083 00:34:58,390 --> 00:35:00,150 1084 00:35:00,150 --> 00:35:01,670 1085 00:35:01,670 --> 00:35:04,870 1086 00:35:04,870 --> 00:35:07,589 1087 00:35:07,589 --> 00:35:09,670 1088 00:35:09,670 --> 00:35:11,910 1089 00:35:11,910 --> 00:35:11,920 1090 00:35:11,920 --> 00:35:12,710 1091 00:35:12,710 --> 00:35:14,710 1092 00:35:14,710 --> 00:35:16,630 1093 00:35:16,630 --> 00:35:19,270 1094 00:35:19,270 --> 00:35:21,990 1095 00:35:21,990 --> 00:35:23,349 1096 00:35:23,349 --> 00:35:25,030 1097 00:35:25,030 --> 00:35:27,030 1098 00:35:27,030 --> 00:35:29,349 1099 00:35:29,349 --> 00:35:31,990 1100 00:35:31,990 --> 00:35:33,829 1101 00:35:33,829 --> 00:35:35,270 1102 00:35:35,270 --> 00:35:37,190 1103 00:35:37,190 --> 00:35:38,950 1104 00:35:38,950 --> 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