1 00:00:24,880 --> 00:00:26,630 hi everybody can you hear me yes good um welcome back uh to um to the course and um so we are now at lecture 10 and um uh i let's see what have we been doing we've been talking about um undecidability um so we introduced last time uh reducibilities and mapping reducibilities for proving various problems are undecidable and this time we're going to and today we're going to introduce a a more sophisticated method for proving undecidability uh using reduceabilities and that's called the computation history method and and that's pretty much what's used in all of the you know more serious cases where people have proved problems undecidable it's it's pretty much a widespread uh uh method for uh especially you know it's not an active area of research these days but at times when people were proving problems undecidable this was frequently a method that was used um so we'll go over that today and we'll prove um a few examples of problems undecidable using that method all right so why don't we get started um uh so first of all most importantly you should remember how we you how we prove um problems are undecidable in the first place and that is by showing um some other undecidable problem that we already know is undecidable is reducible to the problem of interest so typically you might be reducing atm um which is a problem that we started with and showed to be undecidable by the diagonalization method um or might be some other problem that we've subsequently shown to be undecidable and you take that problem and you reduce it to the problem you're trying to show undecidable okay oops let me fix that sorry um so um so this is the thing to keep in mind to prove a language is undecidable show that atm or some other known undecidable language is reducible to the problem b that you have at hand that you're trying to show undecidable all right so we are now going to um um [Music] first i'm going to start off by remembering um hilbert's 10th problem which we discussed briefly a few lectures back um and as you may remember so that's a problem where uh you're given say some polynomial you know over several variables you know like three x squared y minus 15z plus 2xz equals 5. and you want to solve that problem you want to find a solution that solves that equation but you require that the solution involves only integers so those are so called diaphantine problems because the requirement is that the solution being integers and hilbert's problem was to ask for a an algorithm not the language that he used but doesn't matter he essentially he asked for an algorithm to test whether a polynomial has a solution in integers or the polynomial equation has a solution in integers and um as uh um uh um as we now know that problem was posed back in 1900 it took 71 years to find the solution but the answer is no there is no there is no such algorithm we talked about this when we were discussing the church turing thesis um and um the method showing that there is no algorithm is by a reduction from atm to that problem so one can use exactly the method that we're using today to prove that problem is undecidable about polynomials the only thing is is that the reduction that single reduction would take the entire semester um and i know that's in fact correct because when i was a graduate student um there was a woman in the math department at berkeley where i where i studied and she uh the whole the whole course was to go through the proof of hilbert's tenth problem um of the solution to hillary problem the the proof of the undecidability of the uh of the solution uh of the undecidability of testing whether a polynomial has an integral solution um so um the thing is that involves some fairly hairy number theory in order to um to come up with the right polynomials to basically simulate the turing machine which so that's kind of getting ahead of ourselves that's what we're going to be doing today um but we're going to instead of looking at that problem because we don't have a whole semester to spend on it um we're going to look at a toy problem instead where there's no number theory but it still has the same basic underlying idea um to that that was employed in um the solution to hilbert's tenth problem um so that toy problem is called the post correspondence problem and it's going to have some other utility for us well the main reason we're studying it is just to illustrate the method um uh so and the method is going to be as i've been saying is this computation history method okay so one we will first talk about uh this post-correspondence problem because that's very easy to understand and then we'll spend a little time introducing that method um all right so um the uh the post the post correspondence problem is as follows you're given uh a bunch of pairs of strings um so here is one pair t1 and b1 and i'm writing them as dominoes i'm calling them dominoes because i'm writing one pair one one of the strings on top of the other string in fact i'm using t and b for top and bottom here so this is t1 is the top string b1 is the bottom string in the first domino and then in the second domino we have two other strings to a top and a bottom and so on um so we have these here uh this collection of dominoes um as sort of the legal dominoes um and what the goal is to find a sequence of those dominoes um uh so you want to pick from these dominoes here you're allowed to repeat dominoes so you want to pick a sequence of those dominoes line them up next to each other and so that the string that you get by reading along the top all of the t's together is going to be exactly the same as the string you get by reading along all the bottoms i'll give you an example um so here to write it down a little bit more formally so uh what i'm calling a match is a sequence of these dominoes so it's t i1 ti2 as well it's it's i1 is the first domino i2 is the second domino and so on and then you what you're going to have is all the top strings concatenated concatenated together is exactly the same as what you get by concatenating together all the bottom strings um so here's an example and i think it'll become clear if you didn't understand it so far but it'll be completely clear from the example i hope so here is a bunch of dominoes um four um and um what i wanna know is is there some selection of these dominoes again you can repeat dominoes otherwise it would be not an interesting problem so you can repeat these as many times as you like and i want to pick up pick these dominoes and line them up so that what you get by reading along the top is the same as what you get by reading along the bottom so first of all if you just stare at this and you're trying to make a match you'll see that there's only one possibility possible way to start this um you know for example if you started with the third domino here b a over a a it would fail immediately because um you know the b and the a are different um and and if you you use the second domino as your starting point that would fail the match as well because a a has to would have to e um be the the the top string would start with a a the bottom string would start with a b and they could never be equal um so by looking at this you can see that the only possible choice that you have is to start with the first domino okay so i'm going to start to build the domino for you just so you can see it so here's the very the the match as um so you can see um the process so the match starts with the very first domino and the way i'm going to write it is i'm going to take the dominoes and kind of skew them so that you can see how the top and bottom strings are lining up um but they're just mouth the same dominoes they've just been i changed the shape so this is the first domino a b over aba now uh the second domino in my and my match it's gotta start with an a as the leftmost symbol on the top string so that would rule out this one for example um but there might be several choices and so it's not exactly obvious which one you should take what i'm going to suggest is we take this one here so we take a a over a b a you see it a a over a b a it's just you know written i'm just skewing uh skewing it so that um i can line up the a over a the b over b the a over a and so on so i'm starting to build this concatenation of the top being equal to the concatenation of the bottom following me so far i hope so now um what's what's next so i need to have something where the top string is going to start with a ba in fact there's only one choice for that so it's clearly going to have to be this domino is going to be next so it's going to put a ba up here but then it's going to a force an a a down below because that's what this domino does um so this ba matches with that now we have the aaa to deal with um so we're going to reuse this domino because that's the only one that starts with an aa so we're going to reuse that one now we have a a and aba now we need something that starts with aba i see two choices it could be this one because this is consistent with a it's at least it captures the a b we could we could try putting this one over here but a better choice would it be would be to use this one because if we use this one we finish the match and therefore we have found what we're looking for um and this this this collection of dominoes here has a mat has a match now it's not always obvious some some collections of dominoes may not have a match and so the computational problem is is there a match um and the theorem that we're going to prove today is that this problem is undecidable simple as it is simple little combinatorial problem of you know uh trying to put together different um uh different tiles like that to form form a match uh there was no algorithm for solving that problem um and and i think this is kind of interesting because it's the first example of a problem that we have where the underlying question does not really seem to have anything to do with computation you know you might argue that atm being undecidable is perhaps a little unsurprising after the fact because you know it's a problem about touring machines so you can imagine turing machines are going to have a tough time answering it um but here's a problem just about strings um and we're going to show it's undecidable so just to formulate this as a language i'm going to call the pcp language is the collection of these um uh pcp uh problems pcp instances which are just collect themselves each one is a collection of dominoes so it's the language of all uh collections of these dominoes where there is a match so it's all of the pcp problems where you can solve them with a match and uh we're going to prove that it's undecidable by showing that atm is reducible to that pcp language um now uh before we do that i'm going to have to i'm going to explain to you what the computation history method is in the first place that we're going to be using uh so before we do that here's our first check-in for the day just to make sure you're following me about uh what the um what this pcp problem is i am going to give you a question to uh test is there a match with these three dominoes so what you think about that i mean the main thing i'm i want to make sure you understand is what a match is in the first place this is not a super hard problem uh to tell whether or not there's a match okay um five seconds already okay closing it out um okay um so uh for those of you who said there is a match i want you to exhibit exhibit the match because in fact there is no such map there is no match um and uh i mean you could get by fiddling around with it i think if you try to find a match you'll pretty quickly uh see that it's you're gonna get um i think what happens is you kind of get into if you try to build build a match with this particular setup you're just going to get stuck sort of repeating yourself and um one point here that you may have missed is that a match has to be a finite sequence so if you were thinking about there there is there is a kind of an infinite match uh that you can build with this uh this this set of dominoes but that's not allowed we're only allowing finite matches and so that may they might change your answer as a result but too late um so um and one way to see that you can't have a uh a match in this problem is that you know you're going to have to start you you you know neither of these two are possible starting points so this one is going to be clearly the only chance that you're going to have as a match uh the first domino here but then once you put this one down um the bottom one the bottom string the the bottom b's that you're going to get the bottom string is going to be longer than the top string and then the other two are going to be have the same length um so you're never going to have the bottom uh the top be the same length as the bottom the bottom is always going to be longer no matter how many uh more dominoes you lay out so this there's no chance of there being a match here um okay so now we're gonna um uh work our way up to introducing this computation history method [Music] um um first let me define something for a turing machine that i'm going to call a configuration um configuration is just a snapshot of the turing machine in the middle of its computation so if you're running the turing machine and you just stop it at some moment it's going to be in a certain state the head is going to be on a certain position and the tape is going to have a certain contents and with that information you can then um continue the copy the computation of the turing machine that's a full set of the information that you need about the turing machine that tells you everything about its cut its computation at that moment in time the state head position and tape contents and so that we're calling that that information together uh state head position tape contents we're calling that a configuration it's a fairly um basic notion i mean if you if you program if you have something if you have like the states of all the variables and where the current execution of the program is at a moment in time same idea um okay so in terms of a picture here here is the turing machine imagine it's in state q3 um the head position is in the sixth position on the tape so p would be six um because it's in the sixth position and the tape contents is going to be this bunch of a's followed by this bunch of b's so i would write it down like this states in q3 head position number six this is the contents of the tape um now what we're going to often do for convenience in using this notion in proofs is that we're going to want to represent a con a configuration as a string um in a particular way that's going to be that's going to this is this concept is going to come up kind of again and again during the during during uh the course and so it's going to be handy to have a particular way of writing down configurations um you know for doing um proofs about them and so the way we're going to write them down is i think it's just maybe i can just put it right up here like this um we're going to write down the the symbols of the tape but what we're going to do is stick in the middle of those symbols the the state of the machine immediately to the left of the position that the machine is currently reading so you can imagine you know the the head here which is coming you know imagine the state is kind of where the head is it's pointing at the symbol immediately to its right so this is just another way of writing down a configuration here's the tape contents which i'm i'm sort of writing formally here i'm breaking the the the the tape contents into two parts which i'm calling t1 and t2 um this is the t1 part the t2 part and i'm putting the uh the the state in between t1 and t2 where i have in mind that the machine's head position is right at the beginning of t2 but maybe this picture just says it all and it's clear enough um so just keep in mind that when when we're going to be writing down configurations of machines we're typically going to write it down this way um okay so i think this is a good moment to take uh to take a moment uh for questions oh i see there's a lot of questions already um um so one's question is how does the encoding differentiate between the string and the state um i if i'm understanding you correctly i'm going to be assuming that um the state the symbols that represent states and the symbols that appear on the tape are distinct from one another so you can always tell just like usually the way we write things down when you have a state symbol or whether you have a tape symbol and so in a configuration you're going to have a bunch of uh tape symbols and a single state symbol represent you know in the position where the head is um somebody says 6 here is just you know this is one two three four five six we're in head the head is in position six that's all i had in mind for six and that's why that six is appearing over there because that's the position of the head um good all right so why don't we continue um now we're all ready to start defining computation histories um which themselves is not very difficult concept computation history for a turing machine m on an input w is just the sequence of configurations you go through when you start the machine at the beginning um with uh m on the um on the tape with w on the tape i'm sorry um and then and the the st the head at the beginning is in position one and the uh state in q zero or whatever the start state is of the machine so that's going to be the very the starting configuration here and these are the sequence of configuration that mean machines go through that the machine goes through step by step every step of the machine is getting represented here until it ends up at an accept that's what we mean by a computation history or sometimes we're going to emphasize that by calling and accepting computation history represent meaning that the machine has accepted its input and these are the sequence of configurations that the machine goes through that's what i mean by a computation history you know i'm sure in i mean i think the terminology changes over the years but the there's a notion similar to that for any kind of um you know if you're you know running a program and you just want to keep track of how the variables are changing um as you're you know in this particular run of the machine and you're writing them all down um it's like a log of the uh history of the settings of the internal states of the machine and variables and so on um so this is just all of the configurations the machine goes through are on the way to accepting its input if the machine does not accept its input there is no computation history okay so there is no accepting computation history at least to emphasize that point um because that can only occur obviously if the machine has accepted its input okay so just as we had with configurations we're going to might want to be able to write down computation histories as strings and it'll be convenient to have a particular format for doing that as well and it's kind of the obvious thing we're just going to take [Music] the sequence of configurations in the computation history and write them down as a string where each configuration is going to be the string for that configuration i'll show you kind of a little example in a second but just to you know focus on the concept here we're just going to write down the string for this configuration the starting configuration and then the next configuration and so on until you get to the accepting configuration and separate each of those strings by some pound sign okay so here is a computation history for m on w um [Music] uh where w is the string w on w2 dot wn so here is the uh here is the first uh configuration so that this part here is that is the computation history encoded as a string i'm just i'm giving you this extra um side information just to make it clearer what's what's going on there so this is the first configuration this is the second configuration and so on and i'm i'm trying to even give you a little bit more you know um detail of how it might look uh by kind of you know making the example a little bit more concrete so let's say that the turing machine when it's in the starting state q0 looking at the very first symbol of of the end of the the input tape in this particular input say w1 um it goes from q0 to q7 and writes an a on the tape and moves its head to the right so that's why the second configuration reflects that see here it went from q0 to q7 now the head is in the second position so that's why that state symbol has moved over one and um the very first position now has become an a because the machine has changed whatever was on that input tape there on the first position to an a and then the next step it goes from q7 reading w2 which is where the head is uh now it's looking at the w2 and goes to q8 writing a c and again moves right okay so that's why i drew these in the way i did and so you go through a sequence of those you get to cue accept and that's the encoded form of the computation history all right um i think that's is that all i wanted to say yeah um so you can uh feel free to ask another question if you want um i'm just gonna relaunch this yeah good so if you for all to for all together on configurations computation histories and how we're going to be writing them down as strings that's what we've done so far no questions so i'm going to move on all right so let me define a new um automaton that we're going to mainly use as uh just to provide an example for us today i'm going to call this a linearly bounded automaton and all it is is a turing machine [Music] where the turing machine is going to be restricted in where it can uh the tape is not going to be infinite anymore the tape is just going to be big enough to hold the input so the machine no longer has the ability to move into the portion of the tape to the right of the input because there is no tape out there it just has the tape sitting here um that contains the input which can the tape itself can vary in size however big the input is that's how big the tape is so the tape adjusts to the length of the input but once you've started the machine with some particular input that's as big as the tape is there's no more um the reason why it's called linearly bounded is because the amount of memory is a linear function of the of the size of the input because you can effectively get somewhat more memory by enlarging the tape alphabet but that's going to be fixed for any given machine so that's where the linear comes from if that's helpful but if you don't get that that's sort of a side remark uh but um what's important to me is that your understand what i mean by an uh a linearly bounded automaton or an lba it's just like a turing machine but this that portion of the tape that originally had blanks is just not there the machine tries to move its head off the off the right end of the input it just sticks there just as if it tried to move its head off the left end of the input doesn't go anywhere okay so now i'm gonna we're gonna ask the same kinds of questions about lbas that we asked for other automata so the acceptance problem if i give you a an input and some particular lba and i want to know does the lba accept that input well you know and now the question is is that decidable um or not uh so it's um at first glance you might think well um an lba is like a turing machine and the atm problem is undecidable so uh you know that that might be a good first guess um [Music] and also if you try to uh to simulate them if you try to figure out how you would go about simulating the machine um you know if you if a given b and w if you actually try to simulate the machine to get the answer so you run b on w well of course if you run it for a while and it eventually halts either accepting or rejecting then you know the answer and you're finished but you know this machine might get into a loop um you know nothing to prevent the machine from looping on that finite amount of on that limited amount of tape that it has um and then you know you might be in trouble but in fact that's not the case because when you start out with a limited amount of tape if you run the machine for a long time and it's not holding it's going and going and going inevitably it's going to have to repeat um get into exactly the same configuration that it did before because there's only a limited number of configurations that the machine has and once it repeats a configuration it's going to be repeating that configuration forever and it's going to be in a loop so this problem in fact is decidable um because the idea is if bmw runs for for a very long time and an amount that you can calculate then you know it's got to be you know cycling you know more than just looping it's got to be repeating itself um and so therefore once it starts repeating itself it's going to be going forever so um and here's the actual calculation which is something i'm sure you could do on your own but just to spell it out so if you have an input of length n um that you're providing to uh b so if w is of length n um the the lba can only go uh for uh this number of different it can only have this number of different configurations the number of states times the number of head positions which is n the number of head positions on the tape times the number of different tape contents you know if if the tape was only one long you know this is the this is the size of the tape alphabet so if the tape were too long it would you know had the tape had two uh two cells on it the number of possible tape contents is would be um the square of the of the alphabet and then and if if the tape is going to be n uh symbols long it's going to be the tape alphabet size to the nth power uh so therefore if a turning machine runs for longer it's got to repeat some configuration and it'll never hold so the decider is going to be hopefully uh clear at this point you're given b and w says that this is the decider for alba it's going to run bmw for this number of steps if it's accepted by then then you accept and if it hasn't if it's rejected or it's still running then you can reject and you know if it's still running at this point it's never going to accept all right any questions on this okay let's move on um all right so now um but what's different now or what's it perhaps interesting now is that even though the acceptance problem for lbas is the is is decidable the emptiness problem is undecidable and that's where the computation history method is going to come in so this is where we're going to be starting to do something new in terms of a proof technique um because so we're going to reduce atm to elba the emptiness problem for lbas um so given an lba does it accept anything or not and this is going to use the computation history method so this will be a chance to illustrate that method for the first time so this the setup initially is just like before we're going to assume we have a turing machine that decides the cb elba problem that of interest and we're going to use that to construct turing machine deciding atm for our contradiction okay so here's s supposedly deciding atm what it's going to do and this is going to be the tricky part s is going to use m and w to design an llba that lba is going to be built with the knowledge of m w in fact it's going to have m and w built into it so that's why i'm calling that lba b sub mw because it depends on mw um as i'll describe and what that lba does it takes its input let's call it the x the input to the the lba and examines that input to see if that input is an accepting computation history for m on w then that lba is just looking for computation histories for mw and that's the only thing it's going to ever accept if you feed it something which is not a computation history you know that sequence of configurations uh for mmw leading to an accept if you feed it something else the lba is just going to reject it it only accepts the accepting computation history for m w and now if you can build such a thing then you can use that machine in your emptiness tester to see if it accepts any strings at all because the only thing that could possibly be accepting is an accepting computation history you don't even know if there exists one because you don't know if m accepts w so you're going to build this lba um and uh which is looking for accepting computation histories and then see if its language is empty or not if its language is not empty the only thing it could be accepting is this computation history so you know m except w whereas if the language is empty you know that there is no computation history for m w and so m does not accept the leaf so that's the whole idea the trick is how do you build this uh this lba um so first you're going to make this lba which tests its input to see if it's an accepting computation history for m on w and you have to build this um uh lba without even knowing whether there is a computation history um it's not like you can take that string and just build the string into the lba because they don't even know if that string exists but what you can do is you can make the lba um follow the rules of m it knows w and knows m so it can test take the its input and using w and the rules of m see if that input is a computation history for m and w um and that's what it's going to do so here is um this machine i'm going to construct i'll i'll give it to you written down and with a little bit more um details of its procedure but just to kind of illustrate it so here is a proposed input to b sub mw this would be the x here this would be an x that would be accepted but anything else that would be provided would not be accepted so this is the sequence of configurations written down as a string that would be the x that i'm providing to the b c b b sub mw and it's supposed to be checking this to make sure it's legit um so so on on input x the way it's gonna the way it's going to proceed it's going to first check to see whether x begins in the right way because you know this lba it knows m and it knows w so the very first thing is it takes a look at the first part of the string up to the pound sign you know there's no pound sign if you're going to just feed junk in it's going to be easily identifiable as junk um so uh you know if you're going to feed something so it's going to the lba is going to take everything up to the first pound sign and just confirm that that thing is the first configuration the starting configuration of um of m on w which means it has to start with the start state and here is w so just going to check that um then it's going to check that each each one of these guys follows legally from the previous one according to the rules of f it's going to first checks that this one is correct and then this one leads to that one according to the rules of m then this one leads to that one according to the rules of m all of that you know the knowledge of m and w is built in so it can do that um and then it just follows along checking this computation you know it's easy to check that a computation is correct that's all that's really going on here um it's going to check that the computation is correct until it gets to the end and then make sure that the last uh configuration that it's been given as input is an accepting configuration that there's an accept state in it um and if everything that passes it it accepts otherwise it uh rejects okay um and the point is that this is an lba you don't oh i just kind of jumped ahead of myself you don't need any additional information so how does it actually do this so you know how do you actually do this on the tape so i claim that the lba doesn't need any extra space to do this um to do this check you know it's just going to be zigzagging back and forth here on the on on the on the input you know checking that the corresponding symbols match up except around the the head position where it gets updated or correctly and it seems it may need to mark it will need to mark on the tape it's allowed to write on the tape just to make sure it keeps track of where it is but this is a very simple i mean i'm not trying to uh if you're not following i'm not trying to uh alarm you but i'm just trying to help you understand that this is not a complicated procedure here to do this check that the that the actual computation of the that's written down of the turing machine is valid um but you know we can deal with a question here uh oh good now we got some questions oops you can be thinking about while i'm thinking about this you can think about that uh check-in over there so here's a good question going from each configuration to the next configuration is it a unique uh next step um well i'm assuming that the turing machine we're starting with is deterministic so this should be only one way to go so the answer to that question is yes this is going to be a unique um string here there's really going to be only one um computation history not that it really matters in a sense um the answer to that question um as long as that accepting computation history corresponds to the machine actually accepting uh oh boy we're all all over the place here uh okay are you ready to end this um in this um in this bowl uh all right well this is where we are uh everybody who said they'd rather be in six or four fig six or four six i know who you are you're all going to be expelled no i'm joking uh i don't know who you are so um uh yeah good so um uh wha what so we are here at the at the break um i and i made this poll on purpose uh at this moment so we can spend a little time let me just start our clock going and i can try to help you those of you who are who answered see that you're baffled um i'll i can try to uh help you understand what's going on all right so this is a this is a good question here um uh okay well okay there's several good questions here so um somebody asks why don't we just test all possible strings uh for b remember if we're trying to test emptiness for b's language um that's the e-lba problem nor do we just try all possible strings um and uh that would work if we had enough time but there's infinitely many strings and so that's not going to be good if we're trying to be a decider so so that is that's why we don't just try all strings so but here's this other question here this is a okay so um the question is how do we find the input x so this is an important question um because uh we don't find the input x there's no you know we the input x um at least the accepted input x would be the accepting computation history um for m on w we're never going to find it you know we're never going to find an x we're not we're building this um this lba not we're never going to run that lba we're designing that lba not for the purposes of running it we're building that lba only for the purpose of uh using the lba language emptiness tester we're going to build that end lba and feed it into the machine turing machine r which is going to tell us whether that lba's language is empty we ourselves is not never going to run that machine we're never going to come up with an x we're just having we're we're we are designing a computation checker and then um the emptiness tester for that that we assumed to exist for for elba is going to say yes there is some computation which this machine accepts or no there is no computation which this machine accepts and that's going to be a computation for m on w and so that's going to tell us whether or not m accepts w so we're never going to actually be finding an x we're never going to be running that lba we're just feeding using that lba as as an input to r um does the computation history method always use lbas no as you'll see uh right after the break because that's the lbas is just kind of an easy place to get started but we're going to use the computation history method and computation histories next on the on the post correspondence problem yes the the computation histories and the input x are always going to be finite so that that was an answer to a question about whether these histories or the inputs are going to be finite or not so the the they're all those strings are always going to be finite and the computation history is going to be finite um so we are at the end of our five minutes um let me just see if there was another question i wanted to answer um let me let's move on okay now getting coming back to the undecidability of this post-correspondence problem so remember the post course upon you know this this problem you're given those dominoes you want to know if there's a match here is a little mini version of the the diagram if that helps you remember it um and we're going to prove that this language is undecidable and so it's undecidable to test whether you have a match given a set of dominoes whether a match is possible and we're going to use we're going to reduce atm to to this language using the computation history method so how in the world are we going to do that um first of all there's a little detail here i want to mention uh just uh um don't focus on this but i'm going to assume the matches always start with the very first domino on the list um the very first domino in the collection so this is going to be like a starting domino this is going to make my proof a little simpler to do it that way and then you can fix that assumption uh later and if we have time at the end i'll do that or maybe after the lecture is over if there are questions i can show you how to fix that assumption but for now um to simplify the proof we're going to assume that there is a starting domino that the matches always have to start with that particular domino um if you didn't follow that point don't worry you can just ignore it you'll see where it comes up later so um now we're going to reduce atm to the pcp so assuming we have a machine that decides the pcp we're going to use that to make a machine that decides the atm as before so here's the turing machine s which is going to decide atm um and the way we're going to do it is like this we're given m and w we want to know as always does m accept w what we're going to do is we're going to build an instance of the pcp problem so we're going to build a bun a collection of dominoes [Music] uh which are going to dip and that collection is going to be built knowing m w so that's going to affect the dominoes we're going to create so this collection of dominoes is going to be called p sub mw depends on an m w and finding a match for this set of w is is going to force you to simulate m on w because the match is going to correspond to a computation history for m on w so uh we're going to use once we do that um once we build uh the set of dominoes who's where a match corresponds to a computation history we're going to use r to determine whether or not there is a match or in other words whether or not there is a computation history which is whether or not m accepts w um so if there is a match we're going to accept because we know m accepts w and if and if there is no uh match we're going to reject because we know m does not accept w okay um this is the plan i haven't told you how to do this yet i mean i'm worried about the significant number of you who are feeling um confused by the method i mean you should you guys should be uh texting me in the tas um to try to at least get a sense of how this is working i mean that we're going to be going we're going to be doing the method again um it's just going to be a little more complicated because we have to also deal with these dominoes and all this all that stuff and that's why i presented it to the fir to you the first time in the setting of the lbas which where in a sense the method comes out perhaps more simply so a match is going to correspond to an accepting computation history um sometimes i if i don't use the word accepting um you know i'm just being a little sloppy computation history and accepting computation history for us right now they're the same i mean later on we may actually talk about rejecting computation histories but let's not get ourselves confused all the except computation histories always have to end with the machine accepting okay uh somebody's asking what does it mean for a match to correspond to a computation history you'll see that's going to be on my next slide oh so this is a good question um is my step two trying to use r to determine whether m accepts w well yes but i'm doing that by testing whether whether this these dominoes have a match because r is besides pcp i can only use r to test whether things have a match because someone asked should step two be to use r to determine whether m accepts w well in effect that's what it's doing but it's doing indirectly through um testing whether this um pcp whether these dominoes have a match because those dominoes are going to force you to simulate mlw okay i think i'm repeating myself here so let's uh avoid getting into a loop on the lecture and um uh see how we actually build piece of mw so now you understand what you have to under have the plan in your mind uh um we are given m w we're trying to make a set of dominoes where a match is going to be a computation history for m w so the string you're going to get in that match the top string and the bottom string which are going to be you have to have to have to match um there they're going to be computation histories but they're going to be a computation history of m w so i want to figure out how to make my dominoes force that okay so my starting domino as i told you there's going to be a special starting domino in my collection so which is going to require the match to start with that domino it's going to be this one here it's going to have these two strings in it and you can see already it's starting to look like a a computation history the dominoes going to have that feature so uh it's the pair of strings and i've written it kind of to help you see what's kind of going on it's a pound sign on the top and the the starting configuration for mmw on the bottom and what i'm going to do for you here is at the bottom of the slide i'm going to take the dominoes i've written down so far and try to be building a match and you'll see how that match is forcing a simulation of the machine okay so let's take this as an illustration let's assume the input to m so i'm running m on w now i'm trying to see does m accept w um w is a string two two three so the start configuration for m on w is q zero that's the starting state for m and then w following is 2 2 3 that's what it's going to appear on the on the tape of the turing machine to start off so this is the start configuration for on w and assuming that i had this particular string is input string to w and so uh given the match the dominoes that i've given you so far just one um this is how the match is going to start uh now there's going to be more dominoes coming so for every possible tape symbol and state don't get confused by the language here this is the important thing if if in the turing machine um and i'll just read this in english to you if the turing machine when it's in state q and the head is reading an a it moves to state r writes a b at that point and its head moves to the right i'm focusing on right rightward moving uh turing machine steps right now but for every possible uh state and and tape symbol and looking at where what happens i'm going to have a domino that's going to capture this information and it's going to be this domino here q a on top br on the bottom so just the string qa on the top br on the bottom so let's see why with that sort of arcane looking why is that a good thing to why is that a useful domino to put in here well if you take a look um let's assume from my turing machine when it's in state q0 which is the starting state and the head is reading a two which it will happen to be reading because the uh this the string w starts with a two if it goes into state q4 no q7 and writes a four and then moves right i'm going to have in this in this domino i'm going to have q02 on top and 4 q7 on the bottom because 4 is what i wrote right here that's you know that's the b and this and the new state that i'm going into is q7 so that's going to be this the domino that i'm going to get and here it is here's that domino appearing in the match so um it's going to take match up with q q 0 2 um which you know you know don't forget the the the top string has to equal the bottom string so and this is going to be the only choice that i have for extending the match so q02 um is going to be on the top when i put that there but that's going to force um 427 to appear at the bottom because that's what's going to be the bottom string corresponding to the q02 on the top okay so if you're looking down here this is the the beginning of the second configuration of the machine that's what i want to be happening i want to be for that match should look like a computation history so now um so this is going to be all possible right moves are going to be handled in this way it's going to be a similar process for the left moves but let me not you know i'll leave that to your imagination now how do i continue on from there what does the rest of this configuration look like well that should just be a copying over of the remaining symbols that were on the tape because they don't change things only change around the head the rest of those symbols which is the two and the three those should just get copied over here so i'm gonna have two additional i'm gonna have additional uh symbols in my dominoes in my uh in my collection so for every um tape symbol i'm gonna have a a be a domino so and for every type symbol a i'm going to have a a be a domino in my collection and so that says uh i can have uh so that there's going to be a 2 2 domino so i can match up this 2 over here but that forces me to put a 2 down over there there's also going to be a 3 3 domino which is going to match up with that 3 but it's going to force me to push it put a 3 down over there that's the only way i can extend this um match that i've built so far i have no choice those are the only dominoes that i'm going to be given and by and so doing i'm forcing you to basically simulate the machine now what i want to have happen next is a pound sign to appear here and that will conclude my second configuration so there's going to be a pound sign pound sign domino as well because there's a pound sign here it's going to get matched with a pound sign up top forces a pound sign to come down um on the bottom and if you look at where we are right now we're exactly like where we were at the beginning when we had just this first domino appearing but now we're one configuration later so if you understood that and i admit it it's it's you know this is a there's just one idea here um [Music] and once you get the idea it's all trivial it's all very simple but there's just you have to get that idea i'm trying to figure out how to get that idea into your head once you get the idea you can write all this stuff yourself um so uh so following the this uh the this description of how uh the you know the the transition function of the machine works and copying over uh the symbols from the tape to the next configuration i'm going to be able to get configuration after configuration going until i get to a point when there's an accept and now from the machine's perspective we're done the machine has accepted its input this is our computation history but is it a match well you know up until this this bottom thing um is the computation history but it's not a match because the top doesn't equal the bottom there's still extra stuff at the bottom which the top is not uh you know it doesn't have so what i'm going to need to do now is is add some additional kind of pseudo steps of the machine where i'm going to allow the top to catch up to the bottom and the way i'm going to be thinking about that and this is not real for a turning machine as much as the turn machine's real anyway but you're going to imagine i'm going to add a new kind of uh move to the machine which is going to allow the head to eat the symbols off the tape like pac-man okay and the way i'm gonna get that effect um on the top if i have any tape symbol next to a cue except on either side on the top what i'm going to get you to write on the bottom is just to accept with that taste with that tape symbol gone and by repeating that you know move after move the the the actual tape is going to be shrinking the symbols on the tape are going to be going down one by one move by move until there's nothing left except just the um you accept all by itself okay so i have this sort of pac-man idea of eating the symbols on the tape um and then finally i get to a point when there's just the cue except alone on the tape there's no symbols left to consume and then i'm going to add one last domino here which is q accept pound pound matching with just pound which just conveniently finishes off the um uh the uh the match and so the match is completed there's actually a little detail here i i hesitate even to bring it up because i i think it's a kind of thing where if you understand everything up to this point you could fill it in yourself but just for completeness sake uh you have to deal with the situation when the the uh the head um of the turing machine might move into the blank portion of the tape which is not taken into account here and the way we get that effect is by having another domino here which allows me to add blank symbols at the bottom as needed that's just a detail you know i'm more worried that you understand the underlying concept um so here's going to be um uh a check-in um trying to remember um okay but you know so okay so what else can we conclude from this information we know so at this point we know that pcp is undecidable that's what we just finished proving what else do we know if anything or do we even know that uh see my picture is a little bit covering the check in but it's still readable so i mean this is not an easy concept to get the to get the idea but once you get it you know it's um you know it's you'll see it's not that bad okay another 15 seconds so this portion doesn't this question is not really relying so much on the computation history method it's just relying on the fact about pcp being undecidable um [Music] okay everybody almost done five seconds all right um so i can see that there is uh some level of confusion um uh so the reason why b is correct is that we know pcp is undecidable but we also know that it's recognizable because you could try all possible uh ways of combining dominoes and one after the next um and you accept if you ever find a match so they might go forever but if there is a match uh possible you'll find it eventually so pcp is a recognizable language undecidable and so therefore we know it's complement it has to be unrecognizable because you know that's something we've shown before if a language and its complement are both recognizable then it's decidable but we know pcp is not decidable so they both both sides can't be recognizable um okay so let's let me prove to you one last theorem that's going to be useful for your homework uh involving the computation history method so you'll have one last chance to get the way we're going to use this though this one is going to be a little bit more similar in spirit to the lba version uh than to the pcp version which has this extra kind of complication about coding the computations of the machine into dominoes here um you know we're going to operate with another automaton which where it's going to be more straightforward but anyway okay getting ahead of myself so if you remember for the for context-free grammars we had the problem the emptiness problem we showed that was decidable so testing whether a context-free grammar's language is empty is decidable however testing whether a context-free grammar's language is everything whether it's equal to sigma star that turns out to be um uh that turns out to be undecidable okay so emptiness testing for context-free grammars decidable sigma-star testing undecidable um okay so we'll show that atm is reducible to all pda via the computation history method um and uh so uh assume we have same patterns assuming we have a decider for all pda and make a decider for ulti decider for atm here's the atm decider and now similar to what we did for the uh lba case but with a with a twist we're going to make a push down automaton that's going to check its input to see whether it's an accepting computation history of m w just like the lba did if you think back to how that worked remember remember you know we um the lba tested its input to see whether it's uh an accepting computation history accept it if it did and then we test the lba's language for emptiness to see if there are any computation histories so we're doing the same kind of thing except now uh the pda is going to test its input to see whether it's a an accepting computation history and if it is it's going to reject it's going to do the reverse of what the lba did and that's going to turn out to be necessary but for the for the moment let's just go with it and maybe we'll see in the proof why the proof needs it to be this way so um this p pushdown automaton is going to accept its input if it's not an accepting computation history for m on w otherwise it will accept so you can think of this pda as accepting all the junk it just doesn't accept um the good stuff the accepting computation history okay it's it's just it it's uh accepting all the things which fail to be an accepting computation history okay and then once we have that we we test whether ours language is everything but whether bmw's languages everything using r because it if if that pda's language is everything then we know um uh there could not be an accepting computation history because that's the one thing that gets um not accepted that's the one thing that gets rejected so if it's accepting everything there is not going to be an accepting computation history so if there is no uh if this is equal to sigma star then there couldn't be an accepting computation history and so we're going to accept okay so how is this going to work so what's different now about uh this case uh from the lba case is remember the lba uh got you know in the the computation history on its input and it used its ability to write on the tape to check that each configuration follows the next one we don't have the ability to write on the tape and push on automaton and by the way i hope you're all comfortable with my using pdas instead of grammars because we can interchange one to the other um i should have mentioned that when when i did it but um so the the pda is going to be using its stack to compare each configuration with the next one okay so the way that's going to happen is it's going to non-deterministically take one of these so the very first thing it does is as before it checks to make sure that the beginning of the input is the start configuration but then once that's that's the easy part but once we get going with that we can we push each configuration well we first of all we now deterministically choose which configuration might be the bet the one that fails um where where one fails to go to the next one because those are the ones we're trying to accept when there's a failure so we now deterministically look for the place that there's a failure we push that onto the stack um and then we pop it off the stack and compare it with the next configuration so that's how we're using the stack instead of being able to mark on the input now there's a so i'm kind of going to illustrate that here so you know we're going as we're going to read this thing here and we're going to put it onto the stack so this thing moves over to here and this input here got put onto the stack you know q 0 w 1 w 2 it's q 0 w 1 w 2 you see that that first configuration is now sitting on the stack now as we're going to pop it off we're going to match it with the second configuration now if you're following me you'll realize that there's a there's a difficulty here because it's coming out in reverse you know it comes out in the reverse order that we put it in and that's not what we need to do so what we're going to do here and here's a little sort of a twist we're going to change the way we're writing down computation histories by making them by writing them uh reversing the even numbered ones so this one here is going to be written down in reverse and that's perfectly okay we can write down configuration computation histories in any way we want to um to meet our needs so we're going to reverse we're going to reverse the alternate ones and now when we push one it's going to come off in the right order to compare with the next one and so that's how the the procedure works okay we're running a little short on time um you need to be let's see so let let me just go to um uh quick recap the computation history method is useful for showing the undecidability of problems when you're testing for the existence of an object each of those four cases that we showed today involving testing whether something exists so is there an integer solution to the polynomial is there some string in the language is there a string that's not in the language um or is there a match so that's a typical case when um this computation history method comes up and so as a quick review uh these are the things we showed today okay so why don't we uh uh we're we're just out of time um and so i'm i'm gonna let you go um but i will stick around uh for another five minutes or ten minutes uh happy to answer any questions if you if you have them um but that's that's officially the end of the lecture um okay so let's let me try to get to there's a lot of questions in the chat don't forget you know write to the tas too so if m does not accept w what will happen to the computation history so if m does not accept w um uh then there is no computation there is no accepting computation history that's the only kind of computation histories we're considering so if m does not accept w there is no accepting computation history there's no computation history um so i don't know what it means what will happen to it it just doesn't doesn't exist um so uh i hope that's helpful um don't we need to be able to add states to the end of the tape i don't know what that means it's taped to the end of the tape i i don't know you'll have to repeat that one sorry or explain that better um ah so this is a good this is a very good question here where does the previous proof fail when trying to reduce atm to ecfg it's got to fail because ecfg is decidable um so that that's a very good question maybe we can just go back to that last slide here because i was running a little short on time i didn't really focus on why we have to accept um the non-computation history strings we why are we accepting all the junk strings and only rejecting um the strings which are the uh the computation histories so you know we're looking for strings that fail you know if the string could fail because it starts wrong string it doesn't have the start configurat configuration or it doesn't maybe doesn't end with the accepting configuration or maybe one configuration doesn't lead properly to the next one any of those cases are a failure and there are going to be reason to accept the reason why we can do do it do that um is because the we're taking advantage of the push downs non-determinism we don't know where the um the failure might occur so we're going to non-deterministically guess where that failure occurs if we were going to flip this around and say let's accept only the good ones and reject everything else we would have to test that each configuration led to the next one so we have to check them all so if something fails it only has to fail in one place and then you know we can accept if it um if it's a good computation history we have to it has to be good everywhere and so the push down automaton really if you're going to get down to the nitty-gritty the push on automaton can check that this configuration leads to that configuration using pushing on the stack and then popping it but now by the time you get to here you want to check that this second configuration leads to the third one you would need to push the second configuration onto the stack but at this point you're already at after the second configuration so we cannot back up and push the second one on the stack so the the push on tom dot is not able to check um in the positive sense that you have a computation history it's only able to and accept them it's only to check in the negative sense that you don't have a con a computation history and uh accept all the ones that fail and that's just because it only has to fail in one place i hope that helps um so that's why we couldn't flip this around and make this proof work for the emptiness problem and only have to work for the the everything problem the old problem all right um okay so bye bye everybody and i will see you on tuesday 2 00:00:26,630 --> 00:00:29,589 hi everybody can you hear me yes good um welcome back uh to um to the course and um so we are now at lecture 10 and um uh i let's see what have we been doing we've been talking about um undecidability um so we introduced last time uh reducibilities and mapping reducibilities for proving various problems are undecidable and this time we're going to and today we're going to introduce a a more sophisticated method for proving undecidability uh using reduceabilities and that's called the computation history method and and that's pretty much what's used in all of the you know more serious cases where people have proved problems undecidable it's it's pretty much a widespread uh uh method for uh especially you know it's not an active area of research these days but at times when people were proving problems undecidable this was frequently a method that was used um so we'll go over that today and we'll prove um a few examples of problems undecidable using that method all right so why don't we get started um uh so first of all most importantly you should remember how we you how we prove um problems are undecidable in the first place and that is by showing um some other undecidable problem that we already know is undecidable is reducible to the problem of interest so typically you might be reducing atm um which is a problem that we started with and showed to be undecidable by the diagonalization method um or might be some other problem that we've subsequently shown to be undecidable and you take that problem and you reduce it to the problem you're trying to show undecidable okay oops let me fix that sorry um so um so this is the thing to keep in mind to prove a language is undecidable show that atm or some other known undecidable language is reducible to the problem b that you have at hand that you're trying to show undecidable all right so we are now going to um um [Music] first i'm going to start off by remembering um hilbert's 10th problem which we discussed briefly a few lectures back um and as you may remember so that's a problem where uh you're given say some polynomial you know over several variables you know like three x squared y minus 15z plus 2xz equals 5. and you want to solve that problem you want to find a solution that solves that equation but you require that the solution involves only integers so those are so called diaphantine problems because the requirement is that the solution being integers and hilbert's problem was to ask for a an algorithm not the language that he used but doesn't matter he essentially he asked for an algorithm to test whether a polynomial has a solution in integers or the polynomial equation has a solution in integers and um as uh um uh um as we now know that problem was posed back in 1900 it took 71 years to find the solution but the answer is no there is no there is no such algorithm we talked about this when we were discussing the church turing thesis um and um the method showing that there is no algorithm is by a reduction from atm to that problem so one can use exactly the method that we're using today to prove that problem is undecidable about polynomials the only thing is is that the reduction that single reduction would take the entire semester um and i know that's in fact correct because when i was a graduate student um there was a woman in the math department at berkeley where i where i studied and she uh the whole the whole course was to go through the proof of hilbert's tenth problem um of the solution to hillary problem the the proof of the undecidability of the uh of the solution uh of the undecidability of testing whether a polynomial has an integral solution um so um the thing is that involves some fairly hairy number theory in order to um to come up with the right polynomials to basically simulate the turing machine which so that's kind of getting ahead of ourselves that's what we're going to be doing today um but we're going to instead of looking at that problem because we don't have a whole semester to spend on it um we're going to look at a toy problem instead where there's no number theory but it still has the same basic underlying idea um to that that was employed in um the solution to hilbert's tenth problem um so that toy problem is called the post correspondence problem and it's going to have some other utility for us well the main reason we're studying it is just to illustrate the method um uh so and the method is going to be as i've been saying is this computation history method okay so one we will first talk about uh this post-correspondence problem because that's very easy to understand and then we'll spend a little time introducing that method um all right so um the uh the post the post correspondence problem is as follows you're given uh a bunch of pairs of strings um so here is one pair t1 and b1 and i'm writing them as dominoes i'm calling them dominoes because i'm writing one pair one one of the strings on top of the other string in fact i'm using t and b for top and bottom here so this is t1 is the top string b1 is the bottom string in the first domino and then in the second domino we have two other strings to a top and a bottom and so on um so we have these here uh this collection of dominoes um as sort of the legal dominoes um and what the goal is to find a sequence of those dominoes um uh so you want to pick from these dominoes here you're allowed to repeat dominoes so you want to pick a sequence of those dominoes line them up next to each other and so that the string that you get by reading along the top all of the t's together is going to be exactly the same as the string you get by reading along all the bottoms i'll give you an example um so here to write it down a little bit more formally so uh what i'm calling a match is a sequence of these dominoes so it's t i1 ti2 as well it's it's i1 is the first domino i2 is the second domino and so on and then you what you're going to have is all the top strings concatenated concatenated together is exactly the same as what you get by concatenating together all the bottom strings um so here's an example and i think it'll become clear if you didn't understand it so far but it'll be completely clear from the example i hope so here is a bunch of dominoes um four um and um what i wanna know is is there some selection of these dominoes again you can repeat dominoes otherwise it would be not an interesting problem so you can repeat these as many times as you like and i want to pick up pick these dominoes and line them up so that what you get by reading along the top is the same as what you get by reading along the bottom so first of all if you just stare at this and you're trying to make a match you'll see that there's only one possibility possible way to start this um you know for example if you started with the third domino here b a over a a it would fail immediately because um you know the b and the a are different um and and if you you use the second domino as your starting point that would fail the match as well because a a has to would have to e um be the the the top string would start with a a the bottom string would start with a b and they could never be equal um so by looking at this you can see that the only possible choice that you have is to start with the first domino okay so i'm going to start to build the domino for you just so you can see it so here's the very the the match as um so you can see um the process so the match starts with the very first domino and the way i'm going to write it is i'm going to take the dominoes and kind of skew them so that you can see how the top and bottom strings are lining up um but they're just mouth the same dominoes they've just been i changed the shape so this is the first domino a b over aba now uh the second domino in my and my match it's gotta start with an a as the leftmost symbol on the top string so that would rule out this one for example um but there might be several choices and so it's not exactly obvious which one you should take what i'm going to suggest is we take this one here so we take a a over a b a you see it a a over a b a it's just you know written i'm just skewing uh skewing it so that um i can line up the a over a the b over b the a over a and so on so i'm starting to build this concatenation of the top being equal to the concatenation of the bottom following me so far i hope so now um what's what's next so i need to have something where the top string is going to start with a ba in fact there's only one choice for that so it's clearly going to have to be this domino is going to be next so it's going to put a ba up here but then it's going to a force an a a down below because that's what this domino does um so this ba matches with that now we have the aaa to deal with um so we're going to reuse this domino because that's the only one that starts with an aa so we're going to reuse that one now we have a a and aba now we need something that starts with aba i see two choices it could be this one because this is consistent with a it's at least it captures the a b we could we could try putting this one over here but a better choice would it be would be to use this one because if we use this one we finish the match and therefore we have found what we're looking for um and this this this collection of dominoes here has a mat has a match now it's not always obvious some some collections of dominoes may not have a match and so the computational problem is is there a match um and the theorem that we're going to prove today is that this problem is undecidable simple as it is simple little combinatorial problem of you know uh trying to put together different um uh different tiles like that to form form a match uh there was no algorithm for solving that problem um and and i think this is kind of interesting because it's the first example of a problem that we have where the underlying question does not really seem to have anything to do with computation you know you might argue that atm being undecidable is perhaps a little unsurprising after the fact because you know it's a problem about touring machines so you can imagine turing machines are going to have a tough time answering it um but here's a problem just about strings um and we're going to show it's undecidable so just to formulate this as a language i'm going to call the pcp language is the collection of these um uh pcp uh problems pcp instances which are just collect themselves each one is a collection of dominoes so it's the language of all uh collections of these dominoes where there is a match so it's all of the pcp problems where you can solve them with a match and uh we're going to prove that it's undecidable by showing that atm is reducible to that pcp language um now uh before we do that i'm going to have to i'm going to explain to you what the computation history method is in the first place that we're going to be using uh so before we do that here's our first check-in for the day just to make sure you're following me about uh what the um what this pcp problem is i am going to give you a question to uh test is there a match with these three dominoes so what you think about that i mean the main thing i'm i want to make sure you understand is what a match is in the first place this is not a super hard problem uh to tell whether or not there's a match okay um five seconds already okay closing it out um okay um so uh for those of you who said there is a match i want you to exhibit exhibit the match because in fact there is no such map there is no match um and uh i mean you could get by fiddling around with it i think if you try to find a match you'll pretty quickly uh see that it's you're gonna get um i think what happens is you kind of get into if you try to build build a match with this particular setup you're just going to get stuck sort of repeating yourself and um one point here that you may have missed is that a match has to be a finite sequence so if you were thinking about there there is there is a kind of an infinite match uh that you can build with this uh this this set of dominoes but that's not allowed we're only allowing finite matches and so that may they might change your answer as a result but too late um so um and one way to see that you can't have a uh a match in this problem is that you know you're going to have to start you you you know neither of these two are possible starting points so this one is going to be clearly the only chance that you're going to have as a match uh the first domino here but then once you put this one down um the bottom one the bottom string the the bottom b's that you're going to get the bottom string is going to be longer than the top string and then the other two are going to be have the same length um so you're never going to have the bottom uh the top be the same length as the bottom the bottom is always going to be longer no matter how many uh more dominoes you lay out so this there's no chance of there being a match here um okay so now we're gonna um uh work our way up to introducing this computation history method [Music] um um first let me define something for a turing machine that i'm going to call a configuration um configuration is just a snapshot of the turing machine in the middle of its computation so if you're running the turing machine and you just stop it at some moment it's going to be in a certain state the head is going to be on a certain position and the tape is going to have a certain contents and with that information you can then um continue the copy the computation of the turing machine that's a full set of the information that you need about the turing machine that tells you everything about its cut its computation at that moment in time the state head position and tape contents and so that we're calling that that information together uh state head position tape contents we're calling that a configuration it's a fairly um basic notion i mean if you if you program if you have something if you have like the states of all the variables and where the current execution of the program is at a moment in time same idea um okay so in terms of a picture here here is the turing machine imagine it's in state q3 um the head position is in the sixth position on the tape so p would be six um because it's in the sixth position and the tape contents is going to be this bunch of a's followed by this bunch of b's so i would write it down like this states in q3 head position number six this is the contents of the tape um now what we're going to often do for convenience in using this notion in proofs is that we're going to want to represent a con a configuration as a string um in a particular way that's going to be that's going to this is this concept is going to come up kind of again and again during the during during uh the course and so it's going to be handy to have a particular way of writing down configurations um you know for doing um proofs about them and so the way we're going to write them down is i think it's just maybe i can just put it right up here like this um we're going to write down the the symbols of the tape but what we're going to do is stick in the middle of those symbols the the state of the machine immediately to the left of the position that the machine is currently reading so you can imagine you know the the head here which is coming you know imagine the state is kind of where the head is it's pointing at the symbol immediately to its right so this is just another way of writing down a configuration here's the tape contents which i'm i'm sort of writing formally here i'm breaking the the the the tape contents into two parts which i'm calling t1 and t2 um this is the t1 part the t2 part and i'm putting the uh the the state in between t1 and t2 where i have in mind that the machine's head position is right at the beginning of t2 but maybe this picture just says it all and it's clear enough um so just keep in mind that when when we're going to be writing down configurations of machines we're typically going to write it down this way um okay so i think this is a good moment to take uh to take a moment uh for questions oh i see there's a lot of questions already um um so one's question is how does the encoding differentiate between the string and the state um i if i'm understanding you correctly i'm going to be assuming that um the state the symbols that represent states and the symbols that appear on the tape are distinct from one another so you can always tell just like usually the way we write things down when you have a state symbol or whether you have a tape symbol and so in a configuration you're going to have a bunch of uh tape symbols and a single state symbol represent you know in the position where the head is um somebody says 6 here is just you know this is one two three four five six we're in head the head is in position six that's all i had in mind for six and that's why that six is appearing over there because that's the position of the head um good all right so why don't we continue um now we're all ready to start defining computation histories um which themselves is not very difficult concept computation history for a turing machine m on an input w is just the sequence of configurations you go through when you start the machine at the beginning um with uh m on the um on the tape with w on the tape i'm sorry um and then and the the st the head at the beginning is in position one and the uh state in q zero or whatever the start state is of the machine so that's going to be the very the starting configuration here and these are the sequence of configuration that mean machines go through that the machine goes through step by step every step of the machine is getting represented here until it ends up at an accept that's what we mean by a computation history or sometimes we're going to emphasize that by calling and accepting computation history represent meaning that the machine has accepted its input and these are the sequence of configurations that the machine goes through that's what i mean by a computation history you know i'm sure in i mean i think the terminology changes over the years but the there's a notion similar to that for any kind of um you know if you're you know running a program and you just want to keep track of how the variables are changing um as you're you know in this particular run of the machine and you're writing them all down um it's like a log of the uh history of the settings of the internal states of the machine and variables and so on um so this is just all of the configurations the machine goes through are on the way to accepting its input if the machine does not accept its input there is no computation history okay so there is no accepting computation history at least to emphasize that point um because that can only occur obviously if the machine has accepted its input okay so just as we had with configurations we're going to might want to be able to write down computation histories as strings and it'll be convenient to have a particular format for doing that as well and it's kind of the obvious thing we're just going to take [Music] the sequence of configurations in the computation history and write them down as a string where each configuration is going to be the string for that configuration i'll show you kind of a little example in a second but just to you know focus on the concept here we're just going to write down the string for this configuration the starting configuration and then the next configuration and so on until you get to the accepting configuration and separate each of those strings by some pound sign okay so here is a computation history for m on w um [Music] uh where w is the string w on w2 dot wn so here is the uh here is the first uh configuration so that this part here is that is the computation history encoded as a string i'm just i'm giving you this extra um side information just to make it clearer what's what's going on there so this is the first configuration this is the second configuration and so on and i'm i'm trying to even give you a little bit more you know um detail of how it might look uh by kind of you know making the example a little bit more concrete so let's say that the turing machine when it's in the starting state q0 looking at the very first symbol of of the end of the the input tape in this particular input say w1 um it goes from q0 to q7 and writes an a on the tape and moves its head to the right so that's why the second configuration reflects that see here it went from q0 to q7 now the head is in the second position so that's why that state symbol has moved over one and um the very first position now has become an a because the machine has changed whatever was on that input tape there on the first position to an a and then the next step it goes from q7 reading w2 which is where the head is uh now it's looking at the w2 and goes to q8 writing a c and again moves right okay so that's why i drew these in the way i did and so you go through a sequence of those you get to cue accept and that's the encoded form of the computation history all right um i think that's is that all i wanted to say yeah um so you can uh feel free to ask another question if you want um i'm just gonna relaunch this yeah good so if you for all to for all together on configurations computation histories and how we're going to be writing them down as strings that's what we've done so far no questions so i'm going to move on all right so let me define a new um automaton that we're going to mainly use as uh just to provide an example for us today i'm going to call this a linearly bounded automaton and all it is is a turing machine [Music] where the turing machine is going to be restricted in where it can uh the tape is not going to be infinite anymore the tape is just going to be big enough to hold the input so the machine no longer has the ability to move into the portion of the tape to the right of the input because there is no tape out there it just has the tape sitting here um that contains the input which can the tape itself can vary in size however big the input is that's how big the tape is so the tape adjusts to the length of the input but once you've started the machine with some particular input that's as big as the tape is there's no more um the reason why it's called linearly bounded is because the amount of memory is a linear function of the of the size of the input because you can effectively get somewhat more memory by enlarging the tape alphabet but that's going to be fixed for any given machine so that's where the linear comes from if that's helpful but if you don't get that that's sort of a side remark uh but um what's important to me is that your understand what i mean by an uh a linearly bounded automaton or an lba it's just like a turing machine but this that portion of the tape that originally had blanks is just not there the machine tries to move its head off the off the right end of the input it just sticks there just as if it tried to move its head off the left end of the input doesn't go anywhere okay so now i'm gonna we're gonna ask the same kinds of questions about lbas that we asked for other automata so the acceptance problem if i give you a an input and some particular lba and i want to know does the lba accept that input well you know and now the question is is that decidable um or not uh so it's um at first glance you might think well um an lba is like a turing machine and the atm problem is undecidable so uh you know that that might be a good first guess um [Music] and also if you try to uh to simulate them if you try to figure out how you would go about simulating the machine um you know if you if a given b and w if you actually try to simulate the machine to get the answer so you run b on w well of course if you run it for a while and it eventually halts either accepting or rejecting then you know the answer and you're finished but you know this machine might get into a loop um you know nothing to prevent the machine from looping on that finite amount of on that limited amount of tape that it has um and then you know you might be in trouble but in fact that's not the case because when you start out with a limited amount of tape if you run the machine for a long time and it's not holding it's going and going and going inevitably it's going to have to repeat um get into exactly the same configuration that it did before because there's only a limited number of configurations that the machine has and once it repeats a configuration it's going to be repeating that configuration forever and it's going to be in a loop so this problem in fact is decidable um because the idea is if bmw runs for for a very long time and an amount that you can calculate then you know it's got to be you know cycling you know more than just looping it's got to be repeating itself um and so therefore once it starts repeating itself it's going to be going forever so um and here's the actual calculation which is something i'm sure you could do on your own but just to spell it out so if you have an input of length n um that you're providing to uh b so if w is of length n um the the lba can only go uh for uh this number of different it can only have this number of different configurations the number of states times the number of head positions which is n the number of head positions on the tape times the number of different tape contents you know if if the tape was only one long you know this is the this is the size of the tape alphabet so if the tape were too long it would you know had the tape had two uh two cells on it the number of possible tape contents is would be um the square of the of the alphabet and then and if if the tape is going to be n uh symbols long it's going to be the tape alphabet size to the nth power uh so therefore if a turning machine runs for longer it's got to repeat some configuration and it'll never hold so the decider is going to be hopefully uh clear at this point you're given b and w says that this is the decider for alba it's going to run bmw for this number of steps if it's accepted by then then you accept and if it hasn't if it's rejected or it's still running then you can reject and you know if it's still running at this point it's never going to accept all right any questions on this okay let's move on um all right so now um but what's different now or what's it perhaps interesting now is that even though the acceptance problem for lbas is the is is decidable the emptiness problem is undecidable and that's where the computation history method is going to come in so this is where we're going to be starting to do something new in terms of a proof technique um because so we're going to reduce atm to elba the emptiness problem for lbas um so given an lba does it accept anything or not and this is going to use the computation history method so this will be a chance to illustrate that method for the first time so this the setup initially is just like before we're going to assume we have a turing machine that decides the cb elba problem that of interest and we're going to use that to construct turing machine deciding atm for our contradiction okay so here's s supposedly deciding atm what it's going to do and this is going to be the tricky part s is going to use m and w to design an llba that lba is going to be built with the knowledge of m w in fact it's going to have m and w built into it so that's why i'm calling that lba b sub mw because it depends on mw um as i'll describe and what that lba does it takes its input let's call it the x the input to the the lba and examines that input to see if that input is an accepting computation history for m on w then that lba is just looking for computation histories for mw and that's the only thing it's going to ever accept if you feed it something which is not a computation history you know that sequence of configurations uh for mmw leading to an accept if you feed it something else the lba is just going to reject it it only accepts the accepting computation history for m w and now if you can build such a thing then you can use that machine in your emptiness tester to see if it accepts any strings at all because the only thing that could possibly be accepting is an accepting computation history you don't even know if there exists one because you don't know if m accepts w so you're going to build this lba um and uh which is looking for accepting computation histories and then see if its language is empty or not if its language is not empty the only thing it could be accepting is this computation history so you know m except w whereas if the language is empty you know that there is no computation history for m w and so m does not accept the leaf so that's the whole idea the trick is how do you build this uh this lba um so first you're going to make this lba which tests its input to see if it's an accepting computation history for m on w and you have to build this um uh lba without even knowing whether there is a computation history um it's not like you can take that string and just build the string into the lba because they don't even know if that string exists but what you can do is you can make the lba um follow the rules of m it knows w and knows m so it can test take the its input and using w and the rules of m see if that input is a computation history for m and w um and that's what it's going to do so here is um this machine i'm going to construct i'll i'll give it to you written down and with a little bit more um details of its procedure but just to kind of illustrate it so here is a proposed input to b sub mw this would be the x here this would be an x that would be accepted but anything else that would be provided would not be accepted so this is the sequence of configurations written down as a string that would be the x that i'm providing to the b c b b sub mw and it's supposed to be checking this to make sure it's legit um so so on on input x the way it's gonna the way it's going to proceed it's going to first check to see whether x begins in the right way because you know this lba it knows m and it knows w so the very first thing is it takes a look at the first part of the string up to the pound sign you know there's no pound sign if you're going to just feed junk in it's going to be easily identifiable as junk um so uh you know if you're going to feed something so it's going to the lba is going to take everything up to the first pound sign and just confirm that that thing is the first configuration the starting configuration of um of m on w which means it has to start with the start state and here is w so just going to check that um then it's going to check that each each one of these guys follows legally from the previous one according to the rules of f it's going to first checks that this one is correct and then this one leads to that one according to the rules of m then this one leads to that one according to the rules of m all of that you know the knowledge of m and w is built in so it can do that um and then it just follows along checking this computation you know it's easy to check that a computation is correct that's all that's really going on here um it's going to check that the computation is correct until it gets to the end and then make sure that the last uh configuration that it's been given as input is an accepting configuration that there's an accept state in it um and if everything that passes it it accepts otherwise it uh rejects okay um and the point is that this is an lba you don't oh i just kind of jumped ahead of myself you don't need any additional information so how does it actually do this so you know how do you actually do this on the tape so i claim that the lba doesn't need any extra space to do this um to do this check you know it's just going to be zigzagging back and forth here on the on on the on the input you know checking that the corresponding symbols match up except around the the head position where it gets updated or correctly and it seems it may need to mark it will need to mark on the tape it's allowed to write on the tape just to make sure it keeps track of where it is but this is a very simple i mean i'm not trying to uh if you're not following i'm not trying to uh alarm you but i'm just trying to help you understand that this is not a complicated procedure here to do this check that the that the actual computation of the that's written down of the turing machine is valid um but you know we can deal with a question here uh oh good now we got some questions oops you can be thinking about while i'm thinking about this you can think about that uh check-in over there so here's a good question going from each configuration to the next configuration is it a unique uh next step um well i'm assuming that the turing machine we're starting with is deterministic so this should be only one way to go so the answer to that question is yes this is going to be a unique um string here there's really going to be only one um computation history not that it really matters in a sense um the answer to that question um as long as that accepting computation history corresponds to the machine actually accepting uh oh boy we're all all over the place here uh okay are you ready to end this um in this um in this bowl uh all right well this is where we are uh everybody who said they'd rather be in six or four fig six or four six i know who you are you're all going to be expelled no i'm joking uh i don't know who you are so um uh yeah good so um uh wha what so we are here at the at the break um i and i made this poll on purpose uh at this moment so we can spend a little time let me just start our clock going and i can try to help you those of you who are who answered see that you're baffled um i'll i can try to uh help you understand what's going on all right so this is a this is a good question here um uh okay well okay there's several good questions here so um somebody asks why don't we just test all possible strings uh for b remember if we're trying to test emptiness for b's language um that's the e-lba problem nor do we just try all possible strings um and uh that would work if we had enough time but there's infinitely many strings and so that's not going to be good if we're trying to be a decider so so that is that's why we don't just try all strings so but here's this other question here this is a okay so um the question is how do we find the input x so this is an important question um because uh we don't find the input x there's no you know we the input x um at least the accepted input x would be the accepting computation history um for m on w we're never going to find it you know we're never going to find an x we're not we're building this um this lba not we're never going to run that lba we're designing that lba not for the purposes of running it we're building that lba only for the purpose of uh using the lba language emptiness tester we're going to build that end lba and feed it into the machine turing machine r which is going to tell us whether that lba's language is empty we ourselves is not never going to run that machine we're never going to come up with an x we're just having we're we're we are designing a computation checker and then um the emptiness tester for that that we assumed to exist for for elba is going to say yes there is some computation which this machine accepts or no there is no computation which this machine accepts and that's going to be a computation for m on w and so that's going to tell us whether or not m accepts w so we're never going to actually be finding an x we're never going to be running that lba we're just feeding using that lba as as an input to r um does the computation history method always use lbas no as you'll see uh right after the break because that's the lbas is just kind of an easy place to get started but we're going to use the computation history method and computation histories next on the on the post correspondence problem yes the the computation histories and the input x are always going to be finite so that that was an answer to a question about whether these histories or the inputs are going to be finite or not so the the they're all those strings are always going to be finite and the computation history is going to be finite um so we are at the end of our five minutes um let me just see if there was another question i wanted to answer um let me let's move on okay now getting coming back to the undecidability of this post-correspondence problem so remember the post course upon you know this this problem you're given those dominoes you want to know if there's a match here is a little mini version of the the diagram if that helps you remember it um and we're going to prove that this language is undecidable and so it's undecidable to test whether you have a match given a set of dominoes whether a match is possible and we're going to use we're going to reduce atm to to this language using the computation history method so how in the world are we going to do that um first of all there's a little detail here i want to mention uh just uh um don't focus on this but i'm going to assume the matches always start with the very first domino on the list um the very first domino in the collection so this is going to be like a starting domino this is going to make my proof a little simpler to do it that way and then you can fix that assumption uh later and if we have time at the end i'll do that or maybe after the lecture is over if there are questions i can show you how to fix that assumption but for now um to simplify the proof we're going to assume that there is a starting domino that the matches always have to start with that particular domino um if you didn't follow that point don't worry you can just ignore it you'll see where it comes up later so um now we're going to reduce atm to the pcp so assuming we have a machine that decides the pcp we're going to use that to make a machine that decides the atm as before so here's the turing machine s which is going to decide atm um and the way we're going to do it is like this we're given m and w we want to know as always does m accept w what we're going to do is we're going to build an instance of the pcp problem so we're going to build a bun a collection of dominoes [Music] uh which are going to dip and that collection is going to be built knowing m w so that's going to affect the dominoes we're going to create so this collection of dominoes is going to be called p sub mw depends on an m w and finding a match for this set of w is is going to force you to simulate m on w because the match is going to correspond to a computation history for m on w so uh we're going to use once we do that um once we build uh the set of dominoes who's where a match corresponds to a computation history we're going to use r to determine whether or not there is a match or in other words whether or not there is a computation history which is whether or not m accepts w um so if there is a match we're going to accept because we know m accepts w and if and if there is no uh match we're going to reject because we know m does not accept w okay um this is the plan i haven't told you how to do this yet i mean i'm worried about the significant number of you who are feeling um confused by the method i mean you should you guys should be uh texting me in the tas um to try to at least get a sense of how this is working i mean that we're going to be going we're going to be doing the method again um it's just going to be a little more complicated because we have to also deal with these dominoes and all this all that stuff and that's why i presented it to the fir to you the first time in the setting of the lbas which where in a sense the method comes out perhaps more simply so a match is going to correspond to an accepting computation history um sometimes i if i don't use the word accepting um you know i'm just being a little sloppy computation history and accepting computation history for us right now they're the same i mean later on we may actually talk about rejecting computation histories but let's not get ourselves confused all the except computation histories always have to end with the machine accepting okay uh somebody's asking what does it mean for a match to correspond to a computation history you'll see that's going to be on my next slide oh so this is a good question um is my step two trying to use r to determine whether m accepts w well yes but i'm doing that by testing whether whether this these dominoes have a match because r is besides pcp i can only use r to test whether things have a match because someone asked should step two be to use r to determine whether m accepts w well in effect that's what it's doing but it's doing indirectly through um testing whether this um pcp whether these dominoes have a match because those dominoes are going to force you to simulate mlw okay i think i'm repeating myself here so let's uh avoid getting into a loop on the lecture and um uh see how we actually build piece of mw so now you understand what you have to under have the plan in your mind uh um we are given m w we're trying to make a set of dominoes where a match is going to be a computation history for m w so the string you're going to get in that match the top string and the bottom string which are going to be you have to have to have to match um there they're going to be computation histories but they're going to be a computation history of m w so i want to figure out how to make my dominoes force that okay so my starting domino as i told you there's going to be a special starting domino in my collection so which is going to require the match to start with that domino it's going to be this one here it's going to have these two strings in it and you can see already it's starting to look like a a computation history the dominoes going to have that feature so uh it's the pair of strings and i've written it kind of to help you see what's kind of going on it's a pound sign on the top and the the starting configuration for mmw on the bottom and what i'm going to do for you here is at the bottom of the slide i'm going to take the dominoes i've written down so far and try to be building a match and you'll see how that match is forcing a simulation of the machine okay so let's take this as an illustration let's assume the input to m so i'm running m on w now i'm trying to see does m accept w um w is a string two two three so the start configuration for m on w is q zero that's the starting state for m and then w following is 2 2 3 that's what it's going to appear on the on the tape of the turing machine to start off so this is the start configuration for on w and assuming that i had this particular string is input string to w and so uh given the match the dominoes that i've given you so far just one um this is how the match is going to start uh now there's going to be more dominoes coming so for every possible tape symbol and state don't get confused by the language here this is the important thing if if in the turing machine um and i'll just read this in english to you if the turing machine when it's in state q and the head is reading an a it moves to state r writes a b at that point and its head moves to the right i'm focusing on right rightward moving uh turing machine steps right now but for every possible uh state and and tape symbol and looking at where what happens i'm going to have a domino that's going to capture this information and it's going to be this domino here q a on top br on the bottom so just the string qa on the top br on the bottom so let's see why with that sort of arcane looking why is that a good thing to why is that a useful domino to put in here well if you take a look um let's assume from my turing machine when it's in state q0 which is the starting state and the head is reading a two which it will happen to be reading because the uh this the string w starts with a two if it goes into state q4 no q7 and writes a four and then moves right i'm going to have in this in this domino i'm going to have q02 on top and 4 q7 on the bottom because 4 is what i wrote right here that's you know that's the b and this and the new state that i'm going into is q7 so that's going to be this the domino that i'm going to get and here it is here's that domino appearing in the match so um it's going to take match up with q q 0 2 um which you know you know don't forget the the the top string has to equal the bottom string so and this is going to be the only choice that i have for extending the match so q02 um is going to be on the top when i put that there but that's going to force um 427 to appear at the bottom because that's what's going to be the bottom string corresponding to the q02 on the top okay so if you're looking down here this is the the beginning of the second configuration of the machine that's what i want to be happening i want to be for that match should look like a computation history so now um so this is going to be all possible right moves are going to be handled in this way it's going to be a similar process for the left moves but let me not you know i'll leave that to your imagination now how do i continue on from there what does the rest of this configuration look like well that should just be a copying over of the remaining symbols that were on the tape because they don't change things only change around the head the rest of those symbols which is the two and the three those should just get copied over here so i'm gonna have two additional i'm gonna have additional uh symbols in my dominoes in my uh in my collection so for every um tape symbol i'm gonna have a a be a domino so and for every type symbol a i'm going to have a a be a domino in my collection and so that says uh i can have uh so that there's going to be a 2 2 domino so i can match up this 2 over here but that forces me to put a 2 down over there there's also going to be a 3 3 domino which is going to match up with that 3 but it's going to force me to push it put a 3 down over there that's the only way i can extend this um match that i've built so far i have no choice those are the only dominoes that i'm going to be given and by and so doing i'm forcing you to basically simulate the machine now what i want to have happen next is a pound sign to appear here and that will conclude my second configuration so there's going to be a pound sign pound sign domino as well because there's a pound sign here it's going to get matched with a pound sign up top forces a pound sign to come down um on the bottom and if you look at where we are right now we're exactly like where we were at the beginning when we had just this first domino appearing but now we're one configuration later so if you understood that and i admit it it's it's you know this is a there's just one idea here um [Music] and once you get the idea it's all trivial it's all very simple but there's just you have to get that idea i'm trying to figure out how to get that idea into your head once you get the idea you can write all this stuff yourself um so uh so following the this uh the this description of how uh the you know the the transition function of the machine works and copying over uh the symbols from the tape to the next configuration i'm going to be able to get configuration after configuration going until i get to a point when there's an accept and now from the machine's perspective we're done the machine has accepted its input this is our computation history but is it a match well you know up until this this bottom thing um is the computation history but it's not a match because the top doesn't equal the bottom there's still extra stuff at the bottom which the top is not uh you know it doesn't have so what i'm going to need to do now is is add some additional kind of pseudo steps of the machine where i'm going to allow the top to catch up to the bottom and the way i'm going to be thinking about that and this is not real for a turning machine as much as the turn machine's real anyway but you're going to imagine i'm going to add a new kind of uh move to the machine which is going to allow the head to eat the symbols off the tape like pac-man okay and the way i'm gonna get that effect um on the top if i have any tape symbol next to a cue except on either side on the top what i'm going to get you to write on the bottom is just to accept with that taste with that tape symbol gone and by repeating that you know move after move the the the actual tape is going to be shrinking the symbols on the tape are going to be going down one by one move by move until there's nothing left except just the um you accept all by itself okay so i have this sort of pac-man idea of eating the symbols on the tape um and then finally i get to a point when there's just the cue except alone on the tape there's no symbols left to consume and then i'm going to add one last domino here which is q accept pound pound matching with just pound which just conveniently finishes off the um uh the uh the match and so the match is completed there's actually a little detail here i i hesitate even to bring it up because i i think it's a kind of thing where if you understand everything up to this point you could fill it in yourself but just for completeness sake uh you have to deal with the situation when the the uh the head um of the turing machine might move into the blank portion of the tape which is not taken into account here and the way we get that effect is by having another domino here which allows me to add blank symbols at the bottom as needed that's just a detail you know i'm more worried that you understand the underlying concept um so here's going to be um uh a check-in um trying to remember um okay but you know so okay so what else can we conclude from this information we know so at this point we know that pcp is undecidable that's what we just finished proving what else do we know if anything or do we even know that uh see my picture is a little bit covering the check in but it's still readable so i mean this is not an easy concept to get the to get the idea but once you get it you know it's um you know it's you'll see it's not that bad okay another 15 seconds so this portion doesn't this question is not really relying so much on the computation history method it's just relying on the fact about pcp being undecidable um [Music] okay everybody almost done five seconds all right um so i can see that there is uh some level of confusion um uh so the reason why b is correct is that we know pcp is undecidable but we also know that it's recognizable because you could try all possible uh ways of combining dominoes and one after the next um and you accept if you ever find a match so they might go forever but if there is a match uh possible you'll find it eventually so pcp is a recognizable language undecidable and so therefore we know it's complement it has to be unrecognizable because you know that's something we've shown before if a language and its complement are both recognizable then it's decidable but we know pcp is not decidable so they both both sides can't be recognizable um okay so let's let me prove to you one last theorem that's going to be useful for your homework uh involving the computation history method so you'll have one last chance to get the way we're going to use this though this one is going to be a little bit more similar in spirit to the lba version uh than to the pcp version which has this extra kind of complication about coding the computations of the machine into dominoes here um you know we're going to operate with another automaton which where it's going to be more straightforward but anyway okay getting ahead of myself so if you remember for the for context-free grammars we had the problem the emptiness problem we showed that was decidable so testing whether a context-free grammar's language is empty is decidable however testing whether a context-free grammar's language is everything whether it's equal to sigma star that turns out to be um uh that turns out to be undecidable okay so emptiness testing for context-free grammars decidable sigma-star testing undecidable um okay so we'll show that atm is reducible to all pda via the computation history method um and uh so uh assume we have same patterns assuming we have a decider for all pda and make a decider for ulti decider for atm here's the atm decider and now similar to what we did for the uh lba case but with a with a twist we're going to make a push down automaton that's going to check its input to see whether it's an accepting computation history of m w just like the lba did if you think back to how that worked remember remember you know we um the lba tested its input to see whether it's uh an accepting computation history accept it if it did and then we test the lba's language for emptiness to see if there are any computation histories so we're doing the same kind of thing except now uh the pda is going to test its input to see whether it's a an accepting computation history and if it is it's going to reject it's going to do the reverse of what the lba did and that's going to turn out to be necessary but for the for the moment let's just go with it and maybe we'll see in the proof why the proof needs it to be this way so um this p pushdown automaton is going to accept its input if it's not an accepting computation history for m on w otherwise it will accept so you can think of this pda as accepting all the junk it just doesn't accept um the good stuff the accepting computation history okay it's it's just it it's uh accepting all the things which fail to be an accepting computation history okay and then once we have that we we test whether ours language is everything but whether bmw's languages everything using r because it if if that pda's language is everything then we know um uh there could not be an accepting computation history because that's the one thing that gets um not accepted that's the one thing that gets rejected so if it's accepting everything there is not going to be an accepting computation history so if there is no uh if this is equal to sigma star then there couldn't be an accepting computation history and so we're going to accept okay so how is this going to work so what's different now about uh this case uh from the lba case is remember the lba uh got you know in the the computation history on its input and it used its ability to write on the tape to check that each configuration follows the next one we don't have the ability to write on the tape and push on automaton and by the way i hope you're all comfortable with my using pdas instead of grammars because we can interchange one to the other um i should have mentioned that when when i did it but um so the the pda is going to be using its stack to compare each configuration with the next one okay so the way that's going to happen is it's going to non-deterministically take one of these so the very first thing it does is as before it checks to make sure that the beginning of the input is the start configuration but then once that's that's the easy part but once we get going with that we can we push each configuration well we first of all we now deterministically choose which configuration might be the bet the one that fails um where where one fails to go to the next one because those are the ones we're trying to accept when there's a failure so we now deterministically look for the place that there's a failure we push that onto the stack um and then we pop it off the stack and compare it with the next configuration so that's how we're using the stack instead of being able to mark on the input now there's a so i'm kind of going to illustrate that here so you know we're going as we're going to read this thing here and we're going to put it onto the stack so this thing moves over to here and this input here got put onto the stack you know q 0 w 1 w 2 it's q 0 w 1 w 2 you see that that first configuration is now sitting on the stack now as we're going to pop it off we're going to match it with the second configuration now if you're following me you'll realize that there's a there's a difficulty here because it's coming out in reverse you know it comes out in the reverse order that we put it in and that's not what we need to do so what we're going to do here and here's a little sort of a twist we're going to change the way we're writing down computation histories by making them by writing them uh reversing the even numbered ones so this one here is going to be written down in reverse and that's perfectly okay we can write down configuration computation histories in any way we want to um to meet our needs so we're going to reverse we're going to reverse the alternate ones and now when we push one it's going to come off in the right order to compare with the next one and so that's how the the procedure works okay we're running a little short on time um you need to be let's see so let let me just go to um uh quick recap the computation history method is useful for showing the undecidability of problems when you're testing for the existence of an object each of those four cases that we showed today involving testing whether something exists so is there an integer solution to the polynomial is there some string in the language is there a string that's not in the language um or is there a match so that's a typical case when um this computation history method comes up and so as a quick review uh these are the things we showed today okay so why don't we uh uh we're we're just out of time um and so i'm i'm gonna let you go um but i will stick around uh for another five minutes or ten minutes uh happy to answer any questions if you if you have them um but that's that's officially the end of the lecture um okay so let's let me try to get to there's a lot of questions in the chat don't forget you know write to the tas too so if m does not accept w what will happen to the computation history so if m does not accept w um uh then there is no computation there is no accepting computation history that's the only kind of computation histories we're considering so if m does not accept w there is no accepting computation history there's no computation history um so i don't know what it means what will happen to it it just doesn't doesn't exist um so uh i hope that's helpful um don't we need to be able to add states to the end of the tape i don't know what that means it's taped to the end of the tape i i don't know you'll have to repeat that one sorry or explain that better um ah so this is a good this is a very good question here where does the previous proof fail when trying to reduce atm to ecfg it's got to fail because ecfg is decidable um so that that's a very good question maybe we can just go back to that last slide here because i was running a little short on time i didn't really focus on why we have to accept um the non-computation history strings we why are we accepting all the junk strings and only rejecting um the strings which are the uh the computation histories so you know we're looking for strings that fail you know if the string could fail because it starts wrong string it doesn't have the start configurat configuration or it doesn't maybe doesn't end with the accepting configuration or maybe one configuration doesn't lead properly to the next one any of those cases are a failure and there are going to be reason to accept the reason why we can do do it do that um is because the we're taking advantage of the push downs non-determinism we don't know where the um the failure might occur so we're going to non-deterministically guess where that failure occurs if we were going to flip this around and say let's accept only the good ones and reject everything else we would have to test that each configuration led to the next one so we have to check them all so if something fails it only has to fail in one place and then you know we can accept if it um if it's a good computation history we have to it has to be good everywhere and so the push down automaton really if you're going to get down to the nitty-gritty the push on automaton can check that this configuration leads to that configuration using pushing on the stack and then popping it but now by the time you get to here you want to check that this second configuration leads to the third one you would need to push the second configuration onto the stack but at this point you're already at after the second configuration so we cannot back up and push the second one on the stack so the the push on tom dot is not able to check um in the positive sense that you have a computation history it's only able to and accept them it's only to check in the negative sense that you don't have a con a computation history and uh accept all the ones that fail and that's just because it only has to fail in one place i hope that helps um so that's why we couldn't flip this around and make this proof work for the emptiness problem and only have to work for the the everything problem the old problem all right um okay so bye bye everybody and i will see you on tuesday 3 00:00:29,589 --> 00:00:31,189 4 00:00:31,189 --> 00:00:31,199 5 00:00:31,199 --> 00:00:32,790 6 00:00:32,790 --> 00:00:35,030 7 00:00:35,030 --> 00:00:36,630 8 00:00:36,630 --> 00:00:39,430 9 00:00:39,430 --> 00:00:45,750 10 00:00:45,750 --> 00:00:45,760 11 00:00:45,760 --> 00:00:47,110 12 00:00:47,110 --> 00:00:48,389 13 00:00:48,389 --> 00:00:50,950 14 00:00:50,950 --> 00:00:50,960 15 00:00:50,960 --> 00:00:53,350 16 00:00:53,350 --> 00:00:54,790 17 00:00:54,790 --> 00:00:56,790 18 00:00:56,790 --> 00:00:58,950 19 00:00:58,950 --> 00:01:03,029 20 00:01:03,029 --> 00:01:06,230 21 00:01:06,230 --> 00:01:06,240 22 00:01:06,240 --> 00:01:07,190 23 00:01:07,190 --> 00:01:08,950 24 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348 00:11:06,790 --> 00:11:08,230 349 00:11:08,230 --> 00:11:11,269 350 00:11:11,269 --> 00:11:13,509 351 00:11:13,509 --> 00:11:15,590 352 00:11:15,590 --> 00:11:18,710 353 00:11:18,710 --> 00:11:21,910 354 00:11:21,910 --> 00:11:24,630 355 00:11:24,630 --> 00:11:26,790 356 00:11:26,790 --> 00:11:31,829 357 00:11:31,829 --> 00:11:34,230 358 00:11:34,230 --> 00:11:37,190 359 00:11:37,190 --> 00:11:39,110 360 00:11:39,110 --> 00:11:40,949 361 00:11:40,949 --> 00:11:40,959 362 00:11:40,959 --> 00:11:42,630 363 00:11:42,630 --> 00:11:44,230 364 00:11:44,230 --> 00:11:46,069 365 00:11:46,069 --> 00:11:47,990 366 00:11:47,990 --> 00:11:50,230 367 00:11:50,230 --> 00:11:53,030 368 00:11:53,030 --> 00:11:56,710 369 00:11:56,710 --> 00:11:59,030 370 00:11:59,030 --> 00:12:00,310 371 00:12:00,310 --> 00:12:03,030 372 00:12:03,030 --> 00:12:03,040 373 00:12:03,040 --> 00:12:04,389 374 00:12:04,389 --> 00:12:06,230 375 00:12:06,230 --> 00:12:08,310 376 00:12:08,310 --> 00:12:11,990 377 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--> 00:13:02,230 407 00:13:02,230 --> 00:13:04,230 408 00:13:04,230 --> 00:13:07,030 409 00:13:07,030 --> 00:13:09,590 410 00:13:09,590 --> 00:13:12,150 411 00:13:12,150 --> 00:13:13,430 412 00:13:13,430 --> 00:13:14,949 413 00:13:14,949 --> 00:13:16,150 414 00:13:16,150 --> 00:13:18,230 415 00:13:18,230 --> 00:13:20,550 416 00:13:20,550 --> 00:13:22,790 417 00:13:22,790 --> 00:13:25,190 418 00:13:25,190 --> 00:13:25,200 419 00:13:25,200 --> 00:13:27,269 420 00:13:27,269 --> 00:13:28,949 421 00:13:28,949 --> 00:13:30,870 422 00:13:30,870 --> 00:13:33,190 423 00:13:33,190 --> 00:13:33,200 424 00:13:33,200 --> 00:13:34,629 425 00:13:34,629 --> 00:13:35,670 426 00:13:35,670 --> 00:13:37,750 427 00:13:37,750 --> 00:13:39,750 428 00:13:39,750 --> 00:13:41,110 429 00:13:41,110 --> 00:13:42,470 430 00:13:42,470 --> 00:13:45,030 431 00:13:45,030 --> 00:13:46,829 432 00:13:46,829 --> 00:13:48,870 433 00:13:48,870 --> 00:13:48,880 434 00:13:48,880 --> 00:13:50,949 435 00:13:50,949 --> 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465 00:14:50,160 --> 00:14:51,030 466 00:14:51,030 --> 00:14:54,790 467 00:14:54,790 --> 00:14:58,150 468 00:14:58,150 --> 00:14:59,750 469 00:14:59,750 --> 00:15:01,350 470 00:15:01,350 --> 00:15:03,110 471 00:15:03,110 --> 00:15:03,120 472 00:15:03,120 --> 00:15:04,550 473 00:15:04,550 --> 00:15:04,560 474 00:15:04,560 --> 00:15:05,509 475 00:15:05,509 --> 00:15:07,590 476 00:15:07,590 --> 00:15:09,990 477 00:15:09,990 --> 00:15:14,389 478 00:15:14,389 --> 00:15:16,310 479 00:15:16,310 --> 00:15:18,069 480 00:15:18,069 --> 00:15:19,590 481 00:15:19,590 --> 00:15:20,870 482 00:15:20,870 --> 00:15:22,389 483 00:15:22,389 --> 00:15:22,399 484 00:15:22,399 --> 00:15:27,110 485 00:15:27,110 --> 00:15:27,120 486 00:15:27,120 --> 00:15:27,910 487 00:15:27,910 --> 00:15:27,920 488 00:15:27,920 --> 00:15:30,230 489 00:15:30,230 --> 00:15:33,189 490 00:15:33,189 --> 00:15:33,199 491 00:15:33,199 --> 00:15:34,550 492 00:15:34,550 --> 00:15:34,560 493 00:15:34,560 --> 00:15:35,350 494 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--> 00:16:22,880 524 00:16:22,880 --> 00:16:25,269 525 00:16:25,269 --> 00:16:27,269 526 00:16:27,269 --> 00:16:28,949 527 00:16:28,949 --> 00:16:32,389 528 00:16:32,389 --> 00:16:34,629 529 00:16:34,629 --> 00:16:36,389 530 00:16:36,389 --> 00:16:36,399 531 00:16:36,399 --> 00:16:38,389 532 00:16:38,389 --> 00:16:39,990 533 00:16:39,990 --> 00:16:43,110 534 00:16:43,110 --> 00:16:43,120 535 00:16:43,120 --> 00:16:43,990 536 00:16:43,990 --> 00:16:44,000 537 00:16:44,000 --> 00:16:46,069 538 00:16:46,069 --> 00:16:48,949 539 00:16:48,949 --> 00:16:52,470 540 00:16:52,470 --> 00:16:54,629 541 00:16:54,629 --> 00:16:55,749 542 00:16:55,749 --> 00:16:57,110 543 00:16:57,110 --> 00:16:59,350 544 00:16:59,350 --> 00:17:01,110 545 00:17:01,110 --> 00:17:02,790 546 00:17:02,790 --> 00:17:05,110 547 00:17:05,110 --> 00:17:06,549 548 00:17:06,549 --> 00:17:09,829 549 00:17:09,829 --> 00:17:11,429 550 00:17:11,429 --> 00:17:13,429 551 00:17:13,429 --> 00:17:15,110 552 00:17:15,110 --> 00:17:16,470 553 00:17:16,470 --> 00:17:18,789 554 00:17:18,789 --> 00:17:21,510 555 00:17:21,510 --> 00:17:23,350 556 00:17:23,350 --> 00:17:24,710 557 00:17:24,710 --> 00:17:26,710 558 00:17:26,710 --> 00:17:29,510 559 00:17:29,510 --> 00:17:31,110 560 00:17:31,110 --> 00:17:32,870 561 00:17:32,870 --> 00:17:32,880 562 00:17:32,880 --> 00:17:34,870 563 00:17:34,870 --> 00:17:36,230 564 00:17:36,230 --> 00:17:36,240 565 00:17:36,240 --> 00:17:38,230 566 00:17:38,230 --> 00:17:38,240 567 00:17:38,240 --> 00:17:39,110 568 00:17:39,110 --> 00:17:41,590 569 00:17:41,590 --> 00:17:43,620 570 00:17:43,620 --> 00:17:43,630 571 00:17:43,630 --> 00:17:44,830 572 00:17:44,830 --> 00:17:48,150 573 00:17:48,150 --> 00:17:48,160 574 00:17:48,160 --> 00:17:50,150 575 00:17:50,150 --> 00:17:51,669 576 00:17:51,669 --> 00:17:53,830 577 00:17:53,830 --> 00:17:56,070 578 00:17:56,070 --> 00:17:56,080 579 00:17:56,080 --> 00:17:57,190 580 00:17:57,190 --> 00:17:59,590 581 00:17:59,590 --> 00:18:01,029 582 00:18:01,029 --> 00:18:01,039 583 00:18:01,039 --> 00:18:03,430 584 00:18:03,430 --> 00:18:03,440 585 00:18:03,440 --> 00:18:04,230 586 00:18:04,230 --> 00:18:05,750 587 00:18:05,750 --> 00:18:06,710 588 00:18:06,710 --> 00:18:08,789 589 00:18:08,789 --> 00:18:11,110 590 00:18:11,110 --> 00:18:12,470 591 00:18:12,470 --> 00:18:12,480 592 00:18:12,480 --> 00:18:13,510 593 00:18:13,510 --> 00:18:14,950 594 00:18:14,950 --> 00:18:14,960 595 00:18:14,960 --> 00:18:16,549 596 00:18:16,549 --> 00:18:18,310 597 00:18:18,310 --> 00:18:21,830 598 00:18:21,830 --> 00:18:23,430 599 00:18:23,430 --> 00:18:25,669 600 00:18:25,669 --> 00:18:28,070 601 00:18:28,070 --> 00:18:29,909 602 00:18:29,909 --> 00:18:32,710 603 00:18:32,710 --> 00:18:34,390 604 00:18:34,390 --> 00:18:36,710 605 00:18:36,710 --> 00:18:38,549 606 00:18:38,549 --> 00:18:41,029 607 00:18:41,029 --> 00:18:42,710 608 00:18:42,710 --> 00:18:42,720 609 00:18:42,720 --> 00:18:43,350 610 00:18:43,350 --> 00:18:43,360 611 00:18:43,360 --> 00:18:44,310 612 00:18:44,310 --> 00:18:44,320 613 00:18:44,320 --> 00:18:45,430 614 00:18:45,430 --> 00:18:48,070 615 00:18:48,070 --> 00:18:49,990 616 00:18:49,990 --> 00:18:51,350 617 00:18:51,350 --> 00:18:53,270 618 00:18:53,270 --> 00:18:55,270 619 00:18:55,270 --> 00:18:57,510 620 00:18:57,510 --> 00:18:57,520 621 00:18:57,520 --> 00:18:59,909 622 00:18:59,909 --> 00:19:02,310 623 00:19:02,310 --> 00:19:04,070 624 00:19:04,070 --> 00:19:06,710 625 00:19:06,710 --> 00:19:08,950 626 00:19:08,950 --> 00:19:10,950 627 00:19:10,950 --> 00:19:12,549 628 00:19:12,549 --> 00:19:14,710 629 00:19:14,710 --> 00:19:16,549 630 00:19:16,549 --> 00:19:18,470 631 00:19:18,470 --> 00:19:20,230 632 00:19:20,230 --> 00:19:20,240 633 00:19:20,240 --> 00:19:21,270 634 00:19:21,270 --> 00:19:23,029 635 00:19:23,029 --> 00:19:25,029 636 00:19:25,029 --> 00:19:26,950 637 00:19:26,950 --> 00:19:29,510 638 00:19:29,510 --> 00:19:32,310 639 00:19:32,310 --> 00:19:32,320 640 00:19:32,320 --> 00:19:33,510 641 00:19:33,510 --> 00:19:36,390 642 00:19:36,390 --> 00:19:37,669 643 00:19:37,669 --> 00:19:40,549 644 00:19:40,549 --> 00:19:40,559 645 00:19:40,559 --> 00:19:42,710 646 00:19:42,710 --> 00:19:44,470 647 00:19:44,470 --> 00:19:45,830 648 00:19:45,830 --> 00:19:48,710 649 00:19:48,710 --> 00:19:49,510 650 00:19:49,510 --> 00:19:51,669 651 00:19:51,669 --> 00:19:54,310 652 00:19:54,310 --> 00:19:56,230 653 00:19:56,230 --> 00:19:58,310 654 00:19:58,310 --> 00:19:58,320 655 00:19:58,320 --> 00:19:59,350 656 00:19:59,350 --> 00:20:01,110 657 00:20:01,110 --> 00:20:03,270 658 00:20:03,270 --> 00:20:05,430 659 00:20:05,430 --> 00:20:07,110 660 00:20:07,110 --> 00:20:08,789 661 00:20:08,789 --> 00:20:08,799 662 00:20:08,799 --> 00:20:09,669 663 00:20:09,669 --> 00:20:11,350 664 00:20:11,350 --> 00:20:11,360 665 00:20:11,360 --> 00:20:12,149 666 00:20:12,149 --> 00:20:14,950 667 00:20:14,950 --> 00:20:17,350 668 00:20:17,350 --> 00:20:19,750 669 00:20:19,750 --> 00:20:19,760 670 00:20:19,760 --> 00:20:21,110 671 00:20:21,110 --> 00:20:22,870 672 00:20:22,870 --> 00:20:25,029 673 00:20:25,029 --> 00:20:26,149 674 00:20:26,149 --> 00:20:28,310 675 00:20:28,310 --> 00:20:28,320 676 00:20:28,320 --> 00:20:29,270 677 00:20:29,270 --> 00:20:31,110 678 00:20:31,110 --> 00:20:33,510 679 00:20:33,510 --> 00:20:35,669 680 00:20:35,669 --> 00:20:37,909 681 00:20:37,909 --> 00:20:41,990 682 00:20:41,990 --> 00:20:43,510 683 00:20:43,510 --> 00:20:45,750 684 00:20:45,750 --> 00:20:47,990 685 00:20:47,990 --> 00:20:50,470 686 00:20:50,470 --> 00:20:52,950 687 00:20:52,950 --> 00:20:55,190 688 00:20:55,190 --> 00:20:58,630 689 00:20:58,630 --> 00:21:00,549 690 00:21:00,549 --> 00:21:02,950 691 00:21:02,950 --> 00:21:02,960 692 00:21:02,960 --> 00:21:04,070 693 00:21:04,070 --> 00:21:05,510 694 00:21:05,510 --> 00:21:07,750 695 00:21:07,750 --> 00:21:11,590 696 00:21:11,590 --> 00:21:13,669 697 00:21:13,669 --> 00:21:15,270 698 00:21:15,270 --> 00:21:17,190 699 00:21:17,190 --> 00:21:19,350 700 00:21:19,350 --> 00:21:19,360 701 00:21:19,360 --> 00:21:20,310 702 00:21:20,310 --> 00:21:21,909 703 00:21:21,909 --> 00:21:23,270 704 00:21:23,270 --> 00:21:25,110 705 00:21:25,110 --> 00:21:26,470 706 00:21:26,470 --> 00:21:26,480 707 00:21:26,480 --> 00:21:28,870 708 00:21:28,870 --> 00:21:28,880 709 00:21:28,880 --> 00:21:31,029 710 00:21:31,029 --> 00:21:33,270 711 00:21:33,270 --> 00:21:35,350 712 00:21:35,350 --> 00:21:35,360 713 00:21:35,360 --> 00:21:36,630 714 00:21:36,630 --> 00:21:38,230 715 00:21:38,230 --> 00:21:41,669 716 00:21:41,669 --> 00:21:43,990 717 00:21:43,990 --> 00:21:45,830 718 00:21:45,830 --> 00:21:47,909 719 00:21:47,909 --> 00:21:47,919 720 00:21:47,919 --> 00:21:48,710 721 00:21:48,710 --> 00:21:51,270 722 00:21:51,270 --> 00:21:52,950 723 00:21:52,950 --> 00:21:55,270 724 00:21:55,270 --> 00:21:57,270 725 00:21:57,270 --> 00:21:59,669 726 00:21:59,669 --> 00:22:00,710 727 00:22:00,710 --> 00:22:03,190 728 00:22:03,190 --> 00:22:05,669 729 00:22:05,669 --> 00:22:07,590 730 00:22:07,590 --> 00:22:11,190 731 00:22:11,190 --> 00:22:12,710 732 00:22:12,710 --> 00:22:13,909 733 00:22:13,909 --> 00:22:16,870 734 00:22:16,870 --> 00:22:18,390 735 00:22:18,390 --> 00:22:20,870 736 00:22:20,870 --> 00:22:20,880 737 00:22:20,880 --> 00:22:22,789 738 00:22:22,789 --> 00:22:23,909 739 00:22:23,909 --> 00:22:26,870 740 00:22:26,870 --> 00:22:29,510 741 00:22:29,510 --> 00:22:31,430 742 00:22:31,430 --> 00:22:33,590 743 00:22:33,590 --> 00:22:34,789 744 00:22:34,789 --> 00:22:36,710 745 00:22:36,710 --> 00:22:36,720 746 00:22:36,720 --> 00:22:38,230 747 00:22:38,230 --> 00:22:38,240 748 00:22:38,240 --> 00:22:39,510 749 00:22:39,510 --> 00:22:40,470 750 00:22:40,470 --> 00:22:45,510 751 00:22:45,510 --> 00:22:47,909 752 00:22:47,909 --> 00:22:49,669 753 00:22:49,669 --> 00:22:51,830 754 00:22:51,830 --> 00:22:54,310 755 00:22:54,310 --> 00:22:57,350 756 00:22:57,350 --> 00:22:59,669 757 00:22:59,669 --> 00:23:01,909 758 00:23:01,909 --> 00:23:03,270 759 00:23:03,270 --> 00:23:05,510 760 00:23:05,510 --> 00:23:06,789 761 00:23:06,789 --> 00:23:10,149 762 00:23:10,149 --> 00:23:11,350 763 00:23:11,350 --> 00:23:14,070 764 00:23:14,070 --> 00:23:15,029 765 00:23:15,029 --> 00:23:17,350 766 00:23:17,350 --> 00:23:18,789 767 00:23:18,789 --> 00:23:21,510 768 00:23:21,510 --> 00:23:24,149 769 00:23:24,149 --> 00:23:24,159 770 00:23:24,159 --> 00:23:24,870 771 00:23:24,870 --> 00:23:26,390 772 00:23:26,390 --> 00:23:28,390 773 00:23:28,390 --> 00:23:30,230 774 00:23:30,230 --> 00:23:31,669 775 00:23:31,669 --> 00:23:34,230 776 00:23:34,230 --> 00:23:36,390 777 00:23:36,390 --> 00:23:38,870 778 00:23:38,870 --> 00:23:38,880 779 00:23:38,880 --> 00:23:40,549 780 00:23:40,549 --> 00:23:42,230 781 00:23:42,230 --> 00:23:43,590 782 00:23:43,590 --> 00:23:46,070 783 00:23:46,070 --> 00:23:48,390 784 00:23:48,390 --> 00:23:50,310 785 00:23:50,310 --> 00:23:52,390 786 00:23:52,390 --> 00:23:54,310 787 00:23:54,310 --> 00:23:56,630 788 00:23:56,630 --> 00:23:58,070 789 00:23:58,070 --> 00:24:00,950 790 00:24:00,950 --> 00:24:02,789 791 00:24:02,789 --> 00:24:04,870 792 00:24:04,870 --> 00:24:05,830 793 00:24:05,830 --> 00:24:08,310 794 00:24:08,310 --> 00:24:10,230 795 00:24:10,230 --> 00:24:11,750 796 00:24:11,750 --> 00:24:14,230 797 00:24:14,230 --> 00:24:15,669 798 00:24:15,669 --> 00:24:17,990 799 00:24:17,990 --> 00:24:21,590 800 00:24:21,590 --> 00:24:23,750 801 00:24:23,750 --> 00:24:25,830 802 00:24:25,830 --> 00:24:27,029 803 00:24:27,029 --> 00:24:27,039 804 00:24:27,039 --> 00:24:28,710 805 00:24:28,710 --> 00:24:30,230 806 00:24:30,230 --> 00:24:32,230 807 00:24:32,230 --> 00:24:34,789 808 00:24:34,789 --> 00:24:36,950 809 00:24:36,950 --> 00:24:41,190 810 00:24:41,190 --> 00:24:42,789 811 00:24:42,789 --> 00:24:44,149 812 00:24:44,149 --> 00:24:46,950 813 00:24:46,950 --> 00:24:48,710 814 00:24:48,710 --> 00:24:53,190 815 00:24:53,190 --> 00:24:55,110 816 00:24:55,110 --> 00:24:55,120 817 00:24:55,120 --> 00:24:56,789 818 00:24:56,789 --> 00:24:58,310 819 00:24:58,310 --> 00:25:01,110 820 00:25:01,110 --> 00:25:01,120 821 00:25:01,120 --> 00:25:02,149 822 00:25:02,149 --> 00:25:03,350 823 00:25:03,350 --> 00:25:05,510 824 00:25:05,510 --> 00:25:07,350 825 00:25:07,350 --> 00:25:09,480 826 00:25:09,480 --> 00:25:09,490 827 00:25:09,490 --> 00:25:11,750 828 00:25:11,750 --> 00:25:13,669 829 00:25:13,669 --> 00:25:16,630 830 00:25:16,630 --> 00:25:18,950 831 00:25:18,950 --> 00:25:21,190 832 00:25:21,190 --> 00:25:23,510 833 00:25:23,510 --> 00:25:24,870 834 00:25:24,870 --> 00:25:26,870 835 00:25:26,870 --> 00:25:29,430 836 00:25:29,430 --> 00:25:30,950 837 00:25:30,950 --> 00:25:32,310 838 00:25:32,310 --> 00:25:34,070 839 00:25:34,070 --> 00:25:35,830 840 00:25:35,830 --> 00:25:38,070 841 00:25:38,070 --> 00:25:41,750 842 00:25:41,750 --> 00:25:45,110 843 00:25:45,110 --> 00:25:47,100 844 00:25:47,100 --> 00:25:47,110 845 00:25:47,110 --> 00:25:48,630 846 00:25:48,630 --> 00:25:48,640 847 00:25:48,640 --> 00:25:49,430 848 00:25:49,430 --> 00:25:53,350 849 00:25:53,350 --> 00:25:57,909 850 00:25:57,909 --> 00:26:00,390 851 00:26:00,390 --> 00:26:02,630 852 00:26:02,630 --> 00:26:05,909 853 00:26:05,909 --> 00:26:08,549 854 00:26:08,549 --> 00:26:10,950 855 00:26:10,950 --> 00:26:10,960 856 00:26:10,960 --> 00:26:12,070 857 00:26:12,070 --> 00:26:14,070 858 00:26:14,070 --> 00:26:15,909 859 00:26:15,909 --> 00:26:17,830 860 00:26:17,830 --> 00:26:21,029 861 00:26:21,029 --> 00:26:22,830 862 00:26:22,830 --> 00:26:26,390 863 00:26:26,390 --> 00:26:28,789 864 00:26:28,789 --> 00:26:30,630 865 00:26:30,630 --> 00:26:33,510 866 00:26:33,510 --> 00:26:35,029 867 00:26:35,029 --> 00:26:38,310 868 00:26:38,310 --> 00:26:40,230 869 00:26:40,230 --> 00:26:42,230 870 00:26:42,230 --> 00:26:44,870 871 00:26:44,870 --> 00:26:48,310 872 00:26:48,310 --> 00:26:49,590 873 00:26:49,590 --> 00:26:49,600 874 00:26:49,600 --> 00:26:50,630 875 00:26:50,630 --> 00:26:52,630 876 00:26:52,630 --> 00:26:53,990 877 00:26:53,990 --> 00:26:57,430 878 00:26:57,430 --> 00:26:59,190 879 00:26:59,190 --> 00:27:00,630 880 00:27:00,630 --> 00:27:02,149 881 00:27:02,149 --> 00:27:02,159 882 00:27:02,159 --> 00:27:03,190 883 00:27:03,190 --> 00:27:03,200 884 00:27:03,200 --> 00:27:04,549 885 00:27:04,549 --> 00:27:06,789 886 00:27:06,789 --> 00:27:07,750 887 00:27:07,750 --> 00:27:09,590 888 00:27:09,590 --> 00:27:10,710 889 00:27:10,710 --> 00:27:13,510 890 00:27:13,510 --> 00:27:14,870 891 00:27:14,870 --> 00:27:17,350 892 00:27:17,350 --> 00:27:20,549 893 00:27:20,549 --> 00:27:23,350 894 00:27:23,350 --> 00:27:25,830 895 00:27:25,830 --> 00:27:27,590 896 00:27:27,590 --> 00:27:30,710 897 00:27:30,710 --> 00:27:32,389 898 00:27:32,389 --> 00:27:35,029 899 00:27:35,029 --> 00:27:36,070 900 00:27:36,070 --> 00:27:38,470 901 00:27:38,470 --> 00:27:38,480 902 00:27:38,480 --> 00:27:40,549 903 00:27:40,549 --> 00:27:41,750 904 00:27:41,750 --> 00:27:41,760 905 00:27:41,760 --> 00:27:43,510 906 00:27:43,510 --> 00:27:45,669 907 00:27:45,669 --> 00:27:47,350 908 00:27:47,350 --> 00:27:50,149 909 00:27:50,149 --> 00:27:52,870 910 00:27:52,870 --> 00:27:59,990 911 00:27:59,990 --> 00:28:03,669 912 00:28:03,669 --> 00:28:08,230 913 00:28:08,230 --> 00:28:08,240 914 00:28:08,240 --> 00:28:10,149 915 00:28:10,149 --> 00:28:11,990 916 00:28:11,990 --> 00:28:13,430 917 00:28:13,430 --> 00:28:15,029 918 00:28:15,029 --> 00:28:19,190 919 00:28:19,190 --> 00:28:21,990 920 00:28:21,990 --> 00:28:23,110 921 00:28:23,110 --> 00:28:25,590 922 00:28:25,590 --> 00:28:27,669 923 00:28:27,669 --> 00:28:30,070 924 00:28:30,070 --> 00:28:31,909 925 00:28:31,909 --> 00:28:33,830 926 00:28:33,830 --> 00:28:33,840 927 00:28:33,840 --> 00:28:34,710 928 00:28:34,710 --> 00:28:37,269 929 00:28:37,269 --> 00:28:39,430 930 00:28:39,430 --> 00:28:39,440 931 00:28:39,440 --> 00:28:41,830 932 00:28:41,830 --> 00:28:43,830 933 00:28:43,830 --> 00:28:46,389 934 00:28:46,389 --> 00:28:50,870 935 00:28:50,870 --> 00:28:50,880 936 00:28:50,880 --> 00:28:52,149 937 00:28:52,149 --> 00:28:54,070 938 00:28:54,070 --> 00:28:54,080 939 00:28:54,080 --> 00:28:55,510 940 00:28:55,510 --> 00:28:58,549 941 00:28:58,549 --> 00:29:00,149 942 00:29:00,149 --> 00:29:02,230 943 00:29:02,230 --> 00:29:05,909 944 00:29:05,909 --> 00:29:08,389 945 00:29:08,389 --> 00:29:11,190 946 00:29:11,190 --> 00:29:12,630 947 00:29:12,630 --> 00:29:15,110 948 00:29:15,110 --> 00:29:18,630 949 00:29:18,630 --> 00:29:20,470 950 00:29:20,470 --> 00:29:22,950 951 00:29:22,950 --> 00:29:25,110 952 00:29:25,110 --> 00:29:26,789 953 00:29:26,789 --> 00:29:26,799 954 00:29:26,799 --> 00:29:27,830 955 00:29:27,830 --> 00:29:29,990 956 00:29:29,990 --> 00:29:32,230 957 00:29:32,230 --> 00:29:34,950 958 00:29:34,950 --> 00:29:36,789 959 00:29:36,789 --> 00:29:39,110 960 00:29:39,110 --> 00:29:41,269 961 00:29:41,269 --> 00:29:43,350 962 00:29:43,350 --> 00:29:43,360 963 00:29:43,360 --> 00:29:44,230 964 00:29:44,230 --> 00:29:46,070 965 00:29:46,070 --> 00:29:48,149 966 00:29:48,149 --> 00:29:49,510 967 00:29:49,510 --> 00:29:51,669 968 00:29:51,669 --> 00:29:53,110 969 00:29:53,110 --> 00:29:55,190 970 00:29:55,190 --> 00:29:55,200 971 00:29:55,200 --> 00:29:56,549 972 00:29:56,549 --> 00:29:58,389 973 00:29:58,389 --> 00:29:58,399 974 00:29:58,399 --> 00:30:01,510 975 00:30:01,510 --> 00:30:03,350 976 00:30:03,350 --> 00:30:05,430 977 00:30:05,430 --> 00:30:08,310 978 00:30:08,310 --> 00:30:11,750 979 00:30:11,750 --> 00:30:14,230 980 00:30:14,230 --> 00:30:16,389 981 00:30:16,389 --> 00:30:18,549 982 00:30:18,549 --> 00:30:20,389 983 00:30:20,389 --> 00:30:22,950 984 00:30:22,950 --> 00:30:24,549 985 00:30:24,549 --> 00:30:26,230 986 00:30:26,230 --> 00:30:26,240 987 00:30:26,240 --> 00:30:27,190 988 00:30:27,190 --> 00:30:29,590 989 00:30:29,590 --> 00:30:31,830 990 00:30:31,830 --> 00:30:33,669 991 00:30:33,669 --> 00:30:35,669 992 00:30:35,669 --> 00:30:37,830 993 00:30:37,830 --> 00:30:39,590 994 00:30:39,590 --> 00:30:40,710 995 00:30:40,710 --> 00:30:42,389 996 00:30:42,389 --> 00:30:42,399 997 00:30:42,399 --> 00:30:43,669 998 00:30:43,669 --> 00:30:45,909 999 00:30:45,909 --> 00:30:49,669 1000 00:30:49,669 --> 00:30:51,430 1001 00:30:51,430 --> 00:30:55,430 1002 00:30:55,430 --> 00:30:56,950 1003 00:30:56,950 --> 00:30:56,960 1004 00:30:56,960 --> 00:30:59,750 1005 00:30:59,750 --> 00:31:03,350 1006 00:31:03,350 --> 00:31:06,389 1007 00:31:06,389 --> 00:31:09,590 1008 00:31:09,590 --> 00:31:11,990 1009 00:31:11,990 --> 00:31:14,389 1010 00:31:14,389 --> 00:31:14,399 1011 00:31:14,399 --> 00:31:15,190 1012 00:31:15,190 --> 00:31:15,200 1013 00:31:15,200 --> 00:31:15,440 1014 00:31:15,440 --> 00:31:15,450 1015 00:31:15,450 --> 00:31:16,950 1016 00:31:16,950 --> 00:31:21,430 1017 00:31:21,430 --> 00:31:23,190 1018 00:31:23,190 --> 00:31:24,870 1019 00:31:24,870 --> 00:31:27,509 1020 00:31:27,509 --> 00:31:29,830 1021 00:31:29,830 --> 00:31:31,750 1022 00:31:31,750 --> 00:31:34,470 1023 00:31:34,470 --> 00:31:36,230 1024 00:31:36,230 --> 00:31:37,990 1025 00:31:37,990 --> 00:31:39,990 1026 00:31:39,990 --> 00:31:42,070 1027 00:31:42,070 --> 00:31:44,630 1028 00:31:44,630 --> 00:31:44,640 1029 00:31:44,640 --> 00:31:45,669 1030 00:31:45,669 --> 00:31:46,950 1031 00:31:46,950 --> 00:31:48,950 1032 00:31:48,950 --> 00:31:51,669 1033 00:31:51,669 --> 00:31:53,990 1034 00:31:53,990 --> 00:31:55,110 1035 00:31:55,110 --> 00:31:56,630 1036 00:31:56,630 --> 00:31:59,509 1037 00:31:59,509 --> 00:31:59,519 1038 00:31:59,519 --> 00:32:00,789 1039 00:32:00,789 --> 00:32:02,870 1040 00:32:02,870 --> 00:32:03,990 1041 00:32:03,990 --> 00:32:06,950 1042 00:32:06,950 --> 00:32:08,389 1043 00:32:08,389 --> 00:32:09,750 1044 00:32:09,750 --> 00:32:14,470 1045 00:32:14,470 --> 00:32:14,480 1046 00:32:14,480 --> 00:32:16,710 1047 00:32:16,710 --> 00:32:19,750 1048 00:32:19,750 --> 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