1 00:00:24,880 --> 00:00:28,070 why don't we get started so as i like to do let's just review where we um have been recently um which is to um discuss context-free languages we talked about the context-free grammars and the push-down automata as a way of describing the context-free languages as you remember the context-free languages are a larger class of languages than the regular languages which is where we started the languages of the finite automata so when you add a stack you get more power you get more languages that you can do and we're very rapidly going to be moving on today to our main model for the semester which is called the turing machine so let's just take a look at what we're going to be covering today and that is first we're going to show that a technique analogous to the one we use for um [Music] proving that languages are not regular but this time for proving languages are not context free so the push down automate and the grammar still have their limitations in terms of what we normally think a computer can do and with that we're going to use that as a kind of a lead into our general purpose model which is the turing machine and uh so we're going to talk about turing machines and of and aspects of of that um and i do i would want to make comment uh so i have posted the solutions for the first problem set i know you're starting to think about the second problem set now which i have posted as well if you want to get a sense of what i'm looking for in terms of the level of detail you can look at the solutions to problem set one because i consider those to be model solutions uh that's part of the reason why i post them just to give you a sense of the level of detail that i'm looking for which is not a whole lot but i do want to make sure you're capturing the main ideas um of what of what's involved in solving the problem so have a look at those um and for problem set two which i'll talk about um in a second so i'll just say a few words if you want to pull that up um you know you can do that but just to get you started on a few of the problems if you're uh finding some challenges there i i don't want you to get stuck really before you even understand what the problem is saying um so for problem number one if you looked at that so that's a problem where you're asked to prove um a certain language is not context-free so that's um and by the way all of the problems in this problem set except uh perhaps for the last one for number six you'll be able to solve we'll have enough material at the end of today's lecture to solve all of them um i believe that's right uh yeah so number six you should have enough as of thursday's lecture to solve that um um uh problem number one um uh it's proving a language is not context-free so we're going to introduce a method for doing that that method is going to come in handy for parts b and c uh if you look at the problem set it has this strange looking thing sigma sigma sigma and parenthesis star it's really just a regular expression that's very simple um you know you should just make sure you understand that that's a way of representing all strings whose length is a multiple of three um and if i stick a sigma in front of that it's all strings whose length is one plus a multiple of three um so um once you understand that and if you think about what kinds of strings are in the language c2 it'll help you want to understand what happens when you take those unions um and um parts b and c are not intended to be very hard but you just have to understand what's going on uh problem number two is about ambiguous grammars i touched on that briefly in lecture um it's enough to solve the problem um the book has a little bit more detail about uh um ambiguous languages uh ambiguous grammars uh ambiguous grammars i should say um and uh so this is a grammar that's supposed to represent a fragment of a programming language um you know with uh if thens um and if then else's uh you know i'm sure you're all familiar with uh those kinds of constructs in programming languages um and there is a natural ambiguity that comes up in a programming language you know if you have you know if some condition then statement one else statement two i presume you understand what the semantics of that is what that means um and the tricky thing is that if you have a you know those statements can themselves be if statements and so if you have the situation where you have if then and if then else is the uh what follows that the question is where does the else attach is it to the second if foot to the first if um so that's kind of a big hint on this problem but that's okay um i think you uh should uh um uh you know you need to take that and figure out how to get a uh an actual member of the language which is ambiguously generated um and then show that it has that show that it is by showing two parse trees uh now or or two left-most derivations if you read the book you'll see that's an alternative way of of representing comparison so um and then what you're supposed to do is give a grammar for the same language which is unambiguous um you don't have to prove that it's unambiguous because that's a bit of a chore but as long as you understand what's going on you should be should be able to come up with an unambiguous grammar which resolves that ambiguity and i don't have in mind changing the language by introducing new uh programming languages contra constructs like a begin end uh that's not in the spirit of this problem because that's a different it's grammar for a different language so you need to be generating the same language um without any other extraneous things going on that are going to resolve the ambiguity the ambiguity needs to be resolved within the structure of the grammar itself so keep that in mind uh for problem number three about the key automata um you know that came up actually as a suggestion um uh last lecture i believe or two lectures back what happens if you take a push down automaton but instead of a push down uh as a instead of a stack you add a q um what happens then well actually it turns out that the model you get is very powerful and turns out to be equivalent in power to a turing machine so you'll see arguments of that kind today how you show that um other models are equivalent no not today that's going to be on um uh that's going to so i apologize the key this is going to be something that you'll also you know i i i i'm confusing myself here uh for problems number problem number three you actually need thursday thursday's lecture as well uh to really at least see examples of how you do that kind of thing um and so uh so thursday uh yeah so i i i'll try to send out a note clarifying this by the end of thursday you'll be able to do everything except for problem six and for problem six you'll need tuesday's lecture a week from today's lecture to do um so uh um all right uh so um problem number four that one you'll be able to do at the end of today we'll that's also gonna you know the problem is i'm i'm working on preparing thursday's lecture too so i'm getting a little i'm confusing myself uh problem number four you'll be able to do after thursday's lecture maybe we should talk about that um next lecture um problem number five um you can do today um but maybe i'm not gonna say anything about that and problem number six i won't say anything about either okay so why don't we just jump in then um and uh look at today's material um what about seven seven is uh oh seven is an optional problem oh i should have mentioned that whenever there's seven is always going to be an optional i indicate that with a star i should have made that clear um on the actual description here but seven is optional uh it's just like we had for problem set one okay let's move let's move on then to uh what we're going to talk about today um uh and just a little bit of review um so we talked about the equivalence of context-free grammars and push down automata as you remember oops let me get myself out of the picture here um as we mentioned last time we actually proved one direction but the other direction of that you just have to know it's true but you don't have to know the proof the proof is a little bit lengthy i would say it's a nice proof but it's pretty long um and uh there are two uh important corollaries to that um if you know what a corollary is just a simple consequence which doesn't need much of a proof uh sort of a very straightforward consequence first of all uh as i think we pointed out at last time one one conclusion one corollary you get is that every regular language is a context-free language because um a finite automaton is a push down automaton that just happens not to use its stack so immediately you get that every regular language is context-free and second of all um you also immediately get that whenever you have a context-free language and a and a regular language and you take their intersection you get back a context-free language so context-free intersect regular is context-free okay that that's actually mentioned in your homework as well as one of the zero-point x problems um which i give to try to get you um they're not you don't have to turn those in but i suggest you look at them i don't know how many of you are looking at them but uh this is a useful fact and and some of those other facts and in zero point x problems are useful so i encourage you to look at them but anyway intersection of context-free and regular is context-free you might ask what about intersection of context-free and context-free do we have closure under intersection the answer is no we do not have close closure under intersection we'll talk about that shortly uh so here's the proof sketch for um i should i wanted to say that the intersection of context free and regular why do we know that's still context-free because the push down automaton for a can be simulating the finite automaton for b inside its finite control inside its finite memory the problem is if you have two context-free languages you have two push-down automata you can't simulate that with one push-down automaton because it has only a single stack so if you're trying to take the intersection of two context-free language languages with only a single stack you're going to be in trouble because it's hard to um anyway i mean that's not a proof but at least it shows you what goes wrong if you try to do the obvious thing um okay uh so uh so if if then just here's an important point that i was trying to make before if a and b are both context-free and you take the intersection uh the result may not necessarily be a context-free language um so the class of context-free languages is not closed under the intersection we'll comment down that uh uh in a bit um the context-free languages are closed under the regular operations however union intersection a union concatenation and star so you should um feel comfortable that you know how to prove that it's in the again it's one of the uh i think it's problem 0.2 and i think the solution is even given in the book for it so you just should know how to prove that um uh not very pretty straightforward um okay so let's move on then um to uh to really basically conclude our um work on context-free languages and to understand the limitations of uh content context-free grammars and what kinds of languages may not be context-free and how do you prove that so how do you prove that for some language there is no grammar um again you know it's not enough just to say you know uh given informal uh comment that you will i i couldn't think of a grammar that or some things of that kind that's not going to be good enough we need to have a proof uh so if we take the language here 0 to the k1 to the k2 to the k so those are strings which are runs of zeros followed by an equal number of ones followed by an equal number of twos so just for zeros then ones then twos all the same length um uh that's a language which is not going to be a context-free language and we'll we'll give a method for proving that you know if you had a stack you can match the ones with the zeros but then once you're done with that the stack is empty and how do you now make sure that the number of twos corresponds to the number of ones that you had so again that's an informal argument that's not good enough to uh to be a proof but it sort of gives an intuition um so we're going to give a method for proving non-context-free languages are not context-free using again a pumping lemma but this is going to be a pumping lemma that applies to context-free languages not to regular languages okay it looks very similar but it has some extra wrinkles thrown in because the other older pumping dilemma was specific to the regular languages and this is going to be something that applies to the context free languages okay so now um let's just read it and then we'll um try to interpret it again it's very similar in spirit basically it says that whenever you have a context-free language all long strings in the language can be pumped in some kind of way so it's going to be a little different kind of pumping than we had before and you stay in the language okay so before we broke the string into three pieces where we could repeat that center piece as many times as as you like um and you stay in the language here we're going to end up breaking the um the string into five pieces so s is going to be broken up into u v x y z um uh and the and the way it's gonna work here so here is a picture so all long strings again there's gonna be a threshold um for whenever you have a language there's going to be some cutoff length so that all the longer strings in that language can be pumped and you stay in the language but the shorter strings there's no guarantee um so if you have a long string in the language length at least this pumping length p then you can break it up into five pieces but now it's that second and fourth string that are going to play that special uh pumping roll um which means that what you can do is you can repeat those um and you stay in the language and it's important that you repeat them both that v and that y the same number of times so you're gonna have a picture that looks something like this um and that is going to you you repeat uh so you can get you if you repeat the v and repeat the y you get u v v x y y z um or if you look at over here it would be u v squared x y squared is e um and that's going to still be in the language um and then we have so that's one condition we'll have to look at all of these conditions when we do the proof but we just understand what the statement is right now so the second condition is that v and y together cannot be empty and really that's another way of saying they can't both be the empty string because if they were both the empty string then repeating them wouldn't change um s and then of course it would stay in the language so it would be kind of meaningless if they were allowed to be empty and the last thing is again going to be there as a matter of convenience for proving str proving uh languages are not context-free because you have to make sure there's no possible way of cutting up the string when you're trying to prove a language is not context-free you have to show the pumping fails and then you it's it's going to be helpful sometimes to limit the ways in which the string can be cut up because then you have it's an easier job for you to work with it um so here it's a little different than before but sort of similar that v x y combined um as a substring so i think i show that over here v x y together is a um is not too long so the v vxy maybe is better seen up here is going to be is going to be at most p um we'll we'll do an example in a minute of using this okay so again here's our pumping limit i just restated it so um you know we have it in front of us and we're going to do a proof i'm just going to give you the idea of the proof first and then we'll go through some of the details the idea is actually pretty simple okay i give it a call proof by picture again remember what we're trying to do we're trying to show that we have this context-free language a and now all long strings in a have this pumping quality that you can break them up into five pieces so that the second and the fourth piece can be repeated and you stay in the language so how do we know that that's going to be true let's take a look at the proof here and how why is that going to be true so first of all i i i'd like to do it qualitatively rather than quantitatively so let's just imagine instead of what thinking we'll calculate what p is later but just imagine the s is some really really long string that's the way i like to think about it so s is just really long what is that going to tell us it's going to tell us something important about the way the grammar produces s which is going to be useful in in getting a way of pumping pumping s so if s is really long um we're going to look at the parse tree for s and we're going to conclude that the parse tree has to be really tall because it's impossible for a very shallow parse tree to generate a very long string and again we'll quantify that in a second but intuitively i think that's not too hard to see why that ought to be true so if you have a long s the pole the parse tree has to be really tall because the parse tree can't generate very many it can't expand by very much at each level so we'll look at how much it can expand but it depends on the grammar how much expansion can we have at each level and it's going to be you can't have just in three levels you know some small grammar generating a string of length a million just you know you'll see that that's just impossible so once you know that the parse tree is really tall here um then you're actually almost done because what does it mean to be really tall it means that there's some path starting at the start variable e i'm calling it uh in this parse tree which goes down to some um terminal symbol in s which goes through many steps that's what it means for the tree to be very tall and each one of those steps is a variable until you get down to the very end okay so that's the way parse trees look you keep expanding variables until you get to a terminal so here's that here you get some some path that's really a long path and once you have a long path that have many many variables appearing on here well the grammar itself has only some fixed number of variables in it so you're going to have to have a repetition coming among the variables that occur on that long path got that so a long string forces a tall parse tree forces a repetition on some path coming out of the start variable uh of some of some other variable that comes out now that's going to tell us how to cut up s because if you look at the subtrees of s that those two r variables are generating shown like this i'm going to use that so if you have to follow what i'm saying here so you know the r here is generating this portion of s and the lower r is generating a smaller portion of s just looking at the subtree that you get here um and that's going to tell us that we can cut up s accordingly so u is that very first part out here generated by e but not by the first r r is generated v is generated by the first r but not by the second r the second r generates exactly x and then we have y and z similarly so that all follows from having a lo a tall parse tree and now we're finished now we know how to cut up s how do we know we can repeat uh v and y and still be in the language well i'll actually show your and show you that you're in the language by exhibiting a parse tree for the string u v v x y y z here it is what i'm gonna get that parse tree by when i expand this lower r instead of expanding it to get x i'm going to follow the same substitutions that i i had when i expanded the upper r so it's as if i took this larger subtree here and i substituted it in for the smaller uh subtree under under the second r and so that i get a picture that looks like this so here i'm substituting under the second r the same subtree that i had originally coming out of the upper r the first r and so now this parse tree is generating the string u v v x y y z which is what i'm looking for and of course you can do that again and again and you're going to keep getting higher and higher exponents of v and y um and in fact you can even get the zero exponent which means that v and y both disappear altogether and for that you do something slightly different which is that you replace the larger subtree by the smaller subtree okay so here which was originally that larger subtree generating v x y i stick instead the smaller subtree i do the substitutions from the smaller sub tree and i just get x uh there and so now the string i generated is u xz which is the same as u v to the zero x y to the zero z okay that's that is the idea of the proof now i think uh you could work out um [Music] the quantities that you need in order to drive this proof um i'm gonna do that for you i actually hate writing down lots of inequalities and um equations and so on on the board because i think they're just almost incomprehensible to follow at least they would be for me but i'm going to put them up there just for completeness sake um so here we're going to give the details of this proof on the next uh the next slide here oh yeah so i just want to give a name to this i'm going to call this the cutting and pasting argument because i'm cutting apart pieces of this parse tree and i'm pasting them into other places within the parse tree to get new strings being generated so this is a cutting and pasting argument so okay let's let's take a look at the details here um uh just so we have to understand well how big does p actually need to be in order for this thing to kick in um well first of all we have to understand how fast that parse tree can be growing um as we go level to level and that's going to be dependent on how big the right-hand side sides of rules are i mean that's expand that really tells you how many what's the fan out you know of each node what's the maximum fanout and that's going to be the maximum uh length of a right-hand side of any rule so for example in that other grammar we had seen last time for arithmetic expressions we had this e goes to e plus t and uh this rule here and in terms of the parse tree that would look uh like an a little element like that and that's actually the longest right hand side that you can get and so the parse tree can be growing by a factor of three each time now that's going to tell us how big the string uh needs to be that's being generated what is the value of p in order to get a high enough parse tree so that you're going to get a repeated variable okay um let's call the height of the parse tree for sh okay so now if you this is just repeating what i just said if you have a tree of height h and the maximum branching is b then you get it most b to the h leaves because each level you get another factor of b coming up because that's how much branching you have so each node at one level can become b nodes at the next level down so you're multiplying by b each time and if you have h levels you're going to have b to the h leaves so the length of s which are really the leaves here that it is at most b to the a the reason why i said most and not exactly is you might be doing substitutions which are shorter right hand sides um okay so to try to show this as a picture here uh pulling that same picture we had before we want h the height to be bigger than the number of variables to force a repetition so the number of variables is going to be written this way v is the variables uh v with bars around it is going to be the number of variables and we want that height to be greater than the number of variables so once you know how high you want that tree to be in order to force the repetition then it tells you how big s has to be so b has to be bigger than b to the v b to the size of v um because then um the the the height that you're going to get is going to be greater than uh uh the size of v which is uh so that's what you want you want h to be greater than the size of v uh so you're gonna set p to be one more than b to the v and so if s is at least that length this whole thing is going to kick in and you're going to get that repeated variable um so we let p to be that value where v is the number of variables in the grammar and so if s is at least p which is greater than b to the v then the length of s is going to be greater than um b to the v so h is going to be what you want to make this thing work i you know if you don't follow that those inequalities i sympathize with you i would never follow that either in a lecture so um but i hope you get the idea um but you know we're not quite finished yet because what i want to now circle back as and look at these three conditions and make sure that we've captured them all because actually it's not totally uh obvious um in each of those cases that we've got them um so there's a few extra things we need to do um um okay so this is concluding the argument you know there are going to be at least three plus one variables in the longest path so there's going to be a repetition so now let's go back here and see now that we have this picture with a repeated uh variable how do we know we can get condition one well that's just the cutting and pacing argument from the previous slide how do we know that v and y are not both empty well actually that's not totally obvious because it's possible that when you generated v here and you generated y um maybe going from this r to that r you got nothing new you know it could have been that r got replaced by t another variable with nothing new coming out and then t got replaced by r so you substituted uh you know t for r and then r for t and you've got nothing new coming out and in that case uh v and y would both be the empty string and that would violate what we want um the way you get around that and these are details here if you're not totally following these points don't worry uh but i do want to just try to it's easy they're easy to describe so i figure let me present the whole thing in full detail so if going from this r to that r doesn't generate anything new you're getting exactly the same things coming out v and y are just the empty string um how do we avoid that from happening well what there's a simple way to address that which is to say uh if you have this string s when you take a parse tree make sure you take a smallest possible parse tree you cannot you're not allowed to start off with an inefficient parse tree that can be shortened and still generate s i want the smallest possible parse tree and that smallest possible parser can't have an r going to another r which is generating nothing new because then you could always have eliminated that step and you would still have a parse tree for s but it would be a smaller parse tree so that would be i want you to start off with the smallest possible parser tree and then you're going to be guaranteed that v or y is going to be something uh not empty so that takes care of uh condition two condition three uh you know how do we know that v x y together is not very long and basically it's the same argument all over again um you just want to make sure that when you're picking the repetition are the two r's here you pick the lowest possible repetitions that occur if you have many many choices and those lowest two those lowest repetitions there's not going to be any lower repetition here and then by the same argument um the the uh since once you have that very first r there's no more repetitions occurring below the v x y can't be very long um because that would again force another repetition to occur um so anyway those are the three conditions and that's the proof of the pumping lemma for context-free languages let's see how we use that um okay so let's do an example of proving a language not context free using the pumping line how do you gonna do a go about doing that because that's the kind of thing you you know at the very least you need to know how to do this in order to do the homework um uh i'd like to motivate you that the stuff is so interesting and fun but it doesn't work for everybody so uh for you practical people out there um pay attention so you can do the homework okay let's go back to that language we had a couple of slides back zero to the g one to the k two to the k it's not a contact three language we're gonna show that now using the pumping lemma for context-free languages um uh so it's gonna do similar to the proofs using for non-regular non-regul not non-regular languages um proof by contradiction so you first you assume the language is context-free and then we're going to apply the pumping lemma um and then we're going to get a contradiction uh so the pumping limit gives that pumping length as we subscribed above and now we just want to pick a longer string in the language and show that that longer string which is supposed to be pumpable and stay in the language in fact is not pumpable um so the pumping dilemma says that you can divide it into five pieces satisfying the three conditions uh condition three implies uh that so now i'm gonna i'm gonna work through the i'm gonna uh show you get a contradiction so condition three implies that you cannot contain both zeros and let's pull up a picture here so here's s uh zeros ones and then twos all the same length condition three uh so if you break it up condition three says v x y together cannot be too long well if v x y together is not too long uh what you know how could it be that when you're repeating v and y you stay in the language uh for one thing you can't have zeros ones and twos all occurring within v x and y um some uh some symbol is going to get left out so then when you pump up you're going to have unequal numbers of symbols and so you're going to be out of the language um okay so no matter how you try to cut it up uh following condition three which is one of the things that restricts the the ways to cut it up you're going to end up when you pump up uh going out of the language and so therefore it's not in d d is wrong uh b shouldn't say b can knife can i nice supposed to be able to write on this thing i guess not i didn't test that oh well that's supposed to be a b um uh so b is a context free language we close so that's the assumption that b is a context-free language that's false and we conclude that's not a context-free language okay let's do uh oh yeah i have a check-in here um so let's see um what we're going to we're going to i'm going to ask you to think about okay my head is blocking part of the text oh that was a while ago um okay uh yes so with just one question by the way in in terms of applying the um in in the pumping dilemma either v or y can be uh empty but not both but anyway let's get to this check in here um so let's look at these two languages a1 and a2 which look very similar to b but a little different um so it's a1 is 0 to the k1 to the gate 2 to the l where k and l could be any numbers any positive you know uh non-negative numbers so basically what this says saying is that the number of zeros and ones are going to be equal but the number of twos can be anything whereas a2 similar but here we're requiring the number of two ones and twos to be equal and the number of zeros can be anything okay now you can easily make i hope uh you should make sure you you can push down automata that can recognize a1 and a2 um because let's just take a1 the push down automaton can push the zeros as it's reading them pop them as it's reading the ones to match them off and make sure that they're the same number of them and then the twos it doesn't care how many there are it just has to make sure that there are no strings there are no letters coming out of order but any number of twos is fine so you can easily make a presentation recognize a1 similarly for a2 um so what can we conclude from that here are the three possibilities let me uh so look at that the class of content language is not closing intersection you know you can read it um so i i want to pull up the uh poll and launch that please fill that out um 10 seconds again just if you don't know the answer just give any answers so that uh because we're not counting correctness um still a few dribbling in okay uh five seconds okay end polling uh most of you got that right um i know is it okay to share these things i don't want to make people who didn't get the right feel bad you know but you should understand i think you know if you're missing something you should understand what is what you're missing the pumping limit shows that a1 union a2 is not a context-free language no as i mentioned at the beginning the reg the context-free languages are closed under union so the pumping alum had better not show that these we already know that these two languages are context-free because we get them from push down on top and we said at the beginning that contact free language is closed under union so we know that these two are context-free so the pumping alignment better not show it they're not context-free something would be terribly gone have gone terribly wrong uh if that were true and also um we know also from a little bit of further reasoning that the context-free languages is not closed under complement by what we've already what we've already discussed uh because they are closed under union as i pointed out they're not closed under intersection and so if they were closed under complement de morgan's laws would say that closure under union and closure under complement would give you closure under intersection but we don't have closer intersections so in fact they're not closed under complement okay uh so in fact this does show us that the class of context-free language is not closed under intersection because the intersection of a1 and a2 two context three languages is b and b is not context-free so it shows that um this is not the the um closure under intersection does not hold okay um so let us continue then we have one more uh example then we'll take a break uh so the pumping limit for context-free languages again uh here's a second example uh here's the language f we have actually seen what's before ww uh two copies of a string um in in um that two copies of any string um i'm gonna show that's not a context free language uh assume that it is context-free the pumping limit gives pumping length now here you have to do a little bit more often the challenge in applying the pumping lemma in either case uh that we've seen uh involves with choosing that string that you need to pump that you're gonna pump so you have to choose s uh in f which is longer than p which has to go with so you might try this one first glance here isn't string it's in the language because it's two copies of the string zero to the p1 zero to and then zero to the p1 that's in the language but it's a bad choice uh let's say let's before i get ahead of myself let's let's draw a picture of this which i think is always helpful to see um so here is runs of zeros and then a one runs of zeros and then a one um why is this a bad choice because you can pump that string and you remain in the language um there is a way to cut that string up um and you'll stay in the language and the way to cut it up is to let the x be just that substring which is just the one and the v and the y can be a couple of zeros or a single zero on either side of that one and now that's going to be a small v x y but if you repeat v and y you're going to um if you repeat v and y you're going to stay in the language because you'll just be adding zeros here you'll be adding same number of zeros there and then you're going to have a string which still looks like ww and you'll still be in the language so that that means of cutting it up doesn't get you out of the language under pumping and the fact is that that's a bad choice for s because there is that way of cutting it up so you have to show there's no way you can't you don't get to pick the way to cut it up um you have to show that there is no way to cut it up um uh in order to um to violate the uh the pumping lever so if instead you use the string 0 to the p1 to the p0 to the p1 to the p so this is zeros followed by one followed by zeros followed by ones all the same number of them that can't be pumped satisfying to the three conditions um and just going through that uh um now if you try to break it up you're going to lose or the or the lemon is going to lose you're going to be happy but the lemon is not going to be happy because it's not going it's going to violate the conditions um condition 3 says vxy is not doesn't span too much and in fact can't span uh two runs of zeros or two runs of ones it's just not big enough because they're more than p things they're p things apart and this one string this string v x y is only p long and so therefore if you repeat of uh v and y you're going to have two runs of zeros or two runs of ones that have unequal length and now that's not going to be of the form w and you're going to be out of the language okay uh so i hope you know that's uh you've got a little practice with that um i think we're at our uh break and i will see you back here in uh five uh minutes um i can get my timer launched here okay so see you soon you can this is a good time by the way to to um uh message me or the tas and i'll i'll try to be uh looking for if you have any questions in the pumping limit can x x can be epsilon in the pumping lemma uh but it's not possible x can be epsilon y can be epsilon but x and y cannot be both cannot both be epsilon because then when you pump you will get nothing new technically v and y can include both zeros and ones yeah v and y can conclude both zeros and ones um so let me uh try to put that back if that's will um [Music] so v and y can have both zeros and ones but they can't have zeros from two different blocks and you can't have ones from two different blocks so what's going to happen is either you're going to get things out of order when you repeat like a v has both zeros and ones in it when you repeat v you're going to have uh zeros and ones and zeros and ones and zeros and ones that's clearly out of the language so that's so that's no good your only hope is to have v to be sticking only inside the zeros and y to be sticking only inside zeros are only inside once but now if you repeat that if just look at what you're going to get you're going to have a string which is going to be if you try to cut that string in half it's not going to be of the right form it's not going to be two copies of the same string because you know it's going to have a run of zeros um followed by a longer or shorter run of zeros or one of ones followed by another run of ones of unequal length so there's no way this can be two strings um two copies of the same string because that that's what you're required f has to be two copies of the same string to be in the language okay uh let me just see where we're running out of time here let me just put my uh my my timer here we've only got 30 seconds okay sorry i'm not getting to answer all the questions here um okay we are done um uh with um our break it's going to come back um and now we're shifting gears in a major way because in a sense everything we've done so far has been kind of a warm up these limited computational models um really are kind of get helping us to set our understanding of automata and the definitions and the notation and they're also going to be helpful in providing examples later on in the term um but really in terms of a model of computation they don't cut it because uh they cannot do very simple things that we normally think of a computer as being able to do so here we're introducing another model of computation called the turing machine and that's really going to be the model that we're going to stick with for the rest of the semester because that's going to be our model of a general purpose uh computer the way you normally think about it um so let's uh we'll spend a little time introducing it and then we will um uh you know we'll we'll continue that discussion next time um so uh in terms of a schematic actually the turing machine model is pretty simple um it has a uh it has um it's going to have states and all that stuff but you know though so there's going to be a finite control here which is going to include the states and transition function as we'll describe in a minute um the point is is that you know it's going to have the input appearing on a tape uh the key difference now is that the the machine is going to be able to change the symbols on the tape um and so we think of the machine as being able to write as well as read the tape okay so that's really uh the key uh feature um of a turing machine is the ability to write on the tape everything else in a sense follows from that a few other differences but so the fact that he he head can read and write so we can use the tape as storage much as we used a stack as storage but it doesn't it's not limited in the way we can access it the way a stack is so that we kind of have very uh flexible access of the information on the tape now being able to write on the tape doesn't do any good if you can't go back and read what you've written later on so we're going to make the head to be able to be two-way so the head can move left to right as before but it can also move back left and that's going to be under control of the transition function so under under program control essentially the tape is going to be oops sorry the tape is infinite to the right and so we're not going to limit how much storage the machine can have so the tape is going to think of as having instead of just having the input on it it's going to have the input but then the rest is going to have infinitely many blanks blank symbols following the input so the tape is infinite or in the right right hand direction um and uh so there's infinitely many blanks i'm going to use that symbol for the blank to follow the input um you can accept or reject oh yeah so that's another thing that's important normally we think of the machine in the previous machines finding a time when i pushed down automata when you got to the end of the input that's when you the acceptance of rejection was decided if you were in except state at the end of the input then you accept it but you have to be in that location um at the end of the input in order for that to take effect that doesn't make any sense anymore because the machine might go off beyond that and still be computing and come back and read the tape later on so it only really makes sense to let the machine accept or reject upon entering the accept or the reject state so we're going to have a special accept state and a special reject state which is also a little different than before and when the machine enters those states then machine then the action takes effect the machine those machine holds and then accepts or holds and then it rejects so we'll make that absolutely clear in the formal definition in a second but just to get the spirit of it um so i'm going to give you an example of the thing running uh sorry you need to again my powerpoint is having issues um uh okay so here is a turing machine recognizing that language b actually i switched gears on you instead of zeros ones and twos i made them a's b's and c's but same idea [Music] so i'm going to show you how the turing machine operates and then we'll give a formal uh definition i hope that's on here i think it is in a second but let's this is an informal discussion of how the machine is going to operate to do this language a to the kb to the kc to the k using its ability to write on the tape as well as read and move its head in both directions okay so let me just first describe um in english how this um [Music] machine operates and then we will um see it in action on this little uh picture i have over here okay so the way the machine is going to operate is the very first thing is the head is going to start here and the head is going to scan off to the right uh making sure that the symbols appear in the correct order so it's seeing that that there are a's and then b's and then c's without checking the quantities just that the order is correct for that you don't need to write a finite automaton can check that the input is of the form a star b star c star um so uh rioting is not necessary the machine if it uh detects symbols out of order it immediately rejects by going into a special reject state otherwise it's going to return its head back to the left end and um uh let me just show that here so here is it oh no i better before i illustrate it over here let's go through the whole algorithm uh so the next thing that happens is you're going to scan right and now you want to do the counting so you're going to scan right again but this time you're going to make a bunch of passes over the input bunch of scans and each time you make a scan you're going to cross off one symbol of each type so you're going to cross off an a you'll cross off a b you'll cross off a c on a single scan and then repeat that crossing off the next a the next b the next c and you want to make sure that you've crossed off all of the symbols on the same run and not crossing off some symbols before other symbols uh while other symbols will remain because that would mean that the counts were not equal if you cross them off and they're all you know run out on the same scan same pass then we know that the numbers have to be start off being equal sort of i mean this is sort of baby stuff here i but i hope you get the idea and we'll kind of illustrate in in a second so if at least one of the if you have the last one of each symbol so what i mean by that is you you just crossed off the last a the last b and the last c then you know that you were originally had an equal number and so you accept because you're crossing off one of each on each scan so if you cross the on the last scan each one of them gets crossed off um then you accept but if uh it was the last of some symbol but not of other symbols so you know you crossed off the last a but there um there were several b's remaining then you started off with an unequal number of a's b's and c's and you can reject or if all symbols still remain after you've crossed them uh one of each off then you haven't done enough passes and you're gonna repeat from stage three and do that uh again another scan okay so here's a little um animation which shows this happening on this diagram so here is the very first stage where you're scanning across making sure things are in the right order um didn't have to write on the tape and now you're gonna reset the the head back to the beginning this is by the way not the most efficient procedure for doing this um so um now that now we're going to do a scan crossing off a single a a single b and a single c so here i'm going to show that here single a single b single c and now as soon as you've crossed out that last c we can return back to the beginning um so scan right cross uh so if all symbols remain so there are still symbols remaining of each type we're going to return to the left and repeat and repeat now we'll get another pass single a single b single c get crossed off we haven't crossed them all off yet no there's of each type there's still our remaining ones so again we return back to the beginning now we have a last pass crossover the last a the last be the last c the last one of each type was crossed off so now we know we can accept because the original string was in the language okay so that's uh to give you at least some idea um that how the turing machine can operate you know more like the way you would think of a computer operating you know maybe it's very primitive you know you you could imagine counting also and the termination can count as well but this is the simplest uh descript uh procedure that i can just describe on um you know uh for you without getting making too complicated um okay so let's do a little checking on that um okay so the way i'm describing this how do we go how do you think and you in a sense you don't quite know enough yet but um [Music] uh how do you think we're going to get this effect of crossing off with the turing machine um are we going to get that by changing the model and adding that ability to cross off to the model um are we going to use a tape alphabet that includes those crossed off symbols among them or um we'll just assume that all turing machines come with the eraser and they can always cross our stuff um so what do you think is the nice way sort of mathematically to describe this this ability to cross things off yeah again again most of your again i think are getting this um uh so like 10 laggards here so please wrap it up um so we can close the poll uh five seconds to go um okay polling ending get your last last call uh all right share the results so most of you uh got that right all turn machines come with an eraser i don't know that was thrown in there as a joke but it got came in second so i don't feel bad if you got it but that's not what i had in mind um the the way the turing machine is going to be riding on the tape is to write a crossed off symbol uh instead of the uh the symbol that was originally there so we're going to add these new crosstalk symbols and that's going to be a common thing for us to do in um you know when we design turing machines we're not going to get down to the implementation level for very long we're going to very quickly shift to a higher level of discussion about the machines but anyway that's how you would do it if you were going to actually build a machine okay so um let us then look at the formal definition of the cursing and maybe i should have done that checking after the formal definition that might have been clear but oh well uh okay here's the formal definition this time a turing machine is a seven tuple um and there is a uh um now here uh we have sigma which is the input alphabet gamma is the tape alphabet so now you're it's sort of a little bit analogous to the stack from before where gamma was the stack alphabet but these are the symbols that you're allowed to write on the tape um or that are allowed to be on the tape um so uh obviously all of the input symbols are among the uh tape uh symbols because they can appear on the tape so you have sigma is a subset of gamma um one thing i didn't mention here is that the input alphabet we don't allow the blank symbol to be uh in the input alphabet so that you can actually use the blank symbol as a delimiter for the end of the input a marker for the end of the input um so uh in fact and blank symbol is always going to be in the tape alphabet uh so we have um you know the this is actually always going to be a proper subset uh because of the blank symbol um what we're just allowing doesn't really matter we're allowing the tape alphabet to have other symbols for convenience so for example these crossed off symbols now let's look at what the transition function how that operates so the transition function remember tells how the machine is actually doing its computation and it says that if you're in a certain state and the head is looking at a certain tape symbol um then you can go to a new state you write a new symbol on the uh at that location on the tape and you can move the head either left or right so that's how we get the effect of the head being able to be bi-directional and here's the writing on the tape comes up right here so you know just an example here we says if we're in state q and the head is looking at an a currently on the tape then we can move to state r we change it a to a b and we move the head right once one square okay um uh uh now here this is important to on when you're giving a certain input here to the turing machine it may gr compute around for a while moving its head back and forth as we were showing and it may eventually halt by either entering the q accept state or the q reject state which i didn't uh bring out here but that's important these are the accepting rejecting special states of the machine um or the machine may never enter one of those it may just go on and on and on and never halt we call that looping a little bit of a misnomer because it's the looping implies some sort of a repetition for us looping just means not holding um and um so therefore m has three possible outcomes for each um input w it might accept w by uh entering uh the except state it could reject w by entering the reject state which means it's going to reject it by holding or we also say we can reject by looping you can reject the string by running forever um okay that's just the terminology that's common in the subject so you either accept it um by halting and accepting or rejecting it by either holding and rejecting or by just going forever that's also considered to be rejecting sort of sort of rejecting the instance by default if you never actually have accepted it then it's going to be rejected okay uh checking three here all right so now um our last check in for the day we say this turing machine model is deterministic um i'm just saying that but if you look at the way we set it up if you've been following these formal definitions so far you would understand why it's deterministic so let's just as a way of checking that um how would we change this definition because we will look on at the next lecture at non-deterministic turing machines so a little bit of a lead into that how would we change this uh definition to make it a non-deterministic turing machine which of those three options would we use so here i'll launch that poll uh got about 10 people left let's give them another 10 seconds um okay i think that's everybody who's answered it from before um here there is so here i think you pretty much almost all of you got the right idea uh it is b in fact uh because when we have the power set symbol here that means there might be several there's a subset of possibilities so that indicates several different ways to go um uh and that's what the that's the essence of non-determinism okay so uh i think we're uh whoops um okay all right so look this is also kind of setting us up for next lecture and where we're going to be going with this um so these are uh basically two in example two or three important definitions here um one is um we have we talked about the regular languages from fine automata we talked about the context free languages from the grammars and the push down automata what are the languages of that the touring machines can do those are called in this course anyway turing recognizable languages is very orti recognizable um those are the lang those are the languages that the turing machine can recognize and so what just to make sure we were on the same page on this the language of the machine is the collection of strings that the machine accepts so the things that are not in the language are the things that are rejected either by looping or by halting and projecting so all the only the ones that are accepted are the light is the language every machine has just a single language it's a language of all strings that that machine accepts um and we'll say that m recognize that language if that language is a collection of such strings that are accepted and uh we will call that language a touring recognizable language okay if there's some turing machine that can recognize it um now uh this feature of being able to reject by running forever is a little bit weird perhaps um uh and uh for from the standpoint of practicality uh it's more convenient if the machine always makes a decision to accept or reject in a finite time and doesn't just reject by going forever and so we're going to bring out a special class of turing machines that have that feature that they always halt they always go to the holding states by the way maybe didn't say this explicitly are the q accept and the q eject states the accept and reject states are the holding states so the machine halts that means it ends up in one of those two so it has made a decision of accepting or rejecting in in by the at the point in which it has halted um so we'll say a machine is a decider if it always holds on every input uh so for every w feed in the machine is eventually going to come to a q accept or to reject we call such a machine a decider and now we're going to say a language uh is that so it will say that the machine decides a language if if it's the language of the machines of the collection of accepted strings uh and the machine is a decider so we'll say that instead of just recognizing the language we'll say that it decides the language um and a turing decidable language is a language um that the machine of all strings the machine accepts uh for some turing machine which is is a decider which is a touring machine that always holds so if a turing machine may sometimes reject by looping then it's only recognizing this language if the turing machine is always holding so it's always rejecting by explicitly coming to a reject state and halting then we'll say it's actually deciding the language so in a sense that's better and we're going to distinguish between those two because um they're not the same uh there are some languages which can be recognized but not decided and so in fact we're going to get the following picture here the deteuring recognizable languages are a proper subset they include all this everything that's decidable certainly is going to be recognizable because um uh you know this being a decider is an additional restriction to impose additional requirement so everything that's decidable is going to be automatically recognizable but there are things which are recognizable but you're not decidable as we'll see um i'll actually give an example of that but not prove it next lecture and just for just to complete out this picture um i'm going to uh also point out we haven't proven this yet but uh we will prove it um that the decidable languages also include all the context-free languages which in turn include the regular languages as it was already seen so we haven't shown this inclusion yet but actually this is the picture we we get um so there's actually a hierarchy of containments here regular languages are a subset of the context-free languages which are in turn a subset of the decidable languages which are in turn are a subset of the recognizer turing recognizable languages um and so with that i think we're going to move to our little bit of a review of what we've done today so we proved that pumping lemma as a tool for showing that languages are not context-free languages we define turing machines which is going to be our model um that we're going to be focusing on for the rest of the term uh not forgetting the other models because they're going to be useful examples for us and we define touring deciders turing machine deciders that halt and all inputs okay so i think with that um we have come to the end of today's lecture um i will stick around a little bit and answer questions in the chat um i will try to share them with everybody um as i'm answering them so i'm not just talking to one person how was the concept of pride applied in so i'm getting one question about the practicality of all this uh bunches of questions are coming in um uh so look uh is this is this stuff all practical um uh i would say um uh yes and no um uh i i don't know which concept you have in mind we're going to introduce lots of concepts in this course and um uh the you know the concept of um uh of you know the fine art automata and the push down automata and context-free languages definitely used in other subjects in other fields in computer science and elsewhere these are very basic and fundamental notions um and so yes and turing machines well i mean that's a model of a general computer if you want to understand computation you're going to need to understand some model and a turing machine is a particularly simple model and that's why we use it as it turns out it doesn't really matter what model you use but we'll talk about that next next time but yeah i would say there's a lot of applied material in this course as time has shown whether it's led to things like public key cryptography which is used on the internet or you know understanding various algorithms i mean that's not the reason i study it i'd study it because i'm more of them coming at it from more of a mathematical perspective i just find the material very uh beautiful and interesting and challenging um but it does have applications um any other questions here i think i'm going to sign off then so get myself set up for my office hours which is in a different on a different zoom like okay so thank you everybody and see you uh on thursday bye 2 00:00:28,070 --> 00:00:30,470 why don't we get started so as i like to do let's just review where we um have been recently um which is to um discuss context-free languages we talked about the context-free grammars and the push-down automata as a way of describing the context-free languages as you remember the context-free languages are a larger class of languages than the regular languages which is where we started the languages of the finite automata so when you add a stack you get more power you get more languages that you can do and we're very rapidly going to be moving on today to our main model for the semester which is called the turing machine so let's just take a look at what we're going to be covering today and that is first we're going to show that a technique analogous to the one we use for um [Music] proving that languages are not regular but this time for proving languages are not context free so the push down automate and the grammar still have their limitations in terms of what we normally think a computer can do and with that we're going to use that as a kind of a lead into our general purpose model which is the turing machine and uh so we're going to talk about turing machines and of and aspects of of that um and i do i would want to make comment uh so i have posted the solutions for the first problem set i know you're starting to think about the second problem set now which i have posted as well if you want to get a sense of what i'm looking for in terms of the level of detail you can look at the solutions to problem set one because i consider those to be model solutions uh that's part of the reason why i post them just to give you a sense of the level of detail that i'm looking for which is not a whole lot but i do want to make sure you're capturing the main ideas um of what of what's involved in solving the problem so have a look at those um and for problem set two which i'll talk about um in a second so i'll just say a few words if you want to pull that up um you know you can do that but just to get you started on a few of the problems if you're uh finding some challenges there i i don't want you to get stuck really before you even understand what the problem is saying um so for problem number one if you looked at that so that's a problem where you're asked to prove um a certain language is not context-free so that's um and by the way all of the problems in this problem set except uh perhaps for the last one for number six you'll be able to solve we'll have enough material at the end of today's lecture to solve all of them um i believe that's right uh yeah so number six you should have enough as of thursday's lecture to solve that um um uh problem number one um uh it's proving a language is not context-free so we're going to introduce a method for doing that that method is going to come in handy for parts b and c uh if you look at the problem set it has this strange looking thing sigma sigma sigma and parenthesis star it's really just a regular expression that's very simple um you know you should just make sure you understand that that's a way of representing all strings whose length is a multiple of three um and if i stick a sigma in front of that it's all strings whose length is one plus a multiple of three um so um once you understand that and if you think about what kinds of strings are in the language c2 it'll help you want to understand what happens when you take those unions um and um parts b and c are not intended to be very hard but you just have to understand what's going on uh problem number two is about ambiguous grammars i touched on that briefly in lecture um it's enough to solve the problem um the book has a little bit more detail about uh um ambiguous languages uh ambiguous grammars uh ambiguous grammars i should say um and uh so this is a grammar that's supposed to represent a fragment of a programming language um you know with uh if thens um and if then else's uh you know i'm sure you're all familiar with uh those kinds of constructs in programming languages um and there is a natural ambiguity that comes up in a programming language you know if you have you know if some condition then statement one else statement two i presume you understand what the semantics of that is what that means um and the tricky thing is that if you have a you know those statements can themselves be if statements and so if you have the situation where you have if then and if then else is the uh what follows that the question is where does the else attach is it to the second if foot to the first if um so that's kind of a big hint on this problem but that's okay um i think you uh should uh um uh you know you need to take that and figure out how to get a uh an actual member of the language which is ambiguously generated um and then show that it has that show that it is by showing two parse trees uh now or or two left-most derivations if you read the book you'll see that's an alternative way of of representing comparison so um and then what you're supposed to do is give a grammar for the same language which is unambiguous um you don't have to prove that it's unambiguous because that's a bit of a chore but as long as you understand what's going on you should be should be able to come up with an unambiguous grammar which resolves that ambiguity and i don't have in mind changing the language by introducing new uh programming languages contra constructs like a begin end uh that's not in the spirit of this problem because that's a different it's grammar for a different language so you need to be generating the same language um without any other extraneous things going on that are going to resolve the ambiguity the ambiguity needs to be resolved within the structure of the grammar itself so keep that in mind uh for problem number three about the key automata um you know that came up actually as a suggestion um uh last lecture i believe or two lectures back what happens if you take a push down automaton but instead of a push down uh as a instead of a stack you add a q um what happens then well actually it turns out that the model you get is very powerful and turns out to be equivalent in power to a turing machine so you'll see arguments of that kind today how you show that um other models are equivalent no not today that's going to be on um uh that's going to so i apologize the key this is going to be something that you'll also you know i i i i'm confusing myself here uh for problems number problem number three you actually need thursday thursday's lecture as well uh to really at least see examples of how you do that kind of thing um and so uh so thursday uh yeah so i i i'll try to send out a note clarifying this by the end of thursday you'll be able to do everything except for problem six and for problem six you'll need tuesday's lecture a week from today's lecture to do um so uh um all right uh so um problem number four that one you'll be able to do at the end of today we'll that's also gonna you know the problem is i'm i'm working on preparing thursday's lecture too so i'm getting a little i'm confusing myself uh problem number four you'll be able to do after thursday's lecture maybe we should talk about that um next lecture um problem number five um you can do today um but maybe i'm not gonna say anything about that and problem number six i won't say anything about either okay so why don't we just jump in then um and uh look at today's material um what about seven seven is uh oh seven is an optional problem oh i should have mentioned that whenever there's seven is always going to be an optional i indicate that with a star i should have made that clear um on the actual description here but seven is optional uh it's just like we had for problem set one okay let's move let's move on then to uh what we're going to talk about today um uh and just a little bit of review um so we talked about the equivalence of context-free grammars and push down automata as you remember oops let me get myself out of the picture here um as we mentioned last time we actually proved one direction but the other direction of that you just have to know it's true but you don't have to know the proof the proof is a little bit lengthy i would say it's a nice proof but it's pretty long um and uh there are two uh important corollaries to that um if you know what a corollary is just a simple consequence which doesn't need much of a proof uh sort of a very straightforward consequence first of all uh as i think we pointed out at last time one one conclusion one corollary you get is that every regular language is a context-free language because um a finite automaton is a push down automaton that just happens not to use its stack so immediately you get that every regular language is context-free and second of all um you also immediately get that whenever you have a context-free language and a and a regular language and you take their intersection you get back a context-free language so context-free intersect regular is context-free okay that that's actually mentioned in your homework as well as one of the zero-point x problems um which i give to try to get you um they're not you don't have to turn those in but i suggest you look at them i don't know how many of you are looking at them but uh this is a useful fact and and some of those other facts and in zero point x problems are useful so i encourage you to look at them but anyway intersection of context-free and regular is context-free you might ask what about intersection of context-free and context-free do we have closure under intersection the answer is no we do not have close closure under intersection we'll talk about that shortly uh so here's the proof sketch for um i should i wanted to say that the intersection of context free and regular why do we know that's still context-free because the push down automaton for a can be simulating the finite automaton for b inside its finite control inside its finite memory the problem is if you have two context-free languages you have two push-down automata you can't simulate that with one push-down automaton because it has only a single stack so if you're trying to take the intersection of two context-free language languages with only a single stack you're going to be in trouble because it's hard to um anyway i mean that's not a proof but at least it shows you what goes wrong if you try to do the obvious thing um okay uh so uh so if if then just here's an important point that i was trying to make before if a and b are both context-free and you take the intersection uh the result may not necessarily be a context-free language um so the class of context-free languages is not closed under the intersection we'll comment down that uh uh in a bit um the context-free languages are closed under the regular operations however union intersection a union concatenation and star so you should um feel comfortable that you know how to prove that it's in the again it's one of the uh i think it's problem 0.2 and i think the solution is even given in the book for it so you just should know how to prove that um uh not very pretty straightforward um okay so let's move on then um to uh to really basically conclude our um work on context-free languages and to understand the limitations of uh content context-free grammars and what kinds of languages may not be context-free and how do you prove that so how do you prove that for some language there is no grammar um again you know it's not enough just to say you know uh given informal uh comment that you will i i couldn't think of a grammar that or some things of that kind that's not going to be good enough we need to have a proof uh so if we take the language here 0 to the k1 to the k2 to the k so those are strings which are runs of zeros followed by an equal number of ones followed by an equal number of twos so just for zeros then ones then twos all the same length um uh that's a language which is not going to be a context-free language and we'll we'll give a method for proving that you know if you had a stack you can match the ones with the zeros but then once you're done with that the stack is empty and how do you now make sure that the number of twos corresponds to the number of ones that you had so again that's an informal argument that's not good enough to uh to be a proof but it sort of gives an intuition um so we're going to give a method for proving non-context-free languages are not context-free using again a pumping lemma but this is going to be a pumping lemma that applies to context-free languages not to regular languages okay it looks very similar but it has some extra wrinkles thrown in because the other older pumping dilemma was specific to the regular languages and this is going to be something that applies to the context free languages okay so now um let's just read it and then we'll um try to interpret it again it's very similar in spirit basically it says that whenever you have a context-free language all long strings in the language can be pumped in some kind of way so it's going to be a little different kind of pumping than we had before and you stay in the language okay so before we broke the string into three pieces where we could repeat that center piece as many times as as you like um and you stay in the language here we're going to end up breaking the um the string into five pieces so s is going to be broken up into u v x y z um uh and the and the way it's gonna work here so here is a picture so all long strings again there's gonna be a threshold um for whenever you have a language there's going to be some cutoff length so that all the longer strings in that language can be pumped and you stay in the language but the shorter strings there's no guarantee um so if you have a long string in the language length at least this pumping length p then you can break it up into five pieces but now it's that second and fourth string that are going to play that special uh pumping roll um which means that what you can do is you can repeat those um and you stay in the language and it's important that you repeat them both that v and that y the same number of times so you're gonna have a picture that looks something like this um and that is going to you you repeat uh so you can get you if you repeat the v and repeat the y you get u v v x y y z um or if you look at over here it would be u v squared x y squared is e um and that's going to still be in the language um and then we have so that's one condition we'll have to look at all of these conditions when we do the proof but we just understand what the statement is right now so the second condition is that v and y together cannot be empty and really that's another way of saying they can't both be the empty string because if they were both the empty string then repeating them wouldn't change um s and then of course it would stay in the language so it would be kind of meaningless if they were allowed to be empty and the last thing is again going to be there as a matter of convenience for proving str proving uh languages are not context-free because you have to make sure there's no possible way of cutting up the string when you're trying to prove a language is not context-free you have to show the pumping fails and then you it's it's going to be helpful sometimes to limit the ways in which the string can be cut up because then you have it's an easier job for you to work with it um so here it's a little different than before but sort of similar that v x y combined um as a substring so i think i show that over here v x y together is a um is not too long so the v vxy maybe is better seen up here is going to be is going to be at most p um we'll we'll do an example in a minute of using this okay so again here's our pumping limit i just restated it so um you know we have it in front of us and we're going to do a proof i'm just going to give you the idea of the proof first and then we'll go through some of the details the idea is actually pretty simple okay i give it a call proof by picture again remember what we're trying to do we're trying to show that we have this context-free language a and now all long strings in a have this pumping quality that you can break them up into five pieces so that the second and the fourth piece can be repeated and you stay in the language so how do we know that that's going to be true let's take a look at the proof here and how why is that going to be true so first of all i i i'd like to do it qualitatively rather than quantitatively so let's just imagine instead of what thinking we'll calculate what p is later but just imagine the s is some really really long string that's the way i like to think about it so s is just really long what is that going to tell us it's going to tell us something important about the way the grammar produces s which is going to be useful in in getting a way of pumping pumping s so if s is really long um we're going to look at the parse tree for s and we're going to conclude that the parse tree has to be really tall because it's impossible for a very shallow parse tree to generate a very long string and again we'll quantify that in a second but intuitively i think that's not too hard to see why that ought to be true so if you have a long s the pole the parse tree has to be really tall because the parse tree can't generate very many it can't expand by very much at each level so we'll look at how much it can expand but it depends on the grammar how much expansion can we have at each level and it's going to be you can't have just in three levels you know some small grammar generating a string of length a million just you know you'll see that that's just impossible so once you know that the parse tree is really tall here um then you're actually almost done because what does it mean to be really tall it means that there's some path starting at the start variable e i'm calling it uh in this parse tree which goes down to some um terminal symbol in s which goes through many steps that's what it means for the tree to be very tall and each one of those steps is a variable until you get down to the very end okay so that's the way parse trees look you keep expanding variables until you get to a terminal so here's that here you get some some path that's really a long path and once you have a long path that have many many variables appearing on here well the grammar itself has only some fixed number of variables in it so you're going to have to have a repetition coming among the variables that occur on that long path got that so a long string forces a tall parse tree forces a repetition on some path coming out of the start variable uh of some of some other variable that comes out now that's going to tell us how to cut up s because if you look at the subtrees of s that those two r variables are generating shown like this i'm going to use that so if you have to follow what i'm saying here so you know the r here is generating this portion of s and the lower r is generating a smaller portion of s just looking at the subtree that you get here um and that's going to tell us that we can cut up s accordingly so u is that very first part out here generated by e but not by the first r r is generated v is generated by the first r but not by the second r the second r generates exactly x and then we have y and z similarly so that all follows from having a lo a tall parse tree and now we're finished now we know how to cut up s how do we know we can repeat uh v and y and still be in the language well i'll actually show your and show you that you're in the language by exhibiting a parse tree for the string u v v x y y z here it is what i'm gonna get that parse tree by when i expand this lower r instead of expanding it to get x i'm going to follow the same substitutions that i i had when i expanded the upper r so it's as if i took this larger subtree here and i substituted it in for the smaller uh subtree under under the second r and so that i get a picture that looks like this so here i'm substituting under the second r the same subtree that i had originally coming out of the upper r the first r and so now this parse tree is generating the string u v v x y y z which is what i'm looking for and of course you can do that again and again and you're going to keep getting higher and higher exponents of v and y um and in fact you can even get the zero exponent which means that v and y both disappear altogether and for that you do something slightly different which is that you replace the larger subtree by the smaller subtree okay so here which was originally that larger subtree generating v x y i stick instead the smaller subtree i do the substitutions from the smaller sub tree and i just get x uh there and so now the string i generated is u xz which is the same as u v to the zero x y to the zero z okay that's that is the idea of the proof now i think uh you could work out um [Music] the quantities that you need in order to drive this proof um i'm gonna do that for you i actually hate writing down lots of inequalities and um equations and so on on the board because i think they're just almost incomprehensible to follow at least they would be for me but i'm going to put them up there just for completeness sake um so here we're going to give the details of this proof on the next uh the next slide here oh yeah so i just want to give a name to this i'm going to call this the cutting and pasting argument because i'm cutting apart pieces of this parse tree and i'm pasting them into other places within the parse tree to get new strings being generated so this is a cutting and pasting argument so okay let's let's take a look at the details here um uh just so we have to understand well how big does p actually need to be in order for this thing to kick in um well first of all we have to understand how fast that parse tree can be growing um as we go level to level and that's going to be dependent on how big the right-hand side sides of rules are i mean that's expand that really tells you how many what's the fan out you know of each node what's the maximum fanout and that's going to be the maximum uh length of a right-hand side of any rule so for example in that other grammar we had seen last time for arithmetic expressions we had this e goes to e plus t and uh this rule here and in terms of the parse tree that would look uh like an a little element like that and that's actually the longest right hand side that you can get and so the parse tree can be growing by a factor of three each time now that's going to tell us how big the string uh needs to be that's being generated what is the value of p in order to get a high enough parse tree so that you're going to get a repeated variable okay um let's call the height of the parse tree for sh okay so now if you this is just repeating what i just said if you have a tree of height h and the maximum branching is b then you get it most b to the h leaves because each level you get another factor of b coming up because that's how much branching you have so each node at one level can become b nodes at the next level down so you're multiplying by b each time and if you have h levels you're going to have b to the h leaves so the length of s which are really the leaves here that it is at most b to the a the reason why i said most and not exactly is you might be doing substitutions which are shorter right hand sides um okay so to try to show this as a picture here uh pulling that same picture we had before we want h the height to be bigger than the number of variables to force a repetition so the number of variables is going to be written this way v is the variables uh v with bars around it is going to be the number of variables and we want that height to be greater than the number of variables so once you know how high you want that tree to be in order to force the repetition then it tells you how big s has to be so b has to be bigger than b to the v b to the size of v um because then um the the the height that you're going to get is going to be greater than uh uh the size of v which is uh so that's what you want you want h to be greater than the size of v uh so you're gonna set p to be one more than b to the v and so if s is at least that length this whole thing is going to kick in and you're going to get that repeated variable um so we let p to be that value where v is the number of variables in the grammar and so if s is at least p which is greater than b to the v then the length of s is going to be greater than um b to the v so h is going to be what you want to make this thing work i you know if you don't follow that those inequalities i sympathize with you i would never follow that either in a lecture so um but i hope you get the idea um but you know we're not quite finished yet because what i want to now circle back as and look at these three conditions and make sure that we've captured them all because actually it's not totally uh obvious um in each of those cases that we've got them um so there's a few extra things we need to do um um okay so this is concluding the argument you know there are going to be at least three plus one variables in the longest path so there's going to be a repetition so now let's go back here and see now that we have this picture with a repeated uh variable how do we know we can get condition one well that's just the cutting and pacing argument from the previous slide how do we know that v and y are not both empty well actually that's not totally obvious because it's possible that when you generated v here and you generated y um maybe going from this r to that r you got nothing new you know it could have been that r got replaced by t another variable with nothing new coming out and then t got replaced by r so you substituted uh you know t for r and then r for t and you've got nothing new coming out and in that case uh v and y would both be the empty string and that would violate what we want um the way you get around that and these are details here if you're not totally following these points don't worry uh but i do want to just try to it's easy they're easy to describe so i figure let me present the whole thing in full detail so if going from this r to that r doesn't generate anything new you're getting exactly the same things coming out v and y are just the empty string um how do we avoid that from happening well what there's a simple way to address that which is to say uh if you have this string s when you take a parse tree make sure you take a smallest possible parse tree you cannot you're not allowed to start off with an inefficient parse tree that can be shortened and still generate s i want the smallest possible parse tree and that smallest possible parser can't have an r going to another r which is generating nothing new because then you could always have eliminated that step and you would still have a parse tree for s but it would be a smaller parse tree so that would be i want you to start off with the smallest possible parser tree and then you're going to be guaranteed that v or y is going to be something uh not empty so that takes care of uh condition two condition three uh you know how do we know that v x y together is not very long and basically it's the same argument all over again um you just want to make sure that when you're picking the repetition are the two r's here you pick the lowest possible repetitions that occur if you have many many choices and those lowest two those lowest repetitions there's not going to be any lower repetition here and then by the same argument um the the uh since once you have that very first r there's no more repetitions occurring below the v x y can't be very long um because that would again force another repetition to occur um so anyway those are the three conditions and that's the proof of the pumping lemma for context-free languages let's see how we use that um okay so let's do an example of proving a language not context free using the pumping line how do you gonna do a go about doing that because that's the kind of thing you you know at the very least you need to know how to do this in order to do the homework um uh i'd like to motivate you that the stuff is so interesting and fun but it doesn't work for everybody so uh for you practical people out there um pay attention so you can do the homework okay let's go back to that language we had a couple of slides back zero to the g one to the k two to the k it's not a contact three language we're gonna show that now using the pumping lemma for context-free languages um uh so it's gonna do similar to the proofs using for non-regular non-regul not non-regular languages um proof by contradiction so you first you assume the language is context-free and then we're going to apply the pumping lemma um and then we're going to get a contradiction uh so the pumping limit gives that pumping length as we subscribed above and now we just want to pick a longer string in the language and show that that longer string which is supposed to be pumpable and stay in the language in fact is not pumpable um so the pumping dilemma says that you can divide it into five pieces satisfying the three conditions uh condition three implies uh that so now i'm gonna i'm gonna work through the i'm gonna uh show you get a contradiction so condition three implies that you cannot contain both zeros and let's pull up a picture here so here's s uh zeros ones and then twos all the same length condition three uh so if you break it up condition three says v x y together cannot be too long well if v x y together is not too long uh what you know how could it be that when you're repeating v and y you stay in the language uh for one thing you can't have zeros ones and twos all occurring within v x and y um some uh some symbol is going to get left out so then when you pump up you're going to have unequal numbers of symbols and so you're going to be out of the language um okay so no matter how you try to cut it up uh following condition three which is one of the things that restricts the the ways to cut it up you're going to end up when you pump up uh going out of the language and so therefore it's not in d d is wrong uh b shouldn't say b can knife can i nice supposed to be able to write on this thing i guess not i didn't test that oh well that's supposed to be a b um uh so b is a context free language we close so that's the assumption that b is a context-free language that's false and we conclude that's not a context-free language okay let's do uh oh yeah i have a check-in here um so let's see um what we're going to we're going to i'm going to ask you to think about okay my head is blocking part of the text oh that was a while ago um okay uh yes so with just one question by the way in in terms of applying the um in in the pumping dilemma either v or y can be uh empty but not both but anyway let's get to this check in here um so let's look at these two languages a1 and a2 which look very similar to b but a little different um so it's a1 is 0 to the k1 to the gate 2 to the l where k and l could be any numbers any positive you know uh non-negative numbers so basically what this says saying is that the number of zeros and ones are going to be equal but the number of twos can be anything whereas a2 similar but here we're requiring the number of two ones and twos to be equal and the number of zeros can be anything okay now you can easily make i hope uh you should make sure you you can push down automata that can recognize a1 and a2 um because let's just take a1 the push down automaton can push the zeros as it's reading them pop them as it's reading the ones to match them off and make sure that they're the same number of them and then the twos it doesn't care how many there are it just has to make sure that there are no strings there are no letters coming out of order but any number of twos is fine so you can easily make a presentation recognize a1 similarly for a2 um so what can we conclude from that here are the three possibilities let me uh so look at that the class of content language is not closing intersection you know you can read it um so i i want to pull up the uh poll and launch that please fill that out um 10 seconds again just if you don't know the answer just give any answers so that uh because we're not counting correctness um still a few dribbling in okay uh five seconds okay end polling uh most of you got that right um i know is it okay to share these things i don't want to make people who didn't get the right feel bad you know but you should understand i think you know if you're missing something you should understand what is what you're missing the pumping limit shows that a1 union a2 is not a context-free language no as i mentioned at the beginning the reg the context-free languages are closed under union so the pumping alum had better not show that these we already know that these two languages are context-free because we get them from push down on top and we said at the beginning that contact free language is closed under union so we know that these two are context-free so the pumping alignment better not show it they're not context-free something would be terribly gone have gone terribly wrong uh if that were true and also um we know also from a little bit of further reasoning that the context-free languages is not closed under complement by what we've already what we've already discussed uh because they are closed under union as i pointed out they're not closed under intersection and so if they were closed under complement de morgan's laws would say that closure under union and closure under complement would give you closure under intersection but we don't have closer intersections so in fact they're not closed under complement okay uh so in fact this does show us that the class of context-free language is not closed under intersection because the intersection of a1 and a2 two context three languages is b and b is not context-free so it shows that um this is not the the um closure under intersection does not hold okay um so let us continue then we have one more uh example then we'll take a break uh so the pumping limit for context-free languages again uh here's a second example uh here's the language f we have actually seen what's before ww uh two copies of a string um in in um that two copies of any string um i'm gonna show that's not a context free language uh assume that it is context-free the pumping limit gives pumping length now here you have to do a little bit more often the challenge in applying the pumping lemma in either case uh that we've seen uh involves with choosing that string that you need to pump that you're gonna pump so you have to choose s uh in f which is longer than p which has to go with so you might try this one first glance here isn't string it's in the language because it's two copies of the string zero to the p1 zero to and then zero to the p1 that's in the language but it's a bad choice uh let's say let's before i get ahead of myself let's let's draw a picture of this which i think is always helpful to see um so here is runs of zeros and then a one runs of zeros and then a one um why is this a bad choice because you can pump that string and you remain in the language um there is a way to cut that string up um and you'll stay in the language and the way to cut it up is to let the x be just that substring which is just the one and the v and the y can be a couple of zeros or a single zero on either side of that one and now that's going to be a small v x y but if you repeat v and y you're going to um if you repeat v and y you're going to stay in the language because you'll just be adding zeros here you'll be adding same number of zeros there and then you're going to have a string which still looks like ww and you'll still be in the language so that that means of cutting it up doesn't get you out of the language under pumping and the fact is that that's a bad choice for s because there is that way of cutting it up so you have to show there's no way you can't you don't get to pick the way to cut it up um you have to show that there is no way to cut it up um uh in order to um to violate the uh the pumping lever so if instead you use the string 0 to the p1 to the p0 to the p1 to the p so this is zeros followed by one followed by zeros followed by ones all the same number of them that can't be pumped satisfying to the three conditions um and just going through that uh um now if you try to break it up you're going to lose or the or the lemon is going to lose you're going to be happy but the lemon is not going to be happy because it's not going it's going to violate the conditions um condition 3 says vxy is not doesn't span too much and in fact can't span uh two runs of zeros or two runs of ones it's just not big enough because they're more than p things they're p things apart and this one string this string v x y is only p long and so therefore if you repeat of uh v and y you're going to have two runs of zeros or two runs of ones that have unequal length and now that's not going to be of the form w and you're going to be out of the language okay uh so i hope you know that's uh you've got a little practice with that um i think we're at our uh break and i will see you back here in uh five uh minutes um i can get my timer launched here okay so see you soon you can this is a good time by the way to to um uh message me or the tas and i'll i'll try to be uh looking for if you have any questions in the pumping limit can x x can be epsilon in the pumping lemma uh but it's not possible x can be epsilon y can be epsilon but x and y cannot be both cannot both be epsilon because then when you pump you will get nothing new technically v and y can include both zeros and ones yeah v and y can conclude both zeros and ones um so let me uh try to put that back if that's will um [Music] so v and y can have both zeros and ones but they can't have zeros from two different blocks and you can't have ones from two different blocks so what's going to happen is either you're going to get things out of order when you repeat like a v has both zeros and ones in it when you repeat v you're going to have uh zeros and ones and zeros and ones and zeros and ones that's clearly out of the language so that's so that's no good your only hope is to have v to be sticking only inside the zeros and y to be sticking only inside zeros are only inside once but now if you repeat that if just look at what you're going to get you're going to have a string which is going to be if you try to cut that string in half it's not going to be of the right form it's not going to be two copies of the same string because you know it's going to have a run of zeros um followed by a longer or shorter run of zeros or one of ones followed by another run of ones of unequal length so there's no way this can be two strings um two copies of the same string because that that's what you're required f has to be two copies of the same string to be in the language okay uh let me just see where we're running out of time here let me just put my uh my my timer here we've only got 30 seconds okay sorry i'm not getting to answer all the questions here um okay we are done um uh with um our break it's going to come back um and now we're shifting gears in a major way because in a sense everything we've done so far has been kind of a warm up these limited computational models um really are kind of get helping us to set our understanding of automata and the definitions and the notation and they're also going to be helpful in providing examples later on in the term um but really in terms of a model of computation they don't cut it because uh they cannot do very simple things that we normally think of a computer as being able to do so here we're introducing another model of computation called the turing machine and that's really going to be the model that we're going to stick with for the rest of the semester because that's going to be our model of a general purpose uh computer the way you normally think about it um so let's uh we'll spend a little time introducing it and then we will um uh you know we'll we'll continue that discussion next time um so uh in terms of a schematic actually the turing machine model is pretty simple um it has a uh it has um it's going to have states and all that stuff but you know though so there's going to be a finite control here which is going to include the states and transition function as we'll describe in a minute um the point is is that you know it's going to have the input appearing on a tape uh the key difference now is that the the machine is going to be able to change the symbols on the tape um and so we think of the machine as being able to write as well as read the tape okay so that's really uh the key uh feature um of a turing machine is the ability to write on the tape everything else in a sense follows from that a few other differences but so the fact that he he head can read and write so we can use the tape as storage much as we used a stack as storage but it doesn't it's not limited in the way we can access it the way a stack is so that we kind of have very uh flexible access of the information on the tape now being able to write on the tape doesn't do any good if you can't go back and read what you've written later on so we're going to make the head to be able to be two-way so the head can move left to right as before but it can also move back left and that's going to be under control of the transition function so under under program control essentially the tape is going to be oops sorry the tape is infinite to the right and so we're not going to limit how much storage the machine can have so the tape is going to think of as having instead of just having the input on it it's going to have the input but then the rest is going to have infinitely many blanks blank symbols following the input so the tape is infinite or in the right right hand direction um and uh so there's infinitely many blanks i'm going to use that symbol for the blank to follow the input um you can accept or reject oh yeah so that's another thing that's important normally we think of the machine in the previous machines finding a time when i pushed down automata when you got to the end of the input that's when you the acceptance of rejection was decided if you were in except state at the end of the input then you accept it but you have to be in that location um at the end of the input in order for that to take effect that doesn't make any sense anymore because the machine might go off beyond that and still be computing and come back and read the tape later on so it only really makes sense to let the machine accept or reject upon entering the accept or the reject state so we're going to have a special accept state and a special reject state which is also a little different than before and when the machine enters those states then machine then the action takes effect the machine those machine holds and then accepts or holds and then it rejects so we'll make that absolutely clear in the formal definition in a second but just to get the spirit of it um so i'm going to give you an example of the thing running uh sorry you need to again my powerpoint is having issues um uh okay so here is a turing machine recognizing that language b actually i switched gears on you instead of zeros ones and twos i made them a's b's and c's but same idea [Music] so i'm going to show you how the turing machine operates and then we'll give a formal uh definition i hope that's on here i think it is in a second but let's this is an informal discussion of how the machine is going to operate to do this language a to the kb to the kc to the k using its ability to write on the tape as well as read and move its head in both directions okay so let me just first describe um in english how this um [Music] machine operates and then we will um see it in action on this little uh picture i have over here okay so the way the machine is going to operate is the very first thing is the head is going to start here and the head is going to scan off to the right uh making sure that the symbols appear in the correct order so it's seeing that that there are a's and then b's and then c's without checking the quantities just that the order is correct for that you don't need to write a finite automaton can check that the input is of the form a star b star c star um so uh rioting is not necessary the machine if it uh detects symbols out of order it immediately rejects by going into a special reject state otherwise it's going to return its head back to the left end and um uh let me just show that here so here is it oh no i better before i illustrate it over here let's go through the whole algorithm uh so the next thing that happens is you're going to scan right and now you want to do the counting so you're going to scan right again but this time you're going to make a bunch of passes over the input bunch of scans and each time you make a scan you're going to cross off one symbol of each type so you're going to cross off an a you'll cross off a b you'll cross off a c on a single scan and then repeat that crossing off the next a the next b the next c and you want to make sure that you've crossed off all of the symbols on the same run and not crossing off some symbols before other symbols uh while other symbols will remain because that would mean that the counts were not equal if you cross them off and they're all you know run out on the same scan same pass then we know that the numbers have to be start off being equal sort of i mean this is sort of baby stuff here i but i hope you get the idea and we'll kind of illustrate in in a second so if at least one of the if you have the last one of each symbol so what i mean by that is you you just crossed off the last a the last b and the last c then you know that you were originally had an equal number and so you accept because you're crossing off one of each on each scan so if you cross the on the last scan each one of them gets crossed off um then you accept but if uh it was the last of some symbol but not of other symbols so you know you crossed off the last a but there um there were several b's remaining then you started off with an unequal number of a's b's and c's and you can reject or if all symbols still remain after you've crossed them uh one of each off then you haven't done enough passes and you're gonna repeat from stage three and do that uh again another scan okay so here's a little um animation which shows this happening on this diagram so here is the very first stage where you're scanning across making sure things are in the right order um didn't have to write on the tape and now you're gonna reset the the head back to the beginning this is by the way not the most efficient procedure for doing this um so um now that now we're going to do a scan crossing off a single a a single b and a single c so here i'm going to show that here single a single b single c and now as soon as you've crossed out that last c we can return back to the beginning um so scan right cross uh so if all symbols remain so there are still symbols remaining of each type we're going to return to the left and repeat and repeat now we'll get another pass single a single b single c get crossed off we haven't crossed them all off yet no there's of each type there's still our remaining ones so again we return back to the beginning now we have a last pass crossover the last a the last be the last c the last one of each type was crossed off so now we know we can accept because the original string was in the language okay so that's uh to give you at least some idea um that how the turing machine can operate you know more like the way you would think of a computer operating you know maybe it's very primitive you know you you could imagine counting also and the termination can count as well but this is the simplest uh descript uh procedure that i can just describe on um you know uh for you without getting making too complicated um okay so let's do a little checking on that um okay so the way i'm describing this how do we go how do you think and you in a sense you don't quite know enough yet but um [Music] uh how do you think we're going to get this effect of crossing off with the turing machine um are we going to get that by changing the model and adding that ability to cross off to the model um are we going to use a tape alphabet that includes those crossed off symbols among them or um we'll just assume that all turing machines come with the eraser and they can always cross our stuff um so what do you think is the nice way sort of mathematically to describe this this ability to cross things off yeah again again most of your again i think are getting this um uh so like 10 laggards here so please wrap it up um so we can close the poll uh five seconds to go um okay polling ending get your last last call uh all right share the results so most of you uh got that right all turn machines come with an eraser i don't know that was thrown in there as a joke but it got came in second so i don't feel bad if you got it but that's not what i had in mind um the the way the turing machine is going to be riding on the tape is to write a crossed off symbol uh instead of the uh the symbol that was originally there so we're going to add these new crosstalk symbols and that's going to be a common thing for us to do in um you know when we design turing machines we're not going to get down to the implementation level for very long we're going to very quickly shift to a higher level of discussion about the machines but anyway that's how you would do it if you were going to actually build a machine okay so um let us then look at the formal definition of the cursing and maybe i should have done that checking after the formal definition that might have been clear but oh well uh okay here's the formal definition this time a turing machine is a seven tuple um and there is a uh um now here uh we have sigma which is the input alphabet gamma is the tape alphabet so now you're it's sort of a little bit analogous to the stack from before where gamma was the stack alphabet but these are the symbols that you're allowed to write on the tape um or that are allowed to be on the tape um so uh obviously all of the input symbols are among the uh tape uh symbols because they can appear on the tape so you have sigma is a subset of gamma um one thing i didn't mention here is that the input alphabet we don't allow the blank symbol to be uh in the input alphabet so that you can actually use the blank symbol as a delimiter for the end of the input a marker for the end of the input um so uh in fact and blank symbol is always going to be in the tape alphabet uh so we have um you know the this is actually always going to be a proper subset uh because of the blank symbol um what we're just allowing doesn't really matter we're allowing the tape alphabet to have other symbols for convenience so for example these crossed off symbols now let's look at what the transition function how that operates so the transition function remember tells how the machine is actually doing its computation and it says that if you're in a certain state and the head is looking at a certain tape symbol um then you can go to a new state you write a new symbol on the uh at that location on the tape and you can move the head either left or right so that's how we get the effect of the head being able to be bi-directional and here's the writing on the tape comes up right here so you know just an example here we says if we're in state q and the head is looking at an a currently on the tape then we can move to state r we change it a to a b and we move the head right once one square okay um uh uh now here this is important to on when you're giving a certain input here to the turing machine it may gr compute around for a while moving its head back and forth as we were showing and it may eventually halt by either entering the q accept state or the q reject state which i didn't uh bring out here but that's important these are the accepting rejecting special states of the machine um or the machine may never enter one of those it may just go on and on and on and never halt we call that looping a little bit of a misnomer because it's the looping implies some sort of a repetition for us looping just means not holding um and um so therefore m has three possible outcomes for each um input w it might accept w by uh entering uh the except state it could reject w by entering the reject state which means it's going to reject it by holding or we also say we can reject by looping you can reject the string by running forever um okay that's just the terminology that's common in the subject so you either accept it um by halting and accepting or rejecting it by either holding and rejecting or by just going forever that's also considered to be rejecting sort of sort of rejecting the instance by default if you never actually have accepted it then it's going to be rejected okay uh checking three here all right so now um our last check in for the day we say this turing machine model is deterministic um i'm just saying that but if you look at the way we set it up if you've been following these formal definitions so far you would understand why it's deterministic so let's just as a way of checking that um how would we change this definition because we will look on at the next lecture at non-deterministic turing machines so a little bit of a lead into that how would we change this uh definition to make it a non-deterministic turing machine which of those three options would we use so here i'll launch that poll uh got about 10 people left let's give them another 10 seconds um okay i think that's everybody who's answered it from before um here there is so here i think you pretty much almost all of you got the right idea uh it is b in fact uh because when we have the power set symbol here that means there might be several there's a subset of possibilities so that indicates several different ways to go um uh and that's what the that's the essence of non-determinism okay so uh i think we're uh whoops um okay all right so look this is also kind of setting us up for next lecture and where we're going to be going with this um so these are uh basically two in example two or three important definitions here um one is um we have we talked about the regular languages from fine automata we talked about the context free languages from the grammars and the push down automata what are the languages of that the touring machines can do those are called in this course anyway turing recognizable languages is very orti recognizable um those are the lang those are the languages that the turing machine can recognize and so what just to make sure we were on the same page on this the language of the machine is the collection of strings that the machine accepts so the things that are not in the language are the things that are rejected either by looping or by halting and projecting so all the only the ones that are accepted are the light is the language every machine has just a single language it's a language of all strings that that machine accepts um and we'll say that m recognize that language if that language is a collection of such strings that are accepted and uh we will call that language a touring recognizable language okay if there's some turing machine that can recognize it um now uh this feature of being able to reject by running forever is a little bit weird perhaps um uh and uh for from the standpoint of practicality uh it's more convenient if the machine always makes a decision to accept or reject in a finite time and doesn't just reject by going forever and so we're going to bring out a special class of turing machines that have that feature that they always halt they always go to the holding states by the way maybe didn't say this explicitly are the q accept and the q eject states the accept and reject states are the holding states so the machine halts that means it ends up in one of those two so it has made a decision of accepting or rejecting in in by the at the point in which it has halted um so we'll say a machine is a decider if it always holds on every input uh so for every w feed in the machine is eventually going to come to a q accept or to reject we call such a machine a decider and now we're going to say a language uh is that so it will say that the machine decides a language if if it's the language of the machines of the collection of accepted strings uh and the machine is a decider so we'll say that instead of just recognizing the language we'll say that it decides the language um and a turing decidable language is a language um that the machine of all strings the machine accepts uh for some turing machine which is is a decider which is a touring machine that always holds so if a turing machine may sometimes reject by looping then it's only recognizing this language if the turing machine is always holding so it's always rejecting by explicitly coming to a reject state and halting then we'll say it's actually deciding the language so in a sense that's better and we're going to distinguish between those two because um they're not the same uh there are some languages which can be recognized but not decided and so in fact we're going to get the following picture here the deteuring recognizable languages are a proper subset they include all this everything that's decidable certainly is going to be recognizable because um uh you know this being a decider is an additional restriction to impose additional requirement so everything that's decidable is going to be automatically recognizable but there are things which are recognizable but you're not decidable as we'll see um i'll actually give an example of that but not prove it next lecture and just for just to complete out this picture um i'm going to uh also point out we haven't proven this yet but uh we will prove it um that the decidable languages also include all the context-free languages which in turn include the regular languages as it was already seen so we haven't shown this inclusion yet but actually this is the picture we we get um so there's actually a hierarchy of containments here regular languages are a subset of the context-free languages which are in turn a subset of the decidable languages which are in turn are a subset of the recognizer turing recognizable languages um and so with that i think we're going to move to our little bit of a review of what we've done today so we proved that pumping lemma as a tool for showing that languages are not context-free languages we define turing machines which is going to be our model um that we're going to be focusing on for the rest of the term uh not forgetting the other models because they're going to be useful examples for us and we define touring deciders turing machine deciders that halt and all inputs okay so i think with that um we have come to the end of today's lecture um i will stick around a little bit and answer questions in the chat um i will try to share them with everybody um as i'm answering them so i'm not just talking to one person how was the concept of pride applied in so i'm getting one question about the practicality of all this uh bunches of questions are coming in um uh so look uh is this is this stuff all practical um uh i would say um uh yes and no um uh i i don't know which concept you have in mind we're going to introduce lots of concepts in this course and um uh the you know the concept of um uh of you know the fine art automata and the push down automata and context-free languages definitely used in other subjects in other fields in computer science and elsewhere these are very basic and fundamental notions um and so yes and turing machines well i mean that's a model of a general computer if you want to understand computation you're going to need to understand some model and a turing machine is a particularly simple model and that's why we use it as it turns out it doesn't really matter what model you use but we'll talk about that next next time but yeah i would say there's a lot of applied material in this course as time has shown whether it's led to things like public key cryptography which is used on the internet or you know understanding various algorithms i mean that's not the reason i study it i'd study it because i'm more of them coming at it from more of a mathematical perspective i just find the material very uh beautiful and interesting and challenging um but it does have applications um any other questions here i think i'm going to sign off then so get myself set up for my office hours which is in a different on a different zoom like okay so thank you everybody and see you uh on thursday bye 3 00:00:30,470 --> 00:00:34,310 4 00:00:34,310 --> 00:00:36,150 5 00:00:36,150 --> 00:00:36,160 6 00:00:36,160 --> 00:00:37,750 7 00:00:37,750 --> 00:00:37,760 8 00:00:37,760 --> 00:00:39,110 9 00:00:39,110 --> 00:00:41,670 10 00:00:41,670 --> 00:00:43,350 11 00:00:43,350 --> 00:00:45,029 12 00:00:45,029 --> 00:00:47,750 13 00:00:47,750 --> 00:00:47,760 14 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--> 00:11:45,269 368 00:11:45,269 --> 00:11:47,110 369 00:11:47,110 --> 00:11:49,910 370 00:11:49,910 --> 00:11:50,949 371 00:11:50,949 --> 00:11:52,389 372 00:11:52,389 --> 00:11:54,870 373 00:11:54,870 --> 00:11:56,069 374 00:11:56,069 --> 00:11:58,310 375 00:11:58,310 --> 00:12:00,550 376 00:12:00,550 --> 00:12:01,990 377 00:12:01,990 --> 00:12:04,150 378 00:12:04,150 --> 00:12:04,160 379 00:12:04,160 --> 00:12:05,990 380 00:12:05,990 --> 00:12:06,000 381 00:12:06,000 --> 00:12:06,949 382 00:12:06,949 --> 00:12:08,829 383 00:12:08,829 --> 00:12:08,839 384 00:12:08,839 --> 00:12:10,389 385 00:12:10,389 --> 00:12:12,629 386 00:12:12,629 --> 00:12:14,790 387 00:12:14,790 --> 00:12:16,150 388 00:12:16,150 --> 00:12:17,990 389 00:12:17,990 --> 00:12:19,750 390 00:12:19,750 --> 00:12:22,069 391 00:12:22,069 --> 00:12:25,670 392 00:12:25,670 --> 00:12:27,430 393 00:12:27,430 --> 00:12:29,110 394 00:12:29,110 --> 00:12:31,750 395 00:12:31,750 --> 00:12:31,760 396 00:12:31,760 --> 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426 00:13:28,230 --> 00:13:28,240 427 00:13:28,240 --> 00:13:29,110 428 00:13:29,110 --> 00:13:30,230 429 00:13:30,230 --> 00:13:33,829 430 00:13:33,829 --> 00:13:33,839 431 00:13:33,839 --> 00:13:34,870 432 00:13:34,870 --> 00:13:36,790 433 00:13:36,790 --> 00:13:39,269 434 00:13:39,269 --> 00:13:40,310 435 00:13:40,310 --> 00:13:41,990 436 00:13:41,990 --> 00:13:44,949 437 00:13:44,949 --> 00:13:47,670 438 00:13:47,670 --> 00:13:49,590 439 00:13:49,590 --> 00:13:51,110 440 00:13:51,110 --> 00:13:53,269 441 00:13:53,269 --> 00:13:55,670 442 00:13:55,670 --> 00:13:57,030 443 00:13:57,030 --> 00:13:57,040 444 00:13:57,040 --> 00:13:58,230 445 00:13:58,230 --> 00:14:00,870 446 00:14:00,870 --> 00:14:02,629 447 00:14:02,629 --> 00:14:05,110 448 00:14:05,110 --> 00:14:07,829 449 00:14:07,829 --> 00:14:10,629 450 00:14:10,629 --> 00:14:12,230 451 00:14:12,230 --> 00:14:14,310 452 00:14:14,310 --> 00:14:17,030 453 00:14:17,030 --> 00:14:19,670 454 00:14:19,670 --> 00:14:21,350 455 00:14:21,350 --> 00:14:24,150 456 00:14:24,150 --> 00:14:27,110 457 00:14:27,110 --> 00:14:30,230 458 00:14:30,230 --> 00:14:30,240 459 00:14:30,240 --> 00:14:32,470 460 00:14:32,470 --> 00:14:34,870 461 00:14:34,870 --> 00:14:34,880 462 00:14:34,880 --> 00:14:36,230 463 00:14:36,230 --> 00:14:38,069 464 00:14:38,069 --> 00:14:39,910 465 00:14:39,910 --> 00:14:41,670 466 00:14:41,670 --> 00:14:43,829 467 00:14:43,829 --> 00:14:45,750 468 00:14:45,750 --> 00:14:47,509 469 00:14:47,509 --> 00:14:49,670 470 00:14:49,670 --> 00:14:51,110 471 00:14:51,110 --> 00:14:52,629 472 00:14:52,629 --> 00:14:54,550 473 00:14:54,550 --> 00:14:56,710 474 00:14:56,710 --> 00:14:59,990 475 00:14:59,990 --> 00:15:01,829 476 00:15:01,829 --> 00:15:03,590 477 00:15:03,590 --> 00:15:05,910 478 00:15:05,910 --> 00:15:07,269 479 00:15:07,269 --> 00:15:10,069 480 00:15:10,069 --> 00:15:12,470 481 00:15:12,470 --> 00:15:14,230 482 00:15:14,230 --> 00:15:15,509 483 00:15:15,509 --> 00:15:17,829 484 00:15:17,829 --> 00:15:20,069 485 00:15:20,069 --> 00:15:22,710 486 00:15:22,710 --> 00:15:24,710 487 00:15:24,710 --> 00:15:27,269 488 00:15:27,269 --> 00:15:29,749 489 00:15:29,749 --> 00:15:31,670 490 00:15:31,670 --> 00:15:33,829 491 00:15:33,829 --> 00:15:36,710 492 00:15:36,710 --> 00:15:41,670 493 00:15:41,670 --> 00:15:41,680 494 00:15:41,680 --> 00:15:42,870 495 00:15:42,870 --> 00:15:42,880 496 00:15:42,880 --> 00:15:43,670 497 00:15:43,670 --> 00:15:43,680 498 00:15:43,680 --> 00:15:45,670 499 00:15:45,670 --> 00:15:47,910 500 00:15:47,910 --> 00:15:50,310 501 00:15:50,310 --> 00:15:52,069 502 00:15:52,069 --> 00:15:53,189 503 00:15:53,189 --> 00:15:55,430 504 00:15:55,430 --> 00:15:57,110 505 00:15:57,110 --> 00:15:59,030 506 00:15:59,030 --> 00:16:00,710 507 00:16:00,710 --> 00:16:02,230 508 00:16:02,230 --> 00:16:04,389 509 00:16:04,389 --> 00:16:05,910 510 00:16:05,910 --> 00:16:07,350 511 00:16:07,350 --> 00:16:08,790 512 00:16:08,790 --> 00:16:12,230 513 00:16:12,230 --> 00:16:14,790 514 00:16:14,790 --> 00:16:18,550 515 00:16:18,550 --> 00:16:20,790 516 00:16:20,790 --> 00:16:23,430 517 00:16:23,430 --> 00:16:23,440 518 00:16:23,440 --> 00:16:24,230 519 00:16:24,230 --> 00:16:26,629 520 00:16:26,629 --> 00:16:28,310 521 00:16:28,310 --> 00:16:31,110 522 00:16:31,110 --> 00:16:32,870 523 00:16:32,870 --> 00:16:34,870 524 00:16:34,870 --> 00:16:38,470 525 00:16:38,470 --> 00:16:40,790 526 00:16:40,790 --> 00:16:42,870 527 00:16:42,870 --> 00:16:45,269 528 00:16:45,269 --> 00:16:49,910 529 00:16:49,910 --> 00:16:53,189 530 00:16:53,189 --> 00:16:53,990 531 00:16:53,990 --> 00:16:54,000 532 00:16:54,000 --> 00:16:54,870 533 00:16:54,870 --> 00:16:56,470 534 00:16:56,470 --> 00:16:56,480 535 00:16:56,480 --> 00:16:57,990 536 00:16:57,990 --> 00:17:00,389 537 00:17:00,389 --> 00:17:01,829 538 00:17:01,829 --> 00:17:03,749 539 00:17:03,749 --> 00:17:05,029 540 00:17:05,029 --> 00:17:07,110 541 00:17:07,110 --> 00:17:09,750 542 00:17:09,750 --> 00:17:11,829 543 00:17:11,829 --> 00:17:13,829 544 00:17:13,829 --> 00:17:15,909 545 00:17:15,909 --> 00:17:17,669 546 00:17:17,669 --> 00:17:19,429 547 00:17:19,429 --> 00:17:19,439 548 00:17:19,439 --> 00:17:20,549 549 00:17:20,549 --> 00:17:20,559 550 00:17:20,559 --> 00:17:21,590 551 00:17:21,590 --> 00:17:22,789 552 00:17:22,789 --> 00:17:23,990 553 00:17:23,990 --> 00:17:25,750 554 00:17:25,750 --> 00:17:25,760 555 00:17:25,760 --> 00:17:26,789 556 00:17:26,789 --> 00:17:28,870 557 00:17:28,870 --> 00:17:31,029 558 00:17:31,029 --> 00:17:33,510 559 00:17:33,510 --> 00:17:36,070 560 00:17:36,070 --> 00:17:36,080 561 00:17:36,080 --> 00:17:37,430 562 00:17:37,430 --> 00:17:38,870 563 00:17:38,870 --> 00:17:41,590 564 00:17:41,590 --> 00:17:42,789 565 00:17:42,789 --> 00:17:44,390 566 00:17:44,390 --> 00:17:48,549 567 00:17:48,549 --> 00:17:51,110 568 00:17:51,110 --> 00:17:53,029 569 00:17:53,029 --> 00:17:55,510 570 00:17:55,510 --> 00:17:58,549 571 00:17:58,549 --> 00:18:00,150 572 00:18:00,150 --> 00:18:03,510 573 00:18:03,510 --> 00:18:06,310 574 00:18:06,310 --> 00:18:07,669 575 00:18:07,669 --> 00:18:09,590 576 00:18:09,590 --> 00:18:11,590 577 00:18:11,590 --> 00:18:14,230 578 00:18:14,230 --> 00:18:16,150 579 00:18:16,150 --> 00:18:17,590 580 00:18:17,590 --> 00:18:19,590 581 00:18:19,590 --> 00:18:22,150 582 00:18:22,150 --> 00:18:23,750 583 00:18:23,750 --> 00:18:26,870 584 00:18:26,870 --> 00:18:29,029 585 00:18:29,029 --> 00:18:31,029 586 00:18:31,029 --> 00:18:32,390 587 00:18:32,390 --> 00:18:33,669 588 00:18:33,669 --> 00:18:35,909 589 00:18:35,909 --> 00:18:38,310 590 00:18:38,310 --> 00:18:38,320 591 00:18:38,320 --> 00:18:39,190 592 00:18:39,190 --> 00:18:43,909 593 00:18:43,909 --> 00:18:45,270 594 00:18:45,270 --> 00:18:47,430 595 00:18:47,430 --> 00:18:49,750 596 00:18:49,750 --> 00:18:50,950 597 00:18:50,950 --> 00:18:54,470 598 00:18:54,470 --> 00:18:56,710 599 00:18:56,710 --> 00:18:58,310 600 00:18:58,310 --> 00:19:00,630 601 00:19:00,630 --> 00:19:02,390 602 00:19:02,390 --> 00:19:02,400 603 00:19:02,400 --> 00:19:04,549 604 00:19:04,549 --> 00:19:06,150 605 00:19:06,150 --> 00:19:07,430 606 00:19:07,430 --> 00:19:09,110 607 00:19:09,110 --> 00:19:11,110 608 00:19:11,110 --> 00:19:13,270 609 00:19:13,270 --> 00:19:15,669 610 00:19:15,669 --> 00:19:17,669 611 00:19:17,669 --> 00:19:20,070 612 00:19:20,070 --> 00:19:21,990 613 00:19:21,990 --> 00:19:23,990 614 00:19:23,990 --> 00:19:24,000 615 00:19:24,000 --> 00:19:25,510 616 00:19:25,510 --> 00:19:26,950 617 00:19:26,950 --> 00:19:29,029 618 00:19:29,029 --> 00:19:31,190 619 00:19:31,190 --> 00:19:32,390 620 00:19:32,390 --> 00:19:32,400 621 00:19:32,400 --> 00:19:33,430 622 00:19:33,430 --> 00:19:33,440 623 00:19:33,440 --> 00:19:36,470 624 00:19:36,470 --> 00:19:39,909 625 00:19:39,909 --> 00:19:41,430 626 00:19:41,430 --> 00:19:43,669 627 00:19:43,669 --> 00:19:45,110 628 00:19:45,110 --> 00:19:47,270 629 00:19:47,270 --> 00:19:47,280 630 00:19:47,280 --> 00:19:48,630 631 00:19:48,630 --> 00:19:49,830 632 00:19:49,830 --> 00:19:50,870 633 00:19:50,870 --> 00:19:52,630 634 00:19:52,630 --> 00:19:55,590 635 00:19:55,590 --> 00:19:58,230 636 00:19:58,230 --> 00:19:58,240 637 00:19:58,240 --> 00:19:59,590 638 00:19:59,590 --> 00:20:02,710 639 00:20:02,710 --> 00:20:02,720 640 00:20:02,720 --> 00:20:03,830 641 00:20:03,830 --> 00:20:05,350 642 00:20:05,350 --> 00:20:07,430 643 00:20:07,430 --> 00:20:09,669 644 00:20:09,669 --> 00:20:12,549 645 00:20:12,549 --> 00:20:14,470 646 00:20:14,470 --> 00:20:16,390 647 00:20:16,390 --> 00:20:18,870 648 00:20:18,870 --> 00:20:20,390 649 00:20:20,390 --> 00:20:22,070 650 00:20:22,070 --> 00:20:24,230 651 00:20:24,230 --> 00:20:25,909 652 00:20:25,909 --> 00:20:27,669 653 00:20:27,669 --> 00:20:30,070 654 00:20:30,070 --> 00:20:31,909 655 00:20:31,909 --> 00:20:34,149 656 00:20:34,149 --> 00:20:37,270 657 00:20:37,270 --> 00:20:40,630 658 00:20:40,630 --> 00:20:42,870 659 00:20:42,870 --> 00:20:42,880 660 00:20:42,880 --> 00:20:43,669 661 00:20:43,669 --> 00:20:46,630 662 00:20:46,630 --> 00:20:48,710 663 00:20:48,710 --> 00:20:50,950 664 00:20:50,950 --> 00:20:53,270 665 00:20:53,270 --> 00:20:56,149 666 00:20:56,149 --> 00:20:58,230 667 00:20:58,230 --> 00:20:58,240 668 00:20:58,240 --> 00:20:59,190 669 00:20:59,190 --> 00:20:59,200 670 00:20:59,200 --> 00:21:00,390 671 00:21:00,390 --> 00:21:00,400 672 00:21:00,400 --> 00:21:01,430 673 00:21:01,430 --> 00:21:02,630 674 00:21:02,630 --> 00:21:04,549 675 00:21:04,549 --> 00:21:06,070 676 00:21:06,070 --> 00:21:06,080 677 00:21:06,080 --> 00:21:06,870 678 00:21:06,870 --> 00:21:09,029 679 00:21:09,029 --> 00:21:11,029 680 00:21:11,029 --> 00:21:11,039 681 00:21:11,039 --> 00:21:13,669 682 00:21:13,669 --> 00:21:15,750 683 00:21:15,750 --> 00:21:17,750 684 00:21:17,750 --> 00:21:19,350 685 00:21:19,350 --> 00:21:21,510 686 00:21:21,510 --> 00:21:25,029 687 00:21:25,029 --> 00:21:27,590 688 00:21:27,590 --> 00:21:30,149 689 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--> 00:22:24,950 719 00:22:24,950 --> 00:22:27,350 720 00:22:27,350 --> 00:22:29,029 721 00:22:29,029 --> 00:22:31,990 722 00:22:31,990 --> 00:22:32,000 723 00:22:32,000 --> 00:22:33,990 724 00:22:33,990 --> 00:22:37,669 725 00:22:37,669 --> 00:22:40,549 726 00:22:40,549 --> 00:22:42,950 727 00:22:42,950 --> 00:22:45,430 728 00:22:45,430 --> 00:22:47,270 729 00:22:47,270 --> 00:22:51,909 730 00:22:51,909 --> 00:22:54,789 731 00:22:54,789 --> 00:22:54,799 732 00:22:54,799 --> 00:22:55,830 733 00:22:55,830 --> 00:22:57,110 734 00:22:57,110 --> 00:22:58,630 735 00:22:58,630 --> 00:23:00,789 736 00:23:00,789 --> 00:23:00,799 737 00:23:00,799 --> 00:23:01,590 738 00:23:01,590 --> 00:23:03,430 739 00:23:03,430 --> 00:23:06,070 740 00:23:06,070 --> 00:23:07,909 741 00:23:07,909 --> 00:23:09,029 742 00:23:09,029 --> 00:23:11,990 743 00:23:11,990 --> 00:23:17,350 744 00:23:17,350 --> 00:23:20,230 745 00:23:20,230 --> 00:23:22,789 746 00:23:22,789 --> 00:23:25,110 747 00:23:25,110 --> 00:23:27,350 748 00:23:27,350 --> 00:23:30,630 749 00:23:30,630 --> 00:23:33,830 750 00:23:33,830 --> 00:23:36,390 751 00:23:36,390 --> 00:23:39,190 752 00:23:39,190 --> 00:23:41,110 753 00:23:41,110 --> 00:23:45,190 754 00:23:45,190 --> 00:23:47,590 755 00:23:47,590 --> 00:23:50,070 756 00:23:50,070 --> 00:23:52,230 757 00:23:52,230 --> 00:23:54,070 758 00:23:54,070 --> 00:23:56,310 759 00:23:56,310 --> 00:23:59,590 760 00:23:59,590 --> 00:24:00,950 761 00:24:00,950 --> 00:24:03,110 762 00:24:03,110 --> 00:24:04,870 763 00:24:04,870 --> 00:24:08,070 764 00:24:08,070 --> 00:24:08,080 765 00:24:08,080 --> 00:24:09,029 766 00:24:09,029 --> 00:24:11,029 767 00:24:11,029 --> 00:24:13,750 768 00:24:13,750 --> 00:24:15,430 769 00:24:15,430 --> 00:24:17,029 770 00:24:17,029 --> 00:24:20,390 771 00:24:20,390 --> 00:24:22,149 772 00:24:22,149 --> 00:24:22,159 773 00:24:22,159 --> 00:24:25,190 774 00:24:25,190 --> 00:24:27,590 775 00:24:27,590 --> 00:24:30,710 776 00:24:30,710 --> 00:24:32,710 777 00:24:32,710 --> 00:24:34,710 778 00:24:34,710 --> 00:24:37,510 779 00:24:37,510 --> 00:24:40,149 780 00:24:40,149 --> 00:24:42,630 781 00:24:42,630 --> 00:24:42,640 782 00:24:42,640 --> 00:24:44,230 783 00:24:44,230 --> 00:24:47,350 784 00:24:47,350 --> 00:24:47,360 785 00:24:47,360 --> 00:24:48,470 786 00:24:48,470 --> 00:24:49,909 787 00:24:49,909 --> 00:24:52,150 788 00:24:52,150 --> 00:24:52,160 789 00:24:52,160 --> 00:24:54,149 790 00:24:54,149 --> 00:24:56,390 791 00:24:56,390 --> 00:24:58,710 792 00:24:58,710 --> 00:25:01,269 793 00:25:01,269 --> 00:25:05,269 794 00:25:05,269 --> 00:25:06,950 795 00:25:06,950 --> 00:25:08,149 796 00:25:08,149 --> 00:25:09,830 797 00:25:09,830 --> 00:25:11,269 798 00:25:11,269 --> 00:25:13,750 799 00:25:13,750 --> 00:25:13,760 800 00:25:13,760 --> 00:25:15,190 801 00:25:15,190 --> 00:25:16,549 802 00:25:16,549 --> 00:25:18,310 803 00:25:18,310 --> 00:25:19,590 804 00:25:19,590 --> 00:25:21,190 805 00:25:21,190 --> 00:25:22,630 806 00:25:22,630 --> 00:25:24,630 807 00:25:24,630 --> 00:25:24,640 808 00:25:24,640 --> 00:25:25,669 809 00:25:25,669 --> 00:25:28,070 810 00:25:28,070 --> 00:25:30,070 811 00:25:30,070 --> 00:25:32,230 812 00:25:32,230 --> 00:25:32,240 813 00:25:32,240 --> 00:25:34,630 814 00:25:34,630 --> 00:25:35,830 815 00:25:35,830 --> 00:25:37,590 816 00:25:37,590 --> 00:25:39,909 817 00:25:39,909 --> 00:25:41,990 818 00:25:41,990 --> 00:25:44,950 819 00:25:44,950 --> 00:25:47,909 820 00:25:47,909 --> 00:25:47,919 821 00:25:47,919 --> 00:25:49,669 822 00:25:49,669 --> 00:25:51,909 823 00:25:51,909 --> 00:25:55,029 824 00:25:55,029 --> 00:25:57,350 825 00:25:57,350 --> 00:25:59,909 826 00:25:59,909 --> 00:26:02,710 827 00:26:02,710 --> 00:26:05,110 828 00:26:05,110 --> 00:26:07,669 829 00:26:07,669 --> 00:26:09,350 830 00:26:09,350 --> 00:26:11,110 831 00:26:11,110 --> 00:26:13,510 832 00:26:13,510 --> 00:26:17,110 833 00:26:17,110 --> 00:26:19,029 834 00:26:19,029 --> 00:26:21,190 835 00:26:21,190 --> 00:26:23,590 836 00:26:23,590 --> 00:26:25,669 837 00:26:25,669 --> 00:26:27,750 838 00:26:27,750 --> 00:26:30,789 839 00:26:30,789 --> 00:26:32,549 840 00:26:32,549 --> 00:26:34,549 841 00:26:34,549 --> 00:26:37,110 842 00:26:37,110 --> 00:26:39,269 843 00:26:39,269 --> 00:26:41,269 844 00:26:41,269 --> 00:26:42,950 845 00:26:42,950 --> 00:26:42,960 846 00:26:42,960 --> 00:26:43,830 847 00:26:43,830 --> 00:26:45,909 848 00:26:45,909 --> 00:26:47,909 849 00:26:47,909 --> 00:26:50,870 850 00:26:50,870 --> 00:26:52,870 851 00:26:52,870 --> 00:26:55,669 852 00:26:55,669 --> 00:26:57,269 853 00:26:57,269 --> 00:26:58,950 854 00:26:58,950 --> 00:27:00,630 855 00:27:00,630 --> 00:27:00,640 856 00:27:00,640 --> 00:27:01,830 857 00:27:01,830 --> 00:27:03,430 858 00:27:03,430 --> 00:27:05,269 859 00:27:05,269 --> 00:27:08,149 860 00:27:08,149 --> 00:27:10,070 861 00:27:10,070 --> 00:27:12,549 862 00:27:12,549 --> 00:27:14,870 863 00:27:14,870 --> 00:27:17,350 864 00:27:17,350 --> 00:27:19,830 865 00:27:19,830 --> 00:27:21,590 866 00:27:21,590 --> 00:27:23,590 867 00:27:23,590 --> 00:27:25,190 868 00:27:25,190 --> 00:27:28,149 869 00:27:28,149 --> 00:27:30,230 870 00:27:30,230 --> 00:27:32,789 871 00:27:32,789 --> 00:27:34,310 872 00:27:34,310 --> 00:27:35,750 873 00:27:35,750 --> 00:27:37,990 874 00:27:37,990 --> 00:27:39,190 875 00:27:39,190 --> 00:27:39,200 876 00:27:39,200 --> 00:27:40,149 877 00:27:40,149 --> 00:27:42,310 878 00:27:42,310 --> 00:27:43,909 879 00:27:43,909 --> 00:27:46,230 880 00:27:46,230 --> 00:27:49,269 881 00:27:49,269 --> 00:27:51,510 882 00:27:51,510 --> 00:27:53,269 883 00:27:53,269 --> 00:27:55,909 884 00:27:55,909 --> 00:27:57,590 885 00:27:57,590 --> 00:28:00,870 886 00:28:00,870 --> 00:28:03,110 887 00:28:03,110 --> 00:28:04,950 888 00:28:04,950 --> 00:28:06,389 889 00:28:06,389 --> 00:28:06,399 890 00:28:06,399 --> 00:28:07,669 891 00:28:07,669 --> 00:28:09,750 892 00:28:09,750 --> 00:28:11,110 893 00:28:11,110 --> 00:28:11,120 894 00:28:11,120 --> 00:28:12,470 895 00:28:12,470 --> 00:28:15,190 896 00:28:15,190 --> 00:28:17,669 897 00:28:17,669 --> 00:28:21,830 898 00:28:21,830 --> 00:28:23,750 899 00:28:23,750 --> 00:28:26,149 900 00:28:26,149 --> 00:28:28,470 901 00:28:28,470 --> 00:28:30,710 902 00:28:30,710 --> 00:28:33,830 903 00:28:33,830 --> 00:28:35,269 904 00:28:35,269 --> 00:28:36,549 905 00:28:36,549 --> 00:28:39,510 906 00:28:39,510 --> 00:28:41,190 907 00:28:41,190 --> 00:28:43,669 908 00:28:43,669 --> 00:28:45,110 909 00:28:45,110 --> 00:28:46,230 910 00:28:46,230 --> 00:28:48,230 911 00:28:48,230 --> 00:28:51,190 912 00:28:51,190 --> 00:28:53,110 913 00:28:53,110 --> 00:28:54,870 914 00:28:54,870 --> 00:28:54,880 915 00:28:54,880 --> 00:28:55,909 916 00:28:55,909 --> 00:28:57,350 917 00:28:57,350 --> 00:28:59,430 918 00:28:59,430 --> 00:29:01,990 919 00:29:01,990 --> 00:29:03,269 920 00:29:03,269 --> 00:29:05,510 921 00:29:05,510 --> 00:29:08,389 922 00:29:08,389 --> 00:29:08,399 923 00:29:08,399 --> 00:29:09,269 924 00:29:09,269 --> 00:29:11,190 925 00:29:11,190 --> 00:29:13,510 926 00:29:13,510 --> 00:29:14,870 927 00:29:14,870 --> 00:29:16,549 928 00:29:16,549 --> 00:29:19,909 929 00:29:19,909 --> 00:29:21,510 930 00:29:21,510 --> 00:29:23,590 931 00:29:23,590 --> 00:29:25,909 932 00:29:25,909 --> 00:29:27,830 933 00:29:27,830 --> 00:29:29,750 934 00:29:29,750 --> 00:29:32,230 935 00:29:32,230 --> 00:29:34,470 936 00:29:34,470 --> 00:29:35,830 937 00:29:35,830 --> 00:29:38,310 938 00:29:38,310 --> 00:29:41,269 939 00:29:41,269 --> 00:29:43,830 940 00:29:43,830 --> 00:29:44,789 941 00:29:44,789 --> 00:29:46,149 942 00:29:46,149 --> 00:29:47,990 943 00:29:47,990 --> 00:29:50,230 944 00:29:50,230 --> 00:29:53,190 945 00:29:53,190 --> 00:29:56,070 946 00:29:56,070 --> 00:29:58,789 947 00:29:58,789 --> 00:30:00,310 948 00:30:00,310 --> 00:30:04,950 949 00:30:04,950 --> 00:30:07,269 950 00:30:07,269 --> 00:30:07,279 951 00:30:07,279 --> 00:30:08,630 952 00:30:08,630 --> 00:30:11,750 953 00:30:11,750 --> 00:30:13,990 954 00:30:13,990 --> 00:30:16,230 955 00:30:16,230 --> 00:30:18,070 956 00:30:18,070 --> 00:30:20,950 957 00:30:20,950 --> 00:30:23,590 958 00:30:23,590 --> 00:30:25,110 959 00:30:25,110 --> 00:30:26,950 960 00:30:26,950 --> 00:30:28,789 961 00:30:28,789 --> 00:30:31,669 962 00:30:31,669 --> 00:30:33,190 963 00:30:33,190 --> 00:30:35,110 964 00:30:35,110 --> 00:30:36,950 965 00:30:36,950 --> 00:30:39,590 966 00:30:39,590 --> 00:30:41,350 967 00:30:41,350 --> 00:30:42,710 968 00:30:42,710 --> 00:30:42,720 969 00:30:42,720 --> 00:30:43,990 970 00:30:43,990 --> 00:30:45,269 971 00:30:45,269 --> 00:30:47,269 972 00:30:47,269 --> 00:30:48,389 973 00:30:48,389 --> 00:30:50,950 974 00:30:50,950 --> 00:30:52,470 975 00:30:52,470 --> 00:30:54,310 976 00:30:54,310 --> 00:30:59,590 977 00:30:59,590 --> 00:31:01,830 978 00:31:01,830 --> 00:31:03,990 979 00:31:03,990 --> 00:31:06,070 980 00:31:06,070 --> 00:31:07,990 981 00:31:07,990 --> 00:31:09,269 982 00:31:09,269 --> 00:31:11,830 983 00:31:11,830 --> 00:31:14,630 984 00:31:14,630 --> 00:31:16,870 985 00:31:16,870 --> 00:31:20,310 986 00:31:20,310 --> 00:31:22,070 987 00:31:22,070 --> 00:31:24,389 988 00:31:24,389 --> 00:31:27,110 989 00:31:27,110 --> 00:31:28,950 990 00:31:28,950 --> 00:31:31,750 991 00:31:31,750 --> 00:31:34,310 992 00:31:34,310 --> 00:31:36,549 993 00:31:36,549 --> 00:31:38,950 994 00:31:38,950 --> 00:31:41,590 995 00:31:41,590 --> 00:31:43,269 996 00:31:43,269 --> 00:31:45,590 997 00:31:45,590 --> 00:31:47,269 998 00:31:47,269 --> 00:31:49,029 999 00:31:49,029 --> 00:31:50,789 1000 00:31:50,789 --> 00:31:53,110 1001 00:31:53,110 --> 00:31:54,789 1002 00:31:54,789 --> 00:31:57,430 1003 00:31:57,430 --> 00:32:00,789 1004 00:32:00,789 --> 00:32:03,590 1005 00:32:03,590 --> 00:32:05,590 1006 00:32:05,590 --> 00:32:07,269 1007 00:32:07,269 --> 00:32:07,279 1008 00:32:07,279 --> 00:32:08,549 1009 00:32:08,549 --> 00:32:11,909 1010 00:32:11,909 --> 00:32:15,190 1011 00:32:15,190 --> 00:32:17,269 1012 00:32:17,269 --> 00:32:19,590 1013 00:32:19,590 --> 00:32:21,750 1014 00:32:21,750 --> 00:32:24,310 1015 00:32:24,310 --> 00:32:26,149 1016 00:32:26,149 --> 00:32:28,389 1017 00:32:28,389 --> 00:32:30,070 1018 00:32:30,070 --> 00:32:31,350 1019 00:32:31,350 --> 00:32:33,669 1020 00:32:33,669 --> 00:32:37,590 1021 00:32:37,590 --> 00:32:39,350 1022 00:32:39,350 --> 00:32:41,029 1023 00:32:41,029 --> 00:32:41,039 1024 00:32:41,039 --> 00:32:42,710 1025 00:32:42,710 --> 00:32:46,950 1026 00:32:46,950 --> 00:32:48,549 1027 00:32:48,549 --> 00:32:50,950 1028 00:32:50,950 --> 00:32:53,029 1029 00:32:53,029 --> 00:32:55,750 1030 00:32:55,750 --> 00:32:57,590 1031 00:32:57,590 --> 00:32:59,269 1032 00:32:59,269 --> 00:32:59,279 1033 00:32:59,279 --> 00:33:00,789 1034 00:33:00,789 --> 00:33:02,630 1035 00:33:02,630 --> 00:33:04,549 1036 00:33:04,549 --> 00:33:05,909 1037 00:33:05,909 --> 00:33:07,269 1038 00:33:07,269 --> 00:33:08,310 1039 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01:10:23,510 --> 01:10:25,270 2237 01:10:25,270 --> 01:10:29,350 2238 01:10:29,350 --> 01:10:32,070 2239 01:10:32,070 --> 01:10:34,550 2240 01:10:34,550 --> 01:10:36,149 2241 01:10:36,149 --> 01:10:37,750 2242 01:10:37,750 --> 01:10:40,229 2243 01:10:40,229 --> 01:10:42,709 2244 01:10:42,709 --> 01:10:44,709 2245 01:10:44,709 --> 01:10:44,719 2246 01:10:44,719 --> 01:10:45,510 2247 01:10:45,510 --> 01:10:47,350 2248 01:10:47,350 --> 01:10:48,950 2249 01:10:48,950 --> 01:10:50,550 2250 01:10:50,550 --> 01:10:52,470 2251 01:10:52,470 --> 01:10:54,470 2252 01:10:54,470 --> 01:10:56,830 2253 01:10:56,830 --> 01:10:58,790 2254 01:10:58,790 --> 01:11:00,470 2255 01:11:00,470 --> 01:11:02,070 2256 01:11:02,070 --> 01:11:03,990 2257 01:11:03,990 --> 01:11:05,990 2258 01:11:05,990 --> 01:11:07,669 2259 01:11:07,669 --> 01:11:10,709 2260 01:11:10,709 --> 01:11:12,550 2261 01:11:12,550 --> 01:11:15,830 2262 01:11:15,830 --> 01:11:17,590 2263 01:11:17,590 --> 01:11:19,910 2264 01:11:19,910 --> 01:11:20,950 2265 01:11:20,950 --> 01:11:23,990 2266 01:11:23,990 --> 01:11:26,390 2267 01:11:26,390 --> 01:11:28,229 2268 01:11:28,229 --> 01:11:28,239 2269 01:11:28,239 --> 01:11:29,590 2270 01:11:29,590 --> 01:11:32,630 2271 01:11:32,630 --> 01:11:34,870 2272 01:11:34,870 --> 01:11:36,630 2273 01:11:36,630 --> 01:11:38,950 2274 01:11:38,950 --> 01:11:41,110 2275 01:11:41,110 --> 01:11:42,229 2276 01:11:42,229 --> 01:11:46,310 2277 01:11:46,310 --> 01:11:49,270 2278 01:11:49,270 --> 01:11:51,510 2279 01:11:51,510 --> 01:11:54,550 2280 01:11:54,550 --> 01:11:56,149 2281 01:11:56,149 --> 01:11:56,159 2282 01:11:56,159 --> 01:11:58,870 2283 01:11:58,870 --> 01:11:58,880 2284 01:11:58,880 --> 01:12:00,310 2285 01:12:00,310 --> 01:12:02,229 2286 01:12:02,229 --> 01:12:05,590 2287 01:12:05,590 --> 01:12:13,270 2288 01:12:13,270 --> 01:12:13,280 2289 01:12:13,280 --> 01:12:15,189 2290 01:12:15,189 --> 01:12:16,709 2291 01:12:16,709 --> 01:12:18,390 2292 01:12:18,390 --> 01:12:20,830 2293 01:12:20,830 --> 01:12:24,870 2294 01:12:24,870 --> 01:12:24,880 2295 01:12:24,880 --> 01:12:26,630 2296 01:12:26,630 --> 01:12:29,750 2297 01:12:29,750 --> 01:12:32,709 2298 01:12:32,709 --> 01:12:32,719 2299 01:12:32,719 --> 01:12:33,830 2300 01:12:33,830 --> 01:12:36,630 2301 01:12:36,630 --> 01:12:38,790 2302 01:12:38,790 --> 01:12:40,870 2303 01:12:40,870 --> 01:12:42,870 2304 01:12:42,870 --> 01:12:46,229 2305 01:12:46,229 --> 01:12:47,430 2306 01:12:47,430 --> 01:12:48,870 2307 01:12:48,870 --> 01:12:50,950 2308 01:12:50,950 --> 01:12:52,630 2309 01:12:52,630 --> 01:12:54,070 2310 01:12:54,070 --> 01:12:56,070 2311 01:12:56,070 --> 01:12:57,270 2312 01:12:57,270 --> 01:12:59,750 2313 01:12:59,750 --> 01:13:01,830 2314 01:13:01,830 --> 01:13:03,830 2315 01:13:03,830 --> 01:13:06,310 2316 01:13:06,310 --> 01:13:07,510 2317 01:13:07,510 --> 01:13:09,270 2318 01:13:09,270 --> 01:13:11,110 2319 01:13:11,110 --> 01:13:13,590 2320 01:13:13,590 --> 01:13:15,590 2321 01:13:15,590 --> 01:13:15,600 2322 01:13:15,600 --> 01:13:16,870 2323 01:13:16,870 --> 01:13:17,750 2324 01:13:17,750 --> 01:13:19,510 2325 01:13:19,510 --> 01:13:21,510 2326 01:13:21,510 --> 01:13:22,790 2327 01:13:22,790 --> 01:13:24,709 2328 01:13:24,709 --> 01:13:26,709 2329 01:13:26,709 --> 01:13:29,350 2330 01:13:29,350 --> 01:13:32,709 2331 01:13:32,709 --> 01:13:34,310 2332 01:13:34,310 --> 01:13:36,550 2333 01:13:36,550 --> 01:13:38,229 2334 01:13:38,229 --> 01:13:40,390 2335 01:13:40,390 --> 01:13:43,189 2336 01:13:43,189 --> 01:13:47,480