1 00:00:25,199 --> 00:00:26,470 hi folks um why don't we get started um welcome back good to see you you all here so um i am going to first we'll we'll recap what we did last time and what we're going to do today i'll talk a little bit about the problem set and we'll also have a break as requested halfway through so why don't we jump in um what we did last time um was we introduced finite besides introduce introducing the course we introduced finite automata and regular languages which are the languages that the finite automata can recognize we talked about these regular operations um and the and then building um those those allow us to build what we call our regular expressions these are ways of describing languages so find we have finite automata can describe languages and regular expressions can describe languages and one of our goals is to show that those two systems are equivalent to one another even though they look rather different um at first glance so um to move in that direction we first proved uh we're going to prove closure properties for all of the for the class of regular languages over these regular operations so we'll show that well we already showed that any two regular languages have their union also being regular and we'll show that for the other two operations as well so let's just look ahead to what we're going to do today we're going to introduce an important concept which is going to be a theme throughout the course called non-determinism and we uh having that as a tool that we can use will be able to show closure under concatenation and star finishing up what we started to do last time and then we'll use those closure constructions to show how to convert regular expressions to finite finite automata um and that's going to be halfway to our goal of showing that the two systems are equivalent to one one another and the following lecture we will show how to do the conversion in the other direction so i thought we would just jump in then and look at um return to the material of the course um as you remember we were looking at the closure properties for the class of regular languages um we started doing that um and if you recall hopefully uh we did closure under union and then we tried to do closure under concatenation um which i have shown here um on the slide um you know the what the proof attempt that we tried to do last time and let's just review that quickly because i think that's going to be uh helpful to see how to fix the problem that came up um so if you remember um we had a uh you know we're given two regular languages a1 and a2 and we're trying to show that the concatenation language a1 a2 is also regular and so the way we you know the way we go about all of these things is you know we assume that a1 and a2 are regular so that means we have machines finite automata for a1 and a2 we'll call them m1 and m2 that recognize a1 and a2 respectively and then what we need to do in order to show that concatenation is regular is to make a finite automaton which recognizes the concat nation and we tried to do that last time um so uh if you remember that concatenation machine m we're calling it uh what is it supposed to do it's supposed to accept its input if it's in the concatenation language and that means that the input can be split into two parts x and y where x is in the a language and y is in the b line and in the end y is accepted by m1 and x is accepted by m1 and y's accepted by m2 sorry garbled that uh so x should be in a1 and y should be in a2 um if you can split w that way then m should accept it so m has to make figure out if there's some way to split the input so that the first machine accepts the first part the second machine accepts the second part and um the idea that we came up with for doing that was to um take these two machines build them in to a a new machine m and then connect the accepting states for m to the start state connect the the accepting states for m1 to the start state for m2 um because the idea would be that if m1 has accepted an initial part you want then you want to pass control to m2 to accept the rest but as we observed that doesn't quite work because um the first place to split w after you found an initial part that's accepted by m1 may not be the right place because the remainder may not be accepted by m2 you might have been better off waiting until you found another place that m1 accepted later on in the stream say over here and then uh by splitting it over there then maybe you do get successfully find that the remainder is accepted by m2 whereas if you tried to split it you know the first place the remainder wouldn't have been accepted by m2 so all you need to do you know m has to know is there someplace to split the input so that you know you can get both parts accepted by the respective machines the problem is is that m might need to know the future in order to know where to make the split and it doesn't have access to the future so what do we do um so what we're going to do is introduce a new concept that will allow us to uh basically get the effect of m1 in the sort of being able to see the future um and that new concept is going to be very important for us throughout the term it's called non-determinism and so we're going to introduce a new kind of finite automaton called a non-deterministic finite automaton and first we'll look at that and then we'll see how that fits in with the previous deterministic finite automaton that we introduced last time so here's an example we're always good to start off with an example um here is an a picture of a non-deterministic fine art automaton looks very similar at first glance to a to the former kind the deterministic finite automaton but if you look a little carefully you see that there are some key differences um the most important difference is that um [Music] the in state q1 for example whereas in the machines that we introduced last time there had to be exactly one way to go on each possible input symbol so you knew how to you know how to follow along through the machine as its computing here there are two ways to go in q1 you can either stay in q1 or you can go to q2 that's the essence of non-determinism there could be many ways to proceed um and furthermore on q1 if you get a b then there's nowhere to go so um that's uh also possible within non-determinism there could be um so let's let's just start looking at these features there are multiple paths forward but multiple paths possible you might be able to have uh one as we had before or many ways to go at each step or maybe zero ways to go at each step those are all legitimate for a non-deterministic machine which doing a non-deterministic computation another difference if you look carefully is that we're allowing here the empty string to appear on a transition that's perhaps a little less uh essential to the spirit of non-determinism but it's going to turn out to be a convenience when we actually apply non-determinism to build machines as you'll see very shortly now if there are many different ways to go and some of those ways to go might have different outcomes you know as we remember from before we accepted the input if you end up in an accept state and we rejected the input if you end up not in an except state in a non-accepting state then you reject but now there might be several different ways to go um so and we'll do an example in a minute but there might be several different ways to go and they might disagree some of them might accept other ones might reject so then what do you do well in that case um acceptance always overrules rejection that's the essence of non-determinism the way we're def the way we're setting it up um you may ask why that is and you know if um sort of the spirit of that will become clear in due course but right now just take it as a rule when we're having a non-deterministic machine acceptance overrules rejection so as long as there is one accepting copy one one of the possible ways to go ends up at an accept we say the whole thing is accepted um the only way could possibly reject if all of the possible ways to go end up at rejection end up at a non-accept state so we'll see example we'll i think we're going to do an example right now yeah so we take for example um this machine n1 now on an input a b and we're going to process the symbols one by one just like we did before but now to follow along there might be several different ways to go so if we take um the uh the first symbol a and we run the machine so the machine is starts at the start state um as before but now an a comes in and there might be two now there are two different ways to go so we're going to keep track of both of them after the machine reads an a you can think of it as think of it as being in two states now simultaneously it can be in state q one and it can be in state q two so those are two different possible places it could be at this moment okay now um now we read the next symbol the b and from a b you take all of each of the places where the machine could be um at the end of the previous symbol and you then apply reading a b the next symbol to each of those uh from each of those states where the machine could be in uh from the previous symbol so the machine could be in q1 and q2 after reading an a now we apply a b well q one on a b goes nowhere so you think of that uh a branch if you will of the computation as just dying off it has nowhere to go it just vanishes however when you're in the the other possibility which was state q2 on a b does allow does have a place to go so that the machine is now going to go from q2 to q3 um uh when it re on that branch of the computation which reading on reading a b and then it has um [Music] uh coming out of q3 there are two symbols there's an a and an empty string symbol now on an a the machine would have to read an a in order to transition along that arrow but on when there's an empty symbol on the arrow that means the machine can go along that arrow for free without even reading anything as long as it gets to q3 it can automatically jump to q4 and so once it has read a b and gone to q3 now um it it can either stay in q3 or it can re it can go along the empty transition and go to q4 so again it is going to be a non-deterministic step at this point um [Music] the essence of having a empty transition is that there's going to be non-determinism that's why we didn't introduce that for deterministic automata because um you don't have to transition along an empty string transition you can stay where you are or you can go along the empty string transition without reading any input and go over to the next state which in this case is q4 so let's just see where we are after reading an a we're in states q1 q2 but now after reading a b we're in states q3 and q4 as possibilities and then we and now we're at the end of the input and we look and see what we got if any one of the states as possibilities that we are right now at the end of the string is an accept state then we say overall the machine has accepted so that corresponds to what we said over here before accept the input if some path leads to an accept so if any way of proceeding um through these you know non-deterministic choices will lead you to an accept then you will say we're going to accept the input okay so this input here is accepted let's do another example suppose we have the input a a instead of a b so a a you know after the first a as before we're in states q one and q two as possibilities now we read in a again now the one that's on state q one that possibility q one possibility after reading an a it again branches to q1 and q2 um so when we know after reading the second a we're going to be in at least q1 and q2 now how about the the state that had been you know on q2 on reading in a the one from before after reading the first day you were in q2 now reading the second a there's nowhere to go so that one just gets removed so after reading a a we remain in states q1 and q2 as possibilities neither of those are accept states so therefore on input a a the machine rejects okay let's just do a couple more and then i'll ask you to do one so we have aba as an input let's see what happens then so um we have a uh remember after reading a b a b the machine is in the two states q three q four as possibilities that's what we had from from um the first example so after reading a b we're in states two three q four now we read another a the confu the q4 on an a has nowhere to go in fact it has nowhere to go on any input so no matter what comes in after you're in state q4 that branch dies but on q3 which is another one of the possibilities reading in a um it can follow along just transition because that's one of the labels on that transition is a so you can follow along just a transition reading in a which is the last uh symbol in the string and so now after aba you are in only state q4 as a possibility but that happens to be an accept state so the machine accepts um okay and now lastly what happens let's take our final examples uh what happens if we have a b b so you know as you remember before when we um were after reading a b that was the first example we were in states q3 q4 as possibilities now we read a b well neither of those states have anywhere to go on a b so now all threads all branches of the computation die off and at this point the machine is totally dead it has no no active possibilities left so certainly it's going to reject this input because none of the active states there are no active states are accepting states and in fact if you looked at anything that came later they would anything that extended the string a b b would also be rejected because once the machine had all all possibilities have died off there's no way for them to uh come back to life on any extensions so with that um oh here's an important point before i i'm going to jump to a check-in on you uh but um uh i think what one thing that you know might be on your mind about this non-determinism is how does how do we actually what is that how does that correspond to reality well it doesn't um we're not intending for non-determinism as we're defining it to correspond to a physical device but nevertheless as you'll see it's a very mathematically useful concept uh this non-determinism and it's going to be playing a big role throughout the subject as will uh as will experience it during the rest of the term so with that uh i'm going to have a little check in it's going to give i'm going to ask you to uh consider what happens on one of the inputs um so here we go what does it do on input a a b so here's the machine you can look at it and i'm gonna i'm supposed to hopefully there's a poll here for me to give to you um so you can uh give me your input so what does the machine do on input a a b most of you have answered again you're not going to be penalized for getting the wrong answer um but uh hopefully you you'll get the right answer anyway let's just take a look here uh so time is up um let's end the polling and share the results so the majority of you majority have gotten the correct answer which is a the machine does accept a a b uh because you know when you have a so if you if you like i'll show you the path that corresponds to accepting you go a a b and then empty string and so that uh uh sequence of steps is one of the non-deterministic possibilities that the machine can follow and that shows that the machine does accept the input a a b you know you could think about it the way we did it before um also you know if you read an a it's in the two possibilities q one q two you read a second a again in the possibilities q one q two now you read a b it's into possibilities q three q four and um that's it a a b so now you read the b it's in possibility q three q four q four is an accepting state that overrules the non-accepting state and so the machine accepts you have to understand this so if you didn't get it right go back and think about where you slipped up okay because this is um you know the you know this is just getting we're just getting warmed up here it's going to get it's going to get a lot harder okay so stop sharing the results um and so let's continue so now we just as we did last time we can formally uh define a fine non-deterministic finite automata here's the picture again um okay uh so it looks a lot like the the the case we had before the deterministic finite automata or dfa as we'll call them um it's a five tuple so i've sort of uh written down little reminders for what those components of that five tuple are that list of five uh five components um so they're all the same as before states alphabet transition function start state and accepting states so that the formal definition definition looks exactly the same except the structure of the transition function so now before if you remember you had a state and an input symbol and you got back a state uh now we have something more complicated looking we have a state and an input symbol but it you know it's instead of just sigma it's sigma sub epsilon and that that's a shorthand for sigma union epsilon and that's a way my way of saying that you're allowed to have on your transition um arrows either an input symbol or an empty string so the transition function has to tell you what to do when you have empty string coming in as well um so that would be part of your table for the transition function now over here what's going on over here well now instead of just producing a single state you know when you've read for example an a on from q1 there's a whole set of possibilities so here we have what's called the power set that's the set of subsets of the collection queue so here we're going to produce an entire subset of states instead of just one state coming out there might be a subset of possible states that you can go to so the power set of q is the set of subsets of um of q so that's what this notation means um again this is something that i'm hopefully presenting to you as a bit of a reminder you've seen this somewhere else before but um please make sure you understand the notation uh going forward because we'll we'll be doing less hand-holding as we start moving forward um okay so just let's take a look at the n1 example here um just to illustrate the illustrate what's going on when you're in state q1 reading an a now you get a whole set of possibilities which in this case is q1 and q2 um whereas if you're reading a b what would be that set coming out of q1 what's the set of possible successor states well there are none so it's the empty set okay so this is hopefully your understanding no notation here so now here's i think really important how do we understand non-determinism intuitively speaking and there are multiple different ways which each has their value under different circumstances so one way is thinking about non-determinism as a kind of parallelism so every time the machine has a non-deterministic choice to make where there is more than one outcome you think of the machine as um [Music] branching forking new threads of the parallel computation at that stage where it makes entire copy of itself when there's a choice of possibilities and then each of those independently proceeds to read the rest of the input as separate threads of the computation so if you're familiar with parallel computing this should be reasonably familiar to you the only key thing to remember is that as this thing forks number possibilities the acceptance rule is that if any one of those possibilities gets to an accept at the end of the input it raises a flag and says accept and that overrules everybody else so acceptance dominates so another way of looking at it is the mathematical view where you can imagine and we're going to use all these so it's you really need to understand them all the mathematical view is if you can think of the computation as kind of a tree of possibilities so you start off at the very beginning sort of at the root of the computation which is that when it really begins but every time there's a non-deterministic branching that occurs um the tree has [Music] that note of the tree has uh multiple children coming out of that node and so the different threads of the computation correspond to different branches of that tree and now you're going to accept if any one of those branches leads to an accepting state okay very obviously somewhat similar to what we had before but it's i think it's a little bit of a different perspective on what not how to think about non-determinism and the last one is going to sound a little weird but actually i think for people who are in the business it's the one they use the most and that's the magical way of thinking about non-determinism and that is um when the machine has non-deterministic choices to make you think of the machine as magically guessing the correct one at every stage and the correct one being the one that will eventually lead to accept okay so you can think of the machine as guessing which is the right way to go and if there is some way right way to go it always guesses right of course if the machine ends up rejecting because there is no right way to go then it doesn't matter there is no good guess but if there is some good guess we'll think of the machine as taking that good guess and going that way okay so um now here is very important thing we introduced this new model uh the uh the non-deterministic foundation nfa it turns out to be even though it looks more powerful because has this non-determinism it isn't any more powerful it can do exactly the same class of languages the regular languages and we'll show that um with the theorem here that if an nfi and if an nfa recognizes a then a is regular um [Music] so um we'll prove that by showing how to convert an nfa to an equivalent dfa which does the same language so we can take an nfa that has a non-determinism and find another dfa which doesn't have non-determinism but does the same language it accepts exactly the same strengths even though it lacks that non-deterministic capability it's this is going to be extremely useful by the way and for example in showing that closure under concatenation okay so uh you know in this presentation here i'm going to ignore the epsilon transitions because once you get the idea for how to do this you could figure out how to incorporate them they just make things a little more complicated so let's just focus on the key aspect of non-determinism which is that the machine could have several ways to go at any point in time there could be several next states on an input okay now the idea for the construction so we're going to take we're going to start with a non-deterministic machine m and we're going to build a deterministic machine m prime which does exactly the same thing and the way m prime works is it's going to do what you would do if you were simulating m what would you do with me this is what we were doing as we were as i was explaining it to to you if you are simulating m every time you get an input symbol you just keep track of what is the set of possible states at that point in time that's what the dfa is going to do it's going to have uh it's going to have to keep track of which possible set of states the nfa could be in at the point on that input where we are right now and then as you get the next symbol the dfa is going to have to update things to keep track of the next set of states the nfa could be in at this point just like you would do okay and so um so here's a kind of a picture um and how do we implement that in in so here's the nfa that we're starting with m and we're going to make here the dfa but um in order to remember which set of states that dfa could be in at a given point so maybe it's the set of states that it that m could be in did i say that wrong which set of states the nfa could be at a given time so maybe m the nfa could be in at some point state q3 and q7 the way the dfa keeps track of that it's going to have a state for every possible subset of states of the nfa that's how it re remembers which subset of states the nfa is in that's the way dfas work they have a separate state for each possibility that they need to keep track of and the possibilities here are the different subsets of states that the nfa could be in at a given point okay so corresponding to this subset to these two possibilities q three q seven the dfa is going to have the sub a state with the subset q3 q7 and it's going to for every possible subset here there's going to be a different state of m prime so n prime is going to be bigger um okay so quickly the construction of m the states of m prime now um q prime are going to be the power set the set of subsets of states uh from the original machine m um and uh the uh now we have to look at how the transition function of the dfa when you make the primed machines are the dfas as a dfa machine so these are the prime the the deterministic component so delta prime um when it when it has a subset that you know something like this um has a has one of its states which corresponds to a subset of states of m and it reads an input symbol you just have to do the updating the way you would naturally do you're going to look at every state in r look at where that can go under a so there's a bunch of sets there and look at all the possible states that could be in one of those subsets and that's the set of states that you you could be in that's going to be the new set of states and that's going to be the new state of m prime okay so it's going to be the subset corresponding to all of the states that could be in uh when you apply the transition function of the non-deterministic machine um to one of the states in that in the subset of states that the not a terminus machine could be in okay it's a little bit of a mouthful you know you you could i suggest you look at this if you didn't quite get it after the fact good to understand the starting state for the nfa for the dfa i'm sorry is going to be which subset now we're going to start off with it's going to be the subset corresponding to just the start state of um uh of of m um and the accepting states are going to be uh of the deterministic machine are going to be all of the subsets that have at least one accepting state from the nfa so i hope you got that because i'm going to be another little check in here um which is i'm going to ask you how big is m prime how many states does n prime have uh i told you what those states are so just go think about that um so check in two if m has n states how many states does m prime have by this construction okay so let's launch the next poll um okay five seconds and i think we're almost done here good all right share results i know if sharing results is a good thing i'm not trying to make you if you didn't get the right answer um because most of the people did get the right answer we didn't get the right answer trying to make you feel bad but i do think you should it's a little bit of suggestion you know that you you need to review some basic concepts so the basic concept here is if you have a collection you have a set of states how many subsets are there and the number of subsets is going to be exponential so if you have a collection of n elements the number of subsets of those n elements is 2 to the n um that's the fact we're using here and that's why m prime has two to the n states if m had n states um and that's you know you should make sure you understand why that is all right so um with that um as requested we're gonna have a little break um and that break is going to last us exactly five minutes so we'll we will return in five minutes i'm going to be prompt so i give you a little timer here so please um uh i'm gonna begin it write it right right when this is over okay almost almost ready i hope you're all refreshed and ready for the second half so now now that we have non-determinism we're going to use uh that as a tool to prove the uh closure properties that we were aiming for for um starting from last lecture okay um so remember we let's look at closure under a union now we already did that but i'm going to do it again but this time using non-determinism and you'll see how powerful non-determinate non-determinism now determinism is because it's going to allow us to do it kind of almost with no effort we'll start off the way we did before i'm going to start off with two dfas but actually these could be um nfas even but let's say we started with the two dfas for the two languages a1 and a2 and we're going to construct the nfa rec now we're going to construct an nfa recognizing the union um and that's good enough because we already know that we can convert nfas to dfas and therefore they do they do regulate languages too um okay so now here are the two dfas that do the languages a1 and a2 and what i'm going to do is i'm going to put them together into a bag of states um which is going to be m the nfa that's going to do the union language so remember what does what does m supposed to do m is supposed to accept its input if either m1 or m2 accept so how is it going to do that what it's going to do we're going to add a new state to m which is going to branch under epsilon transitions and now you can start to see how useful these epsilon transitions are going to be for us going to branch under epsilon transitions to the two original start states of m1 and m2 and we're done y well now non-deterministically as we get an input w coming in to m um m at the very beginning even just right after it gets going the very first thing that happens is it's going to branch to m1 and also branch to m2 non-deterministically as two possibilities and then inside m1 and m2 it's going to actually start reading the input and each one is going to be now following along as it would have originally uh the states corresponding to um reading those input symbols um and m as a combination of m1 and m2 is going to have a possibility for e for one state in m1 and once one stated m2 um and so m is going to have those combined into one in into one um package and now if either at the end of the input if either um of these end up at an accepting state then m [Music] um is going to accept as a non-deterministic fighter automaton because that's how non-determinism works you accept if either if any one of the branches ended up accepting which is just what you need for non-determinate for union so we're doing you and we want either one of these to be accepting and then the non-determinism just is built conveniently to allow us to do the union almost you know for free um so you can again thinking about non-determinism as terms of parallelism you could think of it the non-deterministic machine is running in parallel m1 and m2 on the input and if either one of them ends up accepting m will accept or you can think about it in terms of that guessing that i referred to before which means that as m is getting when it's just about to read these the first symbols of its input it guesses whether um that's going to be an input accepted by m1 or an input accepted by m2 and the magic of non-determinism is that it always guesses right so that input happens to be some an input that's going to be wrecked you're going to be accepted by m2 m is going to guess that m2 is the right way to follow and it's going to go in the m2 direction um because non-determinism the magic is you always guess right wish that was true in real life uh would make exams a lot easier um anyway so now let's see how we can use that to do closure under concatenation okay so now um we're gonna actually have a picture very similar to the one we had originally but now using non-determinism we can make it work so here we have the two machines doing the two languages a1 and a2 and we're going to combine them into one bigger machine m as shown remember what m is supposed to do is accept its input if there's some way of splitting that input such that the first half is accepted by the m1 and the second part is accepted by m2 way we're going to get that effect is by um putting in a uh transit empty empty transitions epsilon transitions going from the starts for the except states of m1 to the start state of m2 just as i've shown here in this diagram so these were the original accepting states of m1 and now um they're going to be declassified as accepting states but they're going to have new transitions empty transitions um attached to them which allow them to branch to m2 without reading any input and so intuitively speaking this is going to do the right thing because once m1 has accepted some part of w then you can non-deterministically branch to m2 and um and you're going to be start processing inside m2 and the the point is as i kind of uh jumped ahead of myself um is that the reason why it fixes the problem we had before is that the epsilon transitions are don't machine does not have to take them it can stay where it is as one non-deterministic option or it can move along the epsilon transition without reading any input as another non-deterministic option so it's using this non-determinism now to both stay in m1 to continue reading more of the input and to jump into m2 to start processing what might be the second half or the second part of the input which m2 accepts and you can think of it in terms of the guessing is that the machine is guessing where to make that split once it found an initial part that's accepted by m1 it guesses that this is the right split point and that passes to m2 but there might be other guesses that it could make you know corresponding to other possibilities and so with non-determinism it always guesses right if there is some way to split the string into two parts uh accepted by m1 and m2 the machine will make that good guess and then m1 will accept the first part and m2 will accept us with the second part and we'll get m accepting that whole uh string uh altogether and so that is the uh solution um to our uh puzzle for how do we do closure under concatenation okay i hope that came through because this is um this is you know we're just getting going with non-determinism we're going to be using non-determinism a lot and you it's going to you're going to need to get very comfortable with it okay um now let's do closure under star um and closure under star works uh very similarly but now we're just gonna have a single language if a is regular so is a star so they're not a pair of languages because a star is a urinary operation applying to just a single uh language so if we have a dfa recognizing a in order to show that a star is regular we have to construct a machine that recognizes a star and a machine we're going to construct is as before and then an nfa okay so um here is m uh the dfa for a and we're going to build an nfa m prime that recognizes a star and let's think now what does it mean to recognize a star so when should m you know if i'm going to give you an input when is it in the star language what does m prime have to do um so m remember what star is star means you can take as many copies of you lot as you like of strings in the original language and that's in the star language so to determine if something's in the star language you have to see can i break it up into pieces which are all in the original language um [Music] so uh um so you want to see can i take my input w and cut it up into a bunch of pieces four in this case um where each of those pieces are members of a they're members of the original language so that's what that's what m prime's job is it's getting it has its input and wants to know can i cut that input up into pieces each of which are accepted by the original machine m that's what m prime does and if you think about it you know um a little bit really what's happening is that as soon as m so you know m prime is going to be simulating m that's why i like to think about this as having m inside so you know you're going to if you were going to be doing this yourself you're going to take w you're going to run it for a while you'll see oh m is accepted now i have to start m over again to see if it accepts uh on the next segment so every time m accepts you're going to restart m to see if it accepts another segment and so you by doing that you're going to be cutting w up into different segments which are each of which is accepted by m of course it's never totally clear whether you should and for any given segment you should cut it there or you should wait a little longer and find another play a later place to cut but that's exactly the same problem that we had before with concatenation and we saw that using non-determinism and we're going to solve it again using non-determinism so the way we're going to get that effect of starting the machine over again once once it's accepted is by adding in epsilon transitions that go from the start states back to the from the accept state back to the start state so now every time m has accepted it has an option not not a requirement but it has an option it can either stay continuing to process or it can restart making a cut at that point and trying to see if there's a yet a different a second another segment of the input that's going to be ex that it's going to accept and this is basically the whole thing with one little um problem that we need to deal with and that is we need to make sure um that m prime accepts the empty string because remember the x empty string is always a member of the starred language and as it's as it's written right now we're going to be requiring there to be at least one copy of one at least one segment we're not taking into account the possibility of no segments which is the empty string and the way we're going to get that is well i mean one thing one way to get to it so you know we're missing the empty string right now so how do we fix it basically you know we're just going to take the construction we have get uh you know on on the screen and we're going to uh adjust it to to add in the empty string because it's possibly missing one way to do that which is tempting but wrong is to make the start state of m an accepting state for m prime so we could have and we could have made this an uh an accepting state too and now m prime is also going to accept the empty string that's the good news the problem is that the start state might be playing some other role in m besides just being the start there might be times when m comes back to the start state you know later on and if we make the start state the accept and accept state it's going to suddenly start accepting a bunch of other things too which might not be intended so it's a bad idea to make the start state and accept state instead we'll take the simples uh solution alternative of adding a new start state which will never be returned to under any circumstances and make that a new start and accept state as well so here we'll have to make this additional modification so as i'm saying this is what we need to do and we will do that is by adding a new start state which is also an accept state to make sure it accepts the empty string and then that also can branch um to start off m as before um if the string that's input is not the empty string and so then m you know we m prime is actually going to have to do some work to see if it can be cut up as it was going as it was doing before um so that's the proof of closure under star i'm not going to do it anything beyond what i've just described these proofs by picture are convincing enough i hope um and if not you know you should they are explained in somewhat more detail somewhat more formally in the textbook but uh for the lecture this is where i'm going to stop with these these two arguments um and so now uh oh yeah one uh quick check in on this so if m has n states how many states does m prime have by this construction um so let's do really these are i'm not intending these to be very hard more just to keep you awake uh so how many states does m prime have okay maybe a little too easy even for a check-in yeah everybody's getting this one um because all you did was we added one new state um so the answer is as you have um i think should pretty much everybody is uh observing that it's just the um it's number b um so i'm gonna end the polling i'm gonna share the results and everybody got that one and so let's continue on um and uh so um the very last thing we're going to do today and show you how to convert regular expressions to nfas thereby showing that every language that you can describe with a regular expression is a regular language on tuesday we'll show how to do the conversion in the other direction and show thereby showing that these two um methods of describing languages are equivalent to one another so here is our theorem if r is or is a regular expression and a is the language is instead of strings that like that regular expression describes then a is regular okay um so um we're going to show how to convert [Music] the the strategy is to convert r to an equivalent nfam and so we have to think about remember these regular expressions uh that we introduced last time these are these expressions that look like um you know a b union b star something like that um so built up out of the regular operations from um from the from the primitive regular expressions that don't have any operations that we're calling atomic so if r is an atomic regular expression it just looks like either just a single symbol or an empty string symbol or a an empty language symbol [Music] or r can be a composite regular expression um whoops we're having a little yeah so so we have two possibilities here um r is either atomic or composite and so let's look at what the equivalent expression is um in each case so if r is just the single letter regular expression that's a totally legitimate regular expression just a regular expression one so that just describes the language of the string one so we have to make a an nfa which accepts which recognizes just that language accepts only the string one so it's a very simple nfa it just starts uh in the start state and on that single symbol it branches to an accept state and there were no other transitions allowed so if you get anything else coming in besides that one that that string which is just that one symbol the nfa is going to reject it like if it goes if it's too long if if you know if it gets a a coming in well there's nowhere to go on from this um accepting state on an a so the machine is just going to die it has to be in an accept state at the end of the input um how about you know why don't you think for yourself for a minute how do we make an nfa which uh accepts only the empty string and no other strings you can do that with just one state with an nfa just this one here um the machine is going to start off in the start state which is also immediately an accept state so it accepts the empty string but if anything else comes in there's nowhere to go when the machine does so this machine accepts just the empty string or its language is the language with one element the empty string how about the empty language well here's a here's an nfa which has no accepting state so it can't be accepting anything now if we have a composite regular expression we're already finished because we showed how to build up we showed constructions which uh give us closure under union uh concatenation and star and those constructions are going to enable us to build up uh the nfas that do the language of these more complex regular expressions built up out of the anaphase that do the individual parts so if we already have anaphase that do r1 and r2 then the closure under union construction gives us an nfa that does r1 union r2 as a language so i'm going to do i hope that's clear but i'm going to do an example which will help hopefully illustrate it and it's going to show you you know basically what i'm giving you is an automatic procedure for converting a regular expression into an equivalent nfa so let's just see if that that procedure in opera in action which is really just following this recipe that i described for you so here is a regular expression a union a b star so this is a regular expression it's some language whatever it is i don't care but i want to make an nfa which recognizes that same language and the way i'm going to do that is first build nfa for the [Music] components um the sub-expressions of this regular expression and then combine them using our closure constructions to be um to be anaphase for larger and larger sub-expressions until i get the n of a that's the equivalent of the entire expression so let's just see how that goes so the the very most primitive parts the the the smallest sub-expressions here are just the expressions for a and for b so here's the one for a just just for a so this is the nfa which um recognizes the language which is just the one string a here's the nfa for the just whose language is just the one string b um and now i want an nfa which is which accepts only the string a b now of course you could just do that by hand um yourself it's simple enough but what i'm arguing is that you we can do this automatically using the closure construction for concatenation because really there's an there's a hidden concatenate concatenation symbol this is the the you know a concatenate b so now for a b i'm going to take um the the thing from a and the part from b so these these two things that i had from before and use the concatenation construction to combine them you see that so now i have automatically a an nfa which does the language whose string is just a b th just the a b string and it's not the simplest nfa you can make a simpler one but the virtue of this one is that i got it automatically just by following the closure construction so now i'm going to do a more complex one just the inside here a union a b um so the way i'm going to build that is from the two parts the a part and the a b part you know the a part and the a b part so here is the a part here's the a b part a b part i've already got those from before it's really kind of a proof by induction here but you know i think it's simple enough we don't have to use that language um so we have um you know the a part the a b part and now we are going to apply the uh the closure under union construction to combine those into one machine and remember how that worked we add a new symbol here which branches under empty string to the previous uh i'm adding a new start state which branches to the original start states under empty empty transition and now this is a an nfa for this language a union a b and lastly now we're one step away from getting uh the star of this and how we're going to do that we're going to take this thing here and apply the construction for the star closure and that's going to be an nfa which does a union a b star which is what we wanted in the first place so first we're going to bring that one down because we've already built that one and now remember how we built um the uh closure under star we made this the accepting states return back to the um the start state and we added a new start state to make sure we got the empty string in there that transitioned to the original star state under epsilon okay so that's all um i wanted to say for today's lecture let's do a quick review um we very important concept non-determinism and and non-deterministic finite automata we proved their equivalent in power uh showed the class of regular languages closed under concatenation and star we showed how to do conversion of regular expressions to nfas so um i think that is it for today's lecture and um thank you all for being here let me i'll try to answer a few of these why does concatenation have order well because it's an ordered construction um is there a simple way to prove closure under concatenation without using non-determinism no um why are the empty strings at the accept state can't they be at any state doesn't star make copies of any part of the input no it's only you have to you have to think about what's going on having you you don't want to you have to branch back to the beginning only on an accept because that means you found a piece that's in the original language um is there an automaton that can add some some or subtract memory automata well depends on what you mean by all that but um certainly there are more powerful machines that we're going to study than find out automata um so um but yes there is um and even finite automata can add and subtract if you present the input in the right way i re refer you to the first problem on the homework um so i think i'm gonna uh check out then um take care everybody bye you 2 00:00:26,470 --> 00:00:28,470 hi folks um why don't we get started um welcome back good to see you you all here so um i am going to first we'll we'll recap what we did last time and what we're going to do today i'll talk a little bit about the problem set and we'll also have a break as requested halfway through so why don't we jump in um what we did last time um was we introduced finite besides introduce introducing the course we introduced finite automata and regular languages which are the languages that the finite automata can recognize we talked about these regular operations um and the and then building um those those allow us to build what we call our regular expressions these are ways of describing languages so find we have finite automata can describe languages and regular expressions can describe languages and one of our goals is to show that those two systems are equivalent to one another even though they look rather different um at first glance so um to move in that direction we first proved uh we're going to prove closure properties for all of the for the class of regular languages over these regular operations so we'll show that well we already showed that any two regular languages have their union also being regular and we'll show that for the other two operations as well so let's just look ahead to what we're going to do today we're going to introduce an important concept which is going to be a theme throughout the course called non-determinism and we uh having that as a tool that we can use will be able to show closure under concatenation and star finishing up what we started to do last time and then we'll use those closure constructions to show how to convert regular expressions to finite finite automata um and that's going to be halfway to our goal of showing that the two systems are equivalent to one one another and the following lecture we will show how to do the conversion in the other direction so i thought we would just jump in then and look at um return to the material of the course um as you remember we were looking at the closure properties for the class of regular languages um we started doing that um and if you recall hopefully uh we did closure under union and then we tried to do closure under concatenation um which i have shown here um on the slide um you know the what the proof attempt that we tried to do last time and let's just review that quickly because i think that's going to be uh helpful to see how to fix the problem that came up um so if you remember um we had a uh you know we're given two regular languages a1 and a2 and we're trying to show that the concatenation language a1 a2 is also regular and so the way we you know the way we go about all of these things is you know we assume that a1 and a2 are regular so that means we have machines finite automata for a1 and a2 we'll call them m1 and m2 that recognize a1 and a2 respectively and then what we need to do in order to show that concatenation is regular is to make a finite automaton which recognizes the concat nation and we tried to do that last time um so uh if you remember that concatenation machine m we're calling it uh what is it supposed to do it's supposed to accept its input if it's in the concatenation language and that means that the input can be split into two parts x and y where x is in the a language and y is in the b line and in the end y is accepted by m1 and x is accepted by m1 and y's accepted by m2 sorry garbled that uh so x should be in a1 and y should be in a2 um if you can split w that way then m should accept it so m has to make figure out if there's some way to split the input so that the first machine accepts the first part the second machine accepts the second part and um the idea that we came up with for doing that was to um take these two machines build them in to a a new machine m and then connect the accepting states for m to the start state connect the the accepting states for m1 to the start state for m2 um because the idea would be that if m1 has accepted an initial part you want then you want to pass control to m2 to accept the rest but as we observed that doesn't quite work because um the first place to split w after you found an initial part that's accepted by m1 may not be the right place because the remainder may not be accepted by m2 you might have been better off waiting until you found another place that m1 accepted later on in the stream say over here and then uh by splitting it over there then maybe you do get successfully find that the remainder is accepted by m2 whereas if you tried to split it you know the first place the remainder wouldn't have been accepted by m2 so all you need to do you know m has to know is there someplace to split the input so that you know you can get both parts accepted by the respective machines the problem is is that m might need to know the future in order to know where to make the split and it doesn't have access to the future so what do we do um so what we're going to do is introduce a new concept that will allow us to uh basically get the effect of m1 in the sort of being able to see the future um and that new concept is going to be very important for us throughout the term it's called non-determinism and so we're going to introduce a new kind of finite automaton called a non-deterministic finite automaton and first we'll look at that and then we'll see how that fits in with the previous deterministic finite automaton that we introduced last time so here's an example we're always good to start off with an example um here is an a picture of a non-deterministic fine art automaton looks very similar at first glance to a to the former kind the deterministic finite automaton but if you look a little carefully you see that there are some key differences um the most important difference is that um [Music] the in state q1 for example whereas in the machines that we introduced last time there had to be exactly one way to go on each possible input symbol so you knew how to you know how to follow along through the machine as its computing here there are two ways to go in q1 you can either stay in q1 or you can go to q2 that's the essence of non-determinism there could be many ways to proceed um and furthermore on q1 if you get a b then there's nowhere to go so um that's uh also possible within non-determinism there could be um so let's let's just start looking at these features there are multiple paths forward but multiple paths possible you might be able to have uh one as we had before or many ways to go at each step or maybe zero ways to go at each step those are all legitimate for a non-deterministic machine which doing a non-deterministic computation another difference if you look carefully is that we're allowing here the empty string to appear on a transition that's perhaps a little less uh essential to the spirit of non-determinism but it's going to turn out to be a convenience when we actually apply non-determinism to build machines as you'll see very shortly now if there are many different ways to go and some of those ways to go might have different outcomes you know as we remember from before we accepted the input if you end up in an accept state and we rejected the input if you end up not in an except state in a non-accepting state then you reject but now there might be several different ways to go um so and we'll do an example in a minute but there might be several different ways to go and they might disagree some of them might accept other ones might reject so then what do you do well in that case um acceptance always overrules rejection that's the essence of non-determinism the way we're def the way we're setting it up um you may ask why that is and you know if um sort of the spirit of that will become clear in due course but right now just take it as a rule when we're having a non-deterministic machine acceptance overrules rejection so as long as there is one accepting copy one one of the possible ways to go ends up at an accept we say the whole thing is accepted um the only way could possibly reject if all of the possible ways to go end up at rejection end up at a non-accept state so we'll see example we'll i think we're going to do an example right now yeah so we take for example um this machine n1 now on an input a b and we're going to process the symbols one by one just like we did before but now to follow along there might be several different ways to go so if we take um the uh the first symbol a and we run the machine so the machine is starts at the start state um as before but now an a comes in and there might be two now there are two different ways to go so we're going to keep track of both of them after the machine reads an a you can think of it as think of it as being in two states now simultaneously it can be in state q one and it can be in state q two so those are two different possible places it could be at this moment okay now um now we read the next symbol the b and from a b you take all of each of the places where the machine could be um at the end of the previous symbol and you then apply reading a b the next symbol to each of those uh from each of those states where the machine could be in uh from the previous symbol so the machine could be in q1 and q2 after reading an a now we apply a b well q one on a b goes nowhere so you think of that uh a branch if you will of the computation as just dying off it has nowhere to go it just vanishes however when you're in the the other possibility which was state q2 on a b does allow does have a place to go so that the machine is now going to go from q2 to q3 um uh when it re on that branch of the computation which reading on reading a b and then it has um [Music] uh coming out of q3 there are two symbols there's an a and an empty string symbol now on an a the machine would have to read an a in order to transition along that arrow but on when there's an empty symbol on the arrow that means the machine can go along that arrow for free without even reading anything as long as it gets to q3 it can automatically jump to q4 and so once it has read a b and gone to q3 now um it it can either stay in q3 or it can re it can go along the empty transition and go to q4 so again it is going to be a non-deterministic step at this point um [Music] the essence of having a empty transition is that there's going to be non-determinism that's why we didn't introduce that for deterministic automata because um you don't have to transition along an empty string transition you can stay where you are or you can go along the empty string transition without reading any input and go over to the next state which in this case is q4 so let's just see where we are after reading an a we're in states q1 q2 but now after reading a b we're in states q3 and q4 as possibilities and then we and now we're at the end of the input and we look and see what we got if any one of the states as possibilities that we are right now at the end of the string is an accept state then we say overall the machine has accepted so that corresponds to what we said over here before accept the input if some path leads to an accept so if any way of proceeding um through these you know non-deterministic choices will lead you to an accept then you will say we're going to accept the input okay so this input here is accepted let's do another example suppose we have the input a a instead of a b so a a you know after the first a as before we're in states q one and q two as possibilities now we read in a again now the one that's on state q one that possibility q one possibility after reading an a it again branches to q1 and q2 um so when we know after reading the second a we're going to be in at least q1 and q2 now how about the the state that had been you know on q2 on reading in a the one from before after reading the first day you were in q2 now reading the second a there's nowhere to go so that one just gets removed so after reading a a we remain in states q1 and q2 as possibilities neither of those are accept states so therefore on input a a the machine rejects okay let's just do a couple more and then i'll ask you to do one so we have aba as an input let's see what happens then so um we have a uh remember after reading a b a b the machine is in the two states q three q four as possibilities that's what we had from from um the first example so after reading a b we're in states two three q four now we read another a the confu the q4 on an a has nowhere to go in fact it has nowhere to go on any input so no matter what comes in after you're in state q4 that branch dies but on q3 which is another one of the possibilities reading in a um it can follow along just transition because that's one of the labels on that transition is a so you can follow along just a transition reading in a which is the last uh symbol in the string and so now after aba you are in only state q4 as a possibility but that happens to be an accept state so the machine accepts um okay and now lastly what happens let's take our final examples uh what happens if we have a b b so you know as you remember before when we um were after reading a b that was the first example we were in states q3 q4 as possibilities now we read a b well neither of those states have anywhere to go on a b so now all threads all branches of the computation die off and at this point the machine is totally dead it has no no active possibilities left so certainly it's going to reject this input because none of the active states there are no active states are accepting states and in fact if you looked at anything that came later they would anything that extended the string a b b would also be rejected because once the machine had all all possibilities have died off there's no way for them to uh come back to life on any extensions so with that um oh here's an important point before i i'm going to jump to a check-in on you uh but um uh i think what one thing that you know might be on your mind about this non-determinism is how does how do we actually what is that how does that correspond to reality well it doesn't um we're not intending for non-determinism as we're defining it to correspond to a physical device but nevertheless as you'll see it's a very mathematically useful concept uh this non-determinism and it's going to be playing a big role throughout the subject as will uh as will experience it during the rest of the term so with that uh i'm going to have a little check in it's going to give i'm going to ask you to uh consider what happens on one of the inputs um so here we go what does it do on input a a b so here's the machine you can look at it and i'm gonna i'm supposed to hopefully there's a poll here for me to give to you um so you can uh give me your input so what does the machine do on input a a b most of you have answered again you're not going to be penalized for getting the wrong answer um but uh hopefully you you'll get the right answer anyway let's just take a look here uh so time is up um let's end the polling and share the results so the majority of you majority have gotten the correct answer which is a the machine does accept a a b uh because you know when you have a so if you if you like i'll show you the path that corresponds to accepting you go a a b and then empty string and so that uh uh sequence of steps is one of the non-deterministic possibilities that the machine can follow and that shows that the machine does accept the input a a b you know you could think about it the way we did it before um also you know if you read an a it's in the two possibilities q one q two you read a second a again in the possibilities q one q two now you read a b it's into possibilities q three q four and um that's it a a b so now you read the b it's in possibility q three q four q four is an accepting state that overrules the non-accepting state and so the machine accepts you have to understand this so if you didn't get it right go back and think about where you slipped up okay because this is um you know the you know this is just getting we're just getting warmed up here it's going to get it's going to get a lot harder okay so stop sharing the results um and so let's continue so now we just as we did last time we can formally uh define a fine non-deterministic finite automata here's the picture again um okay uh so it looks a lot like the the the case we had before the deterministic finite automata or dfa as we'll call them um it's a five tuple so i've sort of uh written down little reminders for what those components of that five tuple are that list of five uh five components um so they're all the same as before states alphabet transition function start state and accepting states so that the formal definition definition looks exactly the same except the structure of the transition function so now before if you remember you had a state and an input symbol and you got back a state uh now we have something more complicated looking we have a state and an input symbol but it you know it's instead of just sigma it's sigma sub epsilon and that that's a shorthand for sigma union epsilon and that's a way my way of saying that you're allowed to have on your transition um arrows either an input symbol or an empty string so the transition function has to tell you what to do when you have empty string coming in as well um so that would be part of your table for the transition function now over here what's going on over here well now instead of just producing a single state you know when you've read for example an a on from q1 there's a whole set of possibilities so here we have what's called the power set that's the set of subsets of the collection queue so here we're going to produce an entire subset of states instead of just one state coming out there might be a subset of possible states that you can go to so the power set of q is the set of subsets of um of q so that's what this notation means um again this is something that i'm hopefully presenting to you as a bit of a reminder you've seen this somewhere else before but um please make sure you understand the notation uh going forward because we'll we'll be doing less hand-holding as we start moving forward um okay so just let's take a look at the n1 example here um just to illustrate the illustrate what's going on when you're in state q1 reading an a now you get a whole set of possibilities which in this case is q1 and q2 um whereas if you're reading a b what would be that set coming out of q1 what's the set of possible successor states well there are none so it's the empty set okay so this is hopefully your understanding no notation here so now here's i think really important how do we understand non-determinism intuitively speaking and there are multiple different ways which each has their value under different circumstances so one way is thinking about non-determinism as a kind of parallelism so every time the machine has a non-deterministic choice to make where there is more than one outcome you think of the machine as um [Music] branching forking new threads of the parallel computation at that stage where it makes entire copy of itself when there's a choice of possibilities and then each of those independently proceeds to read the rest of the input as separate threads of the computation so if you're familiar with parallel computing this should be reasonably familiar to you the only key thing to remember is that as this thing forks number possibilities the acceptance rule is that if any one of those possibilities gets to an accept at the end of the input it raises a flag and says accept and that overrules everybody else so acceptance dominates so another way of looking at it is the mathematical view where you can imagine and we're going to use all these so it's you really need to understand them all the mathematical view is if you can think of the computation as kind of a tree of possibilities so you start off at the very beginning sort of at the root of the computation which is that when it really begins but every time there's a non-deterministic branching that occurs um the tree has [Music] that note of the tree has uh multiple children coming out of that node and so the different threads of the computation correspond to different branches of that tree and now you're going to accept if any one of those branches leads to an accepting state okay very obviously somewhat similar to what we had before but it's i think it's a little bit of a different perspective on what not how to think about non-determinism and the last one is going to sound a little weird but actually i think for people who are in the business it's the one they use the most and that's the magical way of thinking about non-determinism and that is um when the machine has non-deterministic choices to make you think of the machine as magically guessing the correct one at every stage and the correct one being the one that will eventually lead to accept okay so you can think of the machine as guessing which is the right way to go and if there is some way right way to go it always guesses right of course if the machine ends up rejecting because there is no right way to go then it doesn't matter there is no good guess but if there is some good guess we'll think of the machine as taking that good guess and going that way okay so um now here is very important thing we introduced this new model uh the uh the non-deterministic foundation nfa it turns out to be even though it looks more powerful because has this non-determinism it isn't any more powerful it can do exactly the same class of languages the regular languages and we'll show that um with the theorem here that if an nfi and if an nfa recognizes a then a is regular um [Music] so um we'll prove that by showing how to convert an nfa to an equivalent dfa which does the same language so we can take an nfa that has a non-determinism and find another dfa which doesn't have non-determinism but does the same language it accepts exactly the same strengths even though it lacks that non-deterministic capability it's this is going to be extremely useful by the way and for example in showing that closure under concatenation okay so uh you know in this presentation here i'm going to ignore the epsilon transitions because once you get the idea for how to do this you could figure out how to incorporate them they just make things a little more complicated so let's just focus on the key aspect of non-determinism which is that the machine could have several ways to go at any point in time there could be several next states on an input okay now the idea for the construction so we're going to take we're going to start with a non-deterministic machine m and we're going to build a deterministic machine m prime which does exactly the same thing and the way m prime works is it's going to do what you would do if you were simulating m what would you do with me this is what we were doing as we were as i was explaining it to to you if you are simulating m every time you get an input symbol you just keep track of what is the set of possible states at that point in time that's what the dfa is going to do it's going to have uh it's going to have to keep track of which possible set of states the nfa could be in at the point on that input where we are right now and then as you get the next symbol the dfa is going to have to update things to keep track of the next set of states the nfa could be in at this point just like you would do okay and so um so here's a kind of a picture um and how do we implement that in in so here's the nfa that we're starting with m and we're going to make here the dfa but um in order to remember which set of states that dfa could be in at a given point so maybe it's the set of states that it that m could be in did i say that wrong which set of states the nfa could be at a given time so maybe m the nfa could be in at some point state q3 and q7 the way the dfa keeps track of that it's going to have a state for every possible subset of states of the nfa that's how it re remembers which subset of states the nfa is in that's the way dfas work they have a separate state for each possibility that they need to keep track of and the possibilities here are the different subsets of states that the nfa could be in at a given point okay so corresponding to this subset to these two possibilities q three q seven the dfa is going to have the sub a state with the subset q3 q7 and it's going to for every possible subset here there's going to be a different state of m prime so n prime is going to be bigger um okay so quickly the construction of m the states of m prime now um q prime are going to be the power set the set of subsets of states uh from the original machine m um and uh the uh now we have to look at how the transition function of the dfa when you make the primed machines are the dfas as a dfa machine so these are the prime the the deterministic component so delta prime um when it when it has a subset that you know something like this um has a has one of its states which corresponds to a subset of states of m and it reads an input symbol you just have to do the updating the way you would naturally do you're going to look at every state in r look at where that can go under a so there's a bunch of sets there and look at all the possible states that could be in one of those subsets and that's the set of states that you you could be in that's going to be the new set of states and that's going to be the new state of m prime okay so it's going to be the subset corresponding to all of the states that could be in uh when you apply the transition function of the non-deterministic machine um to one of the states in that in the subset of states that the not a terminus machine could be in okay it's a little bit of a mouthful you know you you could i suggest you look at this if you didn't quite get it after the fact good to understand the starting state for the nfa for the dfa i'm sorry is going to be which subset now we're going to start off with it's going to be the subset corresponding to just the start state of um uh of of m um and the accepting states are going to be uh of the deterministic machine are going to be all of the subsets that have at least one accepting state from the nfa so i hope you got that because i'm going to be another little check in here um which is i'm going to ask you how big is m prime how many states does n prime have uh i told you what those states are so just go think about that um so check in two if m has n states how many states does m prime have by this construction okay so let's launch the next poll um okay five seconds and i think we're almost done here good all right share results i know if sharing results is a good thing i'm not trying to make you if you didn't get the right answer um because most of the people did get the right answer we didn't get the right answer trying to make you feel bad but i do think you should it's a little bit of suggestion you know that you you need to review some basic concepts so the basic concept here is if you have a collection you have a set of states how many subsets are there and the number of subsets is going to be exponential so if you have a collection of n elements the number of subsets of those n elements is 2 to the n um that's the fact we're using here and that's why m prime has two to the n states if m had n states um and that's you know you should make sure you understand why that is all right so um with that um as requested we're gonna have a little break um and that break is going to last us exactly five minutes so we'll we will return in five minutes i'm going to be prompt so i give you a little timer here so please um uh i'm gonna begin it write it right right when this is over okay almost almost ready i hope you're all refreshed and ready for the second half so now now that we have non-determinism we're going to use uh that as a tool to prove the uh closure properties that we were aiming for for um starting from last lecture okay um so remember we let's look at closure under a union now we already did that but i'm going to do it again but this time using non-determinism and you'll see how powerful non-determinate non-determinism now determinism is because it's going to allow us to do it kind of almost with no effort we'll start off the way we did before i'm going to start off with two dfas but actually these could be um nfas even but let's say we started with the two dfas for the two languages a1 and a2 and we're going to construct the nfa rec now we're going to construct an nfa recognizing the union um and that's good enough because we already know that we can convert nfas to dfas and therefore they do they do regulate languages too um okay so now here are the two dfas that do the languages a1 and a2 and what i'm going to do is i'm going to put them together into a bag of states um which is going to be m the nfa that's going to do the union language so remember what does what does m supposed to do m is supposed to accept its input if either m1 or m2 accept so how is it going to do that what it's going to do we're going to add a new state to m which is going to branch under epsilon transitions and now you can start to see how useful these epsilon transitions are going to be for us going to branch under epsilon transitions to the two original start states of m1 and m2 and we're done y well now non-deterministically as we get an input w coming in to m um m at the very beginning even just right after it gets going the very first thing that happens is it's going to branch to m1 and also branch to m2 non-deterministically as two possibilities and then inside m1 and m2 it's going to actually start reading the input and each one is going to be now following along as it would have originally uh the states corresponding to um reading those input symbols um and m as a combination of m1 and m2 is going to have a possibility for e for one state in m1 and once one stated m2 um and so m is going to have those combined into one in into one um package and now if either at the end of the input if either um of these end up at an accepting state then m [Music] um is going to accept as a non-deterministic fighter automaton because that's how non-determinism works you accept if either if any one of the branches ended up accepting which is just what you need for non-determinate for union so we're doing you and we want either one of these to be accepting and then the non-determinism just is built conveniently to allow us to do the union almost you know for free um so you can again thinking about non-determinism as terms of parallelism you could think of it the non-deterministic machine is running in parallel m1 and m2 on the input and if either one of them ends up accepting m will accept or you can think about it in terms of that guessing that i referred to before which means that as m is getting when it's just about to read these the first symbols of its input it guesses whether um that's going to be an input accepted by m1 or an input accepted by m2 and the magic of non-determinism is that it always guesses right so that input happens to be some an input that's going to be wrecked you're going to be accepted by m2 m is going to guess that m2 is the right way to follow and it's going to go in the m2 direction um because non-determinism the magic is you always guess right wish that was true in real life uh would make exams a lot easier um anyway so now let's see how we can use that to do closure under concatenation okay so now um we're gonna actually have a picture very similar to the one we had originally but now using non-determinism we can make it work so here we have the two machines doing the two languages a1 and a2 and we're going to combine them into one bigger machine m as shown remember what m is supposed to do is accept its input if there's some way of splitting that input such that the first half is accepted by the m1 and the second part is accepted by m2 way we're going to get that effect is by um putting in a uh transit empty empty transitions epsilon transitions going from the starts for the except states of m1 to the start state of m2 just as i've shown here in this diagram so these were the original accepting states of m1 and now um they're going to be declassified as accepting states but they're going to have new transitions empty transitions um attached to them which allow them to branch to m2 without reading any input and so intuitively speaking this is going to do the right thing because once m1 has accepted some part of w then you can non-deterministically branch to m2 and um and you're going to be start processing inside m2 and the the point is as i kind of uh jumped ahead of myself um is that the reason why it fixes the problem we had before is that the epsilon transitions are don't machine does not have to take them it can stay where it is as one non-deterministic option or it can move along the epsilon transition without reading any input as another non-deterministic option so it's using this non-determinism now to both stay in m1 to continue reading more of the input and to jump into m2 to start processing what might be the second half or the second part of the input which m2 accepts and you can think of it in terms of the guessing is that the machine is guessing where to make that split once it found an initial part that's accepted by m1 it guesses that this is the right split point and that passes to m2 but there might be other guesses that it could make you know corresponding to other possibilities and so with non-determinism it always guesses right if there is some way to split the string into two parts uh accepted by m1 and m2 the machine will make that good guess and then m1 will accept the first part and m2 will accept us with the second part and we'll get m accepting that whole uh string uh altogether and so that is the uh solution um to our uh puzzle for how do we do closure under concatenation okay i hope that came through because this is um this is you know we're just getting going with non-determinism we're going to be using non-determinism a lot and you it's going to you're going to need to get very comfortable with it okay um now let's do closure under star um and closure under star works uh very similarly but now we're just gonna have a single language if a is regular so is a star so they're not a pair of languages because a star is a urinary operation applying to just a single uh language so if we have a dfa recognizing a in order to show that a star is regular we have to construct a machine that recognizes a star and a machine we're going to construct is as before and then an nfa okay so um here is m uh the dfa for a and we're going to build an nfa m prime that recognizes a star and let's think now what does it mean to recognize a star so when should m you know if i'm going to give you an input when is it in the star language what does m prime have to do um so m remember what star is star means you can take as many copies of you lot as you like of strings in the original language and that's in the star language so to determine if something's in the star language you have to see can i break it up into pieces which are all in the original language um [Music] so uh um so you want to see can i take my input w and cut it up into a bunch of pieces four in this case um where each of those pieces are members of a they're members of the original language so that's what that's what m prime's job is it's getting it has its input and wants to know can i cut that input up into pieces each of which are accepted by the original machine m that's what m prime does and if you think about it you know um a little bit really what's happening is that as soon as m so you know m prime is going to be simulating m that's why i like to think about this as having m inside so you know you're going to if you were going to be doing this yourself you're going to take w you're going to run it for a while you'll see oh m is accepted now i have to start m over again to see if it accepts uh on the next segment so every time m accepts you're going to restart m to see if it accepts another segment and so you by doing that you're going to be cutting w up into different segments which are each of which is accepted by m of course it's never totally clear whether you should and for any given segment you should cut it there or you should wait a little longer and find another play a later place to cut but that's exactly the same problem that we had before with concatenation and we saw that using non-determinism and we're going to solve it again using non-determinism so the way we're going to get that effect of starting the machine over again once once it's accepted is by adding in epsilon transitions that go from the start states back to the from the accept state back to the start state so now every time m has accepted it has an option not not a requirement but it has an option it can either stay continuing to process or it can restart making a cut at that point and trying to see if there's a yet a different a second another segment of the input that's going to be ex that it's going to accept and this is basically the whole thing with one little um problem that we need to deal with and that is we need to make sure um that m prime accepts the empty string because remember the x empty string is always a member of the starred language and as it's as it's written right now we're going to be requiring there to be at least one copy of one at least one segment we're not taking into account the possibility of no segments which is the empty string and the way we're going to get that is well i mean one thing one way to get to it so you know we're missing the empty string right now so how do we fix it basically you know we're just going to take the construction we have get uh you know on on the screen and we're going to uh adjust it to to add in the empty string because it's possibly missing one way to do that which is tempting but wrong is to make the start state of m an accepting state for m prime so we could have and we could have made this an uh an accepting state too and now m prime is also going to accept the empty string that's the good news the problem is that the start state might be playing some other role in m besides just being the start there might be times when m comes back to the start state you know later on and if we make the start state the accept and accept state it's going to suddenly start accepting a bunch of other things too which might not be intended so it's a bad idea to make the start state and accept state instead we'll take the simples uh solution alternative of adding a new start state which will never be returned to under any circumstances and make that a new start and accept state as well so here we'll have to make this additional modification so as i'm saying this is what we need to do and we will do that is by adding a new start state which is also an accept state to make sure it accepts the empty string and then that also can branch um to start off m as before um if the string that's input is not the empty string and so then m you know we m prime is actually going to have to do some work to see if it can be cut up as it was going as it was doing before um so that's the proof of closure under star i'm not going to do it anything beyond what i've just described these proofs by picture are convincing enough i hope um and if not you know you should they are explained in somewhat more detail somewhat more formally in the textbook but uh for the lecture this is where i'm going to stop with these these two arguments um and so now uh oh yeah one uh quick check in on this so if m has n states how many states does m prime have by this construction um so let's do really these are i'm not intending these to be very hard more just to keep you awake uh so how many states does m prime have okay maybe a little too easy even for a check-in yeah everybody's getting this one um because all you did was we added one new state um so the answer is as you have um i think should pretty much everybody is uh observing that it's just the um it's number b um so i'm gonna end the polling i'm gonna share the results and everybody got that one and so let's continue on um and uh so um the very last thing we're going to do today and show you how to convert regular expressions to nfas thereby showing that every language that you can describe with a regular expression is a regular language on tuesday we'll show how to do the conversion in the other direction and show thereby showing that these two um methods of describing languages are equivalent to one another so here is our theorem if r is or is a regular expression and a is the language is instead of strings that like that regular expression describes then a is regular okay um so um we're going to show how to convert [Music] the the strategy is to convert r to an equivalent nfam and so we have to think about remember these regular expressions uh that we introduced last time these are these expressions that look like um you know a b union b star something like that um so built up out of the regular operations from um from the from the primitive regular expressions that don't have any operations that we're calling atomic so if r is an atomic regular expression it just looks like either just a single symbol or an empty string symbol or a an empty language symbol [Music] or r can be a composite regular expression um whoops we're having a little yeah so so we have two possibilities here um r is either atomic or composite and so let's look at what the equivalent expression is um in each case so if r is just the single letter regular expression that's a totally legitimate regular expression just a regular expression one so that just describes the language of the string one so we have to make a an nfa which accepts which recognizes just that language accepts only the string one so it's a very simple nfa it just starts uh in the start state and on that single symbol it branches to an accept state and there were no other transitions allowed so if you get anything else coming in besides that one that that string which is just that one symbol the nfa is going to reject it like if it goes if it's too long if if you know if it gets a a coming in well there's nowhere to go on from this um accepting state on an a so the machine is just going to die it has to be in an accept state at the end of the input um how about you know why don't you think for yourself for a minute how do we make an nfa which uh accepts only the empty string and no other strings you can do that with just one state with an nfa just this one here um the machine is going to start off in the start state which is also immediately an accept state so it accepts the empty string but if anything else comes in there's nowhere to go when the machine does so this machine accepts just the empty string or its language is the language with one element the empty string how about the empty language well here's a here's an nfa which has no accepting state so it can't be accepting anything now if we have a composite regular expression we're already finished because we showed how to build up we showed constructions which uh give us closure under union uh concatenation and star and those constructions are going to enable us to build up uh the nfas that do the language of these more complex regular expressions built up out of the anaphase that do the individual parts so if we already have anaphase that do r1 and r2 then the closure under union construction gives us an nfa that does r1 union r2 as a language so i'm going to do i hope that's clear but i'm going to do an example which will help hopefully illustrate it and it's going to show you you know basically what i'm giving you is an automatic procedure for converting a regular expression into an equivalent nfa so let's just see if that that procedure in opera in action which is really just following this recipe that i described for you so here is a regular expression a union a b star so this is a regular expression it's some language whatever it is i don't care but i want to make an nfa which recognizes that same language and the way i'm going to do that is first build nfa for the [Music] components um the sub-expressions of this regular expression and then combine them using our closure constructions to be um to be anaphase for larger and larger sub-expressions until i get the n of a that's the equivalent of the entire expression so let's just see how that goes so the the very most primitive parts the the the smallest sub-expressions here are just the expressions for a and for b so here's the one for a just just for a so this is the nfa which um recognizes the language which is just the one string a here's the nfa for the just whose language is just the one string b um and now i want an nfa which is which accepts only the string a b now of course you could just do that by hand um yourself it's simple enough but what i'm arguing is that you we can do this automatically using the closure construction for concatenation because really there's an there's a hidden concatenate concatenation symbol this is the the you know a concatenate b so now for a b i'm going to take um the the thing from a and the part from b so these these two things that i had from before and use the concatenation construction to combine them you see that so now i have automatically a an nfa which does the language whose string is just a b th just the a b string and it's not the simplest nfa you can make a simpler one but the virtue of this one is that i got it automatically just by following the closure construction so now i'm going to do a more complex one just the inside here a union a b um so the way i'm going to build that is from the two parts the a part and the a b part you know the a part and the a b part so here is the a part here's the a b part a b part i've already got those from before it's really kind of a proof by induction here but you know i think it's simple enough we don't have to use that language um so we have um you know the a part the a b part and now we are going to apply the uh the closure under union construction to combine those into one machine and remember how that worked we add a new symbol here which branches under empty string to the previous uh i'm adding a new start state which branches to the original start states under empty empty transition and now this is a an nfa for this language a union a b and lastly now we're one step away from getting uh the star of this and how we're going to do that we're going to take this thing here and apply the construction for the star closure and that's going to be an nfa which does a union a b star which is what we wanted in the first place so first we're going to bring that one down because we've already built that one and now remember how we built um the uh closure under star we made this the accepting states return back to the um the start state and we added a new start state to make sure we got the empty string in there that transitioned to the original star state under epsilon okay so that's all um i wanted to say for today's lecture let's do a quick review um we very important concept non-determinism and and non-deterministic finite automata we proved their equivalent in power uh showed the class of regular languages closed under concatenation and star we showed how to do conversion of regular expressions to nfas so um i think that is it for today's lecture and um thank you all for being here let me i'll try to answer a few of these why does concatenation have order well because it's an ordered construction um is there a simple way to prove closure under concatenation without using non-determinism no um why are the empty strings at the accept state can't they be at any state doesn't star make copies of any part of the input no it's only you have to you have to think about what's going on having you you don't want to you have to branch back to the beginning only on an accept because that means you found a piece that's in the original language um is there an automaton that can add some some or subtract memory automata well depends on what you mean by all that but um certainly there are more powerful machines that we're going to study than find out automata um so um but yes there is um and even finite automata can add and subtract if you present the input in the right way i re refer you to the first problem on the homework um so i think i'm gonna uh check out then um take care everybody bye you 3 00:00:28,470 --> 00:00:28,480 4 00:00:28,480 --> 00:00:31,029 5 00:00:31,029 --> 00:00:33,430 6 00:00:33,430 --> 00:00:34,790 7 00:00:34,790 --> 00:00:37,030 8 00:00:37,030 --> 00:00:39,110 9 00:00:39,110 --> 00:00:40,869 10 00:00:40,869 --> 00:00:40,879 11 00:00:40,879 --> 00:00:44,069 12 00:00:44,069 --> 00:00:44,079 13 00:00:44,079 --> 00:00:44,869 14 00:00:44,869 --> 00:00:48,150 15 00:00:48,150 --> 00:00:50,150 16 00:00:50,150 --> 00:00:52,150 17 00:00:52,150 --> 00:00:54,470 18 00:00:54,470 --> 00:00:56,069 19 00:00:56,069 --> 00:00:59,029 20 00:00:59,029 --> 00:01:02,150 21 00:01:02,150 --> 00:01:04,869 22 00:01:04,869 --> 00:01:06,310 23 00:01:06,310 --> 00:01:07,670 24 00:01:07,670 --> 00:01:09,510 25 00:01:09,510 --> 00:01:09,520 26 00:01:09,520 --> 00:01:11,030 27 00:01:11,030 --> 00:01:14,550 28 00:01:14,550 --> 00:01:17,830 29 00:01:17,830 --> 00:01:19,590 30 00:01:19,590 --> 00:01:22,310 31 00:01:22,310 --> 00:01:24,789 32 00:01:24,789 --> 00:01:26,789 33 00:01:26,789 --> 00:01:28,630 34 00:01:28,630 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--> 00:05:41,510 154 00:05:41,510 --> 00:05:43,909 155 00:05:43,909 --> 00:05:46,390 156 00:05:46,390 --> 00:05:48,310 157 00:05:48,310 --> 00:05:50,310 158 00:05:50,310 --> 00:05:52,710 159 00:05:52,710 --> 00:05:55,670 160 00:05:55,670 --> 00:05:57,749 161 00:05:57,749 --> 00:05:58,790 162 00:05:58,790 --> 00:06:01,189 163 00:06:01,189 --> 00:06:03,110 164 00:06:03,110 --> 00:06:05,270 165 00:06:05,270 --> 00:06:06,629 166 00:06:06,629 --> 00:06:08,309 167 00:06:08,309 --> 00:06:10,710 168 00:06:10,710 --> 00:06:12,629 169 00:06:12,629 --> 00:06:15,270 170 00:06:15,270 --> 00:06:17,430 171 00:06:17,430 --> 00:06:20,150 172 00:06:20,150 --> 00:06:22,790 173 00:06:22,790 --> 00:06:24,629 174 00:06:24,629 --> 00:06:26,150 175 00:06:26,150 --> 00:06:27,749 176 00:06:27,749 --> 00:06:29,350 177 00:06:29,350 --> 00:06:31,749 178 00:06:31,749 --> 00:06:33,510 179 00:06:33,510 --> 00:06:36,390 180 00:06:36,390 --> 00:06:39,990 181 00:06:39,990 --> 00:06:41,189 182 00:06:41,189 --> 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212 00:07:34,950 --> 00:07:37,589 213 00:07:37,589 --> 00:07:39,589 214 00:07:39,589 --> 00:07:41,589 215 00:07:41,589 --> 00:07:43,589 216 00:07:43,589 --> 00:07:45,589 217 00:07:45,589 --> 00:07:48,869 218 00:07:48,869 --> 00:07:50,390 219 00:07:50,390 --> 00:07:52,390 220 00:07:52,390 --> 00:07:54,869 221 00:07:54,869 --> 00:07:54,879 222 00:07:54,879 --> 00:07:55,749 223 00:07:55,749 --> 00:07:59,749 224 00:07:59,749 --> 00:08:02,230 225 00:08:02,230 --> 00:08:02,240 226 00:08:02,240 --> 00:08:03,749 227 00:08:03,749 --> 00:08:07,189 228 00:08:07,189 --> 00:08:09,589 229 00:08:09,589 --> 00:08:11,110 230 00:08:11,110 --> 00:08:11,120 231 00:08:11,120 --> 00:08:12,710 232 00:08:12,710 --> 00:08:15,110 233 00:08:15,110 --> 00:08:16,869 234 00:08:16,869 --> 00:08:20,790 235 00:08:20,790 --> 00:08:22,710 236 00:08:22,710 --> 00:08:24,790 237 00:08:24,790 --> 00:08:27,670 238 00:08:27,670 --> 00:08:29,589 239 00:08:29,589 --> 00:08:29,599 240 00:08:29,599 --> 00:08:31,909 241 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--> 00:09:28,630 271 00:09:28,630 --> 00:09:31,190 272 00:09:31,190 --> 00:09:33,190 273 00:09:33,190 --> 00:09:37,190 274 00:09:37,190 --> 00:09:39,430 275 00:09:39,430 --> 00:09:41,269 276 00:09:41,269 --> 00:09:42,389 277 00:09:42,389 --> 00:09:45,110 278 00:09:45,110 --> 00:09:47,590 279 00:09:47,590 --> 00:09:50,630 280 00:09:50,630 --> 00:09:53,110 281 00:09:53,110 --> 00:09:55,509 282 00:09:55,509 --> 00:09:57,269 283 00:09:57,269 --> 00:09:59,430 284 00:09:59,430 --> 00:09:59,440 285 00:09:59,440 --> 00:10:00,389 286 00:10:00,389 --> 00:10:02,790 287 00:10:02,790 --> 00:10:04,790 288 00:10:04,790 --> 00:10:06,069 289 00:10:06,069 --> 00:10:09,190 290 00:10:09,190 --> 00:10:11,110 291 00:10:11,110 --> 00:10:13,750 292 00:10:13,750 --> 00:10:15,750 293 00:10:15,750 --> 00:10:18,949 294 00:10:18,949 --> 00:10:22,470 295 00:10:22,470 --> 00:10:25,430 296 00:10:25,430 --> 00:10:26,949 297 00:10:26,949 --> 00:10:30,230 298 00:10:30,230 --> 00:10:33,110 299 00:10:33,110 --> 00:10:35,190 300 00:10:35,190 --> 00:10:37,670 301 00:10:37,670 --> 00:10:37,680 302 00:10:37,680 --> 00:10:39,430 303 00:10:39,430 --> 00:10:39,440 304 00:10:39,440 --> 00:10:40,310 305 00:10:40,310 --> 00:10:41,829 306 00:10:41,829 --> 00:10:43,430 307 00:10:43,430 --> 00:10:45,829 308 00:10:45,829 --> 00:10:48,949 309 00:10:48,949 --> 00:10:51,509 310 00:10:51,509 --> 00:10:53,030 311 00:10:53,030 --> 00:10:55,269 312 00:10:55,269 --> 00:10:56,230 313 00:10:56,230 --> 00:10:58,870 314 00:10:58,870 --> 00:11:01,110 315 00:11:01,110 --> 00:11:03,750 316 00:11:03,750 --> 00:11:05,990 317 00:11:05,990 --> 00:11:07,990 318 00:11:07,990 --> 00:11:11,430 319 00:11:11,430 --> 00:11:11,440 320 00:11:11,440 --> 00:11:12,550 321 00:11:12,550 --> 00:11:16,230 322 00:11:16,230 --> 00:11:19,670 323 00:11:19,670 --> 00:11:22,630 324 00:11:22,630 --> 00:11:26,069 325 00:11:26,069 --> 00:11:28,870 326 00:11:28,870 --> 00:11:31,590 327 00:11:31,590 --> 00:11:34,389 328 00:11:34,389 --> 00:11:35,910 329 00:11:35,910 --> 00:11:37,269 330 00:11:37,269 --> 00:11:39,509 331 00:11:39,509 --> 00:11:42,550 332 00:11:42,550 --> 00:11:44,870 333 00:11:44,870 --> 00:11:46,550 334 00:11:46,550 --> 00:11:46,560 335 00:11:46,560 --> 00:11:47,829 336 00:11:47,829 --> 00:11:49,670 337 00:11:49,670 --> 00:11:50,949 338 00:11:50,949 --> 00:11:52,790 339 00:11:52,790 --> 00:11:55,190 340 00:11:55,190 --> 00:11:55,200 341 00:11:55,200 --> 00:11:57,269 342 00:11:57,269 --> 00:11:59,990 343 00:11:59,990 --> 00:12:03,190 344 00:12:03,190 --> 00:12:05,670 345 00:12:05,670 --> 00:12:07,430 346 00:12:07,430 --> 00:12:11,269 347 00:12:11,269 --> 00:12:12,949 348 00:12:12,949 --> 00:12:15,430 349 00:12:15,430 --> 00:12:17,190 350 00:12:17,190 --> 00:12:18,310 351 00:12:18,310 --> 00:12:18,320 352 00:12:18,320 --> 00:12:18,680 353 00:12:18,680 --> 00:12:18,690 354 00:12:18,690 --> 00:12:20,069 355 00:12:20,069 --> 00:12:20,079 356 00:12:20,079 --> 00:12:22,310 357 00:12:22,310 --> 00:12:25,190 358 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--> 00:13:22,710 388 00:13:22,710 --> 00:13:25,670 389 00:13:25,670 --> 00:13:27,750 390 00:13:27,750 --> 00:13:30,550 391 00:13:30,550 --> 00:13:32,389 392 00:13:32,389 --> 00:13:34,550 393 00:13:34,550 --> 00:13:36,949 394 00:13:36,949 --> 00:13:39,350 395 00:13:39,350 --> 00:13:42,550 396 00:13:42,550 --> 00:13:44,710 397 00:13:44,710 --> 00:13:47,670 398 00:13:47,670 --> 00:13:49,590 399 00:13:49,590 --> 00:13:53,829 400 00:13:53,829 --> 00:13:55,670 401 00:13:55,670 --> 00:13:57,509 402 00:13:57,509 --> 00:13:59,910 403 00:13:59,910 --> 00:14:02,069 404 00:14:02,069 --> 00:14:02,079 405 00:14:02,079 --> 00:14:04,150 406 00:14:04,150 --> 00:14:06,150 407 00:14:06,150 --> 00:14:08,629 408 00:14:08,629 --> 00:14:12,069 409 00:14:12,069 --> 00:14:14,470 410 00:14:14,470 --> 00:14:16,069 411 00:14:16,069 --> 00:14:18,150 412 00:14:18,150 --> 00:14:19,829 413 00:14:19,829 --> 00:14:21,750 414 00:14:21,750 --> 00:14:25,030 415 00:14:25,030 --> 00:14:26,870 416 00:14:26,870 --> 00:14:28,710 417 00:14:28,710 --> 00:14:30,949 418 00:14:30,949 --> 00:14:32,310 419 00:14:32,310 --> 00:14:33,990 420 00:14:33,990 --> 00:14:35,750 421 00:14:35,750 --> 00:14:35,760 422 00:14:35,760 --> 00:14:37,110 423 00:14:37,110 --> 00:14:39,030 424 00:14:39,030 --> 00:14:41,189 425 00:14:41,189 --> 00:14:43,670 426 00:14:43,670 --> 00:14:46,069 427 00:14:46,069 --> 00:14:46,079 428 00:14:46,079 --> 00:14:47,269 429 00:14:47,269 --> 00:14:47,279 430 00:14:47,279 --> 00:14:48,230 431 00:14:48,230 --> 00:14:49,910 432 00:14:49,910 --> 00:14:51,509 433 00:14:51,509 --> 00:14:53,670 434 00:14:53,670 --> 00:14:56,230 435 00:14:56,230 --> 00:14:58,389 436 00:14:58,389 --> 00:14:59,910 437 00:14:59,910 --> 00:15:02,069 438 00:15:02,069 --> 00:15:04,069 439 00:15:04,069 --> 00:15:06,470 440 00:15:06,470 --> 00:15:08,790 441 00:15:08,790 --> 00:15:11,189 442 00:15:11,189 --> 00:15:11,199 443 00:15:11,199 --> 00:15:14,629 444 00:15:14,629 --> 00:15:17,509 445 00:15:17,509 --> 00:15:20,629 446 00:15:20,629 --> 00:15:23,750 447 00:15:23,750 --> 00:15:23,760 448 00:15:23,760 --> 00:15:25,509 449 00:15:25,509 --> 00:15:27,110 450 00:15:27,110 --> 00:15:29,749 451 00:15:29,749 --> 00:15:29,759 452 00:15:29,759 --> 00:15:31,030 453 00:15:31,030 --> 00:15:33,509 454 00:15:33,509 --> 00:15:35,269 455 00:15:35,269 --> 00:15:37,350 456 00:15:37,350 --> 00:15:37,360 457 00:15:37,360 --> 00:15:38,230 458 00:15:38,230 --> 00:15:42,790 459 00:15:42,790 --> 00:15:45,829 460 00:15:45,829 --> 00:15:47,749 461 00:15:47,749 --> 00:15:51,110 462 00:15:51,110 --> 00:15:53,110 463 00:15:53,110 --> 00:15:55,990 464 00:15:55,990 --> 00:15:58,150 465 00:15:58,150 --> 00:16:00,230 466 00:16:00,230 --> 00:16:01,910 467 00:16:01,910 --> 00:16:04,629 468 00:16:04,629 --> 00:16:06,870 469 00:16:06,870 --> 00:16:09,670 470 00:16:09,670 --> 00:16:12,230 471 00:16:12,230 --> 00:16:13,829 472 00:16:13,829 --> 00:16:15,110 473 00:16:15,110 --> 00:16:17,430 474 00:16:17,430 --> 00:16:19,749 475 00:16:19,749 --> 00:16:21,189 476 00:16:21,189 --> 00:16:23,910 477 00:16:23,910 --> 00:16:27,430 478 00:16:27,430 --> 00:16:29,430 479 00:16:29,430 --> 00:16:34,829 480 00:16:34,829 --> 00:16:34,839 481 00:16:34,839 --> 00:16:36,629 482 00:16:36,629 --> 00:16:39,590 483 00:16:39,590 --> 00:16:42,310 484 00:16:42,310 --> 00:16:45,430 485 00:16:45,430 --> 00:16:45,440 486 00:16:45,440 --> 00:16:46,470 487 00:16:46,470 --> 00:16:49,189 488 00:16:49,189 --> 00:16:52,230 489 00:16:52,230 --> 00:16:55,030 490 00:16:55,030 --> 00:16:57,509 491 00:16:57,509 --> 00:16:58,870 492 00:16:58,870 --> 00:17:00,790 493 00:17:00,790 --> 00:17:02,550 494 00:17:02,550 --> 00:17:05,350 495 00:17:05,350 --> 00:17:07,669 496 00:17:07,669 --> 00:17:10,069 497 00:17:10,069 --> 00:17:13,110 498 00:17:13,110 --> 00:17:13,120 499 00:17:13,120 --> 00:17:13,990 500 00:17:13,990 --> 00:17:15,990 501 00:17:15,990 --> 00:17:17,990 502 00:17:17,990 --> 00:17:19,590 503 00:17:19,590 --> 00:17:21,750 504 00:17:21,750 --> 00:17:23,669 505 00:17:23,669 --> 00:17:25,510 506 00:17:25,510 --> 00:17:27,990 507 00:17:27,990 --> 00:17:30,390 508 00:17:30,390 --> 00:17:32,789 509 00:17:32,789 --> 00:17:35,270 510 00:17:35,270 --> 00:17:36,950 511 00:17:36,950 --> 00:17:39,110 512 00:17:39,110 --> 00:17:40,950 513 00:17:40,950 --> 00:17:43,190 514 00:17:43,190 --> 00:17:44,950 515 00:17:44,950 --> 00:17:46,870 516 00:17:46,870 --> 00:17:48,230 517 00:17:48,230 --> 00:17:50,230 518 00:17:50,230 --> 00:17:51,669 519 00:17:51,669 --> 00:17:53,590 520 00:17:53,590 --> 00:17:55,029 521 00:17:55,029 --> 00:17:55,039 522 00:17:55,039 --> 00:17:57,430 523 00:17:57,430 --> 00:17:59,750 524 00:17:59,750 --> 00:18:01,909 525 00:18:01,909 --> 00:18:03,830 526 00:18:03,830 --> 00:18:05,830 527 00:18:05,830 --> 00:18:08,310 528 00:18:08,310 --> 00:18:09,990 529 00:18:09,990 --> 00:18:11,830 530 00:18:11,830 --> 00:18:14,390 531 00:18:14,390 --> 00:18:16,630 532 00:18:16,630 --> 00:18:17,510 533 00:18:17,510 --> 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563 00:19:13,750 --> 00:19:17,590 564 00:19:17,590 --> 00:19:19,830 565 00:19:19,830 --> 00:19:22,070 566 00:19:22,070 --> 00:19:25,110 567 00:19:25,110 --> 00:19:27,510 568 00:19:27,510 --> 00:19:27,520 569 00:19:27,520 --> 00:19:28,470 570 00:19:28,470 --> 00:19:30,950 571 00:19:30,950 --> 00:19:33,430 572 00:19:33,430 --> 00:19:35,350 573 00:19:35,350 --> 00:19:35,360 574 00:19:35,360 --> 00:19:36,549 575 00:19:36,549 --> 00:19:36,559 576 00:19:36,559 --> 00:19:37,669 577 00:19:37,669 --> 00:19:41,029 578 00:19:41,029 --> 00:19:42,549 579 00:19:42,549 --> 00:19:42,559 580 00:19:42,559 --> 00:19:44,070 581 00:19:44,070 --> 00:19:44,080 582 00:19:44,080 --> 00:19:45,750 583 00:19:45,750 --> 00:19:48,630 584 00:19:48,630 --> 00:19:50,150 585 00:19:50,150 --> 00:19:51,750 586 00:19:51,750 --> 00:19:51,760 587 00:19:51,760 --> 00:19:52,870 588 00:19:52,870 --> 00:19:54,870 589 00:19:54,870 --> 00:19:57,510 590 00:19:57,510 --> 00:19:59,430 591 00:19:59,430 --> 00:20:02,470 592 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--> 00:20:54,549 622 00:20:54,549 --> 00:20:57,029 623 00:20:57,029 --> 00:20:59,190 624 00:20:59,190 --> 00:21:02,230 625 00:21:02,230 --> 00:21:05,430 626 00:21:05,430 --> 00:21:08,710 627 00:21:08,710 --> 00:21:11,110 628 00:21:11,110 --> 00:21:13,830 629 00:21:13,830 --> 00:21:13,840 630 00:21:13,840 --> 00:21:14,630 631 00:21:14,630 --> 00:21:17,190 632 00:21:17,190 --> 00:21:19,190 633 00:21:19,190 --> 00:21:21,830 634 00:21:21,830 --> 00:21:24,070 635 00:21:24,070 --> 00:21:26,470 636 00:21:26,470 --> 00:21:28,310 637 00:21:28,310 --> 00:21:30,470 638 00:21:30,470 --> 00:21:32,710 639 00:21:32,710 --> 00:21:34,950 640 00:21:34,950 --> 00:21:37,430 641 00:21:37,430 --> 00:21:40,070 642 00:21:40,070 --> 00:21:43,110 643 00:21:43,110 --> 00:21:45,350 644 00:21:45,350 --> 00:21:47,029 645 00:21:47,029 --> 00:21:48,390 646 00:21:48,390 --> 00:21:49,830 647 00:21:49,830 --> 00:21:52,870 648 00:21:52,870 --> 00:21:54,310 649 00:21:54,310 --> 00:21:56,630 650 00:21:56,630 --> 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680 00:22:52,630 --> 00:22:55,669 681 00:22:55,669 --> 00:22:57,669 682 00:22:57,669 --> 00:23:01,190 683 00:23:01,190 --> 00:23:01,200 684 00:23:01,200 --> 00:23:03,110 685 00:23:03,110 --> 00:23:05,270 686 00:23:05,270 --> 00:23:05,280 687 00:23:05,280 --> 00:23:06,230 688 00:23:06,230 --> 00:23:08,789 689 00:23:08,789 --> 00:23:11,270 690 00:23:11,270 --> 00:23:12,950 691 00:23:12,950 --> 00:23:15,830 692 00:23:15,830 --> 00:23:17,750 693 00:23:17,750 --> 00:23:17,760 694 00:23:17,760 --> 00:23:18,549 695 00:23:18,549 --> 00:23:20,390 696 00:23:20,390 --> 00:23:22,630 697 00:23:22,630 --> 00:23:22,640 698 00:23:22,640 --> 00:23:23,750 699 00:23:23,750 --> 00:23:23,760 700 00:23:23,760 --> 00:23:25,029 701 00:23:25,029 --> 00:23:26,870 702 00:23:26,870 --> 00:23:28,310 703 00:23:28,310 --> 00:23:30,149 704 00:23:30,149 --> 00:23:32,789 705 00:23:32,789 --> 00:23:34,630 706 00:23:34,630 --> 00:23:36,549 707 00:23:36,549 --> 00:23:39,590 708 00:23:39,590 --> 00:23:41,990 709 00:23:41,990 --> 00:23:45,269 710 00:23:45,269 --> 00:23:46,870 711 00:23:46,870 --> 00:23:49,510 712 00:23:49,510 --> 00:23:51,269 713 00:23:51,269 --> 00:23:51,279 714 00:23:51,279 --> 00:23:53,190 715 00:23:53,190 --> 00:23:54,549 716 00:23:54,549 --> 00:23:56,149 717 00:23:56,149 --> 00:23:58,630 718 00:23:58,630 --> 00:23:58,640 719 00:23:58,640 --> 00:24:00,310 720 00:24:00,310 --> 00:24:02,549 721 00:24:02,549 --> 00:24:04,789 722 00:24:04,789 --> 00:24:07,029 723 00:24:07,029 --> 00:24:08,390 724 00:24:08,390 --> 00:24:08,400 725 00:24:08,400 --> 00:24:11,350 726 00:24:11,350 --> 00:24:12,710 727 00:24:12,710 --> 00:24:12,720 728 00:24:12,720 --> 00:24:13,590 729 00:24:13,590 --> 00:24:15,430 730 00:24:15,430 --> 00:24:16,950 731 00:24:16,950 --> 00:24:18,549 732 00:24:18,549 --> 00:24:20,310 733 00:24:20,310 --> 00:24:22,789 734 00:24:22,789 --> 00:24:25,100 735 00:24:25,100 --> 00:24:25,110 736 00:24:25,110 --> 00:24:26,390 737 00:24:26,390 --> 00:24:28,549 738 00:24:28,549 --> 00:24:31,590 739 00:24:31,590 --> 00:24:34,230 740 00:24:34,230 --> 00:24:36,390 741 00:24:36,390 --> 00:24:36,400 742 00:24:36,400 --> 00:24:37,830 743 00:24:37,830 --> 00:24:40,789 744 00:24:40,789 --> 00:24:42,630 745 00:24:42,630 --> 00:24:44,789 746 00:24:44,789 --> 00:24:46,149 747 00:24:46,149 --> 00:24:47,990 748 00:24:47,990 --> 00:24:50,149 749 00:24:50,149 --> 00:24:52,870 750 00:24:52,870 --> 00:24:55,430 751 00:24:55,430 --> 00:24:58,310 752 00:24:58,310 --> 00:25:00,470 753 00:25:00,470 --> 00:25:03,029 754 00:25:03,029 --> 00:25:05,190 755 00:25:05,190 --> 00:25:08,870 756 00:25:08,870 --> 00:25:12,470 757 00:25:12,470 --> 00:25:14,549 758 00:25:14,549 --> 00:25:16,630 759 00:25:16,630 --> 00:25:18,070 760 00:25:18,070 --> 00:25:20,310 761 00:25:20,310 --> 00:25:21,909 762 00:25:21,909 --> 00:25:21,919 763 00:25:21,919 --> 00:25:23,029 764 00:25:23,029 --> 00:25:25,510 765 00:25:25,510 --> 00:25:29,190 766 00:25:29,190 --> 00:25:30,789 767 00:25:30,789 --> 00:25:32,789 768 00:25:32,789 --> 00:25:35,029 769 00:25:35,029 --> 00:25:37,269 770 00:25:37,269 --> 00:25:38,870 771 00:25:38,870 --> 00:25:41,330 772 00:25:41,330 --> 00:25:41,340 773 00:25:41,340 --> 00:25:43,110 774 00:25:43,110 --> 00:25:45,909 775 00:25:45,909 --> 00:25:48,630 776 00:25:48,630 --> 00:25:50,789 777 00:25:50,789 --> 00:25:52,789 778 00:25:52,789 --> 00:25:55,750 779 00:25:55,750 --> 00:25:57,190 780 00:25:57,190 --> 00:25:58,549 781 00:25:58,549 --> 00:26:00,390 782 00:26:00,390 --> 00:26:03,029 783 00:26:03,029 --> 00:26:04,950 784 00:26:04,950 --> 00:26:07,430 785 00:26:07,430 --> 00:26:09,350 786 00:26:09,350 --> 00:26:09,360 787 00:26:09,360 --> 00:26:10,950 788 00:26:10,950 --> 00:26:12,310 789 00:26:12,310 --> 00:26:13,510 790 00:26:13,510 --> 00:26:15,190 791 00:26:15,190 --> 00:26:17,350 792 00:26:17,350 --> 00:26:18,789 793 00:26:18,789 --> 00:26:21,029 794 00:26:21,029 --> 00:26:23,350 795 00:26:23,350 --> 00:26:27,510 796 00:26:27,510 --> 00:26:29,350 797 00:26:29,350 --> 00:26:29,360 798 00:26:29,360 --> 00:26:30,230 799 00:26:30,230 --> 00:26:33,029 800 00:26:33,029 --> 00:26:34,870 801 00:26:34,870 --> 00:26:38,070 802 00:26:38,070 --> 00:26:41,350 803 00:26:41,350 --> 00:26:43,510 804 00:26:43,510 --> 00:26:47,590 805 00:26:47,590 --> 00:26:49,990 806 00:26:49,990 --> 00:26:50,000 807 00:26:50,000 --> 00:26:51,590 808 00:26:51,590 --> 00:26:54,310 809 00:26:54,310 --> 00:26:56,870 810 00:26:56,870 --> 00:26:58,470 811 00:26:58,470 --> 00:27:00,310 812 00:27:00,310 --> 00:27:01,669 813 00:27:01,669 --> 00:27:03,909 814 00:27:03,909 --> 00:27:05,750 815 00:27:05,750 --> 00:27:07,430 816 00:27:07,430 --> 00:27:09,590 817 00:27:09,590 --> 00:27:09,600 818 00:27:09,600 --> 00:27:10,950 819 00:27:10,950 --> 00:27:10,960 820 00:27:10,960 --> 00:27:11,750 821 00:27:11,750 --> 00:27:11,760 822 00:27:11,760 --> 00:27:12,789 823 00:27:12,789 --> 00:27:15,190 824 00:27:15,190 --> 00:27:17,029 825 00:27:17,029 --> 00:27:19,549 826 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--> 00:28:06,149 856 00:28:06,149 --> 00:28:07,590 857 00:28:07,590 --> 00:28:10,789 858 00:28:10,789 --> 00:28:13,830 859 00:28:13,830 --> 00:28:13,840 860 00:28:13,840 --> 00:28:14,830 861 00:28:14,830 --> 00:28:14,840 862 00:28:14,840 --> 00:28:16,630 863 00:28:16,630 --> 00:28:18,630 864 00:28:18,630 --> 00:28:20,470 865 00:28:20,470 --> 00:28:22,710 866 00:28:22,710 --> 00:28:24,149 867 00:28:24,149 --> 00:28:25,990 868 00:28:25,990 --> 00:28:27,269 869 00:28:27,269 --> 00:28:30,549 870 00:28:30,549 --> 00:28:33,110 871 00:28:33,110 --> 00:28:35,029 872 00:28:35,029 --> 00:28:37,190 873 00:28:37,190 --> 00:28:39,669 874 00:28:39,669 --> 00:28:41,909 875 00:28:41,909 --> 00:28:41,919 876 00:28:41,919 --> 00:28:42,789 877 00:28:42,789 --> 00:28:45,590 878 00:28:45,590 --> 00:28:47,110 879 00:28:47,110 --> 00:28:49,990 880 00:28:49,990 --> 00:28:51,830 881 00:28:51,830 --> 00:28:54,549 882 00:28:54,549 --> 00:28:55,830 883 00:28:55,830 --> 00:28:58,149 884 00:28:58,149 --> 00:29:00,470 885 00:29:00,470 --> 00:29:02,710 886 00:29:02,710 --> 00:29:04,470 887 00:29:04,470 --> 00:29:06,070 888 00:29:06,070 --> 00:29:07,990 889 00:29:07,990 --> 00:29:10,470 890 00:29:10,470 --> 00:29:12,630 891 00:29:12,630 --> 00:29:14,789 892 00:29:14,789 --> 00:29:16,630 893 00:29:16,630 --> 00:29:19,269 894 00:29:19,269 --> 00:29:20,549 895 00:29:20,549 --> 00:29:20,559 896 00:29:20,559 --> 00:29:21,669 897 00:29:21,669 --> 00:29:23,669 898 00:29:23,669 --> 00:29:27,909 899 00:29:27,909 --> 00:29:30,070 900 00:29:30,070 --> 00:29:31,190 901 00:29:31,190 --> 00:29:33,110 902 00:29:33,110 --> 00:29:33,120 903 00:29:33,120 --> 00:29:34,710 904 00:29:34,710 --> 00:29:37,510 905 00:29:37,510 --> 00:29:39,350 906 00:29:39,350 --> 00:29:41,269 907 00:29:41,269 --> 00:29:43,590 908 00:29:43,590 --> 00:29:47,029 909 00:29:47,029 --> 00:29:49,510 910 00:29:49,510 --> 00:29:51,029 911 00:29:51,029 --> 00:29:51,039 912 00:29:51,039 --> 00:29:52,470 913 00:29:52,470 --> 00:29:54,789 914 00:29:54,789 --> 00:29:54,799 915 00:29:54,799 --> 00:29:55,669 916 00:29:55,669 --> 00:29:55,679 917 00:29:55,679 --> 00:29:57,110 918 00:29:57,110 --> 00:30:00,070 919 00:30:00,070 --> 00:30:02,310 920 00:30:02,310 --> 00:30:04,950 921 00:30:04,950 --> 00:30:07,430 922 00:30:07,430 --> 00:30:09,190 923 00:30:09,190 --> 00:30:11,669 924 00:30:11,669 --> 00:30:13,750 925 00:30:13,750 --> 00:30:16,710 926 00:30:16,710 --> 00:30:18,230 927 00:30:18,230 --> 00:30:19,909 928 00:30:19,909 --> 00:30:22,470 929 00:30:22,470 --> 00:30:25,750 930 00:30:25,750 --> 00:30:27,909 931 00:30:27,909 --> 00:30:30,870 932 00:30:30,870 --> 00:30:32,630 933 00:30:32,630 --> 00:30:34,870 934 00:30:34,870 --> 00:30:38,149 935 00:30:38,149 --> 00:30:40,149 936 00:30:40,149 --> 00:30:43,110 937 00:30:43,110 --> 00:30:45,510 938 00:30:45,510 --> 00:30:46,470 939 00:30:46,470 --> 00:30:48,389 940 00:30:48,389 --> 00:30:51,269 941 00:30:51,269 --> 00:30:53,590 942 00:30:53,590 --> 00:30:54,950 943 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