1 00:00:24,640 --> 00:00:26,550 okay everybody let's begin um welcome back good to see you all here on zoom um so we're going to pick up with what we had been discussing um last week which was um a an introduction to np completeness so we're following on our description of time complexity we we started talking about the time complexity classes the class p the non-deterministic classes the class np p versus np problem and then leading to this discussion of np completeness and today we're going to prove the big theorem in the field uh which really kind of got things going uh back in the early 1970s by cook and levin that uh there was actually an mp complete problem called that sat in particular is an np complete problem and then we'll also talk about threesat which is a useful tool um so just to remember just to uh remember we had this notion of np completeness for languages that be complete if it's in np and everything else in np is polynomial time reducible to it and if an np complete problem turns out to have a polynomial time solution then every np problem has a polynomial time solution and that's part of the importance of np completeness uh because since we uh consider it unlikely that p equals np uh and that there probably are some problems that uh are in np but are not solvable in polynomial time that's would me imply that an np-complete problem would have that property and so proving a problem and being mp complete is evidence very strong evidence that it doesn't have a polynomial time uh solution and so therefore it's we call intractable uh it's a very difficult problem um so um the way we are going to typically show problems np-complete is by reducing a known a previously known np-complete problem to that problem often it's three-sat as we've seen several examples already or it could be some other uh example so let's just survey briefly uh the things that we've already languages that we've already seen which are mp complete so we have the languages sat which is a direct reduction from every np problem and so we're going to show that today that every np language is polynomial time reducible to sat which is in turn reducible to three sat and we showed previously that threesat is reducible to clique and hampf and in recitation if you went to that they showed that the subset sub problem and the undirected hand path problems are also reducible from previously uh shown well from uh from either from 3sat or in the case of their undirected hand path problem it's reducible from the hand path problem and the conclusion once we have these two blue reducibilities shown then we know that all of these problems are np-complete um so the class np basically breaks down into the np complete problems p problems problems that are in p and then there might be problems in between as well so there are some problems that are not known to be in either category and in fact there's a old theorem which shows that if p is different from np then there are actually problems that are in the intermediate state and um of course it's possible that p equals np and then everything collapses down uh to be the same with the tiny exception of the um sigma star and empty set languages which can never be uh complete um okay um so that's our quick review here's a check in that i'm gonna use to get us ready for the big proof that we're gonna be spending most of the lecture on about um uh showing sat as np-complete but just to define a little notation which you might have maybe you've seen already um but i'm going to do it in the form of a check-in so i'm sure you've all seen the big sum notation uh using a sigma to represent a sum over um over some set of possibilities just as there is the big sigma notation we have you can have other operations um that apply to a a set of uh elements so in this case we're going to be seeing the big and and the big or operation uh it's going to a notation which is going to allow us to talk about taking the end of many things or the ore of many things um because we're we're going to be building these boolean formulas and so ands and orders are going to be the uh operations that we're going to be focusing on and so just as an example just to make sure you're we're all understanding this notation if you have uh two strings x and y uh written out in terms of their individual symbols so they're both of length n so x is x1 to xn y is y 1 to y n and now i write the following expression the big and for i ranging between 1 and n of x i equal to y i if that big and is true what does it tell us about x and y and i'm just going to offer you two possibilities that either x and y agree in some symbol or they are equal namely that they agree on every symbol and so let's just pull that up as a quick poll to get us going here um [Music] i just want to make sure you understand the notation because we're going to be using that a lot in describing the uh polynomial time reduction from uh languages in np to the sat language okay i think most of you have got the idea but so let's finish this up quick um see another 15 seconds please just to give you a chance to participate um okay so i'm gonna close this uh close this poll last chance let's call all right so yes the big and as most of you can see the big and says that you know x1 equals y1 and x2 equals y2 and x3 equals y3 so they're all um every symbol in x is equal to the corresponding symbol in y if we had a big or instead of a big and the uh then a would be the correct answer because then just there would be some place where they agree instead of every place where they they agree um okay so let's um then start to launch in to this big theorem the proof itself is a massive details um with one underlying idea and in fact it's an idea we've seen before um but let's before we get ahead of ourselves let's just understand what we're trying to do uh so we want to show that this language sat is np-complete remember what's that is it's boolean formulas that are satisfiable uh so we already showed that the set pro so being mp complete means it has these two features it's in np and everything in np is reducible to it so first of all sat is np as we've already seen the witness that shows a formula is satisfiable is simply the satisfying assignment that evaluates the true so now we're going to pick some language in np a and show that a is polynomial time reducible to okay so this is going to apply to any language a and np so let a be some language in np it's decided by some non-deterministic turing machine m in time into the k that's what it means to be an np um i'm going to make ignoring the constant factors when we could carry that throughout the proof it just would make this the the description a little more cumbersome and wouldn't change any of the ideas but so let's just say it's m runs in time into the k and recognizes or decides this language a in uh uh so it's an it's in and it's a non-deterministic machine so we've got to give a polynomial time reduction from a to sad that's that's what i have to demonstrate to you so what does that reduction look like it's going to map strings which may or may not be an a to formulas which may or may not be satisfied that's what the reduction's got to do so saying saying that out more formally is it's going to take some string w and map it to a formula we'll call f sub m w where m is the machine that um decides a and w is is is the input so f is going to map w to this formula where the string is in the language exactly when the formula is satisfiable and the formula is going to depend upon m and w that's why it just has it's written in this way so we've seen that kind of thing before so basically my my job you know you know we have this language a that's an np and now we have some string w which might be an a or maybe not is an a and i have to quickly in polynomial time now produce a formula which is going to be satisfiable exactly when the string w is in a okay so how is that formula going to work how can i produce such a formula which i mean i of course i don't know whether w is in a or not um because i i'm just a polynomial time reduction and um a is an np language so polynomial time probably is not enough to solve whether w is in a so i've got to make matt do that mapping without knowing the answer and the idea is that um and this is where you know we've seen things like this before the formula we're going to construct simulates the machine on w so in some way it's going to do that simulation the trick is figuring out how to what that means um but the the interpretation of that formula is that in a sense describes and it says though in my informal language that m accepts w very much like you know there's a lot of parallel here between um this construction and for example the construction um the pcp construction where we made a an instant you know given a machine and an input we made a set of those dominoes where finding a match forced you to simulate the machine here finding a satisfying assignment is going to force you to simulate the machine [Music] so this satisfying assignment is going to be a computation history for mmw and i'm just going to write that computation history in a particular way so it's going to have a somewhat different encoding which is going to help us to visualize what's going on better okay um so that's that's the that's the that's the approach that's the basic idea of what we're trying to what we're gonna do so i'm happy to take a minute if that's if you have any questions about this part but um otherwise i'll just move on to start doing the actual construction of what the formula looks like how is it going to do that how is that formula going to work okay so no questions hopefully we are uh all together okay so first of all let me describe what my computation histories are going to look like and and the situation is a little bit different than what we had before because um when we were talking about the post correspondence problem we had a deterministic machine and now our machine is non-deterministic okay so um we're going to call the object instead of an accepting computation or computation history we're just going to we're going to call it a tableau or sometimes an accepting tableau if you want to emphasize the accepting nature of it but generally we're just going to call it a tableau so a tableau is really an accepting computation history for um for for the machine on an accepting branch of its non-deterministic computation so if m accepts w it's got to have some accepting branch and the tableau is going to be the sequence of configurations that the machine goes through on that accepting branch if there are several accepting branches there may be several tableaus there will there will be several tableaus so there's going to be one tableau for each accepting branch of the of the machine's computation on w if the machine does not accept w there won't be any tableaus and so the whole point is that we're going to make our formula and represent the statement that there is a tableau and satisfying that formula is going to correspond to filling out the symbols in in the tableau to make it a tableau okay so here is a tableau um so tableau is just again an accepting computation history on some branch some accepting branch of the machine's computation so and and it the rows of the ins instead of writing the computation history out linearly we're going to represent it in a table form where each configuration is going to be on a separate row all right now the dimensions of that table are going to be n to the k by n to the k because the machine runs for n to the k steps so there's gonna if there's an accepting branch it's going to accept within that number of steps and we'll have enough rows here to write down all of the configurations that the machine goes through one after the next row by row each one having a configuration uh in it and uh and then at the bottom there'll be an accept you know minor detail if the machine accepts earlier we'll just say the machine stays in the set once it's it enters an accept the machine does not change from that point on so the rule of the machine is nothing changes and it just remains in the same configuration from that point on okay um so um important to understand what we're you know if you're if you're not following what i mean by a tableau you're doomed for this lecture so it it makes sense if you know for you to ask a question to understand what we mean by tableau okay so just a few more elements here so this is going to be the start configuration for mw this would be an accepting configuration here down at the very last row um and you might imagine you know i i think i filled out some hypothetical first step of the machine after the start where maybe the machine was a when it's in remember how we can encode our configurations so this is the machine is in state q0 looking at w uh the first symbol of the input w1 and maybe when it's in q0 or looking in that first symbol it moves to state q 7 and goes right and then changes that w 1 to an a and so now here is the head shown moved 1 position to the right in the new state q7 i mean of course that depends on what the machine is designed to do what the what the transition function is but this um you know could possibly is what happens so does the okay so good does a tableau trace all steps of all branches no the tableau corresponds to one accepting branch each different accepting branch is going to have its own tableau so there might be several different ways of filling out that second row even of course the first row is going to be in any tableau it's got to be the same um you know once i know w here i've written down maybe i should unpack this for you you know it's in the start state here then here are the first you know here are the n symbols of w w is of length n so it's w 1 w 2 up to w n and then i'm padding out the rest with blanks so i should have said that too but okay um so i want to make my table into the k by n to the k because that's going to be enough to represent the entire uh um all of these configurations of m uh running for most into the k steps right because the machine um even if it sends its head moving to the right as fast as it possibly can it's never going to have um it's it's never going to go outside this box if it only runs for n to the k steps so this is going to be big enough to represent the entire computation of m assuming it's running for n to the k time and the input w is of like that all right um what is k so k is the running time of the machine so we assumed from the previous slide that m runs in time n to the k okay um so a good question here how can the tableau be a square table if we have a low number of computation histories but a lot of tape um well i mean the low number of computation history that that's a problem you're not saying that well you know there each computation history is in a different table so this i'm representing a single computation history here there may be there are going to be many configurations um so i can't have a small number of configurations using a lot of tape because you know i can only use one more cell in one more one additional tape cell each time i meant you know if each additional step of the machine is yes so there's really no difference between this if you want to think of this as a computation history that's fine this is really just the standard terminology that's used when you're proving um this this particular theorem typically people talk about it as a tableau but it's really just the computation history um the cue state i mean the question about the what are these states queue this is the way we represent configurations so this means the machine is in uh state they're not all going down the diagonal the the states are going to zigzag here through he through this picture here depending upon how the head of the machine moves so you have to go back and review how we represented configurations of the machine remember the configuration is a snapshot of the machine at a given point um how do we know that m runs in polynomial time we're assuming m runs in polynomial we started off with a language that's in np so it's it's a non-deterministic polynomial time machine and we're picking one branch that accepts and writing down all the sequence configurations the machine goes through let's move on and maybe as you know ask more questions as they come to you and i'll pick out some to answer if that's going to be helpful to others um okay um so uh we're going to now construct this um uh formula to say that m accepts w again that was what we uh that's our goal and it says that a tableau for m on w exists um and basically what that means is we want to say that it starts right it ends right and everything in between is right and then we're going to need some other stuff to talk about how we're going to be encoding those symbols using boolean variables so those are going to be the four parts here's the start starts right here it ends right here it moves right and here it talks about the encoding of the symbols into boolean variables so those are the four parts of this formula that i'm going to describe um i hope you got your question answered as to why the total number of columns is into the k because it's just big enough to fit the entire all configurations if the is running for at most end of the case steps which is what we're assuming okay okay so now just getting back to that i'm going to describe these different components now on separate slots um let me start off with with this component fee cell which is sort of the most fundamental one because it talks about how we're going to be encoding those symbols of the tableau into the boolean variables so again here's kind of our the picture to have in mind uh of this tableau this is this into the k by enter the k table representing um you know the some accepting branch of the machine's computation if there is one and so now let me draw a some one of the cells here i'm going to magnify it so this is the i j cell here and le i'm going to there there are going to be a collection of boolean variables associated with each one of the cells so each one of the cells is going to have a bunch of variables all to itself and those are going to be basically indicator variables they're going to indicate which symbol that cell gets to uh have in it okay so again picture here um the in this tableau you know we don't know how it's going to get filled out but however it gets filled out each one of these cells gets some symbol um and that symbol could either be a tape alphabet symbol or it's symbol representing a state right that's the way we do our configurations so it could be a tape alphabet symbol here showing the magnification maybe it's the tape alphabet symbol which represents the blank symbol or maybe it's a represents some state okay now how am i going to encode that with variables so this let me this is the collection of variables that's going to apply to the entire formula fees of mw um each cell is so each cell ij is going to have um a set of variables one for each possible symbol sigma that's um in the configuration alphabet namely a tape symbol or a state symbol okay um [Music] so i'm gonna have well maybe this will become clear as i'm writing it so if i turn the the uh the variable x i j sigma equal to true that's just a way of saying that cell i j contains a sigma so if i have x you know i j a that means the symbol contains an a so i'm going to illustrate that now for you so imagine you have lights representing all the different x i js x i js for the different sigmas i didn't say that well so um we're in cell i j so these are all x i j variables they're x all x i j sigma variables for the different sigmas that can go in that can go in this uh cell so all different possible sigmas so this is gamma union q okay so now um if i have an a here in that cell so then the variable extra a is true and i'm just helping you visualize that i'm going to that's going to correspond to turning the light on there's going to be a light associated with each one of these variables and it's be turned on when that variable is true similarly if i have the blank symbol is the thing that goes in in that cell um then that variable gets turned on i hope you can see it maybe it's a little bit small on your screen um the x i j blank a variable is is true and similarly if it's q seven here the x i j q seven variable is true okay so that that's the way we're going to be encoding the contents of these cells using these indicator variables and now we have to start making some boolean logic to make sure that those variables reasonably represent the cell con you know contents of these cells so for example what would be the first thing that comes to your mind well we better not have two lights going on in any one of the cells because then we have two symbols on that cells and that's not allowed each cell is going we want each cell to have exactly one symbol and that corresponds to each each each cell having exactly one of its lights turned on or equivalently each cell should have exactly one of the variables be true that's the very first part of the formula is just going to say that so let's let me show you how that what that looks like okay so here we're talking about this free cell it says that there's exactly one light on purcell or in other words exactly one of the x ij sigmas is true for each ij um so this is how i'm going to actually express that using my boolean formula uh i'm sort of color coding the different parts of the formula which i'm writing out to you here in english so first i want to say there's a so in every cell there's at least one light that's on and there's almost one light that's on um so here's the green part this is going to say at least one light is on so i'm going to say that by taking all of the very all of the symbols that can appear in that cell and and take taking in an or over all of those different associated variables so it's either got the first symbol on or the second symbol is there or the or dot or the last symbol is there one of those has got to be there i'm going to write this using my big or notation so for sigma appearing in this set of possibilities one of those variables has got to be on at least that's what this or big word tells you now i want to make sure that there's at most one that's on so that there are not there's one at least one on but there are not two that are on so i'm gonna have an additional part of the formula here which says and i hope you can read this a little small um if i have two different symbols sigma and tau that are configuration uh possible configuration symbols where sigma and tau are not equal so that's i'm reading it out to you if you can't if it's too small for your screen then i'm going to say it's not possible so i have the negation of x i j sigma and x i j tau so saying in another way it's not the case that that cell contains both sigma and tau for any two symbols sigma and tau as long as they're different i'm going to take these two formulas and add them together and this tells me in the cell ij is exactly one of the variables is true exactly one of the lights is on and that's going to represent which symbol goes into that cell and then i want to take the and over all possible cells to make sure that i'm going to now apply that everywhere and so i now i do an and for i and j ranging between 1 and n to the k to apply this logic throughout the picture yes sigma union asking sigma union q contains the input output yes this is not a sigma this is a gamma gamma is the tape alphabet this is any symbol including an input symbol um from sigma is going to be in in gamma so this is any any symbol that can appear on the tape um okay and that's this this expression here that is the expression feasible cell okay so here's a little check-in for you but maybe before we jump into the check and let's just make sure maybe better to take some questions and then we can ask a check-in for you you understand i mean if you're not getting this you should you should try to uh try to figure out how to get it because this is really just the foundation it it only gets more it's not it's not a very complicated proof once you sort of get the idea what's going on but if you're not getting this part you won't be able to get the rest okay i have no idea what what that means but i'll read it out to you guys this looks like a one-hot tensor encoding same from common same form commonly used in ml okay um it's just an indicator i would just call them indicator variables um why is it sigma union q for the big and here sigma union q you mean gamma union q this is wrong okay now i understand why everybody's upset this should be a gamma not a sigma uh there's a booboo sorry about that i don't know if i can fix that without wrecking the whole slide but so i'm not going to even try this symbol here should be gamma not sigma it's a typo thanks for catching that um yes so the question is um okay so uh feasible is just trying to make sure that the encoding represents setting a bunch of symbols into the tableau not so each cell is going to have one symbol exactly not two not zero um so that's what pieces so if you've satisfied if you set the variables to satisfy feasible then there's going to be one symbol on in each one in one symbol in each of those cells um now another question this is not a cnf no this is not a cnf that's the second half of the that's going to be if we hope we don't run out of time but i have a way of converting general sat formulas to cnfs um and preserving satisfiability so we're gonna do that reduction afterward okay let's let's is it a little check to see if you understand uh at some some level what's going on how many variables does this formula actually have is it order n order n square enter the k remember this k is the running time of the machine uh or n to the 2k what do you think so i i i mean for how many variables i mean that in all of uh m sub m yeah how many variables do we have all together in this formula if that's what the question is and here though here are the variables so describing them here x i j sigma okay i'm going to close this so pick something all right ending polling one two three okay yeah so um okay so that's a good question um so first of all the the correct answer is in fact d it's order n to the 2k um now uh i'm getting some questions about what about the size of gamma and q well those are going to be fixed they depend only on the machine but they don't depend on n so thinking about it in term functionally in terms of n that's going to be a constant multiplier um and so it's going to be absorbed within the um the big o uh so that's why we have uh you know these are constant relative to n these are fixed um they're not into the k possible symbols there's a fixed number of symbols depends only on the machine so looking at a particular machine and what happens when you look at large inputs so why is it d and not c well don't forget how big is this table um this is n to the k by n to the k so there are n's n to the 2k cell into the k quantity squared or enter the 2k cells here um and so there's a collection of variables for each cell some fixed number variables for each cell so that's why it's order n to the 2k good okay so let's uh let's move on i think we're actually oh no um okay so we have one more slide and i think then we have a break after uh so now let's next uh talk about constructing um two more pieces of the fisa bmw formula so we already got fisa face of cell done let's look at fee sub start and fee sub except and fistart is going to tell us that the start configuration has exactly these symbols and the accept tells us that the bottom configuration contains an accepting state somewhere so how we're going to write that down well first of all i'm going to write these down just cell by cell so first of all fee start is going to say the cell 1 1 contains the q 0. i mean i know what the start configuration is it should be because you know i you know thinking of me or think you know us as the reduction the reduction is given m it's given w so it knows what the start state is it knows what the symbols of w are so it can it knows what that start configuration is it's just q zero followed by the n symbols of w followed by blanks so it wants it wants the very first um cell in the in the left-hand corner here to be a q0 the start state of the machine which it knows so it's going to say x sub 1 1 q 0 that has to be turned on so it's going to be and a bunch of a bunch of variables here and in order to satisfy fee sub start all of those variables have to be set to true so that means we have to have a q0 in that cell so now we're going to do the next the next cell here the one two the next cell of the uh uh start configuration so fee start is going to have x12 so that's the next place contains w1 and so on x13 contains w2 all the way up to wn and then there's going to be a bunch of additional um uh parts which say that the that we have blanks in the rest just spelling out exactly all of the symbols in that top row because that's what the fiestart um uh formula or sub formula of the of the overall formula we're making looks like okay now let's take a look at fee accept if you accept because i'm just looking for cue accept to appear somewhere in that bottom row i'm going to do that in terms of an or so here is the variables now notice have n to the k because it's the last row in the table so rho n to the k here and then i'm going to vary j from 1 to n to the k so j the column number is going to range from 1 to n to the k and i'm looking for that q except so x enter the kj which is where j is varying and q accept one of those has to be true one of those has to be turned on and so that's why it's a it's a big war and that's my um uh fee accept peace okay and now we'll take a little break and feel free to ask me some more questions um let me just start the our clock and go grab yourself some coffee or ask me some questions happy to answer them why don't we check that q accept only appears once is it impossible for q except to appear twice that's a great question um so that would definitely be a broken configuration if that happened because a configuration can ha can have must have exactly one state symbol appearing the way we're going to enforce that is is with the fee move uh part of the formula which we haven't seen yet so p move is going to guarantee that the machine is acting correctly so that all of the rows of the tableau are all legal configurations and they all legally follow from the previous so that really in a sense the hard heavy lifting is coming in fee move but it's really not that bad um somebody says out of curiosity how close is this intuition proof to the actual proof this is the actual proof there's no i'm not i'm not hiding anything i mean you know we're being a little uh loose here but you can turn this this is this not cutting any corners here this is exactly how the proof goes um and so uh not not uh you're getting you're getting the real deal here somebody wanted to see the previous slide so here we are whoops is there something you want to ask um okay so fee cell says there's exactly one symbol per cell and the variables are set in a way in terms of thinking of them as indicator variables there's exactly one variable set to true in each cell so it's exactly one symbol per cell that's what p cell the f cell tells you what you know if you don't have that then you have a mess so you've got to start with that and then with the other other things are going to be additional conditions which when satisfied going to enforce the rest of the properties that we want why would the proof fail if we n to the k with 2 to the n to the k so okay i i presume you know where does would this use the polynomial running time of the machine uh of the of m the nondeterministic machine i mean if you had a an enormous tab we have to show ultimately that this reduction is a polynomial time reduction and that's going to depend on how big the tableau is because that's going to tell us how big the formula we're producing is fees of mw if e sub mw is exponentially big we don't have a prayer of being able to output that formula in polynomial time if there were less than into the k steps do we repeat the last configuration yeah that's what i said if we have if the machine ends early the last configuration just stays there so we're going to modify the definition of the machine slightly so that you know the um so it just stays yeah um okay um yeah let me not take the other there's a bunch of other questions some of them a little on the technical side um let me uh maybe i'll try to address them as they come along if it if it turns out to work to do that okay so uh the word the break is over um why don't we uh con is it possibly that the encoding configuration will not fit in into the k no no so the question is is there a possibility that the encoding of the configuration won't fit into the k if the machine runs for enter the k steps the the configuration has to fit with an end to the k because it can't use anything more within into the k steps to think about it but no the ques the answer is there's the configurations if the machine runs time into the k that whole um tableau is big enough to write down the entire computation history all right so let's continue fees of move um this is in a sense the part which is going to tell us that we started right we ended right um fee cell says every every cell contains one symbol and now we have to say that the whole interior is correct how are we going to do that so these these are the parts we've already done and the way i'm going to describe that is in terms of these kind of little windows i'm calling neighborhoods so imagine here we have a two by three rectangle which i'm going to call a two by three neighborhood and what i'm going to argue but i'm not going to prove uh here i'm just going to really state it but it's it's really just a sort of more or less obvious fact uh but the proof the book has the formal proof that if any every one of these here is legitimate is legal according to the rules of the machine if every single imagine you know you have these um oops let me put myself back on here so you can see me um if you have um if you have here every um two by three window you can take this as a window and you slide that over the entire picture of the tableau and everything here looks okay as far as the running of the machine so i'll say what that means in a second but if if everything looks locally fine everywhere then the whole tableau has to be a valid tableau in terms of the rules of m um maybe it's easier if i if i describe what i mean by these being legal um so these neighborhoods um these two by three neighborhoods are legal if they're consistent with m's transition function so i'm going to describe rather than i mean to do this formally i would have to go through kind of a process that we went like what we did when we went through uh the construction for the post correspondence problem and see if the machine moves left this thing's happens every move is right that's i think that's it's not really necessary you can kind of get the idea very clearly by doing it a little bit at a higher level um so let's look at what i mean by a legal neighborhood so this illegal neighborhood is a setting of the values the six values of this 2x3 neighborhood in a way which doesn't violate m's rules so for example if m when it's in state q7 reading a b goes into state q3 and moves left then this would be a legal neighborhood because it shows the head moving left um the b becoming a c so reading a b i should also say that it right converts that b to a c and and moves his head left into state q3 so this would be a legal neighborhood if that's the way so being legal depends upon um it depends on the transition function of the machine so given the transition function that's going to tell you which of the legal neighborhoods so another legal neighborhood this would always be a legal neighborhood is that if nothing changes so that means the head of the machine was somewhere else and so um you know whatever was on the tape in this step is going to be the same stuff in the in those places one step later um here's another possible legal neighborhood um is uh if the head suddenly appears you know on one of the on one of the cells either in the left or the right that would correspond to the machine to the machine moving its head from somewhere off the nape off the neighborhood into the neighborhood you know in that step um so this could be a legal neighborhood provided the machine actually does move its head left into a state q5 at some point under under some conditions um and here is another kind of a weird illegal neighborhood if you have a b c and then the a changes to a d that could also be a legal neighborhood if the machine transition function allows an a to get converted to a d when there is some machine when there is some state reading that a and that state also moves its head left so it doesn't move into this picture so those are examples of legal neighborhoods let me show you some illegal neighborhoods um just i'm doing this this is kind of a proof by example now this is perhaps the most intuitive part but i claim that this is easy to turn this into something airtight and formal so this would be clearly illegal if you're if you have a piece of the tape in the pre previous step where it's a b c and then suddenly the b changes to a d the symbol on the tape changes out of nowhere without having a head nearby uh to a different uh to something else that could never happen so that would be illegal another thing would be illegal is if a state appears from nowhere that could never happen um or if it just disappears that could never happen and here's another here's an interesting one if a state becomes two states um don't forget the machine is non-deterministic so the machine in principle could move its head left on one branch and move its head right on a different branch but those would have to be in different tableaus they can't be in the same tableau because that doesn't correspond to any of the threads of the computation those are multiple threads um and i say this because if you think about my claim which is going to put down over here that if every 2x3 neighborhood is legal then the tableau overall corresponds to a computation history this illustrates why it's not enough to have a two by two two by two neighborhood where you really need the two by three um because if this was a two by two neighborhood if you just look at these four this left most uh two by two that could be um a legal neighborhood if it was a two by two if the rules of the machine allowed for that and the right four box right four cells could also be a legal neighborhood so you could have something that looks okay from the perspective of two by two neighborhoods but globally in terms of the overall tableau is completely nonsensical because it has multiple heads um but if you have a three two by three neighborhood it's big enough to prevent this situation from occurring and then you know you can check the details uh and i think it's very plausible that it guarantees that the overall tableau is is is legitimate if all of the two by three neighborhoods are legal okay and so that's what we're going to turn into a um a boolean expression we're going to say for each cell that the set for each neighborhood so here's a neighborhood at the iga location that's i'm calling this sort of this position here sort of the home location for that um neighborhood for each um for each neighborhood uh it has to be set to one of the legal uh possibilities and there's again only a fixed number of those because there's a fixed number of possible symbols that can appear in those cells um so this is that fixed number to the sixth power at most and i'm going to say that um you know the cell in the upper left which would be this one is an r and this one and this one here is an s and this one is a t and this and this one is a v if you just trace down what the indices are telling you it says that that piece of the tape that piece of the tableau here is set according to one of the possible legal settings and i'm just going to or over all of those possible fixed number of legal settings and then i take a an and over all possible tape cells over all possible neighborhoods and so that's going to be my fee move and that's that's it okay let's see can i explain again the third example of illegal so this one over here i presume is i'm being asked about well you know if the machine is in a state q7 you know reading a c you know the head has to move either left or right so that at the next you know the next um configuration there's got to be a state symbol appearing either in this cell or in this cell and here the tape you know the the the the head has basically just vanished with nowhere to you know it's it's gone that could not happen in the in the you know according to the rules of the machine the way we talk about turing machines so that's not possible so this would be an illegal um a neighborhood you want to prevent any of the ill bad stuff from happening anywhere in here it's only good stuff can be happening locally and that guarantees the overall picture is okay um do we have to check that the head doesn't leave the tabo from the left most or right yeah there are some little details here like that so the question is you know have to make sure that that the yeah you probably need to mark i think the book probably does this correctly you may have to mark the left and right ends to make sure that i mean the right end can is not a problem because the the the the machine can never go um off the right end and if you design the machine so that it never moves its head off the left end either which you can do then you wouldn't have to worry about that possibility but yeah otherwise you would have to put some sort of delimiter here to uh to enforce the mach the head not moving off the left end so there are some details like that too um there will be two heads in the same row no this can i don't know what you somebody says there will be two heads in the same row please elaborate because this is designed not to allow two heads in the same row could i go over the or for legal again okay the or so the the big or here what i have in mind is i i take there's gonna be first of all i look at the machine and i look at the transition function and based on that i write down the list of all the legal two by three neighborhoods so all the settings which correspond to legal two way two by three neighborhoods there's gonna be some fixed number of those you know a hundred there's a hundred possible legal neighborhoods of which you know i've written down here four but maybe there's some some number say a hundred so now there's going to be an ore over those hundred different possibilities it's either going to be this legal neighborhood or some other legal neighborhood or some other legal neighborhood and for each one of those legal neighborhoods i'm going to say well the variables are set according to that legal neighborhood or the the variables are set according to the next legal neighborhood or the variables are set according to the next legal neighborhood and do that 100 times one of those has got to end up it's an ore so one of them has to work otherwise the formula fails and will be false because you're going to now end that over all of the neighborhoods in the picture is it possible to have a head on the far left of the configuration and one on the far right you mean a head over here and a head over there i mean how did the head get there can't happen um you know the head the head has to come from a head above it um if you're going to be worrying about the you know details of the boundaries here you know all that's fixable so let's not you know um lose sight of the main idea i mean if you understand the main idea you can fix little details so i want to make sure you understand the main idea what's happening okay so let's finish up this proof um so in summary um we gave a reduction from a to sat this is what we needed it was in those four pieces and you really just need to argue that that formula we're building is not too big it's and it's going to be basically the size of the tableau if you look at what we constructed the number of variables is roughly the size of the tableau and the amount of logic that we're putting into the formula is also going to be a fixed amount of logic independent of n for each of the variables in that tableau and now somebody asked me about like this how big the indices are um you know the indices for the you know the x i value the inj values um technically they're going to be numbers between 1 and n to the k so you're going to have to write those down and so that's going to be a slight additional logarithmic cost um to write those things down but it's not really that interesting a point um and so the overall f is going to be computable in polynomial time because the output is not very big and it's also not complicated to write the output down so that's the end of the that's the end of the proof um i can take a couple of questions we why can't we just check that the whole this is a good question why can't we just check that the whole row is legal um you know you can check that you can check that a row actually is a is a configuration but you to check that the row follows from the previous row um you know ultimately the operation of a turing machine is is a local thing i mean hey you know the way the the the way it moves from one configuration to the next depends locally on how the those you know where the head is and so really that's just another way of the way i'm saying it is just really checking the whole configuration but just doing it locally i don't know if that's satisfying to you um okay uh why don't i move on because i just want to make sure we have enough time to get to the very last part which is a little bit i'm afraid a little technical um so we're going to kind of shift gears now and talk about reducing sat to three set and um let's let's see how it goes um i i don't always have the most success with with presenting this uh this little piece because it's slightly a technical argument but um if you don't get it don't worry just you have to accept that it's true um but i'd like to show it to you just to makes the whole presentation uh you know complete in that sense okay um so i'm going to give a reduction that maps general formulas to three cnf formulas so that's how i map sat to if you remember threesat is satisfiability but for three cnfs so it's conjunctive normal form you're in the form of those clauses which are anded together and each clause is an or of a bunch of literals which are variables or negated variables so i want to convert phi to phi prime um which is a three cnf formula but preserve the satisfiability feed prime is not going to be logically equivalent to fee because i you know i could do that too i can convert any formula to a logically equivalent cnf formula maybe not even a three second of it but yeah you won't be able to get a 3c enough but you can get a c and f but um it might be exponentially larger and that's not good enough i have to do the reduction in polynomial time so i can't generate a much larger formula that's exponentially larger and so i'm going to do that by adding additional variables so it won't be logically equivalent because the new formula is going to have additional variables in it okay i'm going to kind of do it by example and let's see how that goes so his fee which is not in 3c and f it's not even in cnf because it's got you know uh orrs of ands appearing which are not allowed to happen in a cnf okay so how we're going to convert that into a 3cnf formula preserving the satisfiability and just working it through with this example i hope at least give you some idea of how you of how to do the conversion in general so first of all i'm going to represent this formula as a tree using its natural tree structure so you understand so a and b becomes a and b written as a tree and then i take or that with c so i get the tree structure here in sort of the natural way and i'm going to label all of these intermediate nodes which are associated now with operations and um i'm assuming also that the formula is fully parenthesized so that each operation i'm only thinking about is applying just a two it's a binary operation um and let's let's ignore negations for the minute because negations you can always push those through down to the leaves but it's just going to make it too complicated so negations turn out not to be a problem so there's only going to be negations at the level of the inputs not at the not negation operations in the middle okay so we have this tree structure here and now i'm going to use these two logical facts you know and i don't know if you know you've probably all seen hands and ors i hope otherwise it's going to be really tough uh but there's also other logical operators such as the implication operator where you have a implies b thought of as a logical operator operation um and so that this is only this requires that if a is true then b is true um however if a is false b can be anything uh and similarly if b is true a can be anything um the only thing that's prohibited is that if a is true and b is false that's the only thing that's uh would be invalid and so if you think about it that's going to be equivalent to saying that either a is false or b is true one of those has to be and that's going to be logically equivalent to saying that a implies b another logical equivalence maybe is more familiar to you is just simply de morgan's laws the mortgage law which says that if you have the a not of a and b that's uh equivalent to saying uh the not of a or the knot of b okay i'm going to make use of both of these um now now here i want to i ran out of room on the slide so i'm going to take myself out of the picture here for for a minute um uh i had no place else to put this so here we have if you're going to think of the and in terms of its truth table so here's a and b in terms of a and b so if one and one is one but all other settings of a and b uh yield zero for the and i'm going to represent those um if you imagine a and b is going to be called c i'm going to represent this information with uh four um small formulas which taken together you and them together kind of force c to have the correct behavior associated with a and b okay um so uh so if a and b are both one and c is one if a is zero and b is one um then that so then it forces c to be zero um and similarly you know every other setting besides a and b being being true uh for c to be false which is what you want when you have and so i'm gonna write this expression here down with z1 being in the place of c by just taking those four uh expressions and ending them together so this is exactly those same four expressions written out linearly now i want to do the same thing for z2 but now that's written in terms of an or so it's a slightly different truth table here up in this corner uh so now if either one is a one we get a one result and so now you know if a a and b are true you get uh c is true however if a is true and b is false that still implies c is true um so i'm going to write down uh those rules for specifying how uh z2 must be set and each one of these things is going to get converted into clauses three clauses with three literals using these uh using these um using these rules over here so i'm going to do that for each zi and lastly to make sure the whole thing is satisfied which means there's an output of one here i'm going to have one clause associated which says that z4 the output is a one now i can convert all of those when i have a and b imply c that's logically equivalent to not a or not b or or c and the way you can see that is really by repeated app by application of these uh rules here running a little low on time so maybe you'll just have to check this offline but quickly a and b implies c using the first equivalence is the not of this part or c and then i can use the morgan to convert that uh knot of an and to an or of the knots and then and then i can remove the parenthesis because uh or is associative and so i get a uh a clause which is what i need so each one of these guys is really equivalent to a clause and so i just get a bunch of clauses and actually technically this needs to be three um copy of three things here should it should be z4 or z4 or z4 which is a lot um so checking numbers so i realize my checking is broken because i only realized that last point just now as i was talking so uh the the actual uh value that you get in terms of oh no the number of clauses is correct no i i i take it back this is fine so if you understood what i was saying hopefully you can see how big the formula fif fee prime is in terms of the number of operations and fee so let's let's see how many people get that i i i acknowledge this maybe a little on the technical side okay i'm gonna close it close it down please enter your value okay yeah the correct answer is 4k plus one um because each one of these operations is going to end up being a row in this picture it's going each operation is going to have a variable associated to it it's going to become a row in this picture and so then and each row is going to have four um clauses which define um you know uh what you need set what you need in order to force the that that that variable to have the right value corresponding to that operation okay and so then you need and you need one extra clause here uh for uh saying that the the this whole thing evaluates to true um okay so that's all i wanted to uh do today um we proved those two main theorems and now we know that there are np-complete problems and all of the other problems that we can get from by reductions from these problems um are also going to be np-complete as long as they're an np okay so um that's that's it feel free to put some questions into the chat or move on to whatever else you're going to be doing next okay so a good question here is why is fee prime not logically equivalent with this construction can't be logically equivalent has logically equivalent means that it gives you exactly the same function as you if you set the variables in the same way you get the same result coming out well v prime has more variables than phi does so it wouldn't even make sense to talk about logical equivalence because there are two functions on different numbers of variables so in that sense it doesn't really make sense you know what you could say is that for every setting of the of the overlapping variables of the variables that appear in both ph and v prime so those are the original variables of fee um there's going to exist some setting of the new variables which is going to make um the uh you know yeah there's going to be there will exist some setting of the new variables which will make the the two formulas agree but that's not the definition of logic logical equivalence okay so why going back to the proof of status the satisfiability proof and the legal neighborhoods could i go over why the number of legal neighborhoods is polynomial the number of legal neighborhoods is not only polynomial it's constant it depends only on the machine it does not depend on n so because each cell can have at most some fixed number of you know can have the number of tape symbols plus the number of state symbols that's you know that depends on the machine only so uh now we have six six tape cells for the six cells in a two by three neighborhood so you're going to have that number to the sixth power but it's still it's a constant to the sixth power still a constant doesn't depend on n so it's not a question of even being polynomial um it's it's a it's a constant value it's a constant multiplier uh if you want to think about it in terms of the um the size of the formula that's going to result don't forget you know we're trying to make a formula um which is uh well the reduction has to be polynomial um it's a polynomial time reduction so that means that as n increases the time to calculate the reduction increases as a polynomial but we're fixing m so m does not change so therefore anything that depends on m only is just going to be a constant um impact on the formula it's not going to be it doesn't depend on it okay everybody bye-bye see you you 2 00:00:26,550 --> 00:00:28,830 okay everybody let's begin um welcome back good to see you all here on zoom um so we're going to pick up with what we had been discussing um last week which was um a an introduction to np completeness so we're following on our description of time complexity we we started talking about the time complexity classes the class p the non-deterministic classes the class np p versus np problem and then leading to this discussion of np completeness and today we're going to prove the big theorem in the field uh which really kind of got things going uh back in the early 1970s by cook and levin that uh there was actually an mp complete problem called that sat in particular is an np complete problem and then we'll also talk about threesat which is a useful tool um so just to remember just to uh remember we had this notion of np completeness for languages that be complete if it's in np and everything else in np is polynomial time reducible to it and if an np complete problem turns out to have a polynomial time solution then every np problem has a polynomial time solution and that's part of the importance of np completeness uh because since we uh consider it unlikely that p equals np uh and that there probably are some problems that uh are in np but are not solvable in polynomial time that's would me imply that an np-complete problem would have that property and so proving a problem and being mp complete is evidence very strong evidence that it doesn't have a polynomial time uh solution and so therefore it's we call intractable uh it's a very difficult problem um so um the way we are going to typically show problems np-complete is by reducing a known a previously known np-complete problem to that problem often it's three-sat as we've seen several examples already or it could be some other uh example so let's just survey briefly uh the things that we've already languages that we've already seen which are mp complete so we have the languages sat which is a direct reduction from every np problem and so we're going to show that today that every np language is polynomial time reducible to sat which is in turn reducible to three sat and we showed previously that threesat is reducible to clique and hampf and in recitation if you went to that they showed that the subset sub problem and the undirected hand path problems are also reducible from previously uh shown well from uh from either from 3sat or in the case of their undirected hand path problem it's reducible from the hand path problem and the conclusion once we have these two blue reducibilities shown then we know that all of these problems are np-complete um so the class np basically breaks down into the np complete problems p problems problems that are in p and then there might be problems in between as well so there are some problems that are not known to be in either category and in fact there's a old theorem which shows that if p is different from np then there are actually problems that are in the intermediate state and um of course it's possible that p equals np and then everything collapses down uh to be the same with the tiny exception of the um sigma star and empty set languages which can never be uh complete um okay um so that's our quick review here's a check in that i'm gonna use to get us ready for the big proof that we're gonna be spending most of the lecture on about um uh showing sat as np-complete but just to define a little notation which you might have maybe you've seen already um but i'm going to do it in the form of a check-in so i'm sure you've all seen the big sum notation uh using a sigma to represent a sum over um over some set of possibilities just as there is the big sigma notation we have you can have other operations um that apply to a a set of uh elements so in this case we're going to be seeing the big and and the big or operation uh it's going to a notation which is going to allow us to talk about taking the end of many things or the ore of many things um because we're we're going to be building these boolean formulas and so ands and orders are going to be the uh operations that we're going to be focusing on and so just as an example just to make sure you're we're all understanding this notation if you have uh two strings x and y uh written out in terms of their individual symbols so they're both of length n so x is x1 to xn y is y 1 to y n and now i write the following expression the big and for i ranging between 1 and n of x i equal to y i if that big and is true what does it tell us about x and y and i'm just going to offer you two possibilities that either x and y agree in some symbol or they are equal namely that they agree on every symbol and so let's just pull that up as a quick poll to get us going here um [Music] i just want to make sure you understand the notation because we're going to be using that a lot in describing the uh polynomial time reduction from uh languages in np to the sat language okay i think most of you have got the idea but so let's finish this up quick um see another 15 seconds please just to give you a chance to participate um okay so i'm gonna close this uh close this poll last chance let's call all right so yes the big and as most of you can see the big and says that you know x1 equals y1 and x2 equals y2 and x3 equals y3 so they're all um every symbol in x is equal to the corresponding symbol in y if we had a big or instead of a big and the uh then a would be the correct answer because then just there would be some place where they agree instead of every place where they they agree um okay so let's um then start to launch in to this big theorem the proof itself is a massive details um with one underlying idea and in fact it's an idea we've seen before um but let's before we get ahead of ourselves let's just understand what we're trying to do uh so we want to show that this language sat is np-complete remember what's that is it's boolean formulas that are satisfiable uh so we already showed that the set pro so being mp complete means it has these two features it's in np and everything in np is reducible to it so first of all sat is np as we've already seen the witness that shows a formula is satisfiable is simply the satisfying assignment that evaluates the true so now we're going to pick some language in np a and show that a is polynomial time reducible to okay so this is going to apply to any language a and np so let a be some language in np it's decided by some non-deterministic turing machine m in time into the k that's what it means to be an np um i'm going to make ignoring the constant factors when we could carry that throughout the proof it just would make this the the description a little more cumbersome and wouldn't change any of the ideas but so let's just say it's m runs in time into the k and recognizes or decides this language a in uh uh so it's an it's in and it's a non-deterministic machine so we've got to give a polynomial time reduction from a to sad that's that's what i have to demonstrate to you so what does that reduction look like it's going to map strings which may or may not be an a to formulas which may or may not be satisfied that's what the reduction's got to do so saying saying that out more formally is it's going to take some string w and map it to a formula we'll call f sub m w where m is the machine that um decides a and w is is is the input so f is going to map w to this formula where the string is in the language exactly when the formula is satisfiable and the formula is going to depend upon m and w that's why it just has it's written in this way so we've seen that kind of thing before so basically my my job you know you know we have this language a that's an np and now we have some string w which might be an a or maybe not is an a and i have to quickly in polynomial time now produce a formula which is going to be satisfiable exactly when the string w is in a okay so how is that formula going to work how can i produce such a formula which i mean i of course i don't know whether w is in a or not um because i i'm just a polynomial time reduction and um a is an np language so polynomial time probably is not enough to solve whether w is in a so i've got to make matt do that mapping without knowing the answer and the idea is that um and this is where you know we've seen things like this before the formula we're going to construct simulates the machine on w so in some way it's going to do that simulation the trick is figuring out how to what that means um but the the interpretation of that formula is that in a sense describes and it says though in my informal language that m accepts w very much like you know there's a lot of parallel here between um this construction and for example the construction um the pcp construction where we made a an instant you know given a machine and an input we made a set of those dominoes where finding a match forced you to simulate the machine here finding a satisfying assignment is going to force you to simulate the machine [Music] so this satisfying assignment is going to be a computation history for mmw and i'm just going to write that computation history in a particular way so it's going to have a somewhat different encoding which is going to help us to visualize what's going on better okay um so that's that's the that's the that's the approach that's the basic idea of what we're trying to what we're gonna do so i'm happy to take a minute if that's if you have any questions about this part but um otherwise i'll just move on to start doing the actual construction of what the formula looks like how is it going to do that how is that formula going to work okay so no questions hopefully we are uh all together okay so first of all let me describe what my computation histories are going to look like and and the situation is a little bit different than what we had before because um when we were talking about the post correspondence problem we had a deterministic machine and now our machine is non-deterministic okay so um we're going to call the object instead of an accepting computation or computation history we're just going to we're going to call it a tableau or sometimes an accepting tableau if you want to emphasize the accepting nature of it but generally we're just going to call it a tableau so a tableau is really an accepting computation history for um for for the machine on an accepting branch of its non-deterministic computation so if m accepts w it's got to have some accepting branch and the tableau is going to be the sequence of configurations that the machine goes through on that accepting branch if there are several accepting branches there may be several tableaus there will there will be several tableaus so there's going to be one tableau for each accepting branch of the of the machine's computation on w if the machine does not accept w there won't be any tableaus and so the whole point is that we're going to make our formula and represent the statement that there is a tableau and satisfying that formula is going to correspond to filling out the symbols in in the tableau to make it a tableau okay so here is a tableau um so tableau is just again an accepting computation history on some branch some accepting branch of the machine's computation so and and it the rows of the ins instead of writing the computation history out linearly we're going to represent it in a table form where each configuration is going to be on a separate row all right now the dimensions of that table are going to be n to the k by n to the k because the machine runs for n to the k steps so there's gonna if there's an accepting branch it's going to accept within that number of steps and we'll have enough rows here to write down all of the configurations that the machine goes through one after the next row by row each one having a configuration uh in it and uh and then at the bottom there'll be an accept you know minor detail if the machine accepts earlier we'll just say the machine stays in the set once it's it enters an accept the machine does not change from that point on so the rule of the machine is nothing changes and it just remains in the same configuration from that point on okay um so um important to understand what we're you know if you're if you're not following what i mean by a tableau you're doomed for this lecture so it it makes sense if you know for you to ask a question to understand what we mean by tableau okay so just a few more elements here so this is going to be the start configuration for mw this would be an accepting configuration here down at the very last row um and you might imagine you know i i think i filled out some hypothetical first step of the machine after the start where maybe the machine was a when it's in remember how we can encode our configurations so this is the machine is in state q0 looking at w uh the first symbol of the input w1 and maybe when it's in q0 or looking in that first symbol it moves to state q 7 and goes right and then changes that w 1 to an a and so now here is the head shown moved 1 position to the right in the new state q7 i mean of course that depends on what the machine is designed to do what the what the transition function is but this um you know could possibly is what happens so does the okay so good does a tableau trace all steps of all branches no the tableau corresponds to one accepting branch each different accepting branch is going to have its own tableau so there might be several different ways of filling out that second row even of course the first row is going to be in any tableau it's got to be the same um you know once i know w here i've written down maybe i should unpack this for you you know it's in the start state here then here are the first you know here are the n symbols of w w is of length n so it's w 1 w 2 up to w n and then i'm padding out the rest with blanks so i should have said that too but okay um so i want to make my table into the k by n to the k because that's going to be enough to represent the entire uh um all of these configurations of m uh running for most into the k steps right because the machine um even if it sends its head moving to the right as fast as it possibly can it's never going to have um it's it's never going to go outside this box if it only runs for n to the k steps so this is going to be big enough to represent the entire computation of m assuming it's running for n to the k time and the input w is of like that all right um what is k so k is the running time of the machine so we assumed from the previous slide that m runs in time n to the k okay um so a good question here how can the tableau be a square table if we have a low number of computation histories but a lot of tape um well i mean the low number of computation history that that's a problem you're not saying that well you know there each computation history is in a different table so this i'm representing a single computation history here there may be there are going to be many configurations um so i can't have a small number of configurations using a lot of tape because you know i can only use one more cell in one more one additional tape cell each time i meant you know if each additional step of the machine is yes so there's really no difference between this if you want to think of this as a computation history that's fine this is really just the standard terminology that's used when you're proving um this this particular theorem typically people talk about it as a tableau but it's really just the computation history um the cue state i mean the question about the what are these states queue this is the way we represent configurations so this means the machine is in uh state they're not all going down the diagonal the the states are going to zigzag here through he through this picture here depending upon how the head of the machine moves so you have to go back and review how we represented configurations of the machine remember the configuration is a snapshot of the machine at a given point um how do we know that m runs in polynomial time we're assuming m runs in polynomial we started off with a language that's in np so it's it's a non-deterministic polynomial time machine and we're picking one branch that accepts and writing down all the sequence configurations the machine goes through let's move on and maybe as you know ask more questions as they come to you and i'll pick out some to answer if that's going to be helpful to others um okay um so uh we're going to now construct this um uh formula to say that m accepts w again that was what we uh that's our goal and it says that a tableau for m on w exists um and basically what that means is we want to say that it starts right it ends right and everything in between is right and then we're going to need some other stuff to talk about how we're going to be encoding those symbols using boolean variables so those are going to be the four parts here's the start starts right here it ends right here it moves right and here it talks about the encoding of the symbols into boolean variables so those are the four parts of this formula that i'm going to describe um i hope you got your question answered as to why the total number of columns is into the k because it's just big enough to fit the entire all configurations if the is running for at most end of the case steps which is what we're assuming okay okay so now just getting back to that i'm going to describe these different components now on separate slots um let me start off with with this component fee cell which is sort of the most fundamental one because it talks about how we're going to be encoding those symbols of the tableau into the boolean variables so again here's kind of our the picture to have in mind uh of this tableau this is this into the k by enter the k table representing um you know the some accepting branch of the machine's computation if there is one and so now let me draw a some one of the cells here i'm going to magnify it so this is the i j cell here and le i'm going to there there are going to be a collection of boolean variables associated with each one of the cells so each one of the cells is going to have a bunch of variables all to itself and those are going to be basically indicator variables they're going to indicate which symbol that cell gets to uh have in it okay so again picture here um the in this tableau you know we don't know how it's going to get filled out but however it gets filled out each one of these cells gets some symbol um and that symbol could either be a tape alphabet symbol or it's symbol representing a state right that's the way we do our configurations so it could be a tape alphabet symbol here showing the magnification maybe it's the tape alphabet symbol which represents the blank symbol or maybe it's a represents some state okay now how am i going to encode that with variables so this let me this is the collection of variables that's going to apply to the entire formula fees of mw um each cell is so each cell ij is going to have um a set of variables one for each possible symbol sigma that's um in the configuration alphabet namely a tape symbol or a state symbol okay um [Music] so i'm gonna have well maybe this will become clear as i'm writing it so if i turn the the uh the variable x i j sigma equal to true that's just a way of saying that cell i j contains a sigma so if i have x you know i j a that means the symbol contains an a so i'm going to illustrate that now for you so imagine you have lights representing all the different x i js x i js for the different sigmas i didn't say that well so um we're in cell i j so these are all x i j variables they're x all x i j sigma variables for the different sigmas that can go in that can go in this uh cell so all different possible sigmas so this is gamma union q okay so now um if i have an a here in that cell so then the variable extra a is true and i'm just helping you visualize that i'm going to that's going to correspond to turning the light on there's going to be a light associated with each one of these variables and it's be turned on when that variable is true similarly if i have the blank symbol is the thing that goes in in that cell um then that variable gets turned on i hope you can see it maybe it's a little bit small on your screen um the x i j blank a variable is is true and similarly if it's q seven here the x i j q seven variable is true okay so that that's the way we're going to be encoding the contents of these cells using these indicator variables and now we have to start making some boolean logic to make sure that those variables reasonably represent the cell con you know contents of these cells so for example what would be the first thing that comes to your mind well we better not have two lights going on in any one of the cells because then we have two symbols on that cells and that's not allowed each cell is going we want each cell to have exactly one symbol and that corresponds to each each each cell having exactly one of its lights turned on or equivalently each cell should have exactly one of the variables be true that's the very first part of the formula is just going to say that so let's let me show you how that what that looks like okay so here we're talking about this free cell it says that there's exactly one light on purcell or in other words exactly one of the x ij sigmas is true for each ij um so this is how i'm going to actually express that using my boolean formula uh i'm sort of color coding the different parts of the formula which i'm writing out to you here in english so first i want to say there's a so in every cell there's at least one light that's on and there's almost one light that's on um so here's the green part this is going to say at least one light is on so i'm going to say that by taking all of the very all of the symbols that can appear in that cell and and take taking in an or over all of those different associated variables so it's either got the first symbol on or the second symbol is there or the or dot or the last symbol is there one of those has got to be there i'm going to write this using my big or notation so for sigma appearing in this set of possibilities one of those variables has got to be on at least that's what this or big word tells you now i want to make sure that there's at most one that's on so that there are not there's one at least one on but there are not two that are on so i'm gonna have an additional part of the formula here which says and i hope you can read this a little small um if i have two different symbols sigma and tau that are configuration uh possible configuration symbols where sigma and tau are not equal so that's i'm reading it out to you if you can't if it's too small for your screen then i'm going to say it's not possible so i have the negation of x i j sigma and x i j tau so saying in another way it's not the case that that cell contains both sigma and tau for any two symbols sigma and tau as long as they're different i'm going to take these two formulas and add them together and this tells me in the cell ij is exactly one of the variables is true exactly one of the lights is on and that's going to represent which symbol goes into that cell and then i want to take the and over all possible cells to make sure that i'm going to now apply that everywhere and so i now i do an and for i and j ranging between 1 and n to the k to apply this logic throughout the picture yes sigma union asking sigma union q contains the input output yes this is not a sigma this is a gamma gamma is the tape alphabet this is any symbol including an input symbol um from sigma is going to be in in gamma so this is any any symbol that can appear on the tape um okay and that's this this expression here that is the expression feasible cell okay so here's a little check-in for you but maybe before we jump into the check and let's just make sure maybe better to take some questions and then we can ask a check-in for you you understand i mean if you're not getting this you should you should try to uh try to figure out how to get it because this is really just the foundation it it only gets more it's not it's not a very complicated proof once you sort of get the idea what's going on but if you're not getting this part you won't be able to get the rest okay i have no idea what what that means but i'll read it out to you guys this looks like a one-hot tensor encoding same from common same form commonly used in ml okay um it's just an indicator i would just call them indicator variables um why is it sigma union q for the big and here sigma union q you mean gamma union q this is wrong okay now i understand why everybody's upset this should be a gamma not a sigma uh there's a booboo sorry about that i don't know if i can fix that without wrecking the whole slide but so i'm not going to even try this symbol here should be gamma not sigma it's a typo thanks for catching that um yes so the question is um okay so uh feasible is just trying to make sure that the encoding represents setting a bunch of symbols into the tableau not so each cell is going to have one symbol exactly not two not zero um so that's what pieces so if you've satisfied if you set the variables to satisfy feasible then there's going to be one symbol on in each one in one symbol in each of those cells um now another question this is not a cnf no this is not a cnf that's the second half of the that's going to be if we hope we don't run out of time but i have a way of converting general sat formulas to cnfs um and preserving satisfiability so we're gonna do that reduction afterward okay let's let's is it a little check to see if you understand uh at some some level what's going on how many variables does this formula actually have is it order n order n square enter the k remember this k is the running time of the machine uh or n to the 2k what do you think so i i i mean for how many variables i mean that in all of uh m sub m yeah how many variables do we have all together in this formula if that's what the question is and here though here are the variables so describing them here x i j sigma okay i'm going to close this so pick something all right ending polling one two three okay yeah so um okay so that's a good question um so first of all the the correct answer is in fact d it's order n to the 2k um now uh i'm getting some questions about what about the size of gamma and q well those are going to be fixed they depend only on the machine but they don't depend on n so thinking about it in term functionally in terms of n that's going to be a constant multiplier um and so it's going to be absorbed within the um the big o uh so that's why we have uh you know these are constant relative to n these are fixed um they're not into the k possible symbols there's a fixed number of symbols depends only on the machine so looking at a particular machine and what happens when you look at large inputs so why is it d and not c well don't forget how big is this table um this is n to the k by n to the k so there are n's n to the 2k cell into the k quantity squared or enter the 2k cells here um and so there's a collection of variables for each cell some fixed number variables for each cell so that's why it's order n to the 2k good okay so let's uh let's move on i think we're actually oh no um okay so we have one more slide and i think then we have a break after uh so now let's next uh talk about constructing um two more pieces of the fisa bmw formula so we already got fisa face of cell done let's look at fee sub start and fee sub except and fistart is going to tell us that the start configuration has exactly these symbols and the accept tells us that the bottom configuration contains an accepting state somewhere so how we're going to write that down well first of all i'm going to write these down just cell by cell so first of all fee start is going to say the cell 1 1 contains the q 0. i mean i know what the start configuration is it should be because you know i you know thinking of me or think you know us as the reduction the reduction is given m it's given w so it knows what the start state is it knows what the symbols of w are so it can it knows what that start configuration is it's just q zero followed by the n symbols of w followed by blanks so it wants it wants the very first um cell in the in the left-hand corner here to be a q0 the start state of the machine which it knows so it's going to say x sub 1 1 q 0 that has to be turned on so it's going to be and a bunch of a bunch of variables here and in order to satisfy fee sub start all of those variables have to be set to true so that means we have to have a q0 in that cell so now we're going to do the next the next cell here the one two the next cell of the uh uh start configuration so fee start is going to have x12 so that's the next place contains w1 and so on x13 contains w2 all the way up to wn and then there's going to be a bunch of additional um uh parts which say that the that we have blanks in the rest just spelling out exactly all of the symbols in that top row because that's what the fiestart um uh formula or sub formula of the of the overall formula we're making looks like okay now let's take a look at fee accept if you accept because i'm just looking for cue accept to appear somewhere in that bottom row i'm going to do that in terms of an or so here is the variables now notice have n to the k because it's the last row in the table so rho n to the k here and then i'm going to vary j from 1 to n to the k so j the column number is going to range from 1 to n to the k and i'm looking for that q except so x enter the kj which is where j is varying and q accept one of those has to be true one of those has to be turned on and so that's why it's a it's a big war and that's my um uh fee accept peace okay and now we'll take a little break and feel free to ask me some more questions um let me just start the our clock and go grab yourself some coffee or ask me some questions happy to answer them why don't we check that q accept only appears once is it impossible for q except to appear twice that's a great question um so that would definitely be a broken configuration if that happened because a configuration can ha can have must have exactly one state symbol appearing the way we're going to enforce that is is with the fee move uh part of the formula which we haven't seen yet so p move is going to guarantee that the machine is acting correctly so that all of the rows of the tableau are all legal configurations and they all legally follow from the previous so that really in a sense the hard heavy lifting is coming in fee move but it's really not that bad um somebody says out of curiosity how close is this intuition proof to the actual proof this is the actual proof there's no i'm not i'm not hiding anything i mean you know we're being a little uh loose here but you can turn this this is this not cutting any corners here this is exactly how the proof goes um and so uh not not uh you're getting you're getting the real deal here somebody wanted to see the previous slide so here we are whoops is there something you want to ask um okay so fee cell says there's exactly one symbol per cell and the variables are set in a way in terms of thinking of them as indicator variables there's exactly one variable set to true in each cell so it's exactly one symbol per cell that's what p cell the f cell tells you what you know if you don't have that then you have a mess so you've got to start with that and then with the other other things are going to be additional conditions which when satisfied going to enforce the rest of the properties that we want why would the proof fail if we n to the k with 2 to the n to the k so okay i i presume you know where does would this use the polynomial running time of the machine uh of the of m the nondeterministic machine i mean if you had a an enormous tab we have to show ultimately that this reduction is a polynomial time reduction and that's going to depend on how big the tableau is because that's going to tell us how big the formula we're producing is fees of mw if e sub mw is exponentially big we don't have a prayer of being able to output that formula in polynomial time if there were less than into the k steps do we repeat the last configuration yeah that's what i said if we have if the machine ends early the last configuration just stays there so we're going to modify the definition of the machine slightly so that you know the um so it just stays yeah um okay um yeah let me not take the other there's a bunch of other questions some of them a little on the technical side um let me uh maybe i'll try to address them as they come along if it if it turns out to work to do that okay so uh the word the break is over um why don't we uh con is it possibly that the encoding configuration will not fit in into the k no no so the question is is there a possibility that the encoding of the configuration won't fit into the k if the machine runs for enter the k steps the the configuration has to fit with an end to the k because it can't use anything more within into the k steps to think about it but no the ques the answer is there's the configurations if the machine runs time into the k that whole um tableau is big enough to write down the entire computation history all right so let's continue fees of move um this is in a sense the part which is going to tell us that we started right we ended right um fee cell says every every cell contains one symbol and now we have to say that the whole interior is correct how are we going to do that so these these are the parts we've already done and the way i'm going to describe that is in terms of these kind of little windows i'm calling neighborhoods so imagine here we have a two by three rectangle which i'm going to call a two by three neighborhood and what i'm going to argue but i'm not going to prove uh here i'm just going to really state it but it's it's really just a sort of more or less obvious fact uh but the proof the book has the formal proof that if any every one of these here is legitimate is legal according to the rules of the machine if every single imagine you know you have these um oops let me put myself back on here so you can see me um if you have um if you have here every um two by three window you can take this as a window and you slide that over the entire picture of the tableau and everything here looks okay as far as the running of the machine so i'll say what that means in a second but if if everything looks locally fine everywhere then the whole tableau has to be a valid tableau in terms of the rules of m um maybe it's easier if i if i describe what i mean by these being legal um so these neighborhoods um these two by three neighborhoods are legal if they're consistent with m's transition function so i'm going to describe rather than i mean to do this formally i would have to go through kind of a process that we went like what we did when we went through uh the construction for the post correspondence problem and see if the machine moves left this thing's happens every move is right that's i think that's it's not really necessary you can kind of get the idea very clearly by doing it a little bit at a higher level um so let's look at what i mean by a legal neighborhood so this illegal neighborhood is a setting of the values the six values of this 2x3 neighborhood in a way which doesn't violate m's rules so for example if m when it's in state q7 reading a b goes into state q3 and moves left then this would be a legal neighborhood because it shows the head moving left um the b becoming a c so reading a b i should also say that it right converts that b to a c and and moves his head left into state q3 so this would be a legal neighborhood if that's the way so being legal depends upon um it depends on the transition function of the machine so given the transition function that's going to tell you which of the legal neighborhoods so another legal neighborhood this would always be a legal neighborhood is that if nothing changes so that means the head of the machine was somewhere else and so um you know whatever was on the tape in this step is going to be the same stuff in the in those places one step later um here's another possible legal neighborhood um is uh if the head suddenly appears you know on one of the on one of the cells either in the left or the right that would correspond to the machine to the machine moving its head from somewhere off the nape off the neighborhood into the neighborhood you know in that step um so this could be a legal neighborhood provided the machine actually does move its head left into a state q5 at some point under under some conditions um and here is another kind of a weird illegal neighborhood if you have a b c and then the a changes to a d that could also be a legal neighborhood if the machine transition function allows an a to get converted to a d when there is some machine when there is some state reading that a and that state also moves its head left so it doesn't move into this picture so those are examples of legal neighborhoods let me show you some illegal neighborhoods um just i'm doing this this is kind of a proof by example now this is perhaps the most intuitive part but i claim that this is easy to turn this into something airtight and formal so this would be clearly illegal if you're if you have a piece of the tape in the pre previous step where it's a b c and then suddenly the b changes to a d the symbol on the tape changes out of nowhere without having a head nearby uh to a different uh to something else that could never happen so that would be illegal another thing would be illegal is if a state appears from nowhere that could never happen um or if it just disappears that could never happen and here's another here's an interesting one if a state becomes two states um don't forget the machine is non-deterministic so the machine in principle could move its head left on one branch and move its head right on a different branch but those would have to be in different tableaus they can't be in the same tableau because that doesn't correspond to any of the threads of the computation those are multiple threads um and i say this because if you think about my claim which is going to put down over here that if every 2x3 neighborhood is legal then the tableau overall corresponds to a computation history this illustrates why it's not enough to have a two by two two by two neighborhood where you really need the two by three um because if this was a two by two neighborhood if you just look at these four this left most uh two by two that could be um a legal neighborhood if it was a two by two if the rules of the machine allowed for that and the right four box right four cells could also be a legal neighborhood so you could have something that looks okay from the perspective of two by two neighborhoods but globally in terms of the overall tableau is completely nonsensical because it has multiple heads um but if you have a three two by three neighborhood it's big enough to prevent this situation from occurring and then you know you can check the details uh and i think it's very plausible that it guarantees that the overall tableau is is is legitimate if all of the two by three neighborhoods are legal okay and so that's what we're going to turn into a um a boolean expression we're going to say for each cell that the set for each neighborhood so here's a neighborhood at the iga location that's i'm calling this sort of this position here sort of the home location for that um neighborhood for each um for each neighborhood uh it has to be set to one of the legal uh possibilities and there's again only a fixed number of those because there's a fixed number of possible symbols that can appear in those cells um so this is that fixed number to the sixth power at most and i'm going to say that um you know the cell in the upper left which would be this one is an r and this one and this one here is an s and this one is a t and this and this one is a v if you just trace down what the indices are telling you it says that that piece of the tape that piece of the tableau here is set according to one of the possible legal settings and i'm just going to or over all of those possible fixed number of legal settings and then i take a an and over all possible tape cells over all possible neighborhoods and so that's going to be my fee move and that's that's it okay let's see can i explain again the third example of illegal so this one over here i presume is i'm being asked about well you know if the machine is in a state q7 you know reading a c you know the head has to move either left or right so that at the next you know the next um configuration there's got to be a state symbol appearing either in this cell or in this cell and here the tape you know the the the the head has basically just vanished with nowhere to you know it's it's gone that could not happen in the in the you know according to the rules of the machine the way we talk about turing machines so that's not possible so this would be an illegal um a neighborhood you want to prevent any of the ill bad stuff from happening anywhere in here it's only good stuff can be happening locally and that guarantees the overall picture is okay um do we have to check that the head doesn't leave the tabo from the left most or right yeah there are some little details here like that so the question is you know have to make sure that that the yeah you probably need to mark i think the book probably does this correctly you may have to mark the left and right ends to make sure that i mean the right end can is not a problem because the the the the machine can never go um off the right end and if you design the machine so that it never moves its head off the left end either which you can do then you wouldn't have to worry about that possibility but yeah otherwise you would have to put some sort of delimiter here to uh to enforce the mach the head not moving off the left end so there are some details like that too um there will be two heads in the same row no this can i don't know what you somebody says there will be two heads in the same row please elaborate because this is designed not to allow two heads in the same row could i go over the or for legal again okay the or so the the big or here what i have in mind is i i take there's gonna be first of all i look at the machine and i look at the transition function and based on that i write down the list of all the legal two by three neighborhoods so all the settings which correspond to legal two way two by three neighborhoods there's gonna be some fixed number of those you know a hundred there's a hundred possible legal neighborhoods of which you know i've written down here four but maybe there's some some number say a hundred so now there's going to be an ore over those hundred different possibilities it's either going to be this legal neighborhood or some other legal neighborhood or some other legal neighborhood and for each one of those legal neighborhoods i'm going to say well the variables are set according to that legal neighborhood or the the variables are set according to the next legal neighborhood or the variables are set according to the next legal neighborhood and do that 100 times one of those has got to end up it's an ore so one of them has to work otherwise the formula fails and will be false because you're going to now end that over all of the neighborhoods in the picture is it possible to have a head on the far left of the configuration and one on the far right you mean a head over here and a head over there i mean how did the head get there can't happen um you know the head the head has to come from a head above it um if you're going to be worrying about the you know details of the boundaries here you know all that's fixable so let's not you know um lose sight of the main idea i mean if you understand the main idea you can fix little details so i want to make sure you understand the main idea what's happening okay so let's finish up this proof um so in summary um we gave a reduction from a to sat this is what we needed it was in those four pieces and you really just need to argue that that formula we're building is not too big it's and it's going to be basically the size of the tableau if you look at what we constructed the number of variables is roughly the size of the tableau and the amount of logic that we're putting into the formula is also going to be a fixed amount of logic independent of n for each of the variables in that tableau and now somebody asked me about like this how big the indices are um you know the indices for the you know the x i value the inj values um technically they're going to be numbers between 1 and n to the k so you're going to have to write those down and so that's going to be a slight additional logarithmic cost um to write those things down but it's not really that interesting a point um and so the overall f is going to be computable in polynomial time because the output is not very big and it's also not complicated to write the output down so that's the end of the that's the end of the proof um i can take a couple of questions we why can't we just check that the whole this is a good question why can't we just check that the whole row is legal um you know you can check that you can check that a row actually is a is a configuration but you to check that the row follows from the previous row um you know ultimately the operation of a turing machine is is a local thing i mean hey you know the way the the the way it moves from one configuration to the next depends locally on how the those you know where the head is and so really that's just another way of the way i'm saying it is just really checking the whole configuration but just doing it locally i don't know if that's satisfying to you um okay uh why don't i move on because i just want to make sure we have enough time to get to the very last part which is a little bit i'm afraid a little technical um so we're going to kind of shift gears now and talk about reducing sat to three set and um let's let's see how it goes um i i don't always have the most success with with presenting this uh this little piece because it's slightly a technical argument but um if you don't get it don't worry just you have to accept that it's true um but i'd like to show it to you just to makes the whole presentation uh you know complete in that sense okay um so i'm going to give a reduction that maps general formulas to three cnf formulas so that's how i map sat to if you remember threesat is satisfiability but for three cnfs so it's conjunctive normal form you're in the form of those clauses which are anded together and each clause is an or of a bunch of literals which are variables or negated variables so i want to convert phi to phi prime um which is a three cnf formula but preserve the satisfiability feed prime is not going to be logically equivalent to fee because i you know i could do that too i can convert any formula to a logically equivalent cnf formula maybe not even a three second of it but yeah you won't be able to get a 3c enough but you can get a c and f but um it might be exponentially larger and that's not good enough i have to do the reduction in polynomial time so i can't generate a much larger formula that's exponentially larger and so i'm going to do that by adding additional variables so it won't be logically equivalent because the new formula is going to have additional variables in it okay i'm going to kind of do it by example and let's see how that goes so his fee which is not in 3c and f it's not even in cnf because it's got you know uh orrs of ands appearing which are not allowed to happen in a cnf okay so how we're going to convert that into a 3cnf formula preserving the satisfiability and just working it through with this example i hope at least give you some idea of how you of how to do the conversion in general so first of all i'm going to represent this formula as a tree using its natural tree structure so you understand so a and b becomes a and b written as a tree and then i take or that with c so i get the tree structure here in sort of the natural way and i'm going to label all of these intermediate nodes which are associated now with operations and um i'm assuming also that the formula is fully parenthesized so that each operation i'm only thinking about is applying just a two it's a binary operation um and let's let's ignore negations for the minute because negations you can always push those through down to the leaves but it's just going to make it too complicated so negations turn out not to be a problem so there's only going to be negations at the level of the inputs not at the not negation operations in the middle okay so we have this tree structure here and now i'm going to use these two logical facts you know and i don't know if you know you've probably all seen hands and ors i hope otherwise it's going to be really tough uh but there's also other logical operators such as the implication operator where you have a implies b thought of as a logical operator operation um and so that this is only this requires that if a is true then b is true um however if a is false b can be anything uh and similarly if b is true a can be anything um the only thing that's prohibited is that if a is true and b is false that's the only thing that's uh would be invalid and so if you think about it that's going to be equivalent to saying that either a is false or b is true one of those has to be and that's going to be logically equivalent to saying that a implies b another logical equivalence maybe is more familiar to you is just simply de morgan's laws the mortgage law which says that if you have the a not of a and b that's uh equivalent to saying uh the not of a or the knot of b okay i'm going to make use of both of these um now now here i want to i ran out of room on the slide so i'm going to take myself out of the picture here for for a minute um uh i had no place else to put this so here we have if you're going to think of the and in terms of its truth table so here's a and b in terms of a and b so if one and one is one but all other settings of a and b uh yield zero for the and i'm going to represent those um if you imagine a and b is going to be called c i'm going to represent this information with uh four um small formulas which taken together you and them together kind of force c to have the correct behavior associated with a and b okay um so uh so if a and b are both one and c is one if a is zero and b is one um then that so then it forces c to be zero um and similarly you know every other setting besides a and b being being true uh for c to be false which is what you want when you have and so i'm gonna write this expression here down with z1 being in the place of c by just taking those four uh expressions and ending them together so this is exactly those same four expressions written out linearly now i want to do the same thing for z2 but now that's written in terms of an or so it's a slightly different truth table here up in this corner uh so now if either one is a one we get a one result and so now you know if a a and b are true you get uh c is true however if a is true and b is false that still implies c is true um so i'm going to write down uh those rules for specifying how uh z2 must be set and each one of these things is going to get converted into clauses three clauses with three literals using these uh using these um using these rules over here so i'm going to do that for each zi and lastly to make sure the whole thing is satisfied which means there's an output of one here i'm going to have one clause associated which says that z4 the output is a one now i can convert all of those when i have a and b imply c that's logically equivalent to not a or not b or or c and the way you can see that is really by repeated app by application of these uh rules here running a little low on time so maybe you'll just have to check this offline but quickly a and b implies c using the first equivalence is the not of this part or c and then i can use the morgan to convert that uh knot of an and to an or of the knots and then and then i can remove the parenthesis because uh or is associative and so i get a uh a clause which is what i need so each one of these guys is really equivalent to a clause and so i just get a bunch of clauses and actually technically this needs to be three um copy of three things here should it should be z4 or z4 or z4 which is a lot um so checking numbers so i realize my checking is broken because i only realized that last point just now as i was talking so uh the the actual uh value that you get in terms of oh no the number of clauses is correct no i i i take it back this is fine so if you understood what i was saying hopefully you can see how big the formula fif fee prime is in terms of the number of operations and fee so let's let's see how many people get that i i i acknowledge this maybe a little on the technical side okay i'm gonna close it close it down please enter your value okay yeah the correct answer is 4k plus one um because each one of these operations is going to end up being a row in this picture it's going each operation is going to have a variable associated to it it's going to become a row in this picture and so then and each row is going to have four um clauses which define um you know uh what you need set what you need in order to force the that that that variable to have the right value corresponding to that operation okay and so then you need and you need one extra clause here uh for uh saying that the the this whole thing evaluates to true um okay so that's all i wanted to uh do today um we proved those two main theorems and now we know that there are np-complete problems and all of the other problems that we can get from by reductions from these problems um are also going to be np-complete as long as they're an np okay so um that's that's it feel free to put some questions into the chat or move on to whatever else you're going to be doing next okay so a good question here is why is fee prime not logically equivalent with this construction can't be logically equivalent has logically equivalent means that it gives you exactly the same function as you if you set the variables in the same way you get the same result coming out well v prime has more variables than phi does so it wouldn't even make sense to talk about logical equivalence because there are two functions on different numbers of variables so in that sense it doesn't really make sense you know what you could say is that for every setting of the of the overlapping variables of the variables that appear in both ph and v prime so those are the original variables of fee um there's going to exist some setting of the new variables which is going to make um the uh you know yeah there's going to be there will exist some setting of the new variables which will make the the two formulas agree but that's not the definition of logic logical equivalence okay so why going back to the proof of status the satisfiability proof and the legal neighborhoods could i go over why the number of legal neighborhoods is polynomial the number of legal neighborhoods is not only polynomial it's constant it depends only on the machine it does not depend on n so because each cell can have at most some fixed number of you know can have the number of tape symbols plus the number of state symbols that's you know that depends on the machine only so uh now we have six six tape cells for the six cells in a two by three neighborhood so you're going to have that number to the sixth power but it's still it's a constant to the sixth power still a constant doesn't depend on n so it's not a question of even being polynomial um it's it's a it's a constant value it's a constant multiplier uh if you want to think about it in terms of the um the size of the formula that's going to result don't forget you know we're trying to make a formula um which is uh well the reduction has to be polynomial um it's a polynomial time reduction so that means that as n increases the time to calculate the reduction increases as a polynomial but we're fixing m so m does not change so therefore anything that depends on m only is just going to be a constant um impact on the formula it's not going to be it doesn't depend on it okay everybody bye-bye see you you 3 00:00:28,830 --> 00:00:31,669 4 00:00:31,669 --> 00:00:34,950 5 00:00:34,950 --> 00:00:34,960 6 00:00:34,960 --> 00:00:36,069 7 00:00:36,069 --> 00:00:38,549 8 00:00:38,549 --> 00:00:40,150 9 00:00:40,150 --> 00:00:40,160 10 00:00:40,160 --> 00:00:42,069 11 00:00:42,069 --> 00:00:43,350 12 00:00:43,350 --> 00:00:44,549 13 00:00:44,549 --> 00:00:44,559 14 00:00:44,559 --> 00:00:46,229 15 00:00:46,229 --> 00:00:46,239 16 00:00:46,239 --> 00:00:47,029 17 00:00:47,029 --> 00:00:48,869 18 00:00:48,869 --> 00:00:51,990 19 00:00:51,990 --> 00:00:52,000 20 00:00:52,000 --> 00:00:52,950 21 00:00:52,950 --> 00:00:54,310 22 00:00:54,310 --> 00:00:55,910 23 00:00:55,910 --> 00:00:58,549 24 00:00:58,549 --> 00:01:00,549 25 00:01:00,549 --> 00:01:02,150 26 00:01:02,150 --> 00:01:04,469 27 00:01:04,469 --> 00:01:07,350 28 00:01:07,350 --> 00:01:10,310 29 00:01:10,310 --> 00:01:11,510 30 00:01:11,510 --> 00:01:13,830 31 00:01:13,830 --> 00:01:16,789 32 00:01:16,789 --> 00:01:20,230 33 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--> 00:11:04,310 328 00:11:04,310 --> 00:11:05,590 329 00:11:05,590 --> 00:11:07,910 330 00:11:07,910 --> 00:11:10,710 331 00:11:10,710 --> 00:11:13,990 332 00:11:13,990 --> 00:11:15,509 333 00:11:15,509 --> 00:11:17,269 334 00:11:17,269 --> 00:11:19,910 335 00:11:19,910 --> 00:11:19,920 336 00:11:19,920 --> 00:11:21,829 337 00:11:21,829 --> 00:11:24,310 338 00:11:24,310 --> 00:11:24,320 339 00:11:24,320 --> 00:11:25,829 340 00:11:25,829 --> 00:11:30,150 341 00:11:30,150 --> 00:11:32,069 342 00:11:32,069 --> 00:11:32,079 343 00:11:32,079 --> 00:11:33,110 344 00:11:33,110 --> 00:11:35,590 345 00:11:35,590 --> 00:11:37,430 346 00:11:37,430 --> 00:11:40,630 347 00:11:40,630 --> 00:11:42,470 348 00:11:42,470 --> 00:11:46,069 349 00:11:46,069 --> 00:11:47,910 350 00:11:47,910 --> 00:11:51,030 351 00:11:51,030 --> 00:11:53,509 352 00:11:53,509 --> 00:11:53,519 353 00:11:53,519 --> 00:11:54,790 354 00:11:54,790 --> 00:11:57,670 355 00:11:57,670 --> 00:12:00,230 356 00:12:00,230 --> 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386 00:13:03,530 --> 00:13:05,190 387 00:13:05,190 --> 00:13:07,110 388 00:13:07,110 --> 00:13:11,750 389 00:13:11,750 --> 00:13:12,870 390 00:13:12,870 --> 00:13:15,910 391 00:13:15,910 --> 00:13:17,190 392 00:13:17,190 --> 00:13:18,629 393 00:13:18,629 --> 00:13:20,790 394 00:13:20,790 --> 00:13:20,800 395 00:13:20,800 --> 00:13:22,150 396 00:13:22,150 --> 00:13:22,160 397 00:13:22,160 --> 00:13:22,949 398 00:13:22,949 --> 00:13:22,959 399 00:13:22,959 --> 00:13:23,910 400 00:13:23,910 --> 00:13:26,310 401 00:13:26,310 --> 00:13:27,509 402 00:13:27,509 --> 00:13:28,949 403 00:13:28,949 --> 00:13:30,310 404 00:13:30,310 --> 00:13:31,990 405 00:13:31,990 --> 00:13:33,430 406 00:13:33,430 --> 00:13:35,110 407 00:13:35,110 --> 00:13:37,750 408 00:13:37,750 --> 00:13:39,670 409 00:13:39,670 --> 00:13:41,670 410 00:13:41,670 --> 00:13:43,590 411 00:13:43,590 --> 00:13:45,509 412 00:13:45,509 --> 00:13:45,519 413 00:13:45,519 --> 00:13:47,189 414 00:13:47,189 --> 00:13:49,829 415 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--> 00:14:34,399 445 00:14:34,399 --> 00:14:35,829 446 00:14:35,829 --> 00:14:37,350 447 00:14:37,350 --> 00:14:39,990 448 00:14:39,990 --> 00:14:40,000 449 00:14:40,000 --> 00:14:41,030 450 00:14:41,030 --> 00:14:41,040 451 00:14:41,040 --> 00:14:41,750 452 00:14:41,750 --> 00:14:43,269 453 00:14:43,269 --> 00:14:45,030 454 00:14:45,030 --> 00:14:48,310 455 00:14:48,310 --> 00:14:52,150 456 00:14:52,150 --> 00:14:55,430 457 00:14:55,430 --> 00:14:59,590 458 00:14:59,590 --> 00:15:01,350 459 00:15:01,350 --> 00:15:01,360 460 00:15:01,360 --> 00:15:02,470 461 00:15:02,470 --> 00:15:03,670 462 00:15:03,670 --> 00:15:05,430 463 00:15:05,430 --> 00:15:08,069 464 00:15:08,069 --> 00:15:09,750 465 00:15:09,750 --> 00:15:11,910 466 00:15:11,910 --> 00:15:13,509 467 00:15:13,509 --> 00:15:16,150 468 00:15:16,150 --> 00:15:18,389 469 00:15:18,389 --> 00:15:21,670 470 00:15:21,670 --> 00:15:23,750 471 00:15:23,750 --> 00:15:26,389 472 00:15:26,389 --> 00:15:29,350 473 00:15:29,350 --> 00:15:32,230 474 00:15:32,230 --> 00:15:32,240 475 00:15:32,240 --> 00:15:33,269 476 00:15:33,269 --> 00:15:33,279 477 00:15:33,279 --> 00:15:34,470 478 00:15:34,470 --> 00:15:35,670 479 00:15:35,670 --> 00:15:39,189 480 00:15:39,189 --> 00:15:41,189 481 00:15:41,189 --> 00:15:43,670 482 00:15:43,670 --> 00:15:46,629 483 00:15:46,629 --> 00:15:48,470 484 00:15:48,470 --> 00:15:52,230 485 00:15:52,230 --> 00:15:54,710 486 00:15:54,710 --> 00:15:57,350 487 00:15:57,350 --> 00:15:59,269 488 00:15:59,269 --> 00:15:59,279 489 00:15:59,279 --> 00:16:00,470 490 00:16:00,470 --> 00:16:03,590 491 00:16:03,590 --> 00:16:05,910 492 00:16:05,910 --> 00:16:07,990 493 00:16:07,990 --> 00:16:09,829 494 00:16:09,829 --> 00:16:12,629 495 00:16:12,629 --> 00:16:15,829 496 00:16:15,829 --> 00:16:16,790 497 00:16:16,790 --> 00:16:19,189 498 00:16:19,189 --> 00:16:22,949 499 00:16:22,949 --> 00:16:24,710 500 00:16:24,710 --> 00:16:24,720 501 00:16:24,720 --> 00:16:26,710 502 00:16:26,710 --> 00:16:30,310 503 00:16:30,310 --> 00:16:30,320 504 00:16:30,320 --> 00:16:31,269 505 00:16:31,269 --> 00:16:32,949 506 00:16:32,949 --> 00:16:35,509 507 00:16:35,509 --> 00:16:37,430 508 00:16:37,430 --> 00:16:39,670 509 00:16:39,670 --> 00:16:41,430 510 00:16:41,430 --> 00:16:44,550 511 00:16:44,550 --> 00:16:46,710 512 00:16:46,710 --> 00:16:47,829 513 00:16:47,829 --> 00:16:49,269 514 00:16:49,269 --> 00:16:50,949 515 00:16:50,949 --> 00:16:50,959 516 00:16:50,959 --> 00:16:52,389 517 00:16:52,389 --> 00:16:54,629 518 00:16:54,629 --> 00:16:56,470 519 00:16:56,470 --> 00:16:58,710 520 00:16:58,710 --> 00:17:00,629 521 00:17:00,629 --> 00:17:02,310 522 00:17:02,310 --> 00:17:03,910 523 00:17:03,910 --> 00:17:05,029 524 00:17:05,029 --> 00:17:06,949 525 00:17:06,949 --> 00:17:08,710 526 00:17:08,710 --> 00:17:08,720 527 00:17:08,720 --> 00:17:11,270 528 00:17:11,270 --> 00:17:13,510 529 00:17:13,510 --> 00:17:13,520 530 00:17:13,520 --> 00:17:14,390 531 00:17:14,390 --> 00:17:14,400 532 00:17:14,400 --> 00:17:15,909 533 00:17:15,909 --> 00:17:17,829 534 00:17:17,829 --> 00:17:18,870 535 00:17:18,870 --> 00:17:20,789 536 00:17:20,789 --> 00:17:22,150 537 00:17:22,150 --> 00:17:23,750 538 00:17:23,750 --> 00:17:23,760 539 00:17:23,760 --> 00:17:24,789 540 00:17:24,789 --> 00:17:27,909 541 00:17:27,909 --> 00:17:29,270 542 00:17:29,270 --> 00:17:33,350 543 00:17:33,350 --> 00:17:36,070 544 00:17:36,070 --> 00:17:37,029 545 00:17:37,029 --> 00:17:39,350 546 00:17:39,350 --> 00:17:41,270 547 00:17:41,270 --> 00:17:43,510 548 00:17:43,510 --> 00:17:43,520 549 00:17:43,520 --> 00:17:44,390 550 00:17:44,390 --> 00:17:47,510 551 00:17:47,510 --> 00:17:50,230 552 00:17:50,230 --> 00:17:50,240 553 00:17:50,240 --> 00:17:52,310 554 00:17:52,310 --> 00:17:54,390 555 00:17:54,390 --> 00:17:54,400 556 00:17:54,400 --> 00:17:55,430 557 00:17:55,430 --> 00:17:55,440 558 00:17:55,440 --> 00:17:56,310 559 00:17:56,310 --> 00:17:57,990 560 00:17:57,990 --> 00:18:01,110 561 00:18:01,110 --> 00:18:03,990 562 00:18:03,990 --> 00:18:07,590 563 00:18:07,590 --> 00:18:09,990 564 00:18:09,990 --> 00:18:11,750 565 00:18:11,750 --> 00:18:14,470 566 00:18:14,470 --> 00:18:16,789 567 00:18:16,789 --> 00:18:18,390 568 00:18:18,390 --> 00:18:22,630 569 00:18:22,630 --> 00:18:25,350 570 00:18:25,350 --> 00:18:27,430 571 00:18:27,430 --> 00:18:27,440 572 00:18:27,440 --> 00:18:28,470 573 00:18:28,470 --> 00:18:30,390 574 00:18:30,390 --> 00:18:32,390 575 00:18:32,390 --> 00:18:34,870 576 00:18:34,870 --> 00:18:34,880 577 00:18:34,880 --> 00:18:35,990 578 00:18:35,990 --> 00:18:38,310 579 00:18:38,310 --> 00:18:40,549 580 00:18:40,549 --> 00:18:43,029 581 00:18:43,029 --> 00:18:44,870 582 00:18:44,870 --> 00:18:44,880 583 00:18:44,880 --> 00:18:45,909 584 00:18:45,909 --> 00:18:48,950 585 00:18:48,950 --> 00:18:48,960 586 00:18:48,960 --> 00:18:51,510 587 00:18:51,510 --> 00:18:53,510 588 00:18:53,510 --> 00:18:55,350 589 00:18:55,350 --> 00:18:57,190 590 00:18:57,190 --> 00:18:59,110 591 00:18:59,110 --> 00:19:01,430 592 00:19:01,430 --> 00:19:02,789 593 00:19:02,789 --> 00:19:02,799 594 00:19:02,799 --> 00:19:03,909 595 00:19:03,909 --> 00:19:05,029 596 00:19:05,029 --> 00:19:06,630 597 00:19:06,630 --> 00:19:08,470 598 00:19:08,470 --> 00:19:10,390 599 00:19:10,390 --> 00:19:13,190 600 00:19:13,190 --> 00:19:15,510 601 00:19:15,510 --> 00:19:19,190 602 00:19:19,190 --> 00:19:20,950 603 00:19:20,950 --> 00:19:20,960 604 00:19:20,960 --> 00:19:23,029 605 00:19:23,029 --> 00:19:25,350 606 00:19:25,350 --> 00:19:25,360 607 00:19:25,360 --> 00:19:26,310 608 00:19:26,310 --> 00:19:28,390 609 00:19:28,390 --> 00:19:30,549 610 00:19:30,549 --> 00:19:30,559 611 00:19:30,559 --> 00:19:31,590 612 00:19:31,590 --> 00:19:34,230 613 00:19:34,230 --> 00:19:34,240 614 00:19:34,240 --> 00:19:35,029 615 00:19:35,029 --> 00:19:36,950 616 00:19:36,950 --> 00:19:36,960 617 00:19:36,960 --> 00:19:38,390 618 00:19:38,390 --> 00:19:39,590 619 00:19:39,590 --> 00:19:42,630 620 00:19:42,630 --> 00:19:45,029 621 00:19:45,029 --> 00:19:47,669 622 00:19:47,669 --> 00:19:49,909 623 00:19:49,909 --> 00:19:51,270 624 00:19:51,270 --> 00:19:53,830 625 00:19:53,830 --> 00:19:56,470 626 00:19:56,470 --> 00:19:58,950 627 00:19:58,950 --> 00:20:01,110 628 00:20:01,110 --> 00:20:02,470 629 00:20:02,470 --> 00:20:05,110 630 00:20:05,110 --> 00:20:07,669 631 00:20:07,669 --> 00:20:09,430 632 00:20:09,430 --> 00:20:12,710 633 00:20:12,710 --> 00:20:14,390 634 00:20:14,390 --> 00:20:16,789 635 00:20:16,789 --> 00:20:18,630 636 00:20:18,630 --> 00:20:21,669 637 00:20:21,669 --> 00:20:25,430 638 00:20:25,430 --> 00:20:31,590 639 00:20:31,590 --> 00:20:33,190 640 00:20:33,190 --> 00:20:35,590 641 00:20:35,590 --> 00:20:37,669 642 00:20:37,669 --> 00:20:40,310 643 00:20:40,310 --> 00:20:42,149 644 00:20:42,149 --> 00:20:43,590 645 00:20:43,590 --> 00:20:45,669 646 00:20:45,669 --> 00:20:47,029 647 00:20:47,029 --> 00:20:49,350 648 00:20:49,350 --> 00:20:51,430 649 00:20:51,430 --> 00:20:52,950 650 00:20:52,950 --> 00:20:54,870 651 00:20:54,870 --> 00:20:55,590 652 00:20:55,590 --> 00:20:55,600 653 00:20:55,600 --> 00:20:58,390 654 00:20:58,390 --> 00:21:00,470 655 00:21:00,470 --> 00:21:03,190 656 00:21:03,190 --> 00:21:03,200 657 00:21:03,200 --> 00:21:04,470 658 00:21:04,470 --> 00:21:06,630 659 00:21:06,630 --> 00:21:08,549 660 00:21:08,549 --> 00:21:10,870 661 00:21:10,870 --> 00:21:14,149 662 00:21:14,149 --> 00:21:16,310 663 00:21:16,310 --> 00:21:17,510 664 00:21:17,510 --> 00:21:19,430 665 00:21:19,430 --> 00:21:21,350 666 00:21:21,350 --> 00:21:22,870 667 00:21:22,870 --> 00:21:24,950 668 00:21:24,950 --> 00:21:27,270 669 00:21:27,270 --> 00:21:28,710 670 00:21:28,710 --> 00:21:31,510 671 00:21:31,510 --> 00:21:31,520 672 00:21:31,520 --> 00:21:35,270 673 00:21:35,270 --> 00:21:36,950 674 00:21:36,950 --> 00:21:38,710 675 00:21:38,710 --> 00:21:40,470 676 00:21:40,470 --> 00:21:42,549 677 00:21:42,549 --> 00:21:44,710 678 00:21:44,710 --> 00:21:46,789 679 00:21:46,789 --> 00:21:48,230 680 00:21:48,230 --> 00:21:50,789 681 00:21:50,789 --> 00:21:52,789 682 00:21:52,789 --> 00:21:54,310 683 00:21:54,310 --> 00:21:55,990 684 00:21:55,990 --> 00:21:58,230 685 00:21:58,230 --> 00:21:59,830 686 00:21:59,830 --> 00:22:01,669 687 00:22:01,669 --> 00:22:01,679 688 00:22:01,679 --> 00:22:02,870 689 00:22:02,870 --> 00:22:05,750 690 00:22:05,750 --> 00:22:05,760 691 00:22:05,760 --> 00:22:07,510 692 00:22:07,510 --> 00:22:09,270 693 00:22:09,270 --> 00:22:11,510 694 00:22:11,510 --> 00:22:14,710 695 00:22:14,710 --> 00:22:14,720 696 00:22:14,720 --> 00:22:15,669 697 00:22:15,669 --> 00:22:17,590 698 00:22:17,590 --> 00:22:19,990 699 00:22:19,990 --> 00:22:22,070 700 00:22:22,070 --> 00:22:23,350 701 00:22:23,350 --> 00:22:24,950 702 00:22:24,950 --> 00:22:26,630 703 00:22:26,630 --> 00:22:29,270 704 00:22:29,270 --> 00:22:31,270 705 00:22:31,270 --> 00:22:34,310 706 00:22:34,310 --> 00:22:36,390 707 00:22:36,390 --> 00:22:36,400 708 00:22:36,400 --> 00:22:37,669 709 00:22:37,669 --> 00:22:37,679 710 00:22:37,679 --> 00:22:38,630 711 00:22:38,630 --> 00:22:38,640 712 00:22:38,640 --> 00:22:41,510 713 00:22:41,510 --> 00:22:43,029 714 00:22:43,029 --> 00:22:45,029 715 00:22:45,029 --> 00:22:45,039 716 00:22:45,039 --> 00:22:46,789 717 00:22:46,789 --> 00:22:46,799 718 00:22:46,799 --> 00:22:48,950 719 00:22:48,950 --> 00:22:48,960 720 00:22:48,960 --> 00:22:49,750 721 00:22:49,750 --> 00:22:49,760 722 00:22:49,760 --> 00:22:50,950 723 00:22:50,950 --> 00:22:53,270 724 00:22:53,270 --> 00:22:55,990 725 00:22:55,990 --> 00:22:58,390 726 00:22:58,390 --> 00:22:58,400 727 00:22:58,400 --> 00:23:01,029 728 00:23:01,029 --> 00:23:01,039 729 00:23:01,039 --> 00:23:02,630 730 00:23:02,630 --> 00:23:04,710 731 00:23:04,710 --> 00:23:06,710 732 00:23:06,710 --> 00:23:09,270 733 00:23:09,270 --> 00:23:10,950 734 00:23:10,950 --> 00:23:12,230 735 00:23:12,230 --> 00:23:14,230 736 00:23:14,230 --> 00:23:16,789 737 00:23:16,789 --> 00:23:16,799 738 00:23:16,799 --> 00:23:17,990 739 00:23:17,990 --> 00:23:19,990 740 00:23:19,990 --> 00:23:22,310 741 00:23:22,310 --> 00:23:24,870 742 00:23:24,870 --> 00:23:26,789 743 00:23:26,789 --> 00:23:30,470 744 00:23:30,470 --> 00:23:33,510 745 00:23:33,510 --> 00:23:37,190 746 00:23:37,190 --> 00:23:37,200 747 00:23:37,200 --> 00:23:38,470 748 00:23:38,470 --> 00:23:40,390 749 00:23:40,390 --> 00:23:42,230 750 00:23:42,230 --> 00:23:44,310 751 00:23:44,310 --> 00:23:47,310 752 00:23:47,310 --> 00:23:50,149 753 00:23:50,149 --> 00:23:52,310 754 00:23:52,310 --> 00:23:54,630 755 00:23:54,630 --> 00:23:54,640 756 00:23:54,640 --> 00:23:56,950 757 00:23:56,950 --> 00:23:58,230 758 00:23:58,230 --> 00:23:58,240 759 00:23:58,240 --> 00:23:58,950 760 00:23:58,950 --> 00:24:00,870 761 00:24:00,870 --> 00:24:03,269 762 00:24:03,269 --> 00:24:06,789 763 00:24:06,789 --> 00:24:06,799 764 00:24:06,799 --> 00:24:07,830 765 00:24:07,830 --> 00:24:09,510 766 00:24:09,510 --> 00:24:11,430 767 00:24:11,430 --> 00:24:13,110 768 00:24:13,110 --> 00:24:14,470 769 00:24:14,470 --> 00:24:16,830 770 00:24:16,830 --> 00:24:19,830 771 00:24:19,830 --> 00:24:24,789 772 00:24:24,789 --> 00:24:26,310 773 00:24:26,310 --> 00:24:28,630 774 00:24:28,630 --> 00:24:31,990 775 00:24:31,990 --> 00:24:32,000 776 00:24:32,000 --> 00:24:33,669 777 00:24:33,669 --> 00:24:35,990 778 00:24:35,990 --> 00:24:37,909 779 00:24:37,909 --> 00:24:40,789 780 00:24:40,789 --> 00:24:42,630 781 00:24:42,630 --> 00:24:43,830 782 00:24:43,830 --> 00:24:46,870 783 00:24:46,870 --> 00:24:49,190 784 00:24:49,190 --> 00:24:51,190 785 00:24:51,190 --> 00:24:52,950 786 00:24:52,950 --> 00:24:54,789 787 00:24:54,789 --> 00:24:54,799 788 00:24:54,799 --> 00:24:55,830 789 00:24:55,830 --> 00:24:58,230 790 00:24:58,230 --> 00:25:01,510 791 00:25:01,510 --> 00:25:04,310 792 00:25:04,310 --> 00:25:07,430 793 00:25:07,430 --> 00:25:08,789 794 00:25:08,789 --> 00:25:12,870 795 00:25:12,870 --> 00:25:14,630 796 00:25:14,630 --> 00:25:16,310 797 00:25:16,310 --> 00:25:16,320 798 00:25:16,320 --> 00:25:17,350 799 00:25:17,350 --> 00:25:21,909 800 00:25:21,909 --> 00:25:23,669 801 00:25:23,669 --> 00:25:26,470 802 00:25:26,470 --> 00:25:26,480 803 00:25:26,480 --> 00:25:27,510 804 00:25:27,510 --> 00:25:29,269 805 00:25:29,269 --> 00:25:30,310 806 00:25:30,310 --> 00:25:31,830 807 00:25:31,830 --> 00:25:34,789 808 00:25:34,789 --> 00:25:37,110 809 00:25:37,110 --> 00:25:37,120 810 00:25:37,120 --> 00:25:38,470 811 00:25:38,470 --> 00:25:39,990 812 00:25:39,990 --> 00:25:41,990 813 00:25:41,990 --> 00:25:44,070 814 00:25:44,070 --> 00:25:45,269 815 00:25:45,269 --> 00:25:45,279 816 00:25:45,279 --> 00:25:46,549 817 00:25:46,549 --> 00:25:48,230 818 00:25:48,230 --> 00:25:48,240 819 00:25:48,240 --> 00:25:49,269 820 00:25:49,269 --> 00:25:53,110 821 00:25:53,110 --> 00:25:54,470 822 00:25:54,470 --> 00:25:57,269 823 00:25:57,269 --> 00:25:59,190 824 00:25:59,190 --> 00:26:02,230 825 00:26:02,230 --> 00:26:02,240 826 00:26:02,240 --> 00:26:03,350 827 00:26:03,350 --> 00:26:05,669 828 00:26:05,669 --> 00:26:05,679 829 00:26:05,679 --> 00:26:07,029 830 00:26:07,029 --> 00:26:07,039 831 00:26:07,039 --> 00:26:07,830 832 00:26:07,830 --> 00:26:11,590 833 00:26:11,590 --> 00:26:13,510 834 00:26:13,510 --> 00:26:17,029 835 00:26:17,029 --> 00:26:17,039 836 00:26:17,039 --> 00:26:20,149 837 00:26:20,149 --> 00:26:24,149 838 00:26:24,149 --> 00:26:26,149 839 00:26:26,149 --> 00:26:27,830 840 00:26:27,830 --> 00:26:30,789 841 00:26:30,789 --> 00:26:33,269 842 00:26:33,269 --> 00:26:33,279 843 00:26:33,279 --> 00:26:34,149 844 00:26:34,149 --> 00:26:38,390 845 00:26:38,390 --> 00:26:39,990 846 00:26:39,990 --> 00:26:43,909 847 00:26:43,909 --> 00:26:43,919 848 00:26:43,919 --> 00:26:44,870 849 00:26:44,870 --> 00:26:44,880 850 00:26:44,880 --> 00:26:45,520 851 00:26:45,520 --> 00:26:45,530 852 00:26:45,530 --> 00:26:47,350 853 00:26:47,350 --> 00:26:49,830 854 00:26:49,830 --> 00:26:52,230 855 00:26:52,230 --> 00:26:53,830 856 00:26:53,830 --> 00:26:55,990 857 00:26:55,990 --> 00:26:59,510 858 00:26:59,510 --> 00:27:01,669 859 00:27:01,669 --> 00:27:04,070 860 00:27:04,070 --> 00:27:07,750 861 00:27:07,750 --> 00:27:09,830 862 00:27:09,830 --> 00:27:09,840 863 00:27:09,840 --> 00:27:10,630 864 00:27:10,630 --> 00:27:12,710 865 00:27:12,710 --> 00:27:14,950 866 00:27:14,950 --> 00:27:18,789 867 00:27:18,789 --> 00:27:20,389 868 00:27:20,389 --> 00:27:22,149 869 00:27:22,149 --> 00:27:25,669 870 00:27:25,669 --> 00:27:27,350 871 00:27:27,350 --> 00:27:27,360 872 00:27:27,360 --> 00:27:28,789 873 00:27:28,789 --> 00:27:31,750 874 00:27:31,750 --> 00:27:33,830 875 00:27:33,830 --> 00:27:36,070 876 00:27:36,070 --> 00:27:39,830 877 00:27:39,830 --> 00:27:39,840 878 00:27:39,840 --> 00:27:40,630 879 00:27:40,630 --> 00:27:43,029 880 00:27:43,029 --> 00:27:43,039 881 00:27:43,039 --> 00:27:43,909 882 00:27:43,909 --> 00:27:47,350 883 00:27:47,350 --> 00:27:48,389 884 00:27:48,389 --> 00:27:48,399 885 00:27:48,399 --> 00:27:49,190 886 00:27:49,190 --> 00:27:49,200 887 00:27:49,200 --> 00:27:50,230 888 00:27:50,230 --> 00:27:55,510 889 00:27:55,510 --> 00:27:56,950 890 00:27:56,950 --> 00:28:01,269 891 00:28:01,269 --> 00:28:01,279 892 00:28:01,279 --> 00:28:02,070 893 00:28:02,070 --> 00:28:03,430 894 00:28:03,430 --> 00:28:05,990 895 00:28:05,990 --> 00:28:07,430 896 00:28:07,430 --> 00:28:09,110 897 00:28:09,110 --> 00:28:10,630 898 00:28:10,630 --> 00:28:12,950 899 00:28:12,950 --> 00:28:15,430 900 00:28:15,430 --> 00:28:15,440 901 00:28:15,440 --> 00:28:16,710 902 00:28:16,710 --> 00:28:18,230 903 00:28:18,230 --> 00:28:19,990 904 00:28:19,990 --> 00:28:23,669 905 00:28:23,669 --> 00:28:24,870 906 00:28:24,870 --> 00:28:26,310 907 00:28:26,310 --> 00:28:26,320 908 00:28:26,320 --> 00:28:27,430 909 00:28:27,430 --> 00:28:27,440 910 00:28:27,440 --> 00:28:28,549 911 00:28:28,549 --> 00:28:30,789 912 00:28:30,789 --> 00:28:33,750 913 00:28:33,750 --> 00:28:36,470 914 00:28:36,470 --> 00:28:39,110 915 00:28:39,110 --> 00:28:41,029 916 00:28:41,029 --> 00:28:42,470 917 00:28:42,470 --> 00:28:42,480 918 00:28:42,480 --> 00:28:43,669 919 00:28:43,669 --> 00:28:45,669 920 00:28:45,669 --> 00:28:48,470 921 00:28:48,470 --> 00:28:49,590 922 00:28:49,590 --> 00:28:51,350 923 00:28:51,350 --> 00:28:54,470 924 00:28:54,470 --> 00:28:56,950 925 00:28:56,950 --> 00:28:59,909 926 00:28:59,909 --> 00:29:01,990 927 00:29:01,990 --> 00:29:03,990 928 00:29:03,990 --> 00:29:07,110 929 00:29:07,110 --> 00:29:09,269 930 00:29:09,269 --> 00:29:10,549 931 00:29:10,549 --> 00:29:12,710 932 00:29:12,710 --> 00:29:14,310 933 00:29:14,310 --> 00:29:15,669 934 00:29:15,669 --> 00:29:17,190 935 00:29:17,190 --> 00:29:19,110 936 00:29:19,110 --> 00:29:21,350 937 00:29:21,350 --> 00:29:23,269 938 00:29:23,269 --> 00:29:25,909 939 00:29:25,909 --> 00:29:29,510 940 00:29:29,510 --> 00:29:30,950 941 00:29:30,950 --> 00:29:32,950 942 00:29:32,950 --> 00:29:34,789 943 00:29:34,789 --> 00:29:36,310 944 00:29:36,310 --> 00:29:36,320 945 00:29:36,320 --> 00:29:38,549 946 00:29:38,549 --> 00:29:40,070 947 00:29:40,070 --> 00:29:40,080 948 00:29:40,080 --> 00:29:41,110 949 00:29:41,110 --> 00:29:43,029 950 00:29:43,029 --> 00:29:44,549 951 00:29:44,549 --> 00:29:46,950 952 00:29:46,950 --> 00:29:54,950 953 00:29:54,950 --> 00:29:54,960 954 00:29:54,960 --> 00:29:56,070 955 00:29:56,070 --> 00:29:57,669 956 00:29:57,669 --> 00:30:00,630 957 00:30:00,630 --> 00:30:00,640 958 00:30:00,640 --> 00:30:02,230 959 00:30:02,230 --> 00:30:02,240 960 00:30:02,240 --> 00:30:04,389 961 00:30:04,389 --> 00:30:06,070 962 00:30:06,070 --> 00:30:07,590 963 00:30:07,590 --> 00:30:09,430 964 00:30:09,430 --> 00:30:11,750 965 00:30:11,750 --> 00:30:13,830 966 00:30:13,830 --> 00:30:15,909 967 00:30:15,909 --> 00:30:18,549 968 00:30:18,549 --> 00:30:21,110 969 00:30:21,110 --> 00:30:23,350 970 00:30:23,350 --> 00:30:26,230 971 00:30:26,230 --> 00:30:29,430 972 00:30:29,430 --> 00:30:31,990 973 00:30:31,990 --> 00:30:35,669 974 00:30:35,669 --> 00:30:38,070 975 00:30:38,070 --> 00:30:40,070 976 00:30:40,070 --> 00:30:43,590 977 00:30:43,590 --> 00:30:45,590 978 00:30:45,590 --> 00:30:47,430 979 00:30:47,430 --> 00:30:49,830 980 00:30:49,830 --> 00:30:49,840 981 00:30:49,840 --> 00:30:52,549 982 00:30:52,549 --> 00:30:54,070 983 00:30:54,070 --> 00:30:55,669 984 00:30:55,669 --> 00:30:58,470 985 00:30:58,470 --> 00:31:01,110 986 00:31:01,110 --> 00:31:02,389 987 00:31:02,389 --> 00:31:05,269 988 00:31:05,269 --> 00:31:06,950 989 00:31:06,950 --> 00:31:09,110 990 00:31:09,110 --> 00:31:11,509 991 00:31:11,509 --> 00:31:13,990 992 00:31:13,990 --> 00:31:15,830 993 00:31:15,830 --> 00:31:17,430 994 00:31:17,430 --> 00:31:19,509 995 00:31:19,509 --> 00:31:22,070 996 00:31:22,070 --> 00:31:22,080 997 00:31:22,080 --> 00:31:23,350 998 00:31:23,350 --> 00:31:25,029 999 00:31:25,029 --> 00:31:25,039 1000 00:31:25,039 --> 00:31:26,710 1001 00:31:26,710 --> 00:31:28,389 1002 00:31:28,389 --> 00:31:28,399 1003 00:31:28,399 --> 00:31:29,830 1004 00:31:29,830 --> 00:31:33,590 1005 00:31:33,590 --> 00:31:34,710 1006 00:31:34,710 --> 00:31:34,720 1007 00:31:34,720 --> 00:31:36,149 1008 00:31:36,149 --> 00:31:38,630 1009 00:31:38,630 --> 00:31:40,549 1010 00:31:40,549 --> 00:31:42,070 1011 00:31:42,070 --> 00:31:44,870 1012 00:31:44,870 --> 00:31:47,269 1013 00:31:47,269 --> 00:31:49,110 1014 00:31:49,110 --> 00:31:53,190 1015 00:31:53,190 --> 00:31:54,950 1016 00:31:54,950 --> 00:31:56,710 1017 00:31:56,710 --> 00:31:56,720 1018 00:31:56,720 --> 00:31:57,509 1019 00:31:57,509 --> 00:32:00,630 1020 00:32:00,630 --> 00:32:02,870 1021 00:32:02,870 --> 00:32:06,549 1022 00:32:06,549 --> 00:32:08,230 1023 00:32:08,230 --> 00:32:10,070 1024 00:32:10,070 --> 00:32:14,950 1025 00:32:14,950 --> 00:32:18,310 1026 00:32:18,310 --> 00:32:20,310 1027 00:32:20,310 --> 00:32:22,470 1028 00:32:22,470 --> 00:32:24,389 1029 00:32:24,389 --> 00:32:27,190 1030 00:32:27,190 --> 00:32:28,470 1031 00:32:28,470 --> 00:32:30,310 1032 00:32:30,310 --> 00:32:33,269 1033 00:32:33,269 --> 00:32:35,509 1034 00:32:35,509 --> 00:32:37,909 1035 00:32:37,909 --> 00:32:41,990 1036 00:32:41,990 --> 00:32:42,000 1037 00:32:42,000 --> 00:32:45,590 1038 00:32:45,590 --> 00:32:48,870 1039 00:32:48,870 --> 00:32:51,509 1040 00:32:51,509 --> 00:32:54,470 1041 00:32:54,470 --> 00:32:56,470 1042 00:32:56,470 --> 00:32:59,669 1043 00:32:59,669 --> 00:33:00,870 1044 00:33:00,870 --> 00:33:03,509 1045 00:33:03,509 --> 00:33:05,509 1046 00:33:05,509 --> 00:33:05,519 1047 00:33:05,519 --> 00:33:08,830 1048 00:33:08,830 --> 00:33:11,029 1049 00:33:11,029 --> 00:33:11,039 1050 00:33:11,039 --> 00:33:11,830 1051 00:33:11,830 --> 00:33:13,669 1052 00:33:13,669 --> 00:33:16,389 1053 00:33:16,389 --> 00:33:16,399 1054 00:33:16,399 --> 00:33:17,269 1055 00:33:17,269 --> 00:33:19,909 1056 00:33:19,909 --> 00:33:21,669 1057 00:33:21,669 --> 00:33:23,830 1058 00:33:23,830 --> 00:33:25,909 1059 00:33:25,909 --> 00:33:29,190 1060 00:33:29,190 --> 00:33:32,070 1061 00:33:32,070 --> 00:33:33,750 1062 00:33:33,750 --> 00:33:37,190 1063 00:33:37,190 --> 00:33:40,630 1064 00:33:40,630 --> 00:33:42,950 1065 00:33:42,950 --> 00:33:44,789 1066 00:33:44,789 --> 00:33:47,509 1067 00:33:47,509 --> 00:33:48,950 1068 00:33:48,950 --> 00:33:50,389 1069 00:33:50,389 --> 00:33:52,310 1070 00:33:52,310 --> 00:34:01,190 1071 00:34:01,190 --> 00:34:03,029 1072 00:34:03,029 --> 00:34:05,029 1073 00:34:05,029 --> 00:34:07,509 1074 00:34:07,509 --> 00:34:10,149 1075 00:34:10,149 --> 00:34:12,470 1076 00:34:12,470 --> 00:34:12,480 1077 00:34:12,480 --> 00:34:13,589 1078 00:34:13,589 --> 00:34:13,599 1079 00:34:13,599 --> 00:34:14,869 1080 00:34:14,869 --> 00:34:16,310 1081 00:34:16,310 --> 00:34:19,190 1082 00:34:19,190 --> 00:34:19,200 1083 00:34:19,200 --> 00:34:23,349 1084 00:34:23,349 --> 00:34:25,829 1085 00:34:25,829 --> 00:34:30,149 1086 00:34:30,149 --> 00:34:31,829 1087 00:34:31,829 --> 00:34:34,710 1088 00:34:34,710 --> 00:34:36,710 1089 00:34:36,710 --> 00:34:36,720 1090 00:34:36,720 --> 00:34:37,750 1091 00:34:37,750 --> 00:34:39,990 1092 00:34:39,990 --> 00:34:43,190 1093 00:34:43,190 --> 00:34:45,510 1094 00:34:45,510 --> 00:34:47,109 1095 00:34:47,109 --> 00:34:48,389 1096 00:34:48,389 --> 00:34:50,310 1097 00:34:50,310 --> 00:34:52,310 1098 00:34:52,310 --> 00:34:54,629 1099 00:34:54,629 --> 00:34:56,470 1100 00:34:56,470 --> 00:34:58,390 1101 00:34:58,390 --> 00:34:58,400 1102 00:34:58,400 --> 00:35:04,630 1103 00:35:04,630 --> 00:35:04,640 1104 00:35:04,640 --> 00:35:05,430 1105 00:35:05,430 --> 00:35:07,030 1106 00:35:07,030 --> 00:35:07,040 1107 00:35:07,040 --> 00:35:08,550 1108 00:35:08,550 --> 00:35:10,790 1109 00:35:10,790 --> 00:35:10,800 1110 00:35:10,800 --> 00:35:13,589 1111 00:35:13,589 --> 00:35:13,599 1112 00:35:13,599 --> 00:35:15,750 1113 00:35:15,750 --> 00:35:17,430 1114 00:35:17,430 --> 00:35:17,440 1115 00:35:17,440 --> 00:35:18,870 1116 00:35:18,870 --> 00:35:18,880 1117 00:35:18,880 --> 00:35:20,470 1118 00:35:20,470 --> 00:35:22,150 1119 00:35:22,150 --> 00:35:22,160 1120 00:35:22,160 --> 00:35:23,270 1121 00:35:23,270 --> 00:35:25,109 1122 00:35:25,109 --> 00:35:25,119 1123 00:35:25,119 --> 00:35:26,150 1124 00:35:26,150 --> 00:35:29,349 1125 00:35:29,349 --> 00:35:31,349 1126 00:35:31,349 --> 00:35:34,150 1127 00:35:34,150 --> 00:35:36,470 1128 00:35:36,470 --> 00:35:38,550 1129 00:35:38,550 --> 00:35:39,510 1130 00:35:39,510 --> 00:35:42,630 1131 00:35:42,630 --> 00:35:42,640 1132 00:35:42,640 --> 00:35:43,349 1133 00:35:43,349 --> 00:35:45,829 1134 00:35:45,829 --> 00:35:48,310 1135 00:35:48,310 --> 00:35:49,430 1136 00:35:49,430 --> 00:35:50,790 1137 00:35:50,790 --> 00:35:53,109 1138 00:35:53,109 --> 00:35:55,829 1139 00:35:55,829 --> 00:35:58,550 1140 00:35:58,550 --> 00:36:00,230 1141 00:36:00,230 --> 00:36:04,310 1142 00:36:04,310 --> 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01:15:20,630 --> 01:15:22,870 2369 01:15:22,870 --> 01:15:24,790 2370 01:15:24,790 --> 01:15:26,790 2371 01:15:26,790 --> 01:15:29,030 2372 01:15:29,030 --> 01:15:30,870 2373 01:15:30,870 --> 01:15:34,070 2374 01:15:34,070 --> 01:15:34,080 2375 01:15:34,080 --> 01:15:34,870 2376 01:15:34,870 --> 01:15:37,510 2377 01:15:37,510 --> 01:15:39,510 2378 01:15:39,510 --> 01:15:41,110 2379 01:15:41,110 --> 01:15:42,870 2380 01:15:42,870 --> 01:15:45,110 2381 01:15:45,110 --> 01:15:46,390 2382 01:15:46,390 --> 01:15:49,110 2383 01:15:49,110 --> 01:15:51,590 2384 01:15:51,590 --> 01:15:53,430 2385 01:15:53,430 --> 01:15:55,350 2386 01:15:55,350 --> 01:15:56,709 2387 01:15:56,709 --> 01:15:59,110 2388 01:15:59,110 --> 01:16:02,470 2389 01:16:02,470 --> 01:16:04,830 2390 01:16:04,830 --> 01:16:06,790 2391 01:16:06,790 --> 01:16:11,750 2392 01:16:11,750 --> 01:16:12,950 2393 01:16:12,950 --> 01:16:14,870 2394 01:16:14,870 --> 01:16:17,110 2395 01:16:17,110 --> 01:16:18,709 2396 01:16:18,709 --> 01:16:23,510 2397 01:16:23,510 --> 01:16:25,430 2398 01:16:25,430 --> 01:16:27,510 2399 01:16:27,510 --> 01:16:29,189 2400 01:16:29,189 --> 01:16:29,199 2401 01:16:29,199 --> 01:16:30,310 2402 01:16:30,310 --> 01:16:32,790 2403 01:16:32,790 --> 01:16:34,870 2404 01:16:34,870 --> 01:16:36,950 2405 01:16:36,950 --> 01:16:40,630 2406 01:16:40,630 --> 01:16:42,709 2407 01:16:42,709 --> 01:16:45,110 2408 01:16:45,110 --> 01:16:45,120 2409 01:16:45,120 --> 01:16:47,030 2410 01:16:47,030 --> 01:16:49,270 2411 01:16:49,270 --> 01:16:52,310 2412 01:16:52,310 --> 01:16:54,550 2413 01:16:54,550 --> 01:16:56,790 2414 01:16:56,790 --> 01:16:56,800 2415 01:16:56,800 --> 01:16:57,590 2416 01:16:57,590 --> 01:16:59,030 2417 01:16:59,030 --> 01:16:59,040 2418 01:16:59,040 --> 01:17:00,070 2419 01:17:00,070 --> 01:17:00,080 2420 01:17:00,080 --> 01:17:00,950 2421 01:17:00,950 --> 01:17:04,310 2422 01:17:04,310 --> 01:17:06,470 2423 01:17:06,470 --> 01:17:07,910 2424 01:17:07,910 --> 01:17:09,990 2425 01:17:09,990 --> 01:17:11,830 2426 01:17:11,830 --> 01:17:14,229 2427 01:17:14,229 --> 01:17:15,669 2428 01:17:15,669 --> 01:17:18,390 2429 01:17:18,390 --> 01:17:19,669 2430 01:17:19,669 --> 01:17:21,990 2431 01:17:21,990 --> 01:17:25,030 2432 01:17:25,030 --> 01:17:26,870 2433 01:17:26,870 --> 01:17:28,870 2434 01:17:28,870 --> 01:17:30,709 2435 01:17:30,709 --> 01:17:30,719 2436 01:17:30,719 --> 01:17:32,229 2437 01:17:32,229 --> 01:17:34,950 2438 01:17:34,950 --> 01:17:37,270 2439 01:17:37,270 --> 01:17:37,280 2440 01:17:37,280 --> 01:17:39,350 2441 01:17:39,350 --> 01:17:42,149 2442 01:17:42,149 --> 01:17:45,669 2443 01:17:45,669 --> 01:17:47,669 2444 01:17:47,669 --> 01:17:50,229 2445 01:17:50,229 --> 01:17:50,239 2446 01:17:50,239 --> 01:17:54,310 2447 01:17:54,310 --> 01:17:56,830 2448 01:17:56,830 --> 01:17:59,030 2449 01:17:59,030 --> 01:18:01,110 2450 01:18:01,110 --> 01:18:03,669 2451 01:18:03,669 --> 01:18:07,990 2452 01:18:07,990 --> 01:18:10,790 2453 01:18:10,790 --> 01:18:26,630 2454 01:18:26,630 --> 01:18:26,640 2455 01:18:26,640 --> 01:18:28,719