1 00:00:25,119 --> 00:00:27,509 welcome everyone um back to um theory of computation so we are now at lecture number 15 and if this is an important lecture um well all of our lectures are important but this is i'm going to introduce one of the major topics that we're going to see going on in various forms through the rest of the semester which is the notion of a complete problem in this case it's going to be an np complete problem and i think some of you have probably heard of that concept already maybe you've seen it in some courses or other but we're going to do that in a in a rather careful and formal way over the next couple of lectures so we are following up on our previous discussions about complexity time complexity we defined the time and non-deterministic time complexity classes as you may remember the classes p and np talked about the p versus np problem looked at some interesting algorithm for showing problems in p called dynamic programming and we started to move toward our discussion of np completeness with the introduction of polynomial time reducibility which is related to some of the earlier reducibility notions that we discussed in the computability section so let's [Music] quick review um what it means for one problem to be polynomial time reducible to another this follows our pattern of reducibility concepts where um if a problem is reducible to another problem and that other problem is solvable then the first problem is solvable so if a problem is reducible to an easy problem that problem becomes easy too um and the kind of reduction that we're going to be looking at here are the mapping reductions but now where the reductions are computable in polynomial time and so we had this um result we mentioned last time that if a is polynomial time reducible to b and b is in p then a is also in p um [Music] so that's going to be um critically important for the you know for the for the whole um of discussion that's coming um our intuition about pnnp um repeating that here um the np problems are the ones where you can verify membership easily as opposed to being able to test membership easily those are the problems that are in p the verification typically requires some kind of a certificate that establishes kind of a proof that the input is a member of that language um and the big question of the field which remains an unsolved problem is whether p equals np it's called the p versus np question and we don't know the answer so whether there are problems that are in np um solvable and nondeterministic polynomial time typically these are problems that involve searching whether they can be solved in polynomial time typically without the searching um and uh if p were equal to np then you could always eliminate searching and a p were different from np then there were cases where you need to search um and we don't know the answer to that uh so in the direction of exploring uh this this question and its ramifications we introduced this problem called sat these are the boolean formulas which have an assignment that enable that event makes them evaluate to true um so we call those satisfiable formulas and uh we um mentioned but have not yet proven and we will not prove until the next lecture on tuesday that there is this theorem uh kind of a very remarkable theorem that says that if you have uh you take the satisfiability problem and if it is solvable quickly then all of the np problems are solvable quickly so if sat is in p then p equals np so in a sense sat is kind of you know the uh it's sort of the super np problem in the sense that all of the difficulty of any np problem is is embedded within sat so if sat becomes easy then all of the other np problems become easy um and we'll eventually prove that but right now we're kind of setting up the um terminology to allow us to do that um so uh anyway we'll we'll we'll get there um so the the the key ingredient for proving this kook 11 theorem is polynomial time reducibility what we're going to show is that every problem in np can be polynomial time reduced to set so every np problem can be converted into a set problem and so working toward that because when you think about it you know there are infinitely many problems in np and um being able to show that all of them are reducible to sat as in a sense kind of um you have to prove a kind of a higher level theorem it's not just a single reduction we're kind of exhibiting a schema of reductions which shows that all of these problems can be reduced to set um and kind of developing our um you know our intuition toward that direction we're going to look at some specific polynomial time reductions today um and we'll start out with a reduction between two problems we have not yet seen one of them called threesat and the other one called clique so on this slide we're going to introduce those two problems so remembering again about boolean formulas we're going to consider a special class of boolean formulas a restricted form of boolean form formulas called conjunctive normal form and so this formula here in particular is in that conjunctive normal form um so just remember um i was explaining some of this to someone else uh earlier this morning showing some of these slides and it was pointed out that not all of you may have be familiar with these um the boolean operations of and and or uh so this is the or symbol here this is the and symbol here you know they operate you know hopefully you've seen the just the concept of boolean and or or is in a sense a little bit like a union um and it's like an intersection and the the symbols here are kind of a little bit similar to the union and intersections you know the intersection the union intersection symbols themselves like or the v-shaped is a little bit like a pointy union symbol and the in the um upside-down v-shape is a little like a pointy intersection symbol so maybe that's if you haven't seen those seen that connection before maybe it's interesting to observe that but anyway um what makes this uh formula be in conjunctive normal form well the conjunctive normal form formulas are organized in a certain way they have um these groups called clauses within the parentheses that are added together that are connected by these and and operations um and within those groups which are called clauses the um elements are ored together those elements are going to be either variables or variables with negation negated variables um so uh just to repeat that uh these well i'll just just state those those variables or negated variables are going to be called literals um and the ors of a bunch of literals are going to be called clauses okay this is just the standard terminology in the field okay so um a literal is a variable or a negated variable and a clause is an or of literals okay just a definition all right so we have a bunch of these clauses here each of the clauses has an or of of liter of literals in it now a chomsky conjunctive normal form uh formula i guess they're both c and f's but uh it's conjunctive normal form formula is one that's written as an and of these clauses okay so we take these clauses which themselves are ors we and them together we get a formula that's conjunctive normal form conjunctive stands for and i think uh conjunction is an and um and then a three cnf is a cnf where each clause has exactly three literals so for example this particular formula is a cnf but it's not a three cnf because it has at least two clauses which violate the condition of three literals per clause this one is okay obviously this first clause and the three sat problem so this is the first of the two problems we're going to be talking about the three sad problem is the satisfiability problem restricted to these three cnf formulas so this is the collection of setup of three cnf formulas which are satisfiable so you can think of it as a kind of a special case of the sat problem where we only care about formulas of the special kind um this being a special case you might imagine that this might be an easier problem to solve but in general it turns out not to be as we will see this is just as solving this special case is just as har as hard as solving the general case for satisfiability um now let's turn to the second of these two languages up in the headline uh the clique problem um so for that we gonna turn to graph theory so we're going to consider graphs you know points and lines connecting them and we will say that a clique in a graph is a collection of nodes collection of the points that are all pairwise connected by lines and a k click is one where you have case arch nodes so here we have a three click four click and five click um and the clique problem is to try to find cliques of a certain specified size embedded within a given graph so you'd like to know whether you're going to be given a graph and it targets click-sized k and you want to know is there a subset of the nodes in the graph of size k that are all connected to one another that's the cleat problem um so obviously um the clique problem just as with the satisfiability problem is a decidable problem you can just try every possible subset of k nodes and see whether it constitutes a clique but that's in general going to be an exponential algorithm if you have you know you know k is a large value like it might be half the size of the graph you might be looking for a very large clique then you have to try many many subsets in order to see whether uh any one of them is a clique it's also going to be and i hope you get this intuition this is a problem that's in np the cleat problem is an np problem because you can easily verify that a graph uh has a k click just by exhibiting the click okay um now uh uh so the clique language here is you're given a graph and a k and you want to know does the graph contain a cake league um and what we're going to show is that these two problems are connected that the three set problem and the cleat problem are related in that you can reduce in polynomial time three sat to click okay so this is um you know at you know just standing back at it back for a minute you know that seems kind of surprising there's no real reason you know uh obviously anyway no obvious reason why there should be a connection between three sat and click click they look very different from one another but we do give that uh we will give that reduction and um it then that implies that if you can find a way of solving the clique problem quickly that'll give you a way of solving the the three set problem quickly um and that's going to be the whole point of this um so we're going to show this over the next slide or two um this this polynomial time reduction and this is something i'm going to walk through it slowly but this is one where you it's really important to try to get a sense of how it's working um uh because this is the kind of thing that we're going to be doing a lot of and you're going to be asked to do it also on the homework and and possibly on the final exam i hate to use that as the motivating force here but at least it might be one motivation for some of you that you have to understand how to do uh this kind of a reduction okay so you know i i think at a high level what's going on is we're showing how to recode a boolean formula satisfiability problem into the problem of testing whether a graph has a clique um so let's see if we have any questions here that are coming up in the in the chat um okay so i this is an interesting question here can we always convert a boolean formula into conjunctive into conjunctive normal form first of all yes the answer is you can always convert a boolean formula into an equivalent one in ins into in cnf but in general that might make the formula exponentially larger [Music] so just the mere fact that you can convert formulas into into conjunctive normal form doesn't mean that solving um that you know that solving uh um conjunctive normal form uh formulas for testing satisfiability of the of the cnf formulas is going to be as hard as testing the general case because just the conversion might be exponential um there's something more uh more a little bit more complicated going on um than that so let's uh um let me just see here uh if he is a satisfiable formula which is not in cnf can be so a similar um you know similar question so the questions that i'm getting are about converting formulas uh to cnf um so yeah you can do it but not in polynomial time in general because the resulting formula you get might be much larger um if you if you're looking for an equivalent formula if you're not looking for an equivalent formula then you know of course then depending upon what you're looking for you might be able to find something smaller um ah this is a this is a i guess a good basic question why is clique in np um because isn't doesn't verifying that you have a clique require going through all the possible cliques once i you have to understand what verifying means verifying means you can verify something if you're given a certificate in the case of a clique the click problem the certificate is the clique um so once you have the certificate you can do the verification in polynomial time finding the certificate of course might be difficult so you only think of np in the context of having that certificate so in a case for like compositeness the certificate might be the factor um you know anyway i hope that's you know it's there there are problems sometimes with what the certificate is is not necessarily obvious um but there can be a certificate um for um showing that the that the input is in the language but in the ones that we've done so far maybe you can argue that this certificate is sort of an obvious thing but it's not always an obvious thing um okay why don't we move on then um so let's see how do we do this polynomial time reduction from three sat to click okay here we go um so i'm just going to give a reduction that's that's that's what the definition means i'm going to give a way of converting formulas to pairs a g and a k where the formula is going to be solid satisfiable if and only if the graph has a k click okay so let's kind of a little bit do it by example and in order to do that so here's going to be a formula now it's in three cnf that's what i need you know in order to be doing this reduction i'm converting three cnf formulas into clique problems we have to a little bit understand what it means when we say uh we talk about the satisfiability of a formula like this because it's going to be helpful in doing the reduction you know obviously it's satisfiable means that the form you can find an assignment of the variables so you're going to set each of these variables a b and c and so on to true or false and you want to make the whole formula evaluate to true but what does that actually mean [Music] it means that because of the structure of the formula that making this formula true corresponds to making each clause true because the clauses are all added together so the only way for the formula to be true is to make each clause true and to make a clause true you have to make at least one of the literals true so it's another way of thinking about uh satisfying this formula you need satisfying these and a satisfying assignment makes at least one true literal in every clause it's really important to think about it that way because that's what's going to be um [Music] kind of the basis for doing this reduction and all of the reductions it's kind of what makes three set easy to think about in terms of its satisfiability if you had a sort of a general um satisfiability problem and you had a satisfiable formula there's no sort of obvious way of sort of seeing what the satisfying assignment looks like but here we understand what it looks like it has that very special form making one true literal in every clause at least one true literal in every clause okay so now we're going to do the reduction so i'm going to take from this formula you and you have to understand you know i know for some of you you know you're going to be you know chafing why am i going so slowly but i want to make sure that we're all together in understanding what the rules of the game are and what we're trying to do um we're trying to convert this formula into a graph and a number so right now my job is to do this reduction is to exhibit that graph okay so i'm going to do that in kind of two steps i'll first i'm going to tell you what the nodes of the graph are then then i'll tell you what the edges of that graph finally i'll tell you what the number k is that's the way this polynomial time reduction is going to work and we have to also observe at the very end that the reduction that i'm given giving you this procedure for building this graph can be done in polynomial time but that i think you'll see once i'm done that that's pretty obvious okay so first as i promised the nodes um so the nodes of this graph are going to correspond to the literals of the formula every literal is going to become a node in the graph and it's going to be labeled with the um name of that literal so it's every node is going to be labeled an a or b or or c bar and so on so here it goes okay so those are the nodes of the graph g one for each literal in the formula okay labeled as promised now i have to tell you what the edges look like the edges are going to be um i'm going to tell you what the edges are by first telling you what they are not i'm going to first explain to you what the forbidden edges are what what edges i'm going to promise not to include and then the ones that i will include are going to be all the others okay so what are going to be the forbidden edges first of all um the the forbidden edges are going to be of two types one is the edges between nodes that come from literals in the same clause so i would just call that no edges within a clause okay so for example these three nodes will not be connected to one another and i'm going to indicate that by writing red dashed lines which means there's not an edge there okay those are those are forbidden from having an edge okay so these three have no edge and the same thing for every other uh triple of three nodes that come from uh clauses okay those are no edges there and there's one other category of edge that i'm forbidding and that are edges that can that um go between inconsistent labels and the nodes with inconsistent labels so for example between a and a bar those are inconsistent you know nothing's wrong with a going a to d those are not inconsistent those are just different labels or going from a to a that's okay but from a to a bar that's that's not allowed um so that's going to be another forbidden edge or for example a to a bar here for b a bar to a or for example from this c bar to c forbidden okay so imagine you're going to write down all of those forbidden edges and then once you've done that uh those are all of the the forbidden edges and then after [Music] taking those away you're going to be putting in all the other edges possible so for example as a let me just uh gray those out so they don't interfere with the picture b from a to a to d um there's going to be an edge um because those are not forbid you know a to a not forbidden because they they're in different um they are in different uh clauses and they're not inconsistent they're they're consistent with one another so here are a whole bunch of other i'm not showing them all becomes very messy but uh here are all the other edges between nodes uh which are where there isn't there's no pro within they're not forbidden okay and that's that's the graph that's g g is all those nodes and those edges um which are not forbidden now i just have to tell you what k a is k is going to be the number of clauses that's going to be the size of the clique i'm looking for in this graph g that i just spelled out for you and i'm going to claim that this graph here that i just described will have a k clique k is the number of clauses exactly when fee was satisfiable it's kind of kind of cool you know um if he is satisfiable then there will be a cake leak here and if he was not satisfiable okay clean so we have to we're gonna prove that as a claim on the next slide okay so k is the number of clauses all right um any questions on that on that construction so i've done with the construction what's left is to argue why the construction works so far so good let's move on okay um so um so here is that very same construction i eliminated the red uh forbidden edges these are the ones that were remaining plus you know anything else that was not forbidden that was the same formula that i had from the previous slide now i'm going to claim that that formula is satisfiable exactly when g has a k click now why in the world is that so this is an if and only if it's proved in both directions and this is going to be the typical kind of thing that you're going to want to do when you're exhibiting a uh you know one of these reductions which is you're going to have an opportunity to do that um and we'll do do that too in our examples uh okay so now uh what i'm what i want to show is that uh if fee is satisfiable so let's prove the forward direction if p is satisfiable then g has a k click okay so first of all if fee is satisfiable that means it has a satisfying assignment okay now here's a common confusion i'm not sure whether it's helpful to sprinkle the confusions in along with the discussion but in case you're worried you might ask well how do i find that satisfying assignment i thought that was you know exponential exponentially hard to do this is a proof this is not an algorithm okay we're just trying to argue the con correctness of this construction so there's no longer any concern about how to find an assignment i'm just saying if there isn't a sign if there is a satisfying assignment then something will happen something that then then um something will happen in particular then we will show that g has a cake league so let's say fee is satisfiable so it has some assignment let's take that satisfying assignment and remember that in any satisfying assignment uh to a formula in cnf that makes at least one literal true in every clause so let's just pick one of those there might be several clauses which have multiple true literals in that case just pick one of them arbitrarily okay so i don't know what the i don't have this sort of indicated on this diagram here but imagine you know maybe in the very first clause a was true in the second clause b was true in the third clause e compliment ebar was true and you know not not you know e bar was true which means e itself was false but e bar was true and that's the way each of those clauses in turn uh got to be true you know in this particular satisfying assignment because you're going to pick one true literal in every clause and now from that choice of literals one per clause i'm going to look at the corresponding nodes in g and i'm going to claim that those nodes taken together form a k click okay so first of all do they have the right number of nodes well sure because i'm picking one node per click i already said k is the number of cliques so i'm getting exactly ka nodes so i have at least the right number of nodes but how do i know they're all connected to one another those nodes uh that i just picked well they're all connected to each other because i'm going to say well because there were no forbidden edges among them and remember that we put in all possible edges except for the forbidden ones so how do i know there are no forbidden edges well what were the rules for being forbidden means they two nodes in the same clique in the same clause well i'm picking one node per clause so i can never be having two nodes from the same clause so i'm never going to run into trouble with uh the first condition of being forbidden so what's the second possibility for being forbidden is that i'm picking two nodes which are inconsistent well how do i know that i didn't end up with two nodes with inconsistent labels that would be bad because then they would have a forbidden edge and what my result would not be a clique how do i know i didn't end up picking um uh in this group i pick a and in this group i pick a a bar well because they all came from the same assignment those are all of the true letters they were all true literals in clauses it's not possible that a was true in this clause and a bar is true in that clause because if a is true here a bar has got to be false it can't be the true literal in this clause so i cannot have any inconsistent any inconsistent nodes appearing among the nodes of my clique and so they're not in the same clause they're not inconsistent so the edge has to be there and that's going to be true for every pair of nodes in that clique in that in that group of nodes and that's why it's a click um so that proves one direction now we stop to prove the other direction because we have to say well if g has a k clique how do we know that ph is satisfied so that's the reverse so let's just take any k clique that's in g and how do i know i can get from that a satisfying assignment to the formula um uh good um getting some good questions here in the chat but let me just move on um take any uh so proving the reverse this the reverse direction but we're assuming we have a cake leak to take any such k-click now first of all you observe that it's got to have one node in every clause it can't have two nodes in the same clause or zero nodes in in a clause first of all it can't have two nodes in the same clause why well because those nodes are never connected to one another so that they cannot be both in the same clique because all the nodes in a clique are connected to one another um by construction by the way we construct a g nodes from the same clause are not connected so they cannot appear in a clique together so um there's going to be at most one node from every clause appearing in this clique but and but but appearing in in this clip but we also have we know we have k notes so that means if there is at most one per um per clause and we only have k clauses that means every clause has got to have one you know if some clause is missing or note then there's going to be some other clause that has two so we know that every um that this clique that would that g has has exactly one node in every clause so how do we get from that a satisfying assignment so what we're going to do is take the current corresponding literals that correspond to that one node in the that was in the uh clique take those literals and set those literals to be true which means setting the variable to be true um or setting um you know if the literal was a bar then setting a to be false because you want a bar to be true you want to set the literals to be true well now we're setting um there's one note in each clause we're setting one literal uh the corresponding literal true so that's going to set one true literal in every clause so that means it's going to be satisfying there's one thing that one has to really to be careful here to double check to make sure that we're doing this consistently that we're not being asked to set you know a true and also a bar true because then that would not be possible but a and a bar are not connected so it's we're never going to be trying to set both a and a bar true if so we're going to be trying to set a true we're never going to try to set a bar true because those are not those a and a bar cannot be in the same clique they're not connected to each other okay so that is the argument um and lastly i claim we're not going to look at this in detail that it's kind of obvious that this reduction can be done in polynomial time namely if i give you that formula you can write out that graph in pretty easily there's no hard work to be done in writing out that that graph or counting up the number of clauses so that's that's the proof that i can reduce three statically um and it tells you also that if i could solve click quickly then i can stop threesat quickly and this is the whole point because i can convert now clique problems to three sat problems no no i couldn't i'm sorry i got backwards convert three set problems the clique problem so if i can solve quick click easily then i can solve three set easily right i just showed a way of converting these formulas to graphs so if that's so that says if i can solve the graphs easily then i can solve the formulas okay why don't we uh turn to some oh question checking but you know we can also um why don't i i'll launch a check-in but we'll we let's just see i'm seeing many many questions here which are actually what my check-in is about i'll turn it back to you guys uh okay so where did we use the fact um that we have three uh literals per clause what does this thing just work even if we had any number of literals per clause um what do you think we're having a we've got a tight race here hmm uh truth is losing out unfortunately all right oh it's really close but still okay uh one more okay it's neck and neck okay almost done 10 seconds are we done here okay that's it uh ending polling uh yes truth is thank god um yeah it works for any size clause we didn't use the fact that it's a three c and f it could have been any number here uh so it's well we've got a i guess it's a plural plurality here though that not a majority um no we didn't where did we use three in any of these argument i didn't mention it maybe you were imagining that that is going to be part of it but three does not come into this at this discussion at all if you think about how it's why it's working um even if we had you know one of these clauses had 10 literals in it as long as k is the number of clauses and we don't connect any variables any literals internal to a clause this whole argument is going to still work so please check that make sure you understand what's going on because i can see that a good chunk of you have not got this right um uh so i got a good question here uh what if it's only just one big clause so it's like the whole formula is just one big war so what does happen in that case good question so in that case um uh so the you know suppose has one big clause with a hundred literals that's the whole formula it's just a big war so we know the formula is going to be satisfiable by the way obviously um so if you look at the corresponding g it's going to have 100 nodes because there's one for every literal um none of them are going to be connected to each other because they're all in the same clause um so it looks like we're going to be in tough shape trying to find a clique there because there's no edges at all everything's forbidden but what is k k is going to be 1 in that case because there's only one clause and uh kind of a degenerate case but a clique with just one you know just a single node is it counts as a one clique because it's just one node there's no need for any edges at all it still counts as a clique so it'll still it'll still work out um uh let's see what else here uh that was kind of a fun question that you asked me thank you um uh there are a lot of questions um i think we're actually pretty near the break also we'll just see yeah um so why don't i um oh many many questions here uh so i i think i'm gonna start start the clock going down for our um our coffee break and um and i will take some of these questions i'll try to answer some of these questions oops um try to answer some of these questions um afterward uh all right good all right um so i got a question about making sure why nodes don't have inconsistent labels if i understand the question correctly um so these would be nodes um so i never put edges between nodes with inconsistent labels that's the rule that's my construction i get to i get to say how the construction works so those two nodes with inconsistent labels can never be in the same clique because there's no edge between them so i'm not sure that answered the question there but that was uh um let's see so getting another question here since any boolean formula can be converted to cnf does that mean um is sat polynomial time reducible to clique as well monomial time reducible to clique but not for the reason that any boolean formula can be converted to c and f because that conversion um is going to be an exponential conversion in general the conversion to c and f so you have to be careful about what we mean by converting a formula to cnf um uh so also getting questions what happens in the case of one set well we didn't really talk about one set uh suppose or i guess so the question is suppose there's only a single literal per clause that's an interesting another sort of edge case here what happens under those circumstances so um there's only a single literal per clause um uh so you have to work out what happens but in that case um uh all the clauses just have one literal in them and now they have um uh um the only way uh this if you think about what is a when you have a one cnf so there's only um one literal per clause the only way that can be satisfiable is you have to make each literal true because there's no warring anymore so every literal has to be true in the assignment so that means you can never have inconsistent literals um and that's the only case when you would not have an edge because you're not the forbidden is the forbidden condition won't happen because you don't have more than one uh node per per clause every clause is going to have just a single node in it uh so it really comes down to whether or not you have an inconsistently labeled um inconsistent clauses i i didn't explain that super well but it's the the argument still works there you should check it for yourself um so this is a basic question how do we see that f is polynomial time i i'm not sure i want to expend a lot of you know a lot of discussion on that because the the conversion from the formula to the graph is kind of a one-to-one you know every no every uh literal becomes a node um the rule for when edges are present it's a very simple rule so i guess i would argue i'm not sure what what more to say it seems pretty clear that the conversion that reduction is polynomial time computable i mean if you wrote a program to do that it would operate very quickly um is it possible for g to have a k plus n clique where n is greater than zero and does that matter now if you think about it the biggest possible clique that this graph g could have is k it could never have more than uh uh you know a bigger clique because two nodes in the same uh clause are never connected you only have k clauses so the answer is no you cannot have a bigger a bigger than a k click it's just not going to happen does each clause need to have the same number of literals i don't see why um uh so question is why are we worrying about threesat um since it didn't seem to matter here there are other examples where it does matter i i i'm actually not sure if we'll see one or not but um there there are cases where it does matter um let me just see if there's any quick questions here uh somebody's saying doesn't does do we have to worry about there being a polynomial number of literals so you have to think about what you what that question means the size of the input which includes all of the literals is n so there's going to be at most n literals um because that's the size of the input uh the i mean at most uh and literals appearing so you don't have to worry about that being polynomial because by the way we're we're defining the size of the input is going to be at most n literals i hope that's clear uh okay when we talk about power okay so this is a good question when we talk about polynomial time it's polynomial in the representation of of the input which is if you want to think about it in terms of bits that's fine or whatever it's not going to matter if you use some larger alphabet but the number of symbols you need you know in your fixed size alphabet the number of symbols you need to write down that input in whatever your reasonable encoding is is going to be n and it's going to be polynomial in n polynomial in that length of the input um so another question is does sat reduce the three set yes we that we will see um sat actually does polynomial time reduce the three set not by converting to an equivalent formula but by some more uh somewhat more involved argument uh than that okay why don't we move on some of these are going to come out anyway in the lecture okay so now let's talk about mp completeness because we've kind of set things up we're not going to prove that you know the basic theorem the cook 11 theorem about mp completeness but at least we'll be able to make the definition okay here's our definition you know what it means for a problem to be in be complete um uh so a language b is called np complete if it has two properties number one is that has to be a member of np and number two every language in np has the polynomial time reduced to that to that language to that np complete language in order for it to be np complete okay so simple picture has to be in nnp and everything in np reduces to it um and so that's kind of the magical property that we claimed that that sat has sat for for one thing is obviously in np and as we the cook 11 theorem shows or will show everything in npe is reducible to sat so that's going to be our first example of an np complete problem um uh so now uh and we're going to get what we claimed also for sat that if sat or any other np-complete problem turns out to be solvable in polynomial time then every np problem is solvable in polynomial time and and that's immediate because everything is reducible in polynomial time to the np complete problem so if you can do it easily you can do everything easily just by going through the reduction okay so the quick level theorem as i mentioned is that sad is np-complete and we're going to actually prove it next lecture but let's assume for the remainder of this lecture that we know it to be true because so i'll use the terminology of problems being np complete uh assuming that we know um that we have uh sat as empty complete okay so we're gonna we're gonna be using some of the things that were proving next lecture uh just in the terminology that we're going to be talking today okay so here's the picture here's the class np and everything in np is polynomial time reducible to set sat itself is a member of np but i didn't want to show it that way because it makes the picture kind of hard to uh hard to uh hard to display so just from the perspective of the reduction everything in np is polynomial time reducible to sad we'll show that next lecture another thing that we'll show next lecture is that sat in turn is polynomial time reducible to three sat so three stat as you remember are just those problems that are in conjunction three are in three c and f um and then what we showed today is that three set is polynomial time reducible to clique so now taking the assumption that sat is np complete so everything is polynomial time reducible to sat which is in turn polynomial time reducible to three sat in in turn reducible to click these reductions as we've seen before composed um you can just reply apply one reduction function after the next if each one each one individually is polynomial the whole thing as a combination is going to be polynomial so now we know that three sat is going to be also np complete because we can reduce anything in np to stat and then to three set and then we get a reduction directly to three sat by composing those two reductions and then furthermore at the clique so now we uh have several np complete problems and moving beyond that we have the hand path problem which we are going to talk about next and we'll show another reduction um in addition to the one we just showed to click now one going from three set to hand um okay uh so in general i think the the the the take away message is that to show some language as np complete you want to show that three sat is polynomial time reducible to it um uh okay some good questions coming in i'll try to answer those so you're going to retake 3sat and reduced to c that's the sort of the most typical case or you might have there's going to be other examples too we might start with another problem that you've already shown to be np complete and read and reduce it to your language so it doesn't have to start with reset though often it does um okay why do we why is this concept important so i would say there are two reasons um and this is going to get a little bit at the uh some of the chat questions um first of all if you're faced with some copy new computational problem you've got some robotics problem um that you want to solve in your thesis and you need to some algorithm about whether the robot arm can do such and move in a certain such a way and involve searching you know possibly searching through a space of different kinds of motions and you want to know you know i need a either i'd like to have a i like to find a polynomial algorithm to solve that problem i'm using this as an example because this actually did happen to one of the former students from this class who was working in robotics and they actually end up proving that the problem that they were trying to solve is np complete um so um uh that's useful information because even though knowing a problem is np complete doesn't guarantee that it's not in p because conceivably p equals np what it does tell you that if you show if you have a problem that's np complete and it does turn out to be in p then all then p equals np so the tremend would be tremendous surprising consequences of of your problem if it was known to be np-complete ending up in p so generally people take a proof of np completeness as powerful evidence that the problem is not in p even though it's not quite a proof it's powerful evidence that it's not in p and so you might as well give up working on trying to find a polynomial algorithm for it because if you do you know you don't have to worry about robotics anymore you're going to you know become famous as a complexity theory person um so i wouldn't worry about so if you show problems np complete you can pretty much assume it's almost certainly not in p um now there's another reason related to the to the for the theorists uh to be care about mp completeness and that is if you're trying to prove p is different from np or p equals np as one of the chat questions is raising um you know the problem you know in order to prove p different from np you know the most likely approach is you pick some p problem in np and show it's not in p right that's what it would mean for them to be different you're going to pick some np problem and show it's not not a p problem well um one thing would be terrible is possibly p is different from np and you pick the wrong problem like suppose i'd spent all my time trying to show you know back 20 years ago i'm working really hard trying to show composites is not in pain which is would have been perfectly reasonable to do because composites is an np and was not known to be in p 20 years ago um and then i invested tons of effort to try to prove because you know i like number theory or god knows what to prove that composites is not in p and then it turns out composites wasn't in p it was the wrong problem to pick even though p might be different from np but what np completeness guarantees is that if you work on a problem which is np-complete you could you can't pick the wrong problem because if any problem is in np and not in p an np complete problem is going to be an example of that right because if the np complete problem is in p everything in np is in p so if np is different from p you know the complete problems are not in p so you might as well work on one of those so those are two ways in which np completeness and has turned out to be is an important concept okay so here's a check-in um uh again this you guys are kind of getting maybe you're getting started to think the way i think you get is some at least one of you would ask this question in the chat um which language that we've probably seen is most analogous to sat okay atm etm or zero to the k1 to the k uh obviously this is maybe subjective you may have your own interpretation of what that means this in a sense perhaps no right answer but what do you think uh okay that's it um i don't know i don't know how in the world that you could see this problem is analogous to to zero to the k1 to the k but okay um i'm sure you have your reason um yes this is a lot like eight atm why well because for one thing we showed in a homework problem that all um touring recognizable languages are reducible to atm mapping reducible to atm so that's a little bit like the notion of completeness that we have for satisfiability um because all np problems are going to be reducible to sad um and the other thing too is that once we start once we uh we started we want to show other problems are undecidable we reduced atm to them and that's also very similar we're going to be reducing sat or threesat so it's indirectly from sat to other problems in order to show that they're np complete so that they're hard they're they're that they're hard in a sense and so uh in kind of both those ways atm and sat are kind of playing similar roles one key difference however between atm and sat is in the for atm we can prove that it's undecidable but for sat we don't know how to prove it's outside of p that would those would be the analogous situations um and uh so the the story for sat which is kind of easily solved by a diagonalization argument for atm there's reasons to believe that we will see later that diagonalization is not going to work to prove sat outside of p and besides that we don't really have any good methods um so uh anyway uh let's move on why is etm less analogous uh i know because atm was the first problem we showed undecidable and sad is kind of the first problem that we're going to be showing np complete i guess that would be my answer um okay let's continue so let's show now that handpath is np-complete assuming that we know sat or threesat is np complete okay so we're going to give a reduction from 3sat to handpath that that that that's what that's what this is about it's just like what we did for clique but now for him path um and and this is going to be very typical in in these reductions typically what happens is that you're trying to simulate a formula use you know a boolean formula for satisfiability from the satisfiability perspective you're trying to simulate that formula with some sort of structures inside the target language which would be hand path and so what the lingo that people use is that you're going to build gadgets to simulate the structures in the formula namely the variables the literals and the and the clauses okay these are going to be substructures of the um either you know the graph in this case that you're that you're building um we'll see what that means so let's take a formula here and uh let's again try to imagine how we would reduce that to the hand path problem so the reduction would produce a graph no it would produce a hand path instance so a g s and t want to know is there a hamiltonian path from s to t in the graph and this is going to be not the whole graph but this is going to be a substructure in that graph the next slide is going to have the global structure of the graph graph but here this is going to be a key element and it's we're going to call that the variable gadget okay what does it look like i don't know if you can see it on your um clearly enough but these uh edges so there are four outside notes here the edges connecting them are all kind of pointed downward and then there are these horizontal nodes here and there are edges connecting them both left to right and right to left okay there's a row of these horizontal nodes okay so you get the picture of what this looks like um you have to look carefully to see the arrowheads i maybe should have made those a little bigger okay so now um uh whoops let me let me just see so now if we're trying to get from this node to that node imagine now you're trying to build a hamiltonian path because this is going to be a part of that graph g that i'm constructing now remember for them there to be a hamiltonian graph that means you have to go through every node in the graph so if i want to get from s to t the only way i'm going to be able to to go to hit these horizontal nodes here is um by picking them up uh as i go from s to t so the only possibility is if you think about it is it would be for the sequence to go like this if you could if that comes through for you so the the path would go from s to this node and then through these hops of along these horizontal nodes and then down to the bottom node here okay so that's one way that you can get from here down to there and pick up all the other nodes along the way which is they're not going to have any other possibility possible ways of getting to them so i'm going to call that a zigzag okay but there's another way to get from s to the bottom node okay which is going to be by doing sort of the dual going to the right then going out to the left and then down to the bottom i'm going to call that a zigzig okay and uh with a little kind of little diagram here just to summarize what it means and this is the this is the classic thing for a variable gadget because it's a structure that when you're trying to think about how the object that you're asking whether it exists or not the hamiltonian path how it relates to that object it's going to have two possibilities which are going to correspond to the variable variable being set true or false in the formula so we're showing how to set you know the simulate the variable in this hand path instance so setting the variable is going to correspond to to to to constructing the path okay now we also have to make sure not only that the variable gets set to true or false but that it gets a set to true and false in a way that makes this a satisfying assignment namely that we get one true literal in every clause so i'm going to add another gadget here called a clause gadget which is just a single node and visiting that node here is going to correspond to um satisfying that clause to having a true literal in that clause um and now how am i going to get i'm going to have it enable a detour from these horizontal nodes out to this clause gadget so here it is here's that detour as i'm going from left to right i can instead of jump doing a single jump here i could branch out and visit that clause node and then come back and pick up my left to right uh path as as i was doing before you realize what i'm trying what's i hope you're seeing the big picture because this has to be a hamiltonian path it has to hit every node that's one of the requirements so i've got a this is going to be one of the nodes in my graph it's the path has got to hit that node the only way it's going to be able to hit that node is by taking a detour off of this uh horizontal path here but notice and this is the this is the key if i'm doing a zigzag then i can make the detour and visit that clause gadget that clause node but if i'm doing a zag zig the way this detour is structured does not allow me to visit that node because if i'm going from right to left here by the time i get to to this outgoing edge now i come want to come back that note has already been taken it only works if i'm going zag zig from going from right to left the left to right if i'm going from left to right then i can do it for going right to left no so this is going to correspond if you think of left the left to right direction as true that's going to correspond to that variable appearing in the positive way in that clause um but not in the negative way the negative way i would reverse the in and out of the dot of the detour flip that around so now i could only do the detail detour when i'm going right to left instead of left to right i hope you get the picture this is how the structure is working now what's only left me to do is put it all together but this slide contains the guts um of what's happening um so you wanna is is there any question that i can answer for you um on this let's move on to the next slide and then as questions come up you can be typing them in we can maybe answer it there okay so here is the big picture so imagine we have that formula that we started we're reducing that formula um uh and um uh and let's say it has m variables and k clauses i'm going to call them clause c1 c2 up to ck here the m variables appearing either positively or negated in that formula and this is the way the global structure of g is going to look like now i'm getting a question here what do those horizontal nodes what role do they play there those nodes are there to allow me to visit those clause nodes the the nodes which represent the clause gadgets which i'm going to place over here so imag so here here is that um here is i mean this is almost a hole of g that i'm just missing uh some a few edges but these are all the nodes of g so remember i'm trying to find out is there a hamiltonian path from s to t now if i didn't have to worry about the these c nodes the answer would be just yes in fact there would be many hamiltonian pairs from s to t because i can do a zigzag or zx zig through each of these uh variable gadgets and i would take me from s to t and i'd be good it's just these c nodes these nodes these ci nodes you know i have to hit them too so they're going to be um [Music] visiting them is going to be enabled by detours from here so let me just try to show you what that looks like so i'm going to magnify little pieces here uh from from these gadgets here and show you how these guys are connected up so here is the x1 gadget so x1 appears positively in c1 right here's x1 and c1 and so that means when i'm going left to right i'm going to be a possibility of visiting c1 um but now let's look what happens with um uh uh which was the next one i met i i had here uh okay um so uh right so the next one is oh yeah so that's so x1 appears positively in c1 and that's why i have the connection like this uh now x1 appears negated in c2 so i only want to do enable the detour to visit c2 when i'm going right to left as opposed to left to right so this uh set of horizontal nodes is only going to allow me to take a detour either out to c1 or out to c2 but not to both because the hamiltonian path is either going to go left to right or right to left it can't do both when it's going through x the x one uh gadget okay you need to didn't kind of you have to i'll try to help you through it but you have to try to think about why it's working so what let's think again together so x2 also appears in c1 but now it appears negated so i'm going to have edges from this x2 gadget whoops x the x2 gadget but now look at the look at how i've i've arranged that detour you know i can leave on the left leftward side and return on the rightward side um which means i can only do that detour when i'm going right to left okay um and that's because x2 is negated in c1 okay so there's maybe it's a lot here if you're if you're not quite getting it but the point is let's say try to quickly prove so that's the whole construction you just do that for every single appearance of a literal in a clause you're going to add these detours which allow you possibly to go visit the uh the clause um so the forward direction is um why is this true so you take any satisfying assignment as i suggested you make the corresponding zigzags or zag zigs through the variable gadgets from s to t and then take the detours to visit the cloud the clause nodes um the reverse direction actually is slightly trickier we're not gonna have time to go through the subtlety of it um but uh what you want to make sure here is that you don't have a kind of a a weird path um occurring because i'm going to start with a hamiltonian path now and build an assignment uh and what we want to make sure that the the path that i'm constructing doesn't go from one from this horizontal nodes to a clause node and then back to somebody else's horizontal nodes um and is kind of a hodgepodge of things which don't make any sense in trying to reconstruct a satisfying assignment what you what you really want to have happen is the hamiltonian path should have clear zigzags and zag zigs that allow you to decide how to set the variables and so that's the role of these little nodes here these are spacers that separate the detours from one another and that force um a visit to the clause node to come back to the same uh place from which it left otherwise you would never be able to visit those uh spacer notes you have to look in the book for that uh but there is a little bit of a detail that you have to go through you have to show it must be zig zags and zig zag zigs and then you get the corresponding truth assignment and the most satisfied fee uh for all paths okay let's uh again the reduction is polynomial time computable i'm not going to say more about that we're kind of a little bit low on time um last check in would this construction still work if g was undirected suppose i just eliminated all the directions from the from the edges uh made them lines instead of arrows would that now show that the undirected hamiltonian path problem is np-complete let me see what do this okay what do the c nodes represent here there's one c node for every clause so there are k clauses named one c one to ck and there are k c nodes these are the so called clause gadgets which are going to force there to be one true literal um in every clause for the satisfying assignment you know it's uh you know uh you have to look at it or or maybe we can spend a little bit more time explaining it but uh that's what that's what the purpose is um does that mean we need only does that mean we need only two inside nodes uh do we know that so the the horizontal nodes do we only need two of them you won't be able to reuse the reuse these nodes for multiple detours for one thing you know once you've gone to a detour you come back to the node next over and so you better not overlay multiple detours and and also you need to keep them separated from each other don't forget and this node x1 can appear in many many different clauses um so you would need to have um possibly many of these horizontal nodes um so some now says 2k inside nodes would suffice probably 2k i would say 3k just to be safe for the spacer nodes um you need to look at carefully at the argument which is laid out in the textbook um you may not actually need the spacer nodes but then it makes the argument just more more ugly so the way the construction is done is you have three k inside notes um that okay uh again several questions like that the graph would start looking messy if x9 was in c c1 yeah if x9 down here was in c1 yeah i would it'd be messy it's okay my c is allowed uh all right i think we're let's end the polling you all are you all in all right share results yes the answer is no uh the construction depends on this being directed i mean you can see that all over the place but for one thing uh you know the whole point of these detours is the directions of the of the um edges and so without that the the this construction me is is going to be just a bunch of you know not going to mean anything it's not going to prove anything it's probably always going to be have a have a hamiltonian path without the directions um so i think we're kind of out of time um uh a quick review um these are the topics we've covered i think we're out of time so i should let you go um but i'll stick around for a few minutes um in case any of you have any questions but i need to run off it for myself for another meeting so i don't have much that much time um clarify my comment about picking the wrong problem to tackle p versus np um when i use composites as an example you know if i worked hard to prove that composites is not in p as a way of proving that there is some np language which is not in p um that would that would have been a mistake because composites is in p that would be just i would would have been working hard to prove something which we now know was false so we don't want to spend time working on the wrong language um you know but the nice thing about np complete languages is that we have a guarantee that if p is different from np that that language is not in p are there problems that are not in p that are in np but are not np complete oh that's a good question are there problems in between uh p and np complete so np complete is sort of like the hardest problems in np and the p problems are obviously the easy problems that are in np and how about is everything either np complete or mp so for one thing there are problems that are not known to be in either category [Music] uh so we'll discuss some of those in due course but one of them is like the graph isomorphism problem testing two graphs if they're really just permutations of one another it's clearly a problem in np but not known to bnp um so there are problems that are not known to be either np complete or in p so there are problems that might be in between um but then there's another theorem out there which says that if you if you assume that p is different from np then you can construct problems which are in between which are neither np complete nor in p they're np problems but they're not empty complete not in p um so uh those problems themselves are perhaps somewhat artificial but they at least prove the point that it is possible to have these intermediate problems um so somebody's asking isn't factorization one not no you know we for the case of factorization um or you have to make sure to make a language out of that by the way um uh but um it's a um uh stuff because factorization is a function so we won't really want to be talking about languages as kind of as the homework suggests np and p are classes of languages but okay that's a separate separate note there factorization um is uh not no it could factorization could be in p and it could be np-complete you know both of those are not ruled out so uh you know i think most people would probably venture to guess that it's it's a problem that's in the in between state that's neither p nor mp complete but not known um uh do we who who first thought of this reduction from three set to hand path it's so clever well it wasn't me uh i believe that uh reduction from uh i think that is due to dick carp who was one of my professors at berkeley where i was a graduate student he that was done before i got there it was around 1971 um when so there were two famous papers there's the cook paper there's also the lemon paper that was in russian that took a while for people to discover out here in the west but the cook paper was 1971 and very quickly followed after he just showed sat as np complete but uh after that uh carp was he he had a paper called reducibility among combinatorial problems and he had a list of about 20 problems that he showed were np complete by reduction from sad um include clique include handpath and a bunch of other things um so and that's was also a very famous uh very famous paper um both of those were people who often talk about cook carp as sort of together they uh they they they really sort of show the importance of np completeness and exist and no whole notion of np completeness um yeah 21 problems so yeah so car proof 21 problems wouldn't be complete in 1972 um so that was shortly after coke showed that sat was mp complete um incidentally i think the terminology np complete wasn't around uh uh until a little later and that might have been might be due to knuth i'm not sure i remember he did a big poll of people about what should be the right language to use for the for that term and i think he came up with it um all right i'm gonna head off guys nice uh seeing you all uh so until uh thursday oh until tuesday uh bye-bye 2 00:00:27,509 --> 00:00:31,189 welcome everyone um back to um theory of computation so we are now at lecture number 15 and if this is an important lecture um well all of our lectures are important but this is i'm going to introduce one of the major topics that we're going to see going on in various forms through the rest of the semester which is the notion of a complete problem in this case it's going to be an np complete problem and i think some of you have probably heard of that concept already maybe you've seen it in some courses or other but we're going to do that in a in a rather careful and formal way over the next couple of lectures so we are following up on our previous discussions about complexity time complexity we defined the time and non-deterministic time complexity classes as you may remember the classes p and np talked about the p versus np problem looked at some interesting algorithm for showing problems in p called dynamic programming and we started to move toward our discussion of np completeness with the introduction of polynomial time reducibility which is related to some of the earlier reducibility notions that we discussed in the computability section so let's [Music] quick review um what it means for one problem to be polynomial time reducible to another this follows our pattern of reducibility concepts where um if a problem is reducible to another problem and that other problem is solvable then the first problem is solvable so if a problem is reducible to an easy problem that problem becomes easy too um and the kind of reduction that we're going to be looking at here are the mapping reductions but now where the reductions are computable in polynomial time and so we had this um result we mentioned last time that if a is polynomial time reducible to b and b is in p then a is also in p um [Music] so that's going to be um critically important for the you know for the for the whole um of discussion that's coming um our intuition about pnnp um repeating that here um the np problems are the ones where you can verify membership easily as opposed to being able to test membership easily those are the problems that are in p the verification typically requires some kind of a certificate that establishes kind of a proof that the input is a member of that language um and the big question of the field which remains an unsolved problem is whether p equals np it's called the p versus np question and we don't know the answer so whether there are problems that are in np um solvable and nondeterministic polynomial time typically these are problems that involve searching whether they can be solved in polynomial time typically without the searching um and uh if p were equal to np then you could always eliminate searching and a p were different from np then there were cases where you need to search um and we don't know the answer to that uh so in the direction of exploring uh this this question and its ramifications we introduced this problem called sat these are the boolean formulas which have an assignment that enable that event makes them evaluate to true um so we call those satisfiable formulas and uh we um mentioned but have not yet proven and we will not prove until the next lecture on tuesday that there is this theorem uh kind of a very remarkable theorem that says that if you have uh you take the satisfiability problem and if it is solvable quickly then all of the np problems are solvable quickly so if sat is in p then p equals np so in a sense sat is kind of you know the uh it's sort of the super np problem in the sense that all of the difficulty of any np problem is is embedded within sat so if sat becomes easy then all of the other np problems become easy um and we'll eventually prove that but right now we're kind of setting up the um terminology to allow us to do that um so uh anyway we'll we'll we'll get there um so the the the key ingredient for proving this kook 11 theorem is polynomial time reducibility what we're going to show is that every problem in np can be polynomial time reduced to set so every np problem can be converted into a set problem and so working toward that because when you think about it you know there are infinitely many problems in np and um being able to show that all of them are reducible to sat as in a sense kind of um you have to prove a kind of a higher level theorem it's not just a single reduction we're kind of exhibiting a schema of reductions which shows that all of these problems can be reduced to set um and kind of developing our um you know our intuition toward that direction we're going to look at some specific polynomial time reductions today um and we'll start out with a reduction between two problems we have not yet seen one of them called threesat and the other one called clique so on this slide we're going to introduce those two problems so remembering again about boolean formulas we're going to consider a special class of boolean formulas a restricted form of boolean form formulas called conjunctive normal form and so this formula here in particular is in that conjunctive normal form um so just remember um i was explaining some of this to someone else uh earlier this morning showing some of these slides and it was pointed out that not all of you may have be familiar with these um the boolean operations of and and or uh so this is the or symbol here this is the and symbol here you know they operate you know hopefully you've seen the just the concept of boolean and or or is in a sense a little bit like a union um and it's like an intersection and the the symbols here are kind of a little bit similar to the union and intersections you know the intersection the union intersection symbols themselves like or the v-shaped is a little bit like a pointy union symbol and the in the um upside-down v-shape is a little like a pointy intersection symbol so maybe that's if you haven't seen those seen that connection before maybe it's interesting to observe that but anyway um what makes this uh formula be in conjunctive normal form well the conjunctive normal form formulas are organized in a certain way they have um these groups called clauses within the parentheses that are added together that are connected by these and and operations um and within those groups which are called clauses the um elements are ored together those elements are going to be either variables or variables with negation negated variables um so uh just to repeat that uh these well i'll just just state those those variables or negated variables are going to be called literals um and the ors of a bunch of literals are going to be called clauses okay this is just the standard terminology in the field okay so um a literal is a variable or a negated variable and a clause is an or of literals okay just a definition all right so we have a bunch of these clauses here each of the clauses has an or of of liter of literals in it now a chomsky conjunctive normal form uh formula i guess they're both c and f's but uh it's conjunctive normal form formula is one that's written as an and of these clauses okay so we take these clauses which themselves are ors we and them together we get a formula that's conjunctive normal form conjunctive stands for and i think uh conjunction is an and um and then a three cnf is a cnf where each clause has exactly three literals so for example this particular formula is a cnf but it's not a three cnf because it has at least two clauses which violate the condition of three literals per clause this one is okay obviously this first clause and the three sat problem so this is the first of the two problems we're going to be talking about the three sad problem is the satisfiability problem restricted to these three cnf formulas so this is the collection of setup of three cnf formulas which are satisfiable so you can think of it as a kind of a special case of the sat problem where we only care about formulas of the special kind um this being a special case you might imagine that this might be an easier problem to solve but in general it turns out not to be as we will see this is just as solving this special case is just as har as hard as solving the general case for satisfiability um now let's turn to the second of these two languages up in the headline uh the clique problem um so for that we gonna turn to graph theory so we're going to consider graphs you know points and lines connecting them and we will say that a clique in a graph is a collection of nodes collection of the points that are all pairwise connected by lines and a k click is one where you have case arch nodes so here we have a three click four click and five click um and the clique problem is to try to find cliques of a certain specified size embedded within a given graph so you'd like to know whether you're going to be given a graph and it targets click-sized k and you want to know is there a subset of the nodes in the graph of size k that are all connected to one another that's the cleat problem um so obviously um the clique problem just as with the satisfiability problem is a decidable problem you can just try every possible subset of k nodes and see whether it constitutes a clique but that's in general going to be an exponential algorithm if you have you know you know k is a large value like it might be half the size of the graph you might be looking for a very large clique then you have to try many many subsets in order to see whether uh any one of them is a clique it's also going to be and i hope you get this intuition this is a problem that's in np the cleat problem is an np problem because you can easily verify that a graph uh has a k click just by exhibiting the click okay um now uh uh so the clique language here is you're given a graph and a k and you want to know does the graph contain a cake league um and what we're going to show is that these two problems are connected that the three set problem and the cleat problem are related in that you can reduce in polynomial time three sat to click okay so this is um you know at you know just standing back at it back for a minute you know that seems kind of surprising there's no real reason you know uh obviously anyway no obvious reason why there should be a connection between three sat and click click they look very different from one another but we do give that uh we will give that reduction and um it then that implies that if you can find a way of solving the clique problem quickly that'll give you a way of solving the the three set problem quickly um and that's going to be the whole point of this um so we're going to show this over the next slide or two um this this polynomial time reduction and this is something i'm going to walk through it slowly but this is one where you it's really important to try to get a sense of how it's working um uh because this is the kind of thing that we're going to be doing a lot of and you're going to be asked to do it also on the homework and and possibly on the final exam i hate to use that as the motivating force here but at least it might be one motivation for some of you that you have to understand how to do uh this kind of a reduction okay so you know i i think at a high level what's going on is we're showing how to recode a boolean formula satisfiability problem into the problem of testing whether a graph has a clique um so let's see if we have any questions here that are coming up in the in the chat um okay so i this is an interesting question here can we always convert a boolean formula into conjunctive into conjunctive normal form first of all yes the answer is you can always convert a boolean formula into an equivalent one in ins into in cnf but in general that might make the formula exponentially larger [Music] so just the mere fact that you can convert formulas into into conjunctive normal form doesn't mean that solving um that you know that solving uh um conjunctive normal form uh formulas for testing satisfiability of the of the cnf formulas is going to be as hard as testing the general case because just the conversion might be exponential um there's something more uh more a little bit more complicated going on um than that so let's uh um let me just see here uh if he is a satisfiable formula which is not in cnf can be so a similar um you know similar question so the questions that i'm getting are about converting formulas uh to cnf um so yeah you can do it but not in polynomial time in general because the resulting formula you get might be much larger um if you if you're looking for an equivalent formula if you're not looking for an equivalent formula then you know of course then depending upon what you're looking for you might be able to find something smaller um ah this is a this is a i guess a good basic question why is clique in np um because isn't doesn't verifying that you have a clique require going through all the possible cliques once i you have to understand what verifying means verifying means you can verify something if you're given a certificate in the case of a clique the click problem the certificate is the clique um so once you have the certificate you can do the verification in polynomial time finding the certificate of course might be difficult so you only think of np in the context of having that certificate so in a case for like compositeness the certificate might be the factor um you know anyway i hope that's you know it's there there are problems sometimes with what the certificate is is not necessarily obvious um but there can be a certificate um for um showing that the that the input is in the language but in the ones that we've done so far maybe you can argue that this certificate is sort of an obvious thing but it's not always an obvious thing um okay why don't we move on then um so let's see how do we do this polynomial time reduction from three sat to click okay here we go um so i'm just going to give a reduction that's that's that's what the definition means i'm going to give a way of converting formulas to pairs a g and a k where the formula is going to be solid satisfiable if and only if the graph has a k click okay so let's kind of a little bit do it by example and in order to do that so here's going to be a formula now it's in three cnf that's what i need you know in order to be doing this reduction i'm converting three cnf formulas into clique problems we have to a little bit understand what it means when we say uh we talk about the satisfiability of a formula like this because it's going to be helpful in doing the reduction you know obviously it's satisfiable means that the form you can find an assignment of the variables so you're going to set each of these variables a b and c and so on to true or false and you want to make the whole formula evaluate to true but what does that actually mean [Music] it means that because of the structure of the formula that making this formula true corresponds to making each clause true because the clauses are all added together so the only way for the formula to be true is to make each clause true and to make a clause true you have to make at least one of the literals true so it's another way of thinking about uh satisfying this formula you need satisfying these and a satisfying assignment makes at least one true literal in every clause it's really important to think about it that way because that's what's going to be um [Music] kind of the basis for doing this reduction and all of the reductions it's kind of what makes three set easy to think about in terms of its satisfiability if you had a sort of a general um satisfiability problem and you had a satisfiable formula there's no sort of obvious way of sort of seeing what the satisfying assignment looks like but here we understand what it looks like it has that very special form making one true literal in every clause at least one true literal in every clause okay so now we're going to do the reduction so i'm going to take from this formula you and you have to understand you know i know for some of you you know you're going to be you know chafing why am i going so slowly but i want to make sure that we're all together in understanding what the rules of the game are and what we're trying to do um we're trying to convert this formula into a graph and a number so right now my job is to do this reduction is to exhibit that graph okay so i'm going to do that in kind of two steps i'll first i'm going to tell you what the nodes of the graph are then then i'll tell you what the edges of that graph finally i'll tell you what the number k is that's the way this polynomial time reduction is going to work and we have to also observe at the very end that the reduction that i'm given giving you this procedure for building this graph can be done in polynomial time but that i think you'll see once i'm done that that's pretty obvious okay so first as i promised the nodes um so the nodes of this graph are going to correspond to the literals of the formula every literal is going to become a node in the graph and it's going to be labeled with the um name of that literal so it's every node is going to be labeled an a or b or or c bar and so on so here it goes okay so those are the nodes of the graph g one for each literal in the formula okay labeled as promised now i have to tell you what the edges look like the edges are going to be um i'm going to tell you what the edges are by first telling you what they are not i'm going to first explain to you what the forbidden edges are what what edges i'm going to promise not to include and then the ones that i will include are going to be all the others okay so what are going to be the forbidden edges first of all um the the forbidden edges are going to be of two types one is the edges between nodes that come from literals in the same clause so i would just call that no edges within a clause okay so for example these three nodes will not be connected to one another and i'm going to indicate that by writing red dashed lines which means there's not an edge there okay those are those are forbidden from having an edge okay so these three have no edge and the same thing for every other uh triple of three nodes that come from uh clauses okay those are no edges there and there's one other category of edge that i'm forbidding and that are edges that can that um go between inconsistent labels and the nodes with inconsistent labels so for example between a and a bar those are inconsistent you know nothing's wrong with a going a to d those are not inconsistent those are just different labels or going from a to a that's okay but from a to a bar that's that's not allowed um so that's going to be another forbidden edge or for example a to a bar here for b a bar to a or for example from this c bar to c forbidden okay so imagine you're going to write down all of those forbidden edges and then once you've done that uh those are all of the the forbidden edges and then after [Music] taking those away you're going to be putting in all the other edges possible so for example as a let me just uh gray those out so they don't interfere with the picture b from a to a to d um there's going to be an edge um because those are not forbid you know a to a not forbidden because they they're in different um they are in different uh clauses and they're not inconsistent they're they're consistent with one another so here are a whole bunch of other i'm not showing them all becomes very messy but uh here are all the other edges between nodes uh which are where there isn't there's no pro within they're not forbidden okay and that's that's the graph that's g g is all those nodes and those edges um which are not forbidden now i just have to tell you what k a is k is going to be the number of clauses that's going to be the size of the clique i'm looking for in this graph g that i just spelled out for you and i'm going to claim that this graph here that i just described will have a k clique k is the number of clauses exactly when fee was satisfiable it's kind of kind of cool you know um if he is satisfiable then there will be a cake leak here and if he was not satisfiable okay clean so we have to we're gonna prove that as a claim on the next slide okay so k is the number of clauses all right um any questions on that on that construction so i've done with the construction what's left is to argue why the construction works so far so good let's move on okay um so um so here is that very same construction i eliminated the red uh forbidden edges these are the ones that were remaining plus you know anything else that was not forbidden that was the same formula that i had from the previous slide now i'm going to claim that that formula is satisfiable exactly when g has a k click now why in the world is that so this is an if and only if it's proved in both directions and this is going to be the typical kind of thing that you're going to want to do when you're exhibiting a uh you know one of these reductions which is you're going to have an opportunity to do that um and we'll do do that too in our examples uh okay so now uh what i'm what i want to show is that uh if fee is satisfiable so let's prove the forward direction if p is satisfiable then g has a k click okay so first of all if fee is satisfiable that means it has a satisfying assignment okay now here's a common confusion i'm not sure whether it's helpful to sprinkle the confusions in along with the discussion but in case you're worried you might ask well how do i find that satisfying assignment i thought that was you know exponential exponentially hard to do this is a proof this is not an algorithm okay we're just trying to argue the con correctness of this construction so there's no longer any concern about how to find an assignment i'm just saying if there isn't a sign if there is a satisfying assignment then something will happen something that then then um something will happen in particular then we will show that g has a cake league so let's say fee is satisfiable so it has some assignment let's take that satisfying assignment and remember that in any satisfying assignment uh to a formula in cnf that makes at least one literal true in every clause so let's just pick one of those there might be several clauses which have multiple true literals in that case just pick one of them arbitrarily okay so i don't know what the i don't have this sort of indicated on this diagram here but imagine you know maybe in the very first clause a was true in the second clause b was true in the third clause e compliment ebar was true and you know not not you know e bar was true which means e itself was false but e bar was true and that's the way each of those clauses in turn uh got to be true you know in this particular satisfying assignment because you're going to pick one true literal in every clause and now from that choice of literals one per clause i'm going to look at the corresponding nodes in g and i'm going to claim that those nodes taken together form a k click okay so first of all do they have the right number of nodes well sure because i'm picking one node per click i already said k is the number of cliques so i'm getting exactly ka nodes so i have at least the right number of nodes but how do i know they're all connected to one another those nodes uh that i just picked well they're all connected to each other because i'm going to say well because there were no forbidden edges among them and remember that we put in all possible edges except for the forbidden ones so how do i know there are no forbidden edges well what were the rules for being forbidden means they two nodes in the same clique in the same clause well i'm picking one node per clause so i can never be having two nodes from the same clause so i'm never going to run into trouble with uh the first condition of being forbidden so what's the second possibility for being forbidden is that i'm picking two nodes which are inconsistent well how do i know that i didn't end up with two nodes with inconsistent labels that would be bad because then they would have a forbidden edge and what my result would not be a clique how do i know i didn't end up picking um uh in this group i pick a and in this group i pick a a bar well because they all came from the same assignment those are all of the true letters they were all true literals in clauses it's not possible that a was true in this clause and a bar is true in that clause because if a is true here a bar has got to be false it can't be the true literal in this clause so i cannot have any inconsistent any inconsistent nodes appearing among the nodes of my clique and so they're not in the same clause they're not inconsistent so the edge has to be there and that's going to be true for every pair of nodes in that clique in that in that group of nodes and that's why it's a click um so that proves one direction now we stop to prove the other direction because we have to say well if g has a k clique how do we know that ph is satisfied so that's the reverse so let's just take any k clique that's in g and how do i know i can get from that a satisfying assignment to the formula um uh good um getting some good questions here in the chat but let me just move on um take any uh so proving the reverse this the reverse direction but we're assuming we have a cake leak to take any such k-click now first of all you observe that it's got to have one node in every clause it can't have two nodes in the same clause or zero nodes in in a clause first of all it can't have two nodes in the same clause why well because those nodes are never connected to one another so that they cannot be both in the same clique because all the nodes in a clique are connected to one another um by construction by the way we construct a g nodes from the same clause are not connected so they cannot appear in a clique together so um there's going to be at most one node from every clause appearing in this clique but and but but appearing in in this clip but we also have we know we have k notes so that means if there is at most one per um per clause and we only have k clauses that means every clause has got to have one you know if some clause is missing or note then there's going to be some other clause that has two so we know that every um that this clique that would that g has has exactly one node in every clause so how do we get from that a satisfying assignment so what we're going to do is take the current corresponding literals that correspond to that one node in the that was in the uh clique take those literals and set those literals to be true which means setting the variable to be true um or setting um you know if the literal was a bar then setting a to be false because you want a bar to be true you want to set the literals to be true well now we're setting um there's one note in each clause we're setting one literal uh the corresponding literal true so that's going to set one true literal in every clause so that means it's going to be satisfying there's one thing that one has to really to be careful here to double check to make sure that we're doing this consistently that we're not being asked to set you know a true and also a bar true because then that would not be possible but a and a bar are not connected so it's we're never going to be trying to set both a and a bar true if so we're going to be trying to set a true we're never going to try to set a bar true because those are not those a and a bar cannot be in the same clique they're not connected to each other okay so that is the argument um and lastly i claim we're not going to look at this in detail that it's kind of obvious that this reduction can be done in polynomial time namely if i give you that formula you can write out that graph in pretty easily there's no hard work to be done in writing out that that graph or counting up the number of clauses so that's that's the proof that i can reduce three statically um and it tells you also that if i could solve click quickly then i can stop threesat quickly and this is the whole point because i can convert now clique problems to three sat problems no no i couldn't i'm sorry i got backwards convert three set problems the clique problem so if i can solve quick click easily then i can solve three set easily right i just showed a way of converting these formulas to graphs so if that's so that says if i can solve the graphs easily then i can solve the formulas okay why don't we uh turn to some oh question checking but you know we can also um why don't i i'll launch a check-in but we'll we let's just see i'm seeing many many questions here which are actually what my check-in is about i'll turn it back to you guys uh okay so where did we use the fact um that we have three uh literals per clause what does this thing just work even if we had any number of literals per clause um what do you think we're having a we've got a tight race here hmm uh truth is losing out unfortunately all right oh it's really close but still okay uh one more okay it's neck and neck okay almost done 10 seconds are we done here okay that's it uh ending polling uh yes truth is thank god um yeah it works for any size clause we didn't use the fact that it's a three c and f it could have been any number here uh so it's well we've got a i guess it's a plural plurality here though that not a majority um no we didn't where did we use three in any of these argument i didn't mention it maybe you were imagining that that is going to be part of it but three does not come into this at this discussion at all if you think about how it's why it's working um even if we had you know one of these clauses had 10 literals in it as long as k is the number of clauses and we don't connect any variables any literals internal to a clause this whole argument is going to still work so please check that make sure you understand what's going on because i can see that a good chunk of you have not got this right um uh so i got a good question here uh what if it's only just one big clause so it's like the whole formula is just one big war so what does happen in that case good question so in that case um uh so the you know suppose has one big clause with a hundred literals that's the whole formula it's just a big war so we know the formula is going to be satisfiable by the way obviously um so if you look at the corresponding g it's going to have 100 nodes because there's one for every literal um none of them are going to be connected to each other because they're all in the same clause um so it looks like we're going to be in tough shape trying to find a clique there because there's no edges at all everything's forbidden but what is k k is going to be 1 in that case because there's only one clause and uh kind of a degenerate case but a clique with just one you know just a single node is it counts as a one clique because it's just one node there's no need for any edges at all it still counts as a clique so it'll still it'll still work out um uh let's see what else here uh that was kind of a fun question that you asked me thank you um uh there are a lot of questions um i think we're actually pretty near the break also we'll just see yeah um so why don't i um oh many many questions here uh so i i think i'm gonna start start the clock going down for our um our coffee break and um and i will take some of these questions i'll try to answer some of these questions oops um try to answer some of these questions um afterward uh all right good all right um so i got a question about making sure why nodes don't have inconsistent labels if i understand the question correctly um so these would be nodes um so i never put edges between nodes with inconsistent labels that's the rule that's my construction i get to i get to say how the construction works so those two nodes with inconsistent labels can never be in the same clique because there's no edge between them so i'm not sure that answered the question there but that was uh um let's see so getting another question here since any boolean formula can be converted to cnf does that mean um is sat polynomial time reducible to clique as well monomial time reducible to clique but not for the reason that any boolean formula can be converted to c and f because that conversion um is going to be an exponential conversion in general the conversion to c and f so you have to be careful about what we mean by converting a formula to cnf um uh so also getting questions what happens in the case of one set well we didn't really talk about one set uh suppose or i guess so the question is suppose there's only a single literal per clause that's an interesting another sort of edge case here what happens under those circumstances so um there's only a single literal per clause um uh so you have to work out what happens but in that case um uh all the clauses just have one literal in them and now they have um uh um the only way uh this if you think about what is a when you have a one cnf so there's only um one literal per clause the only way that can be satisfiable is you have to make each literal true because there's no warring anymore so every literal has to be true in the assignment so that means you can never have inconsistent literals um and that's the only case when you would not have an edge because you're not the forbidden is the forbidden condition won't happen because you don't have more than one uh node per per clause every clause is going to have just a single node in it uh so it really comes down to whether or not you have an inconsistently labeled um inconsistent clauses i i didn't explain that super well but it's the the argument still works there you should check it for yourself um so this is a basic question how do we see that f is polynomial time i i'm not sure i want to expend a lot of you know a lot of discussion on that because the the conversion from the formula to the graph is kind of a one-to-one you know every no every uh literal becomes a node um the rule for when edges are present it's a very simple rule so i guess i would argue i'm not sure what what more to say it seems pretty clear that the conversion that reduction is polynomial time computable i mean if you wrote a program to do that it would operate very quickly um is it possible for g to have a k plus n clique where n is greater than zero and does that matter now if you think about it the biggest possible clique that this graph g could have is k it could never have more than uh uh you know a bigger clique because two nodes in the same uh clause are never connected you only have k clauses so the answer is no you cannot have a bigger a bigger than a k click it's just not going to happen does each clause need to have the same number of literals i don't see why um uh so question is why are we worrying about threesat um since it didn't seem to matter here there are other examples where it does matter i i i'm actually not sure if we'll see one or not but um there there are cases where it does matter um let me just see if there's any quick questions here uh somebody's saying doesn't does do we have to worry about there being a polynomial number of literals so you have to think about what you what that question means the size of the input which includes all of the literals is n so there's going to be at most n literals um because that's the size of the input uh the i mean at most uh and literals appearing so you don't have to worry about that being polynomial because by the way we're we're defining the size of the input is going to be at most n literals i hope that's clear uh okay when we talk about power okay so this is a good question when we talk about polynomial time it's polynomial in the representation of of the input which is if you want to think about it in terms of bits that's fine or whatever it's not going to matter if you use some larger alphabet but the number of symbols you need you know in your fixed size alphabet the number of symbols you need to write down that input in whatever your reasonable encoding is is going to be n and it's going to be polynomial in n polynomial in that length of the input um so another question is does sat reduce the three set yes we that we will see um sat actually does polynomial time reduce the three set not by converting to an equivalent formula but by some more uh somewhat more involved argument uh than that okay why don't we move on some of these are going to come out anyway in the lecture okay so now let's talk about mp completeness because we've kind of set things up we're not going to prove that you know the basic theorem the cook 11 theorem about mp completeness but at least we'll be able to make the definition okay here's our definition you know what it means for a problem to be in be complete um uh so a language b is called np complete if it has two properties number one is that has to be a member of np and number two every language in np has the polynomial time reduced to that to that language to that np complete language in order for it to be np complete okay so simple picture has to be in nnp and everything in np reduces to it um and so that's kind of the magical property that we claimed that that sat has sat for for one thing is obviously in np and as we the cook 11 theorem shows or will show everything in npe is reducible to sat so that's going to be our first example of an np complete problem um uh so now uh and we're going to get what we claimed also for sat that if sat or any other np-complete problem turns out to be solvable in polynomial time then every np problem is solvable in polynomial time and and that's immediate because everything is reducible in polynomial time to the np complete problem so if you can do it easily you can do everything easily just by going through the reduction okay so the quick level theorem as i mentioned is that sad is np-complete and we're going to actually prove it next lecture but let's assume for the remainder of this lecture that we know it to be true because so i'll use the terminology of problems being np complete uh assuming that we know um that we have uh sat as empty complete okay so we're gonna we're gonna be using some of the things that were proving next lecture uh just in the terminology that we're going to be talking today okay so here's the picture here's the class np and everything in np is polynomial time reducible to set sat itself is a member of np but i didn't want to show it that way because it makes the picture kind of hard to uh hard to uh hard to display so just from the perspective of the reduction everything in np is polynomial time reducible to sad we'll show that next lecture another thing that we'll show next lecture is that sat in turn is polynomial time reducible to three sat so three stat as you remember are just those problems that are in conjunction three are in three c and f um and then what we showed today is that three set is polynomial time reducible to clique so now taking the assumption that sat is np complete so everything is polynomial time reducible to sat which is in turn polynomial time reducible to three sat in in turn reducible to click these reductions as we've seen before composed um you can just reply apply one reduction function after the next if each one each one individually is polynomial the whole thing as a combination is going to be polynomial so now we know that three sat is going to be also np complete because we can reduce anything in np to stat and then to three set and then we get a reduction directly to three sat by composing those two reductions and then furthermore at the clique so now we uh have several np complete problems and moving beyond that we have the hand path problem which we are going to talk about next and we'll show another reduction um in addition to the one we just showed to click now one going from three set to hand um okay uh so in general i think the the the the take away message is that to show some language as np complete you want to show that three sat is polynomial time reducible to it um uh okay some good questions coming in i'll try to answer those so you're going to retake 3sat and reduced to c that's the sort of the most typical case or you might have there's going to be other examples too we might start with another problem that you've already shown to be np complete and read and reduce it to your language so it doesn't have to start with reset though often it does um okay why do we why is this concept important so i would say there are two reasons um and this is going to get a little bit at the uh some of the chat questions um first of all if you're faced with some copy new computational problem you've got some robotics problem um that you want to solve in your thesis and you need to some algorithm about whether the robot arm can do such and move in a certain such a way and involve searching you know possibly searching through a space of different kinds of motions and you want to know you know i need a either i'd like to have a i like to find a polynomial algorithm to solve that problem i'm using this as an example because this actually did happen to one of the former students from this class who was working in robotics and they actually end up proving that the problem that they were trying to solve is np complete um so um uh that's useful information because even though knowing a problem is np complete doesn't guarantee that it's not in p because conceivably p equals np what it does tell you that if you show if you have a problem that's np complete and it does turn out to be in p then all then p equals np so the tremend would be tremendous surprising consequences of of your problem if it was known to be np-complete ending up in p so generally people take a proof of np completeness as powerful evidence that the problem is not in p even though it's not quite a proof it's powerful evidence that it's not in p and so you might as well give up working on trying to find a polynomial algorithm for it because if you do you know you don't have to worry about robotics anymore you're going to you know become famous as a complexity theory person um so i wouldn't worry about so if you show problems np complete you can pretty much assume it's almost certainly not in p um now there's another reason related to the to the for the theorists uh to be care about mp completeness and that is if you're trying to prove p is different from np or p equals np as one of the chat questions is raising um you know the problem you know in order to prove p different from np you know the most likely approach is you pick some p problem in np and show it's not in p right that's what it would mean for them to be different you're going to pick some np problem and show it's not not a p problem well um one thing would be terrible is possibly p is different from np and you pick the wrong problem like suppose i'd spent all my time trying to show you know back 20 years ago i'm working really hard trying to show composites is not in pain which is would have been perfectly reasonable to do because composites is an np and was not known to be in p 20 years ago um and then i invested tons of effort to try to prove because you know i like number theory or god knows what to prove that composites is not in p and then it turns out composites wasn't in p it was the wrong problem to pick even though p might be different from np but what np completeness guarantees is that if you work on a problem which is np-complete you could you can't pick the wrong problem because if any problem is in np and not in p an np complete problem is going to be an example of that right because if the np complete problem is in p everything in np is in p so if np is different from p you know the complete problems are not in p so you might as well work on one of those so those are two ways in which np completeness and has turned out to be is an important concept okay so here's a check-in um uh again this you guys are kind of getting maybe you're getting started to think the way i think you get is some at least one of you would ask this question in the chat um which language that we've probably seen is most analogous to sat okay atm etm or zero to the k1 to the k uh obviously this is maybe subjective you may have your own interpretation of what that means this in a sense perhaps no right answer but what do you think uh okay that's it um i don't know i don't know how in the world that you could see this problem is analogous to to zero to the k1 to the k but okay um i'm sure you have your reason um yes this is a lot like eight atm why well because for one thing we showed in a homework problem that all um touring recognizable languages are reducible to atm mapping reducible to atm so that's a little bit like the notion of completeness that we have for satisfiability um because all np problems are going to be reducible to sad um and the other thing too is that once we start once we uh we started we want to show other problems are undecidable we reduced atm to them and that's also very similar we're going to be reducing sat or threesat so it's indirectly from sat to other problems in order to show that they're np complete so that they're hard they're they're that they're hard in a sense and so uh in kind of both those ways atm and sat are kind of playing similar roles one key difference however between atm and sat is in the for atm we can prove that it's undecidable but for sat we don't know how to prove it's outside of p that would those would be the analogous situations um and uh so the the story for sat which is kind of easily solved by a diagonalization argument for atm there's reasons to believe that we will see later that diagonalization is not going to work to prove sat outside of p and besides that we don't really have any good methods um so uh anyway uh let's move on why is etm less analogous uh i know because atm was the first problem we showed undecidable and sad is kind of the first problem that we're going to be showing np complete i guess that would be my answer um okay let's continue so let's show now that handpath is np-complete assuming that we know sat or threesat is np complete okay so we're going to give a reduction from 3sat to handpath that that that that's what that's what this is about it's just like what we did for clique but now for him path um and and this is going to be very typical in in these reductions typically what happens is that you're trying to simulate a formula use you know a boolean formula for satisfiability from the satisfiability perspective you're trying to simulate that formula with some sort of structures inside the target language which would be hand path and so what the lingo that people use is that you're going to build gadgets to simulate the structures in the formula namely the variables the literals and the and the clauses okay these are going to be substructures of the um either you know the graph in this case that you're that you're building um we'll see what that means so let's take a formula here and uh let's again try to imagine how we would reduce that to the hand path problem so the reduction would produce a graph no it would produce a hand path instance so a g s and t want to know is there a hamiltonian path from s to t in the graph and this is going to be not the whole graph but this is going to be a substructure in that graph the next slide is going to have the global structure of the graph graph but here this is going to be a key element and it's we're going to call that the variable gadget okay what does it look like i don't know if you can see it on your um clearly enough but these uh edges so there are four outside notes here the edges connecting them are all kind of pointed downward and then there are these horizontal nodes here and there are edges connecting them both left to right and right to left okay there's a row of these horizontal nodes okay so you get the picture of what this looks like um you have to look carefully to see the arrowheads i maybe should have made those a little bigger okay so now um uh whoops let me let me just see so now if we're trying to get from this node to that node imagine now you're trying to build a hamiltonian path because this is going to be a part of that graph g that i'm constructing now remember for them there to be a hamiltonian graph that means you have to go through every node in the graph so if i want to get from s to t the only way i'm going to be able to to go to hit these horizontal nodes here is um by picking them up uh as i go from s to t so the only possibility is if you think about it is it would be for the sequence to go like this if you could if that comes through for you so the the path would go from s to this node and then through these hops of along these horizontal nodes and then down to the bottom node here okay so that's one way that you can get from here down to there and pick up all the other nodes along the way which is they're not going to have any other possibility possible ways of getting to them so i'm going to call that a zigzag okay but there's another way to get from s to the bottom node okay which is going to be by doing sort of the dual going to the right then going out to the left and then down to the bottom i'm going to call that a zigzig okay and uh with a little kind of little diagram here just to summarize what it means and this is the this is the classic thing for a variable gadget because it's a structure that when you're trying to think about how the object that you're asking whether it exists or not the hamiltonian path how it relates to that object it's going to have two possibilities which are going to correspond to the variable variable being set true or false in the formula so we're showing how to set you know the simulate the variable in this hand path instance so setting the variable is going to correspond to to to to constructing the path okay now we also have to make sure not only that the variable gets set to true or false but that it gets a set to true and false in a way that makes this a satisfying assignment namely that we get one true literal in every clause so i'm going to add another gadget here called a clause gadget which is just a single node and visiting that node here is going to correspond to um satisfying that clause to having a true literal in that clause um and now how am i going to get i'm going to have it enable a detour from these horizontal nodes out to this clause gadget so here it is here's that detour as i'm going from left to right i can instead of jump doing a single jump here i could branch out and visit that clause node and then come back and pick up my left to right uh path as as i was doing before you realize what i'm trying what's i hope you're seeing the big picture because this has to be a hamiltonian path it has to hit every node that's one of the requirements so i've got a this is going to be one of the nodes in my graph it's the path has got to hit that node the only way it's going to be able to hit that node is by taking a detour off of this uh horizontal path here but notice and this is the this is the key if i'm doing a zigzag then i can make the detour and visit that clause gadget that clause node but if i'm doing a zag zig the way this detour is structured does not allow me to visit that node because if i'm going from right to left here by the time i get to to this outgoing edge now i come want to come back that note has already been taken it only works if i'm going zag zig from going from right to left the left to right if i'm going from left to right then i can do it for going right to left no so this is going to correspond if you think of left the left to right direction as true that's going to correspond to that variable appearing in the positive way in that clause um but not in the negative way the negative way i would reverse the in and out of the dot of the detour flip that around so now i could only do the detail detour when i'm going right to left instead of left to right i hope you get the picture this is how the structure is working now what's only left me to do is put it all together but this slide contains the guts um of what's happening um so you wanna is is there any question that i can answer for you um on this let's move on to the next slide and then as questions come up you can be typing them in we can maybe answer it there okay so here is the big picture so imagine we have that formula that we started we're reducing that formula um uh and um uh and let's say it has m variables and k clauses i'm going to call them clause c1 c2 up to ck here the m variables appearing either positively or negated in that formula and this is the way the global structure of g is going to look like now i'm getting a question here what do those horizontal nodes what role do they play there those nodes are there to allow me to visit those clause nodes the the nodes which represent the clause gadgets which i'm going to place over here so imag so here here is that um here is i mean this is almost a hole of g that i'm just missing uh some a few edges but these are all the nodes of g so remember i'm trying to find out is there a hamiltonian path from s to t now if i didn't have to worry about the these c nodes the answer would be just yes in fact there would be many hamiltonian pairs from s to t because i can do a zigzag or zx zig through each of these uh variable gadgets and i would take me from s to t and i'd be good it's just these c nodes these nodes these ci nodes you know i have to hit them too so they're going to be um [Music] visiting them is going to be enabled by detours from here so let me just try to show you what that looks like so i'm going to magnify little pieces here uh from from these gadgets here and show you how these guys are connected up so here is the x1 gadget so x1 appears positively in c1 right here's x1 and c1 and so that means when i'm going left to right i'm going to be a possibility of visiting c1 um but now let's look what happens with um uh uh which was the next one i met i i had here uh okay um so uh right so the next one is oh yeah so that's so x1 appears positively in c1 and that's why i have the connection like this uh now x1 appears negated in c2 so i only want to do enable the detour to visit c2 when i'm going right to left as opposed to left to right so this uh set of horizontal nodes is only going to allow me to take a detour either out to c1 or out to c2 but not to both because the hamiltonian path is either going to go left to right or right to left it can't do both when it's going through x the x one uh gadget okay you need to didn't kind of you have to i'll try to help you through it but you have to try to think about why it's working so what let's think again together so x2 also appears in c1 but now it appears negated so i'm going to have edges from this x2 gadget whoops x the x2 gadget but now look at the look at how i've i've arranged that detour you know i can leave on the left leftward side and return on the rightward side um which means i can only do that detour when i'm going right to left okay um and that's because x2 is negated in c1 okay so there's maybe it's a lot here if you're if you're not quite getting it but the point is let's say try to quickly prove so that's the whole construction you just do that for every single appearance of a literal in a clause you're going to add these detours which allow you possibly to go visit the uh the clause um so the forward direction is um why is this true so you take any satisfying assignment as i suggested you make the corresponding zigzags or zag zigs through the variable gadgets from s to t and then take the detours to visit the cloud the clause nodes um the reverse direction actually is slightly trickier we're not gonna have time to go through the subtlety of it um but uh what you want to make sure here is that you don't have a kind of a a weird path um occurring because i'm going to start with a hamiltonian path now and build an assignment uh and what we want to make sure that the the path that i'm constructing doesn't go from one from this horizontal nodes to a clause node and then back to somebody else's horizontal nodes um and is kind of a hodgepodge of things which don't make any sense in trying to reconstruct a satisfying assignment what you what you really want to have happen is the hamiltonian path should have clear zigzags and zag zigs that allow you to decide how to set the variables and so that's the role of these little nodes here these are spacers that separate the detours from one another and that force um a visit to the clause node to come back to the same uh place from which it left otherwise you would never be able to visit those uh spacer notes you have to look in the book for that uh but there is a little bit of a detail that you have to go through you have to show it must be zig zags and zig zag zigs and then you get the corresponding truth assignment and the most satisfied fee uh for all paths okay let's uh again the reduction is polynomial time computable i'm not going to say more about that we're kind of a little bit low on time um last check in would this construction still work if g was undirected suppose i just eliminated all the directions from the from the edges uh made them lines instead of arrows would that now show that the undirected hamiltonian path problem is np-complete let me see what do this okay what do the c nodes represent here there's one c node for every clause so there are k clauses named one c one to ck and there are k c nodes these are the so called clause gadgets which are going to force there to be one true literal um in every clause for the satisfying assignment you know it's uh you know uh you have to look at it or or maybe we can spend a little bit more time explaining it but uh that's what that's what the purpose is um does that mean we need only does that mean we need only two inside nodes uh do we know that so the the horizontal nodes do we only need two of them you won't be able to reuse the reuse these nodes for multiple detours for one thing you know once you've gone to a detour you come back to the node next over and so you better not overlay multiple detours and and also you need to keep them separated from each other don't forget and this node x1 can appear in many many different clauses um so you would need to have um possibly many of these horizontal nodes um so some now says 2k inside nodes would suffice probably 2k i would say 3k just to be safe for the spacer nodes um you need to look at carefully at the argument which is laid out in the textbook um you may not actually need the spacer nodes but then it makes the argument just more more ugly so the way the construction is done is you have three k inside notes um that okay uh again several questions like that the graph would start looking messy if x9 was in c c1 yeah if x9 down here was in c1 yeah i would it'd be messy it's okay my c is allowed uh all right i think we're let's end the polling you all are you all in all right share results yes the answer is no uh the construction depends on this being directed i mean you can see that all over the place but for one thing uh you know the whole point of these detours is the directions of the of the um edges and so without that the the this construction me is is going to be just a bunch of you know not going to mean anything it's not going to prove anything it's probably always going to be have a have a hamiltonian path without the directions um so i think we're kind of out of time um uh a quick review um these are the topics we've covered i think we're out of time so i should let you go um but i'll stick around for a few minutes um in case any of you have any questions but i need to run off it for myself for another meeting so i don't have much that much time um clarify my comment about picking the wrong problem to tackle p versus np um when i use composites as an example you know if i worked hard to prove that composites is not in p as a way of proving that there is some np language which is not in p um that would that would have been a mistake because composites is in p that would be just i would would have been working hard to prove something which we now know was false so we don't want to spend time working on the wrong language um you know but the nice thing about np complete languages is that we have a guarantee that if p is different from np that that language is not in p are there problems that are not in p that are in np but are not np complete oh that's a good question are there problems in between uh p and np complete so np complete is sort of like the hardest problems in np and the p problems are obviously the easy problems that are in np and how about is everything either np complete or mp so for one thing there are problems that are not known to be in either category [Music] uh so we'll discuss some of those in due course but one of them is like the graph isomorphism problem testing two graphs if they're really just permutations of one another it's clearly a problem in np but not known to bnp um so there are problems that are not known to be either np complete or in p so there are problems that might be in between um but then there's another theorem out there which says that if you if you assume that p is different from np then you can construct problems which are in between which are neither np complete nor in p they're np problems but they're not empty complete not in p um so uh those problems themselves are perhaps somewhat artificial but they at least prove the point that it is possible to have these intermediate problems um so somebody's asking isn't factorization one not no you know we for the case of factorization um or you have to make sure to make a language out of that by the way um uh but um it's a um uh stuff because factorization is a function so we won't really want to be talking about languages as kind of as the homework suggests np and p are classes of languages but okay that's a separate separate note there factorization um is uh not no it could factorization could be in p and it could be np-complete you know both of those are not ruled out so uh you know i think most people would probably venture to guess that it's it's a problem that's in the in between state that's neither p nor mp complete but not known um uh do we who who first thought of this reduction from three set to hand path it's so clever well it wasn't me uh i believe that uh reduction from uh i think that is due to dick carp who was one of my professors at berkeley where i was a graduate student he that was done before i got there it was around 1971 um when so there were two famous papers there's the cook paper there's also the lemon paper that was in russian that took a while for people to discover out here in the west but the cook paper was 1971 and very quickly followed after he just showed sat as np complete but uh after that uh carp was he he had a paper called reducibility among combinatorial problems and he had a list of about 20 problems that he showed were np complete by reduction from sad um include clique include handpath and a bunch of other things um so and that's was also a very famous uh very famous paper um both of those were people who often talk about cook carp as sort of together they uh they they they really sort of show the importance of np completeness and exist and no whole notion of np completeness um yeah 21 problems so yeah so car proof 21 problems wouldn't be complete in 1972 um so that was shortly after coke showed that sat was mp complete um incidentally i think the terminology np complete wasn't around uh uh until a little later and that might have been might be due to knuth i'm not sure i remember he did a big poll of people about what should be the right language to use for the for that term and i think he came up with it um all right i'm gonna head off guys nice uh seeing you all uh so until uh thursday oh until tuesday uh bye-bye 3 00:00:31,189 --> 00:00:31,199 4 00:00:31,199 --> 00:00:32,470 5 00:00:32,470 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--> 00:08:57,430 272 00:08:57,430 --> 00:08:59,190 273 00:08:59,190 --> 00:09:01,910 274 00:09:01,910 --> 00:09:04,550 275 00:09:04,550 --> 00:09:08,230 276 00:09:08,230 --> 00:09:10,310 277 00:09:10,310 --> 00:09:13,509 278 00:09:13,509 --> 00:09:13,519 279 00:09:13,519 --> 00:09:14,389 280 00:09:14,389 --> 00:09:14,399 281 00:09:14,399 --> 00:09:15,590 282 00:09:15,590 --> 00:09:18,070 283 00:09:18,070 --> 00:09:20,310 284 00:09:20,310 --> 00:09:21,750 285 00:09:21,750 --> 00:09:23,509 286 00:09:23,509 --> 00:09:24,949 287 00:09:24,949 --> 00:09:28,230 288 00:09:28,230 --> 00:09:28,240 289 00:09:28,240 --> 00:09:30,070 290 00:09:30,070 --> 00:09:32,150 291 00:09:32,150 --> 00:09:34,310 292 00:09:34,310 --> 00:09:36,470 293 00:09:36,470 --> 00:09:37,750 294 00:09:37,750 --> 00:09:40,230 295 00:09:40,230 --> 00:09:43,430 296 00:09:43,430 --> 00:09:43,440 297 00:09:43,440 --> 00:09:44,790 298 00:09:44,790 --> 00:09:47,269 299 00:09:47,269 --> 00:09:49,430 300 00:09:49,430 --> 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330 00:10:48,710 --> 00:10:52,230 331 00:10:52,230 --> 00:10:55,110 332 00:10:55,110 --> 00:10:57,190 333 00:10:57,190 --> 00:10:59,670 334 00:10:59,670 --> 00:10:59,680 335 00:10:59,680 --> 00:11:01,590 336 00:11:01,590 --> 00:11:03,030 337 00:11:03,030 --> 00:11:04,470 338 00:11:04,470 --> 00:11:06,790 339 00:11:06,790 --> 00:11:10,630 340 00:11:10,630 --> 00:11:10,640 341 00:11:10,640 --> 00:11:11,509 342 00:11:11,509 --> 00:11:13,030 343 00:11:13,030 --> 00:11:14,470 344 00:11:14,470 --> 00:11:17,030 345 00:11:17,030 --> 00:11:19,509 346 00:11:19,509 --> 00:11:21,750 347 00:11:21,750 --> 00:11:23,750 348 00:11:23,750 --> 00:11:26,470 349 00:11:26,470 --> 00:11:26,480 350 00:11:26,480 --> 00:11:29,030 351 00:11:29,030 --> 00:11:30,949 352 00:11:30,949 --> 00:11:32,389 353 00:11:32,389 --> 00:11:32,399 354 00:11:32,399 --> 00:11:33,670 355 00:11:33,670 --> 00:11:33,680 356 00:11:33,680 --> 00:11:34,470 357 00:11:34,470 --> 00:11:36,389 358 00:11:36,389 --> 00:11:36,399 359 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--> 00:12:36,470 389 00:12:36,470 --> 00:12:36,480 390 00:12:36,480 --> 00:12:37,590 391 00:12:37,590 --> 00:12:39,350 392 00:12:39,350 --> 00:12:39,360 393 00:12:39,360 --> 00:12:40,870 394 00:12:40,870 --> 00:12:42,629 395 00:12:42,629 --> 00:12:44,629 396 00:12:44,629 --> 00:12:47,590 397 00:12:47,590 --> 00:12:49,509 398 00:12:49,509 --> 00:12:52,550 399 00:12:52,550 --> 00:12:54,069 400 00:12:54,069 --> 00:12:55,910 401 00:12:55,910 --> 00:12:55,920 402 00:12:55,920 --> 00:12:57,670 403 00:12:57,670 --> 00:12:59,990 404 00:12:59,990 --> 00:13:01,910 405 00:13:01,910 --> 00:13:03,990 406 00:13:03,990 --> 00:13:05,350 407 00:13:05,350 --> 00:13:07,190 408 00:13:07,190 --> 00:13:10,069 409 00:13:10,069 --> 00:13:11,509 410 00:13:11,509 --> 00:13:13,670 411 00:13:13,670 --> 00:13:15,030 412 00:13:15,030 --> 00:13:16,790 413 00:13:16,790 --> 00:13:20,389 414 00:13:20,389 --> 00:13:23,990 415 00:13:23,990 --> 00:13:26,629 416 00:13:26,629 --> 00:13:28,310 417 00:13:28,310 --> 00:13:30,389 418 00:13:30,389 --> 00:13:30,399 419 00:13:30,399 --> 00:13:31,910 420 00:13:31,910 --> 00:13:33,829 421 00:13:33,829 --> 00:13:35,670 422 00:13:35,670 --> 00:13:37,190 423 00:13:37,190 --> 00:13:39,670 424 00:13:39,670 --> 00:13:42,710 425 00:13:42,710 --> 00:13:44,710 426 00:13:44,710 --> 00:13:46,870 427 00:13:46,870 --> 00:13:49,910 428 00:13:49,910 --> 00:13:49,920 429 00:13:49,920 --> 00:13:51,910 430 00:13:51,910 --> 00:13:54,150 431 00:13:54,150 --> 00:13:55,990 432 00:13:55,990 --> 00:13:56,000 433 00:13:56,000 --> 00:13:56,949 434 00:13:56,949 --> 00:14:00,310 435 00:14:00,310 --> 00:14:01,829 436 00:14:01,829 --> 00:14:03,910 437 00:14:03,910 --> 00:14:03,920 438 00:14:03,920 --> 00:14:05,670 439 00:14:05,670 --> 00:14:08,389 440 00:14:08,389 --> 00:14:10,310 441 00:14:10,310 --> 00:14:12,230 442 00:14:12,230 --> 00:14:14,389 443 00:14:14,389 --> 00:14:16,629 444 00:14:16,629 --> 00:14:16,639 445 00:14:16,639 --> 00:14:17,750 446 00:14:17,750 --> 00:14:21,590 447 00:14:21,590 --> 00:14:23,110 448 00:14:23,110 --> 00:14:24,949 449 00:14:24,949 --> 00:14:26,870 450 00:14:26,870 --> 00:14:26,880 451 00:14:26,880 --> 00:14:27,750 452 00:14:27,750 --> 00:14:29,030 453 00:14:29,030 --> 00:14:30,710 454 00:14:30,710 --> 00:14:32,870 455 00:14:32,870 --> 00:14:32,880 456 00:14:32,880 --> 00:14:33,829 457 00:14:33,829 --> 00:14:36,790 458 00:14:36,790 --> 00:14:38,790 459 00:14:38,790 --> 00:14:40,710 460 00:14:40,710 --> 00:14:43,590 461 00:14:43,590 --> 00:14:46,150 462 00:14:46,150 --> 00:14:48,150 463 00:14:48,150 --> 00:14:49,829 464 00:14:49,829 --> 00:14:51,110 465 00:14:51,110 --> 00:14:52,230 466 00:14:52,230 --> 00:14:54,150 467 00:14:54,150 --> 00:14:56,230 468 00:14:56,230 --> 00:14:59,030 469 00:14:59,030 --> 00:15:00,790 470 00:15:00,790 --> 00:15:03,670 471 00:15:03,670 --> 00:15:05,750 472 00:15:05,750 --> 00:15:07,030 473 00:15:07,030 --> 00:15:08,949 474 00:15:08,949 --> 00:15:11,430 475 00:15:11,430 --> 00:15:11,440 476 00:15:11,440 --> 00:15:13,030 477 00:15:13,030 --> 00:15:13,040 478 00:15:13,040 --> 00:15:13,829 479 00:15:13,829 --> 00:15:16,790 480 00:15:16,790 --> 00:15:18,310 481 00:15:18,310 --> 00:15:19,829 482 00:15:19,829 --> 00:15:19,839 483 00:15:19,839 --> 00:15:21,030 484 00:15:21,030 --> 00:15:21,040 485 00:15:21,040 --> 00:15:23,269 486 00:15:23,269 --> 00:15:24,870 487 00:15:24,870 --> 00:15:36,550 488 00:15:36,550 --> 00:15:39,350 489 00:15:39,350 --> 00:15:43,670 490 00:15:43,670 --> 00:15:45,110 491 00:15:45,110 --> 00:15:45,120 492 00:15:45,120 --> 00:15:46,550 493 00:15:46,550 --> 00:15:48,790 494 00:15:48,790 --> 00:15:48,800 495 00:15:48,800 --> 00:15:49,670 496 00:15:49,670 --> 00:15:51,749 497 00:15:51,749 --> 00:15:54,550 498 00:15:54,550 --> 00:15:58,949 499 00:15:58,949 --> 00:15:58,959 500 00:15:58,959 --> 00:16:00,470 501 00:16:00,470 --> 00:16:02,910 502 00:16:02,910 --> 00:16:04,970 503 00:16:04,970 --> 00:16:04,980 504 00:16:04,980 --> 00:16:06,310 505 00:16:06,310 --> 00:16:06,320 506 00:16:06,320 --> 00:16:07,189 507 00:16:07,189 --> 00:16:09,350 508 00:16:09,350 --> 00:16:11,350 509 00:16:11,350 --> 00:16:13,269 510 00:16:13,269 --> 00:16:15,430 511 00:16:15,430 --> 00:16:18,069 512 00:16:18,069 --> 00:16:18,079 513 00:16:18,079 --> 00:16:19,110 514 00:16:19,110 --> 00:16:22,389 515 00:16:22,389 --> 00:16:25,030 516 00:16:25,030 --> 00:16:26,949 517 00:16:26,949 --> 00:16:29,030 518 00:16:29,030 --> 00:16:31,910 519 00:16:31,910 --> 00:16:34,949 520 00:16:34,949 --> 00:16:36,790 521 00:16:36,790 --> 00:16:36,800 522 00:16:36,800 --> 00:16:38,389 523 00:16:38,389 --> 00:16:39,590 524 00:16:39,590 --> 00:16:42,870 525 00:16:42,870 --> 00:16:42,880 526 00:16:42,880 --> 00:16:44,710 527 00:16:44,710 --> 00:16:48,829 528 00:16:48,829 --> 00:16:52,389 529 00:16:52,389 --> 00:16:56,150 530 00:16:56,150 --> 00:16:58,069 531 00:16:58,069 --> 00:17:00,150 532 00:17:00,150 --> 00:17:01,990 533 00:17:01,990 --> 00:17:05,829 534 00:17:05,829 --> 00:17:08,630 535 00:17:08,630 --> 00:17:10,789 536 00:17:10,789 --> 00:17:12,470 537 00:17:12,470 --> 00:17:12,480 538 00:17:12,480 --> 00:17:16,829 539 00:17:16,829 --> 00:17:19,510 540 00:17:19,510 --> 00:17:21,990 541 00:17:21,990 --> 00:17:24,390 542 00:17:24,390 --> 00:17:26,309 543 00:17:26,309 --> 00:17:27,510 544 00:17:27,510 --> 00:17:29,590 545 00:17:29,590 --> 00:17:29,600 546 00:17:29,600 --> 00:17:33,029 547 00:17:33,029 --> 00:17:33,039 548 00:17:33,039 --> 00:17:33,909 549 00:17:33,909 --> 00:17:36,470 550 00:17:36,470 --> 00:17:36,480 551 00:17:36,480 --> 00:17:37,430 552 00:17:37,430 --> 00:17:39,990 553 00:17:39,990 --> 00:17:40,000 554 00:17:40,000 --> 00:17:41,110 555 00:17:41,110 --> 00:17:43,190 556 00:17:43,190 --> 00:17:44,870 557 00:17:44,870 --> 00:17:47,669 558 00:17:47,669 --> 00:17:48,789 559 00:17:48,789 --> 00:17:50,310 560 00:17:50,310 --> 00:17:53,270 561 00:17:53,270 --> 00:17:56,470 562 00:17:56,470 --> 00:17:58,470 563 00:17:58,470 --> 00:18:00,789 564 00:18:00,789 --> 00:18:00,799 565 00:18:00,799 --> 00:18:02,470 566 00:18:02,470 --> 00:18:04,950 567 00:18:04,950 --> 00:18:06,950 568 00:18:06,950 --> 00:18:08,950 569 00:18:08,950 --> 00:18:11,110 570 00:18:11,110 --> 00:18:13,110 571 00:18:13,110 --> 00:18:14,950 572 00:18:14,950 --> 00:18:16,710 573 00:18:16,710 --> 00:18:18,549 574 00:18:18,549 --> 00:18:19,669 575 00:18:19,669 --> 00:18:19,679 576 00:18:19,679 --> 00:18:20,789 577 00:18:20,789 --> 00:18:23,669 578 00:18:23,669 --> 00:18:25,590 579 00:18:25,590 --> 00:18:27,029 580 00:18:27,029 --> 00:18:29,350 581 00:18:29,350 --> 00:18:32,070 582 00:18:32,070 --> 00:18:33,990 583 00:18:33,990 --> 00:18:36,630 584 00:18:36,630 --> 00:18:38,710 585 00:18:38,710 --> 00:18:40,470 586 00:18:40,470 --> 00:18:42,230 587 00:18:42,230 --> 00:18:43,830 588 00:18:43,830 --> 00:18:45,350 589 00:18:45,350 --> 00:18:45,360 590 00:18:45,360 --> 00:18:46,390 591 00:18:46,390 --> 00:18:46,400 592 00:18:46,400 --> 00:18:47,830 593 00:18:47,830 --> 00:18:49,510 594 00:18:49,510 --> 00:18:49,520 595 00:18:49,520 --> 00:18:50,630 596 00:18:50,630 --> 00:18:52,549 597 00:18:52,549 --> 00:18:55,190 598 00:18:55,190 --> 00:18:57,110 599 00:18:57,110 --> 00:18:57,120 600 00:18:57,120 --> 00:18:57,909 601 00:18:57,909 --> 00:18:59,029 602 00:18:59,029 --> 00:18:59,039 603 00:18:59,039 --> 00:19:03,190 604 00:19:03,190 --> 00:19:04,470 605 00:19:04,470 --> 00:19:06,230 606 00:19:06,230 --> 00:19:07,350 607 00:19:07,350 --> 00:19:07,360 608 00:19:07,360 --> 00:19:09,110 609 00:19:09,110 --> 00:19:09,120 610 00:19:09,120 --> 00:19:10,549 611 00:19:10,549 --> 00:19:10,559 612 00:19:10,559 --> 00:19:11,430 613 00:19:11,430 --> 00:19:11,440 614 00:19:11,440 --> 00:19:12,549 615 00:19:12,549 --> 00:19:15,029 616 00:19:15,029 --> 00:19:16,470 617 00:19:16,470 --> 00:19:19,430 618 00:19:19,430 --> 00:19:22,390 619 00:19:22,390 --> 00:19:24,630 620 00:19:24,630 --> 00:19:26,789 621 00:19:26,789 --> 00:19:26,799 622 00:19:26,799 --> 00:19:27,590 623 00:19:27,590 --> 00:19:29,430 624 00:19:29,430 --> 00:19:31,590 625 00:19:31,590 --> 00:19:34,230 626 00:19:34,230 --> 00:19:35,510 627 00:19:35,510 --> 00:19:37,990 628 00:19:37,990 --> 00:19:42,470 629 00:19:42,470 --> 00:19:44,230 630 00:19:44,230 --> 00:19:45,430 631 00:19:45,430 --> 00:19:47,350 632 00:19:47,350 --> 00:19:49,909 633 00:19:49,909 --> 00:19:51,590 634 00:19:51,590 --> 00:19:53,669 635 00:19:53,669 --> 00:19:55,510 636 00:19:55,510 --> 00:19:57,350 637 00:19:57,350 --> 00:19:58,710 638 00:19:58,710 --> 00:20:01,430 639 00:20:01,430 --> 00:20:03,190 640 00:20:03,190 --> 00:20:04,789 641 00:20:04,789 --> 00:20:06,789 642 00:20:06,789 --> 00:20:08,080 643 00:20:08,080 --> 00:20:08,090 644 00:20:08,090 --> 00:20:10,390 645 00:20:10,390 --> 00:20:12,149 646 00:20:12,149 --> 00:20:13,750 647 00:20:13,750 --> 00:20:16,310 648 00:20:16,310 --> 00:20:21,990 649 00:20:21,990 --> 00:20:22,000 650 00:20:22,000 --> 00:20:23,590 651 00:20:23,590 --> 00:20:26,310 652 00:20:26,310 --> 00:20:28,230 653 00:20:28,230 --> 00:20:30,470 654 00:20:30,470 --> 00:20:32,470 655 00:20:32,470 --> 00:20:34,230 656 00:20:34,230 --> 00:20:37,190 657 00:20:37,190 --> 00:20:39,590 658 00:20:39,590 --> 00:20:43,190 659 00:20:43,190 --> 00:20:44,789 660 00:20:44,789 --> 00:20:46,470 661 00:20:46,470 --> 00:20:49,909 662 00:20:49,909 --> 00:20:49,919 663 00:20:49,919 --> 00:20:51,190 664 00:20:51,190 --> 00:20:53,830 665 00:20:53,830 --> 00:20:56,549 666 00:20:56,549 --> 00:20:58,580 667 00:20:58,580 --> 00:20:58,590 668 00:20:58,590 --> 00:20:59,830 669 00:20:59,830 --> 00:21:01,590 670 00:21:01,590 --> 00:21:02,789 671 00:21:02,789 --> 00:21:04,710 672 00:21:04,710 --> 00:21:06,950 673 00:21:06,950 --> 00:21:08,310 674 00:21:08,310 --> 00:21:08,320 675 00:21:08,320 --> 00:21:09,909 676 00:21:09,909 --> 00:21:12,390 677 00:21:12,390 --> 00:21:13,990 678 00:21:13,990 --> 00:21:16,070 679 00:21:16,070 --> 00:21:17,750 680 00:21:17,750 --> 00:21:19,830 681 00:21:19,830 --> 00:21:21,590 682 00:21:21,590 --> 00:21:23,669 683 00:21:23,669 --> 00:21:25,830 684 00:21:25,830 --> 00:21:27,270 685 00:21:27,270 --> 00:21:27,280 686 00:21:27,280 --> 00:21:28,830 687 00:21:28,830 --> 00:21:31,510 688 00:21:31,510 --> 00:21:34,230 689 00:21:34,230 --> 00:21:36,470 690 00:21:36,470 --> 00:21:38,870 691 00:21:38,870 --> 00:21:40,470 692 00:21:40,470 --> 00:21:42,549 693 00:21:42,549 --> 00:21:43,669 694 00:21:43,669 --> 00:21:45,750 695 00:21:45,750 --> 00:21:47,510 696 00:21:47,510 --> 00:21:48,950 697 00:21:48,950 --> 00:21:48,960 698 00:21:48,960 --> 00:21:51,750 699 00:21:51,750 --> 00:21:53,669 700 00:21:53,669 --> 00:21:57,990 701 00:21:57,990 --> 00:22:00,710 702 00:22:00,710 --> 00:22:04,630 703 00:22:04,630 --> 00:22:06,230 704 00:22:06,230 --> 00:22:08,070 705 00:22:08,070 --> 00:22:09,990 706 00:22:09,990 --> 00:22:11,590 707 00:22:11,590 --> 00:22:13,590 708 00:22:13,590 --> 00:22:15,430 709 00:22:15,430 --> 00:22:15,440 710 00:22:15,440 --> 00:22:17,750 711 00:22:17,750 --> 00:22:19,990 712 00:22:19,990 --> 00:22:22,630 713 00:22:22,630 --> 00:22:24,230 714 00:22:24,230 --> 00:22:26,390 715 00:22:26,390 --> 00:22:28,549 716 00:22:28,549 --> 00:22:29,590 717 00:22:29,590 --> 00:22:31,909 718 00:22:31,909 --> 00:22:34,390 719 00:22:34,390 --> 00:22:36,070 720 00:22:36,070 --> 00:22:37,830 721 00:22:37,830 --> 00:22:41,830 722 00:22:41,830 --> 00:22:44,070 723 00:22:44,070 --> 00:22:46,870 724 00:22:46,870 --> 00:22:49,350 725 00:22:49,350 --> 00:22:51,190 726 00:22:51,190 --> 00:22:51,200 727 00:22:51,200 --> 00:22:52,310 728 00:22:52,310 --> 00:22:52,320 729 00:22:52,320 --> 00:22:53,110 730 00:22:53,110 --> 00:22:56,310 731 00:22:56,310 --> 00:22:57,990 732 00:22:57,990 --> 00:23:00,470 733 00:23:00,470 --> 00:23:04,710 734 00:23:04,710 --> 00:23:07,029 735 00:23:07,029 --> 00:23:09,590 736 00:23:09,590 --> 00:23:12,710 737 00:23:12,710 --> 00:23:12,720 738 00:23:12,720 --> 00:23:14,710 739 00:23:14,710 --> 00:23:16,710 740 00:23:16,710 --> 00:23:18,230 741 00:23:18,230 --> 00:23:20,149 742 00:23:20,149 --> 00:23:22,390 743 00:23:22,390 --> 00:23:24,950 744 00:23:24,950 --> 00:23:26,870 745 00:23:26,870 --> 00:23:26,880 746 00:23:26,880 --> 00:23:27,990 747 00:23:27,990 --> 00:23:29,669 748 00:23:29,669 --> 00:23:31,750 749 00:23:31,750 --> 00:23:34,070 750 00:23:34,070 --> 00:23:36,470 751 00:23:36,470 --> 00:23:37,909 752 00:23:37,909 --> 00:23:39,190 753 00:23:39,190 --> 00:23:39,200 754 00:23:39,200 --> 00:23:40,870 755 00:23:40,870 --> 00:23:42,549 756 00:23:42,549 --> 00:23:45,909 757 00:23:45,909 --> 00:23:47,669 758 00:23:47,669 --> 00:23:50,230 759 00:23:50,230 --> 00:23:53,190 760 00:23:53,190 --> 00:23:55,110 761 00:23:55,110 --> 00:23:57,430 762 00:23:57,430 --> 00:23:58,789 763 00:23:58,789 --> 00:24:01,590 764 00:24:01,590 --> 00:24:03,990 765 00:24:03,990 --> 00:24:05,909 766 00:24:05,909 --> 00:24:08,310 767 00:24:08,310 --> 00:24:10,950 768 00:24:10,950 --> 00:24:13,590 769 00:24:13,590 --> 00:24:15,190 770 00:24:15,190 --> 00:24:16,870 771 00:24:16,870 --> 00:24:19,269 772 00:24:19,269 --> 00:24:21,990 773 00:24:21,990 --> 00:24:26,230 774 00:24:26,230 --> 00:24:27,750 775 00:24:27,750 --> 00:24:29,430 776 00:24:29,430 --> 00:24:31,990 777 00:24:31,990 --> 00:24:33,510 778 00:24:33,510 --> 00:24:35,510 779 00:24:35,510 --> 00:24:37,789 780 00:24:37,789 --> 00:24:40,549 781 00:24:40,549 --> 00:24:42,870 782 00:24:42,870 --> 00:24:45,269 783 00:24:45,269 --> 00:24:45,279 784 00:24:45,279 --> 00:24:46,789 785 00:24:46,789 --> 00:24:49,029 786 00:24:49,029 --> 00:24:50,870 787 00:24:50,870 --> 00:24:52,710 788 00:24:52,710 --> 00:24:54,470 789 00:24:54,470 --> 00:24:56,549 790 00:24:56,549 --> 00:24:59,510 791 00:24:59,510 --> 00:25:01,110 792 00:25:01,110 --> 00:25:03,750 793 00:25:03,750 --> 00:25:06,310 794 00:25:06,310 --> 00:25:08,950 795 00:25:08,950 --> 00:25:08,960 796 00:25:08,960 --> 00:25:10,310 797 00:25:10,310 --> 00:25:11,990 798 00:25:11,990 --> 00:25:14,470 799 00:25:14,470 --> 00:25:17,029 800 00:25:17,029 --> 00:25:17,039 801 00:25:17,039 --> 00:25:18,230 802 00:25:18,230 --> 00:25:20,950 803 00:25:20,950 --> 00:25:21,670 804 00:25:21,670 --> 00:25:21,680 805 00:25:21,680 --> 00:25:22,789 806 00:25:22,789 --> 00:25:24,870 807 00:25:24,870 --> 00:25:26,789 808 00:25:26,789 --> 00:25:28,390 809 00:25:28,390 --> 00:25:31,029 810 00:25:31,029 --> 00:25:32,950 811 00:25:32,950 --> 00:25:34,310 812 00:25:34,310 --> 00:25:36,390 813 00:25:36,390 --> 00:25:40,310 814 00:25:40,310 --> 00:25:42,789 815 00:25:42,789 --> 00:25:45,350 816 00:25:45,350 --> 00:25:47,350 817 00:25:47,350 --> 00:25:47,360 818 00:25:47,360 --> 00:25:48,630 819 00:25:48,630 --> 00:25:50,470 820 00:25:50,470 --> 00:25:52,870 821 00:25:52,870 --> 00:25:54,230 822 00:25:54,230 --> 00:25:56,070 823 00:25:56,070 --> 00:25:57,669 824 00:25:57,669 --> 00:26:00,070 825 00:26:00,070 --> 00:26:03,350 826 00:26:03,350 --> 00:26:05,350 827 00:26:05,350 --> 00:26:07,029 828 00:26:07,029 --> 00:26:09,190 829 00:26:09,190 --> 00:26:10,789 830 00:26:10,789 --> 00:26:12,630 831 00:26:12,630 --> 00:26:13,830 832 00:26:13,830 --> 00:26:15,190 833 00:26:15,190 --> 00:26:18,390 834 00:26:18,390 --> 00:26:20,630 835 00:26:20,630 --> 00:26:22,870 836 00:26:22,870 --> 00:26:27,909 837 00:26:27,909 --> 00:26:30,470 838 00:26:30,470 --> 00:26:33,029 839 00:26:33,029 --> 00:26:35,269 840 00:26:35,269 --> 00:26:37,590 841 00:26:37,590 --> 00:26:37,600 842 00:26:37,600 --> 00:26:38,390 843 00:26:38,390 --> 00:26:40,950 844 00:26:40,950 --> 00:26:43,350 845 00:26:43,350 --> 00:26:46,070 846 00:26:46,070 --> 00:26:47,750 847 00:26:47,750 --> 00:26:50,149 848 00:26:50,149 --> 00:26:53,830 849 00:26:53,830 --> 00:26:55,669 850 00:26:55,669 --> 00:26:57,110 851 00:26:57,110 --> 00:27:00,710 852 00:27:00,710 --> 00:27:02,630 853 00:27:02,630 --> 00:27:06,149 854 00:27:06,149 --> 00:27:08,870 855 00:27:08,870 --> 00:27:09,750 856 00:27:09,750 --> 00:27:09,760 857 00:27:09,760 --> 00:27:10,789 858 00:27:10,789 --> 00:27:12,789 859 00:27:12,789 --> 00:27:14,230 860 00:27:14,230 --> 00:27:16,710 861 00:27:16,710 --> 00:27:25,750 862 00:27:25,750 --> 00:27:29,430 863 00:27:29,430 --> 00:27:29,440 864 00:27:29,440 --> 00:27:31,750 865 00:27:31,750 --> 00:27:31,760 866 00:27:31,760 --> 00:27:33,350 867 00:27:33,350 --> 00:27:35,350 868 00:27:35,350 --> 00:27:37,350 869 00:27:37,350 --> 00:27:39,269 870 00:27:39,269 --> 00:27:41,190 871 00:27:41,190 --> 00:27:42,389 872 00:27:42,389 --> 00:27:43,750 873 00:27:43,750 --> 00:27:43,760 874 00:27:43,760 --> 00:27:45,750 875 00:27:45,750 --> 00:27:48,149 876 00:27:48,149 --> 00:27:50,710 877 00:27:50,710 --> 00:27:53,110 878 00:27:53,110 --> 00:27:55,830 879 00:27:55,830 --> 00:27:59,590 880 00:27:59,590 --> 00:28:03,269 881 00:28:03,269 --> 00:28:05,110 882 00:28:05,110 --> 00:28:06,630 883 00:28:06,630 --> 00:28:08,070 884 00:28:08,070 --> 00:28:12,310 885 00:28:12,310 --> 00:28:12,320 886 00:28:12,320 --> 00:28:13,190 887 00:28:13,190 --> 00:28:15,029 888 00:28:15,029 --> 00:28:16,470 889 00:28:16,470 --> 00:28:19,750 890 00:28:19,750 --> 00:28:21,990 891 00:28:21,990 --> 00:28:23,590 892 00:28:23,590 --> 00:28:23,600 893 00:28:23,600 --> 00:28:25,190 894 00:28:25,190 --> 00:28:25,200 895 00:28:25,200 --> 00:28:26,310 896 00:28:26,310 --> 00:28:28,549 897 00:28:28,549 --> 00:28:31,029 898 00:28:31,029 --> 00:28:33,029 899 00:28:33,029 --> 00:28:33,039 900 00:28:33,039 --> 00:28:34,230 901 00:28:34,230 --> 00:28:36,549 902 00:28:36,549 --> 00:28:39,430 903 00:28:39,430 --> 00:28:43,190 904 00:28:43,190 --> 00:28:45,430 905 00:28:45,430 --> 00:28:47,830 906 00:28:47,830 --> 00:28:50,710 907 00:28:50,710 --> 00:28:50,720 908 00:28:50,720 --> 00:28:51,510 909 00:28:51,510 --> 00:28:53,909 910 00:28:53,909 --> 00:28:55,830 911 00:28:55,830 --> 00:28:57,669 912 00:28:57,669 --> 00:28:59,430 913 00:28:59,430 --> 00:29:01,029 914 00:29:01,029 --> 00:29:03,110 915 00:29:03,110 --> 00:29:04,789 916 00:29:04,789 --> 00:29:05,909 917 00:29:05,909 --> 00:29:05,919 918 00:29:05,919 --> 00:29:07,110 919 00:29:07,110 --> 00:29:09,830 920 00:29:09,830 --> 00:29:12,789 921 00:29:12,789 --> 00:29:14,789 922 00:29:14,789 --> 00:29:17,350 923 00:29:17,350 --> 00:29:19,750 924 00:29:19,750 --> 00:29:21,669 925 00:29:21,669 --> 00:29:23,510 926 00:29:23,510 --> 00:29:25,590 927 00:29:25,590 --> 00:29:28,310 928 00:29:28,310 --> 00:29:30,310 929 00:29:30,310 --> 00:29:33,750 930 00:29:33,750 --> 00:29:33,760 931 00:29:33,760 --> 00:29:34,870 932 00:29:34,870 --> 00:29:35,830 933 00:29:35,830 --> 00:29:37,430 934 00:29:37,430 --> 00:29:39,430 935 00:29:39,430 --> 00:29:41,430 936 00:29:41,430 --> 00:29:43,430 937 00:29:43,430 --> 00:29:46,710 938 00:29:46,710 --> 00:29:49,990 939 00:29:49,990 --> 00:29:53,830 940 00:29:53,830 --> 00:29:57,909 941 00:29:57,909 --> 00:29:57,919 942 00:29:57,919 --> 00:29:58,789 943 00:29:58,789 --> 00:30:00,310 944 00:30:00,310 --> 00:30:02,950 945 00:30:02,950 --> 00:30:05,190 946 00:30:05,190 --> 00:30:07,110 947 00:30:07,110 --> 00:30:07,120 948 00:30:07,120 --> 00:30:09,750 949 00:30:09,750 --> 00:30:11,350 950 00:30:11,350 --> 00:30:12,950 951 00:30:12,950 --> 00:30:15,350 952 00:30:15,350 --> 00:30:18,310 953 00:30:18,310 --> 00:30:21,110 954 00:30:21,110 --> 00:30:22,710 955 00:30:22,710 --> 00:30:24,870 956 00:30:24,870 --> 00:30:25,909 957 00:30:25,909 --> 00:30:27,510 958 00:30:27,510 --> 00:30:29,669 959 00:30:29,669 --> 00:30:32,310 960 00:30:32,310 --> 00:30:34,870 961 00:30:34,870 --> 00:30:36,870 962 00:30:36,870 --> 00:30:39,110 963 00:30:39,110 --> 00:30:40,470 964 00:30:40,470 --> 00:30:43,990 965 00:30:43,990 --> 00:30:44,000 966 00:30:44,000 --> 00:30:45,590 967 00:30:45,590 --> 00:30:47,110 968 00:30:47,110 --> 00:30:51,029 969 00:30:51,029 --> 00:30:52,710 970 00:30:52,710 --> 00:30:55,430 971 00:30:55,430 --> 00:30:56,950 972 00:30:56,950 --> 00:30:59,669 973 00:30:59,669 --> 00:31:05,990 974 00:31:05,990 --> 00:31:08,310 975 00:31:08,310 --> 00:31:10,710 976 00:31:10,710 --> 00:31:13,269 977 00:31:13,269 --> 00:31:15,909 978 00:31:15,909 --> 00:31:18,310 979 00:31:18,310 --> 00:31:21,269 980 00:31:21,269 --> 00:31:23,430 981 00:31:23,430 --> 00:31:26,230 982 00:31:26,230 --> 00:31:29,509 983 00:31:29,509 --> 00:31:31,990 984 00:31:31,990 --> 00:31:33,990 985 00:31:33,990 --> 00:31:36,789 986 00:31:36,789 --> 00:31:40,630 987 00:31:40,630 --> 00:31:43,029 988 00:31:43,029 --> 00:31:46,789 989 00:31:46,789 --> 00:31:48,230 990 00:31:48,230 --> 00:31:48,240 991 00:31:48,240 --> 00:31:49,350 992 00:31:49,350 --> 00:31:51,269 993 00:31:51,269 --> 00:31:53,909 994 00:31:53,909 --> 00:31:58,710 995 00:31:58,710 --> 00:32:02,870 996 00:32:02,870 --> 00:32:04,630 997 00:32:04,630 --> 00:32:06,470 998 00:32:06,470 --> 00:32:09,269 999 00:32:09,269 --> 00:32:11,269 1000 00:32:11,269 --> 00:32:12,549 1001 00:32:12,549 --> 00:32:14,710 1002 00:32:14,710 --> 00:32:16,710 1003 00:32:16,710 --> 00:32:19,350 1004 00:32:19,350 --> 00:32:23,190 1005 00:32:23,190 --> 00:32:25,590 1006 00:32:25,590 --> 00:32:27,909 1007 00:32:27,909 --> 00:32:30,950 1008 00:32:30,950 --> 00:32:34,789 1009 00:32:34,789 --> 00:32:34,799 1010 00:32:34,799 --> 00:32:35,990 1011 00:32:35,990 --> 00:32:36,000 1012 00:32:36,000 --> 00:32:36,950 1013 00:32:36,950 --> 00:32:39,590 1014 00:32:39,590 --> 00:32:42,230 1015 00:32:42,230 --> 00:32:45,269 1016 00:32:45,269 --> 00:32:45,279 1017 00:32:45,279 --> 00:32:46,470 1018 00:32:46,470 --> 00:32:48,789 1019 00:32:48,789 --> 00:32:50,789 1020 00:32:50,789 --> 00:32:53,509 1021 00:32:53,509 --> 00:32:56,149 1022 00:32:56,149 --> 00:32:58,310 1023 00:32:58,310 --> 00:33:00,149 1024 00:33:00,149 --> 00:33:02,070 1025 00:33:02,070 --> 00:33:05,430 1026 00:33:05,430 --> 00:33:07,269 1027 00:33:07,269 --> 00:33:07,279 1028 00:33:07,279 --> 00:33:09,350 1029 00:33:09,350 --> 00:33:12,630 1030 00:33:12,630 --> 00:33:15,430 1031 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01:22:46,960 2599 01:22:46,960 --> 01:22:49,110 2600 01:22:49,110 --> 01:22:49,120 2601 01:22:49,120 --> 01:22:50,709 2602 01:22:50,709 --> 01:22:53,510 2603 01:22:53,510 --> 01:22:53,520 2604 01:22:53,520 --> 01:22:54,470 2605 01:22:54,470 --> 01:22:55,830 2606 01:22:55,830 --> 01:22:57,270 2607 01:22:57,270 --> 01:22:59,189 2608 01:22:59,189 --> 01:23:01,030 2609 01:23:01,030 --> 01:23:03,030 2610 01:23:03,030 --> 01:23:05,189 2611 01:23:05,189 --> 01:23:08,390 2612 01:23:08,390 --> 01:23:08,400 2613 01:23:08,400 --> 01:23:09,590 2614 01:23:09,590 --> 01:23:11,030 2615 01:23:11,030 --> 01:23:13,270 2616 01:23:13,270 --> 01:23:16,070 2617 01:23:16,070 --> 01:23:18,790 2618 01:23:18,790 --> 01:23:22,229 2619 01:23:22,229 --> 01:23:24,149 2620 01:23:24,149 --> 01:23:26,390 2621 01:23:26,390 --> 01:23:28,950 2622 01:23:28,950 --> 01:23:28,960 2623 01:23:28,960 --> 01:23:31,189 2624 01:23:31,189 --> 01:23:31,199 2625 01:23:31,199 --> 01:23:33,030 2626 01:23:33,030 --> 01:23:34,550 2627 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01:24:26,629 2656 01:24:26,629 --> 01:24:28,070 2657 01:24:28,070 --> 01:24:31,110 2658 01:24:31,110 --> 01:24:33,350 2659 01:24:33,350 --> 01:24:35,510 2660 01:24:35,510 --> 01:24:35,520 2661 01:24:35,520 --> 01:24:36,950 2662 01:24:36,950 --> 01:24:39,189 2663 01:24:39,189 --> 01:24:41,910 2664 01:24:41,910 --> 01:24:44,310 2665 01:24:44,310 --> 01:24:45,750 2666 01:24:45,750 --> 01:24:49,270 2667 01:24:49,270 --> 01:24:51,110 2668 01:24:51,110 --> 01:24:53,510 2669 01:24:53,510 --> 01:24:53,520 2670 01:24:53,520 --> 01:24:55,110 2671 01:24:55,110 --> 01:24:55,120 2672 01:24:55,120 --> 01:24:57,110 2673 01:24:57,110 --> 01:24:59,270 2674 01:24:59,270 --> 01:25:00,950 2675 01:25:00,950 --> 01:25:02,870 2676 01:25:02,870 --> 01:25:05,510 2677 01:25:05,510 --> 01:25:08,390 2678 01:25:08,390 --> 01:25:10,790 2679 01:25:10,790 --> 01:25:13,990 2680 01:25:13,990 --> 01:25:16,390 2681 01:25:16,390 --> 01:25:17,750 2682 01:25:17,750 --> 01:25:17,760 2683 01:25:17,760 --> 01:25:19,590 2684 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