1 00:00:02,700 --> 00:00:05,470 alright so this lecture was about hyperbolic paraboloids and the extent to which they don't exist or exist here is a regular non existing hyperbolic paraboloid with concentric squares no diagonals folded here and so is just a few questions about this what does it mean other things these are all s by and paying i believe and they're all open so they're all good questions we don't know whether the the good triangulation the alternating one works for arbitrary and we only know up to N equals 104 the various fixed angles that we checked we don't have a proof technique to do arbitrary n you could try to make do some amount of alternation but not somewhere in between the two extremes and probably you get something in between goodness but I don't know could certainly play with that and it's very natural to try this with larger than squares the only trouble is the center no longer has a single degree of freedom so the only thing to answer there would be how do you want to initially fold like if you're doing a hexagon how hexagons probably clear what you might want to do but for general Kagan how do you want to arrange that innermost keg on from that you could propagate out just like we did with the square if you use an alternating triangulation say probably works well but we haven't tried it could be a cool project to do that not that hard so those were open problems next we have some sort of math general math questions c1 c2 and semi creases these are related questions so let me do a quick visual review of these terms I'm going to do it for one-dimensional curves in 2d because that's a lot easier to think about then what we really care about which is two-dimensional surfaces in 3d but all the ideas carry over so here is just a function let's say for example this is a parabolic arc then a straight segment this point is missing and instead it's up here these points are present and here we have another parabolic arc so this function i would call piecewise C infinity let me explain what these things mean so we have on the one hand c0 is a term meaning just continuous continuous means no jumps like this so this is not a continuous function this so this function overall is not even c0 c1 is a stronger condition which means that you not only are you continuous but you also have a continuous and existing first derivative so whereas here we're thinking about F here we're thinking of F prime so not only can you talk about the function being continuous c0 c1 but you can talk about individual moments in time so for example you know this moment here is going to be c1 it has a nice derivative that derivative is changing continuously here here however this is a c0 but not see one because the tangent on the left is different from the tangent on the right so this is not a c1 function because the derivative is jumping at that point I'm not going to draw the derivative here it's a little harder to draw but you could draw it at the set with the same x-axis and you'd see a jump from one angle to another this point is not even c0 this point is c0 but not see one this point is c1 because the tangent here is equal on the left is equal to the tangent on the right if I drew this properly where this is the bottom of the parabola okay you can ask for more than see one and we do in plus we talk about c2 and this says you should have a continuous second derivative so acceleration and this this point for example is not see two as well the tangents meet if you take the next derivative you see that it's suddenly here the tangent was completely flat it has a second derivative of zero and then suddenly it starts increasing now you can design functions that are even see infinity see infinity means no matter how many derivatives you take its continuous and defined no jumps but if you do a parabola for example it will not be so well behaved I think yes they're right got you know x y equals x squared you take the derivative you get 2x take the derivative of that you get 12 and then the derivative of that you get zero so the second derivative here is to all along the curve this I guess y prime Y double prime so in particular here we had a second derivative 0 here we have a second derivative of 2 alright so that's quick crash course on individual points being c0 c1 or c2 or C infinities when you can go all the way creases are points on the paper where you're not see one that means that you have two different tangent planes coming together typically so discontinuities in the first derivative that's what we call creases semi creases our discontinuities in the second derivative so semi crease is where your Nazi to crease is where you're not see one according to this particular paper semi crease is not too common a term but crease is very calm this is the usual meaning of crease semi creases maybe a little new to this paper we're talking about non-existence of hype ours and so for example this would be a crease and this would be a semi crease here the derivative changes that looks like a crease if you had a one dimensional piece of paper this doesn't look like a crease it's kind of smooth but not it's a little bit smooth at c1 but it's not see too so there's a semi crease there and part of that paper is worrying about semi creases because we want to deal with see two parts because see two functions are nice I mean ideally we have C infinity parts but we only need see two parts and so we subdivided the semi creases and basically argue most of the time so increases don't really happen so you don't have to worry about them but that's what they are any questions about smoothness these are all different kinds of smoothness yeah all of these things are C infinity except at these points so every polynomial C infinity you can differentiate it all the time eventually get 20 Exponential's or C infinity pretty much all functions you can think of that are can smooth RC infinity yeah all right so next question is about one of the proofs which was that got to remember the proof the proof involving normals basically we want involving tangent planes than the other one is involving normals this is the polygonal implies flat proof so this was if you have a region of paper is bounded by polygonal creases so they're piecewise straight then that region of paper if it son creased must actually be flat and I'm just gonna I want to talk about just one part of the proof so there was a sequence of steps but ultimately what we wanted was some Seng and bf this is a boundary of your region okay somewhere over here and we assume that it's straight so it took a few steps to get here but suppose we have a straight region and locally around this point there's some ruling okay and then we were looking at a point Q here I'm going to define this ruling to be RFQ the rule line from Q and so q is an arbitrary pointing slides along this segment and then we're interested in a normal vector here which we called n of Q that's normal to this the surface at this point and so as q slides and if Q changes and we were claiming a few things but that the definition of this guy being normal is it has to be perpendicular to normal to the surface is perpendicular to the surface so it has to be perpendicular to this sort of axis of the surface and has to be perpendicular to this axis so we know just from definition of normal n of Q is perpendicular to B F and n fq is perpendicular to RFQ the rule line so that's the definition but it wasn't quite what we wanted we wanted to prove for example that the derivative of the normal funny thing is perpendicular to bf that was one of the things we wanted and the other one is the corresponding perpendicular to the rule line I don't know if I use this notation and lecture but this just means they're perpendicular she's like the dot product of those two vectors is zero so how do we go from here to here this is actually the part that was most confusing to me I find this one more confusing because it uses the fact that its torso ruled but the question was actually about this one so unless you ask about this one now I'll just cover this one it's a little simpler both are in the notes so why is this true I guess first the intuitive reason so you've got all these different normals here we're going to end up concluding they're all pointing the same but in general they're changing direction somehow we know that at all times the normal is perpendicular to this segment so it's pointing away straight away from the segments we've got a right angle here no matter so it's kind of spinning around the segment at best as it moves around we claim that the change in the normal the derivative of the normal as you walk along this segment must also be perpendicular to this segment intuitively this is obvious if you think about it enough if you were changing in a direction that was not perpendicular to normal to this segment for example you're going this way or something then while initially your perpendicular your kind of falling over and then you'll no longer be perpendicular so if you change in a first-order way that is not perpendicular to this thing then afterwards you will no longer be perpendicular so that's intuitively why this follows from this okay the form one way to formalize that is to use Taylor expansion so i'll just write that quickly if you look at a point Q and I'm going to use this notation q plus epsilon to mean a point just a little bit over this is Q plus Epsilon this is Q for very small epsilon this is going to be approximately n of Q plus epsilon and prime of Q in Taylor expansion then you have a pep salon square but for small enough epsilon this is a good approximation and so on the one hand you have if you look at n of Q plus epsilon and you compare it to be F you know that these are perpendicular and what you can write that as a product being equal to 0 so the dot product of these two vectors should be equal to 0 that is being perpendicular we know this is true for all points Q so in particular is going to be true for Q plus epsilon so that's what we know but now we can expand this thing and we you know get these two terms so we get n fq dot F plus epsilon and prime of Q dot BF that's just distributing that sum over this product dot products work okay that way now this thing is also zero because this is just another point q dotted with BF and they should always be perpendicular and this is the thing we care about now if this whole sum is equal to 0 and this term is equal to 0 then this better be equal to 0 so right implies this thing equal to 0 and that's what we want it that n prime of Q is perpendicular to bf so that's the the algebra way to see it but if you think about it enough and you believe things are linear to the first order so we say then that has to be true you can get second derivatives and things as well but first derivative better hold in particular any questions about that you can do the same thing with this property but it's messier you need to use the tour cyl ruled property which we didn't really go into much detail on so I think I'll skip it last question is what does it all mean if we're proving things are impossible and yet here i have physical models of them existing what's going on what's the difference between mathematical paper and real paper and of course there's no real mathematical answer to that question but there we have a couple competing theories for what might be happening in this model one is that perhaps the paper is not being isometric perhaps it's stretching or shearing so normally we think of paper as unstretched ball and not sharable the only way to kind of share is to fold it that is almost perfectly true but of course no material is perfect so it might be shearing or changing the geometry the intrinsic geometry just enough to somehow make this possible our theory only says that it's impossible if you're exactly perfect so maybe if interesting to measure how exact paper is question it is possible to imitate this year hold it this is not a sheer I hold it we can add creases to my gate just to simulate a sheer right but only by an increases that's what the that's what we've proved in lecture so yeah you can kind of a cashier but only by n increases now it doesn't look like there's extra creases here theory number two is there are lots of really tiny creases here that are so small I mean here you could detect the crease because it was a big change in angle but if they're super super tiny here and you know never fold these things completely perfect it could be lots of little tiny things maybe you look under a microscope and you'll see that I don't know it could be an interesting project to try to investigate what really happens with paper in these kinds of models in real life and how it differs from from mathematics but the two theories are maybe some stretching or maybe extra creases we know by adding some by adding the diagonals it folds now here you could pretty clearly see the diagonals are not being added but there might be there might be effectively there might be enough added creases that are together so tiny you can't see them there is not any places semi creases are not enough the theory says that if you have these creases and possibly any number of semi creases you can't fold it at all so you need to add actual creases yeah good question all right that's that's it for old material now I want it to show you some cool new material which is based on again later than you don't give you a nice place right the creases here do not look like straight lines and yet we proved they must be straight lines so something's going on now we didn't know initially the straight creases had to stay straight in 3d but once we prove that we were pretty sure this didn't exist it's not obvious that's very Croesus stay straight but they do so yeah it's evidence there's a problem okay I wanted to talk about this paper because it's kind of related it's about how to efficiently make pleat folding happen and it's written by a bunch of people John cardinel Marty our cameraman shinji i'ma horiyoshi ito Musashi Kiyomi Stefan langerman rue Hitler Hara who's here on sabbatical and tacky aki no good practice so here's the kind of setting it's going to be familiar in that we have a 1d piece of paper so that's a segment and we're also going to assume that it's uniformly creased so each of these is a spacing of one and now usually we think of pre assigning mountains and valleys here and then you can only fold mountains on the mountains you can only fold valleys in the valleys all right the model here is that you're allowed to fold and unfold repeatedly on the same segments so you can do some layers simple folds though I think we'll just need all layers here ok but you can also sort of unfold any previous folds and I'll say the unfolds are free we don't you don't pay for the unfolding operations because I mean it's just a constant factor you can only unfold things that have previously been folded so we'll count the number of folds you make each time you do a full it's a mountain or Valley say through all the layers or through some of the layers but our goal is in the end to have our segments creased in a particular pattern and the pleat pattern would be MV MV MV so we want the last time each crease to be folded to be a particular pattern but along the way we could fold it in lots of it could be this one could be mountain for a while as long as the last time we creased this thing its Valley we're happy so we end up with sort of a mountain valley strings and we want to know how many folds how many folding operations do you need to make in order to achieve a particular mountain valley string and in particular we care about strings like MV MV MV because as we'll learn later today when we fold these up leading takes a lot of time it's tedious be nice if you could do things faster now why do we think we can do things faster because I can take a piece of paper and imagine this is one-dimensional and fold it in half and then fold it in half again boom I got how many creases two creases for the price of one operation now I fold again boom I get for creases for the price of one I fold it again well I get eight creases for the price of one so the number of folding operations you do here may be much much smaller than n n is going to be the number of the creases that the length of your strength ok so if I fold a piece of paper like this actually tried this and then I unfold to see what mountain valley assignment I got unfolding is free I get this weird shape and I didn't do it perfectly it's hard to do perfectly because of creep I get a weird thing anyone know what this thing is called I think that's pretty much if I unfold them to 90 degrees dragon curve yeah dragon curve looks like this this is the Wikipedia drawing and the more you fold in half that you keep adding this iteration it's a nice fractal it's kind of cool it doesn't happens you can prove it doesn't self intersect I mean it touches itself at vertices but it doesn't properly cross itself you take the limit it looks something like this if you fill it in very cool wikipedia calls it the I don't want to spoil the surprise but there's actually a book where the the sort of section headings are iterations of this fractal anyone know what this book is Jurassic Park Ian Malcolm right there so pretty cool it's called the Jurassic Park fractal to some at least on Wikipedia um and so that's one particular mountain valley assignment you can get and let me tell you what it is it's pretty easy to figure out and you see why it's a fractal so the first time we make a fold let's say I always make mountains just for simplicity you have a little bit of choice here but not very much okay so first I make a mountain fold so now I've got things folded in half now let's say I make a mountain fold over here of course it's already fold in half which means i get a valley over here so i'm going to yeah so now i make another mountain fold but I don't maybe I should have done it the other way anyway so it is so I make another mountain fold here and on the Left I get this on the right I get the reflection of that and then I make another mountain fold I get this on the left and I get the reflection of that on the right and when I reflect I'm reading backwards and also inverting everything v and so on so you see the kind of fractal nature here and that's what gives the cool curve keep repeating that you have essentially n choices their login folds that you make H could be a mountain or Valley so there n different MV strings you can get in essentially log n folds but sadly none of them is the one we want MV MV MV the pleat fold so that leaves us with the question of how many operations do you need to get a pleat fold this is where things get cool so yeah well why don't I tell you some answers so if you want to fold mm mm or MV MV please turn out to be the same because you can do ms on the odd positions v's on the even positions then you can do it in log squared n folds and you need at least log squared n over log log n this is way way faster than n there's an open problem between log squared n and log squared over log log small gap of log factor but somewhere in between here if you want to do let's say a random pattern then we can prove you need a bow sorry and divided by log in so the obvious step the obvious way to do it is fold each crease unfold that takes n steps so you can always improve by a log factor and that's the best you can do for most strings but these strings are sufficiently special that you can get a huge factor a basically exponential win instead of doing roughly n steps you're only doing poly log steps so there's four results here which ones would you like to see how many people vote for log squared n upper bound for how many people vote for log squared over log log lower bound completely different set of four how many people vote for the n over log n story upper bound for and lower bounds okay i think the upper bound here wins is curious they all these similar ideas let me I'm going to briefly sketch because I want to get to folding stuff how you achieve n over log n upper bound maybe then I'll do log squared upper bound here lower bounds are well they're lower bounds it doesn't say anything so if you have an MV string to get this bound arbitrary string I'm going to consider epsilon sorry 1 minus epsilon log in consecutive letters and split into basically n over log n chunks each of size roughly log in a little bit smaller than log in why do I do that because the number of different chunk values is 2 to the 1 minus epsilon times log n right each can be mountain of valley now to to the log n is n so this is n to the 1 minus epsilon possible different chunk values these are chunks why is that interesting because I have basically n over log and chunks but there's only n to the 1 minus epsilon different trunk values this is much much smaller than this I think of epsilon being a half the square root of n this is n over log n so there's actually most many of the chunks have to be repeated so this means an average chunk is repeated what is it an over log n of them just like n over log n times n to the 1 minus epsilon times which is n to the epsilon over log n right this is the number of chunks here and we're dividing by the number of different ones so typical repetition is going to be n to the epsilon over log in time so I'll just assume all chunks repeated this many times it's all linear so it doesn't matter whether some are more common some are less common okay cool now what so if I look at one of these chunks and all of its repetitions I would like to fold them all somehow more efficiently and there's one key idea in this paper on how to do that which is so here's my strip maybe here's one rep instance of the chunk here's another one they can be kind of spaced out arbitrarily so what I'm going to do is fold here fold here fold here I think that's right all the bisectors so I'll end up with one chunk like this then another chunk like this then another chunk something like this then this one's very tiny okay I get my four repetitions of the chunk all on top of each other what I then do so this is going to take how many repetitions is like n to the epsilon over logins this is about n to the epsilon over log n steps to do all these folds now I fold here fold here fold here fold here simultaneously getting them all correct that's going to take 1 minus epsilon time basically log n steps then I unfold and then I discover darn it half of them were upside down because I got this one right and this one right but these two are flipped so that's no good so then I recurse on the remaining half but this turns out to be a geometric series so if I do all four of these and then I do two of them and then I do one of them I'll be done and it will only cost me another factor of two in time so it's I get log n plus n to the epsilon over log in I have to do this for each of the different chunk values so i need to do n to the 1 minus epsilon times log n plus n to the epsilon over again x like to because of the geometric series and this comes out to andover login tada magic the login is basically coming from the trunk size being log in and 1 minus epsilon log n is as big as we could make it because we needed this to be a lot smaller than n so that's how you save a factor of log in a little crazy but it works in theory that's the upper bound the lower bound is actually pretty obvious because if you think if you have you only use k folds you're doing there's n to the K or maybe to n to the K different things you can do because there's n different places you can make a fold could be a mountain or valley and this better be at least two to the end because they're two to the N different Mountain Valley patterns you could make and you have to somehow make them with K folds these are all the things you can make with K folds and you work that out and K has to be at least about n over log n so that's why it's optimal and because this is an information-theoretic argument this works in the average case as well take a random example you need at least n over log in most examples probability will need and over log in so that's those lower bounds let me briefly tell you about this log squared and upper bound uses the same idea but because everything is repeated in the mmm mmm string you don't it's a lot easier to do this kind of folding oh sorry there's one step I left out here so great you make these folds you line things up you fold these things then you unfold so you've and recurse eventually you fix these guys but you also destroyed this case in this crease in this crease so you kind of mess things up a little you've got to do this repeatedly for every chunk you don't want to mess up previous chunks that you've done so you have to go back and fix this fold unfold fix this fold unfold but that again only you only cost n to the epsilon over log n so not too bad so another factor of two here but I'm ignoring constant factors okay that was an over log in upper bound let's do log squared upper bound for mMmmm so I need to look this up it's a little tricky we're going to use the trick of the Dragon curve essentially which is repeatedly fold in half and we're going to keep doing that until we are left with three creases which haven't yet been folded so this is my folded bundle there's you know many many many layers here and then I'm going to fold mmm on those three creases instead of going all the way to Dragon curve which would just be do em fold here first I'm going to fold this unfold fold this unfold fold this unfold so I get this pattern when I unfold this it actually turns out to be kind of nice and alternating it will be something mmm something Valley Valley Valley something about a mountain them something Valley Valley Valley and so on okay great half of my things roughly our mountains I just need to fix these valleys so I use the same trick which is I'm going to fold use red maybe fold in the middle of these mountain segments I do that I line up all the valley segments so i end up with after folding this is going to look like you almost need a computer or call this out but it that we have done implemented this it works it's going to look like this repeat I mean with many layers like this all the valleys are on top of each other these question marks are still question marks this mountain is that mountain this mountain is that mutton my mountain is your mountain I don't know so now we're going to fold mountain mountain mountain mountain on all five of these things and these two remain mountains and then unfold and what i end up with is sort of twice as good i end up with ? and then i think seven sorry not quite because this part also gets messed up so i end up with Valley Mountain ? then seven mountains then ? then some not so pretty stuff em vvvvv em ? and then this repeats until we get to the end of the string and then the end looks kind of like this so the point is I had 3 m's in a row now i have 7 ms in a row each time i do this i roughly double the number of Em's in a row by the end almost everything will be ms after I do this log n times I have pretty much all Em's and then I can finish it off and each of these steps took log n steps because I had to fold along all these things how did I fold along them not one at a time of course I fold in half and fold in half and fold in half roughly I choose the middle most red line to fold first and about log n steps later everything will be piled up so I get to use this efficiency of dragon curve things I'm messing up these creases of course but it's okay because I get a big chunk that all gets correct at once half of it gets wrong half of its correct but the half that's getting correct gets longer and longer each step doubling so only log squared steps pretty sure this is optimal but the best lower bound we can prove is log squared over log log all right finally let's fold some stuff I thought we would fold some hyperbolic paraboloids and maybe put them together into structures like this so you have your squares if you don't have squares bunch of squares up here you can start folding as many as you like we're going to put them together I will demonstrate up here so the first thing we want to do is fold along the diagonals these aren't perfect squares but it's ok so we fold along on diagonal and the other diagonal this is mainly so we know where the center is but also because we need to fold a diagonal so we kill two birds with one stone whereas we just saw how to kill in this case log n birds with one stone save log in here we're only killing two so now you've got your diagonals near square we want to fold it's kind of cool I color the edges with my chalk hands now where everything's going to be parallel to the sides of the square so first thing we're going to do is construct a square that's half is big in the centered along that point so this involves folding the bottom edge usually do this valley-fold fold the bottom edge to line up with that Center and also line up the edges now very important when you make this fold don't fold it all the way just fold the middle half between the two diagonals that's really key otherwise this will not work it's the only thing you have to be careful of don't fall all the way just in the middle okay that's one quarter do that four times so we go here line up with the center fold the middle once you've got the inner square you're going to make two more there's going to be this square at the one quarter mark and an inner square at the three quarters mark to make the outer square you fold the edge to the thing that you just folded keep keep careful note of which crease you're folding to should be from the previous square not from the current Square to fold the inner square you go all the way up to the opposite side of the first square and you can just look visually are these evenly spaced if yes you did it right if not you're in trouble probably okay if you make a few extra creases but don't try to make as few extra creases as possible so then repeat that four times and you'll get your I guess four squares if you count the outer boundary all right it's a little hard to see once you've got that those squares looks like I do one of them incorrectly got your nice concentric squares four of them now you're halfway done you flip it over and do valley folds on the other side so because you want these to be alternating Mountain Valley Mountain Valley and just fill in all the squares in between those squares and you can figure out what the reference markers are there all the mountain folds well actually every other mountain fold will be one of your references don't go to one of your new Valley folds always go to the mountain once and just check that you're always filling in halfway in between two of your Mountain folds don't forget to flip over very important and skip every other guy otherwise you will think refold existing crease the wrong way this is what it looks like after one quarter of those alright once you've done all this crazy precreasing you have concentric squares alternating mountain valley then the fun part begins this is actually literally fun not its I guess also harder but much more exciting than all that precreasing so in this case you have to fold all the creases at once and the easy way to do that is start at the outside and make the outermost ring fold it go around once or twice make sure everything is folded including that diagonal so here I've Valley folded everything including the diet outer diagonal and then you want to mountain fold the next one and get them to be against each other you're aiming for a kind of X shape which is drawn over here it's going to look like that ideally just keep collapsing square by square making sure you alternating mountain valley as I mean it should just fall into place pretty much but paper likes to miss fold a little bit so you got to fix it all here I've done two squares a little more it's easier and easier because the squares get smaller and smaller we've got three squares done so I'm like halfway to annex fourth square centers a little bit tricky try to get it also alternate so now it's nice and thin it's like repeated sync folds of a waterbomb base and it's like an X and then this is your opportunity to wreak Reese all your crease is really hard just give it a good squeeze on each of the three four legs of the starfish and now you've got your hyperbolic paraboloid to make it you want to take two midpoints here of the squares pull them apart until it's a little bit open and then give it a twist and then open up a little you've got your hyperbolic paraboloid tah-dah if we make enough of these we can assemble it into some cool shapes how many people have folded one at least one few people i will show you you can fold more we're going to assemble something might need a bunch of people this is the algorithm we use for converting a polyhedron into a bunch of hyperbolic paraboloids it's in this paper from 99 and we take each face of the polygons of the polyhedron sorry so if you have a cube you've got a bunch of squares for each square we make what we call a 4 hat which is for hyperbolic paraboloids joined in a cycle like this and you've got to be careful to join it the right way but then these the tips of the high parts that are not joined I mean two of the one tip of each of the hype ours comes together at the center then these tips are going to represent the edge of the polygon these red dots and so these two sides are the sides of one hype of two these two sides of the sides of one hype are they're going to join to an adjacent high power over here so that's the idea yeah so I've got already filled with a bunch of these already so maybe what a hat looks like so tick two of them join them along those edges we're going to use tape or staples to join them don't have a fancy lock I'm afraid and join these edges together and you get this is a 3 hat you can put on your head whatever and that would represent a triangle and then we're going to join along these two sides to an adjacent triangle or whatever so we could make a platonic solid i guess the simplest one would be a tetrahedron that has six edges so it needs 12 hype ours I was thinking we could make this solid it's never been made before should be cool it's the simplest Archimedean solid terms the number of edges the truncated tetrahedron this requires 36 parts so for ambitious we can go for it but I think we're low on time so tetrahedra might be easier bit people want to come up and start assembling all right finished taping that hold these this goes here you're gonna hold this one virtual assembly here so maybe we'll finish this Emily next class 2 00:00:05,470 --> 00:00:08,179 alright so this lecture was about hyperbolic paraboloids and the extent to which they don't exist or exist here is a regular non existing hyperbolic paraboloid with concentric squares no diagonals folded here and so is just a few questions about this what does it mean other things these are all s by and paying i believe and they're all open so they're all good questions we don't know whether the the good triangulation the alternating one works for arbitrary and we only know up to N equals 104 the various fixed angles that we checked we don't have a proof technique to do arbitrary n you could try to make do some amount of alternation but not somewhere in between the two extremes and probably you get something in between goodness but I don't know could certainly play with that and it's very natural to try this with larger than squares the only trouble is the center no longer has a single degree of freedom so the only thing to answer there would be how do you want to initially fold like if you're doing a hexagon how hexagons probably clear what you might want to do but for general Kagan how do you want to arrange that innermost keg on from that you could propagate out just like we did with the square if you use an alternating triangulation say probably works well but we haven't tried it could be a cool project to do that not that hard so those were open problems next we have some sort of math general math questions c1 c2 and semi creases these are related questions so let me do a quick visual review of these terms I'm going to do it for one-dimensional curves in 2d because that's a lot easier to think about then what we really care about which is two-dimensional surfaces in 3d but all the ideas carry over so here is just a function let's say for example this is a parabolic arc then a straight segment this point is missing and instead it's up here these points are present and here we have another parabolic arc so this function i would call piecewise C infinity let me explain what these things mean so we have on the one hand c0 is a term meaning just continuous continuous means no jumps like this so this is not a continuous function this so this function overall is not even c0 c1 is a stronger condition which means that you not only are you continuous but you also have a continuous and existing first derivative so whereas here we're thinking about F here we're thinking of F prime so not only can you talk about the function being continuous c0 c1 but you can talk about individual moments in time so for example you know this moment here is going to be c1 it has a nice derivative that derivative is changing continuously here here however this is a c0 but not see one because the tangent on the left is different from the tangent on the right so this is not a c1 function because the derivative is jumping at that point I'm not going to draw the derivative here it's a little harder to draw but you could draw it at the set with the same x-axis and you'd see a jump from one angle to another this point is not even c0 this point is c0 but not see one this point is c1 because the tangent here is equal on the left is equal to the tangent on the right if I drew this properly where this is the bottom of the parabola okay you can ask for more than see one and we do in plus we talk about c2 and this says you should have a continuous second derivative so acceleration and this this point for example is not see two as well the tangents meet if you take the next derivative you see that it's suddenly here the tangent was completely flat it has a second derivative of zero and then suddenly it starts increasing now you can design functions that are even see infinity see infinity means no matter how many derivatives you take its continuous and defined no jumps but if you do a parabola for example it will not be so well behaved I think yes they're right got you know x y equals x squared you take the derivative you get 2x take the derivative of that you get 12 and then the derivative of that you get zero so the second derivative here is to all along the curve this I guess y prime Y double prime so in particular here we had a second derivative 0 here we have a second derivative of 2 alright so that's quick crash course on individual points being c0 c1 or c2 or C infinities when you can go all the way creases are points on the paper where you're not see one that means that you have two different tangent planes coming together typically so discontinuities in the first derivative that's what we call creases semi creases our discontinuities in the second derivative so semi crease is where your Nazi to crease is where you're not see one according to this particular paper semi crease is not too common a term but crease is very calm this is the usual meaning of crease semi creases maybe a little new to this paper we're talking about non-existence of hype ours and so for example this would be a crease and this would be a semi crease here the derivative changes that looks like a crease if you had a one dimensional piece of paper this doesn't look like a crease it's kind of smooth but not it's a little bit smooth at c1 but it's not see too so there's a semi crease there and part of that paper is worrying about semi creases because we want to deal with see two parts because see two functions are nice I mean ideally we have C infinity parts but we only need see two parts and so we subdivided the semi creases and basically argue most of the time so increases don't really happen so you don't have to worry about them but that's what they are any questions about smoothness these are all different kinds of smoothness yeah all of these things are C infinity except at these points so every polynomial C infinity you can differentiate it all the time eventually get 20 Exponential's or C infinity pretty much all functions you can think of that are can smooth RC infinity yeah all right so next question is about one of the proofs which was that got to remember the proof the proof involving normals basically we want involving tangent planes than the other one is involving normals this is the polygonal implies flat proof so this was if you have a region of paper is bounded by polygonal creases so they're piecewise straight then that region of paper if it son creased must actually be flat and I'm just gonna I want to talk about just one part of the proof so there was a sequence of steps but ultimately what we wanted was some Seng and bf this is a boundary of your region okay somewhere over here and we assume that it's straight so it took a few steps to get here but suppose we have a straight region and locally around this point there's some ruling okay and then we were looking at a point Q here I'm going to define this ruling to be RFQ the rule line from Q and so q is an arbitrary pointing slides along this segment and then we're interested in a normal vector here which we called n of Q that's normal to this the surface at this point and so as q slides and if Q changes and we were claiming a few things but that the definition of this guy being normal is it has to be perpendicular to normal to the surface is perpendicular to the surface so it has to be perpendicular to this sort of axis of the surface and has to be perpendicular to this axis so we know just from definition of normal n of Q is perpendicular to B F and n fq is perpendicular to RFQ the rule line so that's the definition but it wasn't quite what we wanted we wanted to prove for example that the derivative of the normal funny thing is perpendicular to bf that was one of the things we wanted and the other one is the corresponding perpendicular to the rule line I don't know if I use this notation and lecture but this just means they're perpendicular she's like the dot product of those two vectors is zero so how do we go from here to here this is actually the part that was most confusing to me I find this one more confusing because it uses the fact that its torso ruled but the question was actually about this one so unless you ask about this one now I'll just cover this one it's a little simpler both are in the notes so why is this true I guess first the intuitive reason so you've got all these different normals here we're going to end up concluding they're all pointing the same but in general they're changing direction somehow we know that at all times the normal is perpendicular to this segment so it's pointing away straight away from the segments we've got a right angle here no matter so it's kind of spinning around the segment at best as it moves around we claim that the change in the normal the derivative of the normal as you walk along this segment must also be perpendicular to this segment intuitively this is obvious if you think about it enough if you were changing in a direction that was not perpendicular to normal to this segment for example you're going this way or something then while initially your perpendicular your kind of falling over and then you'll no longer be perpendicular so if you change in a first-order way that is not perpendicular to this thing then afterwards you will no longer be perpendicular so that's intuitively why this follows from this okay the form one way to formalize that is to use Taylor expansion so i'll just write that quickly if you look at a point Q and I'm going to use this notation q plus epsilon to mean a point just a little bit over this is Q plus Epsilon this is Q for very small epsilon this is going to be approximately n of Q plus epsilon and prime of Q in Taylor expansion then you have a pep salon square but for small enough epsilon this is a good approximation and so on the one hand you have if you look at n of Q plus epsilon and you compare it to be F you know that these are perpendicular and what you can write that as a product being equal to 0 so the dot product of these two vectors should be equal to 0 that is being perpendicular we know this is true for all points Q so in particular is going to be true for Q plus epsilon so that's what we know but now we can expand this thing and we you know get these two terms so we get n fq dot F plus epsilon and prime of Q dot BF that's just distributing that sum over this product dot products work okay that way now this thing is also zero because this is just another point q dotted with BF and they should always be perpendicular and this is the thing we care about now if this whole sum is equal to 0 and this term is equal to 0 then this better be equal to 0 so right implies this thing equal to 0 and that's what we want it that n prime of Q is perpendicular to bf so that's the the algebra way to see it but if you think about it enough and you believe things are linear to the first order so we say then that has to be true you can get second derivatives and things as well but first derivative better hold in particular any questions about that you can do the same thing with this property but it's messier you need to use the tour cyl ruled property which we didn't really go into much detail on so I think I'll skip it last question is what does it all mean if we're proving things are impossible and yet here i have physical models of them existing what's going on what's the difference between mathematical paper and real paper and of course there's no real mathematical answer to that question but there we have a couple competing theories for what might be happening in this model one is that perhaps the paper is not being isometric perhaps it's stretching or shearing so normally we think of paper as unstretched ball and not sharable the only way to kind of share is to fold it that is almost perfectly true but of course no material is perfect so it might be shearing or changing the geometry the intrinsic geometry just enough to somehow make this possible our theory only says that it's impossible if you're exactly perfect so maybe if interesting to measure how exact paper is question it is possible to imitate this year hold it this is not a sheer I hold it we can add creases to my gate just to simulate a sheer right but only by an increases that's what the that's what we've proved in lecture so yeah you can kind of a cashier but only by n increases now it doesn't look like there's extra creases here theory number two is there are lots of really tiny creases here that are so small I mean here you could detect the crease because it was a big change in angle but if they're super super tiny here and you know never fold these things completely perfect it could be lots of little tiny things maybe you look under a microscope and you'll see that I don't know it could be an interesting project to try to investigate what really happens with paper in these kinds of models in real life and how it differs from from mathematics but the two theories are maybe some stretching or maybe extra creases we know by adding some by adding the diagonals it folds now here you could pretty clearly see the diagonals are not being added but there might be there might be effectively there might be enough added creases that are together so tiny you can't see them there is not any places semi creases are not enough the theory says that if you have these creases and possibly any number of semi creases you can't fold it at all so you need to add actual creases yeah good question all right that's that's it for old material now I want it to show you some cool new material which is based on again later than you don't give you a nice place right the creases here do not look like straight lines and yet we proved they must be straight lines so something's going on now we didn't know initially the straight creases had to stay straight in 3d but once we prove that we were pretty sure this didn't exist it's not obvious that's very Croesus stay straight but they do so yeah it's evidence there's a problem okay I wanted to talk about this paper because it's kind of related it's about how to efficiently make pleat folding happen and it's written by a bunch of people John cardinel Marty our cameraman shinji i'ma horiyoshi ito Musashi Kiyomi Stefan langerman rue Hitler Hara who's here on sabbatical and tacky aki no good practice so here's the kind of setting it's going to be familiar in that we have a 1d piece of paper so that's a segment and we're also going to assume that it's uniformly creased so each of these is a spacing of one and now usually we think of pre assigning mountains and valleys here and then you can only fold mountains on the mountains you can only fold valleys in the valleys all right the model here is that you're allowed to fold and unfold repeatedly on the same segments so you can do some layers simple folds though I think we'll just need all layers here ok but you can also sort of unfold any previous folds and I'll say the unfolds are free we don't you don't pay for the unfolding operations because I mean it's just a constant factor you can only unfold things that have previously been folded so we'll count the number of folds you make each time you do a full it's a mountain or Valley say through all the layers or through some of the layers but our goal is in the end to have our segments creased in a particular pattern and the pleat pattern would be MV MV MV so we want the last time each crease to be folded to be a particular pattern but along the way we could fold it in lots of it could be this one could be mountain for a while as long as the last time we creased this thing its Valley we're happy so we end up with sort of a mountain valley strings and we want to know how many folds how many folding operations do you need to make in order to achieve a particular mountain valley string and in particular we care about strings like MV MV MV because as we'll learn later today when we fold these up leading takes a lot of time it's tedious be nice if you could do things faster now why do we think we can do things faster because I can take a piece of paper and imagine this is one-dimensional and fold it in half and then fold it in half again boom I got how many creases two creases for the price of one operation now I fold again boom I get for creases for the price of one I fold it again well I get eight creases for the price of one so the number of folding operations you do here may be much much smaller than n n is going to be the number of the creases that the length of your strength ok so if I fold a piece of paper like this actually tried this and then I unfold to see what mountain valley assignment I got unfolding is free I get this weird shape and I didn't do it perfectly it's hard to do perfectly because of creep I get a weird thing anyone know what this thing is called I think that's pretty much if I unfold them to 90 degrees dragon curve yeah dragon curve looks like this this is the Wikipedia drawing and the more you fold in half that you keep adding this iteration it's a nice fractal it's kind of cool it doesn't happens you can prove it doesn't self intersect I mean it touches itself at vertices but it doesn't properly cross itself you take the limit it looks something like this if you fill it in very cool wikipedia calls it the I don't want to spoil the surprise but there's actually a book where the the sort of section headings are iterations of this fractal anyone know what this book is Jurassic Park Ian Malcolm right there so pretty cool it's called the Jurassic Park fractal to some at least on Wikipedia um and so that's one particular mountain valley assignment you can get and let me tell you what it is it's pretty easy to figure out and you see why it's a fractal so the first time we make a fold let's say I always make mountains just for simplicity you have a little bit of choice here but not very much okay so first I make a mountain fold so now I've got things folded in half now let's say I make a mountain fold over here of course it's already fold in half which means i get a valley over here so i'm going to yeah so now i make another mountain fold but I don't maybe I should have done it the other way anyway so it is so I make another mountain fold here and on the Left I get this on the right I get the reflection of that and then I make another mountain fold I get this on the left and I get the reflection of that on the right and when I reflect I'm reading backwards and also inverting everything v and so on so you see the kind of fractal nature here and that's what gives the cool curve keep repeating that you have essentially n choices their login folds that you make H could be a mountain or Valley so there n different MV strings you can get in essentially log n folds but sadly none of them is the one we want MV MV MV the pleat fold so that leaves us with the question of how many operations do you need to get a pleat fold this is where things get cool so yeah well why don't I tell you some answers so if you want to fold mm mm or MV MV please turn out to be the same because you can do ms on the odd positions v's on the even positions then you can do it in log squared n folds and you need at least log squared n over log log n this is way way faster than n there's an open problem between log squared n and log squared over log log small gap of log factor but somewhere in between here if you want to do let's say a random pattern then we can prove you need a bow sorry and divided by log in so the obvious step the obvious way to do it is fold each crease unfold that takes n steps so you can always improve by a log factor and that's the best you can do for most strings but these strings are sufficiently special that you can get a huge factor a basically exponential win instead of doing roughly n steps you're only doing poly log steps so there's four results here which ones would you like to see how many people vote for log squared n upper bound for how many people vote for log squared over log log lower bound completely different set of four how many people vote for the n over log n story upper bound for and lower bounds okay i think the upper bound here wins is curious they all these similar ideas let me I'm going to briefly sketch because I want to get to folding stuff how you achieve n over log n upper bound maybe then I'll do log squared upper bound here lower bounds are well they're lower bounds it doesn't say anything so if you have an MV string to get this bound arbitrary string I'm going to consider epsilon sorry 1 minus epsilon log in consecutive letters and split into basically n over log n chunks each of size roughly log in a little bit smaller than log in why do I do that because the number of different chunk values is 2 to the 1 minus epsilon times log n right each can be mountain of valley now to to the log n is n so this is n to the 1 minus epsilon possible different chunk values these are chunks why is that interesting because I have basically n over log and chunks but there's only n to the 1 minus epsilon different trunk values this is much much smaller than this I think of epsilon being a half the square root of n this is n over log n so there's actually most many of the chunks have to be repeated so this means an average chunk is repeated what is it an over log n of them just like n over log n times n to the 1 minus epsilon times which is n to the epsilon over log n right this is the number of chunks here and we're dividing by the number of different ones so typical repetition is going to be n to the epsilon over log in time so I'll just assume all chunks repeated this many times it's all linear so it doesn't matter whether some are more common some are less common okay cool now what so if I look at one of these chunks and all of its repetitions I would like to fold them all somehow more efficiently and there's one key idea in this paper on how to do that which is so here's my strip maybe here's one rep instance of the chunk here's another one they can be kind of spaced out arbitrarily so what I'm going to do is fold here fold here fold here I think that's right all the bisectors so I'll end up with one chunk like this then another chunk like this then another chunk something like this then this one's very tiny okay I get my four repetitions of the chunk all on top of each other what I then do so this is going to take how many repetitions is like n to the epsilon over logins this is about n to the epsilon over log n steps to do all these folds now I fold here fold here fold here fold here simultaneously getting them all correct that's going to take 1 minus epsilon time basically log n steps then I unfold and then I discover darn it half of them were upside down because I got this one right and this one right but these two are flipped so that's no good so then I recurse on the remaining half but this turns out to be a geometric series so if I do all four of these and then I do two of them and then I do one of them I'll be done and it will only cost me another factor of two in time so it's I get log n plus n to the epsilon over log in I have to do this for each of the different chunk values so i need to do n to the 1 minus epsilon times log n plus n to the epsilon over again x like to because of the geometric series and this comes out to andover login tada magic the login is basically coming from the trunk size being log in and 1 minus epsilon log n is as big as we could make it because we needed this to be a lot smaller than n so that's how you save a factor of log in a little crazy but it works in theory that's the upper bound the lower bound is actually pretty obvious because if you think if you have you only use k folds you're doing there's n to the K or maybe to n to the K different things you can do because there's n different places you can make a fold could be a mountain or valley and this better be at least two to the end because they're two to the N different Mountain Valley patterns you could make and you have to somehow make them with K folds these are all the things you can make with K folds and you work that out and K has to be at least about n over log n so that's why it's optimal and because this is an information-theoretic argument this works in the average case as well take a random example you need at least n over log in most examples probability will need and over log in so that's those lower bounds let me briefly tell you about this log squared and upper bound uses the same idea but because everything is repeated in the mmm mmm string you don't it's a lot easier to do this kind of folding oh sorry there's one step I left out here so great you make these folds you line things up you fold these things then you unfold so you've and recurse eventually you fix these guys but you also destroyed this case in this crease in this crease so you kind of mess things up a little you've got to do this repeatedly for every chunk you don't want to mess up previous chunks that you've done so you have to go back and fix this fold unfold fix this fold unfold but that again only you only cost n to the epsilon over log n so not too bad so another factor of two here but I'm ignoring constant factors okay that was an over log in upper bound let's do log squared upper bound for mMmmm so I need to look this up it's a little tricky we're going to use the trick of the Dragon curve essentially which is repeatedly fold in half and we're going to keep doing that until we are left with three creases which haven't yet been folded so this is my folded bundle there's you know many many many layers here and then I'm going to fold mmm on those three creases instead of going all the way to Dragon curve which would just be do em fold here first I'm going to fold this unfold fold this unfold fold this unfold so I get this pattern when I unfold this it actually turns out to be kind of nice and alternating it will be something mmm something Valley Valley Valley something about a mountain them something Valley Valley Valley and so on okay great half of my things roughly our mountains I just need to fix these valleys so I use the same trick which is I'm going to fold use red maybe fold in the middle of these mountain segments I do that I line up all the valley segments so i end up with after folding this is going to look like you almost need a computer or call this out but it that we have done implemented this it works it's going to look like this repeat I mean with many layers like this all the valleys are on top of each other these question marks are still question marks this mountain is that mountain this mountain is that mutton my mountain is your mountain I don't know so now we're going to fold mountain mountain mountain mountain on all five of these things and these two remain mountains and then unfold and what i end up with is sort of twice as good i end up with ? and then i think seven sorry not quite because this part also gets messed up so i end up with Valley Mountain ? then seven mountains then ? then some not so pretty stuff em vvvvv em ? and then this repeats until we get to the end of the string and then the end looks kind of like this so the point is I had 3 m's in a row now i have 7 ms in a row each time i do this i roughly double the number of Em's in a row by the end almost everything will be ms after I do this log n times I have pretty much all Em's and then I can finish it off and each of these steps took log n steps because I had to fold along all these things how did I fold along them not one at a time of course I fold in half and fold in half and fold in half roughly I choose the middle most red line to fold first and about log n steps later everything will be piled up so I get to use this efficiency of dragon curve things I'm messing up these creases of course but it's okay because I get a big chunk that all gets correct at once half of it gets wrong half of its correct but the half that's getting correct gets longer and longer each step doubling so only log squared steps pretty sure this is optimal but the best lower bound we can prove is log squared over log log all right finally let's fold some stuff I thought we would fold some hyperbolic paraboloids and maybe put them together into structures like this so you have your squares if you don't have squares bunch of squares up here you can start folding as many as you like we're going to put them together I will demonstrate up here so the first thing we want to do is fold along the diagonals these aren't perfect squares but it's ok so we fold along on diagonal and the other diagonal this is mainly so we know where the center is but also because we need to fold a diagonal so we kill two birds with one stone whereas we just saw how to kill in this case log n birds with one stone save log in here we're only killing two so now you've got your diagonals near square we want to fold it's kind of cool I color the edges with my chalk hands now where everything's going to be parallel to the sides of the square so first thing we're going to do is construct a square that's half is big in the centered along that point so this involves folding the bottom edge usually do this valley-fold fold the bottom edge to line up with that Center and also line up the edges now very important when you make this fold don't fold it all the way just fold the middle half between the two diagonals that's really key otherwise this will not work it's the only thing you have to be careful of don't fall all the way just in the middle okay that's one quarter do that four times so we go here line up with the center fold the middle once you've got the inner square you're going to make two more there's going to be this square at the one quarter mark and an inner square at the three quarters mark to make the outer square you fold the edge to the thing that you just folded keep keep careful note of which crease you're folding to should be from the previous square not from the current Square to fold the inner square you go all the way up to the opposite side of the first square and you can just look visually are these evenly spaced if yes you did it right if not you're in trouble probably okay if you make a few extra creases but don't try to make as few extra creases as possible so then repeat that four times and you'll get your I guess four squares if you count the outer boundary all right it's a little hard to see once you've got that those squares looks like I do one of them incorrectly got your nice concentric squares four of them now you're halfway done you flip it over and do valley folds on the other side so because you want these to be alternating Mountain Valley Mountain Valley and just fill in all the squares in between those squares and you can figure out what the reference markers are there all the mountain folds well actually every other mountain fold will be one of your references don't go to one of your new Valley folds always go to the mountain once and just check that you're always filling in halfway in between two of your Mountain folds don't forget to flip over very important and skip every other guy otherwise you will think refold existing crease the wrong way this is what it looks like after one quarter of those alright once you've done all this crazy precreasing you have concentric squares alternating mountain valley then the fun part begins this is actually literally fun not its I guess also harder but much more exciting than all that precreasing so in this case you have to fold all the creases at once and the easy way to do that is start at the outside and make the outermost ring fold it go around once or twice make sure everything is folded including that diagonal so here I've Valley folded everything including the diet outer diagonal and then you want to mountain fold the next one and get them to be against each other you're aiming for a kind of X shape which is drawn over here it's going to look like that ideally just keep collapsing square by square making sure you alternating mountain valley as I mean it should just fall into place pretty much but paper likes to miss fold a little bit so you got to fix it all here I've done two squares a little more it's easier and easier because the squares get smaller and smaller we've got three squares done so I'm like halfway to annex fourth square centers a little bit tricky try to get it also alternate so now it's nice and thin it's like repeated sync folds of a waterbomb base and it's like an X and then this is your opportunity to wreak Reese all your crease is really hard just give it a good squeeze on each of the three four legs of the starfish and now you've got your hyperbolic paraboloid to make it you want to take two midpoints here of the squares pull them apart until it's a little bit open and then give it a twist and then open up a little you've got your hyperbolic paraboloid tah-dah if we make enough of these we can assemble it into some cool shapes how many people have folded one at least one few people i will show you you can fold more we're going to assemble something might need a bunch of people this is the algorithm we use for converting a polyhedron into a bunch of hyperbolic paraboloids it's in this paper from 99 and we take each face of the polygons of the polyhedron sorry so if you have a cube you've got a bunch of squares for each square we make what we call a 4 hat which is for hyperbolic paraboloids joined in a cycle like this and you've got to be careful to join it the right way but then these the tips of the high parts that are not joined I mean two of the one tip of each of the hype ours comes together at the center then these tips are going to represent the edge of the polygon these red dots and so these two sides are the sides of one hype of two these two sides of the sides of one hype are they're going to join to an adjacent high power over here so that's the idea yeah so I've got already filled with a bunch of these already so maybe what a hat looks like so tick two of them join them along those edges we're going to use tape or staples to join them don't have a fancy lock I'm afraid and join these edges together and you get this is a 3 hat you can put on your head whatever and that would represent a triangle and then we're going to join along these two sides to an adjacent triangle or whatever so we could make a platonic solid i guess the simplest one would be a tetrahedron that has six edges so it needs 12 hype ours I was thinking we could make this solid it's never been made before should be cool it's the simplest Archimedean solid terms the number of edges the truncated tetrahedron this requires 36 parts so for ambitious we can go for it but I think we're low on time so tetrahedra might be easier bit people want to come up and start assembling all right finished taping that hold these this goes here you're gonna hold this one virtual assembly here so maybe we'll finish this Emily next class 3 00:00:08,179 --> 00:00:11,270 4 00:00:11,270 --> 00:00:13,610 5 00:00:13,610 --> 00:00:17,930 6 00:00:17,930 --> 00:00:21,620 7 00:00:21,620 --> 00:00:24,710 8 00:00:24,710 --> 00:00:28,310 9 00:00:28,310 --> 00:00:31,450 10 00:00:31,450 --> 00:00:35,720 11 00:00:35,720 --> 00:00:38,240 12 00:00:38,240 --> 00:00:40,130 13 00:00:40,130 --> 00:00:43,130 14 00:00:43,130 --> 00:00:45,800 15 00:00:45,800 --> 00:00:46,970 16 00:00:46,970 --> 00:00:50,030 17 00:00:50,030 --> 00:00:52,000 18 00:00:52,000 --> 00:00:53,840 19 00:00:53,840 --> 00:00:55,910 20 00:00:55,910 --> 00:00:59,330 21 00:00:59,330 --> 00:01:01,790 22 00:01:01,790 --> 00:01:03,860 23 00:01:03,860 --> 00:01:06,889 24 00:01:06,889 --> 00:01:08,330 25 00:01:08,330 --> 00:01:11,480 26 00:01:11,480 --> 00:01:13,580 27 00:01:13,580 --> 00:01:15,770 28 00:01:15,770 --> 00:01:17,569 29 00:01:17,569 --> 00:01:21,349 30 00:01:21,349 --> 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--> 00:07:44,060 150 00:07:44,060 --> 00:07:49,760 151 00:07:49,760 --> 00:07:51,650 152 00:07:51,650 --> 00:07:53,719 153 00:07:53,719 --> 00:07:57,409 154 00:07:57,409 --> 00:07:59,120 155 00:07:59,120 --> 00:08:07,060 156 00:08:07,060 --> 00:08:12,110 157 00:08:12,110 --> 00:08:17,659 158 00:08:17,659 --> 00:08:18,890 159 00:08:18,890 --> 00:08:21,110 160 00:08:21,110 --> 00:08:22,370 161 00:08:22,370 --> 00:08:24,380 162 00:08:24,380 --> 00:08:28,159 163 00:08:28,159 --> 00:08:30,440 164 00:08:30,440 --> 00:08:33,200 165 00:08:33,200 --> 00:08:37,190 166 00:08:37,190 --> 00:08:39,850 167 00:08:39,850 --> 00:08:42,440 168 00:08:42,440 --> 00:08:44,840 169 00:08:44,840 --> 00:08:46,790 170 00:08:46,790 --> 00:08:48,170 171 00:08:48,170 --> 00:08:51,760 172 00:08:51,760 --> 00:08:54,920 173 00:08:54,920 --> 00:08:59,180 174 00:08:59,180 --> 00:09:01,040 175 00:09:01,040 --> 00:09:03,470 176 00:09:03,470 --> 00:09:07,220 177 00:09:07,220 --> 00:09:11,269 178 00:09:11,269 --> 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--> 00:13:07,210 267 00:13:07,210 --> 00:13:13,930 268 00:13:13,930 --> 00:13:15,700 269 00:13:15,700 --> 00:13:20,200 270 00:13:20,200 --> 00:13:22,660 271 00:13:22,660 --> 00:13:25,920 272 00:13:25,920 --> 00:13:28,270 273 00:13:28,270 --> 00:13:29,980 274 00:13:29,980 --> 00:13:31,780 275 00:13:31,780 --> 00:13:34,930 276 00:13:34,930 --> 00:13:36,640 277 00:13:36,640 --> 00:13:39,190 278 00:13:39,190 --> 00:13:42,100 279 00:13:42,100 --> 00:13:43,870 280 00:13:43,870 --> 00:13:46,540 281 00:13:46,540 --> 00:13:48,370 282 00:13:48,370 --> 00:13:50,110 283 00:13:50,110 --> 00:13:52,270 284 00:13:52,270 --> 00:13:56,350 285 00:13:56,350 --> 00:13:57,640 286 00:13:57,640 --> 00:13:59,890 287 00:13:59,890 --> 00:14:04,810 288 00:14:04,810 --> 00:14:06,340 289 00:14:06,340 --> 00:14:07,900 290 00:14:07,900 --> 00:14:10,060 291 00:14:10,060 --> 00:14:12,580 292 00:14:12,580 --> 00:14:19,270 293 00:14:19,270 --> 00:14:21,700 294 00:14:21,700 --> 00:14:23,260 295 00:14:23,260 --> 00:14:26,800 296 00:14:26,800 --> 00:14:27,460 297 00:14:27,460 --> 00:14:31,210 298 00:14:31,210 --> 00:14:32,950 299 00:14:32,950 --> 00:14:35,110 300 00:14:35,110 --> 00:14:37,000 301 00:14:37,000 --> 00:14:39,660 302 00:14:39,660 --> 00:14:43,600 303 00:14:43,600 --> 00:14:45,850 304 00:14:45,850 --> 00:14:48,610 305 00:14:48,610 --> 00:14:53,470 306 00:14:53,470 --> 00:14:55,720 307 00:14:55,720 --> 00:15:00,010 308 00:15:00,010 --> 00:15:01,510 309 00:15:01,510 --> 00:15:03,640 310 00:15:03,640 --> 00:15:06,880 311 00:15:06,880 --> 00:15:08,980 312 00:15:08,980 --> 00:15:11,950 313 00:15:11,950 --> 00:15:13,180 314 00:15:13,180 --> 00:15:15,670 315 00:15:15,670 --> 00:15:18,879 316 00:15:18,879 --> 00:15:24,129 317 00:15:24,129 --> 00:15:31,059 318 00:15:31,059 --> 00:15:33,789 319 00:15:33,789 --> 00:15:36,220 320 00:15:36,220 --> 00:15:37,689 321 00:15:37,689 --> 00:15:40,599 322 00:15:40,599 --> 00:15:42,280 323 00:15:42,280 --> 00:15:43,629 324 00:15:43,629 --> 00:15:45,879 325 00:15:45,879 --> 00:15:47,829 326 00:15:47,829 --> 00:15:50,470 327 00:15:50,470 --> 00:15:51,849 328 00:15:51,849 --> 00:15:54,429 329 00:15:54,429 --> 00:15:55,960 330 00:15:55,960 --> 00:15:58,449 331 00:15:58,449 --> 00:16:00,609 332 00:16:00,609 --> 00:16:01,929 333 00:16:01,929 --> 00:16:03,729 334 00:16:03,729 --> 00:16:04,720 335 00:16:04,720 --> 00:16:06,729 336 00:16:06,729 --> 00:16:09,729 337 00:16:09,729 --> 00:16:12,460 338 00:16:12,460 --> 00:16:14,549 339 00:16:14,549 --> 00:16:17,650 340 00:16:17,650 --> 00:16:19,720 341 00:16:19,720 --> 00:16:21,909 342 00:16:21,909 --> 00:16:23,739 343 00:16:23,739 --> 00:16:25,569 344 00:16:25,569 --> 00:16:28,150 345 00:16:28,150 --> 00:16:30,999 346 00:16:30,999 --> 00:16:33,030 347 00:16:33,030 --> 00:16:35,840 348 00:16:35,840 --> 00:16:39,600 349 00:16:39,600 --> 00:16:42,180 350 00:16:42,180 --> 00:16:44,190 351 00:16:44,190 --> 00:16:47,040 352 00:16:47,040 --> 00:16:51,660 353 00:16:51,660 --> 00:16:56,520 354 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--> 00:18:20,790 384 00:18:20,790 --> 00:18:22,230 385 00:18:22,230 --> 00:18:24,660 386 00:18:24,660 --> 00:18:27,150 387 00:18:27,150 --> 00:18:30,390 388 00:18:30,390 --> 00:18:34,140 389 00:18:34,140 --> 00:18:39,900 390 00:18:39,900 --> 00:18:44,280 391 00:18:44,280 --> 00:18:51,870 392 00:18:51,870 --> 00:18:55,380 393 00:18:55,380 --> 00:18:57,180 394 00:18:57,180 --> 00:18:58,680 395 00:18:58,680 --> 00:19:00,090 396 00:19:00,090 --> 00:19:01,590 397 00:19:01,590 --> 00:19:03,750 398 00:19:03,750 --> 00:19:05,370 399 00:19:05,370 --> 00:19:06,630 400 00:19:06,630 --> 00:19:10,050 401 00:19:10,050 --> 00:19:11,790 402 00:19:11,790 --> 00:19:13,800 403 00:19:13,800 --> 00:19:19,170 404 00:19:19,170 --> 00:19:21,210 405 00:19:21,210 --> 00:19:23,400 406 00:19:23,400 --> 00:19:26,640 407 00:19:26,640 --> 00:19:28,110 408 00:19:28,110 --> 00:19:30,030 409 00:19:30,030 --> 00:19:33,840 410 00:19:33,840 --> 00:19:35,430 411 00:19:35,430 --> 00:19:39,890 412 00:19:39,890 --> 00:19:43,050 413 00:19:43,050 --> 00:19:45,270 414 00:19:45,270 --> 00:19:53,190 415 00:19:53,190 --> 00:19:54,990 416 00:19:54,990 --> 00:19:58,530 417 00:19:58,530 --> 00:20:01,080 418 00:20:01,080 --> 00:20:03,300 419 00:20:03,300 --> 00:20:06,360 420 00:20:06,360 --> 00:20:08,160 421 00:20:08,160 --> 00:20:11,700 422 00:20:11,700 --> 00:20:13,470 423 00:20:13,470 --> 00:20:16,890 424 00:20:16,890 --> 00:20:20,100 425 00:20:20,100 --> 00:20:22,440 426 00:20:22,440 --> 00:20:25,080 427 00:20:25,080 --> 00:20:27,630 428 00:20:27,630 --> 00:20:29,820 429 00:20:29,820 --> 00:20:32,250 430 00:20:32,250 --> 00:20:35,160 431 00:20:35,160 --> 00:20:38,880 432 00:20:38,880 --> 00:20:42,300 433 00:20:42,300 --> 00:20:44,210 434 00:20:44,210 --> 00:20:47,340 435 00:20:47,340 --> 00:20:49,020 436 00:20:49,020 --> 00:20:53,139 437 00:20:53,139 --> 00:20:56,469 438 00:20:56,469 --> 00:20:58,779 439 00:20:58,779 --> 00:21:00,399 440 00:21:00,399 --> 00:21:04,029 441 00:21:04,029 --> 00:21:06,839 442 00:21:06,839 --> 00:21:10,839 443 00:21:10,839 --> 00:21:13,499 444 00:21:13,499 --> 00:21:16,959 445 00:21:16,959 --> 00:21:18,549 446 00:21:18,549 --> 00:21:21,759 447 00:21:21,759 --> 00:21:23,200 448 00:21:23,200 --> 00:21:24,879 449 00:21:24,879 --> 00:21:27,430 450 00:21:27,430 --> 00:21:29,769 451 00:21:29,769 --> 00:21:31,289 452 00:21:31,289 --> 00:21:35,109 453 00:21:35,109 --> 00:21:36,430 454 00:21:36,430 --> 00:21:38,950 455 00:21:38,950 --> 00:21:41,409 456 00:21:41,409 --> 00:21:43,259 457 00:21:43,259 --> 00:21:48,549 458 00:21:48,549 --> 00:21:50,560 459 00:21:50,560 --> 00:21:52,180 460 00:21:52,180 --> 00:21:55,989 461 00:21:55,989 --> 00:21:57,310 462 00:21:57,310 --> 00:22:00,609 463 00:22:00,609 --> 00:22:04,930 464 00:22:04,930 --> 00:22:07,149 465 00:22:07,149 --> 00:22:08,379 466 00:22:08,379 --> 00:22:10,450 467 00:22:10,450 --> 00:22:11,739 468 00:22:11,739 --> 00:22:15,369 469 00:22:15,369 --> 00:22:16,690 470 00:22:16,690 --> 00:22:18,339 471 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--> 00:23:56,769 501 00:23:56,769 --> 00:24:00,690 502 00:24:00,690 --> 00:24:04,870 503 00:24:04,870 --> 00:24:09,039 504 00:24:09,039 --> 00:24:11,590 505 00:24:11,590 --> 00:24:13,330 506 00:24:13,330 --> 00:24:16,389 507 00:24:16,389 --> 00:24:19,750 508 00:24:19,750 --> 00:24:27,430 509 00:24:27,430 --> 00:24:34,950 510 00:24:34,950 --> 00:24:40,269 511 00:24:40,269 --> 00:24:41,680 512 00:24:41,680 --> 00:24:44,289 513 00:24:44,289 --> 00:24:46,149 514 00:24:46,149 --> 00:24:48,100 515 00:24:48,100 --> 00:24:50,080 516 00:24:50,080 --> 00:24:52,930 517 00:24:52,930 --> 00:24:54,820 518 00:24:54,820 --> 00:24:56,950 519 00:24:56,950 --> 00:24:59,860 520 00:24:59,860 --> 00:25:02,649 521 00:25:02,649 --> 00:25:05,529 522 00:25:05,529 --> 00:25:10,000 523 00:25:10,000 --> 00:25:12,070 524 00:25:12,070 --> 00:25:14,830 525 00:25:14,830 --> 00:25:17,409 526 00:25:17,409 --> 00:25:19,950 527 00:25:19,950 --> 00:25:24,150 528 00:25:24,150 --> 00:25:26,490 529 00:25:26,490 --> 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559 00:27:20,520 --> 00:27:24,750 560 00:27:24,750 --> 00:27:30,450 561 00:27:30,450 --> 00:27:32,610 562 00:27:32,610 --> 00:27:32,620 563 00:27:32,620 --> 00:27:34,170 564 00:27:34,170 --> 00:27:36,880 565 00:27:36,880 --> 00:27:38,050 566 00:27:38,050 --> 00:27:39,370 567 00:27:39,370 --> 00:27:40,720 568 00:27:40,720 --> 00:27:42,940 569 00:27:42,940 --> 00:27:44,860 570 00:27:44,860 --> 00:27:56,200 571 00:27:56,200 --> 00:28:01,330 572 00:28:01,330 --> 00:28:03,190 573 00:28:03,190 --> 00:28:05,610 574 00:28:05,610 --> 00:28:08,710 575 00:28:08,710 --> 00:28:15,520 576 00:28:15,520 --> 00:28:19,210 577 00:28:19,210 --> 00:28:20,980 578 00:28:20,980 --> 00:28:23,590 579 00:28:23,590 --> 00:28:25,390 580 00:28:25,390 --> 00:28:29,260 581 00:28:29,260 --> 00:28:31,510 582 00:28:31,510 --> 00:28:34,110 583 00:28:34,110 --> 00:28:38,280 584 00:28:38,280 --> 00:28:41,860 585 00:28:41,860 --> 00:28:44,260 586 00:28:44,260 --> 00:28:47,590 587 00:28:47,590 --> 00:28:49,900 588 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--> 00:30:13,210 618 00:30:13,210 --> 00:30:17,320 619 00:30:17,320 --> 00:30:19,090 620 00:30:19,090 --> 00:30:22,120 621 00:30:22,120 --> 00:30:25,150 622 00:30:25,150 --> 00:30:28,870 623 00:30:28,870 --> 00:30:30,790 624 00:30:30,790 --> 00:30:32,200 625 00:30:32,200 --> 00:30:33,580 626 00:30:33,580 --> 00:30:36,880 627 00:30:36,880 --> 00:30:38,380 628 00:30:38,380 --> 00:30:40,510 629 00:30:40,510 --> 00:30:42,250 630 00:30:42,250 --> 00:30:43,660 631 00:30:43,660 --> 00:30:45,790 632 00:30:45,790 --> 00:30:47,620 633 00:30:47,620 --> 00:30:50,590 634 00:30:50,590 --> 00:30:52,300 635 00:30:52,300 --> 00:30:53,980 636 00:30:53,980 --> 00:30:55,840 637 00:30:55,840 --> 00:30:58,210 638 00:30:58,210 --> 00:31:01,750 639 00:31:01,750 --> 00:31:04,300 640 00:31:04,300 --> 00:31:07,810 641 00:31:07,810 --> 00:31:09,940 642 00:31:09,940 --> 00:31:13,050 643 00:31:13,050 --> 00:31:15,790 644 00:31:15,790 --> 00:31:18,850 645 00:31:18,850 --> 00:31:21,100 646 00:31:21,100 --> 00:31:23,920 647 00:31:23,920 --> 00:31:25,660 648 00:31:25,660 --> 00:31:29,260 649 00:31:29,260 --> 00:31:31,720 650 00:31:31,720 --> 00:31:34,240 651 00:31:34,240 --> 00:31:36,010 652 00:31:36,010 --> 00:31:37,210 653 00:31:37,210 --> 00:31:38,980 654 00:31:38,980 --> 00:31:40,300 655 00:31:40,300 --> 00:31:42,100 656 00:31:42,100 --> 00:31:43,840 657 00:31:43,840 --> 00:31:46,360 658 00:31:46,360 --> 00:31:48,820 659 00:31:48,820 --> 00:31:51,730 660 00:31:51,730 --> 00:31:55,120 661 00:31:55,120 --> 00:31:55,130 662 00:31:55,130 --> 00:31:57,210 663 00:31:57,210 --> 00:32:00,810 664 00:32:00,810 --> 00:32:04,789 665 00:32:04,789 --> 00:32:11,789 666 00:32:11,789 --> 00:32:14,640 667 00:32:14,640 --> 00:32:16,500 668 00:32:16,500 --> 00:32:18,649 669 00:32:18,649 --> 00:32:21,990 670 00:32:21,990 --> 00:32:25,649 671 00:32:25,649 --> 00:32:27,960 672 00:32:27,960 --> 00:32:29,669 673 00:32:29,669 --> 00:32:31,919 674 00:32:31,919 --> 00:32:35,909 675 00:32:35,909 --> 00:32:37,320 676 00:32:37,320 --> 00:32:39,720 677 00:32:39,720 --> 00:32:42,120 678 00:32:42,120 --> 00:32:44,970 679 00:32:44,970 --> 00:32:48,000 680 00:32:48,000 --> 00:32:49,409 681 00:32:49,409 --> 00:32:54,169 682 00:32:54,169 --> 00:32:57,480 683 00:32:57,480 --> 00:33:00,390 684 00:33:00,390 --> 00:33:04,770 685 00:33:04,770 --> 00:33:07,549 686 00:33:07,549 --> 00:33:13,020 687 00:33:13,020 --> 00:33:16,440 688 00:33:16,440 --> 00:33:21,419 689 00:33:21,419 --> 00:33:25,500 690 00:33:25,500 --> 00:33:28,200 691 00:33:28,200 --> 00:33:32,640 692 00:33:32,640 --> 00:33:35,430 693 00:33:35,430 --> 00:33:36,720 694 00:33:36,720 --> 00:33:40,140 695 00:33:40,140 --> 00:33:42,299 696 00:33:42,299 --> 00:33:43,799 697 00:33:43,799 --> 00:33:45,810 698 00:33:45,810 --> 00:33:47,549 699 00:33:47,549 --> 00:33:49,080 700 00:33:49,080 --> 00:33:51,090 701 00:33:51,090 --> 00:33:53,460 702 00:33:53,460 --> 00:33:54,810 703 00:33:54,810 --> 00:34:00,029 704 00:34:00,029 --> 00:34:02,640 705 00:34:02,640 --> 00:34:04,590 706 00:34:04,590 --> 00:34:08,899 707 00:34:08,899 --> 00:34:08,909 708 00:34:08,909 --> 00:34:10,139 709 00:34:10,139 --> 00:34:12,480 710 00:34:12,480 --> 00:34:14,210 711 00:34:14,210 --> 00:34:23,089 712 00:34:23,089 --> 00:34:28,740 713 00:34:28,740 --> 00:34:33,240 714 00:34:33,240 --> 00:34:34,529 715 00:34:34,529 --> 00:34:37,289 716 00:34:37,289 --> 00:34:39,750 717 00:34:39,750 --> 00:34:41,129 718 00:34:41,129 --> 00:34:42,779 719 00:34:42,779 --> 00:34:44,309 720 00:34:44,309 --> 00:34:48,480 721 00:34:48,480 --> 00:34:49,919 722 00:34:49,919 --> 00:34:52,619 723 00:34:52,619 --> 00:34:55,349 724 00:34:55,349 --> 00:34:57,059 725 00:34:57,059 --> 00:34:58,799 726 00:34:58,799 --> 00:35:00,450 727 00:35:00,450 --> 00:35:02,519 728 00:35:02,519 --> 00:35:05,339 729 00:35:05,339 --> 00:35:07,260 730 00:35:07,260 --> 00:35:09,120 731 00:35:09,120 --> 00:35:10,799 732 00:35:10,799 --> 00:35:13,589 733 00:35:13,589 --> 00:35:16,140 734 00:35:16,140 --> 00:35:18,240 735 00:35:18,240 --> 00:35:20,130 736 00:35:20,130 --> 00:35:21,720 737 00:35:21,720 --> 00:35:24,089 738 00:35:24,089 --> 00:35:26,490 739 00:35:26,490 --> 00:35:28,589 740 00:35:28,589 --> 00:35:33,480 741 00:35:33,480 --> 00:35:35,519 742 00:35:35,519 --> 00:35:38,010 743 00:35:38,010 --> 00:35:41,190 744 00:35:41,190 --> 00:35:45,990 745 00:35:45,990 --> 00:35:47,549 746 00:35:47,549 --> 00:35:52,019 747 00:35:52,019 --> 00:35:53,760 748 00:35:53,760 --> 00:36:00,210 749 00:36:00,210 --> 00:36:02,460 750 00:36:02,460 --> 00:36:05,339 751 00:36:05,339 --> 00:36:10,529 752 00:36:10,529 --> 00:36:15,269 753 00:36:15,269 --> 00:36:16,470 754 00:36:16,470 --> 00:36:17,849 755 00:36:17,849 --> 00:36:19,529 756 00:36:19,529 --> 00:36:19,539 757 00:36:19,539 --> 00:36:24,470 758 00:36:24,470 --> 00:36:27,539 759 00:36:27,539 --> 00:36:31,140 760 00:36:31,140 --> 00:36:33,420 761 00:36:33,420 --> 00:36:36,240 762 00:36:36,240 --> 00:36:37,950 763 00:36:37,950 --> 00:36:40,980 764 00:36:40,980 --> 00:36:42,599 765 00:36:42,599 --> 00:36:44,490 766 00:36:44,490 --> 00:36:46,230 767 00:36:46,230 --> 00:36:48,960 768 00:36:48,960 --> 00:36:52,019 769 00:36:52,019 --> 00:36:53,940 770 00:36:53,940 --> 00:36:55,829 771 00:36:55,829 --> 00:36:57,980 772 00:36:57,980 --> 00:37:00,809 773 00:37:00,809 --> 00:37:03,480 774 00:37:03,480 --> 00:37:06,359 775 00:37:06,359 --> 00:37:08,250 776 00:37:08,250 --> 00:37:10,200 777 00:37:10,200 --> 00:37:11,880 778 00:37:11,880 --> 00:37:14,309 779 00:37:14,309 --> 00:37:20,670 780 00:37:20,670 --> 00:37:25,740 781 00:37:25,740 --> 00:37:25,750 782 00:37:25,750 --> 00:37:36,570 783 00:37:36,570 --> 00:37:39,180 784 00:37:39,180 --> 00:37:41,790 785 00:37:41,790 --> 00:37:43,440 786 00:37:43,440 --> 00:37:46,500 787 00:37:46,500 --> 00:37:48,810 788 00:37:48,810 --> 00:37:51,570 789 00:37:51,570 --> 00:37:55,590 790 00:37:55,590 --> 00:37:58,410 791 00:37:58,410 --> 00:38:01,290 792 00:38:01,290 --> 00:38:03,660 793 00:38:03,660 --> 00:38:06,090 794 00:38:06,090 --> 00:38:09,720 795 00:38:09,720 --> 00:38:11,100 796 00:38:11,100 --> 00:38:13,340 797 00:38:13,340 --> 00:38:17,460 798 00:38:17,460 --> 00:38:19,410 799 00:38:19,410 --> 00:38:23,300 800 00:38:23,300 --> 00:38:26,340 801 00:38:26,340 --> 00:38:29,880 802 00:38:29,880 --> 00:38:52,800 803 00:38:52,800 --> 00:38:54,960 804 00:38:54,960 --> 00:38:57,390 805 00:38:57,390 --> 00:39:02,800 806 00:39:02,800 --> 00:39:06,340 807 00:39:06,340 --> 00:39:09,970 808 00:39:09,970 --> 00:39:12,940 809 00:39:12,940 --> 00:39:14,320 810 00:39:14,320 --> 00:39:15,550 811 00:39:15,550 --> 00:39:17,980 812 00:39:17,980 --> 00:39:20,980 813 00:39:20,980 --> 00:39:22,330 814 00:39:22,330 --> 00:39:26,020 815 00:39:26,020 --> 00:39:27,520 816 00:39:27,520 --> 00:39:30,310 817 00:39:30,310 --> 00:39:32,350 818 00:39:32,350 --> 00:39:36,310 819 00:39:36,310 --> 00:39:37,540 820 00:39:37,540 --> 00:39:40,750 821 00:39:40,750 --> 00:39:51,100 822 00:39:51,100 --> 00:39:53,080 823 00:39:53,080 --> 00:39:57,340 824 00:39:57,340 --> 00:39:59,770 825 00:39:59,770 --> 00:40:49,810 826 00:40:49,810 --> 00:40:52,370 827 00:40:52,370 --> 00:40:54,410 828 00:40:54,410 --> 00:40:56,510 829 00:40:56,510 --> 00:40:58,460 830 00:40:58,460 --> 00:41:04,520 831 00:41:04,520 --> 00:41:05,960 832 00:41:05,960 --> 00:41:08,030 833 00:41:08,030 --> 00:41:10,310 834 00:41:10,310 --> 00:41:12,650 835 00:41:12,650 --> 00:41:19,760 836 00:41:19,760 --> 00:41:21,770 837 00:41:21,770 --> 00:41:24,710 838 00:41:24,710 --> 00:41:26,510 839 00:41:26,510 --> 00:41:28,640 840 00:41:28,640 --> 00:41:30,110 841 00:41:30,110 --> 00:41:32,360 842 00:41:32,360 --> 00:41:34,850 843 00:41:34,850 --> 00:41:37,640 844 00:41:37,640 --> 00:41:41,330 845 00:41:41,330 --> 00:41:44,750 846 00:41:44,750 --> 00:41:46,310 847 00:41:46,310 --> 00:41:51,080 848 00:41:51,080 --> 00:41:53,750 849 00:41:53,750 --> 00:41:56,000 850 00:41:56,000 --> 00:42:02,230 851 00:42:02,230 --> 00:42:04,120 852 00:42:04,120 --> 00:42:06,609 853 00:42:06,609 --> 00:42:08,440 854 00:42:08,440 --> 00:42:15,820 855 00:42:15,820 --> 00:42:18,280 856 00:42:18,280 --> 00:42:20,620 857 00:42:20,620 --> 00:42:23,050 858 00:42:23,050 --> 00:42:27,400 859 00:42:27,400 --> 00:42:29,020 860 00:42:29,020 --> 00:42:31,450 861 00:42:31,450 --> 00:42:34,240 862 00:42:34,240 --> 00:42:39,340 863 00:42:39,340 --> 00:42:42,640 864 00:42:42,640 --> 00:42:45,460 865 00:42:45,460 --> 00:42:49,720 866 00:42:49,720 --> 00:42:51,520 867 00:42:51,520 --> 00:42:55,870 868 00:42:55,870 --> 00:43:01,359 869 00:43:01,359 --> 00:43:03,220 870 00:43:03,220 --> 00:43:08,349 871 00:43:08,349 --> 00:43:14,080 872 00:43:14,080 --> 00:43:20,440 873 00:43:20,440 --> 00:43:21,820 874 00:43:21,820 --> 00:43:24,940 875 00:43:24,940 --> 00:43:27,730 876 00:43:27,730 --> 00:43:31,300 877 00:43:31,300 --> 00:43:34,690 878 00:43:34,690 --> 00:43:37,270 879 00:43:37,270 --> 00:43:38,560 880 00:43:38,560 --> 00:43:41,380 881 00:43:41,380 --> 00:43:43,930 882 00:43:43,930 --> 00:43:47,349 883 00:43:47,349 --> 00:43:49,510 884 00:43:49,510 --> 00:43:52,540 885 00:43:52,540 --> 00:43:54,550 886 00:43:54,550 --> 00:43:58,150 887 00:43:58,150 --> 00:43:59,560 888 00:43:59,560 --> 00:44:01,420 889 00:44:01,420 --> 00:44:03,580 890 00:44:03,580 --> 00:44:08,380 891 00:44:08,380 --> 00:44:11,859 892 00:44:11,859 --> 00:44:13,840 893 00:44:13,840 --> 00:44:16,360 894 00:44:16,360 --> 00:44:23,320 895 00:44:23,320 --> 00:44:25,190 896 00:44:25,190 --> 00:44:34,870 897 00:44:34,870 --> 00:44:39,800 898 00:44:39,800 --> 00:44:42,320 899 00:44:42,320 --> 00:44:44,360 900 00:44:44,360 --> 00:44:48,140 901 00:44:48,140 --> 00:44:55,250 902 00:44:55,250 --> 00:44:58,150 903 00:44:58,150 --> 00:45:01,100 904 00:45:01,100 --> 00:45:02,300 905 00:45:02,300 --> 00:45:04,670 906 00:45:04,670 --> 00:45:12,140 907 00:45:12,140 --> 00:45:13,610 908 00:45:13,610 --> 00:45:17,270 909 00:45:17,270 --> 00:45:21,260 910 00:45:21,260 --> 00:45:23,030 911 00:45:23,030 --> 00:45:24,530 912 00:45:24,530 --> 00:45:26,180 913 00:45:26,180 --> 00:45:27,290 914 00:45:27,290 --> 00:45:33,140 915 00:45:33,140 --> 00:45:36,050 916 00:45:36,050 --> 00:45:38,840 917 00:45:38,840 --> 00:45:42,860 918 00:45:42,860 --> 00:45:45,740 919 00:45:45,740 --> 00:45:50,360 920 00:45:50,360 --> 00:45:54,580 921 00:45:54,580 --> 00:45:58,010 922 00:45:58,010 --> 00:46:01,190