1 00:00:03,280 --> 00:00:04,630 today we're talking about the local behavior of a crease pattern so you take some crease pattern for some flat folding we're thinking about flat foldability is a foldability question i give you a crease pattern like this i want to know does it fold flat like this one does and we're studying what happens locally right around a single vertex so i didn't mention sort of graph terminology it's probably useful to mention that these corners wherever all the edges come together those are vertices these triangles or in general some regions divided by the creases we call faces all right so uh if we just sort of imagine cutting a little disk around that vertex and seeing how it behaves we get a nice circular piece of paper with some crease pattern and we want to understand when those things fold flat and when they don't sometimes they do sometimes they don't this is kind of like a one-dimensional folding problem so let me write some stuff about that so we get we call single vertex crease patterns so something like this is the example i have made something like that that should be flat foldable so we think of uh having a disc of paper it doesn't really matter but it's it's going to be easier to reason about and think about a disc of paper really we just mean some small region around that vertex there's obviously one vertex in the crease pattern let's say we have n creases emanating from one vertex uh this pattern is going to be defined by some sequence of angles in general theta 1 up to theta n if there are increases there will be n angles between them say in clockwise order and let's see all right when we fold this thing flat like in this picture we take this disc we fold it along all of those creases what i'd like to focus on is what happens to the boundary of the paper so there's this outer circle the boundary is a circle and locally if you look at one of these creases it's like folding the circle in half onto itself and when you make all of these creases if you're successful you end up folding the circle onto a portion of the circle so if you look at this boundary which is a little hard to see why don't i trace it it's gonna do something like this okay i'm gonna so this is what the 3d thing looks like blown up a little bit and i changed the angles a little bit but this is what a flat folding would look like and i'm really just focusing on how the circle maps around this circle so one of the a flat folding of a disc also folds a circle onto a circle so we get we have a flat folding in one of these single vertex crease patterns we get a folding of a circle onto a circle this is nice because circles are one-dimensional and last class we talked about folding one-dimensional line segments a circle is just a little bit more complicated it's a different topology we can use a lot of the same mindset at least as we did for folding line segments onto the line in particular you can see yeah this is kind of circular but you could also imagine unrolling it a little bit and making it straight something like this so that it lies on a straight line if you just unroll that circle so you also get a folding of a circle in some sense onto a line by unrolling now at this point i want to mention a slight issue here i mean this for this unrolling to work if you're taking a circle like this in particular you must make at least one fold for this to be possible because right now if i have if i haven't folded anything i just have a big circle i can't unroll a circle onto a line i haven't collapsed it yet here i've made it small enough that it only occupies a portion of the entire circle and so i can unroll it so this is true as long as i make at least one fold okay technically i need to assume another thing which is that this thing came from a flat piece of paper uh i said there are these angles and what i intended to mean is that the sum of those angles is 360 degrees as it is in a flat piece of paper but i'm not always going to require this assumption it's it's what we care about for thinking about a crease pattern of a flat piece of paper and we look at each vertex this will be true but when we reason about these single vertex crease patterns it's really useful to think about sort of in the middle this could get smaller so for example you have a pattern like this let's say i just take these two creases in the bottom and i fold a crimp because i know a crimp isn't a kind of a good thing to do so now i have what's called a cone of paper the sum of the angles there's two angles here this one and this one each one looks like 270 no not 270 135 in total it's 270. so the total angle there is now less than 360. this is what we call a convex cone so if you relax this constraint to allow less than or equal to 360 then we get a convex cone and today i only want to talk about flat paper and convex cones we need this for the induction essentially for proving things about the flat case in the textbook if you look at this chapter on single vertex crease patterns they als we also prove or study the case where the sum of the angles is greater than 360 which you could never make from flat paper but hey it's fun to think about it's a natural generalization and you can characterize all the things that we do here you can generalize to that situation it's just a little bit harder all right so what does it take to fold a circle onto a line in this situation of a convex cone where we make at least one fold our folding is going to lie along portion of the circle we can unroll it a little bit get something on lying on a straight line it's kind of like folding a line segment onto a straight line that was flat folding of one-dimensional pieces of paper but there's a lot of differences but they are similar but a lot of the things we proved last time do not hold in the case of circular paper so i have a bunch of them here so one problem you may recall before if we had a line segment and we put an arbitrary crease pattern then we could just assign a mountain valley assignment alternating mountain valley and that would always fold flat no collisions okay in this world of having a circular piece of paper that's no longer true for example if you have this kind of crease pattern on a circle or in the disk it would look like this those two creases you can't fold that thing flat it's not going to work okay we'll see exactly what what you need to forbid for that to make sense another example is something like this so if you're all right so if i try to draw this in the the line what's happening is i have a short segment and then i have a long segment and so the problem is those ends don't meet but even when the ends meet like in this picture imagine these are vertically aligned if i do this alternating mountain valley pattern mountain valley assignment uh i can't join these ends and make a circle because that would collide with everything so this thing actually does have a flat folding starting from a circle and i'm not going to try to unmap it to a circle it does fold but not like that not with that alternating mountain valley pattern there are other annoying things let me tell you sort of one more on the proof side last time we had a lemma that said if your crease pattern is mingling which still is a meaningful notion here then you have a crimp or an end fold in that case here of course we don't have any end folds here we also do not get this implication mingling does not imply the existence of a crimp and the where the proof breaks down is if you have so if you remember our parentheses from last time we had this kind of pattern and we said well you keep going and either on the left side you have a beginning open paren or on the right side you have a closing round parenthesis in either case you've got an end fold now this can actually happen in a circle and so you have nothing you can do so you could have this pattern there would be no crimp and probably in that case you would not be flat-foldable but in particular this implication fails so we have to do more work and but we're going to get the same kind of result that in linear time we can tell whether one of these things folds flat so that's good news so uh before last time telling whether a crease pattern folded flat was trivial the answer was always yes now it's not so trivial so let's start with characterizing crease patterns that fold flat then we will go to mountain valley patterns that fold flat so be great if i could just say that but i mean here of course single vertex so i give you one of these sequence of angles let's say it sums to 360 or less when is it going to be flat foldable and the answer is very simple let me write it down this thing is going to be flat foldable if and only if the sum of the odd angles is equal to the sum of the even angles and yeah i'm implicitly assuming and requiring here that n is even that's something we'll prove so n is actually even and minus one is going to be odd and if you happen to have started with a flat piece of paper uh for the sum of all of these things is 360 then of course if you have them evenly divided this sum will be 180 degrees but if you started with the convex cone that it might it'll be smaller but for this reason a lot of people or some people call this the 180 degree property or the pi property if you'd like to think in radians instead of degrees and we can do a little example here this guy this crease pattern uh has a 90 degree angle here and it has a 90 degree angle down here so the sum of those two is 180 those are alternating angles different parity classes this is 45 this is the complement 135 and that was also something 180 of course one does it only if the other does so it's flat foldable we're golden something like this is bad because i mean it's an even number but you the odd angles do not equal the sum of the even angles one's clearly bigger okay and this is a general property this is also called kawasaki's theorem or kawasaki jost jostan's theorem they proved it in 1989 so this is pretty much the beginning of origami mathematics so let's prove it there's two things to prove one is that uh any flat foldable crease pattern has this property which is pretty easy and the other which is a little bit harder is that if you have this property you really are flat foldable that's the full characterization so let's start with the easy direction if you're flat foldable you must have that sum okay so it's all about thinking what happens at a crease so we're thinking about the boundary of this piece of paper thinking about the circle and what's happening on the circle is you travel uh say counterclockwise for some amount of time that would be first angle then you make a fold it's either a mountain or valley at this point we don't care either way you turn around maybe you turn around this way or you turn around this way but in terms of your travel along a circle so we went theta one counterclockwise now we're going to go theta two clockwise then we're gonna go theta three counterclockwise and so on we keep alternating back and forth it's much easier to think draw on a line so we go to the right theta 1 then we go to the left theta 2 then we go to the right theta 3 and i don't know about the layering here i'm just drawing it in terms of horizontal travel the y-coordinate means nothing okay in order for this thing to be a flat folding to be a valid folding of a circle of paper these guys have to have the same x coordinate so i don't know i mean the y coordinates they also have to work out but we're not thinking about the y coordinates we're just thinking about x travel we've got to get back to where we started if we're folding a circle that's it so how much trout how much total travel do we do in a sine sense how much do we go right well we go theta 1 to the right then we go theta 2 to the left which is like going negative theta two to the right and then we go theta three to the right then we go negative theta four to the right and you add up this alternating summation that must equal zero okay the other thing uh that i should say is that n has to be even um this is maybe obvious at this point but we're supposed to alternate every time we hit a crease we have to change direction and so in the end it's important that after theta 4 we change direction to visit theta 1. if they were aligned like if theta 4 went over here and then theta 5 went like that and there was an odd number of creases this would be bad even though they're lined up there's no crease here you just went straight from theta 5 to theta 1. you're supposed to change direction every crease so also n is even to alternate directions question all right all right good clear this is pretty easy once you realize the angles correspond to x coordinates in this linear space so why is this enough in order to show that this property is enough for flat foldability we need to actually worry about the stacking order of these edges we need to worry about the y coordinates we need to make sure that they can avoid collision with some mountain valley assignment the good news is we're free to use whatever mountains and valleys we want we can stack things however we want we just need to find some valid state and our intuition from the one-dimensional case was we should try alternating mountains and valleys that seemed like a good thing it avoided collisions for an open piece of paper for a line segment of course we know that's not enough here somewhere i drew this example so it's alternating in the middle and then that last crease is a problem but that's all we need to fix turns out it's not so hard so we have a nice circle what i'd like to do is cut that circle so here i sort of cut it and thought about it as a line segment and then i wanted to rejoin okay i'm going to be careful about where i cut it okay i want to cut at an extreme left or extreme right it doesn't matter you can see here the cut ended up being in the middle that's bad so how do i fix it what does extreme mean how do i tell given a circle it's a little hard to say what extreme means because it's circular but if i take a picture like this i've already drawn it with nice x coordinates i don't know about the y coordinates yet say oh that's bad my cut point ended up being non-extreme well this is extreme so cut here instead okay so what that means is redraw this picture with this being the cut point so maybe this is the beginning so we go like that again and now we wrap around so now we have this segment and now we have the segment and lo and behold it remained extreme and this is always true you can do this sort of cut and rejoin the ends that were messed up before and you'll preserve the x coordinates x coordinates aren't changing when you do this kind of transformation so there's a clear leftmost and there's a clear rightmost you pick either one you cut there instead and now we have this nice property that it's easy to join the ends because it's that as far left as you can go there can't be anything that penetrates because that would be farther left all right let me write down some words corresponding to that so once we cut at this extreme crease you can fold this 1d segment just like we used to be able to so we can fold it flat using the accordion fold alternating mountain and valley we know that works and then all we need to show is that the two ends can join those ends are the two copies of the cut crease that we did in this first step uh we need two things one is that they are aligned and that's where we need this condition that the sum of the odd angles equals the sum of the even angles which is the same as saying the alternating sum is equal to zero that tells us that whatever we do the x coordinates are lined up at the end so they're aligned by the by the assumption that we made here and you can join without crossing so this is a statement about y-coordinates and we know it's okay because it's at the left extreme so by this choice of the proper cut there could be other things that come right to the boundary here but that's considered okay that's not crossing that's just touching nothing can go farther left because it was the left extreme that's the end of the proof any questions about that it's kawasaki's theorem i'll mention just for fun this sort of classic if you allow one of these a non-convex cone where you have more than 360 degrees of material this statement is not true and what can there's one other situation which can happen which is that the alternating sum of angles is not zero but it's uh plus or minus 360 degrees which is fun and you can read it you can see examples of that in the book all right so obviously if you want to tell whether a crease pattern or a single vertex is flat foldable you just compute this thing in linear time and you know yes or no uh so crease patterns are pretty easy just have one extra condition what about mountain valley assignments last time we looked at mountain valley assignments and we showed that crimps and end folds were always enough to fold any mountain valley assignment that's out there that will continue to be true here except now we don't have end folds now it's crimps all the way what does it mean so that is the flat foldable mountain valley patterns we're going to prove that crimps are enough but before we get there i need a bunch of sort of warm-up facts about valid mountain valley patterns the first thing is just counting mountains and valleys so even if you play with a simple example like this one you see it has three mountains and one valley or three valleys and one mountain in fact if if you have a degree four vertex with four creases emanating from that point those are your only choices you could try for example making it all valleys and it i mean you can't do it uh it's hard to demonstrate uh you could try making it two mountains and two valleys so maybe i try to do mountain mountain here and then somehow it's gonna be valley valley there it's no good okay you can that what's happening here is that the number of mountains and the number of valleys must differ by exactly two it doesn't matter who's bigger so the difference could be plus two or minus two because you can always flip over and negate that difference swapping mountains for valleys but they always have to differ by exactly two why is that well this is uh yeah we're again going to think of a picture like this we have horizontal travel according to the theta eyes we're also going to think about the vertical picture if this thing is flat foldable so i'm assuming it's flat foldable then this must happen this is called maekawa's theorem by the way it's proved by june mayakawa in 1986 a little earlier also proved by jacques christine around the same time back before we were good at internet and communication and whatnot and there were little pockets of mathematical origamis and they found each other basically in 89 when there was the 89 as the first origami science math and education conference started bringing these people together the last one was this summer right so why is this true i don't know flat folding looks like something okay it has no crossings it has horizontal travel we understand it ends up back where it started there's also some notion of layers and vertical travel which is a little tricky to think about but the one thing that's easy to think about i mean this forms essentially a polygon it's like a really squash polygon and these vertical segments are actually really just points okay imagine that's just where you turn around so it's like a polygon just stretched or squashed down onto a line we know some things about polygons okay it's an n vertex polygon you know that like the sum of the interior angles is whatever it is i always get confused with that formula so i don't like to think about it one thing we an easier way to think about that same statement is just think about how much turning the polygon does so you start here you turn right 180 degrees you turn left 180 degrees you turn left 180 right 180 right 180 right 180. how much is it in total oh gosh do i have to add no it's so simple it's 360. right just if you think about you're going around in a circle any polygon not just flat but anything in in two dimensions uh the total amount of turning you do is 360. or if you count backwards it's negative 360. okay so this is the key something you should all know about polygons if you look at the sum of the turn angles how much turning you do at each vertex that will always be plus or minus 360 in any polygon this is equivalent to the sum of the interior angles being whatever pi times n minus 2 which i don't remember but this is much easier to remember and it's useful because we can map mountains and valleys to left and right turns as i said every time this was a valley it was a left turn by 180. every time it was a mountain it was a right turn by 180. valleys mountains right turns so i'm going to think of that as negative 180. now of course technically it could be the other way around could be this is negative this is positive but i already have a plus and minus here so it's symmetric so if we sum up the turn angles which we know is supposed to be plus or minus 360 some of the turn angles is going to be we're going to have 180 for each valley so 180 times the number of valleys and we're going to have negative 180 for each mountain and so the 180s factor out and this thing is supposed to equal 360 plus or minus 360. how could that be well number of valleys minus number of mountains should be either two for plus 360 or negative two for minus 360. so that proves the theorem and that's why in a degree four vertex one of them has to be three the other one has to be one there's no other option because that total number is four it's the only way to get the difference to be plus or minus two so that's kind of nice but it's not enough unfortunately if i gave you a mountain valley pattern that satisfies this condition it still might not be flat foldable this may be no surprise it's not like we had such a simple test for the one dimensional case should be easier we had to give an algorithm we said repeatedly crimp and enfold if you get stuck the answer is no if you don't get stuck you folded it great i love this eraser increases entropy all right let me tell you one case that is easy to think about and because it it's nice it reduces directly to the one dimensional case and that's what i call the generic case generic is this very convenient term we use in mathematics it's actually really tricky to define so i'd like to not define it if i can but maybe i should yeah all right uh the simple version which is not quite enough is to say that all of the angles in the crease pattern are different okay here for example two of them are the same there's two 90 degree angles this is not generic it's just easier to draw but in general you i mean you imagine you just draw some random thing none of those lengths are going to be the same yeah the alternating sum is 0 but that's it and that's what i the generic case is that is the only thing that holds true about these angles you can take the alternating sum that equals zero if you take some alternating sum of some subset of the angles that will not equal zero if you take a random example this is going to be true okay just bear with me suppose never any two angles are the same then look at the globally smallest crease it's the globally smallest theta say it's theta i so the picture is theta i theta i minus 1 theta i plus 1. and what do we know if it's globally smallest and none of the values are equal then we know that these two are bigger no big surprise uh think about what happens if this is small these are bigger and you try to make these both valleys or both mountains that's of course bad this is one of the situations we had in the one dimensional case i mean really this is a circle but i can think just locally about what's happening here so in this situation one of these must be a mountain and the other must be a i don't know which order they are but i know they're different and then i know i can apply one of these crimps incidentally the word crimp jason kuh is asking about like why do you call it crimp and not pleat probably i should call it pleat but it's crimp in my mind it's been crimped in my mind since 1998 when we wrote this the 1d paper before i knew the word pleat and so that's my excuse so pleat or crimp take your pick crimp is usually mini pleats together that's crimping cool so crimps we kind of like why do we like crimps we proved in the one-dimensional case last time that if you make a crimp and your original thing was flat-foldable the new thing will be flat-foldable good news is that is still true i told you all these things were no longer true in the circular case it's still true in the circular case uh should i tell you let me just remind you the same proof works remind you what the proof looked like we had a crimp and then and we're hoping to fold that first we know at some point these creases get folded but there may be other junk in here maybe other junk in here we'd like to get rid of that junk and so what we did is move this junk up to here which was always safe we moved this junk down to here and that was still a folded state a flat folded state okay i remember all this stuff therefore we could have folded the crimp first and then we have a flat folding in this situation where this paper is effectively glued together which means it was safe to fold this crimp and glue these together and never touch them again so that is still true in the circular case that proof didn't really assume anything about the paper would even work for like tree-shaped paper or something weird so if we can find a crimp which we can in the generic case then you just make a crimp make a crimp repeat if you ever get stuck you know that the original thing was not flat foldable because the thing you have is not flat foldable why if there's no crimp in the generic case that means you have two valleys at around the smallest angle and that's clearly bad um so i'm not writing a lot of details here is that clear if i have an angle that's surrounded by strictly larger angles i know there must be one mount in one valley then i can safely make the crimp and repeat the trouble comes in when these angles are equal we never had to think about angles being equal in the one-dimensional case it didn't matter much we just said oh you know it's safe to make a crimp as long as this was greater than or equal to this and this was greater than or equal to this that is still true it's safe to make a crimp but how do we know that there is a crimp how do we know that there's a mountain followed by a valley we don't i mean if they're equal this this would be all right that's two valleys is valid if these three angles are equal but i claim still somewhere there's got to be a crimp there's no longer a notion of the globally smallest angle because some of them are equal we've got to find that crimp but it's out there so we just need to prove that it exists so so to do that we're going to generalize this theorem and kawa just in that number of mountains minus number of valleys is plus or minus two that's true and it's a statement about the sum of all of the or you know all of the angles all the creases what i'd like is something a little bit more localized and in particular we're going to think about when i have a whole bunch of angles that are equal how many mountains and valleys can there be in there so what i call local counts so while everything i've said so far is pretty classic this result was proved in 2001 by tom hall so much more recent over a decade later so let's think about k equal angles and i want that to be a maximal sequence of equal angles so the ones right after and right before are different and furthermore i'm going to assume that they're bigger strictly larger angles okay so something like that a bunch of equal angles k of them bigger angles on either side this almost always exists um i should mention this it's so i'm going to say something about this situation but the situation should exist because i take the smallest angle out there and then i see how many friends it has that are all equal if it's the smallest angle or it's one of the smallest angles then the ones are surrounding it are going to be bigger unless all the angles are equal so there's one case which we'll worry about later it's very easy all the angles are equal in fact i could tell you how to worry about that if all the angles are equal then you know everything is crimpable as long as we can find a switch between mountain and valley somewhere and because the number of mountains and valleys is plus or minus two there's got to be so they're almost in equal balance there's got to be a valley it can't be all valleys can't be all mountains so somewhere there's a transition between mountain and valley so you get a crimp out of that okay but otherwise if they're not all equal let's think about k angle equal angles surrounded by two larger guys forgot how technical this is all right then i want to look at the number of mountains and the number of valleys among these uh k plus one creases so within that range it's going to be zero if k is odd it's going to be plus or minus one if k is even okay it's not plus or minus two intuitively plus or minus two is when you go all the way around here we're just looking at a one you know segment portion of the entire circle remember this is actually a circle somewhere and when you this corresponds to in the odd case you're going to be going in the same direction afterwards and the even case you're going to have turned around but not a full circle so it's going to be plus or minus 1. there's a lot of ways to prove this one way is to think about convex cones so if k is even and you fold this thing see one two three four five six equal angles and then there's the two longer guys they may not be the same length but that's what you get in the even case so what i'd like to do is just in order to just reduce the things that we've already thought about i mean you could talk about turn angles again or you could just say hey i know mayakawa's theorem so as long as i can make this into a circle which i can do by adding that stuff now i have a nice circular it's a flat folded state there's no crossings here presumably because this did something i don't actually know that it's zigzagging it might do other stuff but it's not going to cross what i just added so the number of mountains minus number of valleys in total here must be plus or minus 2. i only added one crease this one so that's going to make it either plus or minus 1 or plus or minus 3. if you think about it a little bit it has to be plus or minus 1 and hopefully that's what i wrote here yes because i mean really we're thinking about turn angles again i'm afraid can't totally reduce it we just turned around here and we're about to finish the circle which would make it plus or minus 2 before we finish the circle is plus or minus 1. k is odd it's again this might go like that whatever but because it's an odd number these guys are going in the same direction and so the total turn angle here must be zero and one way to see that is to extend it into a full polygon you know that it's plus or minus two for this whole thing and i added two guys in the same direction if i remove them it's going to go down to plus or minus zero that was a bit hand wavy but do you buy that yeah plus or minus zero is zero yeah you went from two down to zero because you removed two identical guys so let me let me tie this all together because there's so many little cases so i want to know what mountain what mountain valley patterns are flat foldable okay single vertex and the patterns okay i claim there is a crimp if it's flat foldable there has to be a crimp that you can find to do remember crimp was where this length was less than or equal to this length and less than or equal to this length and one of these was m one of them was v that was the definition of a crimp claims such a thing always exists the proof is two parts if all the angles are equal all the angles are equal we know that these conditions are always satisfied because everything is equal so we just need to find an m followed by a v or v followed by an m and by maekalo's theorem we know it's not all m's or all v's so somewhere there has to be a transition from m to v's so just use that lemma theorem count property otherwise they're not all equal then i can apply this local counts result take the smallest angle take one instance of the smallest angle there could be many of them find all of its neighboring friends that are equal then i know that the next angles before and after have to be strictly bigger then i know that the number of mountains and valleys in here if i look at the difference and it's zero that means the number of mountains equals the number of valleys therefore there's at least one of each therefore there's a transition for mountain valley in there there's a transition from mountain to valley like right here that's a valid crimp this guy's bigger this guy's equal this would also be a fine transition because everybody's equal this would also be a fine transition any any transition from mountain valley in here i've got a crimp the even case is a little i've got to think a little bit more k is even k plus 1 which is the number of creases here is odd and the difference between mountains and valleys among those k plus one creases is plus or minus one think about it for a second that means there's again at least one mountain and at least one valley and uh because cave to be even has to be at least two you've gotta have at least three things so there's gonna be at least one of one of one and two of the other in fact and i just need one mountain in one valley somewhere in there and then i know there has to be a transition from mountain to valley otherwise use local counts and in either case we get either a mountain to valley transition or a valley to mountain transition and i get it in such a way that these these inequalities and the lengths hold so in other words that it's a crimp therefore any flat foldable single vertex mountain valley pattern has a crimp make it and as i was arguing here just like last time that crimp will still after you do the crimp you'll still be flat foldable so repeat cook until done is the i like to put it question what about the trivial case of just floating trivial case of folding a circle in half all right that's a good question it's kind of a technicality uh one version would be to say that's zero vertices that doesn't count we're thinking about one vertex and as soon as you think about it as one vertex which is fine then you actually have two uh creases there so it's two creases and hopefully this works uh oh i see you're saying it's not a crimp that's a good point it is flat foldable and yet there's no crimp damn it all right sorry my college right yeah so it's i made a slight mistake here i said if all the angles are equal then meikawa says there's at least one of each that's not true when n equals two then there's two of one and zero of the other okay as long as you have more than two creases then this is true that's enough so you have to handle this as a special case uh that's true this this this is an important special case to remember because this will be the final picture after you do a sequence of crimps you this is the good case uh if your thing is flat foldable you will always end up with that that's a case where everything is excuse me the same smallest it's it's another example where all the angles are equal yeah so that's why this is the situation it's either maekawa and you get a crimp here because n is greater than 2 and here i should say if n is greater than 2 then this is true of course the n equals 2 case it looks like that it better be both valleys or both mountains otherwise it's not going to be flat foldable i forgot about that i should add in the notes but it's in the textbook right that's true it won't actually look like this it's going to at the end of the algorithm it's going to be a cone the two angles will be equal because of kawasaki's theorem and they should still be both mountains or both valleys but it's going to be we can see it right here so i apply my crimp and now i have a cone and there's two angles there's one here which has been fused together there's one here which is original and they're equal and it's two mountains or two valleys if i turn it upside down and then it finishes so that's the algorithm in action two steps first you find this crimp here this is actually the globally smallest angle it's surrounded by bigger angles so it's really easy then i have a cone with two equal angles which is what has to happen at the end and then i i'm done good this is what always happens oh is this the only way to flat fold these things no there are other ah yes one of the two um i think it depends what you're counting if you're counting mountain valley assignments i mean if you just want to know can every mountain valley assignment be folded by crimping then the answer is yes that's what we proved but if you want to get every possible folded state crimps are not going to be enough the reason they're not enough is because we're using this thing said hey here's some hypothetical folded state we can rip it out and make the crimp have been done but then you're there's a folded state you're not able to reach by crimping i mean we need to work that into an actual example where crimping forbids you from reaching some folded state but i'm pretty sure one exists because of this yeah maybe some kind of spiraling thing right what we're missing here is getting all the possible layer stacking orders so we're just trying to match the mountain valley assignment we're not going to match a target layer ordering because we simplified the layer orders in order to make crimps possible so crimps won't get you some some of the folded states but one thing you can play with over here that has been played with by tom hall is if i give you a crease pattern how many flat foldable mountain valley assignments does it have okay we have this nice linear time algorithm to tell you whether a particular one is doable how many are there and there is if you work through all of the things i've shown you in a little bit more detail you can recover this was finding a crimp you could actually look at what are all the crimps that are possible and actually count how many different ways there are to fold how many crimps you can do and then given those crimps how many crimps you can do after that and in same linear time kind of algorithm you can figure out how many different mountain valley patterns are flat foldable that takes a little bit more care i will tell you the extremes how small could the number of flat foldable mountain valley assignments be for a given crease pattern well do you have any guesses when when do i have the least choice of where to make crimps yeah uh for general n sorry but yeah three three angles are i guess two angles i really can't do any i don't have any choice but for general and how should i disperse the theta eyes in order to make there be very few possible crimps when there are only two transitions between mountains and valleys all the mountains are together all the valleys are together that's right if i was asking the question about mountains and valleys but i was asking a question about theta eyes so i want to count how many mountain valley patterns there are mountain valley assignments that are flat foldable for a given crease pattern all the angles are consecutive increasing yeah that pretty much works in fact what i need is the generic case where all the angles are different and they never become equal by folding which is what generic means so then there was only really one crimp i could do which is the globally smallest angle so actually i still have two choices i could do mountain then valley or valley than mountain but i only have two choices for that crimp then i will have two choices for the next crimp two choices for the next crimp in general i get two to the n possible mountain valley assignments for the generic case okay what about what's the opposite where i get as much choice as possible yeah two to the n no uh two to the n over two thank you right they're two to the end conceivable mountain valley assignments can't be that big uh every time i do a crimp i eat two creases not one that's what i forgot good still pretty big though it's like square root of all the possible things you could imagine are indeed feasible it went from 2 to the n to 2n over 2. when would i get the most choice in crimping all angles are equal that's the opposite extreme and in this case a little messier when all the angles are equal the only property you need and you can see that from the proof the only property you need is that the number of mountains minus number of valleys is plus or minus two all the a and the n is even as long as you have that uh we showed here you're going to have an alternation from mountain valley that's a valid crimp because all the angles are equal and you can keep going so it's just how do you disperse that many how do i disperse n over two minus one mountains among n different positions that's what this represents and positions how do i pick n over two minus one of them to be mountains or could be n over two minus one of them are valleys and that's this factor of two so it could have more mountains or more valleys and then you somehow place those among those and this is if you don't know this notation that's just what it means if you want to know with using other notation it's like this n over 2 minus 1 factorial over 2 plus 1 factorial if you don't know factorials i'll tell you about them later cool so that's kind of the end of the single vertex situation um yeah i think i'll mention one interesting open question here which i would love to explore maybe in our problem session um which looks like it'll be mondays at five i think um this is one vertex and we went through all this work to solve single vertex situation and it's interesting as we'll see to think about locally how each vertex behaves but you would think if you have a crease pattern with two vertices shouldn't be that much harder so here we have linear time how quickly can you tell whether a two vertex crease pattern is flat foldable as far as i know no one has looked at that problem surely we can do it in like quadratic time but maybe even linear time i think can't be that hard i think in general if i have a small number of vertices say k vertices much smaller than n creases is there the algorithms people is there fixed parameter retractable algorithm and k or can i get something even getting something like n to the some function of k would be progress i think this should be doable but ideally we get a running time that's exponential in k and linear in n that would be my hope why do i say it has to be exponential because in general if i give you crease pattern with n vertices lots of vertices this problem is np complete which is there's not going to be a polynomial time algorithm for it nothing good we will prove that next wednesday so i'll hold off on that a little bit but for two two vertices how hard could it be all right i want to talk about one related topic and then we will go to origami design and do a little bit on the tree method but before we get there i want to talk about local foldability which is a cool topic people tend to forget about it i really like it i think it would make a cool project also it's a nice algorithm it goes back to vernon hayes 1996 so it's also right at the beginning of origami mathematics and it's this idea all right i give you a crease pattern now arbitrarily many vertices and i ask you if i ask you does it fold flat that's np-complete intractable same paper but what if i ask you just to give me a mountain valley assignment that might fold flat well that's also mp complete but if if you actually want it to fall flat this is really the same problem but if i ask you give me a mountain valley assignment so that if i checked every vertex according to this algorithm at least every vertex folds flat that would seem nice you know definitely i have to find a mountain valley assignment that satisfies these conditions that as i do successive crimps every vertex if i cut out the vertex separately it would fold flat this is the notion of local foldability and there's a linear time algorithm to tell you to give you a mountain valley assignment that ought to work in that each vertex it works it still may not work globally for some other reason but it's pretty good i think this would be actually pretty practical i think in a lot of real world origami settings when you're doing flat folding anyway locally valid locally foldable is going to be enough to be globally foldable that's a guess i don't know if it's true consistent mountain valley assignment if there is one so that each vertex locally folds flat let me uh try to concoct a small example that's relevant here gotta think yeah i think that works all right here we go i think this is in the textbook i remember drawing it in the past three years ago or whatever all right so here's a crease pattern on a square whatever yes i should have this saying thank you uh so this satisfies kawasaki's theorem that was the hard part because these angles sum to 180 and it's symmetric all around okay this is uh these angles are the smallest they're 60 degrees this is an equilateral triangle so we have this not quite the generic case but we have a the smallest angle is surrounded by two larger angles therefore one of these is a mountain the other is a valley that means that these two creases have to have different assignments i'm going to write a not equal sign one of these is not in the other's value they can't be the same also these two cannot be the same also these two cannot be the same that's not possible because i've got three it has to alternate you can't alternate three times you can only alternate an even number of times okay two two of these mountains one's valley and then you've got a problem okay so this thing is not flat foldable and this algorithm will tell you that because it'll say hey i can't even find a mountain valley assignment that could possibly fold each vertex flat so this is a nice algorithm at least it will detect annoying situations like that so in order to solve this we're going to use and i'm just going to sketch how this algorithm works we're going to use this characterization and this idea that we can really find all possible mountain valley assignments just by trying all the possible crimp sequences now there are exponentially many so it's not like i'm actually going to try them all but i need to explore that space of candidate crimps and see what happens all right uh so the idea is kind of crazy the beginning of the algorithm is fold each vertex flat somehow i don't care how just pick a crimp do it pick a crimp do it okay separately for each vertex they're not going to be compatible and don't look at the mountain valley assignment you get but look at the crimping sequence you get so here let's do a little example a non-trivial example so here we have the peace sign and these two angles are equal they're smallest i could crimp this angle first or i could crimp this angle first i have a choice so i'll draw both of them if i crimp this angle first i know these guys are paired up in the sense that they must be not equal ones mountain ones valley after i do that crimp just like in the real example i had you know this angle will equal that angle and these two guys must have equal assignments they must be both mountain or both valley in this situation these two guys are not equal but if i crimp the this pair first and these two guys must be equal equal now in the sense of being mountain or valley okay so there's still you know exponentially many possibilities in how to do this but just pick one pick one of these ways of pairing up you're going to pair up each of the increases into n over two pairs and they're going to have some not equals and equal signs okay so now if you imagine that the general picture like here i for example i might get in this pattern i would be forced to get not equals and these guys paired up and in general i want to look at these kinds of cycles you know if i come if i come into a vertex here's a vertex it's paired up with somebody so if i come in here i can go out somewhere and i come to this vertex it's paired with somebody so i'm gonna just i can keep wandering around and in general there will be a bunch of these paths that you can follow what could the paths do they either close up on themselves maybe things are paired up in such a way that you make a return trip or it could be some other path let me draw a dotted path it could come in here and maybe it gets paired up with this guy maybe it goes off to infinity it reaches the boundary of the paper those are the two possibilities you get paths which go off to the edge off to infinity on either end or you could get cycles the cycles are the problem because whenever i have a cycle i have a parity constraint for example when they're all not equals uh the length of the cycle must be even if they were yeah what's the general statement it's like the parity of the cycle which is whether it's even or odd should be equal to the parity of the number of not equal signs something like that i'm going to cheat here great i just said parody problems it's something like that it's either the parity of the cycle should equal the parity of the number of equals or the variety of the number not equals i think number of equals anyway one of the two and you can just check i mean you're forced if i say okay let's make this mountain then this is either equals or not equals it'll tell me whether this is mountain or valley just walk around the cycle either you get a contradiction or you don't if you get a contradiction we have a problem how do we how could we possibly fix the problem well when we made you look at each of these crimps i mean in fact you could look at each of these vertices separately but you think about one of these crimps and say well could i have done it another way sometimes there are crimps that have other equal choices maybe they're a bunch of equal angles and i could have done a different pairing what happens when i try a different pairing well instead of this being in one cycle and let's say this being in another cycle if i pair these guys up instead i'll end up merging those two cycles into one bigger thing could be a path or a cycle and the algorithm says just keep doing those cycle merges and if you get stuck your thing is not locally foldable that's the hard part to prove otherwise you will find a one of these patterns that you can actually resolve mountains and valleys all the way through so let's say so start with sum folding say local folding whatever i'm going to say with some pairing of creases at vertices and merge two pads or cycles whenever possible and when i say merge i mean whenever you have what are the possible things you could possibly merge when you have at some point during the algorithm a bunch of equal angles you have a choice which of these you crimp obviously the mountain valley sign is not fixed you can crimp any of them you picked one of them and it's sort of merging whatever this thing is attached to to whatever this thing is attached to if there's something else that's in that's disconnected from that thing i want you to instead merge two of them that combines two different connected components two paths or cycles merging means i decrease the total number of pads or cycles i combine two into one whenever that's possible do it you can prove that if you have a parity problem in the merge thing you had to have had a parity problem originally and merging can only fix parity problems that's the claim that's what i will not prove once you know that and it doesn't matter in what order or how you choose to merge you just merge as much as possible and either the resulting thing is okay or not and accordingly you will tell whether this thing is locally foldable sorry i want to move on to other things but i think this would be a fun thing to actually implement it's an easy algorithm and it's i think a pretty good test for problems like this to prevent flat foldability all right so let's move on from foldability to origami design so a bit of a big transition and we're gonna talk about origami design a lot more next class also but uh just started off today and the particular algorithm for origami design i want to talk about is called tree method this is probably the oldest algorithm for origami design uh in that it's been around and people have been thinking about it and developing it for many years uh through this period called the bug wars and people are trying to design more and more complicated insects it's like well i can make an insect with six legs oh yeah well i can make you know a spider with eight legs oh yeah well i can make an insect that has you know a beetle it has wings and horns and there's thorns on the horns and all these crazy things during that time there are a lot of people thinking about how do i make more and more complicated especially uh sort of more more limbs in my creatures more and very precise uh arrangements of those limbs let's say and that is what the tree method deals with and is really formalized by robert lang who published a paper in 96 describing it as a as a sort of complete algorithm and it's still not known for sure that that algorithm always works but uh that's what we've we've been working on me and marty and rob lang for the last four years or so and soon we will publish that thing and prove that this thing always works but i'm going to describe to the algorithm without the proof that it works and what it does tell you its goal from a mathematical perspective so it's interested in practical origami design so we're going to start from a square piece of paper it would also work for triangular pieces of paper or anything convex but squares are what people usually care about rectangles are sometimes it's used for rectangles also okay the idea is i give to you a stick figure so that's formally it's a tree a graph without any cycles it's a metric tree meaning that i put lengths on the edges i know this length this edge length is maybe twice as long as this one so i really draw it with edge lengths in mind then what i want you to do is find some folding of a piece of paper i should really be looking at what i'm trying to match uh so here here maybe this goes down such that i want to find one some folding of a square paper in fact the smallest square possible so that when i project like this vertically the projection of that uh flat folding is exactly that metric tree okay and this is called a uniaxial this thing is called a uniaxial base let me tell you a little bit why it's called any axial base we're thinking about what are called origami bases these are like the beginning of origami models and most classic origami models like more than 60 years ago start from one of these bases you've got water bomb base in the top left preliminary base fish base bird base windmill base and frog base i remember them all i have a little example here this is the water bomb base so it's just very simple crease pattern and you know why this is useful is it gives you sort of four flaps of paper to work with you know maybe you make one of them uh the head and the other two wings and the back one a tail if you're making a crane it doesn't actually starts not from the space but from the other one but same idea if you're folding a crane one of these would be head other tail and two wings this is great if you're making a four-flap animal and if you think about its projection and it's easier to think about in this other picture the projection of this thing is a four-limbed star right you can see it's a plus sign all of them are the same length and so this is actually something you can get out of the tree method you just give that as your input you will get this 3d thing and this crease pattern as your uh origami as the output let me show you the program in action uh so robert lang implemented this thing it's called tree maker it's freely available open source all that good stuff and uh yeah good and i'm not an expert at using it so bear with me but if we wanted to say i would like that star okay now i i drew it obviously not with all the lengths equal but it's ignoring the lengths that i drew and they're actually specified here so all the lengths here are supposed to be one and then i say okay optimize and then make a crease pattern and then show me the crease pattern there it is exactly the crease pattern i made although actually i can see from the mountain valley assignment because this is not really this is not flat origami it's it made it this way so it's flat of course the projection is the same still for limbs okay and so you know stash is for valleys and solid lines for mountains but you can make anything you want so let's say we want to make uh i don't know a lizard or something so the blue lines are the tree uh so here's can i do this yeah there we go uh so here i have a forearm head four leg another forearm body segment tail and two hind legs maybe i want to make that so i say optimize and then make a crease pattern and then boom you fold that it will have exactly that projection and assuming it did a reasonable job at computation this will be the smaller the best way to fold this thing and it's the smallest square that matches exactly that shape you get the best scale factor between the size of your piece of paper and the target shape but actually doing that optimization the first step i did is mp complete so it's not going to do it perfectly we'll prove that wednesday but it the heuristics are pretty good it finds a local minimum and often it finds a pretty good one and sometimes you can coax it to find better ones but yeah it's not perfect but hey it's mp complete so you can't do it this actually shows you what it would look like in x-ray view and then you can say oh that was nice but let's where's my yeah that was good but maybe i really wanted the head segment to be shorter uh you know length 0.5 and then i wanted the tail to be really long and then you can optimize that and find a crease pattern and it'll complain because it's having trouble oh dear demo effect i should have tried this example before it's not oh gosh it's one of these annoying ones i should say i should add some feature like maybe strain split something add a little bit maybe do that and yeah there we go uh i cheated and i'll explain how i cheated last time but if sometimes the particular method fails but you can fix it by adding in another tiny limb off the edge somewhere and of course you can then get rid of that at the end when you're folding it only makes the problem slightly harder and it'll still find a folding but we'll talk about that next time and i'm way out of time so we will stop there 2 00:00:04,630 --> 00:00:08,950 today we're talking about the local behavior of a crease pattern so you take some crease pattern for some flat folding we're thinking about flat foldability is a foldability question i give you a crease pattern like this i want to know does it fold flat like this one does and we're studying what happens locally right around a single vertex so i didn't mention sort of graph terminology it's probably useful to mention that these corners wherever all the edges come together those are vertices these triangles or in general some regions divided by the creases we call faces all right so uh if we just sort of imagine cutting a little disk around that vertex and seeing how it behaves we get a nice circular piece of paper with some crease pattern and we want to understand when those things fold flat and when they don't sometimes they do sometimes they don't this is kind of like a one-dimensional folding problem so let me write some stuff about that so we get we call single vertex crease patterns so something like this is the example i have made something like that that should be flat foldable so we think of uh having a disc of paper it doesn't really matter but it's it's going to be easier to reason about and think about a disc of paper really we just mean some small region around that vertex there's obviously one vertex in the crease pattern let's say we have n creases emanating from one vertex uh this pattern is going to be defined by some sequence of angles in general theta 1 up to theta n if there are increases there will be n angles between them say in clockwise order and let's see all right when we fold this thing flat like in this picture we take this disc we fold it along all of those creases what i'd like to focus on is what happens to the boundary of the paper so there's this outer circle the boundary is a circle and locally if you look at one of these creases it's like folding the circle in half onto itself and when you make all of these creases if you're successful you end up folding the circle onto a portion of the circle so if you look at this boundary which is a little hard to see why don't i trace it it's gonna do something like this okay i'm gonna so this is what the 3d thing looks like blown up a little bit and i changed the angles a little bit but this is what a flat folding would look like and i'm really just focusing on how the circle maps around this circle so one of the a flat folding of a disc also folds a circle onto a circle so we get we have a flat folding in one of these single vertex crease patterns we get a folding of a circle onto a circle this is nice because circles are one-dimensional and last class we talked about folding one-dimensional line segments a circle is just a little bit more complicated it's a different topology we can use a lot of the same mindset at least as we did for folding line segments onto the line in particular you can see yeah this is kind of circular but you could also imagine unrolling it a little bit and making it straight something like this so that it lies on a straight line if you just unroll that circle so you also get a folding of a circle in some sense onto a line by unrolling now at this point i want to mention a slight issue here i mean this for this unrolling to work if you're taking a circle like this in particular you must make at least one fold for this to be possible because right now if i have if i haven't folded anything i just have a big circle i can't unroll a circle onto a line i haven't collapsed it yet here i've made it small enough that it only occupies a portion of the entire circle and so i can unroll it so this is true as long as i make at least one fold okay technically i need to assume another thing which is that this thing came from a flat piece of paper uh i said there are these angles and what i intended to mean is that the sum of those angles is 360 degrees as it is in a flat piece of paper but i'm not always going to require this assumption it's it's what we care about for thinking about a crease pattern of a flat piece of paper and we look at each vertex this will be true but when we reason about these single vertex crease patterns it's really useful to think about sort of in the middle this could get smaller so for example you have a pattern like this let's say i just take these two creases in the bottom and i fold a crimp because i know a crimp isn't a kind of a good thing to do so now i have what's called a cone of paper the sum of the angles there's two angles here this one and this one each one looks like 270 no not 270 135 in total it's 270. so the total angle there is now less than 360. this is what we call a convex cone so if you relax this constraint to allow less than or equal to 360 then we get a convex cone and today i only want to talk about flat paper and convex cones we need this for the induction essentially for proving things about the flat case in the textbook if you look at this chapter on single vertex crease patterns they als we also prove or study the case where the sum of the angles is greater than 360 which you could never make from flat paper but hey it's fun to think about it's a natural generalization and you can characterize all the things that we do here you can generalize to that situation it's just a little bit harder all right so what does it take to fold a circle onto a line in this situation of a convex cone where we make at least one fold our folding is going to lie along portion of the circle we can unroll it a little bit get something on lying on a straight line it's kind of like folding a line segment onto a straight line that was flat folding of one-dimensional pieces of paper but there's a lot of differences but they are similar but a lot of the things we proved last time do not hold in the case of circular paper so i have a bunch of them here so one problem you may recall before if we had a line segment and we put an arbitrary crease pattern then we could just assign a mountain valley assignment alternating mountain valley and that would always fold flat no collisions okay in this world of having a circular piece of paper that's no longer true for example if you have this kind of crease pattern on a circle or in the disk it would look like this those two creases you can't fold that thing flat it's not going to work okay we'll see exactly what what you need to forbid for that to make sense another example is something like this so if you're all right so if i try to draw this in the the line what's happening is i have a short segment and then i have a long segment and so the problem is those ends don't meet but even when the ends meet like in this picture imagine these are vertically aligned if i do this alternating mountain valley pattern mountain valley assignment uh i can't join these ends and make a circle because that would collide with everything so this thing actually does have a flat folding starting from a circle and i'm not going to try to unmap it to a circle it does fold but not like that not with that alternating mountain valley pattern there are other annoying things let me tell you sort of one more on the proof side last time we had a lemma that said if your crease pattern is mingling which still is a meaningful notion here then you have a crimp or an end fold in that case here of course we don't have any end folds here we also do not get this implication mingling does not imply the existence of a crimp and the where the proof breaks down is if you have so if you remember our parentheses from last time we had this kind of pattern and we said well you keep going and either on the left side you have a beginning open paren or on the right side you have a closing round parenthesis in either case you've got an end fold now this can actually happen in a circle and so you have nothing you can do so you could have this pattern there would be no crimp and probably in that case you would not be flat-foldable but in particular this implication fails so we have to do more work and but we're going to get the same kind of result that in linear time we can tell whether one of these things folds flat so that's good news so uh before last time telling whether a crease pattern folded flat was trivial the answer was always yes now it's not so trivial so let's start with characterizing crease patterns that fold flat then we will go to mountain valley patterns that fold flat so be great if i could just say that but i mean here of course single vertex so i give you one of these sequence of angles let's say it sums to 360 or less when is it going to be flat foldable and the answer is very simple let me write it down this thing is going to be flat foldable if and only if the sum of the odd angles is equal to the sum of the even angles and yeah i'm implicitly assuming and requiring here that n is even that's something we'll prove so n is actually even and minus one is going to be odd and if you happen to have started with a flat piece of paper uh for the sum of all of these things is 360 then of course if you have them evenly divided this sum will be 180 degrees but if you started with the convex cone that it might it'll be smaller but for this reason a lot of people or some people call this the 180 degree property or the pi property if you'd like to think in radians instead of degrees and we can do a little example here this guy this crease pattern uh has a 90 degree angle here and it has a 90 degree angle down here so the sum of those two is 180 those are alternating angles different parity classes this is 45 this is the complement 135 and that was also something 180 of course one does it only if the other does so it's flat foldable we're golden something like this is bad because i mean it's an even number but you the odd angles do not equal the sum of the even angles one's clearly bigger okay and this is a general property this is also called kawasaki's theorem or kawasaki jost jostan's theorem they proved it in 1989 so this is pretty much the beginning of origami mathematics so let's prove it there's two things to prove one is that uh any flat foldable crease pattern has this property which is pretty easy and the other which is a little bit harder is that if you have this property you really are flat foldable that's the full characterization so let's start with the easy direction if you're flat foldable you must have that sum okay so it's all about thinking what happens at a crease so we're thinking about the boundary of this piece of paper thinking about the circle and what's happening on the circle is you travel uh say counterclockwise for some amount of time that would be first angle then you make a fold it's either a mountain or valley at this point we don't care either way you turn around maybe you turn around this way or you turn around this way but in terms of your travel along a circle so we went theta one counterclockwise now we're going to go theta two clockwise then we're gonna go theta three counterclockwise and so on we keep alternating back and forth it's much easier to think draw on a line so we go to the right theta 1 then we go to the left theta 2 then we go to the right theta 3 and i don't know about the layering here i'm just drawing it in terms of horizontal travel the y-coordinate means nothing okay in order for this thing to be a flat folding to be a valid folding of a circle of paper these guys have to have the same x coordinate so i don't know i mean the y coordinates they also have to work out but we're not thinking about the y coordinates we're just thinking about x travel we've got to get back to where we started if we're folding a circle that's it so how much trout how much total travel do we do in a sine sense how much do we go right well we go theta 1 to the right then we go theta 2 to the left which is like going negative theta two to the right and then we go theta three to the right then we go negative theta four to the right and you add up this alternating summation that must equal zero okay the other thing uh that i should say is that n has to be even um this is maybe obvious at this point but we're supposed to alternate every time we hit a crease we have to change direction and so in the end it's important that after theta 4 we change direction to visit theta 1. if they were aligned like if theta 4 went over here and then theta 5 went like that and there was an odd number of creases this would be bad even though they're lined up there's no crease here you just went straight from theta 5 to theta 1. you're supposed to change direction every crease so also n is even to alternate directions question all right all right good clear this is pretty easy once you realize the angles correspond to x coordinates in this linear space so why is this enough in order to show that this property is enough for flat foldability we need to actually worry about the stacking order of these edges we need to worry about the y coordinates we need to make sure that they can avoid collision with some mountain valley assignment the good news is we're free to use whatever mountains and valleys we want we can stack things however we want we just need to find some valid state and our intuition from the one-dimensional case was we should try alternating mountains and valleys that seemed like a good thing it avoided collisions for an open piece of paper for a line segment of course we know that's not enough here somewhere i drew this example so it's alternating in the middle and then that last crease is a problem but that's all we need to fix turns out it's not so hard so we have a nice circle what i'd like to do is cut that circle so here i sort of cut it and thought about it as a line segment and then i wanted to rejoin okay i'm going to be careful about where i cut it okay i want to cut at an extreme left or extreme right it doesn't matter you can see here the cut ended up being in the middle that's bad so how do i fix it what does extreme mean how do i tell given a circle it's a little hard to say what extreme means because it's circular but if i take a picture like this i've already drawn it with nice x coordinates i don't know about the y coordinates yet say oh that's bad my cut point ended up being non-extreme well this is extreme so cut here instead okay so what that means is redraw this picture with this being the cut point so maybe this is the beginning so we go like that again and now we wrap around so now we have this segment and now we have the segment and lo and behold it remained extreme and this is always true you can do this sort of cut and rejoin the ends that were messed up before and you'll preserve the x coordinates x coordinates aren't changing when you do this kind of transformation so there's a clear leftmost and there's a clear rightmost you pick either one you cut there instead and now we have this nice property that it's easy to join the ends because it's that as far left as you can go there can't be anything that penetrates because that would be farther left all right let me write down some words corresponding to that so once we cut at this extreme crease you can fold this 1d segment just like we used to be able to so we can fold it flat using the accordion fold alternating mountain and valley we know that works and then all we need to show is that the two ends can join those ends are the two copies of the cut crease that we did in this first step uh we need two things one is that they are aligned and that's where we need this condition that the sum of the odd angles equals the sum of the even angles which is the same as saying the alternating sum is equal to zero that tells us that whatever we do the x coordinates are lined up at the end so they're aligned by the by the assumption that we made here and you can join without crossing so this is a statement about y-coordinates and we know it's okay because it's at the left extreme so by this choice of the proper cut there could be other things that come right to the boundary here but that's considered okay that's not crossing that's just touching nothing can go farther left because it was the left extreme that's the end of the proof any questions about that it's kawasaki's theorem i'll mention just for fun this sort of classic if you allow one of these a non-convex cone where you have more than 360 degrees of material this statement is not true and what can there's one other situation which can happen which is that the alternating sum of angles is not zero but it's uh plus or minus 360 degrees which is fun and you can read it you can see examples of that in the book all right so obviously if you want to tell whether a crease pattern or a single vertex is flat foldable you just compute this thing in linear time and you know yes or no uh so crease patterns are pretty easy just have one extra condition what about mountain valley assignments last time we looked at mountain valley assignments and we showed that crimps and end folds were always enough to fold any mountain valley assignment that's out there that will continue to be true here except now we don't have end folds now it's crimps all the way what does it mean so that is the flat foldable mountain valley patterns we're going to prove that crimps are enough but before we get there i need a bunch of sort of warm-up facts about valid mountain valley patterns the first thing is just counting mountains and valleys so even if you play with a simple example like this one you see it has three mountains and one valley or three valleys and one mountain in fact if if you have a degree four vertex with four creases emanating from that point those are your only choices you could try for example making it all valleys and it i mean you can't do it uh it's hard to demonstrate uh you could try making it two mountains and two valleys so maybe i try to do mountain mountain here and then somehow it's gonna be valley valley there it's no good okay you can that what's happening here is that the number of mountains and the number of valleys must differ by exactly two it doesn't matter who's bigger so the difference could be plus two or minus two because you can always flip over and negate that difference swapping mountains for valleys but they always have to differ by exactly two why is that well this is uh yeah we're again going to think of a picture like this we have horizontal travel according to the theta eyes we're also going to think about the vertical picture if this thing is flat foldable so i'm assuming it's flat foldable then this must happen this is called maekawa's theorem by the way it's proved by june mayakawa in 1986 a little earlier also proved by jacques christine around the same time back before we were good at internet and communication and whatnot and there were little pockets of mathematical origamis and they found each other basically in 89 when there was the 89 as the first origami science math and education conference started bringing these people together the last one was this summer right so why is this true i don't know flat folding looks like something okay it has no crossings it has horizontal travel we understand it ends up back where it started there's also some notion of layers and vertical travel which is a little tricky to think about but the one thing that's easy to think about i mean this forms essentially a polygon it's like a really squash polygon and these vertical segments are actually really just points okay imagine that's just where you turn around so it's like a polygon just stretched or squashed down onto a line we know some things about polygons okay it's an n vertex polygon you know that like the sum of the interior angles is whatever it is i always get confused with that formula so i don't like to think about it one thing we an easier way to think about that same statement is just think about how much turning the polygon does so you start here you turn right 180 degrees you turn left 180 degrees you turn left 180 right 180 right 180 right 180. how much is it in total oh gosh do i have to add no it's so simple it's 360. right just if you think about you're going around in a circle any polygon not just flat but anything in in two dimensions uh the total amount of turning you do is 360. or if you count backwards it's negative 360. okay so this is the key something you should all know about polygons if you look at the sum of the turn angles how much turning you do at each vertex that will always be plus or minus 360 in any polygon this is equivalent to the sum of the interior angles being whatever pi times n minus 2 which i don't remember but this is much easier to remember and it's useful because we can map mountains and valleys to left and right turns as i said every time this was a valley it was a left turn by 180. every time it was a mountain it was a right turn by 180. valleys mountains right turns so i'm going to think of that as negative 180. now of course technically it could be the other way around could be this is negative this is positive but i already have a plus and minus here so it's symmetric so if we sum up the turn angles which we know is supposed to be plus or minus 360 some of the turn angles is going to be we're going to have 180 for each valley so 180 times the number of valleys and we're going to have negative 180 for each mountain and so the 180s factor out and this thing is supposed to equal 360 plus or minus 360. how could that be well number of valleys minus number of mountains should be either two for plus 360 or negative two for minus 360. so that proves the theorem and that's why in a degree four vertex one of them has to be three the other one has to be one there's no other option because that total number is four it's the only way to get the difference to be plus or minus two so that's kind of nice but it's not enough unfortunately if i gave you a mountain valley pattern that satisfies this condition it still might not be flat foldable this may be no surprise it's not like we had such a simple test for the one dimensional case should be easier we had to give an algorithm we said repeatedly crimp and enfold if you get stuck the answer is no if you don't get stuck you folded it great i love this eraser increases entropy all right let me tell you one case that is easy to think about and because it it's nice it reduces directly to the one dimensional case and that's what i call the generic case generic is this very convenient term we use in mathematics it's actually really tricky to define so i'd like to not define it if i can but maybe i should yeah all right uh the simple version which is not quite enough is to say that all of the angles in the crease pattern are different okay here for example two of them are the same there's two 90 degree angles this is not generic it's just easier to draw but in general you i mean you imagine you just draw some random thing none of those lengths are going to be the same yeah the alternating sum is 0 but that's it and that's what i the generic case is that is the only thing that holds true about these angles you can take the alternating sum that equals zero if you take some alternating sum of some subset of the angles that will not equal zero if you take a random example this is going to be true okay just bear with me suppose never any two angles are the same then look at the globally smallest crease it's the globally smallest theta say it's theta i so the picture is theta i theta i minus 1 theta i plus 1. and what do we know if it's globally smallest and none of the values are equal then we know that these two are bigger no big surprise uh think about what happens if this is small these are bigger and you try to make these both valleys or both mountains that's of course bad this is one of the situations we had in the one dimensional case i mean really this is a circle but i can think just locally about what's happening here so in this situation one of these must be a mountain and the other must be a i don't know which order they are but i know they're different and then i know i can apply one of these crimps incidentally the word crimp jason kuh is asking about like why do you call it crimp and not pleat probably i should call it pleat but it's crimp in my mind it's been crimped in my mind since 1998 when we wrote this the 1d paper before i knew the word pleat and so that's my excuse so pleat or crimp take your pick crimp is usually mini pleats together that's crimping cool so crimps we kind of like why do we like crimps we proved in the one-dimensional case last time that if you make a crimp and your original thing was flat-foldable the new thing will be flat-foldable good news is that is still true i told you all these things were no longer true in the circular case it's still true in the circular case uh should i tell you let me just remind you the same proof works remind you what the proof looked like we had a crimp and then and we're hoping to fold that first we know at some point these creases get folded but there may be other junk in here maybe other junk in here we'd like to get rid of that junk and so what we did is move this junk up to here which was always safe we moved this junk down to here and that was still a folded state a flat folded state okay i remember all this stuff therefore we could have folded the crimp first and then we have a flat folding in this situation where this paper is effectively glued together which means it was safe to fold this crimp and glue these together and never touch them again so that is still true in the circular case that proof didn't really assume anything about the paper would even work for like tree-shaped paper or something weird so if we can find a crimp which we can in the generic case then you just make a crimp make a crimp repeat if you ever get stuck you know that the original thing was not flat foldable because the thing you have is not flat foldable why if there's no crimp in the generic case that means you have two valleys at around the smallest angle and that's clearly bad um so i'm not writing a lot of details here is that clear if i have an angle that's surrounded by strictly larger angles i know there must be one mount in one valley then i can safely make the crimp and repeat the trouble comes in when these angles are equal we never had to think about angles being equal in the one-dimensional case it didn't matter much we just said oh you know it's safe to make a crimp as long as this was greater than or equal to this and this was greater than or equal to this that is still true it's safe to make a crimp but how do we know that there is a crimp how do we know that there's a mountain followed by a valley we don't i mean if they're equal this this would be all right that's two valleys is valid if these three angles are equal but i claim still somewhere there's got to be a crimp there's no longer a notion of the globally smallest angle because some of them are equal we've got to find that crimp but it's out there so we just need to prove that it exists so so to do that we're going to generalize this theorem and kawa just in that number of mountains minus number of valleys is plus or minus two that's true and it's a statement about the sum of all of the or you know all of the angles all the creases what i'd like is something a little bit more localized and in particular we're going to think about when i have a whole bunch of angles that are equal how many mountains and valleys can there be in there so what i call local counts so while everything i've said so far is pretty classic this result was proved in 2001 by tom hall so much more recent over a decade later so let's think about k equal angles and i want that to be a maximal sequence of equal angles so the ones right after and right before are different and furthermore i'm going to assume that they're bigger strictly larger angles okay so something like that a bunch of equal angles k of them bigger angles on either side this almost always exists um i should mention this it's so i'm going to say something about this situation but the situation should exist because i take the smallest angle out there and then i see how many friends it has that are all equal if it's the smallest angle or it's one of the smallest angles then the ones are surrounding it are going to be bigger unless all the angles are equal so there's one case which we'll worry about later it's very easy all the angles are equal in fact i could tell you how to worry about that if all the angles are equal then you know everything is crimpable as long as we can find a switch between mountain and valley somewhere and because the number of mountains and valleys is plus or minus two there's got to be so they're almost in equal balance there's got to be a valley it can't be all valleys can't be all mountains so somewhere there's a transition between mountain and valley so you get a crimp out of that okay but otherwise if they're not all equal let's think about k angle equal angles surrounded by two larger guys forgot how technical this is all right then i want to look at the number of mountains and the number of valleys among these uh k plus one creases so within that range it's going to be zero if k is odd it's going to be plus or minus one if k is even okay it's not plus or minus two intuitively plus or minus two is when you go all the way around here we're just looking at a one you know segment portion of the entire circle remember this is actually a circle somewhere and when you this corresponds to in the odd case you're going to be going in the same direction afterwards and the even case you're going to have turned around but not a full circle so it's going to be plus or minus 1. there's a lot of ways to prove this one way is to think about convex cones so if k is even and you fold this thing see one two three four five six equal angles and then there's the two longer guys they may not be the same length but that's what you get in the even case so what i'd like to do is just in order to just reduce the things that we've already thought about i mean you could talk about turn angles again or you could just say hey i know mayakawa's theorem so as long as i can make this into a circle which i can do by adding that stuff now i have a nice circular it's a flat folded state there's no crossings here presumably because this did something i don't actually know that it's zigzagging it might do other stuff but it's not going to cross what i just added so the number of mountains minus number of valleys in total here must be plus or minus 2. i only added one crease this one so that's going to make it either plus or minus 1 or plus or minus 3. if you think about it a little bit it has to be plus or minus 1 and hopefully that's what i wrote here yes because i mean really we're thinking about turn angles again i'm afraid can't totally reduce it we just turned around here and we're about to finish the circle which would make it plus or minus 2 before we finish the circle is plus or minus 1. k is odd it's again this might go like that whatever but because it's an odd number these guys are going in the same direction and so the total turn angle here must be zero and one way to see that is to extend it into a full polygon you know that it's plus or minus two for this whole thing and i added two guys in the same direction if i remove them it's going to go down to plus or minus zero that was a bit hand wavy but do you buy that yeah plus or minus zero is zero yeah you went from two down to zero because you removed two identical guys so let me let me tie this all together because there's so many little cases so i want to know what mountain what mountain valley patterns are flat foldable okay single vertex and the patterns okay i claim there is a crimp if it's flat foldable there has to be a crimp that you can find to do remember crimp was where this length was less than or equal to this length and less than or equal to this length and one of these was m one of them was v that was the definition of a crimp claims such a thing always exists the proof is two parts if all the angles are equal all the angles are equal we know that these conditions are always satisfied because everything is equal so we just need to find an m followed by a v or v followed by an m and by maekalo's theorem we know it's not all m's or all v's so somewhere there has to be a transition from m to v's so just use that lemma theorem count property otherwise they're not all equal then i can apply this local counts result take the smallest angle take one instance of the smallest angle there could be many of them find all of its neighboring friends that are equal then i know that the next angles before and after have to be strictly bigger then i know that the number of mountains and valleys in here if i look at the difference and it's zero that means the number of mountains equals the number of valleys therefore there's at least one of each therefore there's a transition for mountain valley in there there's a transition from mountain to valley like right here that's a valid crimp this guy's bigger this guy's equal this would also be a fine transition because everybody's equal this would also be a fine transition any any transition from mountain valley in here i've got a crimp the even case is a little i've got to think a little bit more k is even k plus 1 which is the number of creases here is odd and the difference between mountains and valleys among those k plus one creases is plus or minus one think about it for a second that means there's again at least one mountain and at least one valley and uh because cave to be even has to be at least two you've gotta have at least three things so there's gonna be at least one of one of one and two of the other in fact and i just need one mountain in one valley somewhere in there and then i know there has to be a transition from mountain to valley otherwise use local counts and in either case we get either a mountain to valley transition or a valley to mountain transition and i get it in such a way that these these inequalities and the lengths hold so in other words that it's a crimp therefore any flat foldable single vertex mountain valley pattern has a crimp make it and as i was arguing here just like last time that crimp will still after you do the crimp you'll still be flat foldable so repeat cook until done is the i like to put it question what about the trivial case of just floating trivial case of folding a circle in half all right that's a good question it's kind of a technicality uh one version would be to say that's zero vertices that doesn't count we're thinking about one vertex and as soon as you think about it as one vertex which is fine then you actually have two uh creases there so it's two creases and hopefully this works uh oh i see you're saying it's not a crimp that's a good point it is flat foldable and yet there's no crimp damn it all right sorry my college right yeah so it's i made a slight mistake here i said if all the angles are equal then meikawa says there's at least one of each that's not true when n equals two then there's two of one and zero of the other okay as long as you have more than two creases then this is true that's enough so you have to handle this as a special case uh that's true this this this is an important special case to remember because this will be the final picture after you do a sequence of crimps you this is the good case uh if your thing is flat foldable you will always end up with that that's a case where everything is excuse me the same smallest it's it's another example where all the angles are equal yeah so that's why this is the situation it's either maekawa and you get a crimp here because n is greater than 2 and here i should say if n is greater than 2 then this is true of course the n equals 2 case it looks like that it better be both valleys or both mountains otherwise it's not going to be flat foldable i forgot about that i should add in the notes but it's in the textbook right that's true it won't actually look like this it's going to at the end of the algorithm it's going to be a cone the two angles will be equal because of kawasaki's theorem and they should still be both mountains or both valleys but it's going to be we can see it right here so i apply my crimp and now i have a cone and there's two angles there's one here which has been fused together there's one here which is original and they're equal and it's two mountains or two valleys if i turn it upside down and then it finishes so that's the algorithm in action two steps first you find this crimp here this is actually the globally smallest angle it's surrounded by bigger angles so it's really easy then i have a cone with two equal angles which is what has to happen at the end and then i i'm done good this is what always happens oh is this the only way to flat fold these things no there are other ah yes one of the two um i think it depends what you're counting if you're counting mountain valley assignments i mean if you just want to know can every mountain valley assignment be folded by crimping then the answer is yes that's what we proved but if you want to get every possible folded state crimps are not going to be enough the reason they're not enough is because we're using this thing said hey here's some hypothetical folded state we can rip it out and make the crimp have been done but then you're there's a folded state you're not able to reach by crimping i mean we need to work that into an actual example where crimping forbids you from reaching some folded state but i'm pretty sure one exists because of this yeah maybe some kind of spiraling thing right what we're missing here is getting all the possible layer stacking orders so we're just trying to match the mountain valley assignment we're not going to match a target layer ordering because we simplified the layer orders in order to make crimps possible so crimps won't get you some some of the folded states but one thing you can play with over here that has been played with by tom hall is if i give you a crease pattern how many flat foldable mountain valley assignments does it have okay we have this nice linear time algorithm to tell you whether a particular one is doable how many are there and there is if you work through all of the things i've shown you in a little bit more detail you can recover this was finding a crimp you could actually look at what are all the crimps that are possible and actually count how many different ways there are to fold how many crimps you can do and then given those crimps how many crimps you can do after that and in same linear time kind of algorithm you can figure out how many different mountain valley patterns are flat foldable that takes a little bit more care i will tell you the extremes how small could the number of flat foldable mountain valley assignments be for a given crease pattern well do you have any guesses when when do i have the least choice of where to make crimps yeah uh for general n sorry but yeah three three angles are i guess two angles i really can't do any i don't have any choice but for general and how should i disperse the theta eyes in order to make there be very few possible crimps when there are only two transitions between mountains and valleys all the mountains are together all the valleys are together that's right if i was asking the question about mountains and valleys but i was asking a question about theta eyes so i want to count how many mountain valley patterns there are mountain valley assignments that are flat foldable for a given crease pattern all the angles are consecutive increasing yeah that pretty much works in fact what i need is the generic case where all the angles are different and they never become equal by folding which is what generic means so then there was only really one crimp i could do which is the globally smallest angle so actually i still have two choices i could do mountain then valley or valley than mountain but i only have two choices for that crimp then i will have two choices for the next crimp two choices for the next crimp in general i get two to the n possible mountain valley assignments for the generic case okay what about what's the opposite where i get as much choice as possible yeah two to the n no uh two to the n over two thank you right they're two to the end conceivable mountain valley assignments can't be that big uh every time i do a crimp i eat two creases not one that's what i forgot good still pretty big though it's like square root of all the possible things you could imagine are indeed feasible it went from 2 to the n to 2n over 2. when would i get the most choice in crimping all angles are equal that's the opposite extreme and in this case a little messier when all the angles are equal the only property you need and you can see that from the proof the only property you need is that the number of mountains minus number of valleys is plus or minus two all the a and the n is even as long as you have that uh we showed here you're going to have an alternation from mountain valley that's a valid crimp because all the angles are equal and you can keep going so it's just how do you disperse that many how do i disperse n over two minus one mountains among n different positions that's what this represents and positions how do i pick n over two minus one of them to be mountains or could be n over two minus one of them are valleys and that's this factor of two so it could have more mountains or more valleys and then you somehow place those among those and this is if you don't know this notation that's just what it means if you want to know with using other notation it's like this n over 2 minus 1 factorial over 2 plus 1 factorial if you don't know factorials i'll tell you about them later cool so that's kind of the end of the single vertex situation um yeah i think i'll mention one interesting open question here which i would love to explore maybe in our problem session um which looks like it'll be mondays at five i think um this is one vertex and we went through all this work to solve single vertex situation and it's interesting as we'll see to think about locally how each vertex behaves but you would think if you have a crease pattern with two vertices shouldn't be that much harder so here we have linear time how quickly can you tell whether a two vertex crease pattern is flat foldable as far as i know no one has looked at that problem surely we can do it in like quadratic time but maybe even linear time i think can't be that hard i think in general if i have a small number of vertices say k vertices much smaller than n creases is there the algorithms people is there fixed parameter retractable algorithm and k or can i get something even getting something like n to the some function of k would be progress i think this should be doable but ideally we get a running time that's exponential in k and linear in n that would be my hope why do i say it has to be exponential because in general if i give you crease pattern with n vertices lots of vertices this problem is np complete which is there's not going to be a polynomial time algorithm for it nothing good we will prove that next wednesday so i'll hold off on that a little bit but for two two vertices how hard could it be all right i want to talk about one related topic and then we will go to origami design and do a little bit on the tree method but before we get there i want to talk about local foldability which is a cool topic people tend to forget about it i really like it i think it would make a cool project also it's a nice algorithm it goes back to vernon hayes 1996 so it's also right at the beginning of origami mathematics and it's this idea all right i give you a crease pattern now arbitrarily many vertices and i ask you if i ask you does it fold flat that's np-complete intractable same paper but what if i ask you just to give me a mountain valley assignment that might fold flat well that's also mp complete but if if you actually want it to fall flat this is really the same problem but if i ask you give me a mountain valley assignment so that if i checked every vertex according to this algorithm at least every vertex folds flat that would seem nice you know definitely i have to find a mountain valley assignment that satisfies these conditions that as i do successive crimps every vertex if i cut out the vertex separately it would fold flat this is the notion of local foldability and there's a linear time algorithm to tell you to give you a mountain valley assignment that ought to work in that each vertex it works it still may not work globally for some other reason but it's pretty good i think this would be actually pretty practical i think in a lot of real world origami settings when you're doing flat folding anyway locally valid locally foldable is going to be enough to be globally foldable that's a guess i don't know if it's true consistent mountain valley assignment if there is one so that each vertex locally folds flat let me uh try to concoct a small example that's relevant here gotta think yeah i think that works all right here we go i think this is in the textbook i remember drawing it in the past three years ago or whatever all right so here's a crease pattern on a square whatever yes i should have this saying thank you uh so this satisfies kawasaki's theorem that was the hard part because these angles sum to 180 and it's symmetric all around okay this is uh these angles are the smallest they're 60 degrees this is an equilateral triangle so we have this not quite the generic case but we have a the smallest angle is surrounded by two larger angles therefore one of these is a mountain the other is a valley that means that these two creases have to have different assignments i'm going to write a not equal sign one of these is not in the other's value they can't be the same also these two cannot be the same also these two cannot be the same that's not possible because i've got three it has to alternate you can't alternate three times you can only alternate an even number of times okay two two of these mountains one's valley and then you've got a problem okay so this thing is not flat foldable and this algorithm will tell you that because it'll say hey i can't even find a mountain valley assignment that could possibly fold each vertex flat so this is a nice algorithm at least it will detect annoying situations like that so in order to solve this we're going to use and i'm just going to sketch how this algorithm works we're going to use this characterization and this idea that we can really find all possible mountain valley assignments just by trying all the possible crimp sequences now there are exponentially many so it's not like i'm actually going to try them all but i need to explore that space of candidate crimps and see what happens all right uh so the idea is kind of crazy the beginning of the algorithm is fold each vertex flat somehow i don't care how just pick a crimp do it pick a crimp do it okay separately for each vertex they're not going to be compatible and don't look at the mountain valley assignment you get but look at the crimping sequence you get so here let's do a little example a non-trivial example so here we have the peace sign and these two angles are equal they're smallest i could crimp this angle first or i could crimp this angle first i have a choice so i'll draw both of them if i crimp this angle first i know these guys are paired up in the sense that they must be not equal ones mountain ones valley after i do that crimp just like in the real example i had you know this angle will equal that angle and these two guys must have equal assignments they must be both mountain or both valley in this situation these two guys are not equal but if i crimp the this pair first and these two guys must be equal equal now in the sense of being mountain or valley okay so there's still you know exponentially many possibilities in how to do this but just pick one pick one of these ways of pairing up you're going to pair up each of the increases into n over two pairs and they're going to have some not equals and equal signs okay so now if you imagine that the general picture like here i for example i might get in this pattern i would be forced to get not equals and these guys paired up and in general i want to look at these kinds of cycles you know if i come if i come into a vertex here's a vertex it's paired up with somebody so if i come in here i can go out somewhere and i come to this vertex it's paired with somebody so i'm gonna just i can keep wandering around and in general there will be a bunch of these paths that you can follow what could the paths do they either close up on themselves maybe things are paired up in such a way that you make a return trip or it could be some other path let me draw a dotted path it could come in here and maybe it gets paired up with this guy maybe it goes off to infinity it reaches the boundary of the paper those are the two possibilities you get paths which go off to the edge off to infinity on either end or you could get cycles the cycles are the problem because whenever i have a cycle i have a parity constraint for example when they're all not equals uh the length of the cycle must be even if they were yeah what's the general statement it's like the parity of the cycle which is whether it's even or odd should be equal to the parity of the number of not equal signs something like that i'm going to cheat here great i just said parody problems it's something like that it's either the parity of the cycle should equal the parity of the number of equals or the variety of the number not equals i think number of equals anyway one of the two and you can just check i mean you're forced if i say okay let's make this mountain then this is either equals or not equals it'll tell me whether this is mountain or valley just walk around the cycle either you get a contradiction or you don't if you get a contradiction we have a problem how do we how could we possibly fix the problem well when we made you look at each of these crimps i mean in fact you could look at each of these vertices separately but you think about one of these crimps and say well could i have done it another way sometimes there are crimps that have other equal choices maybe they're a bunch of equal angles and i could have done a different pairing what happens when i try a different pairing well instead of this being in one cycle and let's say this being in another cycle if i pair these guys up instead i'll end up merging those two cycles into one bigger thing could be a path or a cycle and the algorithm says just keep doing those cycle merges and if you get stuck your thing is not locally foldable that's the hard part to prove otherwise you will find a one of these patterns that you can actually resolve mountains and valleys all the way through so let's say so start with sum folding say local folding whatever i'm going to say with some pairing of creases at vertices and merge two pads or cycles whenever possible and when i say merge i mean whenever you have what are the possible things you could possibly merge when you have at some point during the algorithm a bunch of equal angles you have a choice which of these you crimp obviously the mountain valley sign is not fixed you can crimp any of them you picked one of them and it's sort of merging whatever this thing is attached to to whatever this thing is attached to if there's something else that's in that's disconnected from that thing i want you to instead merge two of them that combines two different connected components two paths or cycles merging means i decrease the total number of pads or cycles i combine two into one whenever that's possible do it you can prove that if you have a parity problem in the merge thing you had to have had a parity problem originally and merging can only fix parity problems that's the claim that's what i will not prove once you know that and it doesn't matter in what order or how you choose to merge you just merge as much as possible and either the resulting thing is okay or not and accordingly you will tell whether this thing is locally foldable sorry i want to move on to other things but i think this would be a fun thing to actually implement it's an easy algorithm and it's i think a pretty good test for problems like this to prevent flat foldability all right so let's move on from foldability to origami design so a bit of a big transition and we're gonna talk about origami design a lot more next class also but uh just started off today and the particular algorithm for origami design i want to talk about is called tree method this is probably the oldest algorithm for origami design uh in that it's been around and people have been thinking about it and developing it for many years uh through this period called the bug wars and people are trying to design more and more complicated insects it's like well i can make an insect with six legs oh yeah well i can make you know a spider with eight legs oh yeah well i can make an insect that has you know a beetle it has wings and horns and there's thorns on the horns and all these crazy things during that time there are a lot of people thinking about how do i make more and more complicated especially uh sort of more more limbs in my creatures more and very precise uh arrangements of those limbs let's say and that is what the tree method deals with and is really formalized by robert lang who published a paper in 96 describing it as a as a sort of complete algorithm and it's still not known for sure that that algorithm always works but uh that's what we've we've been working on me and marty and rob lang for the last four years or so and soon we will publish that thing and prove that this thing always works but i'm going to describe to the algorithm without the proof that it works and what it does tell you its goal from a mathematical perspective so it's interested in practical origami design so we're going to start from a square piece of paper it would also work for triangular pieces of paper or anything convex but squares are what people usually care about rectangles are sometimes it's used for rectangles also okay the idea is i give to you a stick figure so that's formally it's a tree a graph without any cycles it's a metric tree meaning that i put lengths on the edges i know this length this edge length is maybe twice as long as this one so i really draw it with edge lengths in mind then what i want you to do is find some folding of a piece of paper i should really be looking at what i'm trying to match uh so here here maybe this goes down such that i want to find one some folding of a square paper in fact the smallest square possible so that when i project like this vertically the projection of that uh flat folding is exactly that metric tree okay and this is called a uniaxial this thing is called a uniaxial base let me tell you a little bit why it's called any axial base we're thinking about what are called origami bases these are like the beginning of origami models and most classic origami models like more than 60 years ago start from one of these bases you've got water bomb base in the top left preliminary base fish base bird base windmill base and frog base i remember them all i have a little example here this is the water bomb base so it's just very simple crease pattern and you know why this is useful is it gives you sort of four flaps of paper to work with you know maybe you make one of them uh the head and the other two wings and the back one a tail if you're making a crane it doesn't actually starts not from the space but from the other one but same idea if you're folding a crane one of these would be head other tail and two wings this is great if you're making a four-flap animal and if you think about its projection and it's easier to think about in this other picture the projection of this thing is a four-limbed star right you can see it's a plus sign all of them are the same length and so this is actually something you can get out of the tree method you just give that as your input you will get this 3d thing and this crease pattern as your uh origami as the output let me show you the program in action uh so robert lang implemented this thing it's called tree maker it's freely available open source all that good stuff and uh yeah good and i'm not an expert at using it so bear with me but if we wanted to say i would like that star okay now i i drew it obviously not with all the lengths equal but it's ignoring the lengths that i drew and they're actually specified here so all the lengths here are supposed to be one and then i say okay optimize and then make a crease pattern and then show me the crease pattern there it is exactly the crease pattern i made although actually i can see from the mountain valley assignment because this is not really this is not flat origami it's it made it this way so it's flat of course the projection is the same still for limbs okay and so you know stash is for valleys and solid lines for mountains but you can make anything you want so let's say we want to make uh i don't know a lizard or something so the blue lines are the tree uh so here's can i do this yeah there we go uh so here i have a forearm head four leg another forearm body segment tail and two hind legs maybe i want to make that so i say optimize and then make a crease pattern and then boom you fold that it will have exactly that projection and assuming it did a reasonable job at computation this will be the smaller the best way to fold this thing and it's the smallest square that matches exactly that shape you get the best scale factor between the size of your piece of paper and the target shape but actually doing that optimization the first step i did is mp complete so it's not going to do it perfectly we'll prove that wednesday but it the heuristics are pretty good it finds a local minimum and often it finds a pretty good one and sometimes you can coax it to find better ones but yeah it's not perfect but hey it's mp complete so you can't do it this actually shows you what it would look like in x-ray view and then you can say oh that was nice but let's where's my yeah that was good but maybe i really wanted the head segment to be shorter uh you know length 0.5 and then i wanted the tail to be really long and then you can optimize that and find a crease pattern and it'll complain because it's having trouble oh dear demo effect i should have tried this example before it's not oh gosh it's one of these annoying ones i should say i should add some feature like maybe strain split something add a little bit maybe do that and yeah there we go uh i cheated and i'll explain how i cheated last time but if sometimes the particular method fails but you can fix it by adding in another tiny limb off the edge somewhere and of course you can then get rid of that at the end when you're folding it only makes the problem slightly harder and it'll still find a folding but we'll talk about that next time and i'm way out of time so we will stop there 3 00:00:08,950 --> 00:00:12,470 4 00:00:12,470 --> 00:00:13,990 5 00:00:13,990 --> 00:00:14,000 6 00:00:14,000 --> 00:00:15,589 7 00:00:15,589 --> 00:00:16,870 8 00:00:16,870 --> 00:00:18,310 9 00:00:18,310 --> 00:00:22,230 10 00:00:22,230 --> 00:00:25,589 11 00:00:25,589 --> 00:00:28,070 12 00:00:28,070 --> 00:00:30,150 13 00:00:30,150 --> 00:00:31,589 14 00:00:31,589 --> 00:00:32,310 15 00:00:32,310 --> 00:00:34,310 16 00:00:34,310 --> 00:00:36,549 17 00:00:36,549 --> 00:00:38,869 18 00:00:38,869 --> 00:00:40,470 19 00:00:40,470 --> 00:00:43,670 20 00:00:43,670 --> 00:00:46,470 21 00:00:46,470 --> 00:00:46,480 22 00:00:46,480 --> 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--> 00:13:34,949 347 00:13:34,949 --> 00:13:37,269 348 00:13:37,269 --> 00:13:40,629 349 00:13:40,629 --> 00:13:42,310 350 00:13:42,310 --> 00:13:48,069 351 00:13:48,069 --> 00:13:50,150 352 00:13:50,150 --> 00:13:51,829 353 00:13:51,829 --> 00:13:51,839 354 00:13:51,839 --> 00:13:52,550 355 00:13:52,550 --> 00:13:54,230 356 00:13:54,230 --> 00:13:57,670 357 00:13:57,670 --> 00:14:00,949 358 00:14:00,949 --> 00:14:00,959 359 00:14:00,959 --> 00:14:02,150 360 00:14:02,150 --> 00:14:06,069 361 00:14:06,069 --> 00:14:09,189 362 00:14:09,189 --> 00:14:10,710 363 00:14:10,710 --> 00:14:13,030 364 00:14:13,030 --> 00:14:14,550 365 00:14:14,550 --> 00:14:17,829 366 00:14:17,829 --> 00:14:21,670 367 00:14:21,670 --> 00:14:24,150 368 00:14:24,150 --> 00:14:25,189 369 00:14:25,189 --> 00:14:26,389 370 00:14:26,389 --> 00:14:28,389 371 00:14:28,389 --> 00:14:32,870 372 00:14:32,870 --> 00:14:35,750 373 00:14:35,750 --> 00:14:37,509 374 00:14:37,509 --> 00:14:39,110 375 00:14:39,110 --> 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405 00:15:47,749 --> 00:15:48,949 406 00:15:48,949 --> 00:15:51,350 407 00:15:51,350 --> 00:15:52,710 408 00:15:52,710 --> 00:15:53,670 409 00:15:53,670 --> 00:15:55,749 410 00:15:55,749 --> 00:15:55,759 411 00:15:55,759 --> 00:15:56,870 412 00:15:56,870 --> 00:15:59,030 413 00:15:59,030 --> 00:16:00,790 414 00:16:00,790 --> 00:16:03,189 415 00:16:03,189 --> 00:16:03,199 416 00:16:03,199 --> 00:16:04,230 417 00:16:04,230 --> 00:16:05,749 418 00:16:05,749 --> 00:16:07,749 419 00:16:07,749 --> 00:16:10,150 420 00:16:10,150 --> 00:16:10,949 421 00:16:10,949 --> 00:16:14,150 422 00:16:14,150 --> 00:16:14,160 423 00:16:14,160 --> 00:16:14,710 424 00:16:14,710 --> 00:16:18,150 425 00:16:18,150 --> 00:16:21,670 426 00:16:21,670 --> 00:16:22,790 427 00:16:22,790 --> 00:16:24,470 428 00:16:24,470 --> 00:16:26,310 429 00:16:26,310 --> 00:16:26,320 430 00:16:26,320 --> 00:16:29,749 431 00:16:29,749 --> 00:16:31,829 432 00:16:31,829 --> 00:16:33,910 433 00:16:33,910 --> 00:16:35,829 434 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--> 00:17:27,909 464 00:17:27,909 --> 00:17:29,990 465 00:17:29,990 --> 00:17:32,150 466 00:17:32,150 --> 00:17:35,190 467 00:17:35,190 --> 00:17:37,510 468 00:17:37,510 --> 00:17:39,029 469 00:17:39,029 --> 00:17:42,230 470 00:17:42,230 --> 00:17:43,430 471 00:17:43,430 --> 00:17:47,669 472 00:17:47,669 --> 00:17:49,430 473 00:17:49,430 --> 00:17:50,870 474 00:17:50,870 --> 00:17:52,630 475 00:17:52,630 --> 00:17:54,549 476 00:17:54,549 --> 00:17:56,230 477 00:17:56,230 --> 00:17:57,669 478 00:17:57,669 --> 00:18:01,270 479 00:18:01,270 --> 00:18:09,510 480 00:18:09,510 --> 00:18:14,549 481 00:18:14,549 --> 00:18:17,510 482 00:18:17,510 --> 00:18:19,110 483 00:18:19,110 --> 00:18:26,470 484 00:18:26,470 --> 00:18:29,350 485 00:18:29,350 --> 00:18:29,360 486 00:18:29,360 --> 00:18:29,909 487 00:18:29,909 --> 00:18:31,270 488 00:18:31,270 --> 00:18:31,280 489 00:18:31,280 --> 00:18:33,110 490 00:18:33,110 --> 00:18:34,870 491 00:18:34,870 --> 00:18:36,789 492 00:18:36,789 --> 00:18:38,310 493 00:18:38,310 --> 00:18:40,230 494 00:18:40,230 --> 00:18:41,830 495 00:18:41,830 --> 00:18:43,270 496 00:18:43,270 --> 00:18:44,470 497 00:18:44,470 --> 00:18:45,909 498 00:18:45,909 --> 00:18:48,390 499 00:18:48,390 --> 00:18:49,590 500 00:18:49,590 --> 00:18:51,510 501 00:18:51,510 --> 00:18:53,110 502 00:18:53,110 --> 00:18:55,029 503 00:18:55,029 --> 00:18:57,270 504 00:18:57,270 --> 00:18:59,430 505 00:18:59,430 --> 00:19:01,110 506 00:19:01,110 --> 00:19:01,120 507 00:19:01,120 --> 00:19:02,070 508 00:19:02,070 --> 00:19:05,190 509 00:19:05,190 --> 00:19:07,669 510 00:19:07,669 --> 00:19:07,679 511 00:19:07,679 --> 00:19:09,350 512 00:19:09,350 --> 00:19:12,150 513 00:19:12,150 --> 00:19:13,750 514 00:19:13,750 --> 00:19:17,029 515 00:19:17,029 --> 00:19:20,390 516 00:19:20,390 --> 00:19:23,350 517 00:19:23,350 --> 00:19:24,870 518 00:19:24,870 --> 00:19:26,230 519 00:19:26,230 --> 00:19:27,909 520 00:19:27,909 --> 00:19:29,750 521 00:19:29,750 --> 00:19:49,430 522 00:19:49,430 --> 00:19:51,669 523 00:19:51,669 --> 00:19:53,029 524 00:19:53,029 --> 00:19:54,630 525 00:19:54,630 --> 00:19:57,110 526 00:19:57,110 --> 00:20:01,350 527 00:20:01,350 --> 00:20:03,510 528 00:20:03,510 --> 00:20:04,870 529 00:20:04,870 --> 00:20:06,310 530 00:20:06,310 --> 00:20:08,789 531 00:20:08,789 --> 00:20:10,549 532 00:20:10,549 --> 00:20:11,750 533 00:20:11,750 --> 00:20:13,029 534 00:20:13,029 --> 00:20:14,870 535 00:20:14,870 --> 00:20:16,710 536 00:20:16,710 --> 00:20:21,350 537 00:20:21,350 --> 00:20:23,909 538 00:20:23,909 --> 00:20:23,919 539 00:20:23,919 --> 00:20:24,710 540 00:20:24,710 --> 00:20:27,270 541 00:20:27,270 --> 00:20:29,270 542 00:20:29,270 --> 00:20:32,310 543 00:20:32,310 --> 00:20:32,320 544 00:20:32,320 --> 00:20:33,270 545 00:20:33,270 --> 00:20:36,310 546 00:20:36,310 --> 00:20:38,149 547 00:20:38,149 --> 00:20:40,230 548 00:20:40,230 --> 00:20:41,830 549 00:20:41,830 --> 00:20:42,630 550 00:20:42,630 --> 00:20:45,590 551 00:20:45,590 --> 00:20:46,789 552 00:20:46,789 --> 00:20:48,310 553 00:20:48,310 --> 00:20:49,669 554 00:20:49,669 --> 00:20:51,110 555 00:20:51,110 --> 00:20:52,710 556 00:20:52,710 --> 00:20:54,710 557 00:20:54,710 --> 00:20:55,990 558 00:20:55,990 --> 00:20:58,230 559 00:20:58,230 --> 00:20:59,830 560 00:20:59,830 --> 00:21:02,149 561 00:21:02,149 --> 00:21:03,270 562 00:21:03,270 --> 00:21:05,029 563 00:21:05,029 --> 00:21:08,390 564 00:21:08,390 --> 00:21:10,070 565 00:21:10,070 --> 00:21:12,470 566 00:21:12,470 --> 00:21:15,590 567 00:21:15,590 --> 00:21:20,310 568 00:21:20,310 --> 00:21:26,789 569 00:21:26,789 --> 00:21:29,990 570 00:21:29,990 --> 00:21:31,909 571 00:21:31,909 --> 00:21:34,950 572 00:21:34,950 --> 00:21:38,070 573 00:21:38,070 --> 00:21:38,549 574 00:21:38,549 --> 00:21:44,390 575 00:21:44,390 --> 00:21:47,510 576 00:21:47,510 --> 00:21:54,549 577 00:21:54,549 --> 00:21:59,909 578 00:21:59,909 --> 00:22:02,549 579 00:22:02,549 --> 00:22:04,710 580 00:22:04,710 --> 00:22:07,830 581 00:22:07,830 --> 00:22:10,470 582 00:22:10,470 --> 00:22:12,310 583 00:22:12,310 --> 00:22:14,310 584 00:22:14,310 --> 00:22:16,390 585 00:22:16,390 --> 00:22:17,909 586 00:22:17,909 --> 00:22:20,390 587 00:22:20,390 --> 00:22:22,710 588 00:22:22,710 --> 00:22:24,830 589 00:22:24,830 --> 00:22:24,840 590 00:22:24,840 --> 00:22:27,750 591 00:22:27,750 --> 00:22:34,549 592 00:22:34,549 --> 00:22:35,750 593 00:22:35,750 --> 00:22:38,149 594 00:22:38,149 --> 00:22:40,230 595 00:22:40,230 --> 00:22:43,270 596 00:22:43,270 --> 00:22:44,789 597 00:22:44,789 --> 00:22:46,549 598 00:22:46,549 --> 00:22:48,310 599 00:22:48,310 --> 00:22:50,230 600 00:22:50,230 --> 00:22:52,789 601 00:22:52,789 --> 00:22:56,549 602 00:22:56,549 --> 00:22:57,750 603 00:22:57,750 --> 00:23:00,549 604 00:23:00,549 --> 00:23:05,110 605 00:23:05,110 --> 00:23:07,350 606 00:23:07,350 --> 00:23:07,360 607 00:23:07,360 --> 00:23:09,750 608 00:23:09,750 --> 00:23:13,270 609 00:23:13,270 --> 00:23:15,830 610 00:23:15,830 --> 00:23:17,190 611 00:23:17,190 --> 00:23:20,549 612 00:23:20,549 --> 00:23:23,029 613 00:23:23,029 --> 00:23:23,990 614 00:23:23,990 --> 00:23:25,430 615 00:23:25,430 --> 00:23:27,510 616 00:23:27,510 --> 00:23:29,750 617 00:23:29,750 --> 00:23:31,110 618 00:23:31,110 --> 00:23:36,789 619 00:23:36,789 --> 00:23:38,870 620 00:23:38,870 --> 00:23:40,230 621 00:23:40,230 --> 00:23:42,789 622 00:23:42,789 --> 00:23:44,310 623 00:23:44,310 --> 00:23:45,350 624 00:23:45,350 --> 00:23:49,350 625 00:23:49,350 --> 00:23:51,830 626 00:23:51,830 --> 00:23:51,840 627 00:23:51,840 --> 00:23:53,669 628 00:23:53,669 --> 00:23:55,510 629 00:23:55,510 --> 00:23:56,710 630 00:23:56,710 --> 00:23:58,149 631 00:23:58,149 --> 00:24:00,789 632 00:24:00,789 --> 00:24:02,470 633 00:24:02,470 --> 00:24:05,669 634 00:24:05,669 --> 00:24:08,070 635 00:24:08,070 --> 00:24:09,830 636 00:24:09,830 --> 00:24:13,110 637 00:24:13,110 --> 00:24:16,830 638 00:24:16,830 --> 00:24:16,840 639 00:24:16,840 --> 00:24:23,190 640 00:24:23,190 --> 00:24:24,950 641 00:24:24,950 --> 00:24:34,630 642 00:24:34,630 --> 00:24:35,669 643 00:24:35,669 --> 00:24:37,269 644 00:24:37,269 --> 00:24:38,950 645 00:24:38,950 --> 00:24:41,430 646 00:24:41,430 --> 00:24:41,440 647 00:24:41,440 --> 00:24:43,430 648 00:24:43,430 --> 00:24:46,149 649 00:24:46,149 --> 00:24:49,269 650 00:24:49,269 --> 00:24:50,710 651 00:24:50,710 --> 00:24:52,470 652 00:24:52,470 --> 00:24:55,909 653 00:24:55,909 --> 00:24:55,919 654 00:24:55,919 --> 00:24:56,390 655 00:24:56,390 --> 00:24:58,870 656 00:24:58,870 --> 00:25:01,029 657 00:25:01,029 --> 00:25:02,710 658 00:25:02,710 --> 00:25:04,549 659 00:25:04,549 --> 00:25:07,029 660 00:25:07,029 --> 00:25:09,590 661 00:25:09,590 --> 00:25:13,510 662 00:25:13,510 --> 00:25:16,390 663 00:25:16,390 --> 00:25:16,400 664 00:25:16,400 --> 00:25:17,029 665 00:25:17,029 --> 00:25:18,950 666 00:25:18,950 --> 00:25:20,390 667 00:25:20,390 --> 00:25:20,400 668 00:25:20,400 --> 00:25:21,190 669 00:25:21,190 --> 00:25:22,950 670 00:25:22,950 --> 00:25:24,390 671 00:25:24,390 --> 00:25:27,750 672 00:25:27,750 --> 00:25:30,149 673 00:25:30,149 --> 00:25:30,159 674 00:25:30,159 --> 00:25:33,510 675 00:25:33,510 --> 00:25:37,110 676 00:25:37,110 --> 00:25:40,390 677 00:25:40,390 --> 00:25:42,310 678 00:25:42,310 --> 00:25:43,990 679 00:25:43,990 --> 00:25:45,190 680 00:25:45,190 --> 00:25:47,350 681 00:25:47,350 --> 00:25:50,310 682 00:25:50,310 --> 00:25:54,470 683 00:25:54,470 --> 00:25:58,390 684 00:25:58,390 --> 00:26:01,990 685 00:26:01,990 --> 00:26:04,070 686 00:26:04,070 --> 00:26:04,080 687 00:26:04,080 --> 00:26:05,190 688 00:26:05,190 --> 00:26:07,190 689 00:26:07,190 --> 00:26:08,830 690 00:26:08,830 --> 00:26:08,840 691 00:26:08,840 --> 00:26:10,230 692 00:26:10,230 --> 00:26:11,990 693 00:26:11,990 --> 00:26:13,830 694 00:26:13,830 --> 00:26:17,190 695 00:26:17,190 --> 00:26:19,350 696 00:26:19,350 --> 00:26:24,470 697 00:26:24,470 --> 00:26:29,590 698 00:26:29,590 --> 00:26:31,990 699 00:26:31,990 --> 00:26:34,390 700 00:26:34,390 --> 00:26:36,710 701 00:26:36,710 --> 00:26:39,269 702 00:26:39,269 --> 00:26:41,190 703 00:26:41,190 --> 00:26:42,630 704 00:26:42,630 --> 00:26:45,269 705 00:26:45,269 --> 00:26:45,279 706 00:26:45,279 --> 00:26:46,549 707 00:26:46,549 --> 00:26:50,710 708 00:26:50,710 --> 00:26:52,390 709 00:26:52,390 --> 00:26:52,400 710 00:26:52,400 --> 00:26:53,269 711 00:26:53,269 --> 00:26:55,830 712 00:26:55,830 --> 00:26:59,750 713 00:26:59,750 --> 00:27:06,950 714 00:27:06,950 --> 00:27:08,390 715 00:27:08,390 --> 00:27:11,909 716 00:27:11,909 --> 00:27:15,830 717 00:27:15,830 --> 00:27:17,990 718 00:27:17,990 --> 00:27:20,549 719 00:27:20,549 --> 00:27:22,549 720 00:27:22,549 --> 00:27:22,559 721 00:27:22,559 --> 00:27:23,590 722 00:27:23,590 --> 00:27:25,029 723 00:27:25,029 --> 00:27:26,549 724 00:27:26,549 --> 00:27:26,559 725 00:27:26,559 --> 00:27:27,190 726 00:27:27,190 --> 00:27:29,029 727 00:27:29,029 --> 00:27:30,870 728 00:27:30,870 --> 00:27:31,669 729 00:27:31,669 --> 00:27:35,029 730 00:27:35,029 --> 00:27:37,190 731 00:27:37,190 --> 00:27:39,350 732 00:27:39,350 --> 00:27:42,950 733 00:27:42,950 --> 00:27:46,149 734 00:27:46,149 --> 00:27:47,990 735 00:27:47,990 --> 00:27:49,990 736 00:27:49,990 --> 00:27:50,000 737 00:27:50,000 --> 00:27:50,710 738 00:27:50,710 --> 00:27:52,389 739 00:27:52,389 --> 00:27:53,750 740 00:27:53,750 --> 00:27:54,789 741 00:27:54,789 --> 00:27:56,710 742 00:27:56,710 --> 00:27:58,630 743 00:27:58,630 --> 00:28:00,149 744 00:28:00,149 --> 00:28:02,789 745 00:28:02,789 --> 00:28:05,190 746 00:28:05,190 --> 00:28:06,230 747 00:28:06,230 --> 00:28:09,590 748 00:28:09,590 --> 00:28:12,149 749 00:28:12,149 --> 00:28:12,159 750 00:28:12,159 --> 00:28:12,950 751 00:28:12,950 --> 00:28:16,310 752 00:28:16,310 --> 00:28:16,320 753 00:28:16,320 --> 00:28:17,269 754 00:28:17,269 --> 00:28:19,029 755 00:28:19,029 --> 00:28:20,870 756 00:28:20,870 --> 00:28:22,789 757 00:28:22,789 --> 00:28:24,950 758 00:28:24,950 --> 00:28:26,789 759 00:28:26,789 --> 00:28:28,710 760 00:28:28,710 --> 00:28:30,630 761 00:28:30,630 --> 00:28:32,870 762 00:28:32,870 --> 00:28:33,990 763 00:28:33,990 --> 00:28:36,070 764 00:28:36,070 --> 00:28:37,750 765 00:28:37,750 --> 00:28:38,950 766 00:28:38,950 --> 00:28:42,149 767 00:28:42,149 --> 00:28:47,190 768 00:28:47,190 --> 00:28:48,630 769 00:28:48,630 --> 00:28:50,630 770 00:28:50,630 --> 00:28:51,510 771 00:28:51,510 --> 00:28:54,149 772 00:28:54,149 --> 00:28:56,710 773 00:28:56,710 --> 00:28:59,590 774 00:28:59,590 --> 00:29:00,870 775 00:29:00,870 --> 00:29:02,630 776 00:29:02,630 --> 00:29:04,389 777 00:29:04,389 --> 00:29:08,630 778 00:29:08,630 --> 00:29:09,909 779 00:29:09,909 --> 00:29:12,310 780 00:29:12,310 --> 00:29:12,320 781 00:29:12,320 --> 00:29:24,830 782 00:29:24,830 --> 00:29:24,840 783 00:29:24,840 --> 00:29:31,510 784 00:29:31,510 --> 00:29:33,190 785 00:29:33,190 --> 00:29:35,190 786 00:29:35,190 --> 00:29:36,549 787 00:29:36,549 --> 00:29:37,830 788 00:29:37,830 --> 00:29:40,070 789 00:29:40,070 --> 00:29:41,750 790 00:29:41,750 --> 00:29:44,310 791 00:29:44,310 --> 00:29:46,389 792 00:29:46,389 --> 00:29:49,430 793 00:29:49,430 --> 00:29:55,510 794 00:29:55,510 --> 00:30:00,070 795 00:30:00,070 --> 00:30:03,510 796 00:30:03,510 --> 00:30:05,830 797 00:30:05,830 --> 00:30:08,710 798 00:30:08,710 --> 00:30:15,750 799 00:30:15,750 --> 00:30:30,549 800 00:30:30,549 --> 00:30:34,230 801 00:30:34,230 --> 00:30:37,350 802 00:30:37,350 --> 00:30:38,710 803 00:30:38,710 --> 00:30:40,549 804 00:30:40,549 --> 00:30:43,909 805 00:30:43,909 --> 00:30:49,110 806 00:30:49,110 --> 00:30:51,190 807 00:30:51,190 --> 00:30:52,389 808 00:30:52,389 --> 00:30:54,630 809 00:30:54,630 --> 00:30:56,549 810 00:30:56,549 --> 00:30:58,070 811 00:30:58,070 --> 00:30:59,350 812 00:30:59,350 --> 00:31:02,149 813 00:31:02,149 --> 00:31:05,190 814 00:31:05,190 --> 00:31:05,200 815 00:31:05,200 --> 00:31:05,590 816 00:31:05,590 --> 00:31:07,430 817 00:31:07,430 --> 00:31:08,950 818 00:31:08,950 --> 00:31:08,960 819 00:31:08,960 --> 00:31:09,990 820 00:31:09,990 --> 00:31:13,909 821 00:31:13,909 --> 00:31:15,830 822 00:31:15,830 --> 00:31:17,750 823 00:31:17,750 --> 00:31:19,750 824 00:31:19,750 --> 00:31:21,830 825 00:31:21,830 --> 00:31:23,669 826 00:31:23,669 --> 00:31:25,990 827 00:31:25,990 --> 00:31:26,789 828 00:31:26,789 --> 00:31:42,870 829 00:31:42,870 --> 00:31:49,110 830 00:31:49,110 --> 00:31:52,230 831 00:31:52,230 --> 00:31:53,750 832 00:31:53,750 --> 00:31:56,230 833 00:31:56,230 --> 00:31:57,590 834 00:31:57,590 --> 00:31:58,950 835 00:31:58,950 --> 00:32:06,310 836 00:32:06,310 --> 00:32:08,070 837 00:32:08,070 --> 00:32:09,590 838 00:32:09,590 --> 00:32:12,310 839 00:32:12,310 --> 00:32:15,190 840 00:32:15,190 --> 00:32:15,200 841 00:32:15,200 --> 00:32:15,669 842 00:32:15,669 --> 00:32:18,710 843 00:32:18,710 --> 00:32:22,149 844 00:32:22,149 --> 00:32:23,990 845 00:32:23,990 --> 00:32:26,070 846 00:32:26,070 --> 00:32:28,950 847 00:32:28,950 --> 00:32:30,470 848 00:32:30,470 --> 00:32:31,830 849 00:32:31,830 --> 00:32:32,789 850 00:32:32,789 --> 00:32:35,509 851 00:32:35,509 --> 00:32:35,519 852 00:32:35,519 --> 00:32:35,909 853 00:32:35,909 --> 00:32:37,350 854 00:32:37,350 --> 00:32:38,950 855 00:32:38,950 --> 00:32:40,149 856 00:32:40,149 --> 00:32:41,990 857 00:32:41,990 --> 00:32:45,190 858 00:32:45,190 --> 00:32:46,870 859 00:32:46,870 --> 00:32:48,710 860 00:32:48,710 --> 00:32:50,950 861 00:32:50,950 --> 00:32:52,549 862 00:32:52,549 --> 00:32:54,710 863 00:32:54,710 --> 00:32:56,230 864 00:32:56,230 --> 00:32:57,990 865 00:32:57,990 --> 00:32:59,669 866 00:32:59,669 --> 00:33:01,669 867 00:33:01,669 --> 00:33:04,230 868 00:33:04,230 --> 00:33:06,630 869 00:33:06,630 --> 00:33:10,149 870 00:33:10,149 --> 00:33:10,159 871 00:33:10,159 --> 00:33:11,269 872 00:33:11,269 --> 00:33:18,230 873 00:33:18,230 --> 00:33:22,149 874 00:33:22,149 --> 00:33:25,750 875 00:33:25,750 --> 00:33:28,789 876 00:33:28,789 --> 00:33:30,230 877 00:33:30,230 --> 00:33:31,830 878 00:33:31,830 --> 00:33:38,070 879 00:33:38,070 --> 00:33:40,870 880 00:33:40,870 --> 00:33:40,880 881 00:33:40,880 --> 00:33:42,070 882 00:33:42,070 --> 00:33:44,789 883 00:33:44,789 --> 00:33:47,430 884 00:33:47,430 --> 00:33:50,070 885 00:33:50,070 --> 00:33:53,909 886 00:33:53,909 --> 00:33:55,509 887 00:33:55,509 --> 00:33:57,669 888 00:33:57,669 --> 00:33:58,950 889 00:33:58,950 --> 00:34:00,710 890 00:34:00,710 --> 00:34:03,190 891 00:34:03,190 --> 00:34:03,200 892 00:34:03,200 --> 00:34:03,830 893 00:34:03,830 --> 00:34:06,070 894 00:34:06,070 --> 00:34:08,149 895 00:34:08,149 --> 00:34:10,389 896 00:34:10,389 --> 00:34:12,310 897 00:34:12,310 --> 00:34:15,589 898 00:34:15,589 --> 00:34:18,629 899 00:34:18,629 --> 00:34:19,990 900 00:34:19,990 --> 00:34:21,990 901 00:34:21,990 --> 00:34:24,950 902 00:34:24,950 --> 00:34:26,629 903 00:34:26,629 --> 00:34:28,790 904 00:34:28,790 --> 00:34:30,069 905 00:34:30,069 --> 00:34:31,909 906 00:34:31,909 --> 00:34:33,349 907 00:34:33,349 --> 00:34:35,829 908 00:34:35,829 --> 00:34:37,430 909 00:34:37,430 --> 00:34:41,109 910 00:34:41,109 --> 00:34:44,470 911 00:34:44,470 --> 00:34:44,480 912 00:34:44,480 --> 00:34:46,829 913 00:34:46,829 --> 00:34:46,839 914 00:34:46,839 --> 00:34:48,710 915 00:34:48,710 --> 00:34:50,790 916 00:34:50,790 --> 00:34:52,230 917 00:34:52,230 --> 00:34:53,990 918 00:34:53,990 --> 00:34:55,589 919 00:34:55,589 --> 00:34:57,510 920 00:34:57,510 --> 00:35:00,150 921 00:35:00,150 --> 00:35:01,510 922 00:35:01,510 --> 00:35:02,710 923 00:35:02,710 --> 00:35:06,069 924 00:35:06,069 --> 00:35:08,310 925 00:35:08,310 --> 00:35:09,589 926 00:35:09,589 --> 00:35:11,589 927 00:35:11,589 --> 00:35:13,510 928 00:35:13,510 --> 00:35:15,910 929 00:35:15,910 --> 00:35:15,920 930 00:35:15,920 --> 00:35:16,710 931 00:35:16,710 --> 00:35:18,230 932 00:35:18,230 --> 00:35:20,390 933 00:35:20,390 --> 00:35:20,400 934 00:35:20,400 --> 00:35:21,190 935 00:35:21,190 --> 00:35:23,510 936 00:35:23,510 --> 00:35:24,710 937 00:35:24,710 --> 00:35:26,870 938 00:35:26,870 --> 00:35:28,550 939 00:35:28,550 --> 00:35:30,310 940 00:35:30,310 --> 00:35:33,030 941 00:35:33,030 --> 00:35:35,510 942 00:35:35,510 --> 00:35:38,870 943 00:35:38,870 --> 00:35:40,310 944 00:35:40,310 --> 00:35:42,870 945 00:35:42,870 --> 00:35:44,870 946 00:35:44,870 --> 00:35:47,430 947 00:35:47,430 --> 00:35:49,270 948 00:35:49,270 --> 00:35:51,109 949 00:35:51,109 --> 00:35:52,790 950 00:35:52,790 --> 00:35:54,310 951 00:35:54,310 --> 00:35:55,829 952 00:35:55,829 --> 00:35:58,150 953 00:35:58,150 --> 00:35:59,910 954 00:35:59,910 --> 00:35:59,920 955 00:35:59,920 --> 00:36:02,550 956 00:36:02,550 --> 00:36:04,870 957 00:36:04,870 --> 00:36:06,470 958 00:36:06,470 --> 00:36:08,069 959 00:36:08,069 --> 00:36:09,750 960 00:36:09,750 --> 00:36:11,109 961 00:36:11,109 --> 00:36:12,310 962 00:36:12,310 --> 00:36:15,750 963 00:36:15,750 --> 00:36:17,750 964 00:36:17,750 --> 00:36:19,750 965 00:36:19,750 --> 00:36:22,069 966 00:36:22,069 --> 00:36:24,829 967 00:36:24,829 --> 00:36:28,230 968 00:36:28,230 --> 00:36:30,950 969 00:36:30,950 --> 00:36:32,950 970 00:36:32,950 --> 00:36:34,950 971 00:36:34,950 --> 00:36:36,550 972 00:36:36,550 --> 00:36:38,069 973 00:36:38,069 --> 00:36:38,079 974 00:36:38,079 --> 00:36:39,430 975 00:36:39,430 --> 00:36:41,589 976 00:36:41,589 --> 00:36:43,349 977 00:36:43,349 --> 00:36:45,430 978 00:36:45,430 --> 00:36:46,790 979 00:36:46,790 --> 00:36:48,950 980 00:36:48,950 --> 00:36:51,270 981 00:36:51,270 --> 00:36:52,310 982 00:36:52,310 --> 00:36:54,310 983 00:36:54,310 --> 00:36:55,430 984 00:36:55,430 --> 00:36:56,710 985 00:36:56,710 --> 00:36:58,790 986 00:36:58,790 --> 00:36:59,910 987 00:36:59,910 --> 00:37:01,589 988 00:37:01,589 --> 00:37:02,710 989 00:37:02,710 --> 00:37:05,670 990 00:37:05,670 --> 00:37:07,990 991 00:37:07,990 --> 00:37:11,349 992 00:37:11,349 --> 00:37:13,510 993 00:37:13,510 --> 00:37:14,710 994 00:37:14,710 --> 00:37:16,390 995 00:37:16,390 --> 00:37:18,310 996 00:37:18,310 --> 00:37:19,670 997 00:37:19,670 --> 00:37:21,190 998 00:37:21,190 --> 00:37:22,790 999 00:37:22,790 --> 00:37:24,710 1000 00:37:24,710 --> 00:37:48,829 1001 00:37:48,829 --> 00:37:48,839 1002 00:37:48,839 --> 00:37:57,829 1003 00:37:57,829 --> 00:38:00,470 1004 00:38:00,470 --> 00:38:00,480 1005 00:38:00,480 --> 00:38:01,270 1006 00:38:01,270 --> 00:38:04,630 1007 00:38:04,630 --> 00:38:05,910 1008 00:38:05,910 --> 00:38:08,150 1009 00:38:08,150 --> 00:38:09,670 1010 00:38:09,670 --> 00:38:11,190 1011 00:38:11,190 --> 00:38:13,750 1012 00:38:13,750 --> 00:38:15,270 1013 00:38:15,270 --> 00:38:15,280 1014 00:38:15,280 --> 00:38:16,710 1015 00:38:16,710 --> 00:38:17,990 1016 00:38:17,990 --> 00:38:20,390 1017 00:38:20,390 --> 00:38:23,109 1018 00:38:23,109 --> 00:38:27,430 1019 00:38:27,430 --> 00:38:35,030 1020 00:38:35,030 --> 00:38:36,630 1021 00:38:36,630 --> 00:38:38,630 1022 00:38:38,630 --> 00:38:41,990 1023 00:38:41,990 --> 00:38:44,710 1024 00:38:44,710 --> 00:38:48,470 1025 00:38:48,470 --> 00:38:58,550 1026 00:38:58,550 --> 00:39:00,710 1027 00:39:00,710 --> 00:39:02,710 1028 00:39:02,710 --> 00:39:03,910 1029 00:39:03,910 --> 00:39:07,109 1030 00:39:07,109 --> 00:39:16,069 1031 00:39:16,069 --> 00:39:20,150 1032 00:39:20,150 --> 00:39:27,510 1033 00:39:27,510 --> 00:39:29,589 1034 00:39:29,589 --> 00:39:31,109 1035 00:39:31,109 --> 00:39:34,550 1036 00:39:34,550 --> 00:39:34,560 1037 00:39:34,560 --> 00:39:35,030 1038 00:39:35,030 --> 00:39:37,990 1039 00:39:37,990 --> 00:39:39,190 1040 00:39:39,190 --> 00:39:41,190 1041 00:39:41,190 --> 00:39:41,200 1042 00:39:41,200 --> 00:39:41,829 1043 00:39:41,829 --> 00:39:44,069 1044 00:39:44,069 --> 00:39:45,910 1045 00:39:45,910 --> 00:39:47,109 1046 00:39:47,109 --> 00:39:49,349 1047 00:39:49,349 --> 00:39:50,790 1048 00:39:50,790 --> 00:39:53,030 1049 00:39:53,030 --> 00:39:54,829 1050 00:39:54,829 --> 00:39:58,390 1051 00:39:58,390 --> 00:39:59,990 1052 00:39:59,990 --> 00:40:01,589 1053 00:40:01,589 --> 00:40:04,790 1054 00:40:04,790 --> 00:40:06,150 1055 00:40:06,150 --> 00:40:09,270 1056 00:40:09,270 --> 00:40:11,910 1057 00:40:11,910 --> 00:40:13,430 1058 00:40:13,430 --> 00:40:14,710 1059 00:40:14,710 --> 00:40:17,190 1060 00:40:17,190 --> 00:40:18,710 1061 00:40:18,710 --> 00:40:18,720 1062 00:40:18,720 --> 00:40:19,349 1063 00:40:19,349 --> 00:40:21,750 1064 00:40:21,750 --> 00:40:22,950 1065 00:40:22,950 --> 00:40:26,309 1066 00:40:26,309 --> 00:40:27,270 1067 00:40:27,270 --> 00:40:28,630 1068 00:40:28,630 --> 00:40:29,910 1069 00:40:29,910 --> 00:40:31,030 1070 00:40:31,030 --> 00:40:34,069 1071 00:40:34,069 --> 00:40:36,069 1072 00:40:36,069 --> 00:40:38,309 1073 00:40:38,309 --> 00:40:40,470 1074 00:40:40,470 --> 00:40:42,870 1075 00:40:42,870 --> 00:40:45,750 1076 00:40:45,750 --> 00:40:45,760 1077 00:40:45,760 --> 00:40:50,790 1078 00:40:50,790 --> 00:40:52,150 1079 00:40:52,150 --> 00:40:54,470 1080 00:40:54,470 --> 00:40:58,150 1081 00:40:58,150 --> 00:41:04,069 1082 00:41:04,069 --> 00:41:09,990 1083 00:41:09,990 --> 00:41:12,710 1084 00:41:12,710 --> 00:41:16,550 1085 00:41:16,550 --> 00:41:18,550 1086 00:41:18,550 --> 00:41:20,069 1087 00:41:20,069 --> 00:41:21,990 1088 00:41:21,990 --> 00:41:23,270 1089 00:41:23,270 --> 00:41:25,430 1090 00:41:25,430 --> 00:41:27,190 1091 00:41:27,190 --> 00:41:28,630 1092 00:41:28,630 --> 00:41:28,640 1093 00:41:28,640 --> 00:41:30,470 1094 00:41:30,470 --> 00:41:34,150 1095 00:41:34,150 --> 00:41:36,069 1096 00:41:36,069 --> 00:41:38,150 1097 00:41:38,150 --> 00:41:39,829 1098 00:41:39,829 --> 00:41:41,750 1099 00:41:41,750 --> 00:41:43,750 1100 00:41:43,750 --> 00:41:46,790 1101 00:41:46,790 --> 00:41:48,150 1102 00:41:48,150 --> 00:41:55,190 1103 00:41:55,190 --> 00:42:00,069 1104 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