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alright so today's class we talked about  vertex unfolding unfolding orthogonal  polyhedra and something else which there  weren't any questions about gone already  coaches rigidity theorem so we will go  through those in turn a bunch of small  questions and to cool new things new  updates so first question is about  vertex unfolding why do we fold things  in a chain and kind of a line linear  style would you get any benefit out of a  tree the answer is four lines I mean  chains are easy to avoid intersection  that's the reason we use them and these  kinds of unfoldings we could dedicate a  different slab for every piece and this  is great and incidentally I didn't  mention this in lecture but it's it's  obvious once you think about it for five  minutes if you want general cuts if  you're allowed to cut anywhere on the  surface and you want a vertex unfolding  then anything is possible because you  can take any surface and chain if I it  in the same way we do with hinged  dissections right you subdivide if you  first you triangulate so BB subdivide  ease triangle into three triangles then  you can hinge it together in a chain and  then you can just lay it out like this  so general vertex unfolding is trivial  for this reason for when you're only  allowed to cut along edges things get  much harder the only technique we have  right now is facet path unfolding but of  course vertex unfolding convex polyhedra  is an easier version of edge unfolding  convex polyhedra but we have no idea how  to do that so maybe it's possible and  probably we already know it's not  possible with chains so if it's possible  it's somehow possible the trees I guess  we know tree m street style unfoldings  can be helpful with general cuttings but  this is not directly relevant to vertex  unfolding but this is a source unfolding  in the star unfolding again so yeah who  knows but so far chains are just a  useful proof technique alright next  question is this is vertex unfolding  summarized in the lecture notes on the  right right side so we end up with a  nice or Larian graph in this case every  vertex  happens to have even degree in general  two of them might have odd degree and we  just get a path but the observation in  this question maybe I've got the other  side so you can read both is that we  visit the same vertex multiple times  like we attach these two triangles at  this vertex we also attach these two  triangles at that vertex is that okay  yes we define it to be okay I think I  briefly mentioned it's defined to be  okay why I don't know it's hard to  imagine little dots of paper there but  somehow you imagine that they can stay  connected it is in particular it's the  natural definition of a hinged  dissection you could add hinges at those  two places but also it's impossible  otherwise this was a fun little puzzle  to think about but if you take oh I  actually have a slide for it forgotten  don't have to draw it's hard to draw an  icosahedron if you take an icosahedron  it has 20 faces and the duel is the  dodecahedron so it has 12 vertices so  there aren't enough to go around if you  wanted at least if you wanted a chain  unfolding but actually any tree  unfolding if you have 20 triangles  you're going to need 19 vertices to  connect them together and there's only  12 vertices so you can't do it I mean  unless yeah that literally true i guess  you could imagine in a general tree  unfolding you could use one point to  connect together many many triangles but  I'm pretty sure it's not possible if  you're allowed to reuse vertices so  that's why we allow it I mean we get a  more interesting result when we allow  revisiting vertices could be there's  still an interesting question like this  which polyhedra have vertex unfoldings  we only get to use a vertex once but I  don't know any results on it all right  next question is any progress on vertex  unfolding and there's one new result  before I showed you think one example of  a vertex unfolding polyhedron this guy  box on a box  initially we said oh this is edge on  unfold the ball but it's also vertex on  unfold abul because this face here it's  donut and the five sides of that cube  have to fit in here and there's not  enough area for them and that's true  also if your vertex unfolding no matter  how this stays attached to the donut  face it's gotta lie in here and that's  not possible but this is an unsatisfying  example because you have a face that is  not a disc and in general you're not  topologically convex which you can't  just move the vertices around and make  this a convex polyhedron not strictly  convex anyway so here is a new example  with Zack able froze so yeah it was he  took this class two years ago or no five  years ago I guess this is a  topologically convex polyhedron so you  could move the vertices to make this  convex that's not so obvious and it is  vertex uh none foldable which we will  prove so this is one of the two new  results for today idiot the example is a  dark prism here and a triangular prism  and they light triangular prism just at  right angles sort of to each other then  you take the union of that solid and  then you take the boundary so first you  can check all the faces are discs and  for topological convexity you want to  check that if you look at any two faces  like this light s 1 and this dark one  they should only share one edge at most  they should share nothing a vertex or an  edge that's topological convexity  because that's how things work in convex  if the polygons were convex so like here  they share just that edge so it is  topologically convex let's prove that it  is vertex uh none foldable so in some  sense this is a strengthening of the  witches hats and stuff because this is  also edge unfold edge uh none foldable  but even cool is vertex on unfold a bowl  turns out let's just look at the  connections between the  arc and the light there are 1 2 3 4  vertices on this side and four vertices  on the back side the Kinect dark to  light and shurz make some star wars  reference address all those vertices are  symmetric so we can just look at any of  them some somewhere light the lights  that has to be attached to the dark side  where could that be let's say vertex a  because it's got to be somewhere and  symmetric so vertex a has three angles  incident to it there's alpha that's this  angle which is big it's reflex there's  this beta and there's this gamma beta is  90 degrees here because this is  rectangles to be a rectangle or yeah  rectangle gamma is well let me stick  what I want from these angles there's  some parameters in the construction I  haven't yet told you so we want alpha to  be bigger than 270 beta equals 90 and we  want alpha plus gamma equal to 360 so  this is possible if you angle the these  guys correctly this is the short version  I guess the most interesting is this  alpha plus gamma equals 360 yeah that  obvious not especially but I'll claim is  true in this picture we'll see it in an  unfolding actually on the very next  slide yeah let's go there still have the  same picture but when unfolded is  supposed to look like like this so okay  so what happens at a if you want to  connect the light side which is alpha  that's forced you can you can attach by  a vertex unfolding to beta or gamma well  beta is impossible because ninety plus  greater than 270 is greater than 360  so you would get overlapped locally at  this vertex if you attach these even if  your vertex unfolding so you've got to  attach to gamma alpha plus gamma is  supposed to be exactly 360 and so it's  actually vertex unfolding doesn't buy  you anything in order to not locally  overlap here they must fit right like  that then the only other condition we  need is that these prisms are long  enough so that you get overlap over here  so it's not a purely local overlap but  it's very close by it's the same two  faces overlap here they have an overlap  here and here if this were shorter if  this would like this short that this is  the length of the prism here then you'd  avoid overlap and that's this picture so  this is an unfolding of one of the two  prism halves this is unfolding the other  if you attach them these vertices as is  necessary you get overlap unless the  prism or short than you would get this  kind of unfolding so that's the proof  you can actually make the proof even  simpler kind of nifty although it's less  obvious that this example exists I mean  you give the coordinates and you can  check it exactly and has all the  properties you want so it's kind of each  prism has been bent a little bit more or  less the same example but now it has the  feature that alpha plus beta prime is  greater than 360 and alpha plus gamma  prime is greater than 360 so you just  can't attach anything even locally you  get pure overlap so easy to check if  your computer easier to check and prove  and verify if your computer less hard  visually or are less easy visually so  that's what's known the open problem  which I should probably write down is if  your faces are convex polygons is there  a vertex on unfold the bowl polyhedron  that we still don't know so that the  witch's hat all the faces are triangles  there's no edge unfolding but it has a  vertex unfolding I think these guys are  nice they're topologically convex but  the faces are still very non convex and  we use that because we want a giant  angle reflex angle together with  something to add up to more than 360 if  you're always adding up to convex angles  you don't get that  but maybe you could combine it with a  polyhedron that's not Hamiltonian so you  have to have three things joining  together and then you could get more  than 360 I don't know it could be an  interesting problem to work on but so  far it is open so that is vertex  unfolding next question is about general  unfolding of non convex polyhedra still  it's open whether every polyhedron say  without boundary has a general unfolding  still we know that all orthogonal  polyhedra of gene is zero have an  unfolding that remains the state of the  art but we have a better way to unfold  orthogonal polyhedra of gina's 0 and  this is a new paper it's not yet online  but it will be shortly just final  version just went to the journal  unfolding orthogonal polyhedra with  quadratic refinement so two of the same  authors Marilla Damian and Robin  flatland from the exponential unfolding  and you may recall we had a we saw  solution in lecture that has exponential  refinement the number of cuts in  particular can be exponential in n in  the worst case this requires you take  the grid you refine it at most quadratic  number of times in each dimension so a  number of cuts is polynomial I guess it  most into the seventh if you cut n  squared by N squared by N squared by n I  think it's actually much less than that  probably more like N squared but I need  to be careful so I want to cover the  solutions it's a simple variation  actually on the previous one but also  there are some questions about how the  exponential one work so I want to go  through that one more time we have new  figures for it from this paper so looks  a little different maybe a little bit  clearer as our goal and I need my hands  so remember the general idea was to  slice your polyhedron in one direction  so you end up with these bands strips  and then you visit the strips in a tree  like fashion  and in particular at the leaves like the  closest to you you've got a you've got a  band and then you've got a front face  and so that would look something like  this little box so you've got a band  that goes around oh sorry which way is  it Bend is this way should be an X and Z  yeah and then there's the front face  which here is drawn on the back those  are the five out of six phases of of the  box that are actually present this this  face that's closest to us is actually  attached to another band okay I mean you  could have leaves on this side or that  side okay so what do we do we enter at s  and we exit at T and we are told that  the first turn you make from s should be  a right turn and the last term before  tea should be a left turn and the you  start in the unfolding you will start  going vertical in s that's why you  should turn right because we never want  to turn left we never want to go left in  the unfolding this is here and we also  want to end in a vertical state 40 so  that we can chain these together okay so  we started s we're told to turn right  which we do here then we're basically  going to zigzag and if you check along  the surface we turn right we go here up  there and then we turn left and then  right and then this is a left because  it's we're supposed to be doing it from  below and then a right with that red  thing then a left right down here and  the left and we're done okay and the  number of balances out so we start  vertical we end vertical and then when  we're done with all these things we  thicken it's not drawn here but these  strips get thickened and you end up  attaching these fatter versions but  basically you're following the same path  we don't think in it at the beginning  because we're actually going to do this  many times on the same cell and then we  abstract that picture to this  iconography which says this I didn't  talk about much in  lecturer says you should start by  turning right in s and then you should  end by turning left and then visiting T  so that's one picture but I mean it  depends where you're going so you can  they're actually eight different  versions I showed you this one but all  those combinations have to happen and  the top ones are when you're all the way  forward bottom ones are where you're all  the way back in these slabs so that's  that's slightly different and then  there's weather s is on the left or s is  on the right relative to T and then  there's whether you start by turning to  the right from s or end by turning to  the left t is always the opposite but  those are the eight possibilities from  three different choices are you far  forward are you far back do you turn  right from s or do you turn left and is  that's the left or zest to the right  that's why you have eight combinations  they all look basically the same but  those are your base cases now so this is  still all review of the old version and  then the last part of the old method is  non leaves so in general you have a band  which is here drawn as two separate  pieces but really this is 11 band a  bands are easy to unfold they go  straight but then you have all these  children so you have your parent which  is the S&T part you have to start here  and end here recursively you want to  visit this whole sub tree so you might  have children that are on the front side  children that are on the back side and  what we do again we're told we should  initially turn right say so if we turn  right we visit the first thing we tell  this leaf or this child actually this is  not necessarily a leaf we tell this  child that the next thing we want to do  is turn to the left because we just  turned right actually we just turned  right twice so we've just gone right  right we will actually be unfolding  initially up so we turn right twice now  we better turn left and then we will end  up  pointing I believe down because  initially this is pointing down after  you do this construction you'll be  pointing down again and that's why you  want to turn left here so you unroll  this thing you can verify you are  actually just zigzagging and so you go  back and forth through these verify this  always proceeds to the right but we know  we're using different iconography here  right this is an S on the left that  initially turns left this is an S on the  right that initially turns right that's  why we need two different different  forms of the recursive call this is a  picture where initially we turn right on  s and s is to the right but there are  different versions of this picture just  like in the base cases where asses to  the left and where you initially have to  turn left but they look the same I mean  you just visit them back and forth like  this at some point you've finished the  front children then you go up and here  it's a little bit shifted that these  would normally be aligned then you have  to visit the top children need to do  that by visiting and going back and  forth then of course when you're done  here you have to unwind everything you  did so then we go over here we visit  that guy back and forth eventually we  will end up on the inside here and  having doubled everything we come back  out to tea ok and all the time we're  unfolding to the right now what can we  say about this well what sucks is all of  these children get visited twice and  that's why you're doubling at every  recursive level and so if your recursion  tree looks like a long binary tree then  you get exponential right not quite  we'll come back to that here's an  example of it being exponential with  this kind of tree it's a little hard to  draw too many levels deeps we have one  level deep and then another level deep  but you'll notice that it's not uniform  right this leaf gets visited  exponentially many  times with this one still only got  visited once of course it's shallower  but even these two leaves are at the  same level they did not get visited the  same in overtime so there's some hope of  improving things as you might imagine so  this is where we go so the observation  there's two observations here two ideas  maybe I'll tell you where we started as  I think can never remember years I was  teaching this class either two or five  years ago and I had idea a number one we  worked on at an open problem session but  it was not enough idea what number one  is just part of the story and it's an  idea that comes from the other padua  class I teach which is heavy light  decomposition in advanced data  structures so this is a very common  technique in advanced data structures  goes back to Slater and karjan you've  ever heard of splay trees or link cut  trees it's that era which was the 80s I  believe 83 it's a very useful tool  whenever you have general tree structure  and you want to make it balanced you  should think heavy light decomposition  they're actually a few ways to balance a  tree but heavy light decomposition is  usually the most powerful so essentially  what it does is say if I could just  treat this thing this path as a single  node and then treat these guys as  separate nodes then I'd kind of be  balanced it would be like one super node  and then a bunch of children hanging off  of it one two three four so that's the  picture I'd like to make I'm going to  contract paths only paths of my tree and  it turns out there's always a way to  take paths can track them into single  notes so the resulting thing has height  at most log n no matter what true you  start with and heavy light decomposition  is a way to do this it's very simple to  define for every node  you define the idea of a heavy child  this is a child with more than half the  weight the weight of a subtree is the  number of descendant notes so the number  of nodes in that sub tree so if we look  at this picture you've got this node it  has one two three four five six seven  eight nine nodes below it this subtree  only has one so this subtree has 77 is  bigger than half of nine right most of  the weight is to the left so we call  this edge heavy similarly this edge is  heavy because this is 55 is more than  half of 7 this one's only three nodes  below it that is still more than half as  half of this is 2.5 i believe so these  this edge is heavy this one is not heavy  so actually in the real picture it would  look like this that's the this is a  heavy pass and this guy's all by itself  so is one more note but it has the same  effect of balancing the tree so what's  the point if I generally have some tree  in general i might have you know many  children no idea how many children are  out there so I mean every node can have  many children these correspond to the  various bands that are attached to my my  parent band and right so that's heavy  child in general though every node has  at most one heavy child because you  can't have more than two halves if I  have more than half of the weight in one  child I can't have more than half the  weight in another child so every now it  has at most one heavy child  these edges connecting nodes and they're  heavy children we call heavy edges and  whenever we have another there's at most  one heavy edge going down from it so  heavy edges form heavy paths that's the  point so this red thing is called a  heavy path cool so what do we know about  heavy paths or actually so the the other  edges i should say these guys we call  light edges this is the heavy light  decomposition so every edge is either  heavy or light could be a node only has  light children because it's well  distributed in the worst case it has one  heavy child one heavy edge coming out of  it so we get heavy paths we don't say  heavy edges warm pads and if we look at  what i normally call the light depth of  a node this is if i look at a node in  this tree and i measure how many light  edges do i have to traverse in order to  get from the root of the tree to that  node here it's only one light edge  number of light edges follow on any path  is it most log n log base two of n  because every time i follow a light edge  I know that I started with say n nodes I  follow a light edge I know that I have  at most half n nodes because my weight  goes down by half if I start with M I  can only reduce by a factor of two login  times before I run out of nodes that I'm  in a leaf when I get down to a way to  one so light depth of any node is at  most log in because weight halves every  every step  very light step so this is why if you  contract the heavy pads to single nodes  like death is or the remain the depth of  the remaining tree is only log n so what  does this tell us if I could somehow  deal with paths efficiently and not feel  the recursive pain that we're getting  here from doubling everything along a  path then I just need then I could chain  them together in the usual bad recursive  way I can afford to double on a balanced  tree so if I could somehow treat these  paths as single things we'd be all set  that was idea number one and that's that  for years I knew I knew this should be  possible but what we were missing was  idea number two it's it's funny it's you  stare at this picture long enough it's  actually kind of kind of obvious it  requires more work after the idea okay  so here same pictures before I spread it  out a little bit but so we end up  visiting this guy last and then we end  up recursively visiting this one a  second time then this one a second time  then this one a second time we go down  here we go round here visit this one a  second time unwind is it this one a  second time this one second time this  one a second time and we're out did I  visit them all a second time  no there's one that I didn't visit a  second time which is the last child I  only visited that one once and that's  all we need is one child because I only  need to do a path efficiently so I'm  recursively visiting most of the  children here those are going to be my  light children the one heavy child I  want to aim to visit that won last  because that one's only visited once so  it doesn't get recursively doubled at  this level of course the higher up at  the level it will get double but there's  only log n levels we're going to get  doubled so will only add a factor of n  okay idea number two is that the last  child gets visited only once at this  recursive level and if you've taken an  algorithms class you know recursively  visiting things once versus twice makes  a big difference here it's a matter of  doubling or just stay in the same so  well if we can arrange to visit these  children first then do this one last  then visit this child first then do this  one last then visit this child first and  do this one last and then do all the  children whatever order then we'll  guarantee that this path doesn't get  recursively refined at this level so  it's really being treated as a single  node basically and that lets you tie  together these unfoldings in an  efficient way the only issue is I need  to be able this construction isn't  enough because this construction  basically forced this one to be the last  child to get visited I need to be able  to control which child is last visited  because the heavy property is a property  of the tree one of these guys maybe this  one is heavy maybe that one's heavy  we've got to be able to end on that so  there are two cases first case is that  we always start from the bottom here and  first case is that the on the top side  is where the last child where the heavy  it is so this is the easy case because  the bottom we don't care we just visit  the children whatever orders natural  then we come up and basically we have  the freedom to spin around here we have  to turn left but we could either choose  to immediately go up here or keep going  and go up here or keep going go up here  that would visit that one too early or  go here it turns out this is the right  choice but we had freedom to just spin  around and choose when to go up just try  them all one of the wound up being last  it's it's I mean basically you want this  guy to be in the middle between where  you start and the one just to the left  here because you're kind of rain bowling  back and forth zeroing in on the last  guy and just by I mean by pigeonhole I  suppose one of these choices will work  because there's k different choices up  there k choices you can make a when to  turn up they each result in a different  guy being last and so one of them  corresponds to the correct one it's easy  to figure out which one it is so if it's  on the top side it's really easy you  just pick it this would be great this  would actually give us linear refinement  if we only had this case the other case  is when the last guy the guy we want to  make last the the heavy child is on the  bottom side in this case we need to do  more work so this is basically the old  construction here we don't really have  freedom from s we can't just bypass this  guy because it will actually get to be  hard to get him later so you've got to  visit him and back and forth but the one  thing we could do is just omit a child  just forget about it for a while don't  visit it visit everything else in  whatever order feels natural to you okay  and then we're going to splice this guy  in so we'll see where's where does the  path go at some point it will go above  this last proposed last guy and what  we're going to do which is on the next  slide is on the return trip so when  we're so remember we have this picture  we come out here then we recursively  double everything so  some point we are coming here doubling  and at that point we turn in but if S&T  are aligned like this we are pointing  the wrong way we wanted to get here but  where it doesn't even look like we're in  the same side but we are we just have to  recursively double everything again in  the end when you recursively double  everything again you will be here and we  haven't finished doubling here but in  the end you will come out here and exit  at T so overall these guys which haven't  been visited yet they'll end up being  doubled from this procedure these guys  wound up being quadrupled so we're doing  for recursive calls for most of the  children some of them might get lucky  and only double but all we care about is  that this guy only gets visited once  that's the key is this this will be  recursively here it's hard to unwind but  then you'll end up on the inside of this  channel then you'll end up visiting this  guy then you'll end up visiting this guy  and then you get out of T so you won't  come back to here that's the key that's  all we need you can write down a  recurrence at this point  if you look at the level of refinement  for an N vertex polyhedron or an end  slab polyhedron would be and n let's say  is the number of nodes in the tree of  slabs this picture in general that is  proportional to the number of vertices  in the polyhedron in the worst case it's  going to be the max of two things are n  minus 1 and 4 times are of n over 2 well  I should say and sub I max over I so  what I mean to say here and this is well  ok this is also technically an n sub I  but this is an upper bound certainly so  if we look at the recursion how much  refinement we need I mean this the worst  refinement over the whole tree is is the  max of all of its sub trees right so  there's a max refinement needed here max  refinement needed here and then also of  course what we need to do within the  node and on the one hand there's the  heavy situation the heavy edge that only  decreases the size of the tree by one  but it only gets recursively visited one  time one times are of n minus 1 there's  the other like children these are all  light children and sub I represents the  number of nodes in that child we know  that this is at most and over 2 because  we know that at most half of the weight  lives in the light child now these get  recursively visited at most four times  up to four times worst case it will be  four times so this term and this  recurrence this this first term doesn't  matter our event is at most are of n  minus 1 yeah big deal that's just saying  our event increases with n so really  this is going to be them and then the  max disappear I mean all we need is this  is at most four times R of n over 2 and  this is a recurrence if you've read clrs  is  follows the master theorem and you know  immediately this is order N squared B  theta N squared but this is an upper  bound in the worst case it will actually  be theta N squared so the number amount  of refinement is at most n squared if  this is not obvious to you an easy way  to imagine it is you start with  something of size n you're visiting four  times something of size n over 2 this is  called a recurrence tree recursion tree  each of these visits four times  something of size n over 4 and so on  obviously log n levels here and at the  bottom you're going to have a bunch of  ones that's when you stop recursing and  if you add up everything level by level  how much work am I doing here and how  much work am I doing here 2n how much  work am I doing here it's harder to add  but there are 24 squared 16 things each  and over four so it's for n in general  it goes up by a factor of two every time  which means you are dominated by the  last layer and if you count it's going  to be N squared because this is this is  2 to the 1 so our 220 to the one to the  two after you get to log in it's two to  the log n times n so that is N squared  and this is a geometric series if you  want to add all these numbers up and so  it's it's 2 times n squared so that's  proof of this bound and that's how you  get quadratic refinement so either idea  is kind of easy but two together pretty  powerful questions about unfolding I  think this algorithm should be pretty  clear one of the questions was are they  practical I think the answer is no has  the same issue of strip folding from  origami but I think they would make  great animations this is still a cool  to see virtually but to I mean do it  physically is probably crazy no one has  tried as far as I know another idea this  is an open problem so orthogonal is  great as all these nice properties can  you generalize it not to arbitrary  polyhedra but to some other kind of  orthogonal such as this is a hex  hexagonal lattice in 3d you I mean it's  just extruded in one dimension but then  you have hex grid and another dimension  you could try to do that my guess is if  you set up the band's to be in the hex  dimensions it just works but I haven't  tried so it's potentially some  low-hanging fruit here there should be  other 3d structures that might work a  typical one in computational geometry is  called see oriented polyhedra we're here  we have three oriented polyhedra there's  three different directions that the  faces can be perpendicular to take ten  of them can you apply the same technique  that gets dicey ER but something like  this I feel like maybe possible I should  try it out but as far as I know this is  all open last question is Cochise  rigidity theorem seems intuitively  obvious or as i read in the notes is  Cochise rigidity theorem obvious and my  answer is no not obvious to me I have  two reasons why it's not obvious one is  it's not true in two dimensions it's  kind of weak statement but if I take a  convex polyhedron in 2d so these are  rigid faces this is the equivalent of a  triangle this is not rigid of course but  so that's one statement and then the  other thing is convexity is necessary  there's this thing Stefan's polyhedron  actually originally it was at connolly  polyhedron we've seen some results by  Connolly he's rigidity expert and then  this was simplified by this guy Klaus  Stefan in the 70s this is a drawing of  it flexing in 3d they I mean non-convex  polyhedra can be flexible they happen to  preserve their volume as they flex but  yeah so that's two which that it's hard  to say why a theorem is not obvious to  me it's not obvious of course if you  play with convex polyhedra they do are  rigid so maybe it's obvious from that  but we've got to prove experimental  evidence mathematically other questions  all right the then I have some fun  videos to show you the first one is the  making of this origami piece which I've  probably shown a picture of before it's  by Brian Chan it's one square paper no  cuts men's at manas with a little crane  instead of that's it normally who knows  and then he also made this nice box and  then the glass is etched by Peter Huck  runs the glass slab mighty  you

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alright so today's class we talked about  vertex unfolding unfolding orthogonal  polyhedra and something else which there  weren't any questions about gone already  coaches rigidity theorem so we will go  through those in turn a bunch of small  questions and to cool new things new  updates so first question is about  vertex unfolding why do we fold things  in a chain and kind of a line linear  style would you get any benefit out of a  tree the answer is four lines I mean  chains are easy to avoid intersection  that's the reason we use them and these  kinds of unfoldings we could dedicate a  different slab for every piece and this  is great and incidentally I didn't  mention this in lecture but it's it's  obvious once you think about it for five  minutes if you want general cuts if  you're allowed to cut anywhere on the  surface and you want a vertex unfolding  then anything is possible because you  can take any surface and chain if I it  in the same way we do with hinged  dissections right you subdivide if you  first you triangulate so BB subdivide  ease triangle into three triangles then  you can hinge it together in a chain and  then you can just lay it out like this  so general vertex unfolding is trivial  for this reason for when you're only  allowed to cut along edges things get  much harder the only technique we have  right now is facet path unfolding but of  course vertex unfolding convex polyhedra  is an easier version of edge unfolding  convex polyhedra but we have no idea how  to do that so maybe it's possible and  probably we already know it's not  possible with chains so if it's possible  it's somehow possible the trees I guess  we know tree m street style unfoldings  can be helpful with general cuttings but  this is not directly relevant to vertex  unfolding but this is a source unfolding  in the star unfolding again so yeah who  knows but so far chains are just a  useful proof technique alright next  question is this is vertex unfolding  summarized in the lecture notes on the  right right side so we end up with a  nice or Larian graph in this case every  vertex  happens to have even degree in general  two of them might have odd degree and we  just get a path but the observation in  this question maybe I've got the other  side so you can read both is that we  visit the same vertex multiple times  like we attach these two triangles at  this vertex we also attach these two  triangles at that vertex is that okay  yes we define it to be okay I think I  briefly mentioned it's defined to be  okay why I don't know it's hard to  imagine little dots of paper there but  somehow you imagine that they can stay  connected it is in particular it's the  natural definition of a hinged  dissection you could add hinges at those  two places but also it's impossible  otherwise this was a fun little puzzle  to think about but if you take oh I  actually have a slide for it forgotten  don't have to draw it's hard to draw an  icosahedron if you take an icosahedron  it has 20 faces and the duel is the  dodecahedron so it has 12 vertices so  there aren't enough to go around if you  wanted at least if you wanted a chain  unfolding but actually any tree  unfolding if you have 20 triangles  you're going to need 19 vertices to  connect them together and there's only  12 vertices so you can't do it I mean  unless yeah that literally true i guess  you could imagine in a general tree  unfolding you could use one point to  connect together many many triangles but  I'm pretty sure it's not possible if  you're allowed to reuse vertices so  that's why we allow it I mean we get a  more interesting result when we allow  revisiting vertices could be there's  still an interesting question like this  which polyhedra have vertex unfoldings  we only get to use a vertex once but I  don't know any results on it all right  next question is any progress on vertex  unfolding and there's one new result  before I showed you think one example of  a vertex unfolding polyhedron this guy  box on a box  initially we said oh this is edge on  unfold the ball but it's also vertex on  unfold abul because this face here it's  donut and the five sides of that cube  have to fit in here and there's not  enough area for them and that's true  also if your vertex unfolding no matter  how this stays attached to the donut  face it's gotta lie in here and that's  not possible but this is an unsatisfying  example because you have a face that is  not a disc and in general you're not  topologically convex which you can't  just move the vertices around and make  this a convex polyhedron not strictly  convex anyway so here is a new example  with Zack able froze so yeah it was he  took this class two years ago or no five  years ago I guess this is a  topologically convex polyhedron so you  could move the vertices to make this  convex that's not so obvious and it is  vertex uh none foldable which we will  prove so this is one of the two new  results for today idiot the example is a  dark prism here and a triangular prism  and they light triangular prism just at  right angles sort of to each other then  you take the union of that solid and  then you take the boundary so first you  can check all the faces are discs and  for topological convexity you want to  check that if you look at any two faces  like this light s 1 and this dark one  they should only share one edge at most  they should share nothing a vertex or an  edge that's topological convexity  because that's how things work in convex  if the polygons were convex so like here  they share just that edge so it is  topologically convex let's prove that it  is vertex uh none foldable so in some  sense this is a strengthening of the  witches hats and stuff because this is  also edge unfold edge uh none foldable  but even cool is vertex on unfold a bowl  turns out let's just look at the  connections between the  arc and the light there are 1 2 3 4  vertices on this side and four vertices  on the back side the Kinect dark to  light and shurz make some star wars  reference address all those vertices are  symmetric so we can just look at any of  them some somewhere light the lights  that has to be attached to the dark side  where could that be let's say vertex a  because it's got to be somewhere and  symmetric so vertex a has three angles  incident to it there's alpha that's this  angle which is big it's reflex there's  this beta and there's this gamma beta is  90 degrees here because this is  rectangles to be a rectangle or yeah  rectangle gamma is well let me stick  what I want from these angles there's  some parameters in the construction I  haven't yet told you so we want alpha to  be bigger than 270 beta equals 90 and we  want alpha plus gamma equal to 360 so  this is possible if you angle the these  guys correctly this is the short version  I guess the most interesting is this  alpha plus gamma equals 360 yeah that  obvious not especially but I'll claim is  true in this picture we'll see it in an  unfolding actually on the very next  slide yeah let's go there still have the  same picture but when unfolded is  supposed to look like like this so okay  so what happens at a if you want to  connect the light side which is alpha  that's forced you can you can attach by  a vertex unfolding to beta or gamma well  beta is impossible because ninety plus  greater than 270 is greater than 360  so you would get overlapped locally at  this vertex if you attach these even if  your vertex unfolding so you've got to  attach to gamma alpha plus gamma is  supposed to be exactly 360 and so it's  actually vertex unfolding doesn't buy  you anything in order to not locally  overlap here they must fit right like  that then the only other condition we  need is that these prisms are long  enough so that you get overlap over here  so it's not a purely local overlap but  it's very close by it's the same two  faces overlap here they have an overlap  here and here if this were shorter if  this would like this short that this is  the length of the prism here then you'd  avoid overlap and that's this picture so  this is an unfolding of one of the two  prism halves this is unfolding the other  if you attach them these vertices as is  necessary you get overlap unless the  prism or short than you would get this  kind of unfolding so that's the proof  you can actually make the proof even  simpler kind of nifty although it's less  obvious that this example exists I mean  you give the coordinates and you can  check it exactly and has all the  properties you want so it's kind of each  prism has been bent a little bit more or  less the same example but now it has the  feature that alpha plus beta prime is  greater than 360 and alpha plus gamma  prime is greater than 360 so you just  can't attach anything even locally you  get pure overlap so easy to check if  your computer easier to check and prove  and verify if your computer less hard  visually or are less easy visually so  that's what's known the open problem  which I should probably write down is if  your faces are convex polygons is there  a vertex on unfold the bowl polyhedron  that we still don't know so that the  witch's hat all the faces are triangles  there's no edge unfolding but it has a  vertex unfolding I think these guys are  nice they're topologically convex but  the faces are still very non convex and  we use that because we want a giant  angle reflex angle together with  something to add up to more than 360 if  you're always adding up to convex angles  you don't get that  but maybe you could combine it with a  polyhedron that's not Hamiltonian so you  have to have three things joining  together and then you could get more  than 360 I don't know it could be an  interesting problem to work on but so  far it is open so that is vertex  unfolding next question is about general  unfolding of non convex polyhedra still  it's open whether every polyhedron say  without boundary has a general unfolding  still we know that all orthogonal  polyhedra of gene is zero have an  unfolding that remains the state of the  art but we have a better way to unfold  orthogonal polyhedra of gina's 0 and  this is a new paper it's not yet online  but it will be shortly just final  version just went to the journal  unfolding orthogonal polyhedra with  quadratic refinement so two of the same  authors Marilla Damian and Robin  flatland from the exponential unfolding  and you may recall we had a we saw  solution in lecture that has exponential  refinement the number of cuts in  particular can be exponential in n in  the worst case this requires you take  the grid you refine it at most quadratic  number of times in each dimension so a  number of cuts is polynomial I guess it  most into the seventh if you cut n  squared by N squared by N squared by n I  think it's actually much less than that  probably more like N squared but I need  to be careful so I want to cover the  solutions it's a simple variation  actually on the previous one but also  there are some questions about how the  exponential one work so I want to go  through that one more time we have new  figures for it from this paper so looks  a little different maybe a little bit  clearer as our goal and I need my hands  so remember the general idea was to  slice your polyhedron in one direction  so you end up with these bands strips  and then you visit the strips in a tree  like fashion  and in particular at the leaves like the  closest to you you've got a you've got a  band and then you've got a front face  and so that would look something like  this little box so you've got a band  that goes around oh sorry which way is  it Bend is this way should be an X and Z  yeah and then there's the front face  which here is drawn on the back those  are the five out of six phases of of the  box that are actually present this this  face that's closest to us is actually  attached to another band okay I mean you  could have leaves on this side or that  side okay so what do we do we enter at s  and we exit at T and we are told that  the first turn you make from s should be  a right turn and the last term before  tea should be a left turn and the you  start in the unfolding you will start  going vertical in s that's why you  should turn right because we never want  to turn left we never want to go left in  the unfolding this is here and we also  want to end in a vertical state 40 so  that we can chain these together okay so  we started s we're told to turn right  which we do here then we're basically  going to zigzag and if you check along  the surface we turn right we go here up  there and then we turn left and then  right and then this is a left because  it's we're supposed to be doing it from  below and then a right with that red  thing then a left right down here and  the left and we're done okay and the  number of balances out so we start  vertical we end vertical and then when  we're done with all these things we  thicken it's not drawn here but these  strips get thickened and you end up  attaching these fatter versions but  basically you're following the same path  we don't think in it at the beginning  because we're actually going to do this  many times on the same cell and then we  abstract that picture to this  iconography which says this I didn't  talk about much in  lecturer says you should start by  turning right in s and then you should  end by turning left and then visiting T  so that's one picture but I mean it  depends where you're going so you can  they're actually eight different  versions I showed you this one but all  those combinations have to happen and  the top ones are when you're all the way  forward bottom ones are where you're all  the way back in these slabs so that's  that's slightly different and then  there's weather s is on the left or s is  on the right relative to T and then  there's whether you start by turning to  the right from s or end by turning to  the left t is always the opposite but  those are the eight possibilities from  three different choices are you far  forward are you far back do you turn  right from s or do you turn left and is  that's the left or zest to the right  that's why you have eight combinations  they all look basically the same but  those are your base cases now so this is  still all review of the old version and  then the last part of the old method is  non leaves so in general you have a band  which is here drawn as two separate  pieces but really this is 11 band a  bands are easy to unfold they go  straight but then you have all these  children so you have your parent which  is the S&T part you have to start here  and end here recursively you want to  visit this whole sub tree so you might  have children that are on the front side  children that are on the back side and  what we do again we're told we should  initially turn right say so if we turn  right we visit the first thing we tell  this leaf or this child actually this is  not necessarily a leaf we tell this  child that the next thing we want to do  is turn to the left because we just  turned right actually we just turned  right twice so we've just gone right  right we will actually be unfolding  initially up so we turn right twice now  we better turn left and then we will end  up  pointing I believe down because  initially this is pointing down after  you do this construction you'll be  pointing down again and that's why you  want to turn left here so you unroll  this thing you can verify you are  actually just zigzagging and so you go  back and forth through these verify this  always proceeds to the right but we know  we're using different iconography here  right this is an S on the left that  initially turns left this is an S on the  right that initially turns right that's  why we need two different different  forms of the recursive call this is a  picture where initially we turn right on  s and s is to the right but there are  different versions of this picture just  like in the base cases where asses to  the left and where you initially have to  turn left but they look the same I mean  you just visit them back and forth like  this at some point you've finished the  front children then you go up and here  it's a little bit shifted that these  would normally be aligned then you have  to visit the top children need to do  that by visiting and going back and  forth then of course when you're done  here you have to unwind everything you  did so then we go over here we visit  that guy back and forth eventually we  will end up on the inside here and  having doubled everything we come back  out to tea ok and all the time we're  unfolding to the right now what can we  say about this well what sucks is all of  these children get visited twice and  that's why you're doubling at every  recursive level and so if your recursion  tree looks like a long binary tree then  you get exponential right not quite  we'll come back to that here's an  example of it being exponential with  this kind of tree it's a little hard to  draw too many levels deeps we have one  level deep and then another level deep  but you'll notice that it's not uniform  right this leaf gets visited  exponentially many  times with this one still only got  visited once of course it's shallower  but even these two leaves are at the  same level they did not get visited the  same in overtime so there's some hope of  improving things as you might imagine so  this is where we go so the observation  there's two observations here two ideas  maybe I'll tell you where we started as  I think can never remember years I was  teaching this class either two or five  years ago and I had idea a number one we  worked on at an open problem session but  it was not enough idea what number one  is just part of the story and it's an  idea that comes from the other padua  class I teach which is heavy light  decomposition in advanced data  structures so this is a very common  technique in advanced data structures  goes back to Slater and karjan you've  ever heard of splay trees or link cut  trees it's that era which was the 80s I  believe 83 it's a very useful tool  whenever you have general tree structure  and you want to make it balanced you  should think heavy light decomposition  they're actually a few ways to balance a  tree but heavy light decomposition is  usually the most powerful so essentially  what it does is say if I could just  treat this thing this path as a single  node and then treat these guys as  separate nodes then I'd kind of be  balanced it would be like one super node  and then a bunch of children hanging off  of it one two three four so that's the  picture I'd like to make I'm going to  contract paths only paths of my tree and  it turns out there's always a way to  take paths can track them into single  notes so the resulting thing has height  at most log n no matter what true you  start with and heavy light decomposition  is a way to do this it's very simple to  define for every node  you define the idea of a heavy child  this is a child with more than half the  weight the weight of a subtree is the  number of descendant notes so the number  of nodes in that sub tree so if we look  at this picture you've got this node it  has one two three four five six seven  eight nine nodes below it this subtree  only has one so this subtree has 77 is  bigger than half of nine right most of  the weight is to the left so we call  this edge heavy similarly this edge is  heavy because this is 55 is more than  half of 7 this one's only three nodes  below it that is still more than half as  half of this is 2.5 i believe so these  this edge is heavy this one is not heavy  so actually in the real picture it would  look like this that's the this is a  heavy pass and this guy's all by itself  so is one more note but it has the same  effect of balancing the tree so what's  the point if I generally have some tree  in general i might have you know many  children no idea how many children are  out there so I mean every node can have  many children these correspond to the  various bands that are attached to my my  parent band and right so that's heavy  child in general though every node has  at most one heavy child because you  can't have more than two halves if I  have more than half of the weight in one  child I can't have more than half the  weight in another child so every now it  has at most one heavy child  these edges connecting nodes and they're  heavy children we call heavy edges and  whenever we have another there's at most  one heavy edge going down from it so  heavy edges form heavy paths that's the  point so this red thing is called a  heavy path cool so what do we know about  heavy paths or actually so the the other  edges i should say these guys we call  light edges this is the heavy light  decomposition so every edge is either  heavy or light could be a node only has  light children because it's well  distributed in the worst case it has one  heavy child one heavy edge coming out of  it so we get heavy paths we don't say  heavy edges warm pads and if we look at  what i normally call the light depth of  a node this is if i look at a node in  this tree and i measure how many light  edges do i have to traverse in order to  get from the root of the tree to that  node here it's only one light edge  number of light edges follow on any path  is it most log n log base two of n  because every time i follow a light edge  I know that I started with say n nodes I  follow a light edge I know that I have  at most half n nodes because my weight  goes down by half if I start with M I  can only reduce by a factor of two login  times before I run out of nodes that I'm  in a leaf when I get down to a way to  one so light depth of any node is at  most log in because weight halves every  every step  very light step so this is why if you  contract the heavy pads to single nodes  like death is or the remain the depth of  the remaining tree is only log n so what  does this tell us if I could somehow  deal with paths efficiently and not feel  the recursive pain that we're getting  here from doubling everything along a  path then I just need then I could chain  them together in the usual bad recursive  way I can afford to double on a balanced  tree so if I could somehow treat these  paths as single things we'd be all set  that was idea number one and that's that  for years I knew I knew this should be  possible but what we were missing was  idea number two it's it's funny it's you  stare at this picture long enough it's  actually kind of kind of obvious it  requires more work after the idea okay  so here same pictures before I spread it  out a little bit but so we end up  visiting this guy last and then we end  up recursively visiting this one a  second time then this one a second time  then this one a second time we go down  here we go round here visit this one a  second time unwind is it this one a  second time this one second time this  one a second time and we're out did I  visit them all a second time  no there's one that I didn't visit a  second time which is the last child I  only visited that one once and that's  all we need is one child because I only  need to do a path efficiently so I'm  recursively visiting most of the  children here those are going to be my  light children the one heavy child I  want to aim to visit that won last  because that one's only visited once so  it doesn't get recursively doubled at  this level of course the higher up at  the level it will get double but there's  only log n levels we're going to get  doubled so will only add a factor of n  okay idea number two is that the last  child gets visited only once at this  recursive level and if you've taken an  algorithms class you know recursively  visiting things once versus twice makes  a big difference here it's a matter of  doubling or just stay in the same so  well if we can arrange to visit these  children first then do this one last  then visit this child first then do this  one last then visit this child first and  do this one last and then do all the  children whatever order then we'll  guarantee that this path doesn't get  recursively refined at this level so  it's really being treated as a single  node basically and that lets you tie  together these unfoldings in an  efficient way the only issue is I need  to be able this construction isn't  enough because this construction  basically forced this one to be the last  child to get visited I need to be able  to control which child is last visited  because the heavy property is a property  of the tree one of these guys maybe this  one is heavy maybe that one's heavy  we've got to be able to end on that so  there are two cases first case is that  we always start from the bottom here and  first case is that the on the top side  is where the last child where the heavy  it is so this is the easy case because  the bottom we don't care we just visit  the children whatever orders natural  then we come up and basically we have  the freedom to spin around here we have  to turn left but we could either choose  to immediately go up here or keep going  and go up here or keep going go up here  that would visit that one too early or  go here it turns out this is the right  choice but we had freedom to just spin  around and choose when to go up just try  them all one of the wound up being last  it's it's I mean basically you want this  guy to be in the middle between where  you start and the one just to the left  here because you're kind of rain bowling  back and forth zeroing in on the last  guy and just by I mean by pigeonhole I  suppose one of these choices will work  because there's k different choices up  there k choices you can make a when to  turn up they each result in a different  guy being last and so one of them  corresponds to the correct one it's easy  to figure out which one it is so if it's  on the top side it's really easy you  just pick it this would be great this  would actually give us linear refinement  if we only had this case the other case  is when the last guy the guy we want to  make last the the heavy child is on the  bottom side in this case we need to do  more work so this is basically the old  construction here we don't really have  freedom from s we can't just bypass this  guy because it will actually get to be  hard to get him later so you've got to  visit him and back and forth but the one  thing we could do is just omit a child  just forget about it for a while don't  visit it visit everything else in  whatever order feels natural to you okay  and then we're going to splice this guy  in so we'll see where's where does the  path go at some point it will go above  this last proposed last guy and what  we're going to do which is on the next  slide is on the return trip so when  we're so remember we have this picture  we come out here then we recursively  double everything so  some point we are coming here doubling  and at that point we turn in but if S&T  are aligned like this we are pointing  the wrong way we wanted to get here but  where it doesn't even look like we're in  the same side but we are we just have to  recursively double everything again in  the end when you recursively double  everything again you will be here and we  haven't finished doubling here but in  the end you will come out here and exit  at T so overall these guys which haven't  been visited yet they'll end up being  doubled from this procedure these guys  wound up being quadrupled so we're doing  for recursive calls for most of the  children some of them might get lucky  and only double but all we care about is  that this guy only gets visited once  that's the key is this this will be  recursively here it's hard to unwind but  then you'll end up on the inside of this  channel then you'll end up visiting this  guy then you'll end up visiting this guy  and then you get out of T so you won't  come back to here that's the key that's  all we need you can write down a  recurrence at this point  if you look at the level of refinement  for an N vertex polyhedron or an end  slab polyhedron would be and n let's say  is the number of nodes in the tree of  slabs this picture in general that is  proportional to the number of vertices  in the polyhedron in the worst case it's  going to be the max of two things are n  minus 1 and 4 times are of n over 2 well  I should say and sub I max over I so  what I mean to say here and this is well  ok this is also technically an n sub I  but this is an upper bound certainly so  if we look at the recursion how much  refinement we need I mean this the worst  refinement over the whole tree is is the  max of all of its sub trees right so  there's a max refinement needed here max  refinement needed here and then also of  course what we need to do within the  node and on the one hand there's the  heavy situation the heavy edge that only  decreases the size of the tree by one  but it only gets recursively visited one  time one times are of n minus 1 there's  the other like children these are all  light children and sub I represents the  number of nodes in that child we know  that this is at most and over 2 because  we know that at most half of the weight  lives in the light child now these get  recursively visited at most four times  up to four times worst case it will be  four times so this term and this  recurrence this this first term doesn't  matter our event is at most are of n  minus 1 yeah big deal that's just saying  our event increases with n so really  this is going to be them and then the  max disappear I mean all we need is this  is at most four times R of n over 2 and  this is a recurrence if you've read clrs  is  follows the master theorem and you know  immediately this is order N squared B  theta N squared but this is an upper  bound in the worst case it will actually  be theta N squared so the number amount  of refinement is at most n squared if  this is not obvious to you an easy way  to imagine it is you start with  something of size n you're visiting four  times something of size n over 2 this is  called a recurrence tree recursion tree  each of these visits four times  something of size n over 4 and so on  obviously log n levels here and at the  bottom you're going to have a bunch of  ones that's when you stop recursing and  if you add up everything level by level  how much work am I doing here and how  much work am I doing here 2n how much  work am I doing here it's harder to add  but there are 24 squared 16 things each  and over four so it's for n in general  it goes up by a factor of two every time  which means you are dominated by the  last layer and if you count it's going  to be N squared because this is this is  2 to the 1 so our 220 to the one to the  two after you get to log in it's two to  the log n times n so that is N squared  and this is a geometric series if you  want to add all these numbers up and so  it's it's 2 times n squared so that's  proof of this bound and that's how you  get quadratic refinement so either idea  is kind of easy but two together pretty  powerful questions about unfolding I  think this algorithm should be pretty  clear one of the questions was are they  practical I think the answer is no has  the same issue of strip folding from  origami but I think they would make  great animations this is still a cool  to see virtually but to I mean do it  physically is probably crazy no one has  tried as far as I know another idea this  is an open problem so orthogonal is  great as all these nice properties can  you generalize it not to arbitrary  polyhedra but to some other kind of  orthogonal such as this is a hex  hexagonal lattice in 3d you I mean it's  just extruded in one dimension but then  you have hex grid and another dimension  you could try to do that my guess is if  you set up the band's to be in the hex  dimensions it just works but I haven't  tried so it's potentially some  low-hanging fruit here there should be  other 3d structures that might work a  typical one in computational geometry is  called see oriented polyhedra we're here  we have three oriented polyhedra there's  three different directions that the  faces can be perpendicular to take ten  of them can you apply the same technique  that gets dicey ER but something like  this I feel like maybe possible I should  try it out but as far as I know this is  all open last question is Cochise  rigidity theorem seems intuitively  obvious or as i read in the notes is  Cochise rigidity theorem obvious and my  answer is no not obvious to me I have  two reasons why it's not obvious one is  it's not true in two dimensions it's  kind of weak statement but if I take a  convex polyhedron in 2d so these are  rigid faces this is the equivalent of a  triangle this is not rigid of course but  so that's one statement and then the  other thing is convexity is necessary  there's this thing Stefan's polyhedron  actually originally it was at connolly  polyhedron we've seen some results by  Connolly he's rigidity expert and then  this was simplified by this guy Klaus  Stefan in the 70s this is a drawing of  it flexing in 3d they I mean non-convex  polyhedra can be flexible they happen to  preserve their volume as they flex but  yeah so that's two which that it's hard  to say why a theorem is not obvious to  me it's not obvious of course if you  play with convex polyhedra they do are  rigid so maybe it's obvious from that  but we've got to prove experimental  evidence mathematically other questions  all right the then I have some fun  videos to show you the first one is the  making of this origami piece which I've  probably shown a picture of before it's  by Brian Chan it's one square paper no  cuts men's at manas with a little crane  instead of that's it normally who knows  and then he also made this nice box and  then the glass is etched by Peter Huck  runs the glass slab mighty  you
 

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394
00:17:43,570 --> 00:17:46,030

 

395
00:17:46,030 --> 00:17:46,040

 

396
00:17:46,040 --> 00:17:48,030


397
00:17:48,030 --> 00:17:51,970

 

398
00:17:51,970 --> 00:17:53,980

 

399
00:17:53,980 --> 00:17:54,970

 

400
00:17:54,970 --> 00:17:56,650

 

401
00:17:56,650 --> 00:17:58,750

 

402
00:17:58,750 --> 00:18:00,700

 

403
00:18:00,700 --> 00:18:04,150

 

404
00:18:04,150 --> 00:18:06,310

 

405
00:18:06,310 --> 00:18:08,500

 

406
00:18:08,500 --> 00:18:10,840

 

407
00:18:10,840 --> 00:18:12,610

 

408
00:18:12,610 --> 00:18:15,130

 

409
00:18:15,130 --> 00:18:17,530

 

410
00:18:17,530 --> 00:18:20,230

 

411
00:18:20,230 --> 00:18:22,600

 

412
00:18:22,600 --> 00:18:24,070

 

413
00:18:24,070 --> 00:18:25,870

 

414
00:18:25,870 --> 00:18:28,030

 

415
00:18:28,030 --> 00:18:30,100

 

416
00:18:30,100 --> 00:18:31,450

 

417
00:18:31,450 --> 00:18:33,850

 

418
00:18:33,850 --> 00:18:35,530

 

419
00:18:35,530 --> 00:18:37,660

 

420
00:18:37,660 --> 00:18:40,990

 

421
00:18:40,990 --> 00:18:42,280

 

422
00:18:42,280 --> 00:18:45,580

 

423
00:18:45,580 --> 00:18:47,110

 

424
00:18:47,110 --> 00:18:49,810

 

425
00:18:49,810 --> 00:18:53,020

 

426
00:18:53,020 --> 00:18:55,540

 

427
00:18:55,540 --> 00:18:59,230

 

428
00:18:59,230 --> 00:19:01,450

 

429
00:19:01,450 --> 00:19:03,190

 

430
00:19:03,190 --> 00:19:05,530

 

431
00:19:05,530 --> 00:19:08,440

 

432
00:19:08,440 --> 00:19:12,370

 

433
00:19:12,370 --> 00:19:16,870

 

434
00:19:16,870 --> 00:19:19,510

 

435
00:19:19,510 --> 00:19:20,860

 

436
00:19:20,860 --> 00:19:23,320

 

437
00:19:23,320 --> 00:19:33,490

 

438
00:19:33,490 --> 00:19:40,750

 

439
00:19:40,750 --> 00:19:43,750

 

440
00:19:43,750 --> 00:19:45,430

 

441
00:19:45,430 --> 00:19:46,660

 

442
00:19:46,660 --> 00:19:49,390

 

443
00:19:49,390 --> 00:19:52,590

 

444
00:19:52,590 --> 00:19:55,810

 

445
00:19:55,810 --> 00:19:58,780

 

446
00:19:58,780 --> 00:19:59,680

 

447
00:19:59,680 --> 00:20:00,970

 

448
00:20:00,970 --> 00:20:03,749

 

449
00:20:03,749 --> 00:20:05,919

 

450
00:20:05,919 --> 00:20:07,570

 

451
00:20:07,570 --> 00:20:09,340

 

452
00:20:09,340 --> 00:20:14,460

 

453
00:20:14,460 --> 00:20:17,320

 

454
00:20:17,320 --> 00:20:21,159

 

455
00:20:21,159 --> 00:20:25,840

 

456
00:20:25,840 --> 00:20:30,940

 

457
00:20:30,940 --> 00:20:33,070

 

458
00:20:33,070 --> 00:20:38,950

 

459
00:20:38,950 --> 00:20:40,419

 

460
00:20:40,419 --> 00:20:42,039

 

461
00:20:42,039 --> 00:20:44,649

 

462
00:20:44,649 --> 00:20:46,810

 

463
00:20:46,810 --> 00:20:48,340

 

464
00:20:48,340 --> 00:20:50,470

 

465
00:20:50,470 --> 00:20:52,450

 

466
00:20:52,450 --> 00:20:54,090

 

467
00:20:54,090 --> 00:20:57,850

 

468
00:20:57,850 --> 00:21:00,430

 

469
00:21:00,430 --> 00:21:03,610

 

470
00:21:03,610 --> 00:21:07,720

 

471
00:21:07,720 --> 00:21:10,509

 

472
00:21:10,509 --> 00:21:12,549

 

473
00:21:12,549 --> 00:21:14,110

 

474
00:21:14,110 --> 00:21:16,119

 

475
00:21:16,119 --> 00:21:18,070

 

476
00:21:18,070 --> 00:21:20,980

 

477
00:21:20,980 --> 00:21:27,850

 

478
00:21:27,850 --> 00:21:31,299

 

479
00:21:31,299 --> 00:21:33,129

 

480
00:21:33,129 --> 00:21:36,070

 

481
00:21:36,070 --> 00:21:38,110

 

482
00:21:38,110 --> 00:21:40,299

 

483
00:21:40,299 --> 00:21:44,919

 

484
00:21:44,919 --> 00:21:47,350

 

485
00:21:47,350 --> 00:21:51,100

 

486
00:21:51,100 --> 00:21:52,299

 

487
00:21:52,299 --> 00:21:54,129

 

488
00:21:54,129 --> 00:21:56,710

 

489
00:21:56,710 --> 00:22:02,169

 

490
00:22:02,169 --> 00:22:04,269

 

491
00:22:04,269 --> 00:22:06,220

 

492
00:22:06,220 --> 00:22:12,040

 

493
00:22:12,040 --> 00:22:19,150

 

494
00:22:19,150 --> 00:22:28,520

 

495
00:22:28,520 --> 00:22:34,130

 

496
00:22:34,130 --> 00:22:40,190

 

497
00:22:40,190 --> 00:22:45,080

 

498
00:22:45,080 --> 00:22:47,750

 

499
00:22:47,750 --> 00:22:49,940

 

500
00:22:49,940 --> 00:22:53,990

 

501
00:22:53,990 --> 00:22:58,220

 

502
00:22:58,220 --> 00:23:00,740

 

503
00:23:00,740 --> 00:23:03,230

 

504
00:23:03,230 --> 00:23:06,950

 

505
00:23:06,950 --> 00:23:11,420

 

506
00:23:11,420 --> 00:23:14,180

 

507
00:23:14,180 --> 00:23:17,990

 

508
00:23:17,990 --> 00:23:21,350

 

509
00:23:21,350 --> 00:23:25,190

 

510
00:23:25,190 --> 00:23:30,320

 

511
00:23:30,320 --> 00:23:32,060

 

512
00:23:32,060 --> 00:23:34,940

 

513
00:23:34,940 --> 00:23:37,160

 

514
00:23:37,160 --> 00:23:41,120

 

515
00:23:41,120 --> 00:23:44,570

 

516
00:23:44,570 --> 00:23:46,160

 

517
00:23:46,160 --> 00:23:49,220

 

518
00:23:49,220 --> 00:23:51,920

 

519
00:23:51,920 --> 00:23:53,660

 

520
00:23:53,660 --> 00:23:57,710

 

521
00:23:57,710 --> 00:24:03,200

 

522
00:24:03,200 --> 00:24:07,880

 

523
00:24:07,880 --> 00:24:13,010

 

524
00:24:13,010 --> 00:24:16,310

 

525
00:24:16,310 --> 00:24:17,750

 

526
00:24:17,750 --> 00:24:19,520

 

527
00:24:19,520 --> 00:24:22,550

 

528
00:24:22,550 --> 00:24:24,620

 

529
00:24:24,620 --> 00:24:27,140

 

530
00:24:27,140 --> 00:24:33,790

 

531
00:24:33,790 --> 00:24:36,170

 

532
00:24:36,170 --> 00:24:38,780

 

533
00:24:38,780 --> 00:24:41,690

 

534
00:24:41,690 --> 00:24:45,140

 

535
00:24:45,140 --> 00:24:52,310

 

536
00:24:52,310 --> 00:24:56,180

 

537
00:24:56,180 --> 00:24:59,510

 

538
00:24:59,510 --> 00:25:02,060

 

539
00:25:02,060 --> 00:25:03,740

 

540
00:25:03,740 --> 00:25:06,770

 

541
00:25:06,770 --> 00:25:08,000

 

542
00:25:08,000 --> 00:25:09,980

 

543
00:25:09,980 --> 00:25:12,110

 

544
00:25:12,110 --> 00:25:17,270

 

545
00:25:17,270 --> 00:25:26,330

 

546
00:25:26,330 --> 00:25:28,610

 

547
00:25:28,610 --> 00:25:37,040

 

548
00:25:37,040 --> 00:25:40,280

 

549
00:25:40,280 --> 00:25:42,500

 

550
00:25:42,500 --> 00:25:44,120

 

551
00:25:44,120 --> 00:25:46,060

 

552
00:25:46,060 --> 00:25:49,490

 

553
00:25:49,490 --> 00:25:53,830

 

554
00:25:53,830 --> 00:25:57,050

 

555
00:25:57,050 --> 00:26:00,200

 

556
00:26:00,200 --> 00:26:02,450

 

557
00:26:02,450 --> 00:26:05,780

 

558
00:26:05,780 --> 00:26:08,120

 

559
00:26:08,120 --> 00:26:11,090

 

560
00:26:11,090 --> 00:26:12,950

 

561
00:26:12,950 --> 00:26:14,810

 

562
00:26:14,810 --> 00:26:18,260

 

563
00:26:18,260 --> 00:26:21,500

 

564
00:26:21,500 --> 00:26:28,259

 

565
00:26:28,259 --> 00:26:35,710

 

566
00:26:35,710 --> 00:26:37,779

 

567
00:26:37,779 --> 00:26:41,350

 

568
00:26:41,350 --> 00:26:43,810

 

569
00:26:43,810 --> 00:26:45,549

 

570
00:26:45,549 --> 00:26:50,649

 

571
00:26:50,649 --> 00:26:53,259

 

572
00:26:53,259 --> 00:26:55,090

 

573
00:26:55,090 --> 00:26:58,629

 

574
00:26:58,629 --> 00:27:00,669

 

575
00:27:00,669 --> 00:27:02,680

 

576
00:27:02,680 --> 00:27:05,139

 

577
00:27:05,139 --> 00:27:07,269

 

578
00:27:07,269 --> 00:27:09,489

 

579
00:27:09,489 --> 00:27:11,680

 

580
00:27:11,680 --> 00:27:14,529

 

581
00:27:14,529 --> 00:27:19,869

 

582
00:27:19,869 --> 00:27:21,159

 

583
00:27:21,159 --> 00:27:23,649

 

584
00:27:23,649 --> 00:27:26,769

 

585
00:27:26,769 --> 00:27:30,249

 

586
00:27:30,249 --> 00:27:33,759

 

587
00:27:33,759 --> 00:27:36,430

 

588
00:27:36,430 --> 00:27:38,200

 

589
00:27:38,200 --> 00:27:40,450

 

590
00:27:40,450 --> 00:27:42,940

 

591
00:27:42,940 --> 00:27:46,989

 

592
00:27:46,989 --> 00:27:50,440

 

593
00:27:50,440 --> 00:27:53,169

 

594
00:27:53,169 --> 00:27:55,479

 

595
00:27:55,479 --> 00:28:03,010

 

596
00:28:03,010 --> 00:28:07,630

 

597
00:28:07,630 --> 00:28:11,160

 

598
00:28:11,160 --> 00:28:14,800

 

599
00:28:14,800 --> 00:28:19,240

 

600
00:28:19,240 --> 00:28:22,030

 

601
00:28:22,030 --> 00:28:24,250

 

602
00:28:24,250 --> 00:28:25,510

 

603
00:28:25,510 --> 00:28:28,030

 

604
00:28:28,030 --> 00:28:29,350

 

605
00:28:29,350 --> 00:28:32,590

 

606
00:28:32,590 --> 00:28:34,090

 

607
00:28:34,090 --> 00:28:36,520

 

608
00:28:36,520 --> 00:28:38,200

 

609
00:28:38,200 --> 00:28:39,880

 

610
00:28:39,880 --> 00:28:42,510

 

611
00:28:42,510 --> 00:28:45,220

 

612
00:28:45,220 --> 00:28:55,390

 

613
00:28:55,390 --> 00:29:04,450

 

614
00:29:04,450 --> 00:29:05,800

 

615
00:29:05,800 --> 00:29:07,570

 

616
00:29:07,570 --> 00:29:10,210

 

617
00:29:10,210 --> 00:29:12,940

 

618
00:29:12,940 --> 00:29:15,580

 

619
00:29:15,580 --> 00:29:17,830

 

620
00:29:17,830 --> 00:29:20,020

 

621
00:29:20,020 --> 00:29:22,090

 

622
00:29:22,090 --> 00:29:24,460

 

623
00:29:24,460 --> 00:29:26,620

 

624
00:29:26,620 --> 00:29:29,050

 

625
00:29:29,050 --> 00:29:31,270

 

626
00:29:31,270 --> 00:29:32,770

 

627
00:29:32,770 --> 00:29:35,230

 

628
00:29:35,230 --> 00:29:36,700

 

629
00:29:36,700 --> 00:29:39,310

 

630
00:29:39,310 --> 00:29:41,170

 

631
00:29:41,170 --> 00:29:43,510

 

632
00:29:43,510 --> 00:29:47,230

 

633
00:29:47,230 --> 00:29:49,600

 

634
00:29:49,600 --> 00:29:52,510

 

635
00:29:52,510 --> 00:29:54,460

 

636
00:29:54,460 --> 00:29:56,770

 

637
00:29:56,770 --> 00:29:58,690

 

638
00:29:58,690 --> 00:30:02,410

 

639
00:30:02,410 --> 00:30:04,930

 

640
00:30:04,930 --> 00:30:08,610

 

641
00:30:08,610 --> 00:30:12,610

 

642
00:30:12,610 --> 00:30:16,930

 

643
00:30:16,930 --> 00:30:19,900

 

644
00:30:19,900 --> 00:30:21,520

 

645
00:30:21,520 --> 00:30:23,170

 

646
00:30:23,170 --> 00:30:25,990

 

647
00:30:25,990 --> 00:30:28,660

 

648
00:30:28,660 --> 00:30:30,100

 

649
00:30:30,100 --> 00:30:33,130

 

650
00:30:33,130 --> 00:30:35,800

 

651
00:30:35,800 --> 00:30:37,420

 

652
00:30:37,420 --> 00:30:40,540

 

653
00:30:40,540 --> 00:30:42,490

 

654
00:30:42,490 --> 00:30:44,920

 

655
00:30:44,920 --> 00:30:46,690

 

656
00:30:46,690 --> 00:30:49,540

 

657
00:30:49,540 --> 00:30:51,940

 

658
00:30:51,940 --> 00:30:54,160

 

659
00:30:54,160 --> 00:30:56,350

 

660
00:30:56,350 --> 00:30:59,410

 

661
00:30:59,410 --> 00:31:04,300

 

662
00:31:04,300 --> 00:31:06,330

 

663
00:31:06,330 --> 00:31:09,130

 

664
00:31:09,130 --> 00:31:10,810

 

665
00:31:10,810 --> 00:31:12,670

 

666
00:31:12,670 --> 00:31:14,440

 

667
00:31:14,440 --> 00:31:16,300

 

668
00:31:16,300 --> 00:31:18,880

 

669
00:31:18,880 --> 00:31:20,200

 

670
00:31:20,200 --> 00:31:23,710

 

671
00:31:23,710 --> 00:31:25,240

 

672
00:31:25,240 --> 00:31:28,030

 

673
00:31:28,030 --> 00:31:32,260

 

674
00:31:32,260 --> 00:31:37,510

 

675
00:31:37,510 --> 00:31:39,970

 

676
00:31:39,970 --> 00:31:41,770

 

677
00:31:41,770 --> 00:31:43,300

 

678
00:31:43,300 --> 00:31:46,180

 

679
00:31:46,180 --> 00:31:47,980

 

680
00:31:47,980 --> 00:31:50,140

 

681
00:31:50,140 --> 00:31:53,500

 

682
00:31:53,500 --> 00:31:55,210

 

683
00:31:55,210 --> 00:31:58,420

 

684
00:31:58,420 --> 00:32:00,400

 

685
00:32:00,400 --> 00:32:03,420

 

686
00:32:03,420 --> 00:32:07,270

 

687
00:32:07,270 --> 00:32:09,940

 

688
00:32:09,940 --> 00:32:11,710

 

689
00:32:11,710 --> 00:32:16,960

 

690
00:32:16,960 --> 00:32:18,580

 

691
00:32:18,580 --> 00:32:23,020

 

692
00:32:23,020 --> 00:32:25,600

 

693
00:32:25,600 --> 00:32:29,140

 

694
00:32:29,140 --> 00:32:30,730

 

695
00:32:30,730 --> 00:32:33,640

 

696
00:32:33,640 --> 00:32:41,320

 

697
00:32:41,320 --> 00:32:45,070

 

698
00:32:45,070 --> 00:32:47,620

 

699
00:32:47,620 --> 00:32:48,910

 

700
00:32:48,910 --> 00:32:51,640

 

701
00:32:51,640 --> 00:32:55,419

 

702
00:32:55,419 --> 00:32:58,090

 

703
00:32:58,090 --> 00:33:01,990

 

704
00:33:01,990 --> 00:33:04,000

 

705
00:33:04,000 --> 00:33:06,100

 

706
00:33:06,100 --> 00:33:09,669

 

707
00:33:09,669 --> 00:33:10,930

 

708
00:33:10,930 --> 00:33:13,299

 

709
00:33:13,299 --> 00:33:15,970

 

710
00:33:15,970 --> 00:33:18,549

 

711
00:33:18,549 --> 00:33:21,040

 

712
00:33:21,040 --> 00:33:22,900

 

713
00:33:22,900 --> 00:33:26,400

 

714
00:33:26,400 --> 00:33:30,549

 

715
00:33:30,549 --> 00:33:33,520

 

716
00:33:33,520 --> 00:33:34,900

 

717
00:33:34,900 --> 00:33:36,850

 

718
00:33:36,850 --> 00:33:38,830

 

719
00:33:38,830 --> 00:33:41,350

 

720
00:33:41,350 --> 00:33:44,740

 

721
00:33:44,740 --> 00:33:46,840

 

722
00:33:46,840 --> 00:33:54,320

 

723
00:33:54,320 --> 00:33:56,990

 

724
00:33:56,990 --> 00:34:00,180

 

725
00:34:00,180 --> 00:34:04,050

 

726
00:34:04,050 --> 00:34:05,760

 

727
00:34:05,760 --> 00:34:08,659

 

728
00:34:08,659 --> 00:34:10,380

 

729
00:34:10,380 --> 00:34:14,760

 

730
00:34:14,760 --> 00:34:19,919

 

731
00:34:19,919 --> 00:34:28,590

 

732
00:34:28,590 --> 00:34:34,020

 

733
00:34:34,020 --> 00:34:36,930

 

734
00:34:36,930 --> 00:34:41,490

 

735
00:34:41,490 --> 00:34:46,560

 

736
00:34:46,560 --> 00:34:48,630

 

737
00:34:48,630 --> 00:34:52,260

 

738
00:34:52,260 --> 00:34:54,330

 

739
00:34:54,330 --> 00:34:56,130

 

740
00:34:56,130 --> 00:34:57,630

 

741
00:34:57,630 --> 00:34:59,460

 

742
00:34:59,460 --> 00:35:01,050

 

743
00:35:01,050 --> 00:35:04,110

 

744
00:35:04,110 --> 00:35:07,590

 

745
00:35:07,590 --> 00:35:09,390

 

746
00:35:09,390 --> 00:35:12,570

 

747
00:35:12,570 --> 00:35:15,960

 

748
00:35:15,960 --> 00:35:17,790

 

749
00:35:17,790 --> 00:35:21,780

 

750
00:35:21,780 --> 00:35:23,670

 

751
00:35:23,670 --> 00:35:27,810

 

752
00:35:27,810 --> 00:35:29,550

 

753
00:35:29,550 --> 00:35:32,220

 

754
00:35:32,220 --> 00:35:34,230

 

755
00:35:34,230 --> 00:35:36,690

 

756
00:35:36,690 --> 00:35:40,500

 

757
00:35:40,500 --> 00:35:42,810

 

758
00:35:42,810 --> 00:35:44,910

 

759
00:35:44,910 --> 00:35:47,760

 

760
00:35:47,760 --> 00:35:50,250

 

761
00:35:50,250 --> 00:35:53,010

 

762
00:35:53,010 --> 00:35:55,020

 

763
00:35:55,020 --> 00:35:59,000

 

764
00:35:59,000 --> 00:36:03,060

 

765
00:36:03,060 --> 00:36:03,070

 

766
00:36:03,070 --> 00:36:04,180


767
00:36:04,180 --> 00:36:07,150

 

768
00:36:07,150 --> 00:36:11,890

 

769
00:36:11,890 --> 00:36:13,270

 

770
00:36:13,270 --> 00:36:14,710

 

771
00:36:14,710 --> 00:36:16,390

 

772
00:36:16,390 --> 00:36:20,109

 

773
00:36:20,109 --> 00:36:21,790

 

774
00:36:21,790 --> 00:36:26,010

 

775
00:36:26,010 --> 00:36:29,950

 

776
00:36:29,950 --> 00:36:34,780

 

777
00:36:34,780 --> 00:36:36,790

 

778
00:36:36,790 --> 00:36:38,740

 

779
00:36:38,740 --> 00:36:44,819

 

780
00:36:44,819 --> 00:36:49,780

 

781
00:36:49,780 --> 00:36:50,920

 

782
00:36:50,920 --> 00:36:54,940

 

783
00:36:54,940 --> 00:37:00,180

 

784
00:37:00,180 --> 00:37:03,700

 

785
00:37:03,700 --> 00:37:08,050

 

786
00:37:08,050 --> 00:37:10,240

 

787
00:37:10,240 --> 00:37:14,500

 

788
00:37:14,500 --> 00:37:17,500

 

789
00:37:17,500 --> 00:37:19,170

 

790
00:37:19,170 --> 00:37:22,240

 

791
00:37:22,240 --> 00:37:27,280

 

792
00:37:27,280 --> 00:37:31,599

 

793
00:37:31,599 --> 00:37:34,809

 

794
00:37:34,809 --> 00:37:38,109

 

795
00:37:38,109 --> 00:37:42,790

 

796
00:37:42,790 --> 00:37:45,190

 

797
00:37:45,190 --> 00:37:47,079

 

798
00:37:47,079 --> 00:37:50,859

 

799
00:37:50,859 --> 00:37:54,160

 

800
00:37:54,160 --> 00:37:58,180

 

801
00:37:58,180 --> 00:38:01,420

 

802
00:38:01,420 --> 00:38:05,140

 

803
00:38:05,140 --> 00:38:06,430

 

804
00:38:06,430 --> 00:38:08,980

 

805
00:38:08,980 --> 00:38:11,950

 

806
00:38:11,950 --> 00:38:13,720

 

807
00:38:13,720 --> 00:38:15,520

 

808
00:38:15,520 --> 00:38:17,560

 

809
00:38:17,560 --> 00:38:20,710

 

810
00:38:20,710 --> 00:38:23,350

 

811
00:38:23,350 --> 00:38:27,070

 

812
00:38:27,070 --> 00:38:30,580

 

813
00:38:30,580 --> 00:38:32,500

 

814
00:38:32,500 --> 00:38:34,150

 

815
00:38:34,150 --> 00:38:35,680

 

816
00:38:35,680 --> 00:38:38,200

 

817
00:38:38,200 --> 00:38:41,620

 

818
00:38:41,620 --> 00:38:43,690

 

819
00:38:43,690 --> 00:38:45,520

 

820
00:38:45,520 --> 00:38:47,950

 

821
00:38:47,950 --> 00:38:49,990

 

822
00:38:49,990 --> 00:38:53,470

 

823
00:38:53,470 --> 00:38:55,360

 

824
00:38:55,360 --> 00:38:57,970

 

825
00:38:57,970 --> 00:38:59,710

 

826
00:38:59,710 --> 00:39:01,720

 

827
00:39:01,720 --> 00:39:04,750

 

828
00:39:04,750 --> 00:39:06,430

 

829
00:39:06,430 --> 00:39:07,780

 

830
00:39:07,780 --> 00:39:11,830

 

831
00:39:11,830 --> 00:39:13,750

 

832
00:39:13,750 --> 00:39:16,300

 

833
00:39:16,300 --> 00:39:19,030

 

834
00:39:19,030 --> 00:39:22,000

 

835
00:39:22,000 --> 00:39:26,680

 

836
00:39:26,680 --> 00:39:28,450

 

837
00:39:28,450 --> 00:39:31,810

 

838
00:39:31,810 --> 00:39:34,330

 

839
00:39:34,330 --> 00:39:38,050

 

840
00:39:38,050 --> 00:39:41,010

 

841
00:39:41,010 --> 00:39:44,980

 

842
00:39:44,980 --> 00:39:48,030

 

843
00:39:48,030 --> 00:39:52,540

 

844
00:39:52,540 --> 00:39:54,490

 

845
00:39:54,490 --> 00:40:00,990

 

846
00:40:00,990 --> 00:40:03,310

 

847
00:40:03,310 --> 00:40:05,020

 

848
00:40:05,020 --> 00:40:07,930

 

849
00:40:07,930 --> 00:40:09,430

 

850
00:40:09,430 --> 00:40:11,830

 

851
00:40:11,830 --> 00:40:15,310

 

852
00:40:15,310 --> 00:40:17,320

 

853
00:40:17,320 --> 00:40:22,570

 

854
00:40:22,570 --> 00:40:25,450

 

855
00:40:25,450 --> 00:40:27,940

 

856
00:40:27,940 --> 00:40:30,980

 

857
00:40:30,980 --> 00:40:35,359

 

858
00:40:35,359 --> 00:40:37,010

 

859
00:40:37,010 --> 00:40:38,690

 

860
00:40:38,690 --> 00:40:40,430

 

861
00:40:40,430 --> 00:40:42,590

 

862
00:40:42,590 --> 00:40:45,620

 

863
00:40:45,620 --> 00:40:52,480

 

864
00:40:52,480 --> 00:40:56,090

 

865
00:40:56,090 --> 00:40:58,640

 

866
00:40:58,640 --> 00:41:01,670

 

867
00:41:01,670 --> 00:41:03,170

 

868
00:41:03,170 --> 00:41:05,690

 

869
00:41:05,690 --> 00:41:08,770

 

870
00:41:08,770 --> 00:41:13,760

 

871
00:41:13,760 --> 00:41:16,040

 

872
00:41:16,040 --> 00:41:18,050

 

873
00:41:18,050 --> 00:41:27,710

 

874
00:41:27,710 --> 00:41:27,720

 

875
00:41:27,720 --> 00:41:29,780