1 00:00:02,850 --> 00:00:06,850 PROFESSOR: Today, we are talking about the local behavior 2 00:00:06,850 --> 00:00:08,970 of a crease pattern. 3 00:00:08,970 --> 00:00:13,240 So you take some crease pattern for some flat folding-- 4 00:00:13,240 --> 00:00:15,419 we're thinking about flat foldability. 5 00:00:15,419 --> 00:00:16,710 This is a foldability question. 6 00:00:16,710 --> 00:00:18,293 I give you a crease pattern like this. 7 00:00:18,293 --> 00:00:22,350 I want to know, does it fold flat, like this one does. 8 00:00:22,350 --> 00:00:25,660 And we're studying what happens locally 9 00:00:25,660 --> 00:00:28,220 right around a single vertex. 10 00:00:28,220 --> 00:00:30,870 So I didn't mention graph terminology. 11 00:00:30,870 --> 00:00:32,509 It's probably useful to mention that. 12 00:00:32,509 --> 00:00:35,580 These corners-- where all the edges come together-- those 13 00:00:35,580 --> 00:00:36,270 are vertices. 14 00:00:36,270 --> 00:00:39,040 These triangles-- or in general, some regions 15 00:00:39,040 --> 00:00:41,255 divided by the creases-- we call faces. 16 00:00:44,060 --> 00:00:48,100 If we just sort of imagine cutting a little disk 17 00:00:48,100 --> 00:00:50,770 around that vertex and seeing how it behaves, 18 00:00:50,770 --> 00:00:54,900 we get a nice circular piece of paper with some crease pattern. 19 00:00:54,900 --> 00:00:58,000 And we want to understand when those things fold flat and when 20 00:00:58,000 --> 00:00:58,500 they don't. 21 00:00:58,500 --> 00:01:00,710 Sometimes they do, sometimes they don't. 22 00:01:00,710 --> 00:01:05,700 This is kind of like a one dimensional folding problem. 23 00:01:05,700 --> 00:01:09,280 So let me write some stuff about that. 24 00:01:16,130 --> 00:01:19,770 So we get what we call single vertex crease patterns. 25 00:01:30,860 --> 00:01:34,700 So something like-- this is the example 26 00:01:34,700 --> 00:01:37,660 I have made-- just something like that. 27 00:01:37,660 --> 00:01:38,890 That should be flat foldable. 28 00:01:41,590 --> 00:01:48,270 So we think of having a disk of paper. 29 00:01:48,270 --> 00:01:50,544 It doesn't really matter, but it's 30 00:01:50,544 --> 00:01:51,960 going to be easier to reason about 31 00:01:51,960 --> 00:01:53,790 and think about a disk of paper. 32 00:01:53,790 --> 00:01:56,940 Really, we just mean some small region around that vertex. 33 00:01:56,940 --> 00:01:59,800 There's obviously one vertex in the crease pattern. 34 00:01:59,800 --> 00:02:05,416 Let's say we have n creases emanating from one vertex. 35 00:02:16,510 --> 00:02:18,610 This pattern is going to be defined 36 00:02:18,610 --> 00:02:21,640 by some sequence of angles. 37 00:02:38,560 --> 00:02:41,890 In general, theta 1 up to theta n. 38 00:02:41,890 --> 00:02:46,060 If there are n creases, there will be n angles between them, 39 00:02:46,060 --> 00:02:48,790 say in clockwise order. 40 00:02:48,790 --> 00:02:53,000 And let's see. 41 00:02:53,000 --> 00:03:00,330 When we fold this thing flat-- like in this picture-- 42 00:03:00,330 --> 00:03:02,760 we take this disk. 43 00:03:02,760 --> 00:03:06,330 We fold it along all of those creases. 44 00:03:06,330 --> 00:03:07,860 What I'd like to focus on is what 45 00:03:07,860 --> 00:03:10,590 happens to the boundary of the paper. 46 00:03:10,590 --> 00:03:11,870 So there's this outer circle. 47 00:03:11,870 --> 00:03:13,720 The boundary is a circle. 48 00:03:13,720 --> 00:03:15,840 And locally, if you look at one of these creases, 49 00:03:15,840 --> 00:03:19,950 it's like folding the circle in half onto itself. 50 00:03:19,950 --> 00:03:22,900 And when you make all of these creases, if you're successful, 51 00:03:22,900 --> 00:03:27,716 you end up folding the circle onto a portion of the circle. 52 00:03:27,716 --> 00:03:29,340 So if you look at this boundary-- which 53 00:03:29,340 --> 00:03:31,680 is a little hard to see. 54 00:03:31,680 --> 00:03:32,960 Why don't I trace it? 55 00:03:32,960 --> 00:03:35,980 It's going to do something like this. 56 00:03:43,400 --> 00:03:47,720 So this is what the 3D thing looks like blown up 57 00:03:47,720 --> 00:03:50,851 a little bit-- and I changed the angles a little bit. 58 00:03:50,851 --> 00:03:52,850 But this is what a flat folding would look like. 59 00:03:52,850 --> 00:03:54,830 And I'm really just focusing on how 60 00:03:54,830 --> 00:04:00,270 the circle maps around this circle. 61 00:04:00,270 --> 00:04:08,826 So a flat folding of a disk also folds a circle onto a circle. 62 00:04:08,826 --> 00:04:14,630 So we have a flat folding on one of these single vertex crease 63 00:04:14,630 --> 00:04:28,250 patterns, we get a folding of a circle onto a circle. 64 00:04:28,250 --> 00:04:32,090 This is nice, because circles are one dimensional. 65 00:04:32,090 --> 00:04:34,710 And last class, we talked about folding one dimensional line 66 00:04:34,710 --> 00:04:35,210 segments. 67 00:04:35,210 --> 00:04:37,650 A circle is just a little bit more complicated. 68 00:04:37,650 --> 00:04:39,740 It's different topology. 69 00:04:39,740 --> 00:04:42,060 But we can use a lot of the same mindset, 70 00:04:42,060 --> 00:04:46,180 at least, as we did for folding line segments onto the line. 71 00:04:46,180 --> 00:04:50,940 In particular, you can see, yeah, this is kind of circular. 72 00:04:50,940 --> 00:04:53,860 But you could also imagine unrolling it a little bit 73 00:04:53,860 --> 00:04:55,640 and making it straight. 74 00:04:55,640 --> 00:04:58,090 Something like this. 75 00:04:58,090 --> 00:05:00,310 So that it lies on a straight line, 76 00:05:00,310 --> 00:05:03,160 if you just unroll that circle. 77 00:05:03,160 --> 00:05:06,870 So you also get a folding of a circle in some sense 78 00:05:06,870 --> 00:05:11,705 onto a line by unrolling. 79 00:05:14,720 --> 00:05:18,380 Now, at this point, I want to mention a slight issue here. 80 00:05:20,960 --> 00:05:25,590 I mean for this unrolling to work-- 81 00:05:25,590 --> 00:05:29,500 if you're taking a circle like this-- in particular, 82 00:05:29,500 --> 00:05:32,350 you must make at least one fold for this to be possible. 83 00:05:32,350 --> 00:05:35,080 Because right now, if I haven't folded anything, 84 00:05:35,080 --> 00:05:37,070 I just have a big circle, I can't unroll circle 85 00:05:37,070 --> 00:05:38,000 onto a line. 86 00:05:38,000 --> 00:05:39,650 I haven't collapsed it. 87 00:05:39,650 --> 00:05:41,970 Yet here, I've made it small enough 88 00:05:41,970 --> 00:05:44,510 that it only occupies a portion of the entire circle. 89 00:05:44,510 --> 00:05:45,960 And so I can unroll it. 90 00:05:45,960 --> 00:05:49,140 So this is true, as long as I make at least one fold. 91 00:05:52,560 --> 00:05:53,410 OK. 92 00:05:53,410 --> 00:05:55,840 Technically, I need to assume another thing, which 93 00:05:55,840 --> 00:05:59,920 is that this thing came from a flat piece of paper. 94 00:05:59,920 --> 00:06:01,640 I said there were these angles. 95 00:06:01,640 --> 00:06:06,790 And what I intended to mean is that the sum of those angles 96 00:06:06,790 --> 00:06:13,190 is 360 degrees, as it is in a flat piece of paper. 97 00:06:13,190 --> 00:06:16,340 But I'm not always going to require this assumption. 98 00:06:16,340 --> 00:06:17,940 It's what we care about for thinking 99 00:06:17,940 --> 00:06:20,556 about a crease pattern of a flat piece of paper 100 00:06:20,556 --> 00:06:21,680 and we look at each vertex. 101 00:06:21,680 --> 00:06:23,960 This will be true. 102 00:06:23,960 --> 00:06:27,790 But when we reason about these single vertex crease patterns, 103 00:06:27,790 --> 00:06:31,890 it's really useful to think about, sort of in the middle, 104 00:06:31,890 --> 00:06:33,130 this could get smaller. 105 00:06:33,130 --> 00:06:35,660 So for example, you have a pattern like this. 106 00:06:35,660 --> 00:06:38,880 Let's say I just take these two creases on the bottom 107 00:06:38,880 --> 00:06:40,450 and I fold a crimp. 108 00:06:40,450 --> 00:06:43,580 Because I know a cramp is kind of a good thing to do. 109 00:06:43,580 --> 00:06:46,540 So now I have what's called a cone of paper. 110 00:06:46,540 --> 00:06:49,054 The sum of the angles-- there's two angles here, 111 00:06:49,054 --> 00:06:49,970 this one and this one. 112 00:06:49,970 --> 00:06:55,920 Each one looks like 270-- no, not 270, 135? 113 00:06:55,920 --> 00:06:57,530 In total, it's 270. 114 00:06:57,530 --> 00:06:59,620 So the total angle there is now less than 360. 115 00:06:59,620 --> 00:07:03,660 This is what we call a convex cone. 116 00:07:03,660 --> 00:07:09,640 So if you relax this constraint to allow less than 117 00:07:09,640 --> 00:07:13,770 or equal to 360, then we get a convex cone. 118 00:07:16,730 --> 00:07:19,340 And today, I only want to talk about flat paper and convex 119 00:07:19,340 --> 00:07:19,840 cones. 120 00:07:19,840 --> 00:07:21,990 We need this for the induction, essentially-- 121 00:07:21,990 --> 00:07:24,590 for proving things about the flat case. 122 00:07:24,590 --> 00:07:27,740 In the textbook-- if you look at this chapter on single vertex 123 00:07:27,740 --> 00:07:30,500 crease patterns-- we also prove or study 124 00:07:30,500 --> 00:07:33,170 the case where the sum of the angles is greater than 360, 125 00:07:33,170 --> 00:07:35,300 which you could never make from flat paper. 126 00:07:35,300 --> 00:07:36,680 But hey, it's fun to think about. 127 00:07:36,680 --> 00:07:38,740 It's a natural generalization. 128 00:07:38,740 --> 00:07:41,000 And all the things that we do here, 129 00:07:41,000 --> 00:07:42,721 you can generalize to that situation. 130 00:07:42,721 --> 00:07:43,970 It's just a little bit harder. 131 00:07:51,780 --> 00:07:57,420 So what does it take to fold a circle onto a line? 132 00:07:57,420 --> 00:07:59,390 In this situation of a convex cone, 133 00:07:59,390 --> 00:08:02,454 where we make at least one fold, our folding 134 00:08:02,454 --> 00:08:04,370 is going to lie along a portion of the circle. 135 00:08:04,370 --> 00:08:06,390 We can unroll it a little bit, get 136 00:08:06,390 --> 00:08:08,270 something lying on a straight line. 137 00:08:10,930 --> 00:08:16,480 It's kind of like folding a line segment onto a straight line, 138 00:08:16,480 --> 00:08:21,200 if it was flat folding of one dimensional pieces of paper. 139 00:08:28,220 --> 00:08:30,600 But there's a lot of differences. 140 00:08:30,600 --> 00:08:32,679 I mean, they are similar. 141 00:08:32,679 --> 00:08:34,470 But a lot of the things we proved last time 142 00:08:34,470 --> 00:08:36,500 do not hold in the case of circular paper. 143 00:08:36,500 --> 00:08:40,400 So I have a bunch of them here. 144 00:08:40,400 --> 00:08:43,070 So one problem-- you may recall before, 145 00:08:43,070 --> 00:08:45,650 if we had a line segment and we put an arbitrary crease 146 00:08:45,650 --> 00:08:49,100 pattern, then we could just assign a mountain valley 147 00:08:49,100 --> 00:08:51,090 assignment, alternating mountain valley. 148 00:08:51,090 --> 00:08:54,350 And that would always fold flat, no collisions. 149 00:08:54,350 --> 00:08:54,850 OK. 150 00:08:54,850 --> 00:08:58,480 In this world of having a circular piece of paper, 151 00:08:58,480 --> 00:09:00,670 that's no longer true. 152 00:09:00,670 --> 00:09:04,570 For example, if you have this kind of crease pattern 153 00:09:04,570 --> 00:09:07,070 on a circle or in the disk, it would 154 00:09:07,070 --> 00:09:09,880 look like this, those two creases. 155 00:09:09,880 --> 00:09:11,440 You can't fold that thing flat. 156 00:09:11,440 --> 00:09:13,240 It's not going to work. 157 00:09:13,240 --> 00:09:16,300 We'll see exactly what you need to forbid 158 00:09:16,300 --> 00:09:18,890 for that to make sense. 159 00:09:18,890 --> 00:09:25,500 Another example is something like this. 160 00:09:31,720 --> 00:09:35,320 I mean, if I tried to draw this in the line, what's happening 161 00:09:35,320 --> 00:09:38,550 is I have a short segment and then I have a long segment. 162 00:09:38,550 --> 00:09:40,800 So the problem is those ends don't meet. 163 00:09:40,800 --> 00:09:43,220 But even when the ends meet-- like in this picture, 164 00:09:43,220 --> 00:09:45,660 imagine these are vertically aligned-- 165 00:09:45,660 --> 00:09:48,560 if I do this alternating mountain valley pattern-- 166 00:09:48,560 --> 00:09:52,010 mountain valley assignment-- I can't join these ends 167 00:09:52,010 --> 00:09:55,550 and make a circle, because that would collide with everything. 168 00:09:55,550 --> 00:09:59,220 So this thing actually does have a flat folding 169 00:09:59,220 --> 00:10:00,680 starting from a circle. 170 00:10:00,680 --> 00:10:03,410 And I'm not going to try to un-map it to a circle. 171 00:10:03,410 --> 00:10:05,170 It does fold, but not like that. 172 00:10:08,170 --> 00:10:11,620 Not with that alternating mountain valley pattern. 173 00:10:11,620 --> 00:10:13,440 There are other annoying things. 174 00:10:13,440 --> 00:10:19,660 Let me tell you one more on the proof side. 175 00:10:19,660 --> 00:10:26,150 Last time, we had a lemma that said, 176 00:10:26,150 --> 00:10:27,840 if your crease pattern is mingling-- 177 00:10:27,840 --> 00:10:30,360 which still is a meaningful notion here-- 178 00:10:30,360 --> 00:10:34,336 then you have a crimp, or an end fold in that case. 179 00:10:34,336 --> 00:10:36,210 Here, of course, we don't have any end folds. 180 00:10:36,210 --> 00:10:38,409 Here, we also do not get this implication. 181 00:10:38,409 --> 00:10:40,450 Mingling does not imply the existence of a crimp. 182 00:10:40,450 --> 00:10:45,020 And where the proof breaks down is if you have-- so 183 00:10:45,020 --> 00:10:48,720 if you remember, our parentheses from last time-- 184 00:10:48,720 --> 00:10:50,270 we had this kind of pattern. 185 00:10:50,270 --> 00:10:56,420 And we said, well, you keep going and either 186 00:10:56,420 --> 00:10:58,630 on the left side, you have a beginning open paren 187 00:10:58,630 --> 00:11:02,406 or on the right side you have a closing round parenthesis. 188 00:11:02,406 --> 00:11:04,030 In either case, you've got an end fold. 189 00:11:04,030 --> 00:11:06,090 Now, this can actually happen in a circle. 190 00:11:06,090 --> 00:11:07,949 And so you have nothing you can do. 191 00:11:07,949 --> 00:11:09,240 So you could have this pattern. 192 00:11:09,240 --> 00:11:11,720 There would be no crimp. 193 00:11:11,720 --> 00:11:14,700 And probably, in that case, you would not be flat foldable, 194 00:11:14,700 --> 00:11:17,580 but in particular, this implication fails. 195 00:11:17,580 --> 00:11:19,364 So we have to do more work. 196 00:11:19,364 --> 00:11:21,280 But we're going to get the same kind of result 197 00:11:21,280 --> 00:11:23,360 that, in linear time, we can tell 198 00:11:23,360 --> 00:11:25,020 whether one of these things folds flat. 199 00:11:28,219 --> 00:11:32,490 So that's good news. 200 00:11:32,490 --> 00:11:37,890 So last time, telling whether a crease pattern folded flat 201 00:11:37,890 --> 00:11:38,460 was trivial. 202 00:11:38,460 --> 00:11:40,110 The answer was always yes. 203 00:11:40,110 --> 00:11:41,494 And now, it's not so trivial. 204 00:11:41,494 --> 00:11:42,910 So let's start with characterizing 205 00:11:42,910 --> 00:11:44,600 crease patterns that fold flat. 206 00:11:44,600 --> 00:11:47,430 Then we will go to mountain valley patterns that fold flat. 207 00:12:16,766 --> 00:12:18,390 It'd be great if I could just say that, 208 00:12:18,390 --> 00:12:20,265 but I mean here, of course, single vertex. 209 00:12:26,040 --> 00:12:29,180 So I give you one of these sequence of angles. 210 00:12:29,180 --> 00:12:32,050 Let's say it sums to 360 or less. 211 00:12:32,050 --> 00:12:35,620 When is it going to be flat foldable? 212 00:12:35,620 --> 00:12:36,870 And the answer is very simple. 213 00:12:39,620 --> 00:12:40,540 Let me write it down. 214 00:12:53,970 --> 00:12:57,140 This thing is going to be flat foldable if and only 215 00:12:57,140 --> 00:13:08,424 if the sum of the odd angles is equal to the sum 216 00:13:08,424 --> 00:13:09,215 of the even angles. 217 00:13:16,130 --> 00:13:20,560 And, yeah, I'm implicitly assuming and requiring 218 00:13:20,560 --> 00:13:22,290 here that n is even. 219 00:13:22,290 --> 00:13:23,940 That's something we'll prove. 220 00:13:23,940 --> 00:13:27,760 So n is actually even and minus 1 is going be odd. 221 00:13:27,760 --> 00:13:29,530 And if you happen to have started 222 00:13:29,530 --> 00:13:33,720 with a flat piece of paper-- for the sum of all these things 223 00:13:33,720 --> 00:13:36,920 is 360-- then of course, if you have them evenly divided, 224 00:13:36,920 --> 00:13:40,950 this sum will be 180 degrees. 225 00:13:40,950 --> 00:13:44,780 But if you started with the convex cone, it'll be smaller. 226 00:13:48,370 --> 00:13:50,710 For this reason, a lot of people-- or some people 227 00:13:50,710 --> 00:13:53,850 call this the 180 degree property, or the pie property-- 228 00:13:53,850 --> 00:13:57,740 if you like to think in radians instead of degrees. 229 00:13:57,740 --> 00:14:00,780 And we can do a little example here. 230 00:14:00,780 --> 00:14:08,480 This guy-- this crease pattern-- has a 90 degree angle here. 231 00:14:08,480 --> 00:14:10,870 And it has a 90 degree angle down here. 232 00:14:10,870 --> 00:14:12,790 So the sum of those two is 180. 233 00:14:12,790 --> 00:14:16,210 Those are alternating angles, different parity classes. 234 00:14:16,210 --> 00:14:17,940 This is 45. 235 00:14:17,940 --> 00:14:21,740 This is the complement, 135. 236 00:14:21,740 --> 00:14:24,620 And those also sum to 180, of course. 237 00:14:24,620 --> 00:14:26,080 One does only if the other does. 238 00:14:26,080 --> 00:14:27,760 So it's flat foldable. 239 00:14:27,760 --> 00:14:28,460 We're golden. 240 00:14:28,460 --> 00:14:33,140 Something like this is bad, because I 241 00:14:33,140 --> 00:14:34,280 mean it's an even number. 242 00:14:34,280 --> 00:14:39,240 But the odd angles do not equal the sum of the even angles. 243 00:14:39,240 --> 00:14:41,279 One's clearly bigger. 244 00:14:41,279 --> 00:14:42,570 And this is a general property. 245 00:14:42,570 --> 00:14:45,000 This is also called Kawasaki's theorem 246 00:14:45,000 --> 00:14:48,220 or Kawasaki-Justin's theorem. 247 00:14:48,220 --> 00:14:50,920 They proved it in 1989. 248 00:14:50,920 --> 00:14:52,600 This is pretty much the beginning 249 00:14:52,600 --> 00:14:56,050 of origami mathematics. 250 00:14:56,050 --> 00:14:58,224 So let's prove it. 251 00:14:58,224 --> 00:14:59,390 There's two things to prove. 252 00:14:59,390 --> 00:15:02,390 One is that any flat foldable crease pattern 253 00:15:02,390 --> 00:15:04,487 has this property, which is pretty easy. 254 00:15:04,487 --> 00:15:06,320 And the other, which is a little bit harder, 255 00:15:06,320 --> 00:15:07,380 is that if you have this property, 256 00:15:07,380 --> 00:15:08,760 you really are flat foldable. 257 00:15:08,760 --> 00:15:10,135 That's the full characterization. 258 00:15:14,090 --> 00:15:17,240 So let's start with the easy direction. 259 00:15:17,240 --> 00:15:19,930 If you're flat foldable, you must have that sum. 260 00:15:22,870 --> 00:15:25,100 OK. 261 00:15:25,100 --> 00:15:29,421 So it's all about thinking what happens at a crease. 262 00:15:29,421 --> 00:15:30,920 So we're thinking about the boundary 263 00:15:30,920 --> 00:15:31,878 of this piece of paper. 264 00:15:31,878 --> 00:15:33,590 We're thinking about the circle. 265 00:15:33,590 --> 00:15:40,260 And what's happening on the circle as you travel, 266 00:15:40,260 --> 00:15:42,870 say counterclockwise for some amount 267 00:15:42,870 --> 00:15:45,450 of time, that would be first angle. 268 00:15:45,450 --> 00:15:46,326 Then you make a fold. 269 00:15:46,326 --> 00:15:47,700 It's either a mountain or valley. 270 00:15:47,700 --> 00:15:49,140 At this point, we don't care. 271 00:15:49,140 --> 00:15:52,250 Either way, you turn around-- maybe you turn around this way, 272 00:15:52,250 --> 00:15:53,910 or you turn around this way. 273 00:15:53,910 --> 00:15:57,140 But in terms as your travel along a circle-- 274 00:15:57,140 --> 00:15:59,170 so we went theta 1 counterclockwise. 275 00:15:59,170 --> 00:16:02,110 Now we're going to go theta 2 clockwise; 276 00:16:02,110 --> 00:16:05,460 then we're going to go theta 3 counterclockwise, and so on. 277 00:16:05,460 --> 00:16:07,930 We keep alternating back and forth. 278 00:16:07,930 --> 00:16:10,290 It's much easier to think drawn a line. 279 00:16:10,290 --> 00:16:12,760 So we go to the right, theta 1. 280 00:16:12,760 --> 00:16:16,130 Then we go to the left, theta 2. 281 00:16:16,130 --> 00:16:21,040 Then we go to the right, theta 3. 282 00:16:21,040 --> 00:16:23,000 And I don't know about the layering here. 283 00:16:23,000 --> 00:16:25,419 I'm just drawing it in terms of horizontal travel. 284 00:16:25,419 --> 00:16:26,710 The y-coordinate means nothing. 285 00:16:30,270 --> 00:16:34,050 In order for this thing to be a flat folding-- 286 00:16:34,050 --> 00:16:37,310 to be a valid folding of a circle of paper-- 287 00:16:37,310 --> 00:16:41,150 these guys have to have the same x-coordinate. 288 00:16:41,150 --> 00:16:43,400 So I don't know-- I mean, the y-coordinates, they also 289 00:16:43,400 --> 00:16:44,280 have to work out. 290 00:16:44,280 --> 00:16:46,238 But we're not thinking about the y-coordinates. 291 00:16:46,238 --> 00:16:48,185 We're just thinking about x travel. 292 00:16:48,185 --> 00:16:49,560 We've got to get back to where we 293 00:16:49,560 --> 00:16:51,770 started if we're folding a circle. 294 00:16:51,770 --> 00:16:53,170 That's it. 295 00:16:53,170 --> 00:16:57,170 So how much total travel do we do, in a sign sense? 296 00:16:57,170 --> 00:16:58,490 How much do we go right? 297 00:16:58,490 --> 00:17:00,280 Well, we go theta 1 to the right. 298 00:17:00,280 --> 00:17:02,010 Then we go theta 2 to the left, which 299 00:17:02,010 --> 00:17:05,030 is like going negative theta 2 to the right. 300 00:17:05,030 --> 00:17:07,599 And then we go through theta 3 to the right. 301 00:17:07,599 --> 00:17:10,470 Then we go negative theta 4 to the right. 302 00:17:10,470 --> 00:17:15,119 And you add up this alternating summation that must equal 0. 303 00:17:17,960 --> 00:17:23,619 The other thing that I should say is that n has to be even. 304 00:17:23,619 --> 00:17:26,950 This is maybe obvious at this point. 305 00:17:26,950 --> 00:17:28,840 But we're supposed to alternate. 306 00:17:28,840 --> 00:17:32,340 Every time we hit a crease, we have to change direction. 307 00:17:32,340 --> 00:17:36,390 And so in the end, it's important that after theta 4, 308 00:17:36,390 --> 00:17:39,230 we change direction to visit theta 1. 309 00:17:39,230 --> 00:17:43,490 If they were aligned-- like if theta 4 went over here 310 00:17:43,490 --> 00:17:48,120 and then theta 5 went like that, and there 311 00:17:48,120 --> 00:17:50,232 was an odd number of creases, this would be bad. 312 00:17:50,232 --> 00:17:52,440 Even though they're lined up, there's no crease here. 313 00:17:52,440 --> 00:17:55,419 You just went straight from theta 5 theta 1. 314 00:17:55,419 --> 00:17:57,460 You're supposed to change direction every crease. 315 00:17:57,460 --> 00:18:02,550 So also, n is even to alternate directions. 316 00:18:09,650 --> 00:18:10,494 Question? 317 00:18:10,494 --> 00:18:11,345 AUDIENCE: Oh, no. 318 00:18:11,345 --> 00:18:11,886 You already-- 319 00:18:11,886 --> 00:18:14,670 PROFESSOR: All right, good. 320 00:18:14,670 --> 00:18:15,570 Clear? 321 00:18:15,570 --> 00:18:18,730 This is pretty easy, once you realize the angles correspond 322 00:18:18,730 --> 00:18:20,820 to x-coordinates in this linear space. 323 00:18:26,470 --> 00:18:28,850 So why is this enough? 324 00:18:28,850 --> 00:18:30,680 In order to show that this property is 325 00:18:30,680 --> 00:18:33,610 enough for flat foldability, we need 326 00:18:33,610 --> 00:18:36,422 to actually worry about the stacking order of these edges. 327 00:18:36,422 --> 00:18:38,130 We need to worry about the y-coordinates. 328 00:18:38,130 --> 00:18:40,309 We need to make sure that they can avoid collision 329 00:18:40,309 --> 00:18:41,850 with some mountain valley assignment. 330 00:18:41,850 --> 00:18:43,615 The good news is we're free to use whatever mountains 331 00:18:43,615 --> 00:18:44,530 and valleys we want. 332 00:18:44,530 --> 00:18:46,030 We can stack things however we want. 333 00:18:46,030 --> 00:18:48,470 We just need to find some valid state. 334 00:18:48,470 --> 00:18:50,670 And our intuition, from the one dimensional case, 335 00:18:50,670 --> 00:18:53,790 was we should try alternating mountains and valleys. 336 00:18:53,790 --> 00:18:55,040 That seemed like a good thing. 337 00:18:55,040 --> 00:18:58,570 It avoided collisions for an open piece of paper-- 338 00:18:58,570 --> 00:19:00,220 for a line segment. 339 00:19:00,220 --> 00:19:02,130 Of course, we know that's not enough. 340 00:19:02,130 --> 00:19:06,700 Here-- somewhere-- I drew this example. 341 00:19:06,700 --> 00:19:09,440 So it's alternating in the middle, 342 00:19:09,440 --> 00:19:12,310 and then that last crease is a problem. 343 00:19:12,310 --> 00:19:13,560 But that's all we need to fix. 344 00:19:13,560 --> 00:19:17,220 It turns out it's not so hard. 345 00:19:17,220 --> 00:19:20,460 So we have a nice circle. 346 00:19:20,460 --> 00:19:23,384 What I'd like to do is cut that circle. 347 00:19:23,384 --> 00:19:25,300 So here, I sort of cut it and thought about it 348 00:19:25,300 --> 00:19:27,740 as a line segment and then I wanted to rejoin. 349 00:19:27,740 --> 00:19:28,240 OK. 350 00:19:28,240 --> 00:19:30,380 I'm going to be careful about where I cut it. 351 00:19:49,284 --> 00:19:52,201 OK, I want to cut at an extreme left or extreme right. 352 00:19:52,201 --> 00:19:52,950 It doesn't matter. 353 00:19:52,950 --> 00:19:55,800 We can see here the cut ended up being in the middle. 354 00:19:55,800 --> 00:19:57,320 That's bad. 355 00:19:57,320 --> 00:19:58,310 So how do I fix it? 356 00:20:01,530 --> 00:20:02,900 What does extreme mean? 357 00:20:02,900 --> 00:20:04,400 How do I tell-- given a circle, it's 358 00:20:04,400 --> 00:20:06,066 a little hard to say what extreme means, 359 00:20:06,066 --> 00:20:07,760 because it's circular. 360 00:20:07,760 --> 00:20:09,180 But if I take a picture like this, 361 00:20:09,180 --> 00:20:11,138 I've already drawn it with nice x-coordinates-- 362 00:20:11,138 --> 00:20:13,330 I don't know about the y-coordinates yet-- I say, 363 00:20:13,330 --> 00:20:13,970 oh that's bad. 364 00:20:13,970 --> 00:20:16,900 My cut point ended up being non-extreme. 365 00:20:16,900 --> 00:20:18,440 Well, this is extreme. 366 00:20:18,440 --> 00:20:21,840 So cut here instead. 367 00:20:21,840 --> 00:20:24,810 So what that means is redraw this picture 368 00:20:24,810 --> 00:20:26,820 with this being the cut point. 369 00:20:26,820 --> 00:20:29,490 So maybe, this is the beginning. 370 00:20:29,490 --> 00:20:33,470 So we go like that again and now we wrap around. 371 00:20:33,470 --> 00:20:35,630 So now we have this segment. 372 00:20:35,630 --> 00:20:38,370 And now we have this segment. 373 00:20:38,370 --> 00:20:40,430 And lo and behold, it remained extreme. 374 00:20:40,430 --> 00:20:41,460 And this is always true. 375 00:20:41,460 --> 00:20:44,060 You can do this sort of cut and rejoin 376 00:20:44,060 --> 00:20:46,392 the ends that were messed up before. 377 00:20:46,392 --> 00:20:47,975 And you'll preserve the x-coordinates. 378 00:20:47,975 --> 00:20:49,500 The x-coordinates aren't changing 379 00:20:49,500 --> 00:20:51,340 when you do this kind of transformation. 380 00:20:51,340 --> 00:20:54,030 So there's a clear leftmost and there's a clear rightmost. 381 00:20:54,030 --> 00:20:56,130 You pick either one, you cut there instead. 382 00:20:56,130 --> 00:20:58,210 And now we have this nice property 383 00:20:58,210 --> 00:21:01,330 that it's easy to join the ends, because it's 384 00:21:01,330 --> 00:21:02,470 as far left as you can go. 385 00:21:02,470 --> 00:21:04,140 There can't be anything that penetrates, 386 00:21:04,140 --> 00:21:05,830 because that would be farther left. 387 00:21:08,820 --> 00:21:12,470 Let me write down some words corresponding to that. 388 00:21:12,470 --> 00:21:15,770 So once we cut at this extreme crease, 389 00:21:15,770 --> 00:21:22,310 you can fold this 1D segment just 390 00:21:22,310 --> 00:21:23,520 like we used to be able to. 391 00:21:27,080 --> 00:21:31,550 So we can fold it flat using the accordion fold, 392 00:21:31,550 --> 00:21:35,050 alternating mountain and valley. 393 00:21:35,050 --> 00:21:36,730 We know that works. 394 00:21:36,730 --> 00:21:41,430 And then all we need to show is that the two ends can join. 395 00:21:44,510 --> 00:21:54,819 Those ends are the two copies of the cut crease 396 00:21:54,819 --> 00:21:56,110 that we did in this first step. 397 00:22:00,190 --> 00:22:02,250 We need two things. 398 00:22:02,250 --> 00:22:05,670 One is that they are aligned and that's 399 00:22:05,670 --> 00:22:10,310 where we need this condition-- that the sum of the odd angles 400 00:22:10,310 --> 00:22:12,980 equals the sum of even angles, which is the same as saying 401 00:22:12,980 --> 00:22:15,350 the alternating sum is equal to zero. 402 00:22:15,350 --> 00:22:17,270 That tells us that whatever we do, 403 00:22:17,270 --> 00:22:20,650 the x-coordinates are lined up at the end. 404 00:22:20,650 --> 00:22:24,710 So they are aligned by the assumption that we made here. 405 00:22:24,710 --> 00:22:31,220 And you can join without crossing. 406 00:22:34,730 --> 00:22:37,240 So this is a statement about y-coordinates. 407 00:22:37,240 --> 00:22:40,530 And we know it's OK, because it's at the left extreme. 408 00:22:40,530 --> 00:22:43,292 So by this choice of the proper cut, 409 00:22:43,292 --> 00:22:45,750 there could be other things that come right to the boundary 410 00:22:45,750 --> 00:22:47,570 here, but that's considered OK. 411 00:22:47,570 --> 00:22:50,400 That's not crossing, that just touching. 412 00:22:50,400 --> 00:22:52,866 Nothing can go farther left because it 413 00:22:52,866 --> 00:22:53,740 was the left extreme. 414 00:22:56,504 --> 00:22:57,670 That's the end of the proof. 415 00:22:57,670 --> 00:23:01,020 Any questions about that? 416 00:23:01,020 --> 00:23:05,060 That's Kawasaki's theorem-- Kawasaki-Justin. 417 00:23:05,060 --> 00:23:09,850 I would mention just for fun-- this is sort of classic-- 418 00:23:09,850 --> 00:23:15,410 if you allow one of these-- a non-convex cone, 419 00:23:15,410 --> 00:23:18,250 where you have more than 360 degrees of material-- 420 00:23:18,250 --> 00:23:20,550 this statement is not true. 421 00:23:20,550 --> 00:23:24,197 And there's one other situation which can happen, 422 00:23:24,197 --> 00:23:26,530 which is that the alternating sum of angles is not zero, 423 00:23:26,530 --> 00:23:30,430 but it's plus or minus 360 degrees, which is fun. 424 00:23:30,430 --> 00:23:32,550 And you can see examples of that in the book. 425 00:23:36,930 --> 00:23:37,670 All right. 426 00:23:37,670 --> 00:23:39,260 So obviously, if you want to tell 427 00:23:39,260 --> 00:23:42,661 whether a crease pattern-- or a single vertex is flat foldable, 428 00:23:42,661 --> 00:23:44,410 you just compute this thing in linear time 429 00:23:44,410 --> 00:23:48,220 and you know yes or no. 430 00:23:48,220 --> 00:23:51,290 So crease patterns are pretty easy-- just 431 00:23:51,290 --> 00:23:53,840 have one extra condition. 432 00:23:53,840 --> 00:23:55,580 What about mountain valley assignments? 433 00:23:55,580 --> 00:23:58,100 Last time, we looked at mountain valley assignments 434 00:23:58,100 --> 00:24:01,214 and we showed that crimps and end folds were always 435 00:24:01,214 --> 00:24:03,380 enough to fold any mountain valley assignment that's 436 00:24:03,380 --> 00:24:05,760 out there. 437 00:24:05,760 --> 00:24:08,002 That will continue to be true here, 438 00:24:08,002 --> 00:24:09,460 except now we don't have end folds. 439 00:24:09,460 --> 00:24:13,141 Now, it's crimps all the way. 440 00:24:13,141 --> 00:24:13,890 What does it mean? 441 00:24:16,750 --> 00:24:25,725 So that is the flat foldable mountain valley patterns. 442 00:24:34,267 --> 00:24:36,100 We're going to prove that crimps are enough, 443 00:24:36,100 --> 00:24:37,724 but before we get there, I need a bunch 444 00:24:37,724 --> 00:24:43,570 of sort of warm up facts about valid mountain valley patterns. 445 00:24:43,570 --> 00:24:49,530 The first thing is just counting mountains and valleys. 446 00:24:49,530 --> 00:24:52,650 So even if you play with a simple example like this one, 447 00:24:52,650 --> 00:24:56,420 you see it as three mountains and one valley 448 00:24:56,420 --> 00:24:58,790 or three valleys and one mountain. 449 00:24:58,790 --> 00:25:02,525 In fact, if you have a degree 4 vertex, 450 00:25:02,525 --> 00:25:04,780 with four creases emanating from that point, 451 00:25:04,780 --> 00:25:06,830 those are your only choices. 452 00:25:06,830 --> 00:25:09,770 You could try, for example, making it all valleys, 453 00:25:09,770 --> 00:25:14,350 and it-- I mean, you can't do it. 454 00:25:14,350 --> 00:25:16,110 It's hard to demonstrate. 455 00:25:16,110 --> 00:25:19,200 You could try making it two mountains and two valleys. 456 00:25:19,200 --> 00:25:21,330 Maybe I try to do mountain mountain here 457 00:25:21,330 --> 00:25:23,940 and then somehow this is going to be valley valley there. 458 00:25:23,940 --> 00:25:26,740 It's no good. 459 00:25:26,740 --> 00:25:33,620 But what's happening here is that the number of mountains 460 00:25:33,620 --> 00:25:39,455 and the number valleys must differ by exactly two. 461 00:25:39,455 --> 00:25:41,920 It doesn't matter who's bigger. 462 00:25:41,920 --> 00:25:44,440 So the difference could be plus 2 or minus 2, because you 463 00:25:44,440 --> 00:25:47,880 can always flip over and negate that difference, swapping 464 00:25:47,880 --> 00:25:49,900 mountains for valleys. 465 00:25:49,900 --> 00:25:51,950 But they always have to differ by exactly two. 466 00:25:54,590 --> 00:25:56,510 Why is that? 467 00:25:56,510 --> 00:26:01,580 Well, this is-- yeah, we're again 468 00:26:01,580 --> 00:26:03,990 going to think of a picture like this. 469 00:26:03,990 --> 00:26:07,140 We have horizontal travel according to the theta i's. 470 00:26:07,140 --> 00:26:10,290 We're also going to think about the vertical picture. 471 00:26:10,290 --> 00:26:12,900 If this thing is flat foldable-- so I'm assuming, 472 00:26:12,900 --> 00:26:16,950 it's flat foldable-- then this must happen. 473 00:26:16,950 --> 00:26:19,850 This is called Maekawa's theorem, by the way. 474 00:26:19,850 --> 00:26:29,740 It was proved by Jun Maekawa in 1986-- a little earlier. 475 00:26:29,740 --> 00:26:34,560 Also proved by Jacques Justin around the same time-- 476 00:26:34,560 --> 00:26:37,480 back before we were good at internet and communication 477 00:26:37,480 --> 00:26:39,410 and whatnot. 478 00:26:39,410 --> 00:26:42,650 And there were little pockets of mathematical origamists 479 00:26:42,650 --> 00:26:45,160 and they found each other basically 480 00:26:45,160 --> 00:26:51,730 in '89, when there was the-- '89 was the first origami science 481 00:26:51,730 --> 00:26:53,715 math and education conference-- started 482 00:26:53,715 --> 00:26:56,010 bringing these people together. 483 00:26:56,010 --> 00:26:57,270 The last one was this summer. 484 00:26:59,700 --> 00:27:00,200 Right. 485 00:27:00,200 --> 00:27:01,170 So why is this true? 486 00:27:06,819 --> 00:27:07,360 I don't know. 487 00:27:07,360 --> 00:27:09,730 A flat folding looks like something. 488 00:27:09,730 --> 00:27:11,980 It has no crossings. 489 00:27:11,980 --> 00:27:16,020 It has-- the horizontal travel, we understand. 490 00:27:16,020 --> 00:27:17,810 It ends up back where it started. 491 00:27:17,810 --> 00:27:21,600 There's also some notion of layers and vertical travel, 492 00:27:21,600 --> 00:27:23,451 which is a little tricky to think about. 493 00:27:23,451 --> 00:27:25,450 But the one thing that's easy to think about-- I 494 00:27:25,450 --> 00:27:27,240 mean, this forms essentially a polygon. 495 00:27:27,240 --> 00:27:29,080 It's like a really squashed polygon. 496 00:27:29,080 --> 00:27:32,940 And these vertical segments are actually really just points. 497 00:27:32,940 --> 00:27:35,190 OK, imagine. 498 00:27:35,190 --> 00:27:37,250 That's just where you turn around. 499 00:27:37,250 --> 00:27:41,420 So it's like a polygon, just stretched or squashed 500 00:27:41,420 --> 00:27:43,040 down onto a line. 501 00:27:43,040 --> 00:27:46,470 And we know some things about polygons. 502 00:27:46,470 --> 00:27:47,990 OK, as an end vertex polygon-- you 503 00:27:47,990 --> 00:27:50,070 know that the sum of the interior angles 504 00:27:50,070 --> 00:27:51,664 is-- whatever it is. 505 00:27:51,664 --> 00:27:53,330 I always get confused with that formula. 506 00:27:53,330 --> 00:27:54,746 So I don't like to think about it. 507 00:27:54,746 --> 00:27:56,770 But one thing we-- an easier way to think 508 00:27:56,770 --> 00:27:59,810 about that same statement is just think about how much 509 00:27:59,810 --> 00:28:01,550 turning the polygon does. 510 00:28:01,550 --> 00:28:04,480 So you start here, you turn right 180 degrees, 511 00:28:04,480 --> 00:28:08,720 you turn left 180 degrees, you turn left 180, right 180, 512 00:28:08,720 --> 00:28:10,630 right 180, right 180. 513 00:28:10,630 --> 00:28:12,420 How much is it in total? 514 00:28:12,420 --> 00:28:14,570 Oh gosh, do I have to add? 515 00:28:14,570 --> 00:28:15,550 No. 516 00:28:15,550 --> 00:28:16,250 It's so simple. 517 00:28:16,250 --> 00:28:17,080 It's 360. 518 00:28:17,080 --> 00:28:17,580 Right? 519 00:28:17,580 --> 00:28:19,590 Just, if you think about if you're going around 520 00:28:19,590 --> 00:28:23,780 in a circle, any polygon-- not just flat-- but anything in two 521 00:28:23,780 --> 00:28:27,450 dimensions, the total amount of turning you do is 360. 522 00:28:27,450 --> 00:28:30,960 Or if you count backwards, it's negative 360. 523 00:28:30,960 --> 00:28:33,320 So this is the key-- something you should all 524 00:28:33,320 --> 00:28:35,120 know about polygons. 525 00:28:35,120 --> 00:28:37,740 If you look at sum of the turn angles-- how much turning you 526 00:28:37,740 --> 00:28:39,720 do at each vertex-- that will always 527 00:28:39,720 --> 00:28:42,885 be plus or minus 360 in any polygon. 528 00:28:47,390 --> 00:28:49,880 This is equivalent to the sum of the interior angles being, 529 00:28:49,880 --> 00:28:52,820 whatever, pi times n minus 2, which I don't remember. 530 00:28:52,820 --> 00:28:56,660 But this is much easier to remember. 531 00:28:56,660 --> 00:29:00,980 And it's useful, because we can map mountains and valleys 532 00:29:00,980 --> 00:29:02,040 to left and right turns. 533 00:29:02,040 --> 00:29:04,510 As I said, every time this is a valley, 534 00:29:04,510 --> 00:29:08,750 it was a left turn by 180. 535 00:29:08,750 --> 00:29:11,905 Every time it was a mountain, it was a right turn by 180. 536 00:29:11,905 --> 00:29:32,790 So valleys, mountains, right turns-- 537 00:29:32,790 --> 00:29:34,869 so I'm going to think of that as negative 180. 538 00:29:34,869 --> 00:29:37,410 Now, of course technically, it could be the other way around. 539 00:29:37,410 --> 00:29:39,368 It could be this is negative, this is positive. 540 00:29:39,368 --> 00:29:41,740 But I already have a plus and minus here. 541 00:29:41,740 --> 00:29:44,340 So it's symmetric. 542 00:29:44,340 --> 00:29:46,380 So if we sum up the turn angles-- which 543 00:29:46,380 --> 00:29:49,080 we know is supposed to be plus or minus 360-- 544 00:29:49,080 --> 00:29:57,360 the sum of the turn angles is going to be-- we're 545 00:29:57,360 --> 00:30:03,090 going to have 180 for each valley, so 180 546 00:30:03,090 --> 00:30:06,050 times the number of valleys. 547 00:30:06,050 --> 00:30:09,370 And we're going to have negative 180 for each mountain. 548 00:30:15,860 --> 00:30:17,920 And so the 180's factor out. 549 00:30:30,740 --> 00:30:36,080 And this thing is supposed to equal 360-- plus or minus 360. 550 00:30:36,080 --> 00:30:36,882 How could that be? 551 00:30:36,882 --> 00:30:39,090 Well, the number of valleys minus number of mountains 552 00:30:39,090 --> 00:30:41,810 should be either 2-- for plus 360-- or negative 2-- 553 00:30:41,810 --> 00:30:43,895 for minus 360. 554 00:30:43,895 --> 00:30:45,020 So that proves the theorem. 555 00:30:49,250 --> 00:30:51,670 And that's why, in a degree 4 vertex, one of them 556 00:30:51,670 --> 00:30:54,860 has to be 3, the other one has to be 1. 557 00:30:54,860 --> 00:30:57,905 There's no other option, because the total number is 4. 558 00:30:57,905 --> 00:30:59,530 It's the only way to get the difference 559 00:30:59,530 --> 00:31:02,260 to be plus or minus 2. 560 00:31:02,260 --> 00:31:04,890 So that's kind of nice. 561 00:31:04,890 --> 00:31:06,447 But it's not enough, unfortunately. 562 00:31:06,447 --> 00:31:08,280 If I gave you a mountain valley pattern that 563 00:31:08,280 --> 00:31:10,590 satisfies this condition, it still 564 00:31:10,590 --> 00:31:14,120 might not be flat foldable. 565 00:31:14,120 --> 00:31:15,380 That's maybe no surprise. 566 00:31:15,380 --> 00:31:17,830 It's not like we had such a simple test 567 00:31:17,830 --> 00:31:20,670 for the one dimensional case, which should be easier. 568 00:31:20,670 --> 00:31:21,940 We had to give an algorithm. 569 00:31:21,940 --> 00:31:23,920 We said repeatedly crimp and end fold. 570 00:31:23,920 --> 00:31:25,910 If you get stuck, the answer is no. 571 00:31:25,910 --> 00:31:27,895 If you don't get stuck, you folded it great. 572 00:31:43,280 --> 00:31:45,405 I love this eraser-- increases entropy. 573 00:31:51,110 --> 00:31:55,040 Let me tell you one case that is easy to think about. 574 00:31:55,040 --> 00:31:56,340 Because it's nice. 575 00:31:56,340 --> 00:31:59,140 It reduces directly to the one dimensional case. 576 00:31:59,140 --> 00:32:01,060 And that's what I call the generic case. 577 00:32:06,380 --> 00:32:09,010 Generic is this very convenient term we use in mathematics. 578 00:32:09,010 --> 00:32:12,450 It's actually really tricky to define. 579 00:32:12,450 --> 00:32:15,260 So I'd like to not define it, if I can. 580 00:32:15,260 --> 00:32:16,490 But maybe I should. 581 00:32:22,320 --> 00:32:24,410 The simple version, which is not quite enough, 582 00:32:24,410 --> 00:32:26,920 is to say that all of the angles in the crease pattern 583 00:32:26,920 --> 00:32:28,991 are different. 584 00:32:28,991 --> 00:32:30,990 OK, here, for example, two of them are the same. 585 00:32:30,990 --> 00:32:31,920 There's two 90 degree angles. 586 00:32:31,920 --> 00:32:32,990 This is not generic. 587 00:32:32,990 --> 00:32:35,030 It's just easier to draw. 588 00:32:35,030 --> 00:32:36,770 But in general, I mean, you imagine, 589 00:32:36,770 --> 00:32:39,110 you just draw some random thing. 590 00:32:39,110 --> 00:32:41,080 None of those lengths are going to be the same. 591 00:32:41,080 --> 00:32:42,920 Yeah, the alternating sum is zero. 592 00:32:42,920 --> 00:32:45,290 But that's it. 593 00:32:45,290 --> 00:32:46,790 And that's what I-- the generic case 594 00:32:46,790 --> 00:32:50,870 is that is the only thing that holds true about these angles. 595 00:32:50,870 --> 00:32:52,670 You can take the alternating sum. 596 00:32:52,670 --> 00:32:53,990 That equals zero. 597 00:32:53,990 --> 00:32:56,680 If you take some alternating sum of some subset of the angles, 598 00:32:56,680 --> 00:32:58,190 that will not equal zero. 599 00:32:58,190 --> 00:33:00,760 If you take a random example, this is going to be true. 600 00:33:00,760 --> 00:33:02,700 OK, just bear with me. 601 00:33:02,700 --> 00:33:06,760 Suppose never any two angles are the same. 602 00:33:06,760 --> 00:33:11,230 Then look at the globally smallest crease, 603 00:33:11,230 --> 00:33:13,230 so globally smallest theta. 604 00:33:18,380 --> 00:33:20,000 Say it's theta i. 605 00:33:20,000 --> 00:33:27,961 So the picture is theta i, theta i minus 1, theta i plus 1. 606 00:33:27,961 --> 00:33:29,460 And what do we know if it's globally 607 00:33:29,460 --> 00:33:32,080 smallest and none of the values are equal? 608 00:33:32,080 --> 00:33:38,035 Then we know that these two are bigger. 609 00:33:38,035 --> 00:33:40,250 No big surprise. 610 00:33:40,250 --> 00:33:44,780 Think about what happens if this is small, these are bigger, 611 00:33:44,780 --> 00:33:48,890 and you try to make these both valleys or both mountains, 612 00:33:48,890 --> 00:33:50,200 that's, of course, bad. 613 00:33:50,200 --> 00:33:54,860 This is one of the situations we had 614 00:33:54,860 --> 00:33:56,380 in the one dimensional case. 615 00:33:56,380 --> 00:33:57,930 I mean, really, this is a circle, 616 00:33:57,930 --> 00:34:00,640 but I can think just locally about what's happening here. 617 00:34:00,640 --> 00:34:03,990 So in this situation, one of these must be a mountain 618 00:34:03,990 --> 00:34:05,890 and the other must be a valley. 619 00:34:05,890 --> 00:34:08,360 I don't know which order they are, 620 00:34:08,360 --> 00:34:10,040 but I know they're different. 621 00:34:10,040 --> 00:34:15,670 And then I know I can apply one of these crimps. 622 00:34:15,670 --> 00:34:19,330 Incidentally, the word crimp, Jason Koo was asking about, 623 00:34:19,330 --> 00:34:22,120 like, why do you call it crimp and not pleat? 624 00:34:22,120 --> 00:34:23,639 Probably, I should call it pleat. 625 00:34:23,639 --> 00:34:25,940 But it's crimp in my mind. 626 00:34:25,940 --> 00:34:27,980 It's been crimp in my mind since 1998 627 00:34:27,980 --> 00:34:29,750 when we wrote the 1D paper. 628 00:34:29,750 --> 00:34:33,480 It was before I knew the word pleat, so that's my excuse. 629 00:34:33,480 --> 00:34:35,710 So pleat or crimp, take your pick. 630 00:34:35,710 --> 00:34:37,503 Crimp is usually many pleats together. 631 00:34:37,503 --> 00:34:38,169 That's crimping. 632 00:34:41,190 --> 00:34:41,870 Cool. 633 00:34:41,870 --> 00:34:44,090 So crimps, we kind of like. 634 00:34:44,090 --> 00:34:45,354 Why do we like crimps? 635 00:34:48,820 --> 00:34:51,550 We proved, in the one dimensional case last time, 636 00:34:51,550 --> 00:34:54,320 that if you make a crimp and your original thing 637 00:34:54,320 --> 00:34:57,690 was flat foldable, the new thing will be flat foldable. 638 00:34:57,690 --> 00:34:59,679 Good news is, that is still true. 639 00:34:59,679 --> 00:35:01,470 I told you, all these things were no longer 640 00:35:01,470 --> 00:35:02,840 true in the circular case. 641 00:35:02,840 --> 00:35:06,500 It's still true in the circular case. 642 00:35:06,500 --> 00:35:07,500 Should I tell you-- 643 00:35:07,500 --> 00:35:10,180 Let me just remind you-- the same proof works-- remind you 644 00:35:10,180 --> 00:35:11,620 what the proof looked like. 645 00:35:11,620 --> 00:35:16,900 We had a crimp, and we're hoping to fold that first. 646 00:35:16,900 --> 00:35:19,160 We know at some point, these creases get folded. 647 00:35:19,160 --> 00:35:20,990 But there may be other junk in here, 648 00:35:20,990 --> 00:35:23,110 or maybe other junk in here. 649 00:35:23,110 --> 00:35:24,950 We'd like to get rid of that junk. 650 00:35:24,950 --> 00:35:27,010 And so what we did is move this junk up 651 00:35:27,010 --> 00:35:28,360 to here, which was always safe. 652 00:35:28,360 --> 00:35:30,400 We moved this junk down to here. 653 00:35:30,400 --> 00:35:35,598 And that was still a folded state-- a flat folded state. 654 00:35:35,598 --> 00:35:38,980 OK, remember all this stuff. 655 00:35:38,980 --> 00:35:41,190 Therefore, we could have folded the crimp first. 656 00:35:41,190 --> 00:35:44,010 And then we have a flat folding in this situation, where 657 00:35:44,010 --> 00:35:48,220 this paper is effectively glued together, which means, 658 00:35:48,220 --> 00:35:50,480 it was safe to fold this crimp and glue these 659 00:35:50,480 --> 00:35:52,829 together and never touch them again. 660 00:35:52,829 --> 00:35:54,620 So that is still true in the circular case. 661 00:35:54,620 --> 00:35:57,740 That proof didn't really assume anything about the paper. 662 00:35:57,740 --> 00:35:59,290 Would even work for like tree shaped 663 00:35:59,290 --> 00:36:02,670 paper or something weird. 664 00:36:02,670 --> 00:36:06,610 So if we can find a crimp, which we can in the generic case, 665 00:36:06,610 --> 00:36:08,690 then you just make a crimp, make a crimp, repeat. 666 00:36:08,690 --> 00:36:11,050 If you ever get stuck, you know that the original thing was not 667 00:36:11,050 --> 00:36:13,675 flat foldable, because the thing you have is not flat foldable. 668 00:36:15,810 --> 00:36:16,380 Why? 669 00:36:16,380 --> 00:36:18,620 If there's no cramp in the generic case, 670 00:36:18,620 --> 00:36:21,740 that means you have two valleys around the smallest angle, 671 00:36:21,740 --> 00:36:22,950 and that's clearly bad. 672 00:36:26,020 --> 00:36:28,610 So-- I'm not writing a lot of details here. 673 00:36:28,610 --> 00:36:31,040 Is that clear? 674 00:36:31,040 --> 00:36:32,950 If I have an angle that's surrounded 675 00:36:32,950 --> 00:36:34,911 by strictly larger angles, I know 676 00:36:34,911 --> 00:36:36,660 there must be one mountain and one valley. 677 00:36:36,660 --> 00:36:39,600 Then I can safely make a crimp and repeat. 678 00:36:39,600 --> 00:36:43,500 The trouble comes in when these angles are equal. 679 00:36:43,500 --> 00:36:45,164 We never had to think about angles 680 00:36:45,164 --> 00:36:46,830 being equal in the one dimensional case. 681 00:36:46,830 --> 00:36:49,090 It didn't matter much. 682 00:36:49,090 --> 00:36:52,400 We just said, oh, you know, it's safe to make a crimp, 683 00:36:52,400 --> 00:36:55,490 as long as this was greater than or equal to this 684 00:36:55,490 --> 00:36:57,740 and this was greater than or equal to this. 685 00:36:57,740 --> 00:36:59,992 That is still true; it's safe to make a crimp. 686 00:36:59,992 --> 00:37:01,700 But how do we know that there is a crimp? 687 00:37:01,700 --> 00:37:05,720 How do we know that there's a mountain followed by valley? 688 00:37:05,720 --> 00:37:06,420 We don't. 689 00:37:06,420 --> 00:37:11,400 I mean, if they're equal, this would be all right. 690 00:37:11,400 --> 00:37:14,580 That's two valleys is valid if these three angles are equal. 691 00:37:14,580 --> 00:37:17,144 But I claim still somewhere there's got to be a crimp. 692 00:37:17,144 --> 00:37:19,560 There's no longer a notion of the globally smallest angle, 693 00:37:19,560 --> 00:37:21,380 because some of them are equal. 694 00:37:21,380 --> 00:37:24,770 We've got to find that crimp, but it's out there. 695 00:37:24,770 --> 00:37:27,115 So we just need to prove that it exists. 696 00:37:57,960 --> 00:38:02,670 So to do that, we're going to generalize this theorem-- 697 00:38:02,670 --> 00:38:05,440 Meakawa-Justin-- that number of mountains minus number 698 00:38:05,440 --> 00:38:06,730 of valleys is plus or minus 2. 699 00:38:06,730 --> 00:38:08,930 That's true, and it's a statement 700 00:38:08,930 --> 00:38:13,710 about the sum of all the angles-- all of the creases. 701 00:38:13,710 --> 00:38:16,455 What I'd like is something a little bit more localized. 702 00:38:16,455 --> 00:38:18,080 And in particular, we're going to think 703 00:38:18,080 --> 00:38:22,282 about when I have a whole bunch of angles that are equal, 704 00:38:22,282 --> 00:38:24,490 how many mountains and valleys can there be in there. 705 00:38:27,480 --> 00:38:30,440 So it's what I call local counts. 706 00:38:35,170 --> 00:38:37,770 So while everything I said so far is pretty classic, 707 00:38:37,770 --> 00:38:41,560 this result was proved in 2001 by Tom Hall, 708 00:38:41,560 --> 00:38:45,910 so much more recent, over a decade later. 709 00:38:48,580 --> 00:38:52,140 So let's think about k equal angles. 710 00:38:58,640 --> 00:39:01,400 And I want that to be a maximal sequence of equal angles, 711 00:39:01,400 --> 00:39:05,560 so the ones right after and right before are different. 712 00:39:05,560 --> 00:39:07,550 And furthermore, I'm going to assume 713 00:39:07,550 --> 00:39:20,520 that they're bigger-- strictly larger angles. 714 00:39:23,320 --> 00:39:28,710 OK, so something like that. 715 00:39:28,710 --> 00:39:31,230 A bunch of equal angles-- k of them-- 716 00:39:31,230 --> 00:39:33,960 bigger angles on either side. 717 00:39:33,960 --> 00:39:36,660 This almost always exists. 718 00:39:36,660 --> 00:39:38,890 I should mention-- so I'm going to say something 719 00:39:38,890 --> 00:39:39,930 about this situation. 720 00:39:39,930 --> 00:39:42,820 But this situation should exist, because I take the smallest 721 00:39:42,820 --> 00:39:45,630 angle out there and then I see how many friends 722 00:39:45,630 --> 00:39:47,270 it has that are all equal. 723 00:39:47,270 --> 00:39:50,070 If it's the smallest angle-- or it's one of the smallest 724 00:39:50,070 --> 00:39:52,900 angles-- then the ones surrounding 725 00:39:52,900 --> 00:39:57,790 it are going to be bigger-- unless all the angles are 726 00:39:57,790 --> 00:39:58,466 equal. 727 00:39:58,466 --> 00:40:00,590 So there's one case, which we'll worry about later. 728 00:40:00,590 --> 00:40:01,690 It's very easy. 729 00:40:01,690 --> 00:40:04,197 All the angles are equal. 730 00:40:04,197 --> 00:40:06,280 In fact, I could tell you how to worry about that. 731 00:40:06,280 --> 00:40:11,279 If all the angles are equal, then everything 732 00:40:11,279 --> 00:40:13,820 is crimpable as long as we can find a switch between mountain 733 00:40:13,820 --> 00:40:16,142 and valley-- somewhere. 734 00:40:16,142 --> 00:40:18,100 And because the number of mountains and valleys 735 00:40:18,100 --> 00:40:21,660 is plus or minus 2, there's got to be-- so they're almost 736 00:40:21,660 --> 00:40:25,960 in equal balance-- there's got to be a valley. 737 00:40:25,960 --> 00:40:27,984 It can't be all valleys, can't be all mountains. 738 00:40:27,984 --> 00:40:29,400 So somewhere, there's a transition 739 00:40:29,400 --> 00:40:31,180 between mountain and valley. 740 00:40:31,180 --> 00:40:34,490 So you get a crimp out of that. 741 00:40:34,490 --> 00:40:37,110 But otherwise, if they're not all equal-- let's 742 00:40:37,110 --> 00:40:43,230 think about k equal angles surrounded by two larger guys. 743 00:40:43,230 --> 00:40:45,321 Forgot how technical this is. 744 00:40:45,321 --> 00:40:45,820 All right. 745 00:40:45,820 --> 00:40:52,770 Then I want to look at the number of mountains 746 00:40:52,770 --> 00:40:58,020 and the number valleys among these k plus one creases, 747 00:40:58,020 --> 00:40:59,440 so within that range. 748 00:41:04,130 --> 00:41:07,460 It's going to be 0, if k is odd. 749 00:41:10,160 --> 00:41:13,210 And it's going to be plus or minus 1, if k is even. 750 00:41:17,140 --> 00:41:18,730 It's not plus or minus 2. 751 00:41:18,730 --> 00:41:22,130 Intuitively, plus or minus 2 is when you go all the way around. 752 00:41:22,130 --> 00:41:24,580 Here, we're just looking at a segment-- 753 00:41:24,580 --> 00:41:27,260 a portion of the entire circle. 754 00:41:27,260 --> 00:41:30,730 Remember, this is actually a circle somewhere. 755 00:41:30,730 --> 00:41:35,190 And this corresponds to, in the odd case, 756 00:41:35,190 --> 00:41:38,280 you're going to be going in the same direction afterwards. 757 00:41:38,280 --> 00:41:39,900 In the even case, you're going to have 758 00:41:39,900 --> 00:41:41,870 turned around but not a full circle. 759 00:41:41,870 --> 00:41:43,890 So it's going to be plus or minus 1. 760 00:41:43,890 --> 00:41:46,740 There's a lot of ways to prove this. 761 00:41:46,740 --> 00:41:50,770 One way is to think about convex cones. 762 00:41:50,770 --> 00:42:00,570 So if k is even, and you fold this thing-- let's see, 763 00:42:00,570 --> 00:42:06,409 one, two, three, four, five, six-- equal angles. 764 00:42:06,409 --> 00:42:07,950 And then there's the two longer guys. 765 00:42:07,950 --> 00:42:09,960 They may not be the same length, but that's 766 00:42:09,960 --> 00:42:13,660 what you get in the even case. 767 00:42:13,660 --> 00:42:18,114 So what I'd like to do is just, in order to just reduce 768 00:42:18,114 --> 00:42:20,280 to things that we've already thought about-- I mean, 769 00:42:20,280 --> 00:42:22,020 you could talk about turn angles again, 770 00:42:22,020 --> 00:42:25,040 or you could just say, hey, I know Maekawa's theorem. 771 00:42:25,040 --> 00:42:28,100 So as long as I could make this into a circle, which 772 00:42:28,100 --> 00:42:31,940 I can do by adding that stuff, now I 773 00:42:31,940 --> 00:42:34,620 have a nice circular-- it's a flat folded state. 774 00:42:34,620 --> 00:42:37,980 There's no crossings here, presumably 775 00:42:37,980 --> 00:42:39,210 because this did something. 776 00:42:39,210 --> 00:42:41,001 I don't actually know that it's zigzagging. 777 00:42:41,001 --> 00:42:42,140 It might do other stuff. 778 00:42:42,140 --> 00:42:45,195 But it's not going to cross what I just added. 779 00:42:45,195 --> 00:42:47,320 So the number of mountains minus number of valleys, 780 00:42:47,320 --> 00:42:51,280 in total here, must be plus or minus 2. 781 00:42:51,280 --> 00:42:55,160 I only added one crease-- this one. 782 00:42:55,160 --> 00:42:57,730 So that's going to make it either plus or minus 1 783 00:42:57,730 --> 00:42:59,302 or plus or minus 3. 784 00:42:59,302 --> 00:43:00,760 If you think about it a little bit, 785 00:43:00,760 --> 00:43:02,250 it has to be plus or minus 1. 786 00:43:02,250 --> 00:43:04,124 And hopefully, that's what I wrote here, yes. 787 00:43:06,760 --> 00:43:10,550 Because I mean, really, we're thinking about turn angles 788 00:43:10,550 --> 00:43:11,580 again, I'm afraid. 789 00:43:11,580 --> 00:43:13,330 Can't totally reduce it. 790 00:43:13,330 --> 00:43:15,766 We just turned around here. 791 00:43:15,766 --> 00:43:17,640 And we were about to finish the circle, which 792 00:43:17,640 --> 00:43:18,960 would make it plus or minus 2. 793 00:43:18,960 --> 00:43:23,510 Before we finish the circle, it's plus or minus 1. 794 00:43:23,510 --> 00:43:24,160 k is odd. 795 00:43:28,600 --> 00:43:31,080 Again, this might go like that, whatever. 796 00:43:31,080 --> 00:43:33,220 But, because it's an odd number, these guys 797 00:43:33,220 --> 00:43:36,030 are going in the same direction. 798 00:43:36,030 --> 00:43:39,120 And so the total turn angle here must be 0. 799 00:43:39,120 --> 00:43:44,240 And one way to see that is to extend it into a full polygon. 800 00:43:44,240 --> 00:43:48,250 You know that it's plus or minus 2 for this whole thing. 801 00:43:48,250 --> 00:43:52,100 And I added two guys in the same direction. 802 00:43:52,100 --> 00:43:54,820 If I remove them, it's going to go down to plus or minus 0. 803 00:43:59,929 --> 00:44:00,970 That was a bit hand wavy. 804 00:44:00,970 --> 00:44:02,145 But do you buy that? 805 00:44:05,823 --> 00:44:08,697 AUDIENCE: [INAUDIBLE] plus or minus 0? 806 00:44:08,697 --> 00:44:11,492 PROFESSOR: Plus or minus 0 is 0, yeah. 807 00:44:11,492 --> 00:44:13,200 You went from 2 down to zero, because you 808 00:44:13,200 --> 00:44:14,620 remove two identical guys. 809 00:44:18,730 --> 00:44:22,790 So let me tie this all together, because there's 810 00:44:22,790 --> 00:44:24,255 so many little cases. 811 00:44:34,630 --> 00:44:37,780 So I want to know what mountain valley patterns are 812 00:44:37,780 --> 00:44:49,005 flat foldable, single vertex, m/v patterns. 813 00:44:56,630 --> 00:45:04,710 I claim there is a crimp. 814 00:45:04,710 --> 00:45:06,890 If it's flat foldable, there has to be a crimp 815 00:45:06,890 --> 00:45:08,400 that you can find to do. 816 00:45:08,400 --> 00:45:14,330 Remember, a crimp was where this length 817 00:45:14,330 --> 00:45:18,240 was less than or equal to this length and less than 818 00:45:18,240 --> 00:45:20,250 or equal to this length. 819 00:45:20,250 --> 00:45:21,370 And one of these was m. 820 00:45:21,370 --> 00:45:23,800 One of them was v. That was the definition of a crimp. 821 00:45:23,800 --> 00:45:26,090 I claim such a thing always exists. 822 00:45:26,090 --> 00:45:30,230 The proof is two parts. 823 00:45:30,230 --> 00:45:39,310 If all the angles are equal, we know that these conditions 824 00:45:39,310 --> 00:45:43,080 are always satisfied, because everything is equal. 825 00:45:43,080 --> 00:45:45,180 So we just need to find an m followed by v 826 00:45:45,180 --> 00:45:47,010 or a v followed by an m. 827 00:45:47,010 --> 00:45:52,731 And by Maekawa's theorem, we know it's not all m's or all 828 00:45:52,731 --> 00:45:53,230 v's. 829 00:45:53,230 --> 00:45:56,441 So somewhere, there has to be a transition from m to v's. 830 00:46:04,150 --> 00:46:09,580 So just use that lemma theorem or the count property. 831 00:46:09,580 --> 00:46:12,890 Otherwise, if they're not all equal, 832 00:46:12,890 --> 00:46:16,250 then I can apply this local counts result, 833 00:46:16,250 --> 00:46:20,060 take the smallest angle-- take one instance of the smallest 834 00:46:20,060 --> 00:46:20,560 angle. 835 00:46:20,560 --> 00:46:21,770 There could be many of them. 836 00:46:21,770 --> 00:46:25,540 Find all of its neighboring friends that are equal. 837 00:46:25,540 --> 00:46:27,620 Then I know that the next angles before and after 838 00:46:27,620 --> 00:46:29,770 have to be strictly bigger. 839 00:46:29,770 --> 00:46:33,770 Then I know that the number of mountains and valleys in here-- 840 00:46:33,770 --> 00:46:35,330 if I look at the difference and it's 841 00:46:35,330 --> 00:46:37,762 0, that means the number of mountains 842 00:46:37,762 --> 00:46:38,970 equals the number of valleys. 843 00:46:38,970 --> 00:46:40,669 Therefore, there's at least one of each. 844 00:46:40,669 --> 00:46:42,710 And therefore, there's a transition from mountain 845 00:46:42,710 --> 00:46:44,250 to valley in there. 846 00:46:44,250 --> 00:46:46,390 Because the transition from mountain to valley-- 847 00:46:46,390 --> 00:46:49,020 like right here, that's a valid crimp. 848 00:46:49,020 --> 00:46:52,160 This guy's bigger; this guy's equal. 849 00:46:52,160 --> 00:46:54,030 This would also be a fine transition, 850 00:46:54,030 --> 00:46:55,114 because everybody's equal. 851 00:46:55,114 --> 00:46:56,654 This would also be a fine transition. 852 00:46:56,654 --> 00:46:58,890 Any transition from mountain to valley in here, 853 00:46:58,890 --> 00:47:00,660 I've got a crimp. 854 00:47:00,660 --> 00:47:04,480 The even case is a little-- I got to think a little bit more. 855 00:47:04,480 --> 00:47:07,610 k is even. 856 00:47:07,610 --> 00:47:12,140 k plus 1, which is the number of creases here, is odd. 857 00:47:12,140 --> 00:47:14,310 And the difference between mountains and valleys 858 00:47:14,310 --> 00:47:16,907 among those k plus one creases is plus or minus 1. 859 00:47:16,907 --> 00:47:17,990 Think about it for second. 860 00:47:17,990 --> 00:47:20,531 That means there's, again, at least one mountain and at least 861 00:47:20,531 --> 00:47:22,940 one valley. 862 00:47:22,940 --> 00:47:26,090 Because k, to be even, has to be at least 2. 863 00:47:26,090 --> 00:47:27,907 We've got to have at least three things. 864 00:47:27,907 --> 00:47:29,990 So there's going to be at least one of one and two 865 00:47:29,990 --> 00:47:31,486 of the other, in fact. 866 00:47:31,486 --> 00:47:34,110 And I just need one mountain and one valley somewhere in there. 867 00:47:34,110 --> 00:47:35,890 And then I know there has to be a transition from mountain 868 00:47:35,890 --> 00:47:36,470 to valley. 869 00:47:39,190 --> 00:47:42,140 Otherwise, use local counts. 870 00:47:45,890 --> 00:47:56,200 And in either case, we get either a mountain 871 00:47:56,200 --> 00:48:03,460 to valley transition or a valley to mountain transition. 872 00:48:03,460 --> 00:48:07,360 And I get it in such a way that these inequalities 873 00:48:07,360 --> 00:48:08,939 on the lengths hold. 874 00:48:08,939 --> 00:48:10,480 So in other words, that it's a crimp. 875 00:48:15,590 --> 00:48:18,520 Therefore, any flat foldable, single vertex, mountain valley 876 00:48:18,520 --> 00:48:20,320 pattern has a crimp. 877 00:48:20,320 --> 00:48:21,590 Make it. 878 00:48:21,590 --> 00:48:25,300 And as I was arguing here, just like last time, 879 00:48:25,300 --> 00:48:27,750 that crimp will still-- after you do the crimp, 880 00:48:27,750 --> 00:48:30,120 you'll still be flat foldable, so repeat. 881 00:48:30,120 --> 00:48:34,100 Cook until done is how I like to put it. 882 00:48:34,100 --> 00:48:34,600 Question. 883 00:48:34,600 --> 00:48:37,183 AUDIENCE: What about the trivial case of just folding a circle 884 00:48:37,183 --> 00:48:37,827 in half? 885 00:48:37,827 --> 00:48:39,930 PROFESSOR: The trivial case of folding a circle in half. 886 00:48:39,930 --> 00:48:40,430 All right. 887 00:48:40,430 --> 00:48:41,450 That's a good question. 888 00:48:41,450 --> 00:48:42,616 It's kind of a technicality. 889 00:48:45,650 --> 00:48:48,309 One version would be to say, that's zero vertices. 890 00:48:48,309 --> 00:48:49,100 That doesn't count. 891 00:48:49,100 --> 00:48:50,760 And we're thinking about one vertex. 892 00:48:50,760 --> 00:48:52,134 And as soon as you think about it 893 00:48:52,134 --> 00:48:54,730 as one vertex, which is fine, then you actually 894 00:48:54,730 --> 00:48:58,510 have two creases there. 895 00:48:58,510 --> 00:48:59,570 So it's two creases. 896 00:48:59,570 --> 00:49:02,052 And hopefully, this works. 897 00:49:02,052 --> 00:49:03,260 Oh, I see what you're saying. 898 00:49:03,260 --> 00:49:04,890 It's not a crimp. 899 00:49:04,890 --> 00:49:05,740 That's a good point. 900 00:49:05,740 --> 00:49:08,434 AUDIENCE: But then it's flat foldable. 901 00:49:08,434 --> 00:49:10,850 PROFESSOR: It is flat foldable, and yet, there's no crimp. 902 00:49:10,850 --> 00:49:13,061 Damn it. 903 00:49:13,061 --> 00:49:13,560 All right. 904 00:49:13,560 --> 00:49:14,060 Sorry? 905 00:49:14,060 --> 00:49:17,080 AUDIENCE: But it follows Maekawa's, right? 906 00:49:17,080 --> 00:49:18,260 PROFESSOR: Yeah. 907 00:49:18,260 --> 00:49:21,262 So I made a slight mistake here. 908 00:49:21,262 --> 00:49:22,720 I said if all the angles are equal, 909 00:49:22,720 --> 00:49:24,760 then Maekawa says there's at least one of each. 910 00:49:24,760 --> 00:49:27,230 That's not true when n equals 2. 911 00:49:27,230 --> 00:49:30,450 Then there's two of one and zero of the other. 912 00:49:30,450 --> 00:49:36,060 As long as you have more than two creases, then this is true. 913 00:49:36,060 --> 00:49:36,865 That's enough. 914 00:49:36,865 --> 00:49:38,740 So you have to handle this as a special case. 915 00:49:38,740 --> 00:49:40,365 AUDIENCE: At the end of your algorithm, 916 00:49:40,365 --> 00:49:43,396 you're always left with two of the same-- 917 00:49:43,396 --> 00:49:45,232 PROFESSOR: That's true. 918 00:49:45,232 --> 00:49:47,390 This is an important special case to remember, 919 00:49:47,390 --> 00:49:50,210 because this will be the final picture. 920 00:49:50,210 --> 00:49:54,882 After you do a sequence of crimps, this is the good case. 921 00:49:54,882 --> 00:49:56,340 If your thing is flat foldable, you 922 00:49:56,340 --> 00:49:57,435 will always end up with that. 923 00:49:57,435 --> 00:49:59,200 AUDIENCE: That's a case where everything 924 00:49:59,200 --> 00:50:02,905 is the same, smallest. 925 00:50:02,905 --> 00:50:06,530 PROFESSOR: It's another example where all the angles are equal. 926 00:50:06,530 --> 00:50:09,130 So that's why this is the situation. 927 00:50:09,130 --> 00:50:14,070 It's either Maekawa and you get a cramp here, 928 00:50:14,070 --> 00:50:16,500 because n is greater than 2. 929 00:50:16,500 --> 00:50:20,590 And here, I should say, if n is greater than 2, 930 00:50:20,590 --> 00:50:21,740 then this is true. 931 00:50:21,740 --> 00:50:24,484 Of course the n equals 2 case, it looks like that. 932 00:50:24,484 --> 00:50:26,650 And it had better be both valleys or both mountains. 933 00:50:26,650 --> 00:50:28,120 Otherwise, it's not going to be flat foldable. 934 00:50:28,120 --> 00:50:28,970 I forgot about that. 935 00:50:28,970 --> 00:50:30,200 I should add it in the notes. 936 00:50:30,200 --> 00:50:31,315 But it's in the textbook, right. 937 00:50:31,315 --> 00:50:32,964 AUDIENCE: At the end of the algorithm, 938 00:50:32,964 --> 00:50:34,255 it's not going to be a circle. 939 00:50:34,255 --> 00:50:35,254 It's going to be a cone. 940 00:50:35,254 --> 00:50:36,300 PROFESSOR: That's true. 941 00:50:36,300 --> 00:50:38,650 It won't actually look like this. 942 00:50:38,650 --> 00:50:41,330 At the end of the algorithm, it's going to be a cone. 943 00:50:41,330 --> 00:50:45,590 The two angles will be equal, because of Kawasaki's theorem. 944 00:50:45,590 --> 00:50:48,700 And they should still be both mountains or both valleys. 945 00:50:48,700 --> 00:50:51,170 But it's going to be-- we can see it right here. 946 00:50:51,170 --> 00:50:54,522 So I apply my crimp, and now I have a cone, 947 00:50:54,522 --> 00:50:55,480 and there's two angles. 948 00:50:55,480 --> 00:50:58,850 There's one here, which has been fused together. 949 00:50:58,850 --> 00:51:01,670 And there's one here, which is the original. 950 00:51:01,670 --> 00:51:02,950 They're equal. 951 00:51:02,950 --> 00:51:05,530 And it's two mountains or two valleys, 952 00:51:05,530 --> 00:51:07,290 if I turned it upside down. 953 00:51:07,290 --> 00:51:08,880 And then it finishes. 954 00:51:08,880 --> 00:51:11,550 So that's the algorithm in action-- two steps. 955 00:51:11,550 --> 00:51:13,146 First you find this crimp here. 956 00:51:13,146 --> 00:51:15,020 This is actually the globally smallest angle. 957 00:51:15,020 --> 00:51:16,395 It's surrounded by bigger angles. 958 00:51:16,395 --> 00:51:17,680 So it's really easy. 959 00:51:17,680 --> 00:51:21,360 Then I have a cone with two equal angles, which 960 00:51:21,360 --> 00:51:23,130 is what has to happen at the end. 961 00:51:23,130 --> 00:51:27,241 And then I'm done. 962 00:51:27,241 --> 00:51:27,740 Good? 963 00:51:27,740 --> 00:51:31,170 AUDIENCE: Is that the only way to do it? 964 00:51:31,170 --> 00:51:34,430 PROFESSOR: This is what always happens. 965 00:51:34,430 --> 00:51:37,830 Oh, is this the only way to flat fold these things? 966 00:51:37,830 --> 00:51:38,550 No. 967 00:51:38,550 --> 00:51:42,290 There are other-- yes, one of the two. 968 00:51:53,060 --> 00:51:56,620 I think it depends what you're counting. 969 00:51:56,620 --> 00:51:59,430 If you're counting mountain valley assignments-- I mean, 970 00:51:59,430 --> 00:52:02,100 if you just want to know, can every mountain valley 971 00:52:02,100 --> 00:52:03,812 assignment be folded by crimping. 972 00:52:03,812 --> 00:52:04,770 Then the answer is yes. 973 00:52:04,770 --> 00:52:05,890 That's what we proved. 974 00:52:05,890 --> 00:52:08,694 But if you want to get every possible folded state, 975 00:52:08,694 --> 00:52:10,110 crimps are not going to be enough. 976 00:52:10,110 --> 00:52:13,130 The reason they're not enough is because we're using this thing. 977 00:52:13,130 --> 00:52:15,570 So, hey, here's some hypothetical folded state. 978 00:52:15,570 --> 00:52:17,890 We can rip it out and make the crimp has been done. 979 00:52:17,890 --> 00:52:19,700 But then there's a folded state you're not 980 00:52:19,700 --> 00:52:21,970 able to reach by crimping. 981 00:52:21,970 --> 00:52:24,927 We need to work that into an actual example, where crimping 982 00:52:24,927 --> 00:52:26,760 forbids you from reaching some folded state. 983 00:52:26,760 --> 00:52:28,868 But I'm pretty sure one exists because of this. 984 00:52:28,868 --> 00:52:30,618 AUDIENCE: Could that be more like spirals? 985 00:52:30,618 --> 00:52:32,890 PROFESSOR: Yeah, maybe some kind of spiraling thing. 986 00:52:32,890 --> 00:52:35,102 AUDIENCE: Well, layering takes into account-- 987 00:52:35,102 --> 00:52:35,810 PROFESSOR: Right. 988 00:52:35,810 --> 00:52:38,490 What we're missing here is getting all the possible layer 989 00:52:38,490 --> 00:52:40,020 stacking orders. 990 00:52:40,020 --> 00:52:43,290 So we're just trying to match a mountain valley assignment. 991 00:52:43,290 --> 00:52:45,360 We're not going to match a target layer ordering, 992 00:52:45,360 --> 00:52:48,360 because we simplified the layer orders in order 993 00:52:48,360 --> 00:52:50,790 to make crimps possible. 994 00:52:50,790 --> 00:52:55,980 So crimps won't get you some of the folded states. 995 00:52:55,980 --> 00:53:04,730 But one thing you can play with over here-- 996 00:53:04,730 --> 00:53:08,400 that has been played with by Tom Hall-- 997 00:53:08,400 --> 00:53:15,440 is if I give you a crease pattern, 998 00:53:15,440 --> 00:53:18,840 how many flat foldable mountain valley 999 00:53:18,840 --> 00:53:20,439 assignments does it have? 1000 00:53:20,439 --> 00:53:22,730 OK, we have this nice linear time algorithm to tell you 1001 00:53:22,730 --> 00:53:24,450 whether a particular one is doable. 1002 00:53:24,450 --> 00:53:26,250 How many are there? 1003 00:53:26,250 --> 00:53:30,460 And there is-- if you work through all 1004 00:53:30,460 --> 00:53:33,140 of the things I've shown you in a little bit more detail, 1005 00:53:33,140 --> 00:53:36,055 you can recover-- this was finding a crimp. 1006 00:53:36,055 --> 00:53:37,430 But you can actually look at what 1007 00:53:37,430 --> 00:53:39,580 are all the crimps that are possible 1008 00:53:39,580 --> 00:53:43,340 and actually count how many different ways 1009 00:53:43,340 --> 00:53:44,420 there are to fold. 1010 00:53:44,420 --> 00:53:45,546 How many crimps you can do. 1011 00:53:45,546 --> 00:53:47,545 And then given those crimps, how many crimps you 1012 00:53:47,545 --> 00:53:48,340 can go after that. 1013 00:53:48,340 --> 00:53:51,050 And in the same linear time kind of algorithm, 1014 00:53:51,050 --> 00:53:53,470 you can figure out how many different mountain valley 1015 00:53:53,470 --> 00:53:55,410 patterns are flat foldable. 1016 00:53:55,410 --> 00:53:57,120 That takes a little bit more care. 1017 00:53:57,120 --> 00:53:58,835 I will tell you the extremes. 1018 00:54:02,860 --> 00:54:08,300 How small could the number of flat foldable mountain valley 1019 00:54:08,300 --> 00:54:12,560 assignments be for a given crease pattern? 1020 00:54:12,560 --> 00:54:14,226 Well, do you have any guesses when? 1021 00:54:14,226 --> 00:54:17,005 When do I have the least choice of where to make crimps? 1022 00:54:24,140 --> 00:54:24,680 Yeah. 1023 00:54:24,680 --> 00:54:26,664 AUDIENCE: We have three angles. 1024 00:54:26,664 --> 00:54:28,590 PROFESSOR: For general n. 1025 00:54:28,590 --> 00:54:29,680 Sorry. 1026 00:54:29,680 --> 00:54:32,135 But, yeah, three angles or, I guess, two angles. 1027 00:54:32,135 --> 00:54:34,980 We really can't do any-- I don't have any choice. 1028 00:54:34,980 --> 00:54:38,350 But for general n, how should I disperse 1029 00:54:38,350 --> 00:54:40,630 the theta i's in order to make there 1030 00:54:40,630 --> 00:54:42,870 be very few possible crimps? 1031 00:54:42,870 --> 00:54:44,790 AUDIENCE: When there are only two transitions. 1032 00:54:44,790 --> 00:54:46,706 PROFESSOR: When there are only two transitions 1033 00:54:46,706 --> 00:54:48,090 between mountains and valleys. 1034 00:54:48,090 --> 00:54:49,381 All the mountains are together. 1035 00:54:49,381 --> 00:54:51,127 All the valleys are together. 1036 00:54:51,127 --> 00:54:53,460 That's right, if I was asking a question about mountains 1037 00:54:53,460 --> 00:54:53,990 and valleys. 1038 00:54:53,990 --> 00:54:56,140 But I was asking a question about theta i's. 1039 00:54:56,140 --> 00:54:59,510 So I want to count how many mountain valley patterns there 1040 00:54:59,510 --> 00:55:02,990 are-- mountain valley assignments that 1041 00:55:02,990 --> 00:55:10,453 are flat foldable for a given crease pattern. 1042 00:55:10,453 --> 00:55:12,417 AUDIENCE: So if you have a bunch of angles, 1043 00:55:12,417 --> 00:55:14,872 consecutive theta i's are increasing? 1044 00:55:14,872 --> 00:55:17,300 PROFESSOR: All the angles are consecutive increasing. 1045 00:55:17,300 --> 00:55:18,770 Yeah, that pretty much works. 1046 00:55:18,770 --> 00:55:21,410 In fact, what I need is the generic case, 1047 00:55:21,410 --> 00:55:24,950 where all angles are different and they never 1048 00:55:24,950 --> 00:55:28,840 become equal by folding, which is what generic means. 1049 00:55:28,840 --> 00:55:31,130 So then there was only really one crimp I could do, 1050 00:55:31,130 --> 00:55:34,970 which is the globally smallest angle. 1051 00:55:34,970 --> 00:55:36,590 So actually, I still have two choices. 1052 00:55:36,590 --> 00:55:39,510 I could do mountain then valley or valley then mountain. 1053 00:55:39,510 --> 00:55:42,100 But I only have two choices for that crimp. 1054 00:55:42,100 --> 00:55:44,100 Then I will have two choices for the next crimp, 1055 00:55:44,100 --> 00:55:45,560 two choices for the next crimp. 1056 00:55:45,560 --> 00:55:51,920 In general, I get 2 to the n possible mountain valley 1057 00:55:51,920 --> 00:55:53,910 assignments for the generic case. 1058 00:55:59,500 --> 00:56:02,570 OK, what about, what's the opposite, 1059 00:56:02,570 --> 00:56:04,470 where I get as much choice as possible? 1060 00:56:04,470 --> 00:56:09,070 AUDIENCE: Isn't 2 to the n every mountain valley assignment? 1061 00:56:09,070 --> 00:56:14,401 PROFESSOR: Yes, 2 to the n-- no-- 2 to the n over 2. 1062 00:56:14,401 --> 00:56:14,900 Thank you. 1063 00:56:14,900 --> 00:56:15,399 Right. 1064 00:56:15,399 --> 00:56:17,960 There are 2 to the n conceivable mountain valley assignments. 1065 00:56:17,960 --> 00:56:19,440 It can't be that big. 1066 00:56:19,440 --> 00:56:21,660 Every time I do a crimp, I eat two creases, not one. 1067 00:56:21,660 --> 00:56:23,450 That's what I forgot. 1068 00:56:23,450 --> 00:56:25,090 Good. 1069 00:56:25,090 --> 00:56:26,560 It's still pretty big though. 1070 00:56:26,560 --> 00:56:29,040 It's like square root of all the possible things you 1071 00:56:29,040 --> 00:56:32,220 could imagine are indeed feasible. 1072 00:56:32,220 --> 00:56:34,580 What, from 2 to the n to 2 to the n over 2. 1073 00:56:34,580 --> 00:56:37,578 When would I get the most choice in crimping? 1074 00:56:37,578 --> 00:56:38,977 AUDIENCE: All angles are equal. 1075 00:56:38,977 --> 00:56:40,310 PROFESSOR: All angles are equal. 1076 00:56:40,310 --> 00:56:42,070 That's the opposite extreme. 1077 00:56:42,070 --> 00:56:44,270 And in this case-- a little messier. 1078 00:57:05,300 --> 00:57:08,140 When all the angles are equal, the only property 1079 00:57:08,140 --> 00:57:10,300 you need-- and you can see that from the proof-- 1080 00:57:10,300 --> 00:57:11,780 the only property you need is that the number 1081 00:57:11,780 --> 00:57:13,196 mountains minus the number valleys 1082 00:57:13,196 --> 00:57:17,430 is plus or minus 2, and n is even. 1083 00:57:17,430 --> 00:57:21,280 As long as you have that, we showed here, 1084 00:57:21,280 --> 00:57:22,822 you're going to have-- on alternation 1085 00:57:22,822 --> 00:57:25,155 from mountain valley-- that's a valid crimp, because all 1086 00:57:25,155 --> 00:57:26,050 the angles are equal. 1087 00:57:26,050 --> 00:57:27,340 And you keep going. 1088 00:57:27,340 --> 00:57:31,890 So it's just, how do you disperse that many-- how 1089 00:57:31,890 --> 00:57:36,960 do I disperse n over 2 minus 1 mountains 1090 00:57:36,960 --> 00:57:39,440 among n different positions. 1091 00:57:39,440 --> 00:57:40,740 That's what this represents. 1092 00:57:40,740 --> 00:57:43,420 And positions, how do I pick n over 2 minus 1 of them 1093 00:57:43,420 --> 00:57:44,540 to be mountains. 1094 00:57:44,540 --> 00:57:47,890 Or it could be n over 2 minus 1 of them are valleys. 1095 00:57:47,890 --> 00:57:49,430 And that's this factor of 2. 1096 00:57:49,430 --> 00:57:51,430 So it could have more mountains or more valleys. 1097 00:57:51,430 --> 00:57:54,020 And then you somehow place those among those. 1098 00:57:54,020 --> 00:57:56,430 And if you don't know this notation, 1099 00:57:56,430 --> 00:57:58,320 that's just what it means. 1100 00:57:58,320 --> 00:58:02,170 If you want to know using other notation, 1101 00:58:02,170 --> 00:58:06,460 it's like this, n over 2 minus 1 factorial, 1102 00:58:06,460 --> 00:58:08,750 over 2 plus 1 factorial. 1103 00:58:08,750 --> 00:58:14,160 If you don't know factorials, I'll tell you about them later. 1104 00:58:14,160 --> 00:58:15,910 Cool. 1105 00:58:15,910 --> 00:58:18,480 So that's kind of the end of the single vertex situation. 1106 00:58:22,020 --> 00:58:24,757 Yeah, I think I'll mention one interesting open question here, 1107 00:58:24,757 --> 00:58:26,840 which I would love to explore maybe in our problem 1108 00:58:26,840 --> 00:58:31,980 session-- which looks like it'll be Mondays at 5:00, I think. 1109 00:58:34,898 --> 00:58:37,440 This is one vertex, and we went through all this work 1110 00:58:37,440 --> 00:58:39,000 to solve a single vertex situation. 1111 00:58:39,000 --> 00:58:41,430 And it's interesting, as we'll see, 1112 00:58:41,430 --> 00:58:45,460 to think about locally how each vertex behaves. 1113 00:58:45,460 --> 00:58:47,820 But you would think, if you have a crease pattern 1114 00:58:47,820 --> 00:58:51,280 with two vertices, it shouldn't be that much harder. 1115 00:58:51,280 --> 00:58:52,780 So here we have linear time. 1116 00:58:52,780 --> 00:58:55,370 How quickly can you tell whether a two vertex crease 1117 00:58:55,370 --> 00:58:56,920 pattern is flat foldable? 1118 00:58:56,920 --> 00:58:59,760 As far as I know, no one has looked at that problem. 1119 00:58:59,760 --> 00:59:02,390 Surely, we can do it in quadratic time 1120 00:59:02,390 --> 00:59:04,560 but maybe even linear time, I think. 1121 00:59:04,560 --> 00:59:08,300 It can't be that hard, I think. 1122 00:59:08,300 --> 00:59:11,030 In general, if I have a small number of vertices-- 1123 00:59:11,030 --> 00:59:16,420 say k vertices-- much smaller than n creases, 1124 00:59:16,420 --> 00:59:18,800 is there-- for the algorithms people-- is there 1125 00:59:18,800 --> 00:59:21,360 fixed parameter tractable algorithm in k? 1126 00:59:21,360 --> 00:59:23,850 Or can I get something-- even getting something like n 1127 00:59:23,850 --> 00:59:27,590 to some function of k, would be progress. 1128 00:59:27,590 --> 00:59:29,190 I think this should be doable. 1129 00:59:29,190 --> 00:59:31,200 But ideally, we get a running time 1130 00:59:31,200 --> 00:59:35,800 that's exponential in k and linear in n. 1131 00:59:35,800 --> 00:59:38,220 That would be my hope. 1132 00:59:38,220 --> 00:59:39,960 Why do I say it has to be exponential? 1133 00:59:39,960 --> 00:59:41,460 Because in general, if I give you 1134 00:59:41,460 --> 00:59:43,960 a crease pattern with n vertices-- lots of vertices-- 1135 00:59:43,960 --> 00:59:45,780 this problem is NP-complete, which 1136 00:59:45,780 --> 00:59:48,410 is-- there's not going to be a polynomial time 1137 00:59:48,410 --> 00:59:49,460 algorithm for it. 1138 00:59:49,460 --> 00:59:50,910 Nothing good. 1139 00:59:50,910 --> 00:59:53,610 We will prove that next Wednesday. 1140 00:59:53,610 --> 00:59:56,640 So I'll hold off on that a little bit. 1141 00:59:56,640 --> 00:59:58,930 But for two vertices, how hard could it be? 1142 01:00:01,540 --> 01:00:02,110 All right. 1143 01:00:23,860 --> 01:00:25,610 I want to talk about one related topic. 1144 01:00:25,610 --> 01:00:28,060 And then we will go to origami design 1145 01:00:28,060 --> 01:00:30,960 and do a little bit on the tree method. 1146 01:00:30,960 --> 01:00:33,640 But before we get there, I want to talk 1147 01:00:33,640 --> 01:00:41,800 about local foldability, which is a cool topic. 1148 01:00:41,800 --> 01:00:43,520 People tend to forget about it. 1149 01:00:43,520 --> 01:00:44,230 I really like it. 1150 01:00:44,230 --> 01:00:46,594 I think it would make a cool project also. 1151 01:00:46,594 --> 01:00:47,510 It's a nice algorithm. 1152 01:00:47,510 --> 01:00:50,750 It goes back to Bern and Hayes, 1996. 1153 01:00:50,750 --> 01:00:52,680 So it's also right at the beginning 1154 01:00:52,680 --> 01:00:55,860 of origami mathematics. 1155 01:00:55,860 --> 01:00:57,850 And it's this idea, all right, I give you 1156 01:00:57,850 --> 01:01:01,410 a crease pattern now arbitrarily many vertices. 1157 01:01:01,410 --> 01:01:03,910 And if I ask you, does it fold flat? 1158 01:01:03,910 --> 01:01:07,410 That's NP-complete-- intractable-- same paper. 1159 01:01:07,410 --> 01:01:09,730 But what if I ask you just to give me a mountain valley 1160 01:01:09,730 --> 01:01:13,440 assignment that might fold flat? 1161 01:01:13,440 --> 01:01:15,710 Well, that's also NP-complete. 1162 01:01:15,710 --> 01:01:18,650 But if you actually want it to fold flat, 1163 01:01:18,650 --> 01:01:20,110 this is really the same problem. 1164 01:01:20,110 --> 01:01:23,480 But if I ask you, give me a mountain valley assignment, 1165 01:01:23,480 --> 01:01:25,400 so that if I checked every vertex according 1166 01:01:25,400 --> 01:01:29,082 to this algorithm, at least every vertex folds flat. 1167 01:01:29,082 --> 01:01:32,001 That would seem nice. 1168 01:01:32,001 --> 01:01:34,500 Definitely, I have to find a mountain valley assignment that 1169 01:01:34,500 --> 01:01:36,690 satisfies these conditions, that as I 1170 01:01:36,690 --> 01:01:39,820 do successive crimps, every vertex-- if I cut out 1171 01:01:39,820 --> 01:01:41,570 the vertex separately, it would fold flat. 1172 01:01:41,570 --> 01:01:43,860 This is the notion of local foldability. 1173 01:01:43,860 --> 01:01:46,370 And there's a linear time algorithm 1174 01:01:46,370 --> 01:01:49,810 to give you a mountain valley assignment that ought to work, 1175 01:01:49,810 --> 01:01:51,310 in that at each vertex it works. 1176 01:01:51,310 --> 01:01:53,750 It still may not work globally for some other reason. 1177 01:01:56,360 --> 01:01:59,277 But it's pretty good. 1178 01:01:59,277 --> 01:02:01,360 I think this would be, actually, pretty practical. 1179 01:02:01,360 --> 01:02:05,440 I think in a lot of real world origami settings-- 1180 01:02:05,440 --> 01:02:09,660 when you're doing flat folding anyway-- 1181 01:02:09,660 --> 01:02:13,430 locally foldable is going to be enough to be globally foldable. 1182 01:02:13,430 --> 01:02:14,717 That's a guess. 1183 01:02:14,717 --> 01:02:15,800 I don't know if it's true. 1184 01:02:22,520 --> 01:02:28,290 Consistent mountain valley assignment, 1185 01:02:28,290 --> 01:02:44,510 if there is one, so that each vertex locally folds flat. 1186 01:03:05,470 --> 01:03:11,230 Let me try to concoct a small example that's relevant 1187 01:03:11,230 --> 01:03:21,610 here-- got to think. 1188 01:03:31,890 --> 01:03:32,890 Yes, I think that works. 1189 01:03:32,890 --> 01:03:35,357 All right, here we go. 1190 01:03:35,357 --> 01:03:36,690 I think this is in the textbook. 1191 01:03:36,690 --> 01:03:41,140 I remember drawing it in the past three years 1192 01:03:41,140 --> 01:03:43,590 ago or whatever. 1193 01:03:43,590 --> 01:03:46,958 So here's a crease pattern on a square or whatever. 1194 01:03:46,958 --> 01:03:48,416 AUDIENCE: The top one should have-- 1195 01:03:48,416 --> 01:03:51,440 PROFESSOR: Yes, it should have this. 1196 01:03:51,440 --> 01:03:53,720 Thank you. 1197 01:03:53,720 --> 01:03:55,870 So this satisfies Kawasaki's theorem-- 1198 01:03:55,870 --> 01:03:59,350 that was the hard part-- because these angles sum to 180. 1199 01:03:59,350 --> 01:04:01,500 And it's symmetric all around. 1200 01:04:01,500 --> 01:04:05,080 OK, these angles are the smallest. 1201 01:04:05,080 --> 01:04:07,460 They are 60 degrees. 1202 01:04:07,460 --> 01:04:09,546 This is an equilateral triangle. 1203 01:04:09,546 --> 01:04:11,420 So we have this-- not quite the generic case, 1204 01:04:11,420 --> 01:04:13,180 but we have the smallest angle is 1205 01:04:13,180 --> 01:04:16,257 surrounded by two larger angles. 1206 01:04:16,257 --> 01:04:17,840 Therefore, one of these is a mountain; 1207 01:04:17,840 --> 01:04:19,480 the other is a valley. 1208 01:04:19,480 --> 01:04:22,300 That means that these two creases 1209 01:04:22,300 --> 01:04:24,080 have to have different assignments-- I'm 1210 01:04:24,080 --> 01:04:25,414 going to write a not equal sign. 1211 01:04:25,414 --> 01:04:27,663 One of these is a mountain, and the other is a valley. 1212 01:04:27,663 --> 01:04:28,800 They can't be the same. 1213 01:04:28,800 --> 01:04:31,170 Also, these two cannot be the same. 1214 01:04:31,170 --> 01:04:33,410 Also, these two cannot be the same. 1215 01:04:33,410 --> 01:04:35,340 That's not possible. 1216 01:04:35,340 --> 01:04:37,640 Because I've got three, it has to alternate. 1217 01:04:37,640 --> 01:04:39,570 You can alternate three times. 1218 01:04:39,570 --> 01:04:42,540 You can only alternative at even number of times. 1219 01:04:42,540 --> 01:04:43,960 Two of these are mountains. 1220 01:04:43,960 --> 01:04:44,690 One is valley. 1221 01:04:44,690 --> 01:04:46,450 And then you've got a problem. 1222 01:04:46,450 --> 01:04:48,640 OK, so this thing is not flat foldable. 1223 01:04:48,640 --> 01:04:50,224 And this algorithm will tell you that. 1224 01:04:50,224 --> 01:04:51,806 Because it will say, hey, I can't even 1225 01:04:51,806 --> 01:04:53,400 find a mountain valley assignment 1226 01:04:53,400 --> 01:04:57,420 that could possibly fold each vertex flat. 1227 01:04:57,420 --> 01:05:02,100 So this is a nice algorithm. 1228 01:05:02,100 --> 01:05:06,700 At least, it will detect annoying situations like that. 1229 01:05:06,700 --> 01:05:11,320 So in order to solve this, we're going to use-- 1230 01:05:11,320 --> 01:05:13,810 and I'm just going to sketch how this algorithm works-- 1231 01:05:13,810 --> 01:05:16,490 we're going to use this characterization and this idea 1232 01:05:16,490 --> 01:05:19,040 that we can really find all possible mountain valley 1233 01:05:19,040 --> 01:05:22,634 assignments just by trying all the possible crimp sequences. 1234 01:05:22,634 --> 01:05:24,050 Now, there are exponentially many. 1235 01:05:24,050 --> 01:05:26,270 So it's not like I'm actually going to try them all. 1236 01:05:26,270 --> 01:05:30,045 But I need to explore that space of candidate crimps 1237 01:05:30,045 --> 01:05:30,920 and see what happens. 1238 01:05:36,850 --> 01:05:43,300 So the idea is kind of crazy. 1239 01:05:43,300 --> 01:05:45,290 The beginning of the algorithm is 1240 01:05:45,290 --> 01:05:48,570 fold each vertex flat-- somehow. 1241 01:05:48,570 --> 01:05:50,150 I don't care how. 1242 01:05:50,150 --> 01:05:51,520 Just pick a crimp, do it. 1243 01:05:51,520 --> 01:05:54,400 Pick a crimp, do it-- separately for each vertex. 1244 01:05:54,400 --> 01:05:56,440 They're not going to be compatible. 1245 01:05:56,440 --> 01:05:59,270 And don't look at the mountain valley assignment you get. 1246 01:05:59,270 --> 01:06:02,460 But look at the crimping sequence you get. 1247 01:06:02,460 --> 01:06:06,880 So here, let's do a little example-- 1248 01:06:06,880 --> 01:06:08,650 a non-trivial example. 1249 01:06:08,650 --> 01:06:12,260 So here, we have the peace sign, and these two angles are equal. 1250 01:06:12,260 --> 01:06:13,190 They're smallest. 1251 01:06:13,190 --> 01:06:15,442 I could crimp this angle first, or I 1252 01:06:15,442 --> 01:06:16,650 could crimp this angle first. 1253 01:06:16,650 --> 01:06:17,927 I have a choice. 1254 01:06:17,927 --> 01:06:19,010 So I'll draw both of them. 1255 01:06:23,080 --> 01:06:27,340 If I crimp this angle first, I know these guys 1256 01:06:27,340 --> 01:06:29,810 are paired up in the sense that they must be not equal-- 1257 01:06:29,810 --> 01:06:31,730 one's mountain and one's valley. 1258 01:06:31,730 --> 01:06:34,660 After I do that crimp-- just like in the real example 1259 01:06:34,660 --> 01:06:38,120 I had-- this angle will equal that angle. 1260 01:06:38,120 --> 01:06:42,160 And these two guys must have equal assignments. 1261 01:06:42,160 --> 01:06:44,480 They must be both mountain or both valley. 1262 01:06:44,480 --> 01:06:47,410 In this situation, these two guys 1263 01:06:47,410 --> 01:06:51,100 are not equal, if I crimp this pair first. 1264 01:06:51,100 --> 01:06:53,035 And these two guys must be equal-- equal now 1265 01:06:53,035 --> 01:06:56,500 in the sense of being mountain or valley. 1266 01:06:56,500 --> 01:07:00,170 OK, so there are still exponentially many 1267 01:07:00,170 --> 01:07:01,640 possibilities in how to do this. 1268 01:07:01,640 --> 01:07:05,000 But just pick one-- pick one of these ways of pairing up. 1269 01:07:05,000 --> 01:07:07,540 You're going to pair up each of the n creases 1270 01:07:07,540 --> 01:07:09,830 into n over 2 pairs. 1271 01:07:09,830 --> 01:07:14,550 And they're going to have some not equal and equal signs. 1272 01:07:14,550 --> 01:07:16,850 So now, if you imagine the general picture, 1273 01:07:16,850 --> 01:07:20,630 like here, for example, I might get-- in this pattern, 1274 01:07:20,630 --> 01:07:22,710 I would be forced to get not equals 1275 01:07:22,710 --> 01:07:25,160 and these guys paired up. 1276 01:07:25,160 --> 01:07:28,575 And in general, I want to look at these kinds of cycles. 1277 01:07:31,450 --> 01:07:37,150 If I come into a vertex-- here's a vertex-- 1278 01:07:37,150 --> 01:07:38,970 it's paired up with somebody. 1279 01:07:38,970 --> 01:07:41,900 So if I come in here, I can go out somewhere. 1280 01:07:41,900 --> 01:07:45,260 And I come to this vertex, it's paired with somebody. 1281 01:07:45,260 --> 01:07:48,380 So I'm going to just-- I can keep wandering around. 1282 01:07:48,380 --> 01:07:51,815 And in general, there will be a bunch of these paths 1283 01:07:51,815 --> 01:07:53,650 that you can follow. 1284 01:07:53,650 --> 01:07:54,650 What could the paths do? 1285 01:07:54,650 --> 01:07:58,350 They either close up on themselves-- maybe things 1286 01:07:58,350 --> 01:08:03,660 are paired up in such a way that you make a return trip. 1287 01:08:03,660 --> 01:08:05,880 Or it could be some other path. 1288 01:08:05,880 --> 01:08:08,305 Let me draw a dotted path. 1289 01:08:08,305 --> 01:08:10,245 It could come in here, and maybe it 1290 01:08:10,245 --> 01:08:12,180 gets paired up with this guy. 1291 01:08:12,180 --> 01:08:13,610 Maybe it goes off to infinity. 1292 01:08:13,610 --> 01:08:15,267 It reaches the boundary of the paper. 1293 01:08:15,267 --> 01:08:16,600 Those are the two possibilities. 1294 01:08:16,600 --> 01:08:19,250 You get paths, which go off to the edge-- 1295 01:08:19,250 --> 01:08:20,870 off to infinity on either end. 1296 01:08:20,870 --> 01:08:22,990 Or you could get cycles. 1297 01:08:22,990 --> 01:08:25,000 The cycles are the problem. 1298 01:08:25,000 --> 01:08:29,479 Because whenever I have a cycle, I have a parity constraint. 1299 01:08:29,479 --> 01:08:32,979 For example, when they're all not equals, 1300 01:08:32,979 --> 01:08:36,010 the length of the cycle must be even. 1301 01:08:36,010 --> 01:08:40,700 If there were-- what's the general statement? 1302 01:08:40,700 --> 01:08:42,844 It's like the parity of the cycle, 1303 01:08:42,844 --> 01:08:44,260 which is whether it's even or odd, 1304 01:08:44,260 --> 01:08:48,719 should be equal to the parity of the number of not equal signs. 1305 01:08:52,340 --> 01:08:54,584 Something like that. 1306 01:08:54,584 --> 01:08:55,500 I've got a sheet here. 1307 01:09:03,244 --> 01:09:05,810 Great, I just said "parity problems." 1308 01:09:05,810 --> 01:09:06,870 It's something like that. 1309 01:09:06,870 --> 01:09:08,970 It's either the parity of the cycle 1310 01:09:08,970 --> 01:09:10,990 should equal the parity of the number of equals 1311 01:09:10,990 --> 01:09:13,090 or the parity of the number of not equals. 1312 01:09:13,090 --> 01:09:15,350 I think, number of equals. 1313 01:09:15,350 --> 01:09:18,260 Anyway, one of the two. 1314 01:09:18,260 --> 01:09:19,319 And you can just check. 1315 01:09:19,319 --> 01:09:20,235 I mean, you're forced. 1316 01:09:20,235 --> 01:09:22,250 If I say, OK, let's make this mountain, 1317 01:09:22,250 --> 01:09:24,069 then this is either equals or not equals. 1318 01:09:24,069 --> 01:09:25,640 It'll tell me whether this is mountain or valley. 1319 01:09:25,640 --> 01:09:26,770 Just walk around the cycle. 1320 01:09:26,770 --> 01:09:28,899 Either you get a contradiction or you don't. 1321 01:09:28,899 --> 01:09:32,520 If you get a contradiction, we have a problem. 1322 01:09:32,520 --> 01:09:35,080 How could we possibly fix the problem? 1323 01:09:35,080 --> 01:09:41,395 Well, when we made-- you look at each of these cramps. 1324 01:09:41,395 --> 01:09:43,770 I mean, in fact, you could look at each of these vertices 1325 01:09:43,770 --> 01:09:44,260 separately. 1326 01:09:44,260 --> 01:09:46,510 But you think about one of these crimps and say, well, 1327 01:09:46,510 --> 01:09:48,069 could I have done it another way? 1328 01:09:48,069 --> 01:09:51,912 Sometimes, there are crimps that have other equal choices. 1329 01:09:51,912 --> 01:09:53,620 Maybe, there are a bunch of equal angles. 1330 01:09:53,620 --> 01:09:56,760 And I could have done a different pairing. 1331 01:09:56,760 --> 01:09:59,070 What happens when I try a different pairing? 1332 01:09:59,070 --> 01:10:03,290 Well, instead of this being in one cycle 1333 01:10:03,290 --> 01:10:07,230 and, let's say, this being in another cycle, 1334 01:10:07,230 --> 01:10:09,510 if I pair these guys up instead, I'll 1335 01:10:09,510 --> 01:10:13,150 end up merging those two cycles into one bigger thing. 1336 01:10:13,150 --> 01:10:15,620 It could be a path or a cycle. 1337 01:10:15,620 --> 01:10:19,850 And the algorithm says, just keep doing those cycle merges. 1338 01:10:19,850 --> 01:10:23,820 And if you get stuck, your thing is not locally foldable. 1339 01:10:23,820 --> 01:10:25,830 That's the hard part to prove. 1340 01:10:25,830 --> 01:10:29,336 Otherwise, you will find one of these patterns 1341 01:10:29,336 --> 01:10:31,460 that you can actually resolve mountains and valleys 1342 01:10:31,460 --> 01:10:33,140 all the way through. 1343 01:10:33,140 --> 01:10:51,330 So let's say, so start with some folding-- say local folding, 1344 01:10:51,330 --> 01:10:52,380 whatever. 1345 01:10:52,380 --> 01:10:58,865 I'm going to say with some pairing of creases at vertices. 1346 01:11:06,460 --> 01:11:29,385 And merge two paths or cycles whenever possible. 1347 01:11:33,000 --> 01:11:37,760 And when I say merge, I mean whenever 1348 01:11:37,760 --> 01:11:41,110 you have-- what are the possible things you could possibly 1349 01:11:41,110 --> 01:11:43,720 merge-- when you have at some point during the algorithm 1350 01:11:43,720 --> 01:11:46,640 a bunch of equal angles, you have a choice which of these 1351 01:11:46,640 --> 01:11:47,180 you crimp. 1352 01:11:47,180 --> 01:11:50,320 Obviously, the mountain valley assignment is not fixed. 1353 01:11:50,320 --> 01:11:53,450 You can crimp any of them. 1354 01:11:53,450 --> 01:11:55,920 You picked one of them, and it's sort 1355 01:11:55,920 --> 01:11:58,210 of merging whatever this thing is attached to, 1356 01:11:58,210 --> 01:12:01,380 to whatever this thing is attached to. 1357 01:12:01,380 --> 01:12:05,660 If there's something else that's disconnected from that thing, 1358 01:12:05,660 --> 01:12:09,090 I want you to instead merge two of them that 1359 01:12:09,090 --> 01:12:11,360 combines two different connected components-- two 1360 01:12:11,360 --> 01:12:12,880 paths or cycles. 1361 01:12:12,880 --> 01:12:14,950 Merging means I decrease the total number 1362 01:12:14,950 --> 01:12:15,780 of paths or cycles. 1363 01:12:15,780 --> 01:12:17,220 I combine two into one. 1364 01:12:17,220 --> 01:12:19,310 Whenever that's possible, do it. 1365 01:12:19,310 --> 01:12:21,580 You can prove that if you have parity 1366 01:12:21,580 --> 01:12:23,784 problem in the merge thing, you had 1367 01:12:23,784 --> 01:12:25,450 to have had a parity problem originally. 1368 01:12:25,450 --> 01:12:29,260 And merging can only fix parity problems. 1369 01:12:29,260 --> 01:12:31,940 That's the claim, and that's what I will not prove. 1370 01:12:31,940 --> 01:12:34,470 Once you know that-- and it doesn't matter in what order 1371 01:12:34,470 --> 01:12:37,860 or how you choose to merge-- you just merge as much as possible. 1372 01:12:37,860 --> 01:12:42,229 And either the resulting thing is OK or not. 1373 01:12:42,229 --> 01:12:44,270 And accordingly, you will tell whether this thing 1374 01:12:44,270 --> 01:12:46,650 is locally foldable. 1375 01:12:46,650 --> 01:12:48,700 Sorry, I want to move on to other things. 1376 01:12:48,700 --> 01:12:51,340 But I think this would be a fun thing to actually implement. 1377 01:12:51,340 --> 01:12:53,310 It's an easy algorithm. 1378 01:12:53,310 --> 01:13:01,000 And it's, I think, a pretty good test it for problems like this 1379 01:13:01,000 --> 01:13:02,290 that prevent flat foldability. 1380 01:13:13,510 --> 01:13:17,620 So let's move on from foldability to origami design. 1381 01:13:21,690 --> 01:13:23,595 So a bit of a big transition. 1382 01:13:23,595 --> 01:13:26,880 And we're going to talk about origami design 1383 01:13:26,880 --> 01:13:34,920 a lot more next class also but just start it off today. 1384 01:13:34,920 --> 01:13:38,460 And the particular algorithm for origami design 1385 01:13:38,460 --> 01:13:41,850 I want to talk about is called the tree method. 1386 01:13:41,850 --> 01:13:44,600 This is probably the oldest algorithm 1387 01:13:44,600 --> 01:13:47,940 for origami design, in that people 1388 01:13:47,940 --> 01:13:51,700 have been thinking about it and developing it for many years 1389 01:13:51,700 --> 01:13:54,140 through this period called the Bug Wars, when people were 1390 01:13:54,140 --> 01:13:56,880 trying to design more and more complicated insects. 1391 01:13:56,880 --> 01:14:00,140 It's like, well, I can make an insect with six legs. 1392 01:14:00,140 --> 01:14:03,610 Oh yeah, well I can make a spider with eight legs. 1393 01:14:03,610 --> 01:14:07,660 Oh yeah, well I can make an insect-- a beetle that 1394 01:14:07,660 --> 01:14:10,600 has wings and horns and there's thorns on the horns-- you know, 1395 01:14:10,600 --> 01:14:12,780 all these crazy things. 1396 01:14:12,780 --> 01:14:14,280 During that time, there were a lot 1397 01:14:14,280 --> 01:14:15,654 of people thinking about how do I 1398 01:14:15,654 --> 01:14:19,950 make more and more complicated-- especially, more 1399 01:14:19,950 --> 01:14:23,990 limbs in my creatures and very precise 1400 01:14:23,990 --> 01:14:25,950 arrangements of those limbs, let's say. 1401 01:14:25,950 --> 01:14:28,690 And that is what the tree method deals with and was really 1402 01:14:28,690 --> 01:14:35,420 formalized by Robert Lang, who published a paper in '96, 1403 01:14:35,420 --> 01:14:39,810 describing it as a sort of complete algorithm. 1404 01:14:39,810 --> 01:14:42,940 And it's still not known for sure 1405 01:14:42,940 --> 01:14:44,480 that that algorithm always works. 1406 01:14:44,480 --> 01:14:46,930 But that's what we've been working 1407 01:14:46,930 --> 01:14:52,260 on-- me and Marty and Rob Lang-- for the last four years or so. 1408 01:14:52,260 --> 01:14:54,600 And soon, we will publish that thing 1409 01:14:54,600 --> 01:14:56,645 and prove that this thing always works. 1410 01:14:56,645 --> 01:14:59,020 But I'm going to describe the algorithm without the proof 1411 01:14:59,020 --> 01:15:00,170 that it works. 1412 01:15:00,170 --> 01:15:05,862 And what it does-- I'll tell you its goal 1413 01:15:05,862 --> 01:15:07,195 from a mathematical perspective. 1414 01:15:18,260 --> 01:15:20,572 So it's interested in practical origami design. 1415 01:15:20,572 --> 01:15:22,780 So we're going to start from a square piece of paper. 1416 01:15:22,780 --> 01:15:24,620 It would also work for triangular pieces 1417 01:15:24,620 --> 01:15:27,700 of paper or anything convex. 1418 01:15:27,700 --> 01:15:31,124 But squares are what people usually care about. 1419 01:15:31,124 --> 01:15:32,790 Sometimes it's used for rectangles also. 1420 01:16:06,510 --> 01:16:10,250 The idea is I give to you a stick figure. 1421 01:16:13,110 --> 01:16:14,750 So that's formally, it's a tree-- 1422 01:16:14,750 --> 01:16:17,090 a graph without any cycles. 1423 01:16:17,090 --> 01:16:21,410 It's a metric tree, meaning that I put lengths on the edges. 1424 01:16:21,410 --> 01:16:23,680 I know this length-- this edge length-- is maybe 1425 01:16:23,680 --> 01:16:24,920 twice as long as this one. 1426 01:16:24,920 --> 01:16:28,380 So I really draw it with edge lengths in mind. 1427 01:16:28,380 --> 01:16:32,600 Then what I want you to do is find 1428 01:16:32,600 --> 01:16:38,820 some folding of a piece of paper-- 1429 01:16:38,820 --> 01:16:41,950 I should really be looking at what I'm trying to match. 1430 01:16:41,950 --> 01:17:06,140 So here, here, this goes down, such 1431 01:17:06,140 --> 01:17:09,290 that I want to find some folding of a square paper-- in fact, 1432 01:17:09,290 --> 01:17:12,920 the smallest square possible, so that when 1433 01:17:12,920 --> 01:17:19,100 I project-- like this-- vertically, 1434 01:17:19,100 --> 01:17:26,270 the projection of that folding is exactly that metric tree. 1435 01:17:26,270 --> 01:17:28,940 And this is called a uniaxial-- this thing is 1436 01:17:28,940 --> 01:17:30,880 called a uniaxial base. 1437 01:17:30,880 --> 01:17:33,700 Let me tell you a little bit why it's called a uniaxial base. 1438 01:17:37,280 --> 01:17:40,220 We're thinking about what are called origami bases. 1439 01:17:40,220 --> 01:17:43,570 These there like the beginning of origami models. 1440 01:17:43,570 --> 01:17:46,290 And most classic origami models-- 1441 01:17:46,290 --> 01:17:51,980 like more than 60 years ago-- start from one of these bases. 1442 01:17:51,980 --> 01:17:58,780 You've got waterbomb base on the top left, preliminary base, 1443 01:17:58,780 --> 01:18:05,557 fish base, bird base windmill base, and frog base. 1444 01:18:05,557 --> 01:18:06,640 I can't remember them all. 1445 01:18:06,640 --> 01:18:08,250 I have a little example here. 1446 01:18:08,250 --> 01:18:10,750 This is the waterbomb base. 1447 01:18:10,750 --> 01:18:14,810 So it's just very simple crease pattern. 1448 01:18:14,810 --> 01:18:17,700 And why this is useful is, it gives you 1449 01:18:17,700 --> 01:18:20,370 sort of four flaps of paper to work with. 1450 01:18:20,370 --> 01:18:23,640 Maybe you make one of them the head 1451 01:18:23,640 --> 01:18:26,590 and the other two wings and the back one a tail, 1452 01:18:26,590 --> 01:18:27,930 if you're making a crane. 1453 01:18:27,930 --> 01:18:29,880 If doesn't actually start not from this space 1454 01:18:29,880 --> 01:18:32,610 but from the other one. 1455 01:18:32,610 --> 01:18:33,990 But the same idea. 1456 01:18:33,990 --> 01:18:35,990 If you're folding a crane, one of these 1457 01:18:35,990 --> 01:18:38,860 would be the head, the other tail, and two wings. 1458 01:18:38,860 --> 01:18:41,870 This is great if you're making a four flap animal. 1459 01:18:41,870 --> 01:18:44,020 And if you think about its projection-- 1460 01:18:44,020 --> 01:18:49,520 and it's easier to think about in this other picture-- 1461 01:18:49,520 --> 01:18:54,710 the projection of this thing is a four limbed star. 1462 01:18:54,710 --> 01:18:56,490 You can see it's a plus sign. 1463 01:18:56,490 --> 01:18:58,296 All of them are the same length. 1464 01:18:58,296 --> 01:18:59,670 And so this is actually something 1465 01:18:59,670 --> 01:19:01,610 you can get out of the tree method. 1466 01:19:01,610 --> 01:19:04,420 You just give that as your input. 1467 01:19:04,420 --> 01:19:07,870 You will get this 3D thing and this crease pattern 1468 01:19:07,870 --> 01:19:14,060 as your origami-- as the output. 1469 01:19:14,060 --> 01:19:17,150 Let me show you the program in action. 1470 01:19:17,150 --> 01:19:18,850 So Robert Lang implemented this thing. 1471 01:19:18,850 --> 01:19:20,110 It's called tree maker. 1472 01:19:20,110 --> 01:19:23,025 It's freely available, open source, all that good stuff. 1473 01:19:28,540 --> 01:19:31,970 And I'm not an expert at using it, so bear with me. 1474 01:19:31,970 --> 01:19:39,970 But if we wanted to say, I would like that star. 1475 01:19:39,970 --> 01:19:43,300 OK, now, I drew it obviously not with all the lengths equal, 1476 01:19:43,300 --> 01:19:45,090 but it's ignoring the lengths that I drew. 1477 01:19:45,090 --> 01:19:46,590 And they're actually specified here. 1478 01:19:46,590 --> 01:19:48,560 So all the lengths here are supposed to be 1. 1479 01:19:48,560 --> 01:19:51,500 And then I say, OK, optimize. 1480 01:19:51,500 --> 01:19:53,660 And then make a crease pattern. 1481 01:19:53,660 --> 01:19:55,635 And then show me the crease pattern. 1482 01:19:55,635 --> 01:19:58,060 And there it is, exactly the crease pattern 1483 01:19:58,060 --> 01:20:00,702 I made, although actually I can see from the mountain valley 1484 01:20:00,702 --> 01:20:02,410 assignment-- because this is not really-- 1485 01:20:02,410 --> 01:20:04,450 this is not flat origami. 1486 01:20:04,450 --> 01:20:06,240 It made it this way. 1487 01:20:06,240 --> 01:20:06,840 So it's flat. 1488 01:20:06,840 --> 01:20:09,620 Of course, the projection is the same-- still four limbs. 1489 01:20:12,560 --> 01:20:17,330 And so it's dashes for valleys and solid lines for mountains. 1490 01:20:17,330 --> 01:20:19,540 But you can make anything you want. 1491 01:20:19,540 --> 01:20:25,130 So let's say we want to make-- I don't know-- a lizard 1492 01:20:25,130 --> 01:20:26,350 or something. 1493 01:20:26,350 --> 01:20:29,300 So the blue lines are the tree. 1494 01:20:37,820 --> 01:20:39,170 So here's-- can I do this? 1495 01:20:39,170 --> 01:20:40,970 Yes, there we go. 1496 01:20:40,970 --> 01:20:46,380 So here, I have a forearm, a head, a foreleg, 1497 01:20:46,380 --> 01:20:51,490 another forearm, a body segment, tail, and two hind legs. 1498 01:20:51,490 --> 01:20:53,820 Maybe I want to make that. 1499 01:20:53,820 --> 01:20:58,630 So I say optimize and then make a crease pattern, and then 1500 01:20:58,630 --> 01:20:59,340 boom. 1501 01:20:59,340 --> 01:21:02,000 You fold that, and it will have exactly that projection. 1502 01:21:02,000 --> 01:21:04,750 And assuming it did a reasonable job at computation, 1503 01:21:04,750 --> 01:21:07,036 this will be the best way to fold this thing. 1504 01:21:07,036 --> 01:21:08,410 And it's the smallest square that 1505 01:21:08,410 --> 01:21:10,670 matches exactly that shape. 1506 01:21:10,670 --> 01:21:15,300 You get the best scale factor between the size 1507 01:21:15,300 --> 01:21:17,570 of your piece of paper and the target shape. 1508 01:21:17,570 --> 01:21:19,790 But actually, doing that optimization-- 1509 01:21:19,790 --> 01:21:21,444 the first step I did-- is NP-complete. 1510 01:21:21,444 --> 01:21:22,985 So it's not going to do it perfectly, 1511 01:21:22,985 --> 01:21:25,720 and we'll prove that Wednesday. 1512 01:21:25,720 --> 01:21:27,520 But the heuristics are pretty good. 1513 01:21:27,520 --> 01:21:28,756 AUDIENCE: It finds a local. 1514 01:21:28,756 --> 01:21:30,690 PROFESSOR: It finds a local minimum, 1515 01:21:30,690 --> 01:21:32,710 and often it finds a pretty good one. 1516 01:21:32,710 --> 01:21:36,000 And sometimes you can coax it to find better ones, but, yeah. 1517 01:21:36,000 --> 01:21:38,020 It's not perfect, but hey, it's NP-complete. 1518 01:21:38,020 --> 01:21:39,580 So you can't do it. 1519 01:21:39,580 --> 01:21:43,360 This actually shows you what it would look like in x-ray view. 1520 01:21:46,640 --> 01:21:50,100 And then you can say, oh, that was nice, 1521 01:21:50,100 --> 01:21:55,980 but let's-- where is my-- yeah, that was good. 1522 01:21:55,980 --> 01:21:58,850 But maybe I really wanted the head segment 1523 01:21:58,850 --> 01:22:03,660 to be shorter length of 0.5. 1524 01:22:03,660 --> 01:22:07,910 And then I wanted the tail to be really long. 1525 01:22:07,910 --> 01:22:12,240 And then you can optimize that and find a crease pattern. 1526 01:22:12,240 --> 01:22:13,170 And it'll complain. 1527 01:22:13,170 --> 01:22:14,810 Because it's having trouble. 1528 01:22:14,810 --> 01:22:15,310 Oh, dear. 1529 01:22:18,150 --> 01:22:18,860 Demo effect. 1530 01:22:18,860 --> 01:22:21,540 I should have tried this example before. 1531 01:22:21,540 --> 01:22:23,935 It's not-- oh, gosh. 1532 01:22:23,935 --> 01:22:25,520 It's one of these annoying ones. 1533 01:22:25,520 --> 01:22:42,710 I should say-- I should add some feature like maybe a strain 1534 01:22:42,710 --> 01:22:48,180 split something, add a little bit, maybe do that 1535 01:22:48,180 --> 01:22:54,150 and-- there we go. 1536 01:22:54,150 --> 01:22:57,050 I cheated, and I'll explain how I cheated last time. 1537 01:22:57,050 --> 01:23:00,380 But if sometimes the particular method 1538 01:23:00,380 --> 01:23:05,090 fails but you can fix it by adding in another tiny limb 1539 01:23:05,090 --> 01:23:07,950 off the edge somewhere-- and of course, you 1540 01:23:07,950 --> 01:23:10,510 can then get rid of that at the end when you're folding, 1541 01:23:10,510 --> 01:23:12,300 it only makes the problem slightly harder. 1542 01:23:12,300 --> 01:23:14,050 And it'll still find a folding. 1543 01:23:14,050 --> 01:23:15,750 But we'll talk about that next time. 1544 01:23:15,750 --> 01:23:21,320 And I'm way out of time, so we will stop there.