1 00:00:03,390 --> 00:00:07,280 PROFESSOR: So this lecture was about hyperbolic paraboloids, 2 00:00:07,280 --> 00:00:10,910 and the extent to which they don't exist or exist. 3 00:00:10,910 --> 00:00:13,650 Here is a regular non-existing hyperbolic 4 00:00:13,650 --> 00:00:19,000 paraboloid with the concentric squares, no diagonals, folded 5 00:00:19,000 --> 00:00:20,300 here. 6 00:00:20,300 --> 00:00:24,380 And so, those are just a few questions about this. 7 00:00:24,380 --> 00:00:26,950 What does it mean, other things. 8 00:00:26,950 --> 00:00:29,420 These are all asked by [INAUDIBLE] I believe, 9 00:00:29,420 --> 00:00:35,290 and they're all open, so-- they're all good questions. 10 00:00:35,290 --> 00:00:38,150 We don't know whether the good triangulation, 11 00:00:38,150 --> 00:00:40,050 the alternating one, works for arbitrary, 12 00:00:40,050 --> 00:00:42,920 and we only know up to n equals 100, 13 00:00:42,920 --> 00:00:45,670 for the various fixed angles that we checked. 14 00:00:45,670 --> 00:00:48,630 We don't have a proof technique to do arbitrary in. 15 00:00:48,630 --> 00:00:51,510 You could try to do some amount of alternation, 16 00:00:51,510 --> 00:00:54,500 but not, somewhere in between the two extremes, 17 00:00:54,500 --> 00:00:57,020 and probably you'll get something in between goodness, 18 00:00:57,020 --> 00:00:58,916 but I don't know. 19 00:00:58,916 --> 00:01:01,070 You could certainly play with that. 20 00:01:01,070 --> 00:01:05,209 And it's very natural to try this with larger than squares. 21 00:01:05,209 --> 00:01:07,390 The only trouble is, the center no longer 22 00:01:07,390 --> 00:01:10,530 has a single degree of freedom, so the only thing 23 00:01:10,530 --> 00:01:14,240 to answer there would be, How do you want to initially fold, 24 00:01:14,240 --> 00:01:16,147 like, if you're doing a hexagon, hexagon 25 00:01:16,147 --> 00:01:17,980 is probably clear what you might want to do, 26 00:01:17,980 --> 00:01:21,530 but for general k-gon, how do you 27 00:01:21,530 --> 00:01:23,770 want to arrange that innermost k-gon? 28 00:01:23,770 --> 00:01:26,090 From that, you could propagate out, 29 00:01:26,090 --> 00:01:28,020 just like we did with the square, if you 30 00:01:28,020 --> 00:01:30,126 use an alternating triangulation, say. 31 00:01:30,126 --> 00:01:32,000 Probably works well, but we haven't tried it. 32 00:01:32,000 --> 00:01:37,040 That could be a cool project to do that, not that hard. 33 00:01:37,040 --> 00:01:40,730 So those were open problems. 34 00:01:40,730 --> 00:01:47,510 Next, we have some, sort of, math, general math questions. 35 00:01:47,510 --> 00:01:50,400 c1, c2, and semi-creases. 36 00:01:50,400 --> 00:01:52,720 These are related questions. 37 00:01:52,720 --> 00:01:56,875 So let me do a quick visual review of these terms. 38 00:02:00,720 --> 00:02:03,710 I'm going to do it for one-dimensional curves in 2D, 39 00:02:03,710 --> 00:02:06,420 because that's a lot easier to think about, then 40 00:02:06,420 --> 00:02:08,039 what we really care about, which is 41 00:02:08,039 --> 00:02:09,910 two-dimensional services in 3D. 42 00:02:09,910 --> 00:02:16,740 But all the ideas carry over, so here is just a function. 43 00:02:29,710 --> 00:02:32,390 Let's say, for example, this is a parabolic arc, 44 00:02:32,390 --> 00:02:33,590 than a straight segment. 45 00:02:33,590 --> 00:02:36,250 This point is missing, and instead it's up here. 46 00:02:36,250 --> 00:02:38,000 These points are present, and here we 47 00:02:38,000 --> 00:02:40,230 have another parabolic arc. 48 00:02:40,230 --> 00:02:50,210 So, this function I would call piecewise C infinity. 49 00:02:50,210 --> 00:02:53,330 Let me explain what these things mean. 50 00:02:53,330 --> 00:02:57,490 So we have, on the one hand, c zero is a term, 51 00:02:57,490 --> 00:02:58,540 meaning just continuous. 52 00:03:02,030 --> 00:03:04,450 Continuous means no jumps like this, 53 00:03:04,450 --> 00:03:08,040 so this is not a continuous function. 54 00:03:08,040 --> 00:03:10,460 So this function overall is not even c zero. c1 55 00:03:10,460 --> 00:03:13,960 is a stronger condition, which means that not only are you 56 00:03:13,960 --> 00:03:18,220 continuous, but you also have a continuous and existing 57 00:03:18,220 --> 00:03:18,960 first derivative. 58 00:03:25,847 --> 00:03:27,680 So whereas here we're thinking about f, here 59 00:03:27,680 --> 00:03:30,500 we're thinking of f prime. 60 00:03:30,500 --> 00:03:34,440 So, not only can you talk about the function being continuous, 61 00:03:34,440 --> 00:03:37,700 c0c1, but you can talk about individual moments in time. 62 00:03:37,700 --> 00:03:42,880 So for example, this moment here is going to be c1. 63 00:03:42,880 --> 00:03:44,650 It has a nice derivative. 64 00:03:44,650 --> 00:03:48,030 That derivative is changing continuously here. 65 00:03:48,030 --> 00:03:54,880 Here, however, this is c zero, but not c1, 66 00:03:54,880 --> 00:03:57,730 because the tangent on the left is 67 00:03:57,730 --> 00:03:59,980 different from the tangent on the right. 68 00:03:59,980 --> 00:04:02,910 So this is not a c1 function, because the derivative 69 00:04:02,910 --> 00:04:03,980 is jumping at that point. 70 00:04:03,980 --> 00:04:05,730 I'm not going to draw the derivative here, 71 00:04:05,730 --> 00:04:07,854 it's a little harder to draw, but you could draw it 72 00:04:07,854 --> 00:04:10,500 with the same x-axis, and you'd see a jump 73 00:04:10,500 --> 00:04:13,460 from one angle to another. 74 00:04:13,460 --> 00:04:17,560 This point is not even c zero. 75 00:04:17,560 --> 00:04:22,480 This point is c zero, but not c1. 76 00:04:22,480 --> 00:04:29,500 This point is c1, because the tangent here is equal, 77 00:04:29,500 --> 00:04:31,590 on the left, is equal to the tangent on the right, 78 00:04:31,590 --> 00:04:33,510 if I drew this properly, where this 79 00:04:33,510 --> 00:04:36,400 is the bottom of the parabola. 80 00:04:36,400 --> 00:04:38,990 You can ask for more than c1, and we do in class, 81 00:04:38,990 --> 00:04:42,200 we talk about c2, and this says you 82 00:04:42,200 --> 00:04:47,230 should have a continuous second derivative. 83 00:04:47,230 --> 00:04:48,840 So acceleration. 84 00:04:48,840 --> 00:04:53,156 And this point, for example, is not c2. 85 00:04:53,156 --> 00:04:54,780 Because while the tangents meet, if you 86 00:04:54,780 --> 00:04:58,020 take the next derivative, you see that it's suddenly-- here 87 00:04:58,020 --> 00:04:59,630 the tangent was completely flat, it 88 00:04:59,630 --> 00:05:02,010 has a second derivative of zero, and then suddenly it 89 00:05:02,010 --> 00:05:03,580 starts increasing. 90 00:05:03,580 --> 00:05:06,430 Now you can design functions that are even C infinity. 91 00:05:06,430 --> 00:05:08,610 C infinity means no matter how many derivatives 92 00:05:08,610 --> 00:05:12,320 you take, it's continuous and defined, no jumps. 93 00:05:12,320 --> 00:05:15,080 But if you do a parabola, for example, 94 00:05:15,080 --> 00:05:19,670 it will not be so well-behaved, I think, yes. 95 00:05:22,900 --> 00:05:26,350 Got the x, y equals x squared, you take the derivative, 96 00:05:26,350 --> 00:05:33,240 you get 2x, take the derivative of that, you get 2, and then 97 00:05:33,240 --> 00:05:35,140 the derivative of that, you get zero. 98 00:05:37,680 --> 00:05:44,024 So the second derivative here is 2, all along the curve. 99 00:05:44,024 --> 00:05:46,422 So I guess y prime, y double prime. 100 00:05:46,422 --> 00:05:48,880 So, in particular here, we had a second derivative of zero, 101 00:05:48,880 --> 00:05:52,090 here we have a second derivative of 2. 102 00:05:52,090 --> 00:05:54,470 All right, so that's a quick crash course 103 00:05:54,470 --> 00:05:59,350 on individual points being c zero or c1 or c2, 104 00:05:59,350 --> 00:06:03,340 or C infinity is when you can go all the way. 105 00:06:03,340 --> 00:06:08,210 Creases are points on the paper where you're not c1. 106 00:06:08,210 --> 00:06:11,320 That means that you have two different tangent 107 00:06:11,320 --> 00:06:14,340 plans coming together, typically. 108 00:06:14,340 --> 00:06:16,250 So discontinuity is in the first derivative, 109 00:06:16,250 --> 00:06:18,320 that's what we call creases. 110 00:06:18,320 --> 00:06:24,770 Semi-creases are discontinuities in the second derivative 111 00:06:24,770 --> 00:06:32,970 So semi-crease is where you're not c2, 112 00:06:32,970 --> 00:06:37,870 crease is where you're not c1, according 113 00:06:37,870 --> 00:06:39,680 to this particular paper. 114 00:06:39,680 --> 00:06:41,430 Semi-crease is not too common a term, 115 00:06:41,430 --> 00:06:44,430 but crease is very comm-- this is the usual meaning of crease, 116 00:06:44,430 --> 00:06:47,340 semi-crease is maybe a little new to this paper we're talking 117 00:06:47,340 --> 00:06:50,060 about, nonexistence of pi parse. 118 00:06:50,060 --> 00:06:55,757 And so, for example, this would be a crease, 119 00:06:55,757 --> 00:06:57,090 and this would be a semi-crease. 120 00:07:01,630 --> 00:07:03,920 Here the derivative changes, that looks like a crease, 121 00:07:03,920 --> 00:07:06,120 if you had a one-dimensional piece of paper. 122 00:07:06,120 --> 00:07:08,570 This doesn't look like a crease, it's kind of smooth, 123 00:07:08,570 --> 00:07:11,250 but not-- it's a little bit smooth, it's c1, 124 00:07:11,250 --> 00:07:12,180 but it's not c2. 125 00:07:12,180 --> 00:07:14,420 So there's a semi-crease there. 126 00:07:14,420 --> 00:07:17,310 And part of that paper is sort of worrying about semi-creases, 127 00:07:17,310 --> 00:07:19,570 because we want to deal with c2 parts, 128 00:07:19,570 --> 00:07:22,692 because c2 functions are nice. 129 00:07:22,692 --> 00:07:24,400 I mean, ideally we have C infinity parts, 130 00:07:24,400 --> 00:07:26,690 but we only need c2 parts, and so 131 00:07:26,690 --> 00:07:28,170 we subdivide at the semi-creases. 132 00:07:28,170 --> 00:07:30,670 And basically, argue most the time semi-creases don't really 133 00:07:30,670 --> 00:07:32,586 happen, so you don't have to worry about them. 134 00:07:32,586 --> 00:07:34,860 But that's what they are. 135 00:07:34,860 --> 00:07:36,490 Any questions about smoothness? 136 00:07:36,490 --> 00:07:39,366 These are all different kinds of smoothness. 137 00:07:39,366 --> 00:07:40,830 Yeah. 138 00:07:40,830 --> 00:07:43,270 AUDIENCE: [INAUDIBLE]. 139 00:07:43,270 --> 00:07:47,480 PROFESSOR: All of these things are C infinity, 140 00:07:47,480 --> 00:07:49,770 except at these points. 141 00:07:49,770 --> 00:07:52,260 So every polynomial C infinity, you can differentiate it 142 00:07:52,260 --> 00:07:56,100 all the time, eventually get to zero. 143 00:07:56,100 --> 00:07:58,080 Exponentials are C infinity, pretty 144 00:07:58,080 --> 00:08:00,535 much all functions you can think of that are smooth 145 00:08:00,535 --> 00:08:01,425 are C infinity. 146 00:08:04,070 --> 00:08:04,570 Yeah. 147 00:08:08,080 --> 00:08:12,180 All right, so next question is about one 148 00:08:12,180 --> 00:08:19,560 of the proofs, which was that-- the proof involving normals 149 00:08:19,560 --> 00:08:21,870 basically, the one involving tangent planes, 150 00:08:21,870 --> 00:08:23,560 and the other one is involving normals. 151 00:08:23,560 --> 00:08:26,840 This is the polygonal implies flat proof, 152 00:08:26,840 --> 00:08:29,570 so this was, If you have a region of paper 153 00:08:29,570 --> 00:08:31,980 that is bounded by polygonal creases, 154 00:08:31,980 --> 00:08:37,010 so they're piecewise straight, then that region of paper, 155 00:08:37,010 --> 00:08:40,360 if it's uncreased, must actually be flat. 156 00:08:40,360 --> 00:08:42,145 And I'm just going to, I want to talk 157 00:08:42,145 --> 00:08:43,520 about just one part of the proof. 158 00:08:43,520 --> 00:08:45,790 So there was a sequence of steps, 159 00:08:45,790 --> 00:08:49,470 but ultimately what we wanted was some segment, bf. 160 00:08:49,470 --> 00:08:54,100 This is a boundary of your region, somewhere over here, 161 00:08:54,100 --> 00:08:56,690 and we assume that it's straight. 162 00:08:56,690 --> 00:08:59,970 So, we took a few steps to get here. 163 00:08:59,970 --> 00:09:02,970 But suppose we have a straight region and locally, 164 00:09:02,970 --> 00:09:06,700 around this point, there's some ruling. 165 00:09:06,700 --> 00:09:11,250 And then we're looking at a point, Q, here. 166 00:09:11,250 --> 00:09:16,162 I'm going to define this ruling to be r of q. 167 00:09:16,162 --> 00:09:21,015 The rule line from q, and so q is an arbitrary point, 168 00:09:21,015 --> 00:09:23,310 it slides along this segment. 169 00:09:23,310 --> 00:09:25,580 And then we're interested in a normal vector 170 00:09:25,580 --> 00:09:29,020 here, which we called n of q. 171 00:09:29,020 --> 00:09:32,200 That's normal to the surface at this point, 172 00:09:32,200 --> 00:09:35,230 and so as q slides, n of q changes. 173 00:09:35,230 --> 00:09:37,810 And we were claiming a few things, 174 00:09:37,810 --> 00:09:41,630 but so the definition of this guy being normal 175 00:09:41,630 --> 00:09:45,602 is it has to be perpendicular to-- normal to the surface is 176 00:09:45,602 --> 00:09:46,810 perpendicular to the surface. 177 00:09:46,810 --> 00:09:49,470 It has to be perpendicular to this axis of the surface, 178 00:09:49,470 --> 00:09:51,760 and it has to be perpendicular to this axis. 179 00:09:51,760 --> 00:09:54,620 So we know, just from definition of normal, n of q 180 00:09:54,620 --> 00:10:00,300 is perpendicular to bf, and n of q 181 00:10:00,300 --> 00:10:06,890 is perpendicular to r of q, the rule line. 182 00:10:06,890 --> 00:10:09,900 So that's the definition but it wasn't quite what we wanted. 183 00:10:09,900 --> 00:10:13,770 We wanted to prove, for example, that the derivative 184 00:10:13,770 --> 00:10:19,130 of the normal, funny thing, is perpendicular to bf. 185 00:10:19,130 --> 00:10:20,980 That was one of the things we wanted, 186 00:10:20,980 --> 00:10:24,770 and the other one was the corresponding perpendicular 187 00:10:24,770 --> 00:10:25,810 to the rule line. 188 00:10:25,810 --> 00:10:27,810 I don't know if I used this notation in lecture, 189 00:10:27,810 --> 00:10:29,768 but this just means that they're perpendicular, 190 00:10:29,768 --> 00:10:33,260 which is like the dot product of those two vectors is zero. 191 00:10:33,260 --> 00:10:36,810 So, how do we go from here to here? 192 00:10:36,810 --> 00:10:39,260 This was actually the part that was most confusing to me. 193 00:10:39,260 --> 00:10:42,620 I find this one more confusing, because it uses the fact 194 00:10:42,620 --> 00:10:44,819 that it's [INAUDIBLE], but the question 195 00:10:44,819 --> 00:10:47,360 was actually about this one, so unless you ask about this one 196 00:10:47,360 --> 00:10:48,740 now, I'll just cover this one. 197 00:10:48,740 --> 00:10:49,700 It's a little simpler. 198 00:10:49,700 --> 00:10:51,630 Both are in the notes. 199 00:10:51,630 --> 00:10:54,630 So why is this true? 200 00:10:54,630 --> 00:10:56,810 I guess, first, the intuitive reason. 201 00:10:56,810 --> 00:10:59,297 So, you've got all these different normals here. 202 00:10:59,297 --> 00:11:01,130 We're going to end up concluding they're all 203 00:11:01,130 --> 00:11:03,180 pointed the same way but, in general, 204 00:11:03,180 --> 00:11:07,380 they're changing direction somehow. 205 00:11:07,380 --> 00:11:09,760 We know that at all times the normal 206 00:11:09,760 --> 00:11:11,780 is perpendicular to this segment, 207 00:11:11,780 --> 00:11:15,180 so it is pointing away, straight away from the segments. 208 00:11:15,180 --> 00:11:18,189 So you've got a right angle here. 209 00:11:18,189 --> 00:11:19,980 So it's kind of spinning around the segment 210 00:11:19,980 --> 00:11:23,720 at best, as it moves around. 211 00:11:23,720 --> 00:11:27,410 We claim that the change in the normal, the derivative 212 00:11:27,410 --> 00:11:30,280 of the normal, as you walk along this segment, 213 00:11:30,280 --> 00:11:32,400 must also be perpendicular to the segment. 214 00:11:32,400 --> 00:11:38,020 Intuitively, this is obvious, if you think about it enough. 215 00:11:38,020 --> 00:11:40,520 If you were changing in a direction that was not 216 00:11:40,520 --> 00:11:42,940 perpendicular to normal, to this segment, for example 217 00:11:42,940 --> 00:11:44,870 you're going this way or something, 218 00:11:44,870 --> 00:11:47,220 then, while initially you're perpendicular, 219 00:11:47,220 --> 00:11:49,140 you're kind of falling over, and then you'll 220 00:11:49,140 --> 00:11:50,470 no longer be perpendicular. 221 00:11:50,470 --> 00:11:52,010 So if you change in a first order 222 00:11:52,010 --> 00:11:54,110 way that is not perpendicular to this thing, 223 00:11:54,110 --> 00:11:56,400 then afterwards you will no longer be perpendicular. 224 00:11:56,400 --> 00:12:00,410 So that's intuitively why this follows from this. 225 00:12:03,250 --> 00:12:06,020 One way to formalize that is to use Taylor expansion, 226 00:12:06,020 --> 00:12:08,630 so I'll just write that quickly. 227 00:12:08,630 --> 00:12:10,690 If you look at a point, q, and I'm 228 00:12:10,690 --> 00:12:12,630 going to use this notation, q plus epsilon, 229 00:12:12,630 --> 00:12:14,960 to mean a point just a little bit over. 230 00:12:14,960 --> 00:12:17,490 This is q plus epsilon, this is q. 231 00:12:20,330 --> 00:12:23,140 For a very small epsilon, this is going to be approximately n 232 00:12:23,140 --> 00:12:26,670 of q, plus epsilon, and the prime of q. 233 00:12:29,990 --> 00:12:32,690 In Taylor expansion, then you have epsilon squared, 234 00:12:32,690 --> 00:12:35,950 but for small enough epsilon, this is a good approximation. 235 00:12:35,950 --> 00:12:45,700 And so, on the one hand, you have-- 236 00:12:45,700 --> 00:12:47,810 If you look at n of q plus epsilon, 237 00:12:47,810 --> 00:12:49,286 and you compare it to bf, you know 238 00:12:49,286 --> 00:12:51,660 that these are perpendicular, and what you can write that 239 00:12:51,660 --> 00:12:54,070 as a product being equal to zero. 240 00:12:54,070 --> 00:12:55,730 So the dot product of these two vectors 241 00:12:55,730 --> 00:12:58,097 should be equal to zero, that is, being perpendicular. 242 00:12:58,097 --> 00:12:59,680 We know this is true for all points q, 243 00:12:59,680 --> 00:13:02,390 so in particular, it's going to be true for q plus epsilon. 244 00:13:02,390 --> 00:13:05,730 So that's what we know, but now we can expand this thing, 245 00:13:05,730 --> 00:13:07,650 and we get these two terms. 246 00:13:07,650 --> 00:13:17,820 So we get n of q dot bf, plus epsilon and prime of q, dot bf. 247 00:13:17,820 --> 00:13:21,720 That's just distributing that sum over this product, 248 00:13:21,720 --> 00:13:23,910 dot products work OK that way. 249 00:13:23,910 --> 00:13:27,250 Now this thing is also zero, because this 250 00:13:27,250 --> 00:13:29,204 is just another point q dotted with bf, 251 00:13:29,204 --> 00:13:30,870 and they should always be perpendicular. 252 00:13:30,870 --> 00:13:33,640 And this is the thing we care about. 253 00:13:33,640 --> 00:13:35,560 Now if this whole sum is equal to zero, 254 00:13:35,560 --> 00:13:37,770 and this term is equal to zero, then this better 255 00:13:37,770 --> 00:13:38,870 be equal to zero. 256 00:13:38,870 --> 00:13:41,740 So implies this thing equals zero, 257 00:13:41,740 --> 00:13:43,140 and that's what we wanted. 258 00:13:43,140 --> 00:13:45,040 And prime of q is perpendicular to bf. 259 00:13:45,040 --> 00:13:47,840 So that's the algebra way to see it. 260 00:13:47,840 --> 00:13:50,220 But if you think about it enough, and you believe things 261 00:13:50,220 --> 00:13:52,730 are linear to the first order, as we say, 262 00:13:52,730 --> 00:13:55,657 than that has to be true. 263 00:13:55,657 --> 00:13:57,740 You can get second derivatives and things as well, 264 00:13:57,740 --> 00:14:01,370 but first derivative better hold in, particular. 265 00:14:01,370 --> 00:14:04,696 Any questions about that? 266 00:14:04,696 --> 00:14:06,570 You can do the same thing with this property, 267 00:14:06,570 --> 00:14:07,382 but it's messier. 268 00:14:07,382 --> 00:14:09,340 You need to use the [INAUDIBLE] property, which 269 00:14:09,340 --> 00:14:11,310 we didn't really go into much detail on, 270 00:14:11,310 --> 00:14:13,670 so I think I'll skip it. 271 00:14:16,540 --> 00:14:20,615 Last question is, what does it all mean? 272 00:14:20,615 --> 00:14:22,240 If we're proving things are impossible, 273 00:14:22,240 --> 00:14:26,150 and yet here I have physical models of them existing, 274 00:14:26,150 --> 00:14:26,920 what's going on? 275 00:14:26,920 --> 00:14:28,920 What's the difference between mathematical paper 276 00:14:28,920 --> 00:14:29,940 and real paper? 277 00:14:29,940 --> 00:14:33,300 And, of course, there's no real mathematical answer 278 00:14:33,300 --> 00:14:36,540 to that question, but there-- we have a couple competing 279 00:14:36,540 --> 00:14:40,450 theories for what might be happening in this model. 280 00:14:40,450 --> 00:14:45,080 One is that perhaps the paper is not being isometric. 281 00:14:45,080 --> 00:14:47,660 Perhaps it's stretching or shearing. 282 00:14:47,660 --> 00:14:53,590 So normally we think of paper as unstretchable and not 283 00:14:53,590 --> 00:14:57,630 shearable, the only way to shear is to fold it. 284 00:14:57,630 --> 00:15:00,890 And that is almost perfectly true but, of course, 285 00:15:00,890 --> 00:15:01,990 no material is perfect. 286 00:15:01,990 --> 00:15:05,370 So it might be shearing or changing 287 00:15:05,370 --> 00:15:08,460 the geometry, the intrinsic geometry just enough 288 00:15:08,460 --> 00:15:10,822 to somehow make this possible. 289 00:15:10,822 --> 00:15:12,530 Our theory only says that it's impossible 290 00:15:12,530 --> 00:15:15,050 if you're exactly perfect, so maybe-- it 291 00:15:15,050 --> 00:15:17,960 would be interesting to measure how exact paper is. 292 00:15:17,960 --> 00:15:18,830 Question? 293 00:15:18,830 --> 00:15:21,770 AUDIENCE: If it's possible to imitate a shear, 294 00:15:21,770 --> 00:15:26,019 if it compresses the paper but I fold it, then-- 295 00:15:26,019 --> 00:15:27,310 PROFESSOR: This is not a shear. 296 00:15:27,310 --> 00:15:29,046 AUDIENCE: I know, but if you can fold it, 297 00:15:29,046 --> 00:15:30,760 so that it, like, imitates a shear-- 298 00:15:30,760 --> 00:15:32,910 PROFESSOR: Well, you can add creases to make it, 299 00:15:32,910 --> 00:15:34,374 to simulate a shear. 300 00:15:34,374 --> 00:15:35,040 AUDIENCE: Right. 301 00:15:35,040 --> 00:15:36,623 PROFESSOR: But only by adding creases, 302 00:15:36,623 --> 00:15:38,820 that's what we proved in lecture. 303 00:15:38,820 --> 00:15:41,370 So yeah, you can kind of fake a shear, but only 304 00:15:41,370 --> 00:15:42,120 by adding creases. 305 00:15:42,120 --> 00:15:44,390 Now, it doesn't look like there's extra creases here. 306 00:15:44,390 --> 00:15:46,040 Theory number 2 is there are lots 307 00:15:46,040 --> 00:15:48,860 of really tiny creases here that are so small. 308 00:15:48,860 --> 00:15:50,780 I mean, here you could detect the crease, 309 00:15:50,780 --> 00:15:53,170 because it was a big change in angle, 310 00:15:53,170 --> 00:15:54,732 but if they're super, super tiny here 311 00:15:54,732 --> 00:15:57,070 and, you know, we never fold these things completely 312 00:15:57,070 --> 00:16:00,205 perfect, it could be lots of little tiny things, 313 00:16:00,205 --> 00:16:01,580 maybe you look under a microscope 314 00:16:01,580 --> 00:16:03,380 and you'll see that, I don't know. 315 00:16:03,380 --> 00:16:04,796 It could be an interesting project 316 00:16:04,796 --> 00:16:07,750 to try to investigate what really happens with paper 317 00:16:07,750 --> 00:16:10,600 in these kinds of models in real life, 318 00:16:10,600 --> 00:16:13,000 and how it differs from mathematics. 319 00:16:13,000 --> 00:16:17,370 But the two theories are maybe some stretching or maybe 320 00:16:17,370 --> 00:16:18,220 extra creases. 321 00:16:18,220 --> 00:16:21,550 We know by adding the diagonals, it folds. 322 00:16:21,550 --> 00:16:23,080 Now here, you could pretty clearly 323 00:16:23,080 --> 00:16:25,090 see the diagonals are not being added, 324 00:16:25,090 --> 00:16:27,880 but there might be-- there might be effectively, 325 00:16:27,880 --> 00:16:31,650 there might be enough added creases that are together 326 00:16:31,650 --> 00:16:34,186 so tiny, you can't see them. 327 00:16:34,186 --> 00:16:36,630 AUDIENCE: It's not semi-creases-- 328 00:16:36,630 --> 00:16:39,210 PROFESSOR: Semi-creases are not enough. 329 00:16:39,210 --> 00:16:42,740 The theory says that if you have these creases and possibly 330 00:16:42,740 --> 00:16:46,110 any number of semi-creases, you can't fold it all. 331 00:16:46,110 --> 00:16:48,550 So you need to add actual creases. 332 00:16:48,550 --> 00:16:51,640 Yeah, good question. 333 00:16:51,640 --> 00:16:57,050 All right, that's it for old material. 334 00:16:57,050 --> 00:16:59,950 Now I wanted to show you some cool new material, 335 00:16:59,950 --> 00:17:02,474 which is based on-- 336 00:17:02,474 --> 00:17:03,430 AUDIENCE: [INAUDIBLE]. 337 00:17:03,430 --> 00:17:04,513 PROFESSOR: Yeah, question? 338 00:17:04,513 --> 00:17:06,298 AUDIENCE: [INAUDIBLE] don't really 339 00:17:06,298 --> 00:17:08,192 look like straight lines. 340 00:17:08,192 --> 00:17:08,900 PROFESSOR: Right. 341 00:17:08,900 --> 00:17:10,720 The creases here do not look like straight lines, 342 00:17:10,720 --> 00:17:12,690 and yet we prove they must be straight lines, 343 00:17:12,690 --> 00:17:14,900 so something's going on. 344 00:17:14,900 --> 00:17:17,630 Now, we didn't know initially that straight creases had 345 00:17:17,630 --> 00:17:19,589 to stay straight in 3D, but once we 346 00:17:19,589 --> 00:17:23,172 proved that, we were pretty sure this didn't exist. 347 00:17:23,172 --> 00:17:25,380 It's not obvious that straight creases stay straight, 348 00:17:25,380 --> 00:17:27,803 but they do. 349 00:17:27,803 --> 00:17:31,130 It's evidence there's a problem. 350 00:17:31,130 --> 00:17:32,922 OK, I wanted to talk about this paper, 351 00:17:32,922 --> 00:17:34,130 because it's kind of related. 352 00:17:34,130 --> 00:17:38,550 It's about how to efficiently make pleat folding happen, 353 00:17:38,550 --> 00:17:42,050 and it's written by a bunch of people, Jean Cardinal, Marty, 354 00:17:42,050 --> 00:17:47,900 our cameraman, Shinji Imahori, Yoshi Ito, Messashi Kiomi, 355 00:17:47,900 --> 00:17:50,240 Stefan Langerman, Ryuhei Uehara, who's 356 00:17:50,240 --> 00:17:54,440 here on sabbatical, and Tachiachi Uno. 357 00:17:54,440 --> 00:17:56,510 Good practice. 358 00:17:56,510 --> 00:17:59,940 So, here's the kind of setting. 359 00:17:59,940 --> 00:18:02,390 It's going to be familiar, in that we 360 00:18:02,390 --> 00:18:07,130 have a one-D piece of paper, so it's a segment, 361 00:18:07,130 --> 00:18:10,040 and we're also going to assume that it's uniformly creased, 362 00:18:10,040 --> 00:18:13,890 so each of these is spacing of 1. 363 00:18:13,890 --> 00:18:18,000 And now usually we think of preassigning mountains 364 00:18:18,000 --> 00:18:20,340 and valleys here, and then you could only fold mountains 365 00:18:20,340 --> 00:18:23,390 on the mountains, you can only fold valleys in the valleys. 366 00:18:23,390 --> 00:18:25,040 The model here is that you're allowed 367 00:18:25,040 --> 00:18:29,020 to fold and unfold repeatedly on the same segments. 368 00:18:29,020 --> 00:18:33,510 So you can do some layers, simple folds, 369 00:18:33,510 --> 00:18:37,250 although, I think, we'll just need all layers here today. 370 00:18:40,510 --> 00:18:45,340 But you can also unfold any previous folds. 371 00:18:50,780 --> 00:18:54,599 And I'll say that unfolds are free, 372 00:18:54,599 --> 00:18:56,390 you don't pay for the unfolding operations, 373 00:18:56,390 --> 00:18:58,220 because it's just a constant factor. 374 00:18:58,220 --> 00:19:00,720 You can only unfold things that have previously been folded. 375 00:19:00,720 --> 00:19:02,762 So we'll count the number of folds you make. 376 00:19:02,762 --> 00:19:04,970 Each time you do a fold, it's a mountain or a valley, 377 00:19:04,970 --> 00:19:07,680 say, through all the layers and through some of the layers. 378 00:19:07,680 --> 00:19:11,660 But our goal is in, the end, to have our segment creased 379 00:19:11,660 --> 00:19:12,880 in a particular pattern. 380 00:19:12,880 --> 00:19:17,900 And the pleat pattern would be mv, mv, mv. 381 00:19:17,900 --> 00:19:20,960 So we want the last time each crease to be folded, 382 00:19:20,960 --> 00:19:23,290 to be a particular pattern, but along the way 383 00:19:23,290 --> 00:19:26,840 we could fold it in lots of-- it could be, 384 00:19:26,840 --> 00:19:29,510 this one could be mountain for a while, as long as the last time 385 00:19:29,510 --> 00:19:33,570 we crease this thing, it's a valley, we're happy. 386 00:19:33,570 --> 00:19:37,650 So we end up with mountain valley strings, 387 00:19:37,650 --> 00:19:41,540 and we want to know how many folds, 388 00:19:41,540 --> 00:19:44,110 how many folding operations do you need to make, 389 00:19:44,110 --> 00:19:47,640 in order to achieve a particular mountain valley string? 390 00:19:52,950 --> 00:19:56,370 And, in particular, we care about strings like mv mv mv, 391 00:19:56,370 --> 00:19:59,990 because, as we'll learn later today when we fold these, 392 00:19:59,990 --> 00:20:02,530 pleating takes a lot of time, it's tedious. 393 00:20:02,530 --> 00:20:04,660 It would be nice if you could do things faster. 394 00:20:04,660 --> 00:20:07,600 Now, why do we think we can do things faster? 395 00:20:07,600 --> 00:20:11,840 Because I can take a piece of paper and, imagine this 396 00:20:11,840 --> 00:20:14,780 is one-dimensional, and fold it in half, 397 00:20:14,780 --> 00:20:16,890 and then fold it in half again. 398 00:20:16,890 --> 00:20:19,020 Boom, I got how many creases? 399 00:20:19,020 --> 00:20:21,700 Two creases for the price of one operation. 400 00:20:21,700 --> 00:20:23,530 Now I fold again, Boom! 401 00:20:23,530 --> 00:20:25,710 I get four creases for the price of one. 402 00:20:25,710 --> 00:20:27,180 I fold it again. 403 00:20:27,180 --> 00:20:29,700 Wow, I get eight creases for the price of one. 404 00:20:29,700 --> 00:20:32,620 So the number of folding operations you do here 405 00:20:32,620 --> 00:20:35,260 may be much, much smaller than n. 406 00:20:35,260 --> 00:20:38,970 n is going to be the number of the creases, 407 00:20:38,970 --> 00:20:42,150 the length of your string. 408 00:20:42,150 --> 00:20:44,530 So if I fold a piece of paper like this-- 409 00:20:44,530 --> 00:20:46,600 I actually tried this-- and then I 410 00:20:46,600 --> 00:20:48,970 unfold to see what mountain valley assignment 411 00:20:48,970 --> 00:20:55,060 I got-- unfolding is free-- I get this weird shape. 412 00:20:55,060 --> 00:20:56,390 And I didn't do it perfectly. 413 00:20:56,390 --> 00:20:58,947 It's hard to do perfectly because of creep. 414 00:20:58,947 --> 00:20:59,780 I get a weird thing. 415 00:20:59,780 --> 00:21:03,580 Anyone know what this thing is called? 416 00:21:03,580 --> 00:21:06,917 I think that's pretty much-- if I unfold them to 90 degrees. 417 00:21:06,917 --> 00:21:07,875 AUDIENCE: Dragon curve. 418 00:21:07,875 --> 00:21:08,875 PROFESSOR: Dragon curve. 419 00:21:08,875 --> 00:21:09,700 Yeah. 420 00:21:09,700 --> 00:21:11,880 Dragon curve looks like this. 421 00:21:11,880 --> 00:21:15,110 This is the Wikipedia drawing, and the more 422 00:21:15,110 --> 00:21:18,210 you fold in half, that you keep adding this iteration, 423 00:21:18,210 --> 00:21:20,690 it's a nice fractal. 424 00:21:20,690 --> 00:21:22,190 It's kind of cool, it doesn't have-- 425 00:21:22,190 --> 00:21:23,860 you can prove it doesn't self intersect. 426 00:21:23,860 --> 00:21:27,250 It touches itself at vertices, but it doesn't properly 427 00:21:27,250 --> 00:21:28,541 cross itself. 428 00:21:28,541 --> 00:21:30,540 You take the limit, it looks something like this 429 00:21:30,540 --> 00:21:32,930 if you fill it in, very cool. 430 00:21:32,930 --> 00:21:36,190 Wikipedia calls it the, I don't want to spoil the surprise, 431 00:21:36,190 --> 00:21:40,660 but there's actually a book where the section headings are 432 00:21:40,660 --> 00:21:41,800 iterations of this fractal? 433 00:21:41,800 --> 00:21:43,050 Anyone know what this book is? 434 00:21:43,050 --> 00:21:44,050 AUDIENCE: Jurassic Park. 435 00:21:44,050 --> 00:21:48,370 PROFESSOR: Jurassic Park, yes, Ian Malcolm right there. 436 00:21:48,370 --> 00:21:49,820 So, pretty cool. 437 00:21:49,820 --> 00:21:52,153 It's called the Jurassic Park fractal, to some, at least 438 00:21:52,153 --> 00:21:54,650 on Wikipedia. 439 00:21:54,650 --> 00:21:57,780 And so that's one particular mountain valley assignment 440 00:21:57,780 --> 00:21:58,510 you can get. 441 00:21:58,510 --> 00:22:00,910 And let me tell you what is. 442 00:22:03,610 --> 00:22:06,554 It's pretty easy to figure out, and you see why it's a fractal. 443 00:22:06,554 --> 00:22:07,970 So the first time we make a fold-- 444 00:22:07,970 --> 00:22:10,390 let's say I always make mountains, just for simplicity. 445 00:22:10,390 --> 00:22:14,567 You have a little bit of choice here, but not very much. 446 00:22:14,567 --> 00:22:16,400 So first I make a mountain fold, so now I've 447 00:22:16,400 --> 00:22:17,700 got things folded in half. 448 00:22:17,700 --> 00:22:19,840 Now let's say I make a mountain fold over here. 449 00:22:19,840 --> 00:22:21,465 Of course, it's already folded in half, 450 00:22:21,465 --> 00:22:23,370 which means I get a valley over here. 451 00:22:23,370 --> 00:22:26,370 So I'm going to-- Yeah. 452 00:22:26,370 --> 00:22:31,275 So now I make another mountain fold, maybe 453 00:22:31,275 --> 00:22:34,430 I should have done it the other way, anyway, so it is. 454 00:22:34,430 --> 00:22:38,310 So I make another mountain fold here, and on the left 455 00:22:38,310 --> 00:22:42,090 I get this, on the right I get the reflection of that. 456 00:22:42,090 --> 00:22:43,855 And then I make another mountain fold, 457 00:22:43,855 --> 00:22:45,534 and I get this on the left, and I 458 00:22:45,534 --> 00:22:47,200 get the reflection of that on the right. 459 00:22:47,200 --> 00:22:49,730 And when I reflect, I'm reading backwards and also inverting 460 00:22:49,730 --> 00:22:50,230 everything. 461 00:22:55,780 --> 00:22:57,260 vv and so on. 462 00:22:57,260 --> 00:22:59,870 So you see the kind of fractal nature here, 463 00:22:59,870 --> 00:23:03,090 and that's what gives the cool curve. 464 00:23:03,090 --> 00:23:04,810 Keep repeating that. 465 00:23:04,810 --> 00:23:06,750 You have, essentially, n choices, 466 00:23:06,750 --> 00:23:08,532 because they're log n folds that you make, 467 00:23:08,532 --> 00:23:09,990 each could be a mountain or valley. 468 00:23:09,990 --> 00:23:13,060 So there are n different mv strings 469 00:23:13,060 --> 00:23:19,720 you can get in essentially log n folds. 470 00:23:19,720 --> 00:23:22,000 But sadly, none of them is the one 471 00:23:22,000 --> 00:23:25,490 we want, mvmvmv, the pleat fold. 472 00:23:25,490 --> 00:23:28,405 So that leaves us with the question of how many operations 473 00:23:28,405 --> 00:23:30,300 do you need to get a pleat fold? 474 00:23:36,570 --> 00:23:38,790 This is where things get cool. 475 00:23:38,790 --> 00:23:46,270 So-- why don't I tell you some answers. 476 00:23:46,270 --> 00:23:51,880 So if you want to fold mmmm, or mvmv, 477 00:23:51,880 --> 00:23:53,630 these turn out to be the same, because you 478 00:23:53,630 --> 00:23:57,610 can do m's on the odd positions, v's on the even positions. 479 00:23:57,610 --> 00:24:02,390 Then you can do it in log squared n folds, 480 00:24:02,390 --> 00:24:08,050 and you need at least log squared n over log, log n. 481 00:24:08,050 --> 00:24:10,590 This is way, way faster than n. 482 00:24:10,590 --> 00:24:12,090 There's an open problem between log 483 00:24:12,090 --> 00:24:13,506 squared n and log squared over log 484 00:24:13,506 --> 00:24:16,320 log, small gap of the log log factor. 485 00:24:16,320 --> 00:24:19,150 But somewhere in between here. 486 00:24:19,150 --> 00:24:27,470 If you want to do, let's say, a random pattern, then we 487 00:24:27,470 --> 00:24:38,810 can prove you need about n divided by log n. 488 00:24:38,810 --> 00:24:42,800 So the obvious way to do it is fold each crease unfolds, 489 00:24:42,800 --> 00:24:44,230 that takes n steps. 490 00:24:44,230 --> 00:24:46,600 So you can always improve by a log factor, 491 00:24:46,600 --> 00:24:49,290 and that's the best you could do for most strings. 492 00:24:49,290 --> 00:24:51,630 But these strings are sufficiently special, 493 00:24:51,630 --> 00:24:54,920 that you can get a huge factor, basically exponential, when, 494 00:24:54,920 --> 00:24:56,620 instead of doing roughly n steps, 495 00:24:56,620 --> 00:24:59,290 you're only doing poly log steps. 496 00:24:59,290 --> 00:25:02,050 So, there's four results here. 497 00:25:02,050 --> 00:25:04,100 Which ones would you like to see? 498 00:25:04,100 --> 00:25:06,690 How many people vote for log squared n upper bound? 499 00:25:09,290 --> 00:25:09,790 Four. 500 00:25:09,790 --> 00:25:12,120 How many people vote for log squared over log log 501 00:25:12,120 --> 00:25:13,750 lower bound? 502 00:25:13,750 --> 00:25:15,840 Completely different set of four. 503 00:25:15,840 --> 00:25:19,010 How many people vote for the n over log n story, upper 504 00:25:19,010 --> 00:25:22,461 bound first and lower bound? 505 00:25:22,461 --> 00:25:22,960 OK. 506 00:25:22,960 --> 00:25:26,250 I think the upper bound here wins, just curious. 507 00:25:26,250 --> 00:25:29,447 They all these similar ideas. 508 00:25:29,447 --> 00:25:31,030 I'm going to briefly sketch, because I 509 00:25:31,030 --> 00:25:36,170 want to get to folding stuff, how you achieve n over log n 510 00:25:36,170 --> 00:25:39,090 upper bound, maybe then I'll do log squared upper bound here. 511 00:25:39,090 --> 00:25:45,630 Lower bounds are, well, they're bounds. 512 00:25:45,630 --> 00:25:47,830 That doesn't say anything. 513 00:25:47,830 --> 00:25:52,580 So, if you have an mv string, to get this bound, arbitrary 514 00:25:52,580 --> 00:25:58,840 string, I'm going to consider epsilon-- sorry, 515 00:25:58,840 --> 00:26:04,250 1 minus epsilon log n, consecutive letters, 516 00:26:04,250 --> 00:26:11,340 and split into, basically, n over log n chunks, each of size 517 00:26:11,340 --> 00:26:15,760 roughly log n, a little bit smaller than log n. 518 00:26:15,760 --> 00:26:17,000 Why do I do that? 519 00:26:17,000 --> 00:26:20,270 Because the number of different chunk values 520 00:26:20,270 --> 00:26:23,890 is 2 to the 1 minus epsilon, times log n, right, 521 00:26:23,890 --> 00:26:27,280 each can be mountain or valley. 522 00:26:27,280 --> 00:26:31,090 Now 2 the log n is n, so this is n to the 1 minus 523 00:26:31,090 --> 00:26:34,945 epsilon, possible different chunk values. 524 00:26:42,010 --> 00:26:43,940 These are chunks. 525 00:26:43,940 --> 00:26:45,520 Why is that interesting? 526 00:26:45,520 --> 00:26:48,525 Because I have basically n over log n chunks, 527 00:26:48,525 --> 00:26:50,900 but there's only n to the 1 minus epsilon different chunk 528 00:26:50,900 --> 00:26:51,400 values. 529 00:26:51,400 --> 00:26:53,840 This is much, much smaller than this. 530 00:26:53,840 --> 00:26:56,460 Think of epsilon being a half, as the square root of n, this 531 00:26:56,460 --> 00:26:57,940 is n over log n. 532 00:26:57,940 --> 00:27:02,410 So there's actually most-- many the chunks have to be repeated. 533 00:27:02,410 --> 00:27:13,950 So this means that an average chunk is repeated, What is it? 534 00:27:13,950 --> 00:27:19,620 It's n over log n of them, is n over log n, 535 00:27:19,620 --> 00:27:23,100 times n to the 1 minus epsilon times, 536 00:27:23,100 --> 00:27:26,220 which is n to the epsilon over log n. 537 00:27:29,840 --> 00:27:32,060 This is the number of chunks here, 538 00:27:32,060 --> 00:27:36,030 and we're dividing by the number of different ones, 539 00:27:36,030 --> 00:27:37,844 so typical repetition is going to be 540 00:27:37,844 --> 00:27:39,260 n to the epsilon over log n times. 541 00:27:39,260 --> 00:27:41,689 I'll just assume all chunks are repeated this many times. 542 00:27:41,689 --> 00:27:43,230 It's all linear, so it doesn't matter 543 00:27:43,230 --> 00:27:47,490 whether some are more common and some are less common. 544 00:27:47,490 --> 00:27:51,601 OK, cool. 545 00:27:51,601 --> 00:27:52,100 Now what? 546 00:27:55,780 --> 00:28:01,435 So, if I look at one of these chunks and all of it 547 00:28:01,435 --> 00:28:03,950 is repetitions, I would like to fold them all somehow 548 00:28:03,950 --> 00:28:06,160 more efficiently. 549 00:28:06,160 --> 00:28:08,710 And there's one key idea in this paper 550 00:28:08,710 --> 00:28:17,710 on how to do that, which is-- so here's my strip, maybe here's 551 00:28:17,710 --> 00:28:20,690 one instance of the chunk, here's another one. 552 00:28:20,690 --> 00:28:23,360 They can be kind of spaced out arbitrarily. 553 00:28:23,360 --> 00:28:28,410 So what I'm going to do is fold here, fold here, fold here. 554 00:28:28,410 --> 00:28:30,410 I think that's right, all the bisectors. 555 00:28:30,410 --> 00:28:32,660 So I end up with one chunk like this, 556 00:28:32,660 --> 00:28:37,550 than another chunk like this, than another chunk something 557 00:28:37,550 --> 00:28:39,180 like this. 558 00:28:39,180 --> 00:28:41,410 Then, this one is very tiny. 559 00:28:41,410 --> 00:28:43,390 Can I get my four repetitions of the chunk 560 00:28:43,390 --> 00:28:45,660 all on top of each other? 561 00:28:45,660 --> 00:28:49,810 What I then do, so this is going to take how many repetitions 562 00:28:49,810 --> 00:28:52,060 is, n to the epsilon over log n, so this is about n 563 00:28:52,060 --> 00:28:57,250 to the epsilon over log n steps, to do all these folds. 564 00:28:57,250 --> 00:29:01,240 Now I fold here, fold here, fold here, fold here simultaneously, 565 00:29:01,240 --> 00:29:02,940 getting them all correct. 566 00:29:02,940 --> 00:29:05,240 That's going to take 1 minus epsilon times 567 00:29:05,240 --> 00:29:09,400 basically log n steps, then I unfold. 568 00:29:09,400 --> 00:29:13,580 And then I discover, Darn it, half of them were upside down. 569 00:29:13,580 --> 00:29:15,580 Because I got this one right and this one right, 570 00:29:15,580 --> 00:29:18,990 but these two are flipped, so that's so good, 571 00:29:18,990 --> 00:29:21,967 so then I recurse on the remaining half. 572 00:29:21,967 --> 00:29:23,800 But this turns out to be a geometric series, 573 00:29:23,800 --> 00:29:25,940 so if I do all four of these, and then I do two of them, 574 00:29:25,940 --> 00:29:27,790 and then I do one of them, I'll be done, 575 00:29:27,790 --> 00:29:31,200 and it will only cost me another factor of 2 in time. 576 00:29:31,200 --> 00:29:36,200 So I get log n plus n to the epsilon over log n. 577 00:29:36,200 --> 00:29:39,070 I have to do this for each of the different chunk values, 578 00:29:39,070 --> 00:29:43,190 so I need to do n to the 1 minus epsilon, 579 00:29:43,190 --> 00:29:50,080 times log n plus n to the epsilon over log n, times 2 580 00:29:50,080 --> 00:29:52,630 because of the geometric series. 581 00:29:52,630 --> 00:29:55,225 And this comes out to n over log n. 582 00:29:55,225 --> 00:29:55,724 Ta-da! 583 00:29:59,340 --> 00:30:00,519 Magic. 584 00:30:00,519 --> 00:30:02,810 The log n is basically coming from the chunk size being 585 00:30:02,810 --> 00:30:06,480 log n and 1 minus epsilon log n as big as we could make it, 586 00:30:06,480 --> 00:30:10,810 because we needed this to be a lot smaller than n. 587 00:30:10,810 --> 00:30:14,160 So that's how you save a factor of log n, little crazy, 588 00:30:14,160 --> 00:30:16,986 but it works in theory. 589 00:30:16,986 --> 00:30:18,610 That's the upper bound, the lower bound 590 00:30:18,610 --> 00:30:22,970 is actually pretty obvious, because if you only 591 00:30:22,970 --> 00:30:29,030 use k folds, you're doing-- there's n to the k or maybe 2n 592 00:30:29,030 --> 00:30:31,040 to the k, different things you can do, 593 00:30:31,040 --> 00:30:33,415 because there's n different places you could make a fold, 594 00:30:33,415 --> 00:30:34,770 could be a mountain or a valley. 595 00:30:34,770 --> 00:30:37,490 And this better be, at least, 2 to the n, 596 00:30:37,490 --> 00:30:39,830 because they're 2 the n different mountain valley 597 00:30:39,830 --> 00:30:41,730 patterns you could make, and you have 598 00:30:41,730 --> 00:30:43,180 to somehow make them with k folds. 599 00:30:43,180 --> 00:30:45,570 These are all the things you could make with k folds. 600 00:30:45,570 --> 00:30:47,550 And you work that out and k has to be, 601 00:30:47,550 --> 00:30:50,080 at least, about n over log n. 602 00:30:50,080 --> 00:30:53,130 So that's why it's optimal, and because this is an information 603 00:30:53,130 --> 00:30:55,700 theoretic argument, this works in the average case as well. 604 00:30:55,700 --> 00:30:57,270 Take a random example. 605 00:30:57,270 --> 00:30:59,420 You need at least n over log n. 606 00:30:59,420 --> 00:31:03,190 Most examples, high probability, will need n over log n. 607 00:31:03,190 --> 00:31:05,900 So that's those lower bounds. 608 00:31:05,900 --> 00:31:10,340 Let me briefly tell you about this log squared n upper bound. 609 00:31:10,340 --> 00:31:14,890 It uses the same idea, but because everything is repeated 610 00:31:14,890 --> 00:31:19,170 in the mmmm string, you don't-- it's a lot easier to do this 611 00:31:19,170 --> 00:31:20,570 kind of folding. 612 00:31:20,570 --> 00:31:23,190 Oh sorry, there's one step I left out here. 613 00:31:23,190 --> 00:31:25,600 So great, you make these folds, you line things up, 614 00:31:25,600 --> 00:31:27,820 you fold these things. 615 00:31:27,820 --> 00:31:29,140 Then you unfold. 616 00:31:29,140 --> 00:31:33,010 So in recurse, eventually you fix these guys, 617 00:31:33,010 --> 00:31:35,384 but you also destroyed this crease and this crease 618 00:31:35,384 --> 00:31:37,800 and this crease, so you kind of messed things up a little. 619 00:31:37,800 --> 00:31:39,480 You've got to do this repeatedly for every chunk. 620 00:31:39,480 --> 00:31:41,188 You don't want to mess up previous chunks 621 00:31:41,188 --> 00:31:41,950 that you've done. 622 00:31:41,950 --> 00:31:43,730 So you have to go back and fix this fold, 623 00:31:43,730 --> 00:31:46,000 unfold, fix this fold, unfold, but that, 624 00:31:46,000 --> 00:31:49,350 again, it only costs n to the epsilon over log n, 625 00:31:49,350 --> 00:31:50,850 so not too bad. 626 00:31:50,850 --> 00:31:54,160 So another factor of 2 here. 627 00:31:54,160 --> 00:31:57,480 But I'm ignoring constant factors. 628 00:31:57,480 --> 00:32:00,900 OK, that was n over log n, upper bound. 629 00:32:00,900 --> 00:32:05,340 Let's do log squared upper bound for mmmmm. 630 00:32:08,860 --> 00:32:13,970 So I need to look this up, it's a little tricky. 631 00:32:13,970 --> 00:32:16,120 We're going to use the trick of the dragon curve, 632 00:32:16,120 --> 00:32:18,820 essentially, which is repeatedly fold in half, 633 00:32:18,820 --> 00:32:22,130 and we're going to keep doing that until we 634 00:32:22,130 --> 00:32:27,410 are left with three creases, which haven't yet been folded. 635 00:32:27,410 --> 00:32:28,650 So this is my folded bundle. 636 00:32:28,650 --> 00:32:31,430 There's many, many, many layers here. 637 00:32:31,430 --> 00:32:35,357 And then I'm going to fold mmm on those three creases. 638 00:32:35,357 --> 00:32:37,940 Instead of going all the way to dragon curve, which would just 639 00:32:37,940 --> 00:32:41,960 be, Do m, fold here first, I'm going to fold this, unfold, 640 00:32:41,960 --> 00:32:45,610 fold this, unfold, fold this, unfold, so I get this pattern. 641 00:32:45,610 --> 00:32:48,800 When I unfold this, it actually turns out 642 00:32:48,800 --> 00:32:50,800 to be kind of nice and alternating. 643 00:32:50,800 --> 00:32:56,470 It will be something mmm, something, valley valley 644 00:32:56,470 --> 00:33:00,250 valley, something, mountain mountain mountain, something, 645 00:33:00,250 --> 00:33:03,640 valley valley valley, and so on. 646 00:33:03,640 --> 00:33:08,380 OK great, half of my things, roughly, are mountains. 647 00:33:08,380 --> 00:33:11,020 I just need to fix these valleys. 648 00:33:11,020 --> 00:33:16,905 So I use the same trick, which is I'm going to fold, 649 00:33:16,905 --> 00:33:22,390 let me use red maybe-- fold in the middle of these mountain 650 00:33:22,390 --> 00:33:23,696 segments. 651 00:33:23,696 --> 00:33:27,610 If I do that, I line up all the valley segments. 652 00:33:27,610 --> 00:33:33,680 So I end up with, after folding, this is going to look like, 653 00:33:33,680 --> 00:33:35,880 you almost need a computer to work all this out, 654 00:33:35,880 --> 00:33:39,990 but we have implemented this, it works. 655 00:33:39,990 --> 00:33:42,600 It's going to look like this, I mean, with many layers 656 00:33:42,600 --> 00:33:43,100 like this. 657 00:33:43,100 --> 00:33:45,300 All the values are on top of each other. 658 00:33:45,300 --> 00:33:47,310 These question marks are still question marks. 659 00:33:47,310 --> 00:33:50,100 This mountain is that mountain, this mountain is that mountain, 660 00:33:50,100 --> 00:33:52,182 my mountain is your mountain, I don't know. 661 00:33:52,182 --> 00:33:54,390 So now we're going to fold mountain mountain mountain 662 00:33:54,390 --> 00:33:59,380 mountain, on all five of these things, 663 00:33:59,380 --> 00:34:01,160 and these two remain mountains. 664 00:34:01,160 --> 00:34:05,790 And then unfold, and what I end up with is twice as good. 665 00:34:05,790 --> 00:34:10,840 I end up with question mark, and then, I think, sorry, 666 00:34:10,840 --> 00:34:13,179 not quite-- because this part also gets messed up. 667 00:34:13,179 --> 00:34:17,239 So I end up with valley, mountain, question mark, 668 00:34:17,239 --> 00:34:24,010 then seven mountains, then question mark, 669 00:34:24,010 --> 00:34:30,650 then some not so pretty stuff, m v v v v v m, question mark, 670 00:34:30,650 --> 00:34:34,120 and then this repeats until we get to the end of the string. 671 00:34:34,120 --> 00:34:36,840 And then the end looks kind of like this. 672 00:34:36,840 --> 00:34:38,861 So the point is, I had three m's in a row, 673 00:34:38,861 --> 00:34:40,110 now I have seven m's in a row. 674 00:34:40,110 --> 00:34:41,500 Each time I do this, I roughly double 675 00:34:41,500 --> 00:34:42,690 the number of m's in a row. 676 00:34:42,690 --> 00:34:44,106 By the end, almost everything will 677 00:34:44,106 --> 00:34:48,190 be m's, after I do this log n times. 678 00:34:48,190 --> 00:34:51,800 I have pretty much all m's and then I can finish it off. 679 00:34:51,800 --> 00:34:54,199 And each of these steps took log n steps, 680 00:34:54,199 --> 00:34:56,367 because I had to fold along all these things. 681 00:34:56,367 --> 00:34:57,450 How did I fold along them? 682 00:34:57,450 --> 00:35:00,175 Not one at a time, of course, I fold in half and fold in half, 683 00:35:00,175 --> 00:35:01,570 and fold in half, roughly. 684 00:35:01,570 --> 00:35:04,620 I choose the middlemost red line to fold first, 685 00:35:04,620 --> 00:35:07,600 and about log n steps later, everything will be piled up. 686 00:35:07,600 --> 00:35:10,070 So I get to use this efficiency of dragon curve things. 687 00:35:10,070 --> 00:35:12,360 I'm messing up these creases, of course, 688 00:35:12,360 --> 00:35:14,960 but it's OK, because I get a big chunk 689 00:35:14,960 --> 00:35:17,030 that all gets correct at once. 690 00:35:17,030 --> 00:35:18,922 Half of it gets wrong, half of it's correct, 691 00:35:18,922 --> 00:35:21,380 but the half that's getting correct, gets longer and longer 692 00:35:21,380 --> 00:35:22,800 each step, doubling. 693 00:35:22,800 --> 00:35:24,762 So only log squared steps. 694 00:35:24,762 --> 00:35:26,970 Pretty sure this is optimal, but the best lower bound 695 00:35:26,970 --> 00:35:31,220 we can prove is log squared over log log. 696 00:35:31,220 --> 00:35:34,830 All right, finally, let's fold some stuff. 697 00:35:34,830 --> 00:35:37,362 I thought we would fold some hyperbolic paraboloids 698 00:35:37,362 --> 00:35:43,770 and maybe put them together into structures like this. 699 00:35:43,770 --> 00:35:45,710 So, you have your squares. 700 00:35:45,710 --> 00:35:47,960 If you don't have squares, a bunch of squares up here. 701 00:35:50,560 --> 00:35:53,025 You can start folding as many as you like. 702 00:35:53,025 --> 00:35:54,800 We're going to put them together. 703 00:35:54,800 --> 00:35:57,040 I will demonstrate up here. 704 00:36:00,080 --> 00:36:04,210 So the first thing we want to do is fold along the diagonals. 705 00:36:04,210 --> 00:36:08,510 These aren't perfect squares, but it's OK. 706 00:36:08,510 --> 00:36:14,809 So we fold along on diagonal, and the other diagonal. 707 00:36:14,809 --> 00:36:17,100 This is mainly so we know where the center is, but also 708 00:36:17,100 --> 00:36:18,558 because we need to fold a diagonal, 709 00:36:18,558 --> 00:36:20,075 so we kill two birds with one stone. 710 00:36:25,310 --> 00:36:27,930 Whereas we just saw how to kill, in this case, 711 00:36:27,930 --> 00:36:30,780 log n birds with one stone. 712 00:36:30,780 --> 00:36:33,220 Say log n, here we're only killing two. 713 00:36:33,220 --> 00:36:36,190 So now you've got your diagonals and your square, 714 00:36:36,190 --> 00:36:37,700 we want to fold-- it's kind of cool, 715 00:36:37,700 --> 00:36:40,730 I colored the edges with my chalk hands. 716 00:36:40,730 --> 00:36:42,340 Now everything's going to be parallel 717 00:36:42,340 --> 00:36:43,466 to the sides of the square. 718 00:36:43,466 --> 00:36:44,965 So the first thing we're going to do 719 00:36:44,965 --> 00:36:46,710 is construct a square that's half as 720 00:36:46,710 --> 00:36:49,670 big, centered along that point. 721 00:36:49,670 --> 00:36:53,425 So this involves folding the bottom edge, usually 722 00:36:53,425 --> 00:36:54,550 do this valley fold. 723 00:36:54,550 --> 00:36:56,890 Fold the bottom edge to line up with that center, 724 00:36:56,890 --> 00:36:58,890 and also line up the edges. 725 00:36:58,890 --> 00:37:00,300 Now, very important. 726 00:37:00,300 --> 00:37:03,000 When you make this fold, don't fold it all the way, 727 00:37:03,000 --> 00:37:06,930 just fold the middle half, between the two diagonals. 728 00:37:06,930 --> 00:37:09,470 That's really key, otherwise this will not work. 729 00:37:09,470 --> 00:37:11,490 It's the only thing you have to be careful of. 730 00:37:11,490 --> 00:37:13,810 Don't fall all the way, just in the middle. 731 00:37:13,810 --> 00:37:15,930 OK, that's one quarter. 732 00:37:15,930 --> 00:37:16,940 Do that four times. 733 00:37:20,030 --> 00:37:26,070 So we go here, line up at the center, fold the middle. 734 00:37:37,080 --> 00:37:39,140 Once you've got the inner square, 735 00:37:39,140 --> 00:37:41,267 you're going to make two more. 736 00:37:41,267 --> 00:37:43,600 There's going to be this square at the one quarter mark, 737 00:37:43,600 --> 00:37:48,010 and an inner square at the 3/4 mark. 738 00:37:48,010 --> 00:37:51,390 To make the outer square, you fold the edge to the thing 739 00:37:51,390 --> 00:37:54,310 that you just folded. 740 00:37:54,310 --> 00:37:58,030 Keep careful note of which crease you're folding to, 741 00:37:58,030 --> 00:38:00,800 should be from the previous square, 742 00:38:00,800 --> 00:38:02,850 not from the current square. 743 00:38:02,850 --> 00:38:04,950 To fold the inner square, you go all the way up 744 00:38:04,950 --> 00:38:09,340 to the opposite side of the first square. 745 00:38:09,340 --> 00:38:11,700 And you can just look visually, are these evenly spaced. 746 00:38:11,700 --> 00:38:16,365 If yes, you did it right, if not, you're in trouble. 747 00:38:16,365 --> 00:38:18,530 You're probably OK if you make a few extra creases, 748 00:38:18,530 --> 00:38:21,500 but don't try to make as few extra creases as possible. 749 00:38:24,170 --> 00:38:28,614 So then repeat that four times and you'll get your, 750 00:38:28,614 --> 00:38:30,905 I guess, four squares, if you count the outer boundary. 751 00:38:53,160 --> 00:38:56,732 It's a little hard to see, once you've got that, those squares, 752 00:38:56,732 --> 00:38:58,440 looks like I did one of them incorrectly. 753 00:39:03,570 --> 00:39:07,500 Got your nice concentric squares, four of them. 754 00:39:07,500 --> 00:39:11,040 Now you're halfway down, you flip it over, 755 00:39:11,040 --> 00:39:13,139 and do valley folds on the other side, 756 00:39:13,139 --> 00:39:15,180 because you want these to be alternating mountain 757 00:39:15,180 --> 00:39:18,160 valley, mountain valley, and just fill in all the squares 758 00:39:18,160 --> 00:39:20,430 in between those squares. 759 00:39:20,430 --> 00:39:22,950 And you can figure out what the reference markers are, 760 00:39:22,950 --> 00:39:27,050 they're all the mountain folds, actually every other mountain 761 00:39:27,050 --> 00:39:29,540 fold will be one of your references. 762 00:39:29,540 --> 00:39:31,820 Don't go to one of your new valley folds, 763 00:39:31,820 --> 00:39:35,314 always go to the mountain ones. 764 00:39:35,314 --> 00:39:37,480 And just check that you're always filling in halfway 765 00:39:37,480 --> 00:39:40,590 in between two of your mountain folds. 766 00:39:40,590 --> 00:39:43,765 Don't forget to flip over, very important. 767 00:39:50,734 --> 00:39:52,980 And skip every other guy, otherwise you 768 00:39:52,980 --> 00:39:58,471 will make a refold and just increase the wrong way. 769 00:39:58,471 --> 00:40:00,720 This is what it looks like after one quarter of those. 770 00:40:51,050 --> 00:40:53,700 And once you've done all this crazy pre-creasing, 771 00:40:53,700 --> 00:40:56,040 concentric squares, alternating mountain valley, 772 00:40:56,040 --> 00:40:57,410 then the fun part begins. 773 00:40:57,410 --> 00:40:59,050 This is actually, literally fun. 774 00:41:01,840 --> 00:41:05,170 It's, I guess, also harder, but much more exciting 775 00:41:05,170 --> 00:41:06,660 than all that pre-creasing. 776 00:41:06,660 --> 00:41:09,490 So, in this case, you have to fold all the creases at once, 777 00:41:09,490 --> 00:41:12,340 and the easy way to do that is start at the outside 778 00:41:12,340 --> 00:41:19,340 and make the outermost ring, fold it, just 779 00:41:19,340 --> 00:41:20,890 go around once or twice. 780 00:41:20,890 --> 00:41:24,370 Make sure everything's folded, including that diagonal. 781 00:41:24,370 --> 00:41:26,190 So here I have valley folded everything, 782 00:41:26,190 --> 00:41:28,040 including the outer diagonal. 783 00:41:28,040 --> 00:41:30,520 And then you want to mountain fold the next one, 784 00:41:30,520 --> 00:41:32,250 and get them to be against each other. 785 00:41:32,250 --> 00:41:34,800 You're aiming for a kind of x shape, which 786 00:41:34,800 --> 00:41:37,210 is drawn over here. 787 00:41:37,210 --> 00:41:39,950 It's going to look like that ideally. 788 00:41:39,950 --> 00:41:44,330 Just keep collapsing, square by square, making sure you 789 00:41:44,330 --> 00:41:46,390 alternating mountain valley, it should just 790 00:41:46,390 --> 00:41:50,260 fall into place pretty much. 791 00:41:50,260 --> 00:41:53,620 But paper likes to misfold a little bit, 792 00:41:53,620 --> 00:41:55,430 so you've got to fix it all. 793 00:41:55,430 --> 00:41:58,755 Here I've done two squares, little more. 794 00:42:02,554 --> 00:42:04,470 It gets easier and easier, because the squares 795 00:42:04,470 --> 00:42:06,480 get smaller and smaller. 796 00:42:06,480 --> 00:42:11,060 We've got three squares done, so I'm, like, halfway to an x. 797 00:42:11,060 --> 00:42:11,930 Fourth square. 798 00:42:15,430 --> 00:42:17,020 The center is a little bit tricky, 799 00:42:17,020 --> 00:42:20,530 try to get it so it alternates, so now it's nice and thin. 800 00:42:20,530 --> 00:42:24,760 It's like repeated sync folds of a water bomb base, 801 00:42:24,760 --> 00:42:26,400 and it's like an x. 802 00:42:26,400 --> 00:42:28,100 And then this is your opportunity 803 00:42:28,100 --> 00:42:30,810 to recrease all your creases really hard, 804 00:42:30,810 --> 00:42:34,070 just give it a good squeeze on each of the three, 805 00:42:34,070 --> 00:42:35,185 four legs of the starfish. 806 00:42:38,240 --> 00:42:42,010 And now you've got your hyperbolic paraboloid. 807 00:42:42,010 --> 00:42:46,650 To make it, you want to take two midpoints here of the squares, 808 00:42:46,650 --> 00:42:50,560 pull them apart until it's a little bit open, 809 00:42:50,560 --> 00:42:55,146 and then give it a twist, and then open it up a little, 810 00:42:55,146 --> 00:42:57,760 and you've got your hyperbolic paraboloid. 811 00:43:00,847 --> 00:43:03,260 If we make enough of these, we can assemble it 812 00:43:03,260 --> 00:43:04,760 into some cool shapes. 813 00:43:07,340 --> 00:43:10,390 How many people have folded one? 814 00:43:10,390 --> 00:43:13,450 At least one. 815 00:43:13,450 --> 00:43:16,460 A few people. 816 00:43:16,460 --> 00:43:20,054 I will show you-- you can fold more. 817 00:43:20,054 --> 00:43:21,470 We're going to assemble something, 818 00:43:21,470 --> 00:43:23,840 might need a bunch of people. 819 00:43:23,840 --> 00:43:27,000 This is the algorithm we use for converting a polyhedron 820 00:43:27,000 --> 00:43:31,220 into a bunch of hyperbolic paraboloids. 821 00:43:31,220 --> 00:43:33,900 It's in this paper for '99, and we 822 00:43:33,900 --> 00:43:37,450 take each face of the polygons, of the polyhedron, sorry. 823 00:43:37,450 --> 00:43:39,760 So if you have a cube, you've got a bunch of squares. 824 00:43:39,760 --> 00:43:41,420 For each square, we will make what 825 00:43:41,420 --> 00:43:45,240 we call a four hat, which is four hyperbolic paraboloids, 826 00:43:45,240 --> 00:43:48,170 joined in a cycle, like this. 827 00:43:48,170 --> 00:43:50,790 And you've got to be careful the join it the right way, 828 00:43:50,790 --> 00:43:54,010 but then, these, the tips of the high parts that are not 829 00:43:54,010 --> 00:43:58,740 joined-- I mean, one tip of each of the high parts 830 00:43:58,740 --> 00:44:00,237 comes together at the center. 831 00:44:00,237 --> 00:44:01,820 Than these tips are going to represent 832 00:44:01,820 --> 00:44:04,570 the edge of the polygon, these red dots. 833 00:44:04,570 --> 00:44:11,352 And so these two sides are the sides of-- these two sides 834 00:44:11,352 --> 00:44:12,560 are the sides of [INAUDIBLE]. 835 00:44:12,560 --> 00:44:17,140 They're going to join to an adjacent [INAUDIBLE] over here. 836 00:44:17,140 --> 00:44:18,370 So that's the idea. 837 00:44:22,220 --> 00:44:25,290 So I've got-- I've already folded a bunch of these 838 00:44:25,290 --> 00:44:27,277 already. 839 00:44:27,277 --> 00:44:29,110 I'll show you, maybe, what a hat looks like. 840 00:44:35,410 --> 00:44:41,180 So take two of them, join them along those edges. 841 00:44:41,180 --> 00:44:43,590 We're going to use tape or staples to join them, 842 00:44:43,590 --> 00:44:46,570 I don't have a fancy lock, I'm afraid. 843 00:44:46,570 --> 00:44:51,290 And join these edges together, and you 844 00:44:51,290 --> 00:44:57,330 get-- this is a three-hat and you can put on your head, 845 00:44:57,330 --> 00:45:00,717 whatever, and that would represent a triangle. 846 00:45:00,717 --> 00:45:02,800 And then we're going to join along these two sides 847 00:45:02,800 --> 00:45:05,150 to an adjacent triangle or whatever. 848 00:45:07,760 --> 00:45:12,490 So we could make a platonic solid. 849 00:45:12,490 --> 00:45:17,120 I guess the simplest one would be a tetrahedron has six edges, 850 00:45:17,120 --> 00:45:20,560 so it needs 12 [INAUDIBLE]. 851 00:45:20,560 --> 00:45:22,550 I was thinking we could make this solid, 852 00:45:22,550 --> 00:45:24,020 it's never been made before. 853 00:45:24,020 --> 00:45:24,440 It should be cool. 854 00:45:24,440 --> 00:45:25,940 It's the simplest Archimedean solid, 855 00:45:25,940 --> 00:45:29,130 in terms of number of edges, the truncated tetrahedron. 856 00:45:29,130 --> 00:45:34,580 This requires 36 parts, so if we're ambitious, 857 00:45:34,580 --> 00:45:38,000 we can go for it, but I think we're low on time. 858 00:45:38,000 --> 00:45:41,637 So a tetrahedron might be an easier bet. 859 00:45:41,637 --> 00:45:43,595 If people want to come up and start assembling. 860 00:45:46,440 --> 00:45:47,980 Finish taping that. 861 00:45:47,980 --> 00:45:52,860 Hold these, and then this goes here. 862 00:45:52,860 --> 00:45:55,320 Do you want to hold this one? 863 00:45:55,320 --> 00:45:58,030 Virtual assembly here, so maybe we'll 864 00:45:58,030 --> 00:46:00,520 finish assembling next class.