1 00:00:17,880 --> 00:00:20,820 YUFEI ZHAO: Today we want to look at the sum product 2 00:00:20,820 --> 00:00:21,522 problem. 3 00:00:21,522 --> 00:00:22,980 So for the past few lectures, we've 4 00:00:22,980 --> 00:00:25,260 been discussing the structure of sets 5 00:00:25,260 --> 00:00:28,380 under the addition operation. 6 00:00:28,380 --> 00:00:31,050 Today we're going to throw in one extra operation, so 7 00:00:31,050 --> 00:00:33,310 multiplication, and understand how 8 00:00:33,310 --> 00:00:37,920 sets behave under both addition and multiplication. 9 00:00:37,920 --> 00:00:43,380 And the basic problem here is, can it be the case that A 10 00:00:43,380 --> 00:00:49,980 plus A, A times A, which is, analogously, 11 00:00:49,980 --> 00:00:55,860 the set of all pairwise products of elements from A-- 12 00:00:55,860 --> 00:01:05,760 can these two sets be simultaneously small, 13 00:01:05,760 --> 00:01:08,390 that is, the same for some single A? 14 00:01:11,710 --> 00:01:17,710 Can we have it so that A plus A and A times A 15 00:01:17,710 --> 00:01:20,690 are simultaneously small? 16 00:01:20,690 --> 00:01:24,470 For example, it's easy to make one of them small. 17 00:01:24,470 --> 00:01:29,210 We've seen examples where if you take A to be an arithmetic 18 00:01:29,210 --> 00:01:32,570 progression, then A plus A is more or less as 19 00:01:32,570 --> 00:01:34,400 small as it gets. 20 00:01:34,400 --> 00:01:39,710 But for such an example, you see A times A is pretty large. 21 00:01:39,710 --> 00:01:42,620 It's actually not so clear how to prove how large it 22 00:01:42,620 --> 00:01:43,820 Is. 23 00:01:43,820 --> 00:01:45,320 And there are some very nice proofs. 24 00:01:45,320 --> 00:01:48,620 And this problem has actually been more or less pinned down. 25 00:01:48,620 --> 00:01:54,350 But the short version is that A times A has size close 26 00:01:54,350 --> 00:01:55,790 to its maximum possible. 27 00:01:59,600 --> 00:02:05,690 So it turns out the size of A times A is almost quadratic. 28 00:02:05,690 --> 00:02:08,009 So this number is actually now known fairly precisely. 29 00:02:08,009 --> 00:02:12,410 So this problem of determining the size of A times 30 00:02:12,410 --> 00:02:17,690 A for the interval 1 through N is known as the Erdos 31 00:02:17,690 --> 00:02:19,160 multiplication table problem. 32 00:02:30,190 --> 00:02:32,910 So if you take an N by N multiplication table, 33 00:02:32,910 --> 00:02:35,920 how many numbers do you see in the table? 34 00:02:35,920 --> 00:02:39,570 So that turns out to be sub-quadratic, but not too 35 00:02:39,570 --> 00:02:42,170 sub-quadratic. 36 00:02:42,170 --> 00:02:46,130 So this problem has been more or less solved by Kevin Ford. 37 00:02:46,130 --> 00:02:48,840 And we now know a fairly precise expression, 38 00:02:48,840 --> 00:02:50,840 but I don't want to focus on that. 39 00:02:50,840 --> 00:02:54,200 That's not the topic of today's lecture. 40 00:02:54,200 --> 00:02:55,700 This is just an example. 41 00:02:55,700 --> 00:02:57,950 Alternatively, you can take A times A 42 00:02:57,950 --> 00:03:02,400 to be quite small by taking A to be a geometric progression. 43 00:03:02,400 --> 00:03:04,820 Then it's not too hard to convince yourself 44 00:03:04,820 --> 00:03:08,390 that A plus A must be fairly large in that case. 45 00:03:08,390 --> 00:03:10,310 And the geometric progression doesn't 46 00:03:10,310 --> 00:03:14,630 have so much additive structure, so A plus A will be large. 47 00:03:14,630 --> 00:03:20,380 So can you make A plus A and A times A simultaneously small? 48 00:03:20,380 --> 00:03:24,470 So there's this conjecture that the answer is no. 49 00:03:24,470 --> 00:03:27,480 And this is a famous conjecture in this area, known 50 00:03:27,480 --> 00:03:36,400 as the Erdos similarity conjecture on the sum product 51 00:03:36,400 --> 00:03:40,810 problem, which states that for all finite sets 52 00:03:40,810 --> 00:03:48,730 of real numbers, either A plus A or A times A 53 00:03:48,730 --> 00:03:53,035 has to be close to quadratic size. 54 00:04:00,300 --> 00:04:01,800 So that's the conjecture. 55 00:04:01,800 --> 00:04:03,810 It's still very much open. 56 00:04:03,810 --> 00:04:05,760 Today I want to show you some progress 57 00:04:05,760 --> 00:04:08,810 towards this conjecture via some partial results. 58 00:04:08,810 --> 00:04:12,780 And it will use a nice combination of tools 59 00:04:12,780 --> 00:04:16,519 from graph theory and incidence geometry, 60 00:04:16,519 --> 00:04:18,810 so it nicely ties in together many of the things 61 00:04:18,810 --> 00:04:22,019 that we've seen in this course so far. 62 00:04:22,019 --> 00:04:25,870 So Erdos and Szemeredi proved some bound, 63 00:04:25,870 --> 00:04:30,610 which is like 1 plus c for some constant c. 64 00:04:30,610 --> 00:04:35,387 Today we'll show some bounds for somewhat better c's. 65 00:04:35,387 --> 00:04:35,970 So you'll see. 66 00:04:39,350 --> 00:04:41,480 The first tool that I want to introduce 67 00:04:41,480 --> 00:04:45,150 is a result from graph theory known as the "crossing number 68 00:04:45,150 --> 00:04:46,312 inequality." 69 00:04:56,660 --> 00:04:58,340 So you know that planar graphs are 70 00:04:58,340 --> 00:05:00,100 graphs where you can draw on the planes 71 00:05:00,100 --> 00:05:02,230 so that the edges do not cross. 72 00:05:02,230 --> 00:05:05,000 And there are some famous examples of non-planar graphs, 73 00:05:05,000 --> 00:05:09,070 like K5 and K 3, 3. 74 00:05:09,070 --> 00:05:11,560 But you can ask a more quantitative question. 75 00:05:11,560 --> 00:05:14,320 If I give you a graph, how many crossings 76 00:05:14,320 --> 00:05:18,280 must you have in every drawing of this graph? 77 00:05:18,280 --> 00:05:19,840 And the crossing number inequality 78 00:05:19,840 --> 00:05:22,990 provides some estimate for such a quantity. 79 00:05:22,990 --> 00:05:28,600 So given the graph G, denoted by cr, so crossing of G, 80 00:05:28,600 --> 00:05:40,130 to be the minimum number of crossings in a planar 81 00:05:40,130 --> 00:05:46,570 drawing of G. There is a bit of subtlety 82 00:05:46,570 --> 00:05:50,470 here, where by a planar drawing, do I mean using line segments 83 00:05:50,470 --> 00:05:52,090 or do I mean using curves? 84 00:05:52,090 --> 00:05:55,990 It's actually not clear how it affects this quantity here. 85 00:05:55,990 --> 00:05:57,520 That's a very subtle issue. 86 00:05:57,520 --> 00:06:00,490 So for planar graphs, there's a famous result 87 00:06:00,490 --> 00:06:02,800 that more or less says if a planar graph can 88 00:06:02,800 --> 00:06:04,270 be drawn using continuous curves, 89 00:06:04,270 --> 00:06:06,830 then it can be drawn using straight lines. 90 00:06:06,830 --> 00:06:09,352 But the minimum number of crossings, 91 00:06:09,352 --> 00:06:10,810 the two different ways of drawings, 92 00:06:10,810 --> 00:06:13,400 they might end up with different crossing numbers. 93 00:06:13,400 --> 00:06:15,230 But for the purpose of today's lecture, 94 00:06:15,230 --> 00:06:18,010 we'll use a more general notion, although it doesn't actually 95 00:06:18,010 --> 00:06:19,840 matter for today which one we'll use-- 96 00:06:19,840 --> 00:06:21,745 so planar drawing using curves. 97 00:06:25,870 --> 00:06:28,810 Draw the graph where edges are continuous curves. 98 00:06:28,810 --> 00:06:30,460 How many crossings do you get? 99 00:06:30,460 --> 00:06:35,450 The crossing is a pair of edges that cross. 100 00:06:35,450 --> 00:06:38,260 You can ask-- it's just a cross over point that can-- 101 00:06:38,260 --> 00:06:39,035 it doesn't matter. 102 00:06:39,035 --> 00:06:40,660 So there are many different subtle ways 103 00:06:40,660 --> 00:06:42,040 of defining these things. 104 00:06:42,040 --> 00:06:44,830 They won't really come up for today's lecture. 105 00:06:44,830 --> 00:06:49,510 The crossing number inequality is a result 106 00:06:49,510 --> 00:06:54,010 from the '80s, which give you a lower-bound estimate 107 00:06:54,010 --> 00:06:56,530 on the number of crossings. 108 00:06:56,530 --> 00:07:06,470 If G is a graph with enough edges-- 109 00:07:06,470 --> 00:07:08,220 the number of edges is, let's say, 110 00:07:08,220 --> 00:07:10,790 at least four times the number of vertices-- 111 00:07:13,530 --> 00:07:21,220 then the number of crossings of every drawing of G 112 00:07:21,220 --> 00:07:24,970 is at least the number of edges cubed 113 00:07:24,970 --> 00:07:28,810 divided by the number of vertices squared. 114 00:07:28,810 --> 00:07:32,530 And there's an extra constant factor, which is some constant. 115 00:07:39,380 --> 00:07:43,570 So the constant does not depend on the graph. 116 00:07:43,570 --> 00:07:45,760 In particular, if it has a lot of edges, 117 00:07:45,760 --> 00:07:50,815 then every drawing of G must have a lot of crossings. 118 00:07:50,815 --> 00:07:52,300 So the crossing number inequality 119 00:07:52,300 --> 00:07:56,080 was proved by two separate independent works, 120 00:07:56,080 --> 00:07:59,680 one by Ajtai, Chvatal, Newborn, Szemeredi and the 121 00:07:59,680 --> 00:08:02,470 other by Tom Leighton, our very own Tom Leighton. 122 00:08:06,950 --> 00:08:11,090 So let me first give you some consequences of this theorem, 123 00:08:11,090 --> 00:08:12,690 just for illustration. 124 00:08:12,690 --> 00:08:17,930 So if you have an n-vertex graph with a quadratic number 125 00:08:17,930 --> 00:08:26,460 of edges, then how many crossings must you have? 126 00:08:26,460 --> 00:08:29,070 You plug in these parameters into the theorem. 127 00:08:29,070 --> 00:08:36,270 See that it has necessarily n to the 4th crossings. 128 00:08:39,100 --> 00:08:42,020 But if you just draw the graph in some arbitrary way, 129 00:08:42,020 --> 00:08:44,570 you have at most n to the 4 crossings, 130 00:08:44,570 --> 00:08:49,215 because a crossing involves four points. 131 00:08:49,215 --> 00:08:51,090 So when you have a quadratic number of edges, 132 00:08:51,090 --> 00:08:54,780 you must get basically the maximum number of crossings. 133 00:08:54,780 --> 00:08:58,163 The leading constant term factor is an interesting problem, 134 00:08:58,163 --> 00:08:59,580 which we're not going to get into. 135 00:09:02,540 --> 00:09:06,810 Let's prove the crossing number inequality. 136 00:09:06,810 --> 00:09:11,145 First, the base case of the crossing number inequalities 137 00:09:11,145 --> 00:09:14,065 is when you can draw a graph with no crossings. 138 00:09:14,065 --> 00:09:16,420 And those are planar graphs. 139 00:09:16,420 --> 00:09:29,130 So for every connected planar graph, 140 00:09:29,130 --> 00:09:31,770 if it has at least one cycle-- and you'll see why in a second, 141 00:09:31,770 --> 00:09:33,000 why I say this-- 142 00:09:33,000 --> 00:09:43,380 if with at least one cycle, so that's not a tree, 143 00:09:43,380 --> 00:09:48,150 we must have that 3 times the number of faces 144 00:09:48,150 --> 00:09:53,380 is at most 2 times the number of edges. 145 00:09:53,380 --> 00:09:55,770 So here, we're going to use the key tool 146 00:09:55,770 --> 00:10:01,380 being Euler's formula, which we all 147 00:10:01,380 --> 00:10:05,310 know as the number of vertices minus the number of edges 148 00:10:05,310 --> 00:10:09,080 plus the number of faces equals to 2. 149 00:10:09,080 --> 00:10:12,470 We're here for face, because I draw a planar graph, 150 00:10:12,470 --> 00:10:18,040 and so I count the faces. 151 00:10:18,040 --> 00:10:21,520 Here there are two faces, outer face, inner face, count edges 152 00:10:21,520 --> 00:10:25,380 and vertices, so you have Euler's formula up there. 153 00:10:25,380 --> 00:10:28,390 And plug in Euler's formula for a planar graph 154 00:10:28,390 --> 00:10:32,740 with at least one cycle, so we can obtain this consequence 155 00:10:32,740 --> 00:10:42,030 over here, because every face is adjacent to at least 156 00:10:42,030 --> 00:10:45,330 three edges. 157 00:10:45,330 --> 00:10:47,510 If you go around the face, you see 158 00:10:47,510 --> 00:10:56,350 these three edges, and every edge is counted exactly twice, 159 00:10:56,350 --> 00:11:02,320 is adjacent to exactly two faces. 160 00:11:02,320 --> 00:11:04,785 So you do the double counting, you get that inequality up 161 00:11:04,785 --> 00:11:05,285 there. 162 00:11:07,920 --> 00:11:15,910 So plugging these two into Euler gets you that inequality up 163 00:11:15,910 --> 00:11:16,410 there. 164 00:11:23,730 --> 00:11:31,290 Plugging these two into Euler, we get that the number of edges 165 00:11:31,290 --> 00:11:34,870 is almost 3 times the number of vertices minus 6. 166 00:11:41,662 --> 00:11:43,120 So for this leaves that inequality, 167 00:11:43,120 --> 00:11:48,760 but plug it into Euler, plug in this into Euler, you get this. 168 00:11:48,760 --> 00:11:52,300 So we have that the number of edges 169 00:11:52,300 --> 00:11:57,670 is at most 3 times the number of vertices for every graph G. 170 00:11:57,670 --> 00:12:02,230 So here, we require that the graph is planar 171 00:12:02,230 --> 00:12:07,205 and has at least one cycle, but even if we drop the condition 172 00:12:07,205 --> 00:12:08,830 that it has at least one cycle but just 173 00:12:08,830 --> 00:12:15,760 require that it's planar, every planar graph G 174 00:12:15,760 --> 00:12:18,645 satisfies this inequality over here. 175 00:12:18,645 --> 00:12:20,020 So in other words, you might have 176 00:12:20,020 --> 00:12:24,100 heard before, in a planar graph, the average degree of a vertex 177 00:12:24,100 --> 00:12:25,150 is almost 6. 178 00:12:28,460 --> 00:12:34,790 So in particular, the crossing number of a graph G 179 00:12:34,790 --> 00:12:40,430 is positive if the number of edges 180 00:12:40,430 --> 00:12:42,380 exceeds 3 times the number of vertices. 181 00:12:44,915 --> 00:12:46,810 It's not planar, so it has at least one 182 00:12:46,810 --> 00:12:49,300 crossing every drawing. 183 00:12:49,300 --> 00:13:04,730 And by deleting an edge from each crossing, 184 00:13:04,730 --> 00:13:05,900 we get a planar graph. 185 00:13:11,360 --> 00:13:13,040 You draw the graph. 186 00:13:13,040 --> 00:13:14,040 You have some crossings. 187 00:13:14,040 --> 00:13:17,230 You get rid of an edge associated with each drawing. 188 00:13:17,230 --> 00:13:19,450 Then you get a planar graph. 189 00:13:19,450 --> 00:13:21,610 If you look at this inequality and you 190 00:13:21,610 --> 00:13:24,460 account for the number of edges that you deleted, 191 00:13:24,460 --> 00:13:28,740 we obtain then the inequality that the number 192 00:13:28,740 --> 00:13:31,470 of edges minus the number of crossings 193 00:13:31,470 --> 00:13:34,710 is at least 3 times the number of vertices. 194 00:13:43,130 --> 00:13:50,820 So we obtain the inequality that the lower 195 00:13:50,820 --> 00:13:56,640 bounds in number of crossings as the number of edges 196 00:13:56,640 --> 00:14:01,110 minus 3 times the number of vertices, this one. 197 00:14:06,820 --> 00:14:10,490 So that's some lower bound on the crossing number. 198 00:14:10,490 --> 00:14:12,650 It's not quite the bound that we have over there. 199 00:14:12,650 --> 00:14:14,990 And in fact, if you take a graph with a quadratic number 200 00:14:14,990 --> 00:14:17,840 of edges, this bound here only gives you 201 00:14:17,840 --> 00:14:20,810 quadratic lower bound on the crossing number, some 202 00:14:20,810 --> 00:14:21,480 lower bound. 203 00:14:21,480 --> 00:14:22,860 But it's not a great lower bound. 204 00:14:22,860 --> 00:14:24,750 And we would like to do better. 205 00:14:24,750 --> 00:14:28,710 So here's a trick that is a very nice trick, 206 00:14:28,710 --> 00:14:33,410 where we're going to use this inequality to upgrade it 207 00:14:33,410 --> 00:14:36,170 to a much better inequality, bootstrap it 208 00:14:36,170 --> 00:14:38,430 to a much tighter inequality. 209 00:14:38,430 --> 00:14:40,910 So this involves the use of the probabilistic method. 210 00:14:43,870 --> 00:14:48,410 Let me denote by p some number between 0 and 1, 211 00:14:48,410 --> 00:14:49,760 to be decided later. 212 00:14:54,080 --> 00:14:57,320 And starting with a graph G, let's 213 00:14:57,320 --> 00:15:05,820 let G prime, with vertices and edges being V prime and E 214 00:15:05,820 --> 00:15:16,300 prime, be obtained from G by randomly deleting some 215 00:15:16,300 --> 00:15:19,690 of the vertices, or rather randomly 216 00:15:19,690 --> 00:15:29,600 keeping each vertex with probability p, 217 00:15:29,600 --> 00:15:35,931 independently for each of these vertices. 218 00:15:35,931 --> 00:15:37,940 So you have some graph G. 219 00:15:37,940 --> 00:15:41,330 I keep each vertex with probability p. 220 00:15:41,330 --> 00:15:43,640 And I delete the remaining vertices. 221 00:15:43,640 --> 00:15:45,800 And I get a smaller graph. 222 00:15:45,800 --> 00:15:49,230 I get some induced subgraph. 223 00:15:49,230 --> 00:15:51,440 And I would like to know what can we 224 00:15:51,440 --> 00:15:55,910 say about the crossing number of the smaller graph in comparison 225 00:15:55,910 --> 00:16:00,240 to the crossing number of the original graph? 226 00:16:00,240 --> 00:16:03,860 For the smaller graph, because it's still a planar graph 227 00:16:03,860 --> 00:16:05,770 so G prime-- 228 00:16:05,770 --> 00:16:07,180 so it's still a graph. 229 00:16:07,180 --> 00:16:09,180 It's not a planar graph, but it's still a graph, 230 00:16:09,180 --> 00:16:14,380 so G prime still satisfies this inequality up here. 231 00:16:17,600 --> 00:16:21,950 So G prime still satisfies that the number of crossings 232 00:16:21,950 --> 00:16:24,270 in every drawing of G prime is at least 233 00:16:24,270 --> 00:16:28,500 the number of edges of G prime minus 3 times the number 234 00:16:28,500 --> 00:16:30,520 of vertices of G prime. 235 00:16:33,560 --> 00:16:38,390 But note that G prime is a random graph. 236 00:16:38,390 --> 00:16:40,400 G was fixed, given. 237 00:16:40,400 --> 00:16:44,100 G prime is a random graph. 238 00:16:44,100 --> 00:16:46,520 So let's evaluate the expectation 239 00:16:46,520 --> 00:16:50,270 of both quantities, left-hand side and right-hand side. 240 00:16:53,730 --> 00:16:56,280 If this inequality is true for every G prime, 241 00:16:56,280 --> 00:16:59,370 the same inequality must be true in expectation. 242 00:17:09,369 --> 00:17:13,510 Now what do we know about all the expectations of each 243 00:17:13,510 --> 00:17:17,230 of these quantities? 244 00:17:17,230 --> 00:17:19,940 The number of vertices in expectation-- 245 00:17:19,940 --> 00:17:21,260 that's pretty easy. 246 00:17:21,260 --> 00:17:28,780 So this one here is p times the original number of vertices. 247 00:17:28,780 --> 00:17:30,970 The number of edges is also pretty easy. 248 00:17:30,970 --> 00:17:34,150 Each edge is kept if both endpoints are kept. 249 00:17:34,150 --> 00:17:38,620 So this expectation on the number of edges remaining 250 00:17:38,620 --> 00:17:43,640 is also pretty easy to determine. 251 00:17:43,640 --> 00:17:49,070 The crossing number of the new graph-- 252 00:17:49,070 --> 00:17:51,830 that I have to be a little bit more careful of, 253 00:17:51,830 --> 00:17:54,380 because when you look at the smaller graph, 254 00:17:54,380 --> 00:17:56,780 maybe there's a different way to draw it 255 00:17:56,780 --> 00:18:00,320 that's not just deleting the sum of the vertices 256 00:18:00,320 --> 00:18:02,070 from the original graph. 257 00:18:02,070 --> 00:18:03,560 So even though the original graph 258 00:18:03,560 --> 00:18:06,170 might have a lot of crossings, when you go to a subgraph, 259 00:18:06,170 --> 00:18:09,070 maybe there's a better way to draw it. 260 00:18:09,070 --> 00:18:11,320 But we just need an inequality in the right direction. 261 00:18:11,320 --> 00:18:13,110 So we are still OK. 262 00:18:13,110 --> 00:18:16,170 And I claim that the crossing number of G prime 263 00:18:16,170 --> 00:18:19,740 is in expectation at most p to be 4th 264 00:18:19,740 --> 00:18:23,190 times the crossing number of G. Because if you 265 00:18:23,190 --> 00:18:32,360 keep the same drawing, then the expected number of crossings 266 00:18:32,360 --> 00:18:34,470 that are kept-- 267 00:18:34,470 --> 00:18:38,850 each crossing is kept if all four of its end points 268 00:18:38,850 --> 00:18:40,770 are kept. 269 00:18:40,770 --> 00:18:44,830 So each crossing is kept with probability p to the 4th. 270 00:18:44,830 --> 00:18:47,320 So you can draw it in expectation 271 00:18:47,320 --> 00:18:48,845 with this many crossings. 272 00:18:48,845 --> 00:18:49,720 Maybe it's much less. 273 00:18:49,720 --> 00:18:50,890 Maybe there's a better way to draw it, 274 00:18:50,890 --> 00:18:53,575 but you have an inequality going in the right direction. 275 00:19:01,500 --> 00:19:05,130 Looking at that inequality up there in yellow, 276 00:19:05,130 --> 00:19:08,710 we find that the crossing number of G 277 00:19:08,710 --> 00:19:17,960 is at least p to the minus 2 E minus 3p to the minus 3. 278 00:19:21,090 --> 00:19:25,690 And this is true for every value of p between 0 and 1. 279 00:19:25,690 --> 00:19:30,540 So now you pick a value of p that works most in your favor. 280 00:19:30,540 --> 00:19:33,690 And it turns out you should do this 281 00:19:33,690 --> 00:19:37,710 by setting these two equalities to be 282 00:19:37,710 --> 00:19:39,100 roughly equal to each other. 283 00:19:43,050 --> 00:19:57,090 So setting p between 0 and 1 so that 4 times the-- 284 00:19:57,090 --> 00:19:58,590 basically, set these two terms to be 285 00:19:58,590 --> 00:20:00,020 roughly equal to each other. 286 00:20:03,890 --> 00:20:09,320 And then we get that this quantity here 287 00:20:09,320 --> 00:20:13,910 is at least the claimed quantity, 288 00:20:13,910 --> 00:20:19,270 which is E cubed over V squared up 289 00:20:19,270 --> 00:20:24,140 to some constant factor, which I don't really care about. 290 00:20:24,140 --> 00:20:27,020 In order to set p, I have to be a little bit careful 291 00:20:27,020 --> 00:20:28,560 that p is between 0 and 1. 292 00:20:28,560 --> 00:20:30,500 If you set p to be 1.2, this whole argument 293 00:20:30,500 --> 00:20:33,060 doesn't make any sense. 294 00:20:33,060 --> 00:20:33,980 So this is OK. 295 00:20:36,820 --> 00:20:46,260 So we know p is at most one as long as E is at most 4p. 296 00:20:46,260 --> 00:20:49,470 I mean, the 4 here is not optimal, but if 4 were 2, 297 00:20:49,470 --> 00:20:50,550 then it's not true. 298 00:20:50,550 --> 00:20:54,482 So if E is 2V, you can have a planar graph, 299 00:20:54,482 --> 00:20:56,940 so you shouldn't have a lower bound on the crossing number. 300 00:20:59,550 --> 00:21:02,620 So this is the proof of the crossing number inequality. 301 00:21:02,620 --> 00:21:04,770 As I said, if you have lots of edges, 302 00:21:04,770 --> 00:21:08,960 then you must have lots of crossings. 303 00:21:08,960 --> 00:21:10,002 Any questions? 304 00:21:13,250 --> 00:21:15,200 So let's use the crossing number inequality 305 00:21:15,200 --> 00:21:19,220 to prove a fundamental result in incidence geometry. 306 00:21:26,820 --> 00:21:30,120 Incidence geometry is this area of discrete math 307 00:21:30,120 --> 00:21:33,510 that concerns fairly basic-sounding questions 308 00:21:33,510 --> 00:21:37,530 about incidences between, let's say, points and lines. 309 00:21:37,530 --> 00:21:39,730 And here's an example. 310 00:21:39,730 --> 00:21:52,530 So what's the maximum number of incidences between endpoints 311 00:21:52,530 --> 00:21:57,660 and end lines, where by "incidence" 312 00:21:57,660 --> 00:22:03,150 I mean if p-- so curly p-- is a set of points, 313 00:22:03,150 --> 00:22:09,750 and curly l is a set of lines, then I write I of p and l 314 00:22:09,750 --> 00:22:20,770 to be the number of pairs, one point, one line, such 315 00:22:20,770 --> 00:22:25,380 that the point lies on the line. 316 00:22:25,380 --> 00:22:29,440 So I'm counting incidences between points and lines. 317 00:22:29,440 --> 00:22:30,910 You can view this in many ways. 318 00:22:30,910 --> 00:22:34,000 You can view it as a bipartite graph between points and lines, 319 00:22:34,000 --> 00:22:40,000 and we're counting the number of edges in this bipartite graph. 320 00:22:40,000 --> 00:22:41,730 So I give you end points, end lines. 321 00:22:41,730 --> 00:22:45,350 What's the maximum number of incidences? 322 00:22:45,350 --> 00:22:47,670 It's not such an obvious question. 323 00:22:47,670 --> 00:22:52,560 So let's see how we can approach this question. 324 00:22:52,560 --> 00:22:58,280 But first, let me give you some easy bounds. 325 00:22:58,280 --> 00:23:02,480 So here's a trivial bound-- 326 00:23:07,680 --> 00:23:12,390 so here, I want to know if I give you some number of points, 327 00:23:12,390 --> 00:23:16,250 some number of lines, what's the maximum number of incidences. 328 00:23:16,250 --> 00:23:21,580 So a trivial bound is that the number of incidences 329 00:23:21,580 --> 00:23:26,970 is at most the product between the number of points 330 00:23:26,970 --> 00:23:30,120 and the number of lines. 331 00:23:30,120 --> 00:23:32,220 One point, one line, at most one incidence. 332 00:23:32,220 --> 00:23:34,750 So that's pretty trivial. 333 00:23:34,750 --> 00:23:36,260 We can do better. 334 00:23:36,260 --> 00:23:42,840 So we can do better because, well, you 335 00:23:42,840 --> 00:23:50,050 see, let's use this following fact, that every line-- 336 00:23:52,880 --> 00:24:02,735 so every pair of points determine at most one line. 337 00:24:02,735 --> 00:24:03,810 I have two points. 338 00:24:03,810 --> 00:24:08,780 There's at most one line that contains those two points. 339 00:24:08,780 --> 00:24:15,990 Using this fact, we see that the number of-- 340 00:24:15,990 --> 00:24:23,500 so let's count the number of triples involving 341 00:24:23,500 --> 00:24:34,590 two points and one line such that both points lie 342 00:24:34,590 --> 00:24:35,220 on the line. 343 00:24:39,730 --> 00:24:41,530 So how big can this set be? 344 00:24:41,530 --> 00:24:44,950 So let's try to count it in two different ways. 345 00:24:44,950 --> 00:24:48,900 On one hand, this quantity is at most 346 00:24:48,900 --> 00:24:52,270 the number of points squared, because if I give you 347 00:24:52,270 --> 00:24:56,700 two points, then they determine this line-- 348 00:24:56,700 --> 00:25:01,580 so at most the number of points squared. 349 00:25:01,580 --> 00:25:09,390 But on the other hand, we see that if I give you a line, 350 00:25:09,390 --> 00:25:12,090 I just need to count now the number of-- 351 00:25:15,268 --> 00:25:17,560 let me also require that these two points are distinct. 352 00:25:17,560 --> 00:25:20,860 So if I give you a line, I now need 353 00:25:20,860 --> 00:25:26,990 to count the number of pairs of points on this line. 354 00:25:26,990 --> 00:25:36,750 So I can enumerate over lines and count line 355 00:25:36,750 --> 00:25:42,270 by line how many pairs of points are on that line. 356 00:25:42,270 --> 00:25:45,790 So I get this quantity over here. 357 00:25:45,790 --> 00:25:48,930 On each line, I have that contribution. 358 00:25:48,930 --> 00:25:55,100 And now, using Cauchy-Schwartz inequality, 359 00:25:55,100 --> 00:26:01,130 we find that this squared term is at least 360 00:26:01,130 --> 00:26:11,200 the number of incidences divided by the number of lines. 361 00:26:11,200 --> 00:26:13,660 And the remaining minus 1 term contributes just 362 00:26:13,660 --> 00:26:16,134 to the number of incidences. 363 00:26:20,090 --> 00:26:22,370 So the first is by Cauchy-Schwartz. 364 00:26:26,690 --> 00:26:30,450 So putting these two inequalities together, 365 00:26:30,450 --> 00:26:34,920 we get some upper bound on the number of incidences. 366 00:26:34,920 --> 00:26:37,990 If you have to invert this inequality, 367 00:26:37,990 --> 00:26:42,810 you will get that the number of incidences between points 368 00:26:42,810 --> 00:26:50,360 and lines is upper bounded by the number of points 369 00:26:50,360 --> 00:26:53,930 times the number of lines raised to power 370 00:26:53,930 --> 00:27:00,450 1/2 plus the number of lines. 371 00:27:00,450 --> 00:27:03,290 So that's what you get from this inequality over here. 372 00:27:06,150 --> 00:27:08,720 By considering point-line duality-- 373 00:27:08,720 --> 00:27:12,450 so whenever you have this kind of setup involving points 374 00:27:12,450 --> 00:27:15,870 and lines, you can take the projected duality 375 00:27:15,870 --> 00:27:18,570 and transform the configuration into-- 376 00:27:18,570 --> 00:27:21,150 lines into points and points into lines, and the incidences 377 00:27:21,150 --> 00:27:22,560 are preserved. 378 00:27:22,560 --> 00:27:25,630 So I also have an inequality. 379 00:27:25,630 --> 00:27:29,550 By duality-- I also have an inequality 380 00:27:29,550 --> 00:27:32,760 where I switch the roles of points and lines. 381 00:27:38,610 --> 00:27:40,890 So I is already the numbers. 382 00:27:40,890 --> 00:27:44,470 I don't need to put an extra absolute value sign. 383 00:27:44,470 --> 00:27:46,590 So the number of points and lines 384 00:27:46,590 --> 00:27:50,130 is upper bounded by the number of lines 385 00:27:50,130 --> 00:27:52,740 times the square root of a number of points 386 00:27:52,740 --> 00:27:58,200 plus an extra term, just in case there are very few lines. 387 00:28:02,035 --> 00:28:03,910 So these are the bounds that you have so far. 388 00:28:03,910 --> 00:28:06,160 And the only thing that we have used so far 389 00:28:06,160 --> 00:28:09,340 is the fact that every two points determine at most one 390 00:28:09,340 --> 00:28:12,760 line, and every two lines meet at at most one point. 391 00:28:15,290 --> 00:28:17,200 So these are the bounds that we get. 392 00:28:17,200 --> 00:28:23,720 And in particular, for end points and end lines, 393 00:28:23,720 --> 00:28:26,530 we get the number of incidences is-- 394 00:28:26,530 --> 00:28:29,420 they go off n to the 3/2. 395 00:28:34,440 --> 00:28:36,930 This should remind you of something we've done before. 396 00:28:40,300 --> 00:28:43,560 So in the first part of this course, 397 00:28:43,560 --> 00:28:50,500 when we were looking at extremal numbers, where did 3/2 come up? 398 00:28:50,500 --> 00:28:52,420 AUDIENCE: [INAUDIBLE] like C4? 399 00:28:52,420 --> 00:28:53,980 YUFEI ZHAO: C4, yeah. 400 00:28:53,980 --> 00:29:00,450 So if you compare this quantity to the extremal number of C4, 401 00:29:00,450 --> 00:29:04,690 it's also n to the 3/2. 402 00:29:04,690 --> 00:29:07,750 And in fact, the proof is exactly the same. 403 00:29:07,750 --> 00:29:12,110 All we're using here is that the incidence graph is C4-free 404 00:29:12,110 --> 00:29:17,700 So in fact, this is an argument about C4-free graphs. 405 00:29:17,700 --> 00:29:21,060 So this fact here, every two points determine at most one 406 00:29:21,060 --> 00:29:25,080 line, is saying that if you look at the incidence graph, 407 00:29:25,080 --> 00:29:28,000 there's no C4. 408 00:29:28,000 --> 00:29:30,670 That's all we're using for now. 409 00:29:30,670 --> 00:29:31,712 Any questions? 410 00:29:35,410 --> 00:29:37,000 So is this the truth? 411 00:29:37,000 --> 00:29:39,640 Now, back when we were discussing the extremal number 412 00:29:39,640 --> 00:29:43,650 for C4-free graphs, we saw that, in fact, this 413 00:29:43,650 --> 00:29:45,240 is the correct order. 414 00:29:45,240 --> 00:29:46,740 And what was the construction there? 415 00:29:53,550 --> 00:29:56,790 So the construction also came from incidences, 416 00:29:56,790 --> 00:30:01,290 but incidences of taking all lines 417 00:30:01,290 --> 00:30:07,950 and points in the finite field plain, Fq squared. 418 00:30:07,950 --> 00:30:10,740 If you look at all the lines and all the points 419 00:30:10,740 --> 00:30:13,740 in a finite field plain, then you 420 00:30:13,740 --> 00:30:18,750 get the correct lower bound for C4. 421 00:30:18,750 --> 00:30:23,490 But now we are actually working in the real plane, 422 00:30:23,490 --> 00:30:28,560 so it turns out that the answer is different when you're not 423 00:30:28,560 --> 00:30:30,120 working the finite field. 424 00:30:30,120 --> 00:30:33,615 We're going to be using the topology of the real plane. 425 00:30:33,615 --> 00:30:35,740 And we're going to come up with a different answer. 426 00:30:35,740 --> 00:30:41,070 So it turns out that the truth for the number 427 00:30:41,070 --> 00:30:44,210 of maximum number of incidences in the plane, 428 00:30:44,210 --> 00:30:50,020 for points and lines in the real plane, is not exponent 3/2, 429 00:30:50,020 --> 00:30:53,770 but turns out to be 4/3. 430 00:30:53,770 --> 00:30:56,710 And this is a consequence of an important result 431 00:30:56,710 --> 00:30:58,950 in incidence geometry, a fundamental result, 432 00:30:58,950 --> 00:31:00,800 known as the Szemeredi-Trotter theorem. 433 00:31:07,080 --> 00:31:11,350 So the Szemeredi-Trotter theorem says 434 00:31:11,350 --> 00:31:17,050 that the number of incidences between points and lines 435 00:31:17,050 --> 00:31:19,720 is upper bounded by this function where 436 00:31:19,720 --> 00:31:22,740 you look at the number of points times the number of lines, 437 00:31:22,740 --> 00:31:34,250 and each raised to power 2/3 and plus some additional terms, 438 00:31:34,250 --> 00:31:38,360 just in case there are many more lines compared to points or way 439 00:31:38,360 --> 00:31:41,170 more points compared to lines. 440 00:31:41,170 --> 00:31:44,070 So that's the Szemeredi-Trotter theorem. 441 00:31:44,070 --> 00:31:52,190 And as a corollary, you see that n points, n lines give you 442 00:31:52,190 --> 00:31:58,930 at most n to the 4/3 incidences, in contrast 443 00:31:58,930 --> 00:32:05,950 to the setting of the finite field plain, where you can 444 00:32:05,950 --> 00:32:08,150 get n to the 3/2 incidences. 445 00:32:08,150 --> 00:32:10,870 So somehow, we have to use the topology 446 00:32:10,870 --> 00:32:13,280 of the real plane for this one. 447 00:32:13,280 --> 00:32:15,370 And I want to show you a proof-- 448 00:32:15,370 --> 00:32:17,080 turns out not the original proof, 449 00:32:17,080 --> 00:32:19,490 but it's a proof that uses the crossing number 450 00:32:19,490 --> 00:32:23,525 inequality to prove Szemeredi-Trotter theorem. 451 00:32:23,525 --> 00:32:25,150 You see, in crossing number inequality, 452 00:32:25,150 --> 00:32:29,290 we are using the topology of the real plane. 453 00:32:29,290 --> 00:32:31,657 Where? 454 00:32:31,657 --> 00:32:32,740 AUDIENCE: Euler's formula. 455 00:32:32,740 --> 00:32:34,198 YUFEI ZHAO: Euler's formula, right. 456 00:32:34,198 --> 00:32:36,020 So the very beginning, Euler's formula 457 00:32:36,020 --> 00:32:40,290 has to do with the topology of the real plane. 458 00:32:40,290 --> 00:32:43,790 Now, this bound turns out to be tight. 459 00:32:43,790 --> 00:32:45,530 So let me give you an example showing 460 00:32:45,530 --> 00:32:49,880 that the 4/3 exponent is tight. 461 00:32:49,880 --> 00:32:55,340 And the example is, if you take p 462 00:32:55,340 --> 00:33:06,420 to be this rectangular grid of points, 463 00:33:06,420 --> 00:33:09,750 and L to be a set of lines-- so I'm 464 00:33:09,750 --> 00:33:13,710 going to write the lines by their equation, 465 00:33:13,710 --> 00:33:17,730 where the slope is an integer from 1 through k 466 00:33:17,730 --> 00:33:21,090 and the y-intercept is an integer from 1 467 00:33:21,090 --> 00:33:23,610 through k squared. 468 00:33:23,610 --> 00:33:30,230 And you see here that every line in L 469 00:33:30,230 --> 00:33:42,130 contains exactly k points from P. So we got in total k 470 00:33:42,130 --> 00:33:51,040 to the 4th incidences, which is on the order of n to the 4/3. 471 00:33:53,840 --> 00:33:55,740 So n to the 4/3 third is the right answer. 472 00:33:59,513 --> 00:34:01,930 Now let me show you how to prove Szemeredi-Trotter theorem 473 00:34:01,930 --> 00:34:03,980 from the crossing number inequality. 474 00:34:03,980 --> 00:34:06,700 It turns out to be a very neat application that's 475 00:34:06,700 --> 00:34:10,610 almost a direct consequence once you set up the right graph. 476 00:34:10,610 --> 00:34:15,550 And the idea is that we are going to draw a graph based 477 00:34:15,550 --> 00:34:19,570 on our incidence configuration. 478 00:34:19,570 --> 00:34:26,900 So first, just to clean things up a little bit, 479 00:34:26,900 --> 00:34:41,830 let's get rid of lines in L with 1 or 0 points in P. 480 00:34:41,830 --> 00:34:45,219 So this operation doesn't affect the bounds. 481 00:34:45,219 --> 00:34:46,810 So you can check. 482 00:34:46,810 --> 00:34:50,170 These lines don't contribute much to the incidence bound, 483 00:34:50,170 --> 00:34:53,260 and only contributes to this plus L. 484 00:34:53,260 --> 00:34:55,600 So you can get rid of such lines. 485 00:34:55,600 --> 00:35:03,830 So let's assume that every line in L 486 00:35:03,830 --> 00:35:14,390 contains at least two points from P. 487 00:35:14,390 --> 00:35:19,010 And let's draw a graph based on this incidence structure. 488 00:35:19,010 --> 00:35:19,970 So if I have-- 489 00:35:27,850 --> 00:35:36,980 so suppose these are my points and lines. 490 00:35:36,980 --> 00:35:39,560 I'll just draw a graph where I keep 491 00:35:39,560 --> 00:35:47,560 the points as the vertices, and I put in an edge. 492 00:35:47,560 --> 00:35:55,900 It's a finite edge that connects two adjacent points 493 00:35:55,900 --> 00:35:56,620 on the same line. 494 00:36:02,467 --> 00:36:03,300 So I get some graph. 495 00:36:09,480 --> 00:36:12,938 Let me make this graph a bit more interesting. 496 00:36:23,680 --> 00:36:25,080 So I get some graph. 497 00:36:25,080 --> 00:36:31,340 And how many crossings, at most, does this graph have? 498 00:36:31,340 --> 00:36:42,030 So the number of crossings of G is at most 499 00:36:42,030 --> 00:36:45,920 the number of lines squared, because a crossing 500 00:36:45,920 --> 00:36:47,030 comes from two lines. 501 00:36:49,662 --> 00:36:50,870 So here, you have a crossing. 502 00:36:50,870 --> 00:36:52,448 A crossing comes from two lines. 503 00:36:52,448 --> 00:36:54,740 Number of crossings is at most number of lines squared. 504 00:36:57,380 --> 00:36:59,500 On the other hand, we can give a lower bound 505 00:36:59,500 --> 00:37:04,820 to the number of crossings from the crossing number inequality. 506 00:37:04,820 --> 00:37:07,470 And to do that, I want to estimate the number of edges. 507 00:37:07,470 --> 00:37:09,770 And this is the reason why I assume every line contains 508 00:37:09,770 --> 00:37:16,580 at least two points from P, because a line with now k 509 00:37:16,580 --> 00:37:23,720 incidences gives k minus 1 edges. 510 00:37:26,690 --> 00:37:32,300 And if k is at least 2, then k minus 1 is at least k over 2, 511 00:37:32,300 --> 00:37:33,410 let's say. 512 00:37:33,410 --> 00:37:35,990 I don't care about constant factors. 513 00:37:35,990 --> 00:37:41,270 So by crossing number inequality, 514 00:37:41,270 --> 00:37:46,130 the number of crossings of G is at least 515 00:37:46,130 --> 00:37:50,210 the number of edges cubed over the number of vertices 516 00:37:50,210 --> 00:38:00,320 squared, which is at least the number of incidences 517 00:38:00,320 --> 00:38:05,980 of this configuration cubed over the number of points squared. 518 00:38:05,980 --> 00:38:08,540 Actually, number of vertices is the number of points. 519 00:38:08,540 --> 00:38:12,050 And number of edges, by this argument here, 520 00:38:12,050 --> 00:38:15,428 is on the same order as the number of incidences. 521 00:38:18,500 --> 00:38:24,470 Putting these two facts together, we see-- 522 00:38:24,470 --> 00:38:29,510 there was one extra hypothesis in crossing number inequality. 523 00:38:29,510 --> 00:38:32,780 Provided that this hypothesis holds, 524 00:38:32,780 --> 00:38:37,610 which is that the number of incidences 525 00:38:37,610 --> 00:38:48,720 is at least 8 times the number of points, 526 00:38:48,720 --> 00:38:52,200 so that the original hypothesis holds. 527 00:38:54,810 --> 00:38:58,080 So putting everything together, and rearranging 528 00:38:58,080 --> 00:39:02,190 all of these terms, and using upper and lower bounds 529 00:39:02,190 --> 00:39:08,460 on the crossing number, we find that the number of incidences 530 00:39:08,460 --> 00:39:10,980 is upper bounded by-- 531 00:39:10,980 --> 00:39:22,750 the main term you see is just coming from these two, 532 00:39:22,750 --> 00:39:27,640 but there are a few other terms that we should put in, just 533 00:39:27,640 --> 00:39:31,170 in case this hypothesis is violated, 534 00:39:31,170 --> 00:39:35,080 and also to take care of this assumption over here, 535 00:39:35,080 --> 00:39:39,660 so adding a couple of linear terms corresponding 536 00:39:39,660 --> 00:39:43,200 to the number of points and the number of lines. 537 00:39:43,200 --> 00:39:46,677 If this hypothesis is violated, then the inequality 538 00:39:46,677 --> 00:39:47,260 is still true. 539 00:39:51,650 --> 00:39:55,430 So this proves the crossing numbers inequality. 540 00:39:55,430 --> 00:39:56,638 Any questions? 541 00:40:00,790 --> 00:40:06,450 So we've done these two very neat results. 542 00:40:06,450 --> 00:40:09,760 The question is, what do they have to do with the sum product 543 00:40:09,760 --> 00:40:11,790 problem? 544 00:40:11,790 --> 00:40:15,390 So I want to show you how you can give some lower bound 545 00:40:15,390 --> 00:40:20,010 on the sum product problem using Szemeredi-Trotter theorem. 546 00:40:22,650 --> 00:40:25,750 So it turns out that the sum product problem is intimately 547 00:40:25,750 --> 00:40:28,780 related to incidence geometry. 548 00:40:28,780 --> 00:40:32,500 And the reason-- you'll see in a second precisely why they're 549 00:40:32,500 --> 00:40:36,640 related, but roughly speaking, when you have addition 550 00:40:36,640 --> 00:40:39,580 and multiplication, they're are kind of 551 00:40:39,580 --> 00:40:43,090 like taking slope and y-intercept 552 00:40:43,090 --> 00:40:45,020 of an equation of a line. 553 00:40:45,020 --> 00:40:47,500 So there are two operations that are involved. 554 00:40:47,500 --> 00:40:52,660 So turns out, many incidence geometry problems can be set up 555 00:40:52,660 --> 00:40:53,830 and a way-- 556 00:40:53,830 --> 00:40:55,630 so many sum product problems can be set up 557 00:40:55,630 --> 00:40:58,800 in a way that involves incidence geometry. 558 00:40:58,800 --> 00:41:04,330 And a very short and clever lower bound to the sum product 559 00:41:04,330 --> 00:41:10,270 problem was proved by Elekes in the late '90s. 560 00:41:18,930 --> 00:41:25,470 So he showed the bound that if you have a subset of finite, 561 00:41:25,470 --> 00:41:31,870 subset of reals, then the sum set size times the product set 562 00:41:31,870 --> 00:41:35,665 size is at least A to the 5/2. 563 00:41:39,150 --> 00:41:46,995 As a corollary, one of these two must be fairly large. 564 00:41:46,995 --> 00:41:53,560 The max of the sum set size and the product set size 565 00:41:53,560 --> 00:41:57,495 is at least a to the 5/4. 566 00:42:06,030 --> 00:42:07,440 Let me show you the proof. 567 00:42:07,440 --> 00:42:11,040 I'm going to construct a set of points and a set of lines 568 00:42:11,040 --> 00:42:16,490 based on the set A. And the set of points in R2 569 00:42:16,490 --> 00:42:22,700 is going to be pairs x comma y, where the horizontal coordinate 570 00:42:22,700 --> 00:42:26,740 lies in the sum set, A plus A, and the vertical coordinate 571 00:42:26,740 --> 00:42:38,140 lies in the product set, A times A. And a set of lines 572 00:42:38,140 --> 00:42:40,350 is going to be these lines-- 573 00:42:40,350 --> 00:42:52,350 y equals to a times x minus a prime, where a and a prime 574 00:42:52,350 --> 00:42:56,810 lie in A. 575 00:42:56,810 --> 00:43:00,840 So these are some points and some lines. 576 00:43:00,840 --> 00:43:07,270 And I want to show you that they must have many incidences. 577 00:43:07,270 --> 00:43:09,180 So what are the incidences? 578 00:43:09,180 --> 00:43:17,080 So note that the line y equals to a times x minus a prime-- 579 00:43:17,080 --> 00:43:27,510 it contains the points a prime plus b and ab, 580 00:43:27,510 --> 00:43:34,230 which lies in P for all b in A. You plug it in. 581 00:43:34,230 --> 00:43:39,743 If you plug in a prime plus b into here, you get ab. 582 00:43:39,743 --> 00:43:44,300 And this point lies in P, because the first coordinate 583 00:43:44,300 --> 00:43:45,920 is the sum set. 584 00:43:45,920 --> 00:43:49,550 The second coordinate lies in the product set. 585 00:43:49,550 --> 00:43:57,570 So each line in L contains many incidences. 586 00:43:57,570 --> 00:44:01,960 So each line in L contains a incidents. 587 00:44:01,960 --> 00:44:17,610 So this line, each line in L contains a incidences. 588 00:44:17,610 --> 00:44:23,490 Also, we can easily compute the number of lines 589 00:44:23,490 --> 00:44:26,490 and the number of points. 590 00:44:26,490 --> 00:44:30,870 The number of points is A plus A size 591 00:44:30,870 --> 00:44:36,540 times the size of A times A. And the number of lines 592 00:44:36,540 --> 00:44:41,100 is just the size of A squared. 593 00:44:41,100 --> 00:44:52,541 So by Szemeredi-Trotter, we find that the number of incidences 594 00:44:52,541 --> 00:44:58,250 is lower bounded by noting this fact here. 595 00:44:58,250 --> 00:44:59,880 We have many incidences. 596 00:44:59,880 --> 00:45:07,120 So the number of lines, each line contributes a incidences. 597 00:45:07,120 --> 00:45:09,100 But we also have an upper bound coming 598 00:45:09,100 --> 00:45:11,496 from the Szemeredi-Trotter theorem. 599 00:45:11,496 --> 00:45:18,100 So plugging in the upper bound, we find that you have-- 600 00:45:18,100 --> 00:45:20,080 so now I'm just directly plugging 601 00:45:20,080 --> 00:45:22,895 in the statement of Szemeredi-Trotter. 602 00:45:26,460 --> 00:45:28,317 The main term is the first term. 603 00:45:28,317 --> 00:45:30,150 You should still check the latter two terms, 604 00:45:30,150 --> 00:45:31,650 but the main term is the first term. 605 00:45:34,190 --> 00:45:39,540 So plugging in the values for P and L, 606 00:45:39,540 --> 00:45:55,790 we find this is the case, plus some additional terms, 607 00:45:55,790 --> 00:45:58,910 which you can check are dominated by the first term. 608 00:45:58,910 --> 00:46:00,770 So let me just do a big O over there. 609 00:46:03,530 --> 00:46:06,810 Now you put left and right together, 610 00:46:06,810 --> 00:46:11,070 and we could obtain some lower bound 611 00:46:11,070 --> 00:46:15,090 on the product of the sizes of the sum set and the product 612 00:46:15,090 --> 00:46:18,320 set, thereby yielding allocations. 613 00:46:23,710 --> 00:46:26,460 So this is some lower bound on the sum product problem. 614 00:46:26,460 --> 00:46:30,400 And you see, we went through the crossing number inequality 615 00:46:30,400 --> 00:46:32,875 to prove Szemeredi-Trotter, a basic result 616 00:46:32,875 --> 00:46:34,480 in incidence geometry. 617 00:46:34,480 --> 00:46:39,580 And viewing sum product as an incidence geometry problem, one 618 00:46:39,580 --> 00:46:43,850 can obtain this lower bound over here. 619 00:46:43,850 --> 00:46:44,878 Any questions? 620 00:46:47,950 --> 00:46:51,640 I want to show you a different proof that was found later, 621 00:46:51,640 --> 00:46:54,550 that gives an improvement. 622 00:46:54,550 --> 00:47:01,670 And there's a question, can you do better than 5/4? 623 00:47:01,670 --> 00:47:06,490 So it turns out that there was a very nice result of Solymosi 624 00:47:06,490 --> 00:47:09,914 sometime later that gives you an improvement. 625 00:47:16,210 --> 00:47:21,730 Solymosi proved in 2009 that if A 626 00:47:21,730 --> 00:47:29,030 is a subset of positive reals, then the size of A times 627 00:47:29,030 --> 00:47:33,070 A multiplied by the size of A plus A squared 628 00:47:33,070 --> 00:47:37,540 is at least size of A to the 4th divided 629 00:47:37,540 --> 00:47:45,610 by 4 ceiling log of the size of A, where the log is base 2. 630 00:47:45,610 --> 00:47:48,600 So don't worry about the specific constants. 631 00:47:51,240 --> 00:47:54,640 A being in the positive reals is no big deal, 632 00:47:54,640 --> 00:47:58,630 because you can always separate A as positive and negative 633 00:47:58,630 --> 00:48:00,790 and analyze each part separately. 634 00:48:00,790 --> 00:48:05,260 So as a corollary to Solymosi's theorem, 635 00:48:05,260 --> 00:48:12,730 we obtain that for A, a subset of the reals, the sum 636 00:48:12,730 --> 00:48:17,110 set and the product set, at least one of them 637 00:48:17,110 --> 00:48:24,990 must have size at least A raised to 4/3 638 00:48:24,990 --> 00:48:34,820 divided by 2 times log base 2 size of A raised to 1/3 third. 639 00:48:34,820 --> 00:48:42,210 So basically, A to the 4/3 minus little one in the exponent, 640 00:48:42,210 --> 00:48:43,370 so better than before. 641 00:48:43,370 --> 00:48:44,440 And this is a new bound. 642 00:48:47,580 --> 00:48:52,200 I want to note that in this formulation, where 643 00:48:52,200 --> 00:48:58,320 we are looking at lower bounding this quantity over here, 644 00:48:58,320 --> 00:49:06,160 this is tied up to logarithmic factors, 645 00:49:06,160 --> 00:49:11,710 by considering A to be just the interval from 1 to n. 646 00:49:11,710 --> 00:49:14,140 If A is the interval from 1 to n, 647 00:49:14,140 --> 00:49:17,230 then the left-hand side, A plus A, is around size n. 648 00:49:17,230 --> 00:49:18,660 So you have n squared. 649 00:49:18,660 --> 00:49:20,260 And A times A is also, I mentioned, 650 00:49:20,260 --> 00:49:23,800 around size n squared. 651 00:49:23,800 --> 00:49:26,170 So this inequality here is tight. 652 00:49:26,170 --> 00:49:28,820 The consequence is not tight, but the first inequality 653 00:49:28,820 --> 00:49:29,320 is tight. 654 00:49:34,400 --> 00:49:36,260 So in the remainder of today's lecture, 655 00:49:36,260 --> 00:49:39,960 I want to show you how to prove Solymosi's lower bound. 656 00:49:39,960 --> 00:49:43,550 And it has some similarities to the one 657 00:49:43,550 --> 00:49:50,210 that we've seen, because it also looks at some geometric aspects 658 00:49:50,210 --> 00:49:52,610 of the sum product problem. 659 00:49:52,610 --> 00:49:57,950 But it doesn't use the exact tools that we've seen earlier. 660 00:49:57,950 --> 00:50:00,590 It does use some tools that were related 661 00:50:00,590 --> 00:50:04,520 to the lecture from Monday. 662 00:50:04,520 --> 00:50:07,070 So last time, we discussed this thing 663 00:50:07,070 --> 00:50:09,830 called the "additive energy." 664 00:50:09,830 --> 00:50:12,560 You can come up with a similar notion for the multiplication 665 00:50:12,560 --> 00:50:24,220 operation, so the "multiplicative energy," 666 00:50:24,220 --> 00:50:31,426 which we'll denote by E sub, with the multiplication symbol, 667 00:50:31,426 --> 00:50:35,650 A. So the multiplicative energy is like the additive energy, 668 00:50:35,650 --> 00:50:38,028 except that instead of doing addition, 669 00:50:38,028 --> 00:50:39,820 we're going to do a multiplication instead. 670 00:50:39,820 --> 00:50:45,650 So one way to define it is the number of quadruples such 671 00:50:45,650 --> 00:50:55,530 that there exists some real lambda such that a, comma, 672 00:50:55,530 --> 00:50:58,680 b equals to lambda c, comma, d. 673 00:51:06,400 --> 00:51:08,540 So basically the same as additive energy, 674 00:51:08,540 --> 00:51:12,530 except that we're using multiplications instead. 675 00:51:12,530 --> 00:51:15,330 By the Cauchy-Schwartz inequality-- 676 00:51:15,330 --> 00:51:20,980 and this is a calculation we saw last time, as well-- 677 00:51:20,980 --> 00:51:27,440 we see that if you have a set with small product, then 678 00:51:27,440 --> 00:51:30,280 it must have high multiplicative energy. 679 00:51:30,280 --> 00:51:34,030 So last time, we saw small sum set implies high additive 680 00:51:34,030 --> 00:51:34,570 energy. 681 00:51:34,570 --> 00:51:38,560 Likewise, small product set implies high multiplicative 682 00:51:38,560 --> 00:51:39,570 energy. 683 00:51:39,570 --> 00:51:43,690 In particular, the multiplicative energy of A, 684 00:51:43,690 --> 00:51:50,440 you can rewrite it as sum over all elements 685 00:51:50,440 --> 00:51:55,750 x in the product set of the quantity, which tells you 686 00:51:55,750 --> 00:52:02,730 the number of ways to write x as a product, 687 00:52:02,730 --> 00:52:05,860 this number squared and then summed over all x. 688 00:52:05,860 --> 00:52:09,280 By Cauchy-Schwartz, we find that this quantity here is lower 689 00:52:09,280 --> 00:52:12,430 bounded by the size of A to the 4th divided 690 00:52:12,430 --> 00:52:20,200 by the size of A times A. So to prove Solymosi's theorem, 691 00:52:20,200 --> 00:52:26,710 we are going to actually prove a bound on the energy, 692 00:52:26,710 --> 00:52:28,680 instead of proving it on the set. 693 00:52:28,680 --> 00:52:30,460 We're going to prove it on the energy. 694 00:52:30,460 --> 00:52:42,480 So it suffices to show that the multiplicative energy is 695 00:52:42,480 --> 00:52:49,900 at most 4 times the sum set size times-- 696 00:52:49,900 --> 00:53:01,832 so let me divide the energy by log of A. 697 00:53:01,832 --> 00:53:04,580 So when you plug this into this inequality, 698 00:53:04,580 --> 00:53:05,520 it would imply that. 699 00:53:05,520 --> 00:53:09,590 So it remains to show this inequality 700 00:53:09,590 --> 00:53:12,170 over here upper bounding the multiplicative energy. 701 00:53:20,730 --> 00:53:22,710 There's an important idea that we're 702 00:53:22,710 --> 00:53:25,440 going to use here, which is also pretty common in analysis, 703 00:53:25,440 --> 00:53:32,850 is that instead of considering that energy sum here, 704 00:53:32,850 --> 00:53:35,880 we're going to consider a similar sum, 705 00:53:35,880 --> 00:53:40,110 except we're going to chop up the sum into pieces according 706 00:53:40,110 --> 00:53:44,600 to how big the terms are, so that we're only 707 00:53:44,600 --> 00:53:48,230 looking at contributions of comparable size. 708 00:53:48,230 --> 00:53:50,948 And so this is called a "dyadic decomposition." 709 00:54:02,420 --> 00:54:07,900 The idea is that we can write the multiplicative energy 710 00:54:07,900 --> 00:54:10,400 similar to above, but instead of summing over 711 00:54:10,400 --> 00:54:15,590 x in the product set, let me sum over s in the quotient set. 712 00:54:15,590 --> 00:54:21,670 So you can interpret what this quotient A is. 713 00:54:21,670 --> 00:54:27,140 This is the set of all A divided by B, where A and B are in A. A 714 00:54:27,140 --> 00:54:29,150 is a set of positive reals, so I don't need 715 00:54:29,150 --> 00:54:31,400 to worry about division by 0. 716 00:54:31,400 --> 00:54:36,770 So what remains, then, is the intersection 717 00:54:36,770 --> 00:54:41,700 of s times A and A squared. 718 00:54:41,700 --> 00:54:47,170 Remember, s times A is scaling each element of A by s. 719 00:54:47,170 --> 00:54:49,760 So we have this quantity over here. 720 00:54:49,760 --> 00:54:55,070 So I want to break up the sum into a bunch of smaller 721 00:54:55,070 --> 00:54:59,990 sums, where I want to break up the sum according 722 00:54:59,990 --> 00:55:05,570 to how big the terms are, so that inside each group, 723 00:55:05,570 --> 00:55:08,990 all the terms are roughly of the same size. 724 00:55:08,990 --> 00:55:11,480 And easiest way to do this is to chop them up 725 00:55:11,480 --> 00:55:19,400 into groups where everything inside the same collection 726 00:55:19,400 --> 00:55:21,540 differs by at most a factor of 2. 727 00:55:21,540 --> 00:55:25,100 So that's why it's called a dyadic decomposition, 728 00:55:25,100 --> 00:55:27,740 going from 0 to-- 729 00:55:27,740 --> 00:55:32,960 the maximum possible here is basically A. 730 00:55:32,960 --> 00:55:39,410 So let's look at i going from 0 to log base 2 of A. 731 00:55:39,410 --> 00:55:41,150 So this is the number of bins. 732 00:55:44,350 --> 00:55:49,810 And partition the sum into sub-sums 733 00:55:49,810 --> 00:55:54,190 where I'm looking at the i-th sub-sum consisting 734 00:55:54,190 --> 00:55:58,510 of contributions involving terms with size between 2 735 00:55:58,510 --> 00:56:01,700 to the i and 2 to the i plus 1. 736 00:56:09,530 --> 00:56:15,260 Break up the sum according to the sizes of the summands. 737 00:56:15,260 --> 00:56:18,020 By pigeonhole principle, one of these summands 738 00:56:18,020 --> 00:56:19,760 must be somewhat large. 739 00:56:23,520 --> 00:56:33,730 So by pigeonhole, there exists a k 740 00:56:33,730 --> 00:56:47,920 such that setting D to be the s such that that corresponds 741 00:56:47,920 --> 00:56:52,670 to the k-th term in the sum. 742 00:57:04,250 --> 00:57:17,790 So one has that this sum coming from just contributions from D 743 00:57:17,790 --> 00:57:19,260 is at least-- 744 00:57:23,760 --> 00:57:27,930 so it's at least the multiplicative energy 745 00:57:27,930 --> 00:57:29,760 divided by the number of bins. 746 00:57:36,170 --> 00:57:39,100 All of that many bins-- 747 00:57:39,100 --> 00:57:41,300 by pigeonhole, I can find one bin 748 00:57:41,300 --> 00:57:44,510 that's a pretty large contribution to the sum. 749 00:57:44,510 --> 00:57:52,910 And the right-hand side, we can upper bound each term over here 750 00:57:52,910 --> 00:57:57,470 by 2 to the 2k plus 2, and the number 751 00:57:57,470 --> 00:58:05,590 of terms as the size of D. Let me call the elements of D S1 752 00:58:05,590 --> 00:58:16,820 through Sm, where S1 through Sm are sorted in increasing order. 753 00:58:22,880 --> 00:58:26,960 Now let me draw you a picture of what's going on. 754 00:58:26,960 --> 00:58:40,050 Let's consider for each element of D, so for each i and m, 755 00:58:40,050 --> 00:58:48,230 let's consider the line given by the equation y equals to s 756 00:58:48,230 --> 00:58:51,960 sub i times x. 757 00:58:51,960 --> 00:58:56,290 Let me draw this picture where I'm 758 00:58:56,290 --> 00:59:01,260 looking at the positive quadrant, 759 00:59:01,260 --> 00:59:03,640 so I have a bunch of points in the positive quadrant. 760 00:59:10,180 --> 00:59:13,110 And specifically, I'm interested in these points whose 761 00:59:13,110 --> 00:59:18,205 coordinates, both coordinates are elements of A. 762 00:59:18,205 --> 00:59:26,580 And I want to consider lines through points of A, 763 00:59:26,580 --> 00:59:28,950 but I want to consider lines where 764 00:59:28,950 --> 00:59:34,380 it intersects this A cross A in the desired number of points. 765 00:59:37,030 --> 00:59:39,400 And we find those set, and then let's 766 00:59:39,400 --> 00:59:48,120 draw these lines over here, where this line here, L1 767 00:59:48,120 --> 00:59:53,640 has slope exactly S1, and L2, L3, and so on. 768 00:59:58,990 --> 01:00:03,820 I want to draw one more line, which is somewhat auxiliary, 769 01:00:03,820 --> 01:00:06,770 but just to make our life a bit easier. 770 01:00:06,770 --> 01:00:13,240 Finally, let's let L of m plus 1 be the vertical line, 771 01:00:13,240 --> 01:00:21,030 or rather be the vertical ray, which 772 01:00:21,030 --> 01:00:31,230 goes to the minimum element of A above Lm. 773 01:00:31,230 --> 01:00:34,660 So it's this line over here. 774 01:00:34,660 --> 01:00:36,370 That's Lm plus 1. 775 01:00:40,550 --> 01:00:44,920 So in A cross A, I draw a bunch of lines. 776 01:00:44,920 --> 01:00:46,650 So now all the lines-- 777 01:00:46,650 --> 01:00:50,700 so all these lines involve some point of A and the origin, 778 01:00:50,700 --> 01:00:51,930 but I don't draw all of them. 779 01:00:51,930 --> 01:00:54,600 I draw a select set of them. 780 01:00:54,600 --> 01:00:59,620 And what we said earlier says that the number of lines, 781 01:00:59,620 --> 01:01:02,880 the number of points on each of these strong lines, 782 01:01:02,880 --> 01:01:05,730 is roughly the same for each of these lines. 783 01:01:12,350 --> 01:01:17,240 Let's let capital L sub j denote the set 784 01:01:17,240 --> 01:01:24,350 of points in A cross A that lie on the j-th line. 785 01:01:32,320 --> 01:01:37,930 So that's L1, L2, and so on. 786 01:01:41,230 --> 01:01:49,420 I claim that if you look at two consecutive lines 787 01:01:49,420 --> 01:01:55,390 and look at the sum set of the points in A cross A 788 01:01:55,390 --> 01:02:00,500 that intersect, you're looking at two lines, 789 01:02:00,500 --> 01:02:02,550 and you're adding up points on those two lines. 790 01:02:02,550 --> 01:02:05,680 So you form a grid. 791 01:02:05,680 --> 01:02:13,210 So you end up forming this grid. 792 01:02:13,210 --> 01:02:15,940 And the number of points on this grid 793 01:02:15,940 --> 01:02:19,630 is precisely the product of these two point sets. 794 01:02:28,940 --> 01:02:42,650 Moreover, the sets Lj plus L sub j plus 1 795 01:02:42,650 --> 01:02:48,050 are disjoint for different j. 796 01:02:52,660 --> 01:02:56,440 And this is where we're using the geometry of the plane here. 797 01:02:56,440 --> 01:03:02,960 Because the sum of L1 and L2 lies in the span, 798 01:03:02,960 --> 01:03:06,470 the sum of L2 and L3 in a different span, 799 01:03:06,470 --> 01:03:09,370 so they cannot intersect. 800 01:03:09,370 --> 01:03:11,430 So they lie in-- 801 01:03:11,430 --> 01:03:38,370 so since they span disjoint regions, L1 plus L2 lies here, 802 01:03:38,370 --> 01:03:41,590 L2 plus L3 lies there, and so on. 803 01:03:41,590 --> 01:03:42,780 But they're all disjoint. 804 01:03:51,180 --> 01:03:53,700 Now let's put everything that we know together. 805 01:03:58,720 --> 01:04:06,550 Remember, the goal is to upper bound the multiplicative energy 806 01:04:06,550 --> 01:04:09,840 as a function of the sum set. 807 01:04:09,840 --> 01:04:12,300 So in other words, we want to lower bound the sum set. 808 01:04:12,300 --> 01:04:19,360 So I want to show you that this A plus A has a lot of elements. 809 01:04:19,360 --> 01:04:21,690 There's a lot of sums. 810 01:04:21,690 --> 01:04:25,360 And I have a bunch of disjoint contributions to these sums. 811 01:04:25,360 --> 01:04:28,590 So let's add up those disjoint contributions to the sums. 812 01:04:32,883 --> 01:04:38,220 You see that the size of A plus A squared 813 01:04:38,220 --> 01:04:42,990 is the same as the size of the product set A plus A. 814 01:04:42,990 --> 01:04:45,540 So this is Cartesian product. 815 01:04:45,540 --> 01:04:55,110 Here is-- this is a Cartesian product, 816 01:04:55,110 --> 01:04:58,740 in other words, the grid that is strong up there. 817 01:04:58,740 --> 01:05:01,150 I add this product to itself. 818 01:05:04,753 --> 01:05:06,170 So I should get the same set here. 819 01:05:09,660 --> 01:05:11,800 But how big is this sum set? 820 01:05:11,800 --> 01:05:15,290 That grid, that lattice grid added to itself, 821 01:05:15,290 --> 01:05:16,180 how big should it be? 822 01:05:16,180 --> 01:05:19,220 I want to lower bound the number of sums. 823 01:05:19,220 --> 01:05:23,030 And the key observation is up there. 824 01:05:23,030 --> 01:05:27,880 We can look at contributions coming from distinct spans. 825 01:05:27,880 --> 01:05:34,980 In particular, this sum here, so this sum set here, 826 01:05:34,980 --> 01:05:45,260 size is lower bounded by these distinct Lj plus L j plus 1's. 827 01:05:45,260 --> 01:05:46,210 I threw away a lot. 828 01:05:46,210 --> 01:05:49,430 I only keep the lines on the L's, and I only consider sums 829 01:05:49,430 --> 01:05:52,330 between consecutive L's. 830 01:05:52,330 --> 01:05:54,955 That should be a lower bound to the sum 831 01:05:54,955 --> 01:05:56,280 set of the grid with itself. 832 01:05:58,860 --> 01:06:03,425 But you see, and here, we're using these different-- 833 01:06:03,425 --> 01:06:08,460 for different j's, these contributions are destroyed. 834 01:06:08,460 --> 01:06:15,640 But by what we said up there, Lj plus L j plus 1 is a grid. 835 01:06:15,640 --> 01:06:20,950 So it has size Lj times L j plus 1. 836 01:06:23,900 --> 01:06:32,900 And the size of each Lj is at least 2 to the k. 837 01:06:32,900 --> 01:06:38,150 So the sum here is at least m times 2 to the 2k. 838 01:06:41,410 --> 01:06:48,340 But we saw over here that the energy lower bounds 839 01:06:48,340 --> 01:06:50,290 this 2 to the 2k. 840 01:06:50,290 --> 01:06:58,060 So we have a lower bound that is the multiplicative energy of A 841 01:06:58,060 --> 01:07:05,407 divided by 4 times the log base 2 of the size of A. 842 01:07:05,407 --> 01:07:07,490 So don't worry so much about the constant factors. 843 01:07:07,490 --> 01:07:13,000 That's just the order of magnitude that is important. 844 01:07:13,000 --> 01:07:14,680 And that's it. 845 01:07:14,680 --> 01:07:15,309 Yep. 846 01:07:15,309 --> 01:07:20,130 AUDIENCE: How do you know that the size of big L sub m plus 1? 847 01:07:20,130 --> 01:07:20,880 YUFEI ZHAO: Great. 848 01:07:20,880 --> 01:07:24,200 The question is, what do we know about the size of big L sub 849 01:07:24,200 --> 01:07:25,172 m plus 1? 850 01:07:25,172 --> 01:07:26,130 So that's a good point. 851 01:07:28,940 --> 01:07:30,543 The easiest answer is, if I don't 852 01:07:30,543 --> 01:07:31,960 care about these constant factors, 853 01:07:31,960 --> 01:07:34,660 I don't need to worry about it. 854 01:07:34,660 --> 01:07:40,060 You can think about what is the number of points 855 01:07:40,060 --> 01:07:47,430 on this line above that. 856 01:07:47,430 --> 01:07:53,110 It's essentially the number of elements of A above the biggest 857 01:07:53,110 --> 01:07:57,450 element of s m, above s m. 858 01:08:01,197 --> 01:08:03,123 It's a good question. 859 01:08:03,123 --> 01:08:04,790 I think we don't need to worry about it. 860 01:08:04,790 --> 01:08:08,935 I'm being slightly sloppy here. 861 01:08:08,935 --> 01:08:10,915 Yeah. 862 01:08:10,915 --> 01:08:16,890 AUDIENCE: [INAUDIBLE] 863 01:08:16,890 --> 01:08:18,390 YUFEI ZHAO: I think the question is, 864 01:08:18,390 --> 01:08:22,540 how do we know for j equals to m that you have this bound over 865 01:08:22,540 --> 01:08:23,040 here? 866 01:08:25,866 --> 01:08:33,770 AUDIENCE: [INAUDIBLE] 867 01:08:33,770 --> 01:08:34,520 YUFEI ZHAO: Great. 868 01:08:34,520 --> 01:08:35,260 So yes. 869 01:08:38,224 --> 01:08:48,877 AUDIENCE: [INAUDIBLE] 870 01:08:48,877 --> 01:08:50,710 YUFEI ZHAO: So there are some ways to do it. 871 01:08:50,710 --> 01:08:53,010 You can notice that the vertical line 872 01:08:53,010 --> 01:09:00,340 has at least as many points as the first slanted line. 873 01:09:03,420 --> 01:09:08,700 So details that you can work on. 874 01:09:08,700 --> 01:09:11,109 So this proves Solymosi's theorem, 875 01:09:11,109 --> 01:09:16,470 which gives you a lower bound on the sum set and the product 876 01:09:16,470 --> 01:09:19,600 set sizes and the maximum of those two. 877 01:09:19,600 --> 01:09:21,282 It's based on-- it's very short. 878 01:09:21,282 --> 01:09:21,990 It's very clever. 879 01:09:21,990 --> 01:09:24,540 It took a long time to find. 880 01:09:24,540 --> 01:09:28,649 And it gave a bound on the sum product 881 01:09:28,649 --> 01:09:32,370 problem of 4/3 that actually remained 882 01:09:32,370 --> 01:09:36,810 stuck for a very long time, until just 883 01:09:36,810 --> 01:09:48,420 fairly recently there was an improvement that gives-- 884 01:09:48,420 --> 01:09:52,350 so by Konyagin and Shkredov where 885 01:09:52,350 --> 01:09:58,260 they improved the Solymosi bound from 4/3 to 4/3 886 01:09:58,260 --> 01:10:00,570 plus some really small constant c. 887 01:10:04,300 --> 01:10:06,530 So it's some explicit constant. 888 01:10:06,530 --> 01:10:09,040 I think right now-- so that's being proved over time, 889 01:10:09,040 --> 01:10:12,780 but right now, I think c is around 1 over 1,000 890 01:10:12,780 --> 01:10:13,730 or a few thousand. 891 01:10:13,730 --> 01:10:17,600 So it's some small but explicit constant. 892 01:10:17,600 --> 01:10:22,170 It remains a major open problem to improve this bound 893 01:10:22,170 --> 01:10:25,730 and prove Erdos' similarity conjecture, 894 01:10:25,730 --> 01:10:31,070 that if you have n elements, then one of the sums 895 01:10:31,070 --> 01:10:33,830 or products must be nearly quadratic in size. 896 01:10:33,830 --> 01:10:36,650 And people generally believe that that's the case. 897 01:10:39,320 --> 01:10:40,468 Any questions? 898 01:10:47,950 --> 01:10:51,410 So this concludes all the topics I want to cover in this course. 899 01:10:51,410 --> 01:10:52,470 So we went a long way. 900 01:10:52,470 --> 01:10:54,110 And so the beginning of this course, 901 01:10:54,110 --> 01:10:56,860 we started with extremal graph theory, 902 01:10:56,860 --> 01:11:00,760 looking at the basic problem of if you have a graph that 903 01:11:00,760 --> 01:11:05,830 doesn't contain some subgraph, triangle, C4, what's 904 01:11:05,830 --> 01:11:08,440 the maximum number of edges. 905 01:11:08,440 --> 01:11:11,350 In fact, that showed up even today. 906 01:11:11,350 --> 01:11:13,270 And then we went down to other tools, 907 01:11:13,270 --> 01:11:15,640 like Szemeredi's regularity lemma 908 01:11:15,640 --> 01:11:18,250 that allows us to deduce important arithmetic 909 01:11:18,250 --> 01:11:20,710 consequences, such as Roth's theorem. 910 01:11:20,710 --> 01:11:22,210 It's also an extremal problem if you 911 01:11:22,210 --> 01:11:25,300 have a set without a three-term arithmetic progression, 912 01:11:25,300 --> 01:11:28,880 how many elements can it have? 913 01:11:28,880 --> 01:11:31,900 And so the important tool of Szemeredi's regularity lemma 914 01:11:31,900 --> 01:11:34,270 then later showed up in many different ways 915 01:11:34,270 --> 01:11:36,490 in this course, especially the message 916 01:11:36,490 --> 01:11:38,800 of Szemeredi's regularity lemma, that when 917 01:11:38,800 --> 01:11:41,290 you look at an object, it's important to decompose it 918 01:11:41,290 --> 01:11:44,800 into its structural component and its pseudo-random 919 01:11:44,800 --> 01:11:45,970 component. 920 01:11:45,970 --> 01:11:48,340 So this dichotomy, this interplay 921 01:11:48,340 --> 01:11:50,590 between structure and pseudo randomness, 922 01:11:50,590 --> 01:11:54,190 is a key theme throughout this course. 923 01:11:54,190 --> 01:11:56,170 And it showed up in some of the later topics 924 01:11:56,170 --> 01:11:59,920 as well, when we discussed spectral graph theory, 925 01:11:59,920 --> 01:12:03,250 quasi-randomness, graph limits, and also 926 01:12:03,250 --> 01:12:07,300 in the later Fourier analytic proof of Roth's theorem. 927 01:12:07,300 --> 01:12:09,490 All of these proofs, all of these techniques, 928 01:12:09,490 --> 01:12:12,250 involve some kind of interplay between structure 929 01:12:12,250 --> 01:12:15,990 and pseudo-randomness. 930 01:12:15,990 --> 01:12:17,700 In the past month or, so we've been 931 01:12:17,700 --> 01:12:21,240 looking at Freiman's theorem, this key result 932 01:12:21,240 --> 01:12:24,840 in additive combinatorics concerning the structure 933 01:12:24,840 --> 01:12:27,020 of sets under addition. 934 01:12:27,020 --> 01:12:30,450 And there, we also saw many different tools that came up, 935 01:12:30,450 --> 01:12:33,720 and also connections I mentioned a few lectures ago, 936 01:12:33,720 --> 01:12:35,490 connections to really important results 937 01:12:35,490 --> 01:12:38,220 in geometry to group theory. 938 01:12:38,220 --> 01:12:41,190 And it really extends all around. 939 01:12:41,190 --> 01:12:43,710 And a few takeaways from this course-- 940 01:12:43,710 --> 01:12:47,260 one of them is that graph theory, additive combinatorics, 941 01:12:47,260 --> 01:12:49,030 they are not isolated subjects. 942 01:12:49,030 --> 01:12:52,143 They're connected to a lot within mathematics. 943 01:12:52,143 --> 01:12:54,060 And that's one of the goals I want to show you 944 01:12:54,060 --> 01:12:57,600 in this course, is to show these connections 945 01:12:57,600 --> 01:13:01,170 throughout mathematics and some to analysis, to geometry, 946 01:13:01,170 --> 01:13:02,250 to topology. 947 01:13:02,250 --> 01:13:05,550 And even simple questions can lead 948 01:13:05,550 --> 01:13:08,790 to really deep mathematics. 949 01:13:08,790 --> 01:13:11,280 And some of them I try to show you, try to hint at you, 950 01:13:11,280 --> 01:13:13,980 or at least I mentioned throughout this course. 951 01:13:13,980 --> 01:13:20,220 And what we've seen so far is just the tip of the iceberg. 952 01:13:20,220 --> 01:13:23,430 And there is a lot of still extremely exciting work 953 01:13:23,430 --> 01:13:24,600 that's to be done. 954 01:13:24,600 --> 01:13:28,500 And I've also tried to emphasize many important open problems 955 01:13:28,500 --> 01:13:32,370 that have yet to be better understood. 956 01:13:32,370 --> 01:13:36,373 And I expect in some future iteration of this course, 957 01:13:36,373 --> 01:13:38,040 some of these problems will be resolved, 958 01:13:38,040 --> 01:13:41,460 and I can show the next generation of students 959 01:13:41,460 --> 01:13:43,740 in your seats some new techniques, new methods, 960 01:13:43,740 --> 01:13:44,752 and new theorems. 961 01:13:44,752 --> 01:13:46,210 And I expect that will be the case. 962 01:13:46,210 --> 01:13:47,670 This is a very exciting area. 963 01:13:47,670 --> 01:13:50,200 And it's an area that is very close to my heart. 964 01:13:50,200 --> 01:13:54,170 It's something that I've been thinking about since my PhD. 965 01:13:54,170 --> 01:13:56,730 The bulk of my research work revolves 966 01:13:56,730 --> 01:13:59,460 around better understanding connections between graph 967 01:13:59,460 --> 01:14:02,130 theory, on one hand, and additive combinatorics 968 01:14:02,130 --> 01:14:03,470 on the other hand. 969 01:14:03,470 --> 01:14:05,220 It's been really fun teaching this course, 970 01:14:05,220 --> 01:14:07,680 and happy to have all of you here. 971 01:14:07,680 --> 01:14:08,580 Thank you. 972 01:14:08,580 --> 01:14:11,630 [APPLAUSE]