WEBVTT 00:00:00.000 --> 00:00:02.415 align:middle line:90% [SQUEAKING] 00:00:02.415 --> 00:00:04.347 align:middle line:90% [RUSTLING] 00:00:04.347 --> 00:00:07.245 align:middle line:90% [CLICKING] 00:00:07.245 --> 00:00:10.460 align:middle line:90% 00:00:10.460 --> 00:00:12.870 align:middle line:84% Erik Demaine: All right, welcome back to 6.1200. 00:00:12.870 --> 00:00:16.370 align:middle line:84% Today we are talking about number theory. 00:00:16.370 --> 00:00:20.040 align:middle line:84% And number is, I guess, an old word here for integers, 00:00:20.040 --> 00:00:22.460 align:middle line:90% in modern terminology. 00:00:22.460 --> 00:00:27.170 align:middle line:84% This set black board bold z, which 00:00:27.170 --> 00:00:28.890 align:middle line:84% is all positive and negative integers. 00:00:28.890 --> 00:00:31.280 align:middle line:84% Although mostly we think about natural numbers, 00:00:31.280 --> 00:00:33.470 align:middle line:90% the non-negative integers. 00:00:33.470 --> 00:00:35.970 align:middle line:90% Number theory is a big topic. 00:00:35.970 --> 00:00:38.990 align:middle line:84% We're going to spend three lectures on it, today and two 00:00:38.990 --> 00:00:40.370 align:middle line:90% next week. 00:00:40.370 --> 00:00:43.040 align:middle line:84% And it's a really important branch of-- 00:00:43.040 --> 00:00:46.700 align:middle line:84% subbranch, I guess, or related branch of discrete mathematics 00:00:46.700 --> 00:00:49.310 align:middle line:84% in particular, because it's found a lot of applications 00:00:49.310 --> 00:00:53.060 align:middle line:84% in modern days in cryptography and error correction, 00:00:53.060 --> 00:00:54.950 align:middle line:84% when you use your cell phone, probably 00:00:54.950 --> 00:00:58.610 align:middle line:84% several different things in that software 00:00:58.610 --> 00:01:00.890 align:middle line:84% is using number theory at some level to do crypto, 00:01:00.890 --> 00:01:03.800 align:middle line:84% to access your bank, to do error correction, 00:01:03.800 --> 00:01:07.760 align:middle line:84% to talk over radio waves, all sorts of fun things. 00:01:07.760 --> 00:01:09.380 align:middle line:84% Number theory is also extremely old. 00:01:09.380 --> 00:01:12.220 align:middle line:84% It's probably the oldest branch of mathematics, though, 00:01:12.220 --> 00:01:15.640 align:middle line:84% past 4,000 years it's a little hard to tell exactly what people 00:01:15.640 --> 00:01:16.490 align:middle line:90% were doing. 00:01:16.490 --> 00:01:20.020 align:middle line:84% But in particular, the Babylonians, 3,800 years ago, 00:01:20.020 --> 00:01:22.720 align:middle line:90% were doing number theory. 00:01:22.720 --> 00:01:28.460 align:middle line:84% Pythagoras, 2,500 years ago, is doing number theory. 00:01:28.460 --> 00:01:30.880 align:middle line:84% It had a lot of exciting activity 00:01:30.880 --> 00:01:33.500 align:middle line:90% in the middle of last century. 00:01:33.500 --> 00:01:37.300 align:middle line:84% Hardy is one of the famous number 00:01:37.300 --> 00:01:40.660 align:middle line:90% theorists of near our time. 00:01:40.660 --> 00:01:43.540 align:middle line:84% He wrote, "Number theory is great because it's one science 00:01:43.540 --> 00:01:47.830 align:middle line:84% whose very remoteness from human activities should keep it gentle 00:01:47.830 --> 00:01:48.770 align:middle line:90% and clean." 00:01:48.770 --> 00:01:50.500 align:middle line:84% He was a pacifist, and he wanted to do 00:01:50.500 --> 00:01:53.720 align:middle line:84% math that was completely useless from the perspective of warfare. 00:01:53.720 --> 00:01:57.500 align:middle line:84% Sadly, that's not the case for number theory anymore, 00:01:57.500 --> 00:01:58.520 align:middle line:90% but it used to be. 00:01:58.520 --> 00:02:01.170 align:middle line:90% 00:02:01.170 --> 00:02:04.020 align:middle line:84% But the things we'll be doing here are pretty fundamental. 00:02:04.020 --> 00:02:06.950 align:middle line:84% Today, we're going to talk about divisibility and something 00:02:06.950 --> 00:02:09.180 align:middle line:84% called GCD, which you may have heard of before. 00:02:09.180 --> 00:02:10.680 align:middle line:90% It's a pretty basic notion. 00:02:10.680 --> 00:02:11.850 align:middle line:90% Both of these are. 00:02:11.850 --> 00:02:16.610 align:middle line:84% But we're going to define everything carefully and prove 00:02:16.610 --> 00:02:19.800 align:middle line:84% all sorts of cool properties about divisibility, 00:02:19.800 --> 00:02:23.600 align:middle line:84% and use it to solve at least one fun problem. 00:02:23.600 --> 00:02:29.690 align:middle line:84% So the notation for divisibility is a divides b. 00:02:29.690 --> 00:02:31.670 align:middle line:90% We use a vertical bar. 00:02:31.670 --> 00:02:33.950 align:middle line:84% We use a vertical bar for a few other things 00:02:33.950 --> 00:02:36.300 align:middle line:84% like absolute value and set notation, 00:02:36.300 --> 00:02:38.600 align:middle line:90% so try not to be too confused. 00:02:38.600 --> 00:02:42.320 align:middle line:84% This will only be for a couple of weeks. 00:02:42.320 --> 00:02:46.790 align:middle line:84% Divisibility means a evenly divides b, 00:02:46.790 --> 00:02:50.000 align:middle line:90% or b is evenly divisible by a. 00:02:50.000 --> 00:02:58.200 align:middle line:84% So for example, let's say, 2 divides 4 is an example. 00:02:58.200 --> 00:03:01.900 align:middle line:90% 00:03:01.900 --> 00:03:04.090 align:middle line:84% But in general, what we would like 00:03:04.090 --> 00:03:08.560 align:middle line:90% is a formal definition of this. 00:03:08.560 --> 00:03:12.730 align:middle line:84% So we'll say a divides b If there exists some integer 00:03:12.730 --> 00:03:24.200 align:middle line:84% multiple such that b is that multiple of a, 00:03:24.200 --> 00:03:28.010 align:middle line:90% so b equals k times a. 00:03:28.010 --> 00:03:30.940 align:middle line:84% Very important that this is an integer, otherwise, over reals 00:03:30.940 --> 00:03:34.720 align:middle line:84% this is always the case pretty much. 00:03:34.720 --> 00:03:37.480 align:middle line:84% And we define divisibility in terms of multiplication, 00:03:37.480 --> 00:03:39.500 align:middle line:90% instead of division. 00:03:39.500 --> 00:03:43.910 align:middle line:84% You could divide both sides by a here. 00:03:43.910 --> 00:03:45.430 align:middle line:84% And so, for example, this is saying 00:03:45.430 --> 00:03:47.240 align:middle line:90% b divided by a is an integer. 00:03:47.240 --> 00:03:49.360 align:middle line:84% But that's not such a great definition 00:03:49.360 --> 00:03:51.770 align:middle line:84% because it doesn't work well when you try to divide by 0. 00:03:51.770 --> 00:03:54.920 align:middle line:84% This definition works even when a is zero, 00:03:54.920 --> 00:03:57.810 align:middle line:84% and if you do division, it doesn't. 00:03:57.810 --> 00:04:00.690 align:middle line:84% So that's why we define divisibility this way. 00:04:00.690 --> 00:04:05.960 align:middle line:84% So more interesting examples are 2 divides n means what 00:04:05.960 --> 00:04:10.670 align:middle line:90% we have been saying, n is even. 00:04:10.670 --> 00:04:13.550 align:middle line:84% Whenever we say n is even, what we really mean is 2 divides n. 00:04:13.550 --> 00:04:18.649 align:middle line:84% There's some integer multiple of 2 that equals n. 00:04:18.649 --> 00:04:28.850 align:middle line:84% Some other weirder cases, every natural number negative n 00:04:28.850 --> 00:04:31.010 align:middle line:90% divides n. 00:04:31.010 --> 00:04:33.110 align:middle line:84% It's maybe not so intuitive what divisibility 00:04:33.110 --> 00:04:35.970 align:middle line:84% means for negative numbers, but here are some examples. 00:04:35.970 --> 00:04:38.880 align:middle line:84% We allow this integer multiple k to be negative. 00:04:38.880 --> 00:04:41.150 align:middle line:84% So in particular, k can be minus 1, 00:04:41.150 --> 00:04:45.320 align:middle line:84% and minus 1 times negative n is n. 00:04:45.320 --> 00:04:48.890 align:middle line:84% Also n divides n, but that's a little less exciting. 00:04:48.890 --> 00:04:54.700 align:middle line:84% Another fun fact is that every number divides 0. 00:04:54.700 --> 00:04:57.410 align:middle line:84% This is also maybe not the most intuitive, 00:04:57.410 --> 00:04:59.740 align:middle line:84% but it comes from the definition. k in particular, 00:04:59.740 --> 00:05:00.860 align:middle line:90% can be 0. 00:05:00.860 --> 00:05:04.150 align:middle line:84% And if you take 0 times n you get 0. 00:05:04.150 --> 00:05:06.915 align:middle line:90% So everything divides 0. 00:05:06.915 --> 00:05:10.660 align:middle line:84% A number also divides its negative, things like that. 00:05:10.660 --> 00:05:12.310 align:middle line:84% A consequence of these two things 00:05:12.310 --> 00:05:14.980 align:middle line:90% together is that 0 is even. 00:05:14.980 --> 00:05:17.380 align:middle line:84% If you were ever worried about whether 0 00:05:17.380 --> 00:05:20.920 align:middle line:84% was an exception to evenness in the alternation 00:05:20.920 --> 00:05:22.940 align:middle line:90% of these numbers. 00:05:22.940 --> 00:05:24.970 align:middle line:90% Yes, this is even. 00:05:24.970 --> 00:05:25.610 align:middle line:90% This is even. 00:05:25.610 --> 00:05:26.290 align:middle line:90% This is even. 00:05:26.290 --> 00:05:30.272 align:middle line:84% Because in particular, 2 divides 0. 00:05:30.272 --> 00:05:32.480 align:middle line:84% Everything divides 0, but in particular, 2 divides 0. 00:05:32.480 --> 00:05:37.360 align:middle line:84% That's the meaning of even, OK, great. 00:05:37.360 --> 00:05:43.060 align:middle line:84% Let's state some theorems about divisibility. 00:05:43.060 --> 00:05:45.470 align:middle line:84% I'm going to call these divisibility facts. 00:05:45.470 --> 00:05:49.660 align:middle line:84% We're going to use them several times today. 00:05:49.660 --> 00:05:55.313 align:middle line:84% Let's say we have two numbers divisible by the same thing. 00:05:55.313 --> 00:06:02.430 align:middle line:90% 00:06:02.430 --> 00:06:15.000 align:middle line:84% We call this d is a common divisor of a and b. 00:06:15.000 --> 00:06:21.060 align:middle line:84% So if d divides a and d divides b, we call it a common divisor. 00:06:21.060 --> 00:06:29.340 align:middle line:84% Then I claim, for example, d divides a plus b, their sum. 00:06:29.340 --> 00:06:29.977 align:middle line:90% Why? 00:06:29.977 --> 00:06:31.560 align:middle line:84% This is very easy to prove if you just 00:06:31.560 --> 00:06:32.830 align:middle line:90% work through the definition. 00:06:32.830 --> 00:06:41.640 align:middle line:84% So let's say a equals j times d, and b equals k times d. 00:06:41.640 --> 00:06:44.020 align:middle line:84% For some integers j and k, they might be the same. 00:06:44.020 --> 00:06:45.600 align:middle line:90% They might be different. 00:06:45.600 --> 00:06:49.730 align:middle line:84% Then a plus b, well, it's just the sum of these two things, 00:06:49.730 --> 00:06:50.230 align:middle line:90% right? 00:06:50.230 --> 00:06:55.820 align:middle line:84% It's j plus k times d, not very hard. 00:06:55.820 --> 00:07:03.980 align:middle line:84% More generally, d divides sa plus tb for any integers 00:07:03.980 --> 00:07:04.690 align:middle line:90% s And t. 00:07:04.690 --> 00:07:08.630 align:middle line:90% 00:07:08.630 --> 00:07:13.250 align:middle line:84% This is just a simple extension of this property. 00:07:13.250 --> 00:07:23.750 align:middle line:84% I guess you can write this thing equals sj plus tk times d. 00:07:23.750 --> 00:07:27.750 align:middle line:84% And it turns out this structure is interesting. 00:07:27.750 --> 00:07:32.360 align:middle line:84% It's what we call an integer linear combination. 00:07:32.360 --> 00:07:35.275 align:middle line:84% And this is a concept we will also be using throughout today. 00:07:35.275 --> 00:07:39.470 align:middle line:90% 00:07:39.470 --> 00:07:42.510 align:middle line:84% So much so that I'm going to give an acronym. 00:07:42.510 --> 00:07:46.413 align:middle line:84% I don't usually like acronyms, but this is so long, 00:07:46.413 --> 00:07:48.080 align:middle line:84% that it's going to save me a ton of work 00:07:48.080 --> 00:07:50.800 align:middle line:84% to write ILC for Integer Linear Combination. 00:07:50.800 --> 00:07:54.130 align:middle line:84% If you know linear algebra, this is just linear combination, 00:07:54.130 --> 00:07:56.470 align:middle line:84% but where the coefficients need to be integers. 00:07:56.470 --> 00:07:58.750 align:middle line:90% So it's exactly this notion. 00:07:58.750 --> 00:08:02.460 align:middle line:84% So an integer linear combination of a and b 00:08:02.460 --> 00:08:10.740 align:middle line:84% is some number s times a plus t times b for any s and t 00:08:10.740 --> 00:08:12.660 align:middle line:90% that are integers. 00:08:12.660 --> 00:08:14.340 align:middle line:84% So if I give you two numbers a and b, 00:08:14.340 --> 00:08:16.548 align:middle line:84% we'll only be doing integer linear combinations today 00:08:16.548 --> 00:08:21.430 align:middle line:84% of two numbers, then this is what they always look like. 00:08:21.430 --> 00:08:24.390 align:middle line:84% And given two numbers, there's some numbers you can construct 00:08:24.390 --> 00:08:27.780 align:middle line:84% via integer linear combinations and some you can't. 00:08:27.780 --> 00:08:30.240 align:middle line:84% Actually fun fact, divisibility is 00:08:30.240 --> 00:08:33.150 align:middle line:84% like integer linear combinations of one number. 00:08:33.150 --> 00:08:36.929 align:middle line:84% This is saying b is an integer linear combination of just a. 00:08:36.929 --> 00:08:40.530 align:middle line:84% But that's so special we call it divisibility. 00:08:40.530 --> 00:08:43.260 align:middle line:84% Two numbers is where integer linear combinations really 00:08:43.260 --> 00:08:45.120 align:middle line:90% get interesting. 00:08:45.120 --> 00:08:50.310 align:middle line:84% All right, so to motivate these concepts more, 00:08:50.310 --> 00:08:56.600 align:middle line:84% turns out these are key concepts to solving a famous puzzle, it's 00:08:56.600 --> 00:09:02.760 align:middle line:84% at least 80, 90 years old, called the water-pouring puzzle, 00:09:02.760 --> 00:09:07.610 align:middle line:84% but it became particularly famous through Die Hard 00:09:07.610 --> 00:09:09.360 align:middle line:90% 3, Die Hard with a Vengeance. 00:09:09.360 --> 00:09:11.130 align:middle line:84% So we're going to watch a little clip. 00:09:11.130 --> 00:09:15.240 align:middle line:84% And if this clip is not included in this video, 00:09:15.240 --> 00:09:19.730 align:middle line:84% you can go stream or purchase Die Hard with a Vengeance 00:09:19.730 --> 00:09:23.000 align:middle line:84% and start at the one-hour mark, and you will see an exchange 00:09:23.000 --> 00:09:23.970 align:middle line:90% like the following. 00:09:23.970 --> 00:09:26.150 align:middle line:84% So we have, of course, John McClane 00:09:26.150 --> 00:09:30.980 align:middle line:84% here, who's trying to save Manhattan along 00:09:30.980 --> 00:09:33.470 align:middle line:84% with his friend Zeus, who's an electrician who 00:09:33.470 --> 00:09:37.273 align:middle line:84% just happens to get brought into this whole chaos. 00:09:37.273 --> 00:09:38.690 align:middle line:84% Basically, the bad guys are trying 00:09:38.690 --> 00:09:41.940 align:middle line:84% to distract the good guys by forcing them to solve puzzles, 00:09:41.940 --> 00:09:43.128 align:middle line:90% so here we go. 00:09:43.128 --> 00:09:45.170 align:middle line:84% All right, so we're going to focus on the problem 00:09:45.170 --> 00:09:49.173 align:middle line:84% at hand, which we can formalize as follows. 00:09:49.173 --> 00:09:55.840 align:middle line:90% 00:09:55.840 --> 00:10:04.450 align:middle line:84% So we have two jugs, we'll say, of capacity a and b, 00:10:04.450 --> 00:10:05.990 align:middle line:84% and we'll assume they're integers. 00:10:05.990 --> 00:10:13.330 align:middle line:84% And in the example that saves Manhattan, we have a equals 5, 00:10:13.330 --> 00:10:15.400 align:middle line:90% and B equals 3. 00:10:15.400 --> 00:10:20.950 align:middle line:84% And we'd like to represent some number, in this case 4, 00:10:20.950 --> 00:10:25.780 align:middle line:84% by a sequence of pouring operations. 00:10:25.780 --> 00:10:29.950 align:middle line:84% So how could we represent this water-pouring problem 00:10:29.950 --> 00:10:33.850 align:middle line:84% using techniques that we have from previous lectures? 00:10:33.850 --> 00:10:34.665 align:middle line:90% Any ideas? 00:10:34.665 --> 00:10:40.000 align:middle line:90% 00:10:40.000 --> 00:10:41.600 align:middle line:90% I heard something. 00:10:41.600 --> 00:10:42.100 align:middle line:90% Yeah? 00:10:42.100 --> 00:10:43.142 align:middle line:90% AUDIENCE: State machines. 00:10:43.142 --> 00:10:44.630 align:middle line:84% Erik Demaine: State machines, good, 00:10:44.630 --> 00:10:47.030 align:middle line:90% so we'll use a state machine. 00:10:47.030 --> 00:10:52.130 align:middle line:84% And so we'll have a state x and y, 00:10:52.130 --> 00:10:57.950 align:middle line:84% means that there's currently x gallons in the first jug and y 00:10:57.950 --> 00:11:12.470 align:middle line:84% gallons in the second jug, or let's say the amounts in each 00:11:12.470 --> 00:11:16.400 align:middle line:90% jug/ 00:11:16.400 --> 00:11:24.500 align:middle line:84% So we start at the state 0,0, let's say. 00:11:24.500 --> 00:11:26.270 align:middle line:90% The jug started empty. 00:11:26.270 --> 00:11:28.610 align:middle line:84% We're given infinite water supply 00:11:28.610 --> 00:11:31.490 align:middle line:84% because they're at a fountain, conveniently. 00:11:31.490 --> 00:11:36.830 align:middle line:84% And let's see, we want to finish at a state that 00:11:36.830 --> 00:11:38.760 align:middle line:90% has 4 in one of the two values. 00:11:38.760 --> 00:11:42.350 align:middle line:84% And as John McClane said, it has to be in the bigger one, 00:11:42.350 --> 00:11:45.100 align:middle line:84% in this case, because 4 is bigger than 3. 00:11:45.100 --> 00:11:47.950 align:middle line:90% So we'd like to-- 00:11:47.950 --> 00:11:54.100 align:middle line:84% goal is to get to 4 comma something. 00:11:54.100 --> 00:11:57.080 align:middle line:84% And the tricky part is to write down the transitions here. 00:11:57.080 --> 00:12:06.190 align:middle line:84% So let's write down some of the transitions, transitions. 00:12:06.190 --> 00:12:13.420 align:middle line:84% So given x and y, what are some of the things I could do? 00:12:13.420 --> 00:12:15.580 align:middle line:90% Any suggestions? 00:12:15.580 --> 00:12:18.170 align:middle line:84% I have some partially filled jars. 00:12:18.170 --> 00:12:24.130 align:middle line:84% These are two jars of 5 and 3 pumpkins capacity, respectively. 00:12:24.130 --> 00:12:26.780 align:middle line:90% Maybe I'm at some position here. 00:12:26.780 --> 00:12:29.020 align:middle line:90% What could I do to these drugs? 00:12:29.020 --> 00:12:29.875 align:middle line:90% Yeah, in the back. 00:12:29.875 --> 00:12:30.880 align:middle line:90% AUDIENCE: Fill one to the top. 00:12:30.880 --> 00:12:32.297 align:middle line:84% Erik Demaine: Fill one to the top. 00:12:32.297 --> 00:12:33.350 align:middle line:90% I go to the fountain. 00:12:33.350 --> 00:12:36.100 align:middle line:84% I pour it up until it's completely filled, or this one, 00:12:36.100 --> 00:12:37.910 align:middle line:90% I could go to the filled point. 00:12:37.910 --> 00:12:39.110 align:middle line:90% Other suggestions? 00:12:39.110 --> 00:12:39.966 align:middle line:90% Yeah? 00:12:39.966 --> 00:12:40.758 align:middle line:90% AUDIENCE: Empty it. 00:12:40.758 --> 00:12:43.020 align:middle line:84% Erik Demaine: Empty, I could empty one of the jars. 00:12:43.020 --> 00:12:44.750 align:middle line:84% Well, hopefully I won't need that. 00:12:44.750 --> 00:12:49.460 align:middle line:84% So in the transitions, this corresponds to going to, 00:12:49.460 --> 00:12:53.780 align:middle line:84% I guess, a, y if I fill the first one. 00:12:53.780 --> 00:12:58.100 align:middle line:84% Or I could go to x, b if I fill the second one. 00:12:58.100 --> 00:13:04.260 align:middle line:84% I could empty by going to 0, 7 if I empty the first one. 00:13:04.260 --> 00:13:07.130 align:middle line:90% Or I could go to x, 0. 00:13:07.130 --> 00:13:08.980 align:middle line:90% So these are all-- 00:13:08.980 --> 00:13:09.480 align:middle line:90% sorry. 00:13:09.480 --> 00:13:13.640 align:middle line:84% This is all transitions from x, y. 00:13:13.640 --> 00:13:15.080 align:middle line:90% Anything else? 00:13:15.080 --> 00:13:16.130 align:middle line:90% Yeah? 00:13:16.130 --> 00:13:18.410 align:middle line:90% AUDIENCE: You could put them-- 00:13:18.410 --> 00:13:21.380 align:middle line:84% you could switch one the jugs in one of them to the other one. 00:13:21.380 --> 00:13:23.480 align:middle line:84% Erik Demaine: I could pour the contents of one jug 00:13:23.480 --> 00:13:25.800 align:middle line:90% into the other, cool, yeah. 00:13:25.800 --> 00:13:28.110 align:middle line:84% So here, for example, I filled up a 3. 00:13:28.110 --> 00:13:29.360 align:middle line:90% I poured in 3. 00:13:29.360 --> 00:13:33.090 align:middle line:84% I could fill this one up again and pour it in again. 00:13:33.090 --> 00:13:37.730 align:middle line:84% And at some point, I'll run out of capacity in this jug. 00:13:37.730 --> 00:13:40.120 align:middle line:84% And that's an event that I can detect when 00:13:40.120 --> 00:13:42.550 align:middle line:90% I reach level with this jug. 00:13:42.550 --> 00:13:46.210 align:middle line:84% So to write that down more carefully, given x, y, 00:13:46.210 --> 00:13:49.460 align:middle line:84% we can transition to one of two cases. 00:13:49.460 --> 00:13:53.500 align:middle line:84% Either we get to completely empty the left jug 00:13:53.500 --> 00:13:57.790 align:middle line:84% into the right jug, so this would be x plus y. 00:13:57.790 --> 00:14:01.880 align:middle line:84% But that's only possible if it fits in the second jug 00:14:01.880 --> 00:14:04.600 align:middle line:84% if x plus y is less than or equal to b. 00:14:04.600 --> 00:14:07.100 align:middle line:84% Otherwise-- I have to do arithmetic. 00:14:07.100 --> 00:14:08.440 align:middle line:90% This is too hard. 00:14:08.440 --> 00:14:14.260 align:middle line:84% --we move over b minus y or something. 00:14:14.260 --> 00:14:21.620 align:middle line:84% So we have x minus b minus y, b, otherwise. 00:14:21.620 --> 00:14:23.750 align:middle line:84% If we end up filling up the second one, 00:14:23.750 --> 00:14:27.880 align:middle line:84% then this is how much of the original x that I started with 00:14:27.880 --> 00:14:29.650 align:middle line:90% remains. 00:14:29.650 --> 00:14:33.850 align:middle line:84% And you can do the-- there's one more rule, which is 00:14:33.850 --> 00:14:35.360 align:middle line:90% if I pour from right to left. 00:14:35.360 --> 00:14:40.970 align:middle line:84% Those are available state transitions in this puzzle. 00:14:40.970 --> 00:14:43.140 align:middle line:84% And now maybe I've talked about it long enough. 00:14:43.140 --> 00:14:46.015 align:middle line:84% You have some intuition on how to solve this. 00:14:46.015 --> 00:14:48.650 align:middle line:90% 00:14:48.650 --> 00:14:53.275 align:middle line:84% Start empty, any suggestions on how to construct the number 3? 00:14:53.275 --> 00:14:56.270 align:middle line:90% 00:14:56.270 --> 00:14:57.770 align:middle line:90% Yeah. 00:14:57.770 --> 00:15:01.710 align:middle line:84% AUDIENCE: So you fill up the 5-gallon jug. 00:15:01.710 --> 00:15:04.550 align:middle line:84% Erik Demaine: Just fill up the 5-gallon jug, OK. 00:15:04.550 --> 00:15:07.492 align:middle line:84% AUDIENCE: And transfer it to the 3-gallon jug. 00:15:07.492 --> 00:15:09.950 align:middle line:84% Erik Demaine: OK, we're going to transfer the 3-gallon jug, 00:15:09.950 --> 00:15:11.910 align:middle line:84% until the 3-gallon jug is exactly empty, 00:15:11.910 --> 00:15:13.915 align:middle line:84% which if you're using water, it would be level. 00:15:13.915 --> 00:15:15.290 align:middle line:84% AUDIENCE: Empty the 3-gallon jug. 00:15:15.290 --> 00:15:17.300 align:middle line:84% Erik Demaine: OK, empty the 4-gallon jug. 00:15:17.300 --> 00:15:20.370 align:middle line:84% AUDIENCE: And transfer the 2 into the 4-gallon jug. 00:15:20.370 --> 00:15:21.990 align:middle line:84% Erik Demaine: OK, move the 2 over. 00:15:21.990 --> 00:15:24.530 align:middle line:84% This now feels like Towers of Hanoi, OK. 00:15:24.530 --> 00:15:25.520 align:middle line:90% AUDIENCE: Fill the 5. 00:15:25.520 --> 00:15:27.150 align:middle line:90% Erik Demaine: Fill the 5. 00:15:27.150 --> 00:15:29.060 align:middle line:90% AUDIENCE: And then transfer. 00:15:29.060 --> 00:15:33.600 align:middle line:84% Erik Demaine: And then transfer like this from 5 to-- 00:15:33.600 --> 00:15:35.810 align:middle line:90% did I get the right number? 00:15:35.810 --> 00:15:36.450 align:middle line:90% There we go. 00:15:36.450 --> 00:15:38.780 align:middle line:84% Water's particularly unsettled today. 00:15:38.780 --> 00:15:41.510 align:middle line:84% All right, so we've got 5 on the right if I can count. 00:15:41.510 --> 00:15:44.750 align:middle line:84% And we fill over here, until this one is filled. 00:15:44.750 --> 00:15:46.716 align:middle line:84% And now we've got the number 4, yay. 00:15:46.716 --> 00:15:49.750 align:middle line:90% 00:15:49.750 --> 00:15:55.730 align:middle line:84% That's how we save Manhattan with this water-pouring problem 00:15:55.730 --> 00:16:01.900 align:middle line:84% but while the bad guys are off stealing money from the bank. 00:16:01.900 --> 00:16:04.040 align:middle line:84% But sometimes this puzzle isn't solvable, 00:16:04.040 --> 00:16:05.603 align:middle line:90% so let's take another example. 00:16:05.603 --> 00:16:11.140 align:middle line:90% 00:16:11.140 --> 00:16:16.340 align:middle line:84% Let's say we want a equals 6, b equals 4, 00:16:16.340 --> 00:16:20.530 align:middle line:90% and we want the number 3. 00:16:20.530 --> 00:16:21.377 align:middle line:90% Is this possible? 00:16:21.377 --> 00:16:23.210 align:middle line:84% I already gave away that this is impossible. 00:16:23.210 --> 00:16:25.480 align:middle line:90% Why is it impossible? 00:16:25.480 --> 00:16:28.390 align:middle line:90% Any feelings? 00:16:28.390 --> 00:16:31.540 align:middle line:84% Hint, it has something to do with this stuff. 00:16:31.540 --> 00:16:33.130 align:middle line:90% Yeah. 00:16:33.130 --> 00:16:35.760 align:middle line:84% AUDIENCE: There are no integer linear combinations. 00:16:35.760 --> 00:16:38.390 align:middle line:84% Is it possible that there were 6 and 4 [INAUDIBLE] 00:16:38.390 --> 00:16:39.890 align:middle line:84% Erik Demaine: OK, you said there are 00:16:39.890 --> 00:16:43.770 align:middle line:84% no integer linear combinations where of 6 and 4 that make 3. 00:16:43.770 --> 00:16:44.790 align:middle line:90% That is true. 00:16:44.790 --> 00:16:47.360 align:middle line:84% But this problem isn't directly about integer linear 00:16:47.360 --> 00:16:48.800 align:middle line:90% combinations, is it? 00:16:48.800 --> 00:16:49.435 align:middle line:90% I don't know. 00:16:49.435 --> 00:16:54.260 align:middle line:90% 00:16:54.260 --> 00:16:55.400 align:middle line:90% Yeah? 00:16:55.400 --> 00:16:58.040 align:middle line:84% AUDIENCE: There's no way to get an odd number out of this. 00:16:58.040 --> 00:17:01.683 align:middle line:84% Erik Demaine: There's no way to get an odd number from these two 00:17:01.683 --> 00:17:02.183 align:middle line:90% gallons. 00:17:02.183 --> 00:17:04.849 align:middle line:90% 00:17:04.849 --> 00:17:07.069 align:middle line:90% Both of you are correct. 00:17:07.069 --> 00:17:10.819 align:middle line:84% The simpler reason, in this case, 00:17:10.819 --> 00:17:12.883 align:middle line:84% is it's impossible to get an odd amount. 00:17:12.883 --> 00:17:15.640 align:middle line:90% 00:17:15.640 --> 00:17:17.390 align:middle line:84% I guess, the proof of both of these things 00:17:17.390 --> 00:17:19.829 align:middle line:84% is you look at all the possible transitions, 00:17:19.829 --> 00:17:24.200 align:middle line:84% and you observe that if you start with a and b being even, 00:17:24.200 --> 00:17:26.210 align:middle line:84% you can never construct an odd number 00:17:26.210 --> 00:17:27.960 align:middle line:90% by adding and subtracting. 00:17:27.960 --> 00:17:29.810 align:middle line:84% All we're doing here are taking numbers 00:17:29.810 --> 00:17:32.480 align:middle line:84% that we've constructed, adding them and/or subtracting them 00:17:32.480 --> 00:17:34.370 align:middle line:90% with a and b or each other. 00:17:34.370 --> 00:17:36.080 align:middle line:84% And so if you start with even numbers, 00:17:36.080 --> 00:17:37.660 align:middle line:84% you add and subtract them, you always 00:17:37.660 --> 00:17:39.260 align:middle line:90% stay within the even numbers. 00:17:39.260 --> 00:17:41.050 align:middle line:84% But also, as you were saying, if you 00:17:41.050 --> 00:17:43.570 align:middle line:84% start with an integer linear combination of a and b, 00:17:43.570 --> 00:17:46.480 align:middle line:84% all you can get are integer linear combinations of a and b. 00:17:46.480 --> 00:17:51.610 align:middle line:84% So here is a key lemma about this puzzle 00:17:51.610 --> 00:17:56.110 align:middle line:84% and why that is impossible, more generally, 00:17:56.110 --> 00:18:07.240 align:middle line:84% is all reachable amounts in this puzzle, are integer 00:18:07.240 --> 00:18:12.010 align:middle line:90% linear combinations of a and b. 00:18:12.010 --> 00:18:13.125 align:middle line:90% Let's prove this. 00:18:13.125 --> 00:18:16.450 align:middle line:90% 00:18:16.450 --> 00:18:19.290 align:middle line:84% Any suggestions on how to prove a property like this? 00:18:19.290 --> 00:18:22.495 align:middle line:90% 00:18:22.495 --> 00:18:24.620 align:middle line:84% If you remember all the way back to state machines, 00:18:24.620 --> 00:18:26.480 align:middle line:84% hopefully you've been studying for the test. 00:18:26.480 --> 00:18:27.860 align:middle line:90% State machines are on the test. 00:18:27.860 --> 00:18:29.030 align:middle line:90% This lecture is not. 00:18:29.030 --> 00:18:32.155 align:middle line:90% That will be on quiz 2. 00:18:32.155 --> 00:18:33.030 align:middle line:90% AUDIENCE: Invariance. 00:18:33.030 --> 00:18:35.956 align:middle line:84% Erik Demaine: Invariance, yeah, in particular-- 00:18:35.956 --> 00:18:38.940 align:middle line:90% 00:18:38.940 --> 00:18:41.290 align:middle line:84% this is a statement about invariance. 00:18:41.290 --> 00:18:50.910 align:middle line:84% This is saying, for all reachable x, y, x and y 00:18:50.910 --> 00:18:55.770 align:middle line:84% are integer linear combinations of a and b. 00:18:55.770 --> 00:18:58.750 align:middle line:84% So this is an invariant property. 00:18:58.750 --> 00:19:00.070 align:middle line:90% This is a state predicate. 00:19:00.070 --> 00:19:01.612 align:middle line:84% We want to prove that it's invariant, 00:19:01.612 --> 00:19:04.830 align:middle line:84% that it's true for all reachable states starting from 0, 0. 00:19:04.830 --> 00:19:11.490 align:middle line:84% And we usually do that using the invariant principle which 00:19:11.490 --> 00:19:13.240 align:middle line:90% requires two steps. 00:19:13.240 --> 00:19:15.450 align:middle line:90% One is base case. 00:19:15.450 --> 00:19:19.290 align:middle line:84% We need to check that this predicate holds initially. 00:19:19.290 --> 00:19:22.810 align:middle line:84% And the initial state for us is 0, 0. 00:19:22.810 --> 00:19:26.250 align:middle line:84% And we observe that 0 is indeed, a linear combination 00:19:26.250 --> 00:19:32.530 align:middle line:84% of a and b, namely 0 times a plus 0 times b, great. 00:19:32.530 --> 00:19:33.920 align:middle line:90% Base case is true. 00:19:33.920 --> 00:19:36.280 align:middle line:84% And so now we just need the inductive 00:19:36.280 --> 00:19:39.280 align:middle line:84% step, or the I guess, the preserved, the fact 00:19:39.280 --> 00:19:40.926 align:middle line:84% that this predicate is preserved. 00:19:40.926 --> 00:19:44.230 align:middle line:90% 00:19:44.230 --> 00:19:47.320 align:middle line:84% So suppose that we're at some state 00:19:47.320 --> 00:19:53.890 align:middle line:84% x and y, where the predicate is true, so x equals sa plus tb, 00:19:53.890 --> 00:19:59.590 align:middle line:84% and y equals s prime a plus t prime b. 00:19:59.590 --> 00:20:03.710 align:middle line:84% And then we just look at all of the available transitions. 00:20:03.710 --> 00:20:07.120 align:middle line:84% And we show that every number that you can make, 00:20:07.120 --> 00:20:13.390 align:middle line:84% is indeed an integer linear combination of a and b. 00:20:13.390 --> 00:20:14.870 align:middle line:90% So let's do one example. 00:20:14.870 --> 00:20:17.520 align:middle line:90% We can construct x plus y. 00:20:17.520 --> 00:20:26.650 align:middle line:84% x plus y, we can rewrite as s plus s prime times a and t 00:20:26.650 --> 00:20:29.380 align:middle line:90% plus t prime times b. 00:20:29.380 --> 00:20:33.185 align:middle line:84% So that's if this transition happens to be possible, 00:20:33.185 --> 00:20:35.560 align:middle line:84% it may not be possible because there's this if condition. 00:20:35.560 --> 00:20:38.962 align:middle line:84% But if we show that every number we could possibly build 00:20:38.962 --> 00:20:40.420 align:middle line:84% has this property, then, of course, 00:20:40.420 --> 00:20:42.510 align:middle line:84% all the ones that we can actually reach also 00:20:42.510 --> 00:20:44.250 align:middle line:90% have this property. 00:20:44.250 --> 00:20:51.920 align:middle line:84% So x plus y is indeed an integer linear combination of a and b, 00:20:51.920 --> 00:20:52.420 align:middle line:90% et cetera. 00:20:52.420 --> 00:20:55.890 align:middle line:90% 00:20:55.890 --> 00:20:58.878 align:middle line:84% OK, I don't want to prove it for every single transition 00:20:58.878 --> 00:21:00.670 align:middle line:84% here, just because there's a bunch of them. 00:21:00.670 --> 00:21:03.087 align:middle line:84% But for example, a is indeed an integer linear combination 00:21:03.087 --> 00:21:03.610 align:middle line:90% of a and b. 00:21:03.610 --> 00:21:05.500 align:middle line:90% It's 1 times a plus 0 times b. 00:21:05.500 --> 00:21:06.940 align:middle line:90% This one looks messier. 00:21:06.940 --> 00:21:09.600 align:middle line:84% It's actually just x plus y minus b. 00:21:09.600 --> 00:21:10.960 align:middle line:90% So it's almost the same. 00:21:10.960 --> 00:21:12.550 align:middle line:90% You just add a minus 1 here. 00:21:12.550 --> 00:21:14.370 align:middle line:84% But the general property we're using here 00:21:14.370 --> 00:21:16.410 align:middle line:84% is, that all of the constructible numbers 00:21:16.410 --> 00:21:21.360 align:middle line:84% here are integer linear combinations of x, y a and b. 00:21:21.360 --> 00:21:22.360 align:middle line:90% OK, I lied. 00:21:22.360 --> 00:21:25.060 align:middle line:84% We're using integer linear combinations of four things. 00:21:25.060 --> 00:21:28.430 align:middle line:84% But all of the x and y's are also integer linear combinations 00:21:28.430 --> 00:21:29.010 align:middle line:90% of a and b. 00:21:29.010 --> 00:21:31.610 align:middle line:84% So what we're really taking is integer linear combinations 00:21:31.610 --> 00:21:34.530 align:middle line:84% of integer linear combinations of a and b. 00:21:34.530 --> 00:21:37.100 align:middle line:84% And it turns out that if you take 00:21:37.100 --> 00:21:40.880 align:middle line:84% two integer linear combinations of a and b like, x and y, 00:21:40.880 --> 00:21:43.130 align:middle line:84% and you add them together, or in general 00:21:43.130 --> 00:21:44.940 align:middle line:84% take an integer linear combination of them, 00:21:44.940 --> 00:21:48.150 align:middle line:84% you still get an integer linear combination of a and b. 00:21:48.150 --> 00:21:49.620 align:middle line:90% As you can see from this proof. 00:21:49.620 --> 00:21:51.287 align:middle line:84% It's actually pretty much the same proof 00:21:51.287 --> 00:21:52.700 align:middle line:90% we were doing over here. 00:21:52.700 --> 00:21:55.910 align:middle line:84% This was integer linear combinations of one number, d. 00:21:55.910 --> 00:21:58.070 align:middle line:84% And divisibility was preserved even 00:21:58.070 --> 00:22:00.900 align:middle line:84% when you took integer linear combinations, so same idea. 00:22:00.900 --> 00:22:03.390 align:middle line:84% And for any puzzle where you have that property, 00:22:03.390 --> 00:22:04.912 align:middle line:84% you can only construct numbers that 00:22:04.912 --> 00:22:06.870 align:middle line:84% are integer and linear combinations of a and b. 00:22:06.870 --> 00:22:10.610 align:middle line:84% Now, for this puzzle, you can actually construct all of them 00:22:10.610 --> 00:22:13.010 align:middle line:90% that fit in the jars. 00:22:13.010 --> 00:22:16.860 align:middle line:84% So I'll just mention this as a claim. 00:22:16.860 --> 00:22:21.140 align:middle line:84% And if we have time at the end, we'll come back to it. 00:22:21.140 --> 00:22:38.610 align:middle line:84% We can obtain a value x if and only if, let's say, 00:22:38.610 --> 00:22:44.400 align:middle line:84% x is an integer linear combination of a and b 00:22:44.400 --> 00:22:50.340 align:middle line:84% and 0 is less than or equal to x, is less than 00:22:50.340 --> 00:22:55.090 align:middle line:90% or equal to max of a and b. 00:22:55.090 --> 00:23:00.510 align:middle line:84% So we might try to prove that later. 00:23:00.510 --> 00:23:03.300 align:middle line:84% So now the central question is, how 00:23:03.300 --> 00:23:06.780 align:middle line:84% do we know whether one number is an integer linear combination 00:23:06.780 --> 00:23:07.710 align:middle line:90% of two other numbers? 00:23:07.710 --> 00:23:12.030 align:middle line:84% And when it is, can we actually construct the values s And t, 00:23:12.030 --> 00:23:15.130 align:middle line:90% so that sa plus tb equals x? 00:23:15.130 --> 00:23:17.250 align:middle line:84% Because that's what we needed to solve 00:23:17.250 --> 00:23:19.860 align:middle line:90% this puzzle was some way to-- 00:23:19.860 --> 00:23:21.900 align:middle line:84% we were only given the capacities 00:23:21.900 --> 00:23:23.830 align:middle line:84% of the jars and the target value. 00:23:23.830 --> 00:23:25.735 align:middle line:90% And now we want to represent-- 00:23:25.735 --> 00:23:27.110 align:middle line:84% let's check that this is actually 00:23:27.110 --> 00:23:28.713 align:middle line:90% possible for this example. 00:23:28.713 --> 00:23:36.380 align:middle line:90% 00:23:36.380 --> 00:23:39.030 align:middle line:84% Sorry, not this example, the one that's actually solvable. 00:23:39.030 --> 00:23:45.570 align:middle line:84% So for this solvable puzzle, 4 equals 2 times 5, 00:23:45.570 --> 00:23:48.990 align:middle line:90% that's 10, minus 2 times 3. 00:23:48.990 --> 00:23:51.650 align:middle line:90% That's 6. 00:23:51.650 --> 00:23:53.160 align:middle line:90% 10 minus 6 is 4. 00:23:53.160 --> 00:23:55.740 align:middle line:84% So that's intuitively why this puzzle is solvable. 00:23:55.740 --> 00:23:57.230 align:middle line:90% Although that's not a proof. 00:23:57.230 --> 00:23:58.790 align:middle line:84% I mean, we just saw that it wasn't 00:23:58.790 --> 00:24:00.680 align:middle line:90% limited by this same issue. 00:24:00.680 --> 00:24:03.920 align:middle line:84% So for example, if we started with 6 and 4, 00:24:03.920 --> 00:24:10.140 align:middle line:84% we know by this lemma, because 2 divides 4 and 2 divides 6, 00:24:10.140 --> 00:24:12.380 align:middle line:84% 2 will divide any integer linear combination, 00:24:12.380 --> 00:24:17.400 align:middle line:84% and therefore, any reachable value in this puzzle. 00:24:17.400 --> 00:24:19.190 align:middle line:90% And 2 does not divide 3. 00:24:19.190 --> 00:24:20.580 align:middle line:90% 3 is odd, not even. 00:24:20.580 --> 00:24:22.460 align:middle line:84% And so we will never be able to reach 3 00:24:22.460 --> 00:24:24.310 align:middle line:84% because all the reachable states are even. 00:24:24.310 --> 00:24:28.710 align:middle line:84% That's a corollary now of this lemma, OK. 00:24:28.710 --> 00:24:32.550 align:middle line:84% OK, but in the good case, and the bad case 00:24:32.550 --> 00:24:35.850 align:middle line:84% is easy to see, I guess, how do we know whether we're 00:24:35.850 --> 00:24:37.120 align:middle line:90% in a good case or a bad case? 00:24:37.120 --> 00:24:39.662 align:middle line:84% How do we know whether x is a integer linear combination of a 00:24:39.662 --> 00:24:41.680 align:middle line:90% and b, given x and a and b? 00:24:41.680 --> 00:24:45.690 align:middle line:84% That is the purpose of-- well, some motivation 00:24:45.690 --> 00:24:46.900 align:middle line:90% for the rest of this lecture. 00:24:46.900 --> 00:24:49.710 align:middle line:84% We can use it for all sorts of things. 00:24:49.710 --> 00:24:53.880 align:middle line:84% All right, and I claim that this is going to be true if 00:24:53.880 --> 00:24:59.660 align:middle line:84% and only if this property is true. 00:24:59.660 --> 00:25:05.160 align:middle line:90% 00:25:05.160 --> 00:25:08.020 align:middle line:84% So now I'm going to restate this in terms of divisibility. 00:25:08.020 --> 00:25:10.620 align:middle line:84% This equivalence is not obvious, so we'll 00:25:10.620 --> 00:25:12.750 align:middle line:90% prove it by the end of today. 00:25:12.750 --> 00:25:14.760 align:middle line:84% So what we really want to show is, 00:25:14.760 --> 00:25:16.860 align:middle line:84% that every common divisor of a and b-- 00:25:16.860 --> 00:25:18.780 align:middle line:84% remember this is our definition of d 00:25:18.780 --> 00:25:26.300 align:middle line:84% is a common divisor, a common divisor of a and b, divides a 00:25:26.300 --> 00:25:30.320 align:middle line:84% and it divides b --then it should also divide x. 00:25:30.320 --> 00:25:34.980 align:middle line:84% So in the example, we saw that 2 was a common divisor of 4 and 6, 00:25:34.980 --> 00:25:38.660 align:middle line:84% and so 2 better divide anything that we make. 00:25:38.660 --> 00:25:40.810 align:middle line:84% What are the common divisors of 5 and 3? 00:25:40.810 --> 00:25:43.400 align:middle line:90% 00:25:43.400 --> 00:25:44.185 align:middle line:90% 3 doesn't work. 00:25:44.185 --> 00:25:44.810 align:middle line:90% 5 doesn't work. 00:25:44.810 --> 00:25:45.800 align:middle line:90% 2 doesn't work. 00:25:45.800 --> 00:25:47.010 align:middle line:90% 3 doesn't work. 00:25:47.010 --> 00:25:49.520 align:middle line:90% I guess 1 is the common divisor. 00:25:49.520 --> 00:25:52.700 align:middle line:90% And 1 divides everything. 00:25:52.700 --> 00:25:59.930 align:middle line:84% So yeah, this is true in our good case. 00:25:59.930 --> 00:26:02.750 align:middle line:84% And this, we can rewrite in yet another way, 00:26:02.750 --> 00:26:05.720 align:middle line:90% which motivates our next topic. 00:26:05.720 --> 00:26:10.393 align:middle line:90% The GCD of a and b divides x. 00:26:10.393 --> 00:26:11.810 align:middle line:84% So I claim, in particular, this is 00:26:11.810 --> 00:26:13.520 align:middle line:84% going to be a way to tell whether x 00:26:13.520 --> 00:26:15.380 align:middle line:84% is an integer linear combination of a and b. 00:26:15.380 --> 00:26:17.660 align:middle line:84% You compute something called the GCD of a and b, 00:26:17.660 --> 00:26:20.260 align:middle line:84% and you see whether it divides x. 00:26:20.260 --> 00:26:21.675 align:middle line:90% So let's talk about GCD. 00:26:21.675 --> 00:26:36.640 align:middle line:90% 00:26:36.640 --> 00:26:38.390 align:middle line:90% GCD is another acronym. 00:26:38.390 --> 00:26:41.220 align:middle line:84% It stands for Greatest Common Divisor. 00:26:41.220 --> 00:26:46.240 align:middle line:90% 00:26:46.240 --> 00:26:53.710 align:middle line:84% I assume it comes from Britain, in particular, 00:26:53.710 --> 00:26:57.490 align:middle line:84% where they use "great" to indicate large. 00:26:57.490 --> 00:27:00.700 align:middle line:84% So we might call this the largest common divisor 00:27:00.700 --> 00:27:04.065 align:middle line:84% or the biggest common divisor, all synonyms for the same thing. 00:27:04.065 --> 00:27:11.350 align:middle line:90% 00:27:11.350 --> 00:27:15.520 align:middle line:84% I'm going to give two equivalent definitions of GCD. 00:27:15.520 --> 00:27:18.850 align:middle line:84% To me, the more intuitive one is this. 00:27:18.850 --> 00:27:29.690 align:middle line:90% 00:27:29.690 --> 00:27:33.620 align:middle line:84% But it's a little bit uglier because it has a special case. 00:27:33.620 --> 00:27:36.350 align:middle line:84% So what I'd like to say, is the GCD of a and b, 00:27:36.350 --> 00:27:40.190 align:middle line:84% is the largest integer d such that d divides a and d divides b 00:27:40.190 --> 00:27:43.250 align:middle line:84% And again, this is common divisor. 00:27:43.250 --> 00:27:45.170 align:middle line:84% d is a common divisor of a and b. 00:27:45.170 --> 00:27:47.270 align:middle line:84% And I want it to be the largest common divisor. 00:27:47.270 --> 00:27:48.890 align:middle line:90% That would be GCD. 00:27:48.890 --> 00:27:53.000 align:middle line:84% But there's one annoying special case, which really won't matter 00:27:53.000 --> 00:27:55.530 align:middle line:84% for the rest of the lecture, but to be technically correct, 00:27:55.530 --> 00:27:59.070 align:middle line:84% we should say, what happens when you compute the GCD of 0 and 0? 00:27:59.070 --> 00:28:03.030 align:middle line:84% Because remember, every integer d divides 0. 00:28:03.030 --> 00:28:04.380 align:middle line:90% Everything divides 0. 00:28:04.380 --> 00:28:08.390 align:middle line:84% So what's the largest divisor of 0 and 0? 00:28:08.390 --> 00:28:09.830 align:middle line:90% I mean, infinity? 00:28:09.830 --> 00:28:10.410 align:middle line:90% I don't know. 00:28:10.410 --> 00:28:13.380 align:middle line:84% It's like there is no such integer that's the largest. 00:28:13.380 --> 00:28:15.470 align:middle line:90% So there is no largest integer. 00:28:15.470 --> 00:28:17.590 align:middle line:84% So in this case, it turns out the right thing is 00:28:17.590 --> 00:28:19.880 align:middle line:90% to define GCD of 0, 0 to be 0. 00:28:19.880 --> 00:28:22.450 align:middle line:84% That will be a little counterintuitive maybe, 00:28:22.450 --> 00:28:27.160 align:middle line:84% until you see the next definition. 00:28:27.160 --> 00:28:31.300 align:middle line:90% So here's another definition. 00:28:31.300 --> 00:28:38.260 align:middle line:84% GCD is the only non-negative common divisor 00:28:38.260 --> 00:28:48.090 align:middle line:84% of a and b divisible by all common divisors of a and b. 00:28:48.090 --> 00:29:00.670 align:middle line:90% 00:29:00.670 --> 00:29:05.080 align:middle line:84% So another way to think about the greatest common divisor, 00:29:05.080 --> 00:29:08.620 align:middle line:84% is it's greatest in terms of the divisibility relation, 00:29:08.620 --> 00:29:12.340 align:middle line:84% the one that is the most divisible 00:29:12.340 --> 00:29:15.290 align:middle line:90% by other common divisors. 00:29:15.290 --> 00:29:18.490 align:middle line:84% So it turns out there's going to be one common divisor that's 00:29:18.490 --> 00:29:22.360 align:middle line:84% both non-negative and is divisible by all common divisors 00:29:22.360 --> 00:29:23.920 align:middle line:90% of a and b. 00:29:23.920 --> 00:29:31.390 align:middle line:84% So in these examples, for 4 and 6, GCD of 6 and 4, 00:29:31.390 --> 00:29:32.955 align:middle line:90% is going to be 2. 00:29:32.955 --> 00:29:35.950 align:middle line:90% 00:29:35.950 --> 00:29:37.930 align:middle line:90% Because the only-- let's see. 00:29:37.930 --> 00:29:43.090 align:middle line:84% The common divisors, what are all the common divisors 00:29:43.090 --> 00:29:46.645 align:middle line:90% between 4 and 6? 00:29:46.645 --> 00:29:49.190 align:middle line:84% It may be obvious that there's two. 00:29:49.190 --> 00:29:53.590 align:middle line:84% There's minus 2, there's 1 and minus 1. 00:29:53.590 --> 00:29:56.000 align:middle line:90% 0 divides nothing except itself. 00:29:56.000 --> 00:29:57.790 align:middle line:84% And you can check all the numbers 00:29:57.790 --> 00:29:59.800 align:middle line:90% between negative 6 and 6. 00:29:59.800 --> 00:30:01.990 align:middle line:90% Those are the possible divisors. 00:30:01.990 --> 00:30:05.960 align:middle line:84% And these are the only ones that divide both a and b. 00:30:05.960 --> 00:30:10.490 align:middle line:84% And the largest non-negative 1 here is 2. 00:30:10.490 --> 00:30:13.270 align:middle line:84% And also it has the property that all 00:30:13.270 --> 00:30:17.380 align:middle line:84% of these other common divisors divide 2. 00:30:17.380 --> 00:30:19.890 align:middle line:84% So whichever way you want to think about it, 00:30:19.890 --> 00:30:21.140 align:middle line:90% you can use either definition. 00:30:21.140 --> 00:30:22.600 align:middle line:84% I'm not going to prove that these definitions are 00:30:22.600 --> 00:30:23.060 align:middle line:90% equivalent. 00:30:23.060 --> 00:30:24.727 align:middle line:84% We're just going to take it as an axiom. 00:30:24.727 --> 00:30:26.440 align:middle line:90% But you can prove it. 00:30:26.440 --> 00:30:29.350 align:middle line:90% This may be an exercise. 00:30:29.350 --> 00:30:31.808 align:middle line:84% But when we're proving things about GCD, 00:30:31.808 --> 00:30:33.850 align:middle line:84% we can use either of these definitions, whichever 00:30:33.850 --> 00:30:35.530 align:middle line:90% is more convenient. 00:30:35.530 --> 00:30:44.950 align:middle line:84% So let's do one more interesting example, 00:30:44.950 --> 00:30:54.280 align:middle line:84% which is GCD of, I think, I want to do 0 first, 0, a. 00:30:54.280 --> 00:30:57.637 align:middle line:84% So suppose I want to compute GCD of two numbers. 00:30:57.637 --> 00:30:58.845 align:middle line:90% Maybe I should put a b there. 00:30:58.845 --> 00:31:01.780 align:middle line:90% 00:31:01.780 --> 00:31:04.870 align:middle line:90% And one of them is 0. 00:31:04.870 --> 00:31:08.780 align:middle line:84% Any guesses what the GCD of 0 and b should be? 00:31:08.780 --> 00:31:10.455 align:middle line:90% Remember, everything divides 0. 00:31:10.455 --> 00:31:20.170 align:middle line:90% 00:31:20.170 --> 00:31:20.670 align:middle line:90% AUDIENCE: b. 00:31:20.670 --> 00:31:24.540 align:middle line:84% Erik Demaine: b, it's a good guess, almost right. 00:31:24.540 --> 00:31:26.525 align:middle line:90% What if b was negative 7? 00:31:26.525 --> 00:31:31.350 align:middle line:90% 00:31:31.350 --> 00:31:33.270 align:middle line:84% This should be absolute value of b. 00:31:33.270 --> 00:31:36.540 align:middle line:84% So if we want to handle positive and negative things, 00:31:36.540 --> 00:31:38.610 align:middle line:84% this turns out to be the right answer. 00:31:38.610 --> 00:31:41.927 align:middle line:90% Here is a mini proof. 00:31:41.927 --> 00:31:43.510 align:middle line:84% There are two things we want to check. 00:31:43.510 --> 00:31:48.210 align:middle line:84% One is that absolute value of b is a common divisor of 0 and b. 00:31:48.210 --> 00:31:50.290 align:middle line:84% So well, the absolute value of b-- 00:31:50.290 --> 00:31:53.710 align:middle line:84% this gets notationally pretty gross --divides 0. 00:31:53.710 --> 00:31:56.340 align:middle line:84% That's true because everything divides 0. 00:31:56.340 --> 00:32:00.280 align:middle line:84% And absolute value of b divides b, namely, 00:32:00.280 --> 00:32:03.000 align:middle line:84% when we let k be negative 1 or 1, 00:32:03.000 --> 00:32:07.325 align:middle line:84% depending on whether b was positive or negative or 0. 00:32:07.325 --> 00:32:09.870 align:middle line:90% 00:32:09.870 --> 00:32:12.840 align:middle line:84% Cool, so good it is a common divisor. 00:32:12.840 --> 00:32:17.260 align:middle line:90% 00:32:17.260 --> 00:32:19.048 align:middle line:84% And now I want to claim that it's 00:32:19.048 --> 00:32:20.590 align:middle line:84% the greatest common divisor according 00:32:20.590 --> 00:32:22.340 align:middle line:90% to the second definition. 00:32:22.340 --> 00:32:24.130 align:middle line:84% So let's suppose that we had some number 00:32:24.130 --> 00:32:30.850 align:middle line:84% d which divides 0 and which divides b. 00:32:30.850 --> 00:32:32.750 align:middle line:90% So d divides 0, and d divides b. 00:32:32.750 --> 00:32:36.430 align:middle line:84% Then I claim, in particular, d also 00:32:36.430 --> 00:32:40.930 align:middle line:90% divides the absolute value of b. 00:32:40.930 --> 00:32:43.780 align:middle line:84% Because if you divide a number, you also divide its negation, 00:32:43.780 --> 00:32:47.050 align:middle line:84% just by negating the value k that you're multiplying by. 00:32:47.050 --> 00:32:49.060 align:middle line:84% If I take k times d, and I get b, 00:32:49.060 --> 00:32:52.540 align:middle line:84% then I take negative k times d, And I get absolute value of b 00:32:52.540 --> 00:32:54.790 align:middle line:90% if b happened to be negative. 00:32:54.790 --> 00:32:58.418 align:middle line:84% OK, so that means if you're a common divisor of 0 00:32:58.418 --> 00:33:00.710 align:middle line:84% and b, then you're also a common divisor of this thing, 00:33:00.710 --> 00:33:03.850 align:middle line:84% and that was our second definition of GCD. 00:33:03.850 --> 00:33:07.020 align:middle line:84% So great, this means your greatest. 00:33:07.020 --> 00:33:10.470 align:middle line:90% 00:33:10.470 --> 00:33:16.930 align:middle line:84% OK, now the question arises, how do we compute GCD of a and b? 00:33:16.930 --> 00:33:22.710 align:middle line:84% Because I claimed, over here, that what we need to check 00:33:22.710 --> 00:33:25.420 align:middle line:90% is whether GCD of ab divides x. 00:33:25.420 --> 00:33:28.510 align:middle line:84% So given a and b, we need to be able to compute this thing. 00:33:28.510 --> 00:33:30.310 align:middle line:84% How do we find the greatest common divisor? 00:33:30.310 --> 00:33:32.310 align:middle line:84% It turns out there's a great algorithm to do it. 00:33:32.310 --> 00:33:35.620 align:middle line:84% And the base case of that algorithm is this. 00:33:35.620 --> 00:33:38.320 align:middle line:84% If we can get one of the numbers down to 0, 00:33:38.320 --> 00:33:41.550 align:middle line:90% then we found the GCD. 00:33:41.550 --> 00:33:44.310 align:middle line:84% So our goal now is given some big numbers, 00:33:44.310 --> 00:33:46.980 align:middle line:90% to make them smaller somehow. 00:33:46.980 --> 00:33:57.210 align:middle line:84% And the big idea here to make a GCD algorithm, is subtraction. 00:33:57.210 --> 00:33:59.700 align:middle line:84% How do we make numbers smaller with arithmetic? 00:33:59.700 --> 00:34:01.750 align:middle line:84% Subtraction or division seem like good ideas. 00:34:01.750 --> 00:34:05.760 align:middle line:84% We're going to use both, starting with subtraction. 00:34:05.760 --> 00:34:14.975 align:middle line:84% So I claim GCD of ab equals GCD of a minus b. 00:34:14.975 --> 00:34:21.430 align:middle line:90% b for all integers ab. 00:34:21.430 --> 00:34:24.170 align:middle line:90% 00:34:24.170 --> 00:34:27.137 align:middle line:84% So if a and b are big numbers, this is pretty cool. 00:34:27.137 --> 00:34:29.179 align:middle line:84% This is saying we can subtract one of the numbers 00:34:29.179 --> 00:34:32.780 align:middle line:84% from the other, and we get the same GCD somehow. 00:34:32.780 --> 00:34:37.670 align:middle line:84% We'll prove this in a second, well, now, actually. 00:34:37.670 --> 00:34:39.060 align:middle line:84% I'll leave a little bit of space. 00:34:39.060 --> 00:34:40.944 align:middle line:90% Let's do a proof. 00:34:40.944 --> 00:34:45.650 align:middle line:90% 00:34:45.650 --> 00:34:49.280 align:middle line:84% So then we're going to turn this into an algorithm. 00:34:49.280 --> 00:34:51.210 align:middle line:84% But let's first see why this is the case. 00:34:51.210 --> 00:34:58.050 align:middle line:84% So this is a statement about GCD. 00:34:58.050 --> 00:35:01.670 align:middle line:84% GCD is this complicated definition. 00:35:01.670 --> 00:35:05.760 align:middle line:84% It's a little easier to think about all common divisors. 00:35:05.760 --> 00:35:08.160 align:middle line:84% If you think about the common divisors of a and b, 00:35:08.160 --> 00:35:09.720 align:middle line:84% those are all the numbers d, where 00:35:09.720 --> 00:35:11.220 align:middle line:84% d divides a and d divides b, and you 00:35:11.220 --> 00:35:13.900 align:middle line:84% think of the common divisors of a minus b and b, 00:35:13.900 --> 00:35:16.180 align:middle line:90% I claim these sets are the same. 00:35:16.180 --> 00:35:25.740 align:middle line:84% So the set of common divisors of a and b, 00:35:25.740 --> 00:35:37.650 align:middle line:84% equals the set of common divisors of a minus b and b. 00:35:37.650 --> 00:35:39.420 align:middle line:84% OK, this is a kind of proof we actually 00:35:39.420 --> 00:35:43.170 align:middle line:90% haven't seen in this class yet. 00:35:43.170 --> 00:35:46.380 align:middle line:84% This is the brace, not a minus sign. 00:35:46.380 --> 00:35:48.940 align:middle line:84% When you want to prove two sets are equal, 00:35:48.940 --> 00:35:51.450 align:middle line:84% a very common template for such a proof 00:35:51.450 --> 00:35:54.630 align:middle line:84% is to show that every element of this set 00:35:54.630 --> 00:35:57.750 align:middle line:84% is also an element of this set that, if you recall back 00:35:57.750 --> 00:35:59.830 align:middle line:84% to our set definitions in lecture 1, 00:35:59.830 --> 00:36:02.700 align:middle line:84% is that this set is contained in this set 00:36:02.700 --> 00:36:05.730 align:middle line:84% and also to prove that every element of the second set 00:36:05.730 --> 00:36:07.120 align:middle line:90% is an element of the first set. 00:36:07.120 --> 00:36:10.360 align:middle line:84% That's saying the second set is a subset of the first set. 00:36:10.360 --> 00:36:15.450 align:middle line:84% So we're going to do a proof in two parts. 00:36:15.450 --> 00:36:19.110 align:middle line:84% Part one is to show containment, that the first set is 00:36:19.110 --> 00:36:21.240 align:middle line:90% contained in the second set. 00:36:21.240 --> 00:36:26.190 align:middle line:84% Part two, over here, is to show containment 00:36:26.190 --> 00:36:27.610 align:middle line:90% in the other direction. 00:36:27.610 --> 00:36:30.600 align:middle line:84% This set is a superset of this set. 00:36:30.600 --> 00:36:35.110 align:middle line:84% If I can prove that these sets are equal, then in particular, 00:36:35.110 --> 00:36:37.750 align:middle line:84% the maximum element of these sets, if it exists, 00:36:37.750 --> 00:36:38.968 align:middle line:90% is also going to be equal. 00:36:38.968 --> 00:36:41.010 align:middle line:84% If you have two identical sets, you take the max, 00:36:41.010 --> 00:36:42.468 align:middle line:84% you're going to get the same thing. 00:36:42.468 --> 00:36:45.040 align:middle line:84% Now, I'm using the first definition here. 00:36:45.040 --> 00:36:46.690 align:middle line:84% Of course, there's this one case. 00:36:46.690 --> 00:36:49.560 align:middle line:84% when the two numbers are 0, then there is no maximum. 00:36:49.560 --> 00:36:51.270 align:middle line:84% Because all integers actually, will 00:36:51.270 --> 00:36:53.800 align:middle line:90% be common divisors of 0 and 0. 00:36:53.800 --> 00:36:56.260 align:middle line:84% But in that case, again, if the sets are equal, 00:36:56.260 --> 00:36:57.630 align:middle line:90% that means that a and b-- 00:36:57.630 --> 00:36:59.470 align:middle line:90% or these numbers must be 0. 00:36:59.470 --> 00:37:01.870 align:middle line:84% And then of course, these numbers are also all 0, 00:37:01.870 --> 00:37:06.120 align:middle line:84% And this statement is trivially true when everything is 0. 00:37:06.120 --> 00:37:08.980 align:middle line:84% It's just saying GCD of 0, 0 equals GCD of 0, 0. 00:37:08.980 --> 00:37:09.970 align:middle line:90% That one I'm fine with. 00:37:09.970 --> 00:37:11.740 align:middle line:84% But as long as we're not in the 0, 0 case, 00:37:11.740 --> 00:37:13.210 align:middle line:90% these sets will be finite. 00:37:13.210 --> 00:37:15.670 align:middle line:84% And if they're equal, then their maximums will be the same. 00:37:15.670 --> 00:37:17.610 align:middle line:90% So this is all I need to show. 00:37:17.610 --> 00:37:19.540 align:middle line:90% Let's do it in two parts. 00:37:19.540 --> 00:37:21.690 align:middle line:84% So first, we suppose that we have 00:37:21.690 --> 00:37:23.470 align:middle line:90% a common divisor of a and b. 00:37:23.470 --> 00:37:29.670 align:middle line:84% So that means d divides a and d divides b for some number b. 00:37:29.670 --> 00:37:34.410 align:middle line:84% Well, I claim that also d divides a minus 00:37:34.410 --> 00:37:38.520 align:middle line:84% b because a minus b is an integer linear combination of a 00:37:38.520 --> 00:37:39.310 align:middle line:90% and b. 00:37:39.310 --> 00:37:42.310 align:middle line:84% And we had, over here, in our divisibility facts, 00:37:42.310 --> 00:37:44.160 align:middle line:84% if d divides a and d divides b, then 00:37:44.160 --> 00:37:47.770 align:middle line:84% d divides every integer linear combination of a and b. 00:37:47.770 --> 00:37:51.360 align:middle line:84% So we're plugging in S equals 1, t equals minus 1 here, 00:37:51.360 --> 00:37:56.170 align:middle line:84% and we get that d divides a minus b, great 00:37:56.170 --> 00:38:01.120 align:middle line:84% So we have d divides a, and d divides a minus b. 00:38:01.120 --> 00:38:06.380 align:middle line:84% So in other words, we have divides a and d divides a minus 00:38:06.380 --> 00:38:08.040 align:middle line:90% b and that-- 00:38:08.040 --> 00:38:10.610 align:middle line:90% sorry, wrong one. 00:38:10.610 --> 00:38:11.980 align:middle line:90% I wanted d divides b. 00:38:11.980 --> 00:38:14.870 align:middle line:90% 00:38:14.870 --> 00:38:18.770 align:middle line:84% And this is the meaning of a common divisor of a minus 00:38:18.770 --> 00:38:19.950 align:middle line:90% b and b. 00:38:19.950 --> 00:38:22.040 align:middle line:84% So we assumed that d was a common divisor of this, 00:38:22.040 --> 00:38:24.290 align:middle line:84% and we proved that d was a common divisor of these two 00:38:24.290 --> 00:38:25.020 align:middle line:90% numbers. 00:38:25.020 --> 00:38:26.790 align:middle line:90% Therefore, we have containment. 00:38:26.790 --> 00:38:31.250 align:middle line:84% Every element in this set, is contained in this set. 00:38:31.250 --> 00:38:34.280 align:middle line:84% Now let's prove the reverse, and that will imply 00:38:34.280 --> 00:38:36.150 align:middle line:90% that these sets are equal. 00:38:36.150 --> 00:38:42.380 align:middle line:84% So now suppose that d divides a minus b, and d divides b. 00:38:42.380 --> 00:38:46.360 align:middle line:84% Then I want to show that d divides a. 00:38:46.360 --> 00:38:49.040 align:middle line:90% 00:38:49.040 --> 00:38:51.890 align:middle line:90% Why? 00:38:51.890 --> 00:38:54.680 align:middle line:90% Any suggestions? 00:38:54.680 --> 00:38:56.875 align:middle line:84% Can I use integer linear combinations? 00:38:56.875 --> 00:38:58.518 align:middle line:90% Yeah. 00:38:58.518 --> 00:39:01.390 align:middle line:84% AUDIENCE: [INAUDIBLE] Like t equals 0 in front of it. 00:39:01.390 --> 00:39:03.490 align:middle line:90% Erik Demaine: T equals 0. 00:39:03.490 --> 00:39:06.410 align:middle line:84% I agree, if I put t equals 0 and s equals 1, 00:39:06.410 --> 00:39:09.440 align:middle line:84% then this is an integer linear combination of a and b. 00:39:09.440 --> 00:39:11.900 align:middle line:84% But in this case, I don't know that d divides a. 00:39:11.900 --> 00:39:13.580 align:middle line:84% That's actually what I want to prove. 00:39:13.580 --> 00:39:17.180 align:middle line:84% So I can't use integer linear combinations of a and b. 00:39:17.180 --> 00:39:17.972 align:middle line:90% But-- 00:39:17.972 --> 00:39:20.200 align:middle line:84% AUDIENCE: Yeah, you can't do it with a minus b. 00:39:20.200 --> 00:39:22.450 align:middle line:84% Erik Demaine: I can take an integer linear combination 00:39:22.450 --> 00:39:23.470 align:middle line:90% of a minus b and b. 00:39:23.470 --> 00:39:32.410 align:middle line:84% If I do 1 times this plus 1 times this, a minus b times 1 00:39:32.410 --> 00:39:37.270 align:middle line:90% plus 1 times b is a. 00:39:37.270 --> 00:39:44.110 align:middle line:84% So in other words, because this equals the sum a minus b plus b, 00:39:44.110 --> 00:39:46.870 align:middle line:84% and because I know that d is a common divisor of a minus 00:39:46.870 --> 00:39:50.050 align:middle line:84% b and b, I know that d also divides 00:39:50.050 --> 00:39:53.270 align:middle line:84% any integer linear combination of those two numbers, 00:39:53.270 --> 00:39:57.040 align:middle line:90% and in particular, a, cool. 00:39:57.040 --> 00:40:01.100 align:middle line:84% So both of these are using the divisibility facts 00:40:01.100 --> 00:40:03.120 align:middle line:84% that if you have a common divisor, 00:40:03.120 --> 00:40:07.070 align:middle line:84% you also divide all integer linear combinations of those two 00:40:07.070 --> 00:40:09.320 align:middle line:90% numbers. 00:40:09.320 --> 00:40:11.180 align:middle line:84% So in particular, this means we have 00:40:11.180 --> 00:40:13.320 align:middle line:90% d divides a and d divides b. 00:40:13.320 --> 00:40:15.560 align:middle line:84% I'm just taking this thing that we proved 00:40:15.560 --> 00:40:17.840 align:middle line:90% and this thing that we assumed. 00:40:17.840 --> 00:40:20.270 align:middle line:90% And this says we have a common-- 00:40:20.270 --> 00:40:23.130 align:middle line:84% it's saying if we had a common divisor of a minus b and b, 00:40:23.130 --> 00:40:26.270 align:middle line:84% we also have a common divisor of a and b. 00:40:26.270 --> 00:40:28.670 align:middle line:84% So that says every element in this second set 00:40:28.670 --> 00:40:30.450 align:middle line:90% is an element of the first set. 00:40:30.450 --> 00:40:33.090 align:middle line:84% And because we have containment in both directions, in fact, 00:40:33.090 --> 00:40:35.400 align:middle line:84% these two sets must be exactly the same set. 00:40:35.400 --> 00:40:38.250 align:middle line:84% So if you ever have to prove that two sets are equal, 00:40:38.250 --> 00:40:41.270 align:middle line:84% I recommend a common template is to do it 00:40:41.270 --> 00:40:47.510 align:middle line:84% in two halves like this, OK, great. 00:40:47.510 --> 00:40:49.310 align:middle line:84% So I claim we can get an algorithm out 00:40:49.310 --> 00:40:55.250 align:middle line:84% of this, out of this subtraction idea and GCD of 0, b. 00:40:55.250 --> 00:40:58.090 align:middle line:90% 00:40:58.090 --> 00:41:00.465 align:middle line:84% So let's just do an example of that. 00:41:00.465 --> 00:41:01.840 align:middle line:84% I won't write a careful algorithm 00:41:01.840 --> 00:41:05.020 align:middle line:84% because we're going to do better. 00:41:05.020 --> 00:41:08.530 align:middle line:84% Let's say we want to do GCD of 9, 6. 00:41:08.530 --> 00:41:11.230 align:middle line:90% 00:41:11.230 --> 00:41:17.010 align:middle line:84% OK, oh, and maybe I want to write another fact here. 00:41:17.010 --> 00:41:19.540 align:middle line:90% 00:41:19.540 --> 00:41:22.330 align:middle line:90% Does it matter? 00:41:22.330 --> 00:41:24.970 align:middle line:90% Yeah, I do. 00:41:24.970 --> 00:41:31.270 align:middle line:84% So GCD of a, b is also equal to GCD of b, 00:41:31.270 --> 00:41:35.740 align:middle line:84% a by symmetry of this definition, a and b 00:41:35.740 --> 00:41:38.950 align:middle line:84% can be swapped, and you get exactly the same things 00:41:38.950 --> 00:41:40.480 align:middle line:90% we're talking about. 00:41:40.480 --> 00:41:45.280 align:middle line:84% And if I apply this subtraction lemma with the elements 00:41:45.280 --> 00:41:50.570 align:middle line:84% swapped, what I get is GCD of the first argument, 00:41:50.570 --> 00:41:53.270 align:middle line:84% which is b, minus the second argument, 00:41:53.270 --> 00:41:57.420 align:middle line:84% which is a, comma the second argument, which is a. 00:41:57.420 --> 00:42:02.100 align:middle line:84% So all these things are equal, just from the same subtraction 00:42:02.100 --> 00:42:04.410 align:middle line:90% lemma and using the symmetry. 00:42:04.410 --> 00:42:05.970 align:middle line:84% But what this is saying is either 00:42:05.970 --> 00:42:08.280 align:middle line:84% I can subtract the second argument from the first 00:42:08.280 --> 00:42:13.530 align:middle line:84% and get a minus b, or I can also rewrite this 00:42:13.530 --> 00:42:17.260 align:middle line:90% as GCD of a, b minus a. 00:42:17.260 --> 00:42:19.980 align:middle line:84% Alternatively, I could leave the first argument alone 00:42:19.980 --> 00:42:22.450 align:middle line:84% and subtract it from the second argument. 00:42:22.450 --> 00:42:24.670 align:middle line:84% So I'm free to subtract whichever way I want. 00:42:24.670 --> 00:42:26.160 align:middle line:84% And I find it much easier to think 00:42:26.160 --> 00:42:27.870 align:middle line:84% about non-negative numbers, so I'm always 00:42:27.870 --> 00:42:30.640 align:middle line:84% going to subtract the smaller number from the bigger number. 00:42:30.640 --> 00:42:32.170 align:middle line:90% I can do it either way. 00:42:32.170 --> 00:42:37.470 align:middle line:90% So let's do it for 9, 6. 00:42:37.470 --> 00:42:40.300 align:middle line:84% So I'm going to subtract 6 from 9. 00:42:40.300 --> 00:42:40.990 align:middle line:90% And I get. 00:42:40.990 --> 00:42:42.720 align:middle line:90% This is equal to-- 00:42:42.720 --> 00:42:45.460 align:middle line:90% so this would be a minus b. 00:42:45.460 --> 00:42:48.240 align:middle line:90% This one minus this one is 3. 00:42:48.240 --> 00:42:49.830 align:middle line:90% 9 minus 6 is 3. 00:42:49.830 --> 00:42:52.110 align:middle line:84% The second argument stays the same. 00:42:52.110 --> 00:42:53.800 align:middle line:84% OK, I still don't know the answer, 00:42:53.800 --> 00:42:54.890 align:middle line:84% so I'm going to keep subtracting. 00:42:54.890 --> 00:42:56.682 align:middle line:84% In this case, I'll take the smaller number, 00:42:56.682 --> 00:42:58.340 align:middle line:84% subtract it from the larger number. 00:42:58.340 --> 00:43:03.100 align:middle line:84% So this is b minus a, I guess, replacing b with b minus a, 00:43:03.100 --> 00:43:06.880 align:middle line:90% I get GCD of-- 00:43:06.880 --> 00:43:09.170 align:middle line:90% sorry, I wanted. 00:43:09.170 --> 00:43:10.210 align:middle line:90% Yeah, 3, 3. 00:43:10.210 --> 00:43:13.210 align:middle line:90% 00:43:13.210 --> 00:43:15.310 align:middle line:90% Subtract 3 from 6, I get 3. 00:43:15.310 --> 00:43:16.700 align:middle line:90% OK, that looks pretty easy. 00:43:16.700 --> 00:43:19.070 align:middle line:84% I'm guessing GCD of 3 and 3 is 3. 00:43:19.070 --> 00:43:23.590 align:middle line:84% But if we go all the way, we can subtract one of them 00:43:23.590 --> 00:43:24.500 align:middle line:90% from the other. 00:43:24.500 --> 00:43:28.670 align:middle line:84% I guess, I want to do b minus a again, and we get 3, 0. 00:43:28.670 --> 00:43:30.820 align:middle line:90% And this we actually analyzed. 00:43:30.820 --> 00:43:33.520 align:middle line:90% Over here, we said GCD-- 00:43:33.520 --> 00:43:37.090 align:middle line:84% oh, I got it backwards. --but of 0 comma something, 00:43:37.090 --> 00:43:39.050 align:middle line:84% is the absolute value of something. 00:43:39.050 --> 00:43:41.510 align:middle line:90% So absolute value of 3 is 3. 00:43:41.510 --> 00:43:48.610 align:middle line:84% And so GCD of 3, 0 is 3, OK, great. 00:43:48.610 --> 00:43:54.360 align:middle line:90% Another example, GCD of 1, 100. 00:43:54.360 --> 00:43:56.380 align:middle line:84% OK, well, I'll take the smaller one, 00:43:56.380 --> 00:43:57.790 align:middle line:90% subtract it from the larger one. 00:43:57.790 --> 00:44:02.000 align:middle line:90% I get GCD of 1, 99. 00:44:02.000 --> 00:44:05.170 align:middle line:84% OK, now, I subtract 1 from the larger number, 00:44:05.170 --> 00:44:10.840 align:middle line:90% I get GCD of 1, 98, and so on. 00:44:10.840 --> 00:44:12.280 align:middle line:84% I won't bore you with the details. 00:44:12.280 --> 00:44:14.050 align:middle line:90% In the end, the answer is 1. 00:44:14.050 --> 00:44:19.200 align:middle line:84% But this is a very slow way to arrive at the answer of 1. 00:44:19.200 --> 00:44:23.770 align:middle line:84% I have to do 100 steps to figure out the GCD of 1, 100. 00:44:23.770 --> 00:44:26.320 align:middle line:84% So the number of steps you get in this algorithm, 00:44:26.320 --> 00:44:28.120 align:middle line:84% which I haven't precisely defined, 00:44:28.120 --> 00:44:32.460 align:middle line:84% is going to be roughly, at least in the worst 00:44:32.460 --> 00:44:38.530 align:middle line:84% case, the sum of a and b, or the maximum of a and b, 00:44:38.530 --> 00:44:40.950 align:middle line:84% it doesn't matter, up to constant factors. 00:44:40.950 --> 00:44:46.288 align:middle line:84% And this is what we would call a slow algorithm in number theory. 00:44:46.288 --> 00:44:48.330 align:middle line:84% Because generally, we want to think about numbers 00:44:48.330 --> 00:44:52.070 align:middle line:90% that are huge, 100 bits long. 00:44:52.070 --> 00:44:56.170 align:middle line:84% If you have 100-bit long number, that is as large as 2 00:44:56.170 --> 00:45:00.220 align:middle line:84% to the 100, and that's a very large number. 00:45:00.220 --> 00:45:05.390 align:middle line:84% It's big, roughly the number of particles in the known universe. 00:45:05.390 --> 00:45:07.730 align:middle line:84% I think that's about 2 to the 128 or something. 00:45:07.730 --> 00:45:11.300 align:middle line:84% So we don't want to spend that much time to compute this. 00:45:11.300 --> 00:45:17.830 align:middle line:84% In fact, we can do a lot better using a better technique, which 00:45:17.830 --> 00:45:19.180 align:middle line:90% is division. 00:45:19.180 --> 00:45:23.740 align:middle line:84% So we've seen here, the idea that we can subtract 00:45:23.740 --> 00:45:26.478 align:middle line:84% one number from the other, but if we subtract, 00:45:26.478 --> 00:45:28.270 align:middle line:84% and then we subtract, and then we subtract, 00:45:28.270 --> 00:45:30.860 align:middle line:84% and we subtract, that sounds a lot like division, right? 00:45:30.860 --> 00:45:32.740 align:middle line:84% This is like the bad way to do division 00:45:32.740 --> 00:45:36.470 align:middle line:84% is to repeatedly subtract one number from the other. 00:45:36.470 --> 00:45:39.500 align:middle line:84% And that's exactly what happens in the 100, 1 case. 00:45:39.500 --> 00:45:42.470 align:middle line:84% We have to subtract 1 from 100, subtract it from 99, 00:45:42.470 --> 00:45:48.820 align:middle line:84% keep subtracting 1, until the left-hand side becomes larger 00:45:48.820 --> 00:45:49.990 align:middle line:90% than the right-hand side. 00:45:49.990 --> 00:45:51.100 align:middle line:90% And then we could stop. 00:45:51.100 --> 00:45:55.080 align:middle line:84% In fact, they become equal because 1 divides 100. 00:45:55.080 --> 00:45:58.780 align:middle line:84% But in general, we want to keep subtracting until we go beyond, 00:45:58.780 --> 00:46:01.510 align:middle line:84% or we would go beyond what we can afford. 00:46:01.510 --> 00:46:02.980 align:middle line:90% And that's exactly division. 00:46:02.980 --> 00:46:08.820 align:middle line:84% So let me give you a very precise version of division 00:46:08.820 --> 00:46:19.570 align:middle line:84% with something called the division theorem. 00:46:19.570 --> 00:46:24.490 align:middle line:84% This is one of the core theorems in number theory. 00:46:24.490 --> 00:46:25.720 align:middle line:90% We won't prove it here. 00:46:25.720 --> 00:46:28.950 align:middle line:90% It's kind of tedious. 00:46:28.950 --> 00:46:35.580 align:middle line:84% For every integer n, and for every positive integer d-- 00:46:35.580 --> 00:46:47.040 align:middle line:84% so I'm going to write z plus to mean d is greater than 0 --there 00:46:47.040 --> 00:46:49.163 align:middle line:90% is a unique-- 00:46:49.163 --> 00:46:54.020 align:middle line:90% 00:46:54.020 --> 00:46:58.380 align:middle line:84% and usually we would write this as exists exclamation mark. 00:46:58.380 --> 00:46:59.820 align:middle line:90% There is a unique. 00:46:59.820 --> 00:47:02.520 align:middle line:84% So this means there is, exclamation mark. 00:47:02.520 --> 00:47:03.600 align:middle line:90% It's really exists. 00:47:03.600 --> 00:47:05.120 align:middle line:90% But there's only one of them. 00:47:05.120 --> 00:47:13.970 align:middle line:84% --there exists a unique pair of integers q and r-- 00:47:13.970 --> 00:47:16.520 align:middle line:84% I don't want the parentheses, actually. 00:47:16.520 --> 00:47:27.410 align:middle line:84% I want to say those are both integers --such that n 00:47:27.410 --> 00:47:32.960 align:middle line:90% equals q times d plus r. 00:47:32.960 --> 00:47:38.990 align:middle line:84% and r is between 0 and strictly less than d. 00:47:38.990 --> 00:47:44.390 align:middle line:84% OK, you actually have seen this in other terms probably before. 00:47:44.390 --> 00:47:48.360 align:middle line:84% This q is what we would call the quotient. 00:47:48.360 --> 00:47:52.992 align:middle line:84% And this r is what we call the remainder. 00:47:52.992 --> 00:47:54.700 align:middle line:84% I'm sure you've heard those terms before. 00:47:54.700 --> 00:47:58.590 align:middle line:84% This is like a very precise way to say what division gives us 00:47:58.590 --> 00:48:00.190 align:middle line:90% in all cases. 00:48:00.190 --> 00:48:05.730 align:middle line:84% Given a numerator and a denominator, 00:48:05.730 --> 00:48:10.907 align:middle line:84% that's why they're called n and d here, and as long as d is not 00:48:10.907 --> 00:48:13.240 align:middle line:84% 0, we're going to Furthermore assume that it's positive, 00:48:13.240 --> 00:48:15.990 align:middle line:84% so it's easier to think about, then 00:48:15.990 --> 00:48:18.120 align:middle line:84% there's a unique quotient and remainder you 00:48:18.120 --> 00:48:21.705 align:middle line:90% get from dividing a over b. 00:48:21.705 --> 00:48:27.270 align:middle line:84% A q is how many whole copies of d you can fit inside n, and r 00:48:27.270 --> 00:48:29.400 align:middle line:90% is whatever's left over. 00:48:29.400 --> 00:48:32.250 align:middle line:84% So that's roughly a proof that this is true. 00:48:32.250 --> 00:48:35.490 align:middle line:84% And if you think about what the remainder should be, 00:48:35.490 --> 00:48:36.910 align:middle line:90% it could be 0. 00:48:36.910 --> 00:48:39.640 align:middle line:84% And the 0 case, that's when we have divisibility. 00:48:39.640 --> 00:48:42.750 align:middle line:84% If d divides n, if n equals q times 00:48:42.750 --> 00:48:46.740 align:middle line:84% d, that's exactly divisibility if there's no r. 00:48:46.740 --> 00:48:48.270 align:middle line:90% That's when r equals 0. 00:48:48.270 --> 00:48:49.700 align:middle line:90% That's what we get. 00:48:49.700 --> 00:48:52.790 align:middle line:84% But in general, we know that r should also be less than d. 00:48:52.790 --> 00:48:55.160 align:middle line:84% Because if there was d left over, 00:48:55.160 --> 00:48:57.810 align:middle line:90% we would just increase q by 1. 00:48:57.810 --> 00:48:59.520 align:middle line:90% That would be a whole copy of d. 00:48:59.520 --> 00:49:02.450 align:middle line:84% So the remainder is always going to be between 0 and strictly 00:49:02.450 --> 00:49:03.217 align:middle line:90% less than d. 00:49:03.217 --> 00:49:05.840 align:middle line:90% 00:49:05.840 --> 00:49:09.960 align:middle line:84% I guess the interesting part here, is that there's only one-- 00:49:09.960 --> 00:49:12.830 align:middle line:84% sorry, I should have put an i in unique. 00:49:12.830 --> 00:49:15.560 align:middle line:90% There's only one way to do this. 00:49:15.560 --> 00:49:17.090 align:middle line:84% That with these conditions, there's 00:49:17.090 --> 00:49:21.140 align:middle line:90% only one quotient and remainder, 00:49:21.140 --> 00:49:30.140 align:middle line:84% OK, I also want to give you some definitions, some notation, 00:49:30.140 --> 00:49:35.270 align:middle line:84% which is for the quotient, we write the numerator 00:49:35.270 --> 00:49:38.690 align:middle line:90% div d div for division. 00:49:38.690 --> 00:49:46.410 align:middle line:84% And for the remainder, we write n rem d, rem for remainder. 00:49:46.410 --> 00:49:49.710 align:middle line:84% This is also sometimes called n mod d. 00:49:49.710 --> 00:49:51.540 align:middle line:84% So if you ever see that, we probably 00:49:51.540 --> 00:49:52.690 align:middle line:90% won't use it in this form. 00:49:52.690 --> 00:49:55.148 align:middle line:84% We're going to use mod to mean a slightly related, slightly 00:49:55.148 --> 00:49:56.370 align:middle line:90% different notion. 00:49:56.370 --> 00:49:59.340 align:middle line:84% Also, in the textbook, they write rem 00:49:59.340 --> 00:50:03.690 align:middle line:84% as a two-argument function but same idea. 00:50:03.690 --> 00:50:06.720 align:middle line:84% So we're going to use division theorem a lot 00:50:06.720 --> 00:50:12.450 align:middle line:84% from now on and as a way to reduce our numbers again, 00:50:12.450 --> 00:50:14.950 align:middle line:84% in a more powerful way than subtraction. 00:50:14.950 --> 00:50:18.390 align:middle line:84% So we would like to do subtraction repeatedly. 00:50:18.390 --> 00:50:20.470 align:middle line:90% And what will we be left with? 00:50:20.470 --> 00:50:22.410 align:middle line:84% We'll be left with the remainder. 00:50:22.410 --> 00:50:25.240 align:middle line:84% If you keep subtracting until you can't subtract anymore, 00:50:25.240 --> 00:50:27.660 align:middle line:90% that's remainder. 00:50:27.660 --> 00:50:29.105 align:middle line:90% So let's do that. 00:50:29.105 --> 00:50:32.340 align:middle line:90% 00:50:32.340 --> 00:50:34.522 align:middle line:84% I probably still need these divisibility facts, 00:50:34.522 --> 00:50:35.480 align:middle line:90% so I'll leave it there. 00:50:35.480 --> 00:50:51.431 align:middle line:90% 00:50:51.431 --> 00:50:57.085 align:middle line:90% So this is GCD division theorem. 00:50:57.085 --> 00:51:01.208 align:middle line:90% 00:51:01.208 --> 00:51:03.500 align:middle line:84% Let's suppose a is less than or equal to b is less than 00:51:03.500 --> 00:51:05.083 align:middle line:90% or equal to 1. 00:51:05.083 --> 00:51:07.537 align:middle line:90% 00:51:07.537 --> 00:51:08.370 align:middle line:90% Is that what I want? 00:51:08.370 --> 00:51:09.192 align:middle line:90% No. 00:51:09.192 --> 00:51:10.650 align:middle line:84% Other way around, that seems weird. 00:51:10.650 --> 00:51:12.095 align:middle line:90% It'd be weird to be at most 1. 00:51:12.095 --> 00:51:13.470 align:middle line:84% I want things to be non-negative, 00:51:13.470 --> 00:51:16.393 align:middle line:84% so I'm going to assume a is the bigger one of a and b. 00:51:16.393 --> 00:51:17.810 align:middle line:84% And I want to compute GCD of a, b. 00:51:17.810 --> 00:51:21.263 align:middle line:90% 00:51:21.263 --> 00:51:23.180 align:middle line:84% And I'm going to assume we haven't gotten down 00:51:23.180 --> 00:51:26.930 align:middle line:84% to 0, in particular, because I can't divide by 0, 00:51:26.930 --> 00:51:29.203 align:middle line:84% but also because when we get to 0, we're done. 00:51:29.203 --> 00:51:30.870 align:middle line:84% If one of the numbers is 0, we're happy. 00:51:30.870 --> 00:51:32.537 align:middle line:84% So let's assume they're both at least 1, 00:51:32.537 --> 00:51:35.000 align:middle line:84% and let's assume, by symmetry, that the one on the left 00:51:35.000 --> 00:51:40.130 align:middle line:84% is bigger or equal, then I claim GCD of a, b, 00:51:40.130 --> 00:51:42.550 align:middle line:90% is GCD of a remainder b, b. 00:51:42.550 --> 00:51:47.170 align:middle line:90% 00:51:47.170 --> 00:51:52.420 align:middle line:84% So before, we had a minus b, and now we have a remainder b. 00:51:52.420 --> 00:51:53.810 align:middle line:90% Seems like a small difference. 00:51:53.810 --> 00:51:57.460 align:middle line:84% And indeed, the proof is basically the same. 00:51:57.460 --> 00:51:59.050 align:middle line:84% I'm going to do a proof sketch, which 00:51:59.050 --> 00:52:07.420 align:middle line:84% is to say, well, GCD of a, b is equal to GCD of a minus b, b. 00:52:07.420 --> 00:52:11.080 align:middle line:84% And if we apply by the subtraction theorem, 00:52:11.080 --> 00:52:17.680 align:middle line:84% which we just did over there, and if we apply the subtraction 00:52:17.680 --> 00:52:20.380 align:middle line:84% theorem again, we can subtract the right argument from the left 00:52:20.380 --> 00:52:26.110 align:middle line:84% one, again, a minus 2b, b and so on, eventually, 00:52:26.110 --> 00:52:32.760 align:middle line:84% we get to GCD of a minus a div b times b. 00:52:32.760 --> 00:52:35.410 align:middle line:90% 00:52:35.410 --> 00:52:39.860 align:middle line:84% a div b is how many whole copies of b I can fit into a. 00:52:39.860 --> 00:52:42.260 align:middle line:84% And so if I take that number, multiply it 00:52:42.260 --> 00:52:46.520 align:middle line:84% by b, that's how much I can subtract away before a 00:52:46.520 --> 00:52:49.760 align:middle line:84% would go negative if I subtracted one more time. 00:52:49.760 --> 00:52:53.540 align:middle line:84% And what I'll be left with, is exactly the remainder, 00:52:53.540 --> 00:53:01.010 align:middle line:90% a minus a rem b times-- 00:53:01.010 --> 00:53:04.190 align:middle line:90% oh, no times, comma b. 00:53:04.190 --> 00:53:06.150 align:middle line:84% If I subtract off that many copies of b, 00:53:06.150 --> 00:53:11.140 align:middle line:84% I will be left with a remainder b, sorry, no, a minus. 00:53:11.140 --> 00:53:13.730 align:middle line:90% 00:53:13.730 --> 00:53:20.330 align:middle line:84% That's what we're writing here, in fact. 00:53:20.330 --> 00:53:22.730 align:middle line:90% Let me rewrite this. 00:53:22.730 --> 00:53:31.100 align:middle line:84% We have n, or maybe I'll write it as a or b. 00:53:31.100 --> 00:53:32.450 align:middle line:90% Let me stick with that. 00:53:32.450 --> 00:53:36.170 align:middle line:84% What this over here is saying is, 00:53:36.170 --> 00:53:47.620 align:middle line:84% n div d times d plus n rem d equals to n. 00:53:47.620 --> 00:53:50.920 align:middle line:84% I'm just rewriting this formula using the notations for q and r. 00:53:50.920 --> 00:53:59.050 align:middle line:84% And if I rearrange this, I get that n minus n div d times d 00:53:59.050 --> 00:54:02.800 align:middle line:90% equals n rem d. 00:54:02.800 --> 00:54:04.970 align:middle line:84% And that's exactly what I'm using here. 00:54:04.970 --> 00:54:07.460 align:middle line:84% The remainder is what you get, what's left over. 00:54:07.460 --> 00:54:12.130 align:middle line:84% If you take the numerator and subtract off the div many copies 00:54:12.130 --> 00:54:13.250 align:middle line:90% of the denominator. 00:54:13.250 --> 00:54:14.693 align:middle line:90% So I use dot, dot, dot here. 00:54:14.693 --> 00:54:16.610 align:middle line:84% You can't really use dot, dot, dot in a proof. 00:54:16.610 --> 00:54:18.360 align:middle line:84% Technically, this is a proof by induction. 00:54:18.360 --> 00:54:21.970 align:middle line:84% But I think, hopefully, you're familiar enough with induction 00:54:21.970 --> 00:54:24.230 align:middle line:84% that you could turn this into a formal proof. 00:54:24.230 --> 00:54:29.080 align:middle line:90% It's not hard, OK. 00:54:29.080 --> 00:54:31.420 align:middle line:90% So we can divide. 00:54:31.420 --> 00:54:36.510 align:middle line:84% And this will make this GCD algorithm much faster. 00:54:36.510 --> 00:54:40.750 align:middle line:84% Because now if I want to say, compute GCD of 1, 100, 00:54:40.750 --> 00:54:46.380 align:middle line:84% this is the same thing as GCD of 100, 1. 00:54:46.380 --> 00:54:52.080 align:middle line:84% And what that says is, I can divide 100 by 1 00:54:52.080 --> 00:54:53.590 align:middle line:90% and see what the remainder is. 00:54:53.590 --> 00:54:56.490 align:middle line:84% Well, 1 divides 100, so the remainder is 0. 00:54:56.490 --> 00:55:01.380 align:middle line:84% So I instantly get that this is GCD of 0, 1. 00:55:01.380 --> 00:55:04.330 align:middle line:90% Boom, I'm done. 00:55:04.330 --> 00:55:08.040 align:middle line:84% This GCD of 0 comma anything is the anything. 00:55:08.040 --> 00:55:10.480 align:middle line:84% So in one step now, one division, 00:55:10.480 --> 00:55:12.480 align:middle line:84% instead of having to do a bunch of subtractions, 00:55:12.480 --> 00:55:14.340 align:middle line:90% I compute the GCD. 00:55:14.340 --> 00:55:18.370 align:middle line:84% Great, let's write this down as an algorithm. 00:55:18.370 --> 00:55:40.240 align:middle line:90% 00:55:40.240 --> 00:55:42.160 align:middle line:84% This is called Euclid's algorithm, 00:55:42.160 --> 00:55:43.536 align:middle line:90% or the Euclidean algorithm. 00:55:43.536 --> 00:55:48.430 align:middle line:90% 00:55:48.430 --> 00:55:50.710 align:middle line:84% It's called Euclid's algorithm because Euclid was not 00:55:50.710 --> 00:55:52.960 align:middle line:84% the first to discover it, of course, 00:55:52.960 --> 00:55:57.670 align:middle line:84% but he's still pretty early, around 300 BCE. 00:55:57.670 --> 00:56:01.870 align:middle line:84% But I'll mention an earlier version in a moment. 00:56:01.870 --> 00:56:13.390 align:middle line:84% So what this does is it computes GCD of a, b for a greater 00:56:13.390 --> 00:56:15.918 align:middle line:84% than or equal to b greater than or equal to 0. 00:56:15.918 --> 00:56:17.960 align:middle line:84% OK I'm going to assume everything's non-negative. 00:56:17.960 --> 00:56:20.260 align:middle line:84% If it's negative, just negate it. 00:56:20.260 --> 00:56:22.570 align:middle line:84% I'm going to assume the left argument is bigger 00:56:22.570 --> 00:56:24.250 align:middle line:90% than the right one or equal. 00:56:24.250 --> 00:56:26.960 align:middle line:84% If it's the other way around, swap a and b, 00:56:26.960 --> 00:56:31.540 align:middle line:84% so this is enough for computing GCD of any two integers. 00:56:31.540 --> 00:56:35.580 align:middle line:84% And I'm going to write this, of course, as a state machine. 00:56:35.580 --> 00:56:39.540 align:middle line:84% So in general, I'm going to have a pair of integers, a and b, 00:56:39.540 --> 00:56:42.060 align:middle line:84% where a is greater than or equal to b 00:56:42.060 --> 00:56:44.940 align:middle line:90% is greater than or equal to 0. 00:56:44.940 --> 00:56:49.290 align:middle line:84% So I guess I should say the states are 00:56:49.290 --> 00:56:56.260 align:middle line:84% x, y where x is greater or equal to y, is greater or equal to 0. 00:56:56.260 --> 00:57:12.390 align:middle line:84% And then the transitions, go from x, y to x rem y-- 00:57:12.390 --> 00:57:13.660 align:middle line:90% oh, wrong order. 00:57:13.660 --> 00:57:19.560 align:middle line:90% 00:57:19.560 --> 00:57:24.360 align:middle line:90% --y, x rem y. 00:57:24.360 --> 00:57:26.170 align:middle line:84% OK, almost the same as this theorem, 00:57:26.170 --> 00:57:32.410 align:middle line:84% except I'm going to rewrite this as GCD of b, a rem b, 00:57:32.410 --> 00:57:35.200 align:middle line:84% just flip the arguments because I 00:57:35.200 --> 00:57:36.790 align:middle line:84% would like the left argument to always 00:57:36.790 --> 00:57:38.390 align:middle line:90% be greater than or equal to-- 00:57:38.390 --> 00:57:40.723 align:middle line:84% I would like the left argument to be always greater than 00:57:40.723 --> 00:57:42.970 align:middle line:90% or equal to the right argument. 00:57:42.970 --> 00:57:46.370 align:middle line:84% And if I do them in this order, if x was greater or equal to y, 00:57:46.370 --> 00:57:49.040 align:middle line:84% I do this division, this division will, of course, 00:57:49.040 --> 00:57:51.130 align:middle line:84% be less than y, and so the left one 00:57:51.130 --> 00:57:53.290 align:middle line:84% will be bigger than the right one, actually 00:57:53.290 --> 00:57:56.620 align:middle line:90% strictly in that case. 00:57:56.620 --> 00:58:01.240 align:middle line:84% So that's basically just following this rule repeatedly, 00:58:01.240 --> 00:58:04.390 align:middle line:90% and those are my transitions. 00:58:04.390 --> 00:58:10.450 align:middle line:84% Now, this only works when y is bigger than 0. 00:58:10.450 --> 00:58:22.540 align:middle line:84% If y is 0 at x, 0, we return x as the GCD. 00:58:22.540 --> 00:58:25.270 align:middle line:90% That's the final state. 00:58:25.270 --> 00:58:27.820 align:middle line:84% Or these are various final states, any x, 00:58:27.820 --> 00:58:29.270 align:middle line:90% 0 is going to be a final state. 00:58:29.270 --> 00:58:30.687 align:middle line:84% There's no transition out of there 00:58:30.687 --> 00:58:32.560 align:middle line:84% because we're not allowed to divide by 0. 00:58:32.560 --> 00:58:35.260 align:middle line:84% Because I wrote this for all y greater than 0 rule. 00:58:35.260 --> 00:58:40.800 align:middle line:84% But in that situation, we know the GCD of 0, b or b, 00:58:40.800 --> 00:58:43.770 align:middle line:84% 0 is the absolute value of b because everything here 00:58:43.770 --> 00:58:45.570 align:middle line:90% is non-negative. 00:58:45.570 --> 00:58:48.120 align:middle line:90% The answer is x at that point. 00:58:48.120 --> 00:58:52.110 align:middle line:90% So why does this work? 00:58:52.110 --> 00:58:55.380 align:middle line:84% Well, because we just proved an invariant 00:58:55.380 --> 00:59:01.120 align:middle line:90% that as this machine runs-- 00:59:01.120 --> 00:59:07.140 align:middle line:84% I want to say p of x, y is the state predicate, which 00:59:07.140 --> 00:59:14.010 align:middle line:84% is the GCD of x, y equals the GCD of a, b. 00:59:14.010 --> 00:59:17.130 align:middle line:90% 00:59:17.130 --> 00:59:19.380 align:middle line:84% We designed the change in arguments here. 00:59:19.380 --> 00:59:22.680 align:middle line:84% We started with a, b, so this is true initially. 00:59:22.680 --> 00:59:25.830 align:middle line:84% In our initial state, a, b, of course, GCD of a, b 00:59:25.830 --> 00:59:27.340 align:middle line:90% equals GCD of a, b. 00:59:27.340 --> 00:59:29.780 align:middle line:84% And then every time we do a transition 00:59:29.780 --> 00:59:32.450 align:middle line:84% and we do this division and take the remainder, 00:59:32.450 --> 00:59:39.050 align:middle line:84% we know that the GCDs are equal by this theorem. 00:59:39.050 --> 00:59:40.780 align:middle line:84% And that's exactly what this tells us. 00:59:40.780 --> 00:59:43.820 align:middle line:84% And so this is saying, this theorem 00:59:43.820 --> 00:59:47.010 align:middle line:84% tells us that this state predicate is preserved. 00:59:47.010 --> 00:59:49.130 align:middle line:84% And because it holds initially, that 00:59:49.130 --> 00:59:57.560 align:middle line:84% means it's invariant, OK, true initially, 00:59:57.560 --> 01:00:04.070 align:middle line:84% and it's preserved by this theorem. 01:00:04.070 --> 01:00:06.140 align:middle line:84% And so by the invariant principle, 01:00:06.140 --> 01:00:09.980 align:middle line:84% we know that this will remain true for any reachable state x, 01:00:09.980 --> 01:00:13.490 align:middle line:84% y, we'll have GCD of x, y equal GCD of a, b. 01:00:13.490 --> 01:00:18.300 align:middle line:84% And furthermore, when we get to a final state-- 01:00:18.300 --> 01:00:19.555 align:middle line:90% so this is the final state. 01:00:19.555 --> 01:00:25.550 align:middle line:90% 01:00:25.550 --> 01:00:29.790 align:middle line:84% --we know that this answer is correct because GCD of x, 0 01:00:29.790 --> 01:00:32.140 align:middle line:90% equals x for non-negative x. 01:00:32.140 --> 01:00:36.690 align:middle line:90% So this is partial correctness. 01:00:36.690 --> 01:00:45.720 align:middle line:84% If we get to a final state, GCD of x, 0 01:00:45.720 --> 01:00:51.240 align:middle line:84% equals x for x greater than or equal to 0 by this theorem 01:00:51.240 --> 01:00:54.060 align:middle line:90% that we proved. 01:00:54.060 --> 01:01:00.030 align:middle line:84% So if this machine terminates, we're done. 01:01:00.030 --> 01:01:02.640 align:middle line:84% And I claim this machine does terminate. 01:01:02.640 --> 01:01:07.333 align:middle line:84% So let's prove termination in a couple of ways. 01:01:07.333 --> 01:01:12.360 align:middle line:90% 01:01:12.360 --> 01:01:14.620 align:middle line:84% So how do we prove termination of state machines? 01:01:14.620 --> 01:01:19.840 align:middle line:84% We define some value, some function of states, 01:01:19.840 --> 01:01:23.140 align:middle line:84% that's an integer, a non-negative integer, 01:01:23.140 --> 01:01:25.400 align:middle line:84% and we show that it strictly decreases. 01:01:25.400 --> 01:01:28.820 align:middle line:84% That's our main tool for proving termination of algorithms. 01:01:28.820 --> 01:01:35.570 align:middle line:84% So for example, we could take the function x plus y. 01:01:35.570 --> 01:01:37.556 align:middle line:90% This strictly decreases. 01:01:37.556 --> 01:01:44.300 align:middle line:90% 01:01:44.300 --> 01:01:45.470 align:middle line:90% Why? 01:01:45.470 --> 01:01:47.430 align:middle line:90% Just think about a transition. 01:01:47.430 --> 01:01:49.260 align:middle line:90% We start with x, y. 01:01:49.260 --> 01:01:50.940 align:middle line:90% y doesn't change. 01:01:50.940 --> 01:01:53.250 align:middle line:84% And then the other term in the sum used to be x, 01:01:53.250 --> 01:01:55.580 align:middle line:84% and then it becomes x remainder y. 01:01:55.580 --> 01:02:00.420 align:middle line:84% So we're dividing x by y, so x gets smaller in that process. 01:02:00.420 --> 01:02:01.880 align:middle line:84% And so the sum of these two values, 01:02:01.880 --> 01:02:05.660 align:middle line:84% is going to be smaller than the sum of x and y. 01:02:05.660 --> 01:02:06.690 align:middle line:90% So that's cool. 01:02:06.690 --> 01:02:09.110 align:middle line:84% We now know that this algorithm terminates. 01:02:09.110 --> 01:02:15.530 align:middle line:84% But stating it using this function, 01:02:15.530 --> 01:02:19.880 align:middle line:84% is not so great because it tells us that the number of steps 01:02:19.880 --> 01:02:25.750 align:middle line:90% is at most, a plus b. 01:02:25.750 --> 01:02:27.110 align:middle line:90% But that's not great. 01:02:27.110 --> 01:02:29.080 align:middle line:90% That's what we had before. 01:02:29.080 --> 01:02:33.800 align:middle line:84% From our slow algorithm, we got a running time of a plus b. 01:02:33.800 --> 01:02:34.833 align:middle line:90% So that's still true. 01:02:34.833 --> 01:02:36.500 align:middle line:84% It's an upper bound on the running time. 01:02:36.500 --> 01:02:38.292 align:middle line:84% But I claim that, in fact, our running time 01:02:38.292 --> 01:02:44.410 align:middle line:84% is much smaller by a more clever proof. 01:02:44.410 --> 01:02:46.976 align:middle line:90% Let me erase subtraction. 01:02:46.976 --> 01:03:09.250 align:middle line:90% 01:03:09.250 --> 01:03:19.150 align:middle line:84% So I claim, in fact, If we look at the bit 01:03:19.150 --> 01:03:24.240 align:middle line:84% count of x and the bit count of y in summation, 01:03:24.240 --> 01:03:26.236 align:middle line:90% this also strictly decreases. 01:03:26.236 --> 01:03:31.400 align:middle line:90% 01:03:31.400 --> 01:03:32.640 align:middle line:90% What is the bit count? 01:03:32.640 --> 01:03:35.940 align:middle line:84% This is if you write x in binary, how many bits are there? 01:03:35.940 --> 01:03:38.930 align:middle line:84% What's the length of the binary representation of x? 01:03:38.930 --> 01:03:47.300 align:middle line:84% This is essentially 1 plus log base 2 of x ceiling. 01:03:47.300 --> 01:03:51.530 align:middle line:84% A little gross, but asymptotically, this 01:03:51.530 --> 01:03:55.550 align:middle line:90% is log base 2 of x. 01:03:55.550 --> 01:03:57.360 align:middle line:84% Or when you use asymptotic notation, 01:03:57.360 --> 01:03:59.722 align:middle line:84% you don't even need the base because it only makes 01:03:59.722 --> 01:04:00.930 align:middle line:90% a constant factor difference. 01:04:00.930 --> 01:04:04.770 align:middle line:84% So this is roughly log of x is theta log of x. 01:04:04.770 --> 01:04:06.830 align:middle line:90% This is theta log of y. 01:04:06.830 --> 01:04:17.360 align:middle line:84% And so the number of steps is order log x plus log y. 01:04:17.360 --> 01:04:19.020 align:middle line:90% And that's really good. 01:04:19.020 --> 01:04:20.940 align:middle line:90% That's exponentially faster. 01:04:20.940 --> 01:04:25.210 align:middle line:84% Sorry, I should have written log a plus log b because that's 01:04:25.210 --> 01:04:27.520 align:middle line:90% where we start is a, b. 01:04:27.520 --> 01:04:29.120 align:middle line:84% So if I give you two numbers a and b, 01:04:29.120 --> 01:04:33.790 align:middle line:84% I can compute the GCD in roughly logarithmically many divisions. 01:04:33.790 --> 01:04:38.390 align:middle line:84% Whereas before, we did a plus b, now we have log a plus log b. 01:04:38.390 --> 01:04:40.880 align:middle line:84% So that's what we call an exponential improvement. 01:04:40.880 --> 01:04:44.470 align:middle line:84% And this is what we would call a fast algorithm. 01:04:44.470 --> 01:04:47.500 align:middle line:84% Because now if I have 100-bit integer, it only takes me-- 01:04:47.500 --> 01:04:51.770 align:middle line:84% or two 100-bit integers it only takes me around 100 steps. 01:04:51.770 --> 01:04:53.650 align:middle line:84% Whereas before, it was taking me 2 to the 100 01:04:53.650 --> 01:04:56.380 align:middle line:90% steps, big difference. 01:04:56.380 --> 01:04:58.130 align:middle line:90% OK, why is this true? 01:04:58.130 --> 01:05:02.170 align:middle line:84% Why is bit count of x plus bit count of y strictly decreasing? 01:05:02.170 --> 01:05:13.360 align:middle line:84% Because I claim-- let's just say because x rem y is less than x 01:05:13.360 --> 01:05:14.450 align:middle line:90% over 2. 01:05:14.450 --> 01:05:17.860 align:middle line:84% This is another fact about division. 01:05:17.860 --> 01:05:25.310 align:middle line:84% I think the intuition here is y is smaller than x 01:05:25.310 --> 01:05:26.820 align:middle line:90% when we're doing this division. 01:05:26.820 --> 01:05:31.550 align:middle line:84% And so we can remove at least one whole copy of y from x. 01:05:31.550 --> 01:05:36.030 align:middle line:84% And maybe we remove more, but yeah, 01:05:36.030 --> 01:05:38.910 align:middle line:84% you can think about if y is more than 1/2 x, 01:05:38.910 --> 01:05:41.550 align:middle line:84% then just in the first step, we've removed 1/2 of x. 01:05:41.550 --> 01:05:44.840 align:middle line:84% And so we've gotten down to less than x over 2. 01:05:44.840 --> 01:05:49.290 align:middle line:84% If x is smaller, if y is really small, like 1, 01:05:49.290 --> 01:05:51.360 align:middle line:84% then we're going to be able to eat all of x. 01:05:51.360 --> 01:05:54.710 align:middle line:84% In both cases, you'll end up eating at least 1/2 of x 01:05:54.710 --> 01:05:58.600 align:middle line:84% And so in general, the remainder will be less than 1/2 of x over 01:05:58.600 --> 01:05:59.100 align:middle line:90% 2. 01:05:59.100 --> 01:06:01.850 align:middle line:84% And so if we start with x, and in the next step 01:06:01.850 --> 01:06:03.890 align:middle line:84% we get to x over 2 or smaller, that 01:06:03.890 --> 01:06:06.600 align:middle line:84% means we removed one bit from the binary representation. 01:06:06.600 --> 01:06:10.130 align:middle line:84% Or if you think about it in log base 2, if we halve the number, 01:06:10.130 --> 01:06:11.880 align:middle line:90% log base 2 goes down by 1. 01:06:11.880 --> 01:06:15.620 align:middle line:84% So even with the ceiling, it goes down by 1. 01:06:15.620 --> 01:06:22.480 align:middle line:84% So that's why this Euclid algorithm is fast and good. 01:06:22.480 --> 01:06:23.560 align:middle line:90% That's why we love it. 01:06:23.560 --> 01:06:26.570 align:middle line:84% I have for you a little animation of the algorithm, 01:06:26.570 --> 01:06:28.400 align:middle line:84% and then we'll talk about extensions. 01:06:28.400 --> 01:06:34.700 align:middle line:84% So here, on the screen, we're computing GCD of 189 and 606. 01:06:34.700 --> 01:06:38.890 align:middle line:84% If we think of that as a rectangle, 189 by 606, 01:06:38.890 --> 01:06:45.160 align:middle line:84% then doing a division is like square packing. 01:06:45.160 --> 01:06:50.860 align:middle line:84% So here, if we have an a by b rectangle, 01:06:50.860 --> 01:06:55.570 align:middle line:84% and b is the smaller one, then a b by b square 01:06:55.570 --> 01:06:57.230 align:middle line:90% is like subtracting off b. 01:06:57.230 --> 01:07:00.050 align:middle line:84% And so if you think of the whole rectangle, 01:07:00.050 --> 01:07:02.150 align:middle line:84% it's how many squares can I subtract off? 01:07:02.150 --> 01:07:04.640 align:middle line:84% And what we'll be left with is the remainder. 01:07:04.640 --> 01:07:06.470 align:middle line:84% Because one dimension stays the same, 01:07:06.470 --> 01:07:09.020 align:middle line:84% the other dimension gets reduced to the remainder. 01:07:09.020 --> 01:07:11.620 align:middle line:84% So hopefully that's clear why the geometric view of squares 01:07:11.620 --> 01:07:14.162 align:middle line:84% is the same as what we've been doing with Euclid's algorithm. 01:07:14.162 --> 01:07:15.470 align:middle line:90% Makes for a much nicer picture. 01:07:15.470 --> 01:07:17.670 align:middle line:84% And then you take that rectangle, 01:07:17.670 --> 01:07:20.770 align:middle line:84% and you flip it so that the horizontal dimension is bigger. 01:07:20.770 --> 01:07:22.943 align:middle line:84% You take off as many squares as you can, 01:07:22.943 --> 01:07:25.360 align:middle line:84% and you flip, and you take off as many squares as you can. 01:07:25.360 --> 01:07:26.110 align:middle line:90% Here's just one. 01:07:26.110 --> 01:07:27.390 align:middle line:90% That's just a subtraction. 01:07:27.390 --> 01:07:28.163 align:middle line:90% You flip. 01:07:28.163 --> 01:07:29.830 align:middle line:84% You take off as many squares as you can. 01:07:29.830 --> 01:07:32.550 align:middle line:84% You're left with a remainder of 3 by 6. 01:07:32.550 --> 01:07:34.690 align:middle line:84% And now here we get even division, 01:07:34.690 --> 01:07:36.880 align:middle line:84% and so one of the dimensions becomes 0. 01:07:36.880 --> 01:07:43.390 align:middle line:84% And then the GCD of 3 and 0 is 3, cool. 01:07:43.390 --> 01:07:45.280 align:middle line:90% So if that helps, great. 01:07:45.280 --> 01:07:46.650 align:middle line:90% If not, don't worry. 01:07:46.650 --> 01:07:49.020 align:middle line:84% Hopefully, you can now compute GCD. 01:07:49.020 --> 01:07:50.800 align:middle line:84% We will do another example in a second. 01:07:50.800 --> 01:07:53.760 align:middle line:84% I would like to first strengthen this algorithm 01:07:53.760 --> 01:07:56.370 align:middle line:90% and make it do more. 01:07:56.370 --> 01:07:59.190 align:middle line:84% Because remember, we wanted to compute GCD. 01:07:59.190 --> 01:07:59.940 align:middle line:90% Why? 01:07:59.940 --> 01:08:04.120 align:middle line:84% We wanted to see whether GCD of a, b divided x. 01:08:04.120 --> 01:08:06.240 align:middle line:84% Because if the GCD of a, b divides 01:08:06.240 --> 01:08:10.920 align:middle line:84% x, then all common divisors of a and b will divide x. 01:08:10.920 --> 01:08:13.410 align:middle line:84% And the reason we cared about that is I 01:08:13.410 --> 01:08:15.160 align:middle line:84% claimed it was the same thing as x 01:08:15.160 --> 01:08:17.777 align:middle line:84% being an integer linear combination of a and b. 01:08:17.777 --> 01:08:19.819 align:middle line:84% But where did the integer linear combinations go? 01:08:19.819 --> 01:08:22.609 align:middle line:84% All we've done so far is compute GCD of a, b. 01:08:22.609 --> 01:08:25.180 align:middle line:84% Well, it turns out there's a connection. 01:08:25.180 --> 01:08:32.109 align:middle line:84% And the connection is called Bézout's theorem or lemma. 01:08:32.109 --> 01:08:34.510 align:middle line:84% It's called Bézout's lemma because Bézout's is 01:08:34.510 --> 01:08:37.930 align:middle line:84% like the last person to discover it 01:08:37.930 --> 01:08:43.939 align:middle line:84% in 1779, after another French mathematician, Méziriac in 1624, 01:08:43.939 --> 01:08:49.760 align:middle line:84% and after two Indian mathematicians around 500 ACE. 01:08:49.760 --> 01:08:51.229 align:middle line:90% But we call it Bézout's lemma. 01:08:51.229 --> 01:09:01.090 align:middle line:90% 01:09:01.090 --> 01:09:03.340 align:middle line:90% This is the connection. 01:09:03.340 --> 01:09:07.120 align:middle line:84% GCD of a, b is an integer linear combination of a and b. 01:09:07.120 --> 01:09:11.410 align:middle line:84% And we can prove this using Euclid's algorithm, which 01:09:11.410 --> 01:09:14.279 align:middle line:90% is pretty cool. 01:09:14.279 --> 01:09:18.895 align:middle line:84% We're going to, by Euclid's algorithm. 01:09:18.895 --> 01:09:30.240 align:middle line:90% 01:09:30.240 --> 01:09:34.410 align:middle line:90% OK, let me do this quickly. 01:09:34.410 --> 01:09:37.029 align:middle line:84% Let's assume a is greater or equal to b, again. 01:09:37.029 --> 01:09:38.859 align:middle line:84% If not, we can just swap a and b. 01:09:38.859 --> 01:09:41.350 align:middle line:84% And I guess, we'll assume greater than or equal to 0. 01:09:41.350 --> 01:09:44.880 align:middle line:84% So we can apply Euclid's algorithm. 01:09:44.880 --> 01:09:47.640 align:middle line:84% Euclid's algorithm tells us, or the theorem 01:09:47.640 --> 01:09:50.819 align:middle line:84% we proved before Euclid's algorithm, tells us GCD of a, b 01:09:50.819 --> 01:09:55.535 align:middle line:90% equals GCD of b, a rem b. 01:09:55.535 --> 01:09:58.680 align:middle line:90% 01:09:58.680 --> 01:10:03.325 align:middle line:90% And we can use induction now. 01:10:03.325 --> 01:10:08.010 align:middle line:90% 01:10:08.010 --> 01:10:15.190 align:middle line:90% By strong induction on a plus b. 01:10:15.190 --> 01:10:18.700 align:middle line:84% So this is very similar to what we were saying here. 01:10:18.700 --> 01:10:21.230 align:middle line:90% x plus y strictly decreases. 01:10:21.230 --> 01:10:24.050 align:middle line:84% That's all we need for induction to work here. 01:10:24.050 --> 01:10:27.400 align:middle line:84% So we're assuming by induction, that if a plus 01:10:27.400 --> 01:10:30.730 align:middle line:84% b is smaller than the current value, then GCD of a, b 01:10:30.730 --> 01:10:33.370 align:middle line:84% is an integer linear combination of a and b. 01:10:33.370 --> 01:10:35.650 align:middle line:84% And so here, we have GCD of a and b, 01:10:35.650 --> 01:10:38.780 align:middle line:84% and we've reduced it to GCD of some smaller values, 01:10:38.780 --> 01:10:40.310 align:middle line:90% or whose sum is smaller. 01:10:40.310 --> 01:10:44.620 align:middle line:84% And so we can apply strong induction to this pair. 01:10:44.620 --> 01:10:48.460 align:middle line:84% And so we'll get that this GCD is an integer linear combination 01:10:48.460 --> 01:10:50.320 align:middle line:90% of its arguments. 01:10:50.320 --> 01:10:54.255 align:middle line:90% So let's continue here. 01:10:54.255 --> 01:10:58.120 align:middle line:90% 01:10:58.120 --> 01:11:08.560 align:middle line:84% So by strong induction, we get that GCD of b, a rem b, 01:11:08.560 --> 01:11:16.110 align:middle line:84% is an integer linear combination of b and a rem b. 01:11:16.110 --> 01:11:18.330 align:middle line:84% And this value, of course, is equal to GCD 01:11:18.330 --> 01:11:22.360 align:middle line:90% of a, b, by this equality. 01:11:22.360 --> 01:11:25.110 align:middle line:84% And so we get that the GCD we care about, 01:11:25.110 --> 01:11:27.690 align:middle line:84% is an integer linear combination, not of a and b 01:11:27.690 --> 01:11:32.370 align:middle line:84% but of this reduced version, b and a rem b. 01:11:32.370 --> 01:11:35.910 align:middle line:84% That's kind of annoying, but it turns out 01:11:35.910 --> 01:11:44.212 align:middle line:84% a rem b is an integer linear combination of a and b. 01:11:44.212 --> 01:11:44.712 align:middle line:90% Why? 01:11:44.712 --> 01:11:47.430 align:middle line:90% By the division theorem. 01:11:47.430 --> 01:11:49.750 align:middle line:84% I actually wrote it out over here. 01:11:49.750 --> 01:11:57.690 align:middle line:84% We wrote that n rem d is n minus something times d. 01:11:57.690 --> 01:12:01.470 align:middle line:84% That is, this thing is an integer linear combination 01:12:01.470 --> 01:12:03.810 align:middle line:90% of n And d. 01:12:03.810 --> 01:12:07.380 align:middle line:84% And so if I'm doing a rem b, I get 01:12:07.380 --> 01:12:12.990 align:middle line:84% that it's a minus something, the div, times b. 01:12:12.990 --> 01:12:17.630 align:middle line:84% So that's an integer linear combination of a and b. 01:12:17.630 --> 01:12:21.140 align:middle line:90% So what we're left with is-- 01:12:21.140 --> 01:12:23.750 align:middle line:84% we have that this GCD that we care about 01:12:23.750 --> 01:12:26.060 align:middle line:84% is an integer linear combination of b 01:12:26.060 --> 01:12:28.730 align:middle line:84% and something else that is an integer linear combination of a 01:12:28.730 --> 01:12:29.610 align:middle line:90% and b. 01:12:29.610 --> 01:12:33.140 align:middle line:84% So in other words, we have an integer linear combination 01:12:33.140 --> 01:12:37.580 align:middle line:84% of two integer linear combinations of a and b. 01:12:37.580 --> 01:12:39.800 align:middle line:84% Because b is definitely an integer linear combination 01:12:39.800 --> 01:12:40.320 align:middle line:90% of a and b. 01:12:40.320 --> 01:12:42.560 align:middle line:90% It's 0 times a plus 1 times b. 01:12:42.560 --> 01:12:45.350 align:middle line:84% a rem b is an integer linear combination, as written 01:12:45.350 --> 01:12:46.140 align:middle line:90% over there. 01:12:46.140 --> 01:12:49.100 align:middle line:84% And so now we're just taking an integer linear combination 01:12:49.100 --> 01:12:50.430 align:middle line:90% of those two things. 01:12:50.430 --> 01:12:57.470 align:middle line:84% And that's also an integer linear combination of a and b, 01:12:57.470 --> 01:13:00.770 align:middle line:90% as we showed before. 01:13:00.770 --> 01:13:03.680 align:middle line:84% So that's maybe a little abstract, 01:13:03.680 --> 01:13:05.370 align:middle line:90% but it does prove the theorem. 01:13:05.370 --> 01:13:11.260 align:middle line:84% Now I'm going to turn that into an algorithm, cool algorithm, 01:13:11.260 --> 01:13:22.440 align:middle line:84% maybe over here, which is called the Pulverizer. 01:13:22.440 --> 01:13:25.360 align:middle line:90% 01:13:25.360 --> 01:13:28.088 align:middle line:84% I mentioned Euclid's algorithm was discovered before Euclid. 01:13:28.088 --> 01:13:30.130 align:middle line:84% In fact, a stronger version of Euclid's algorithm 01:13:30.130 --> 01:13:32.560 align:middle line:84% called the Pulverizer, was described 01:13:32.560 --> 01:13:37.190 align:middle line:84% by Aryabhata, an Indian mathematician, around 500 ACE, 01:13:37.190 --> 01:13:40.090 align:middle line:90% so a long time ago. 01:13:40.090 --> 01:13:45.890 align:middle line:84% And he called it Kuttaka, which is Sanskrit for pulverization. 01:13:45.890 --> 01:13:47.900 align:middle line:84% So we're going to call this the Pulverizer. 01:13:47.900 --> 01:13:51.560 align:middle line:84% Most people call this the extended Euclidean algorithm. 01:13:51.560 --> 01:13:58.060 align:middle line:90% 01:13:58.060 --> 01:14:02.210 align:middle line:84% And it's basically, run Euclid's algorithm 01:14:02.210 --> 01:14:05.080 align:middle line:84% but also keep in mind, what's going on in Bézout's lemma 01:14:05.080 --> 01:14:07.490 align:middle line:84% and always keep track of x and y as an integer 01:14:07.490 --> 01:14:10.430 align:middle line:84% and linear combination of a and b. 01:14:10.430 --> 01:14:13.370 align:middle line:90% OK, so that's the intuition. 01:14:13.370 --> 01:14:18.127 align:middle line:84% As a state machine, it's quite simple, kind of simple, a,1, 01:14:18.127 --> 01:14:25.460 align:middle line:84% 0 b, 0, 1 is where we start because a equals 1 times a plus 01:14:25.460 --> 01:14:30.780 align:middle line:84% 0 times b, and b equals 0 times a plus 1 times b. 01:14:30.780 --> 01:14:32.390 align:middle line:84% So that's how we represent a and b 01:14:32.390 --> 01:14:36.120 align:middle line:84% as integer linear combinations of a and b, pretty trivial. 01:14:36.120 --> 01:14:38.410 align:middle line:84% And then our transitions, if we have x, 01:14:38.410 --> 01:14:46.430 align:middle line:84% s, t so this means x equals s times a plus t times b. 01:14:46.430 --> 01:14:51.600 align:middle line:84% And we have y, u, v. So in other words, 01:14:51.600 --> 01:14:56.690 align:middle line:84% we've represented y as u times s plus v times-- 01:14:56.690 --> 01:15:02.990 align:middle line:90% sorry, v times a plus v times b. 01:15:02.990 --> 01:15:05.100 align:middle line:90% Then we transition to-- 01:15:05.100 --> 01:15:09.670 align:middle line:90% 01:15:09.670 --> 01:15:10.820 align:middle line:90% oh, that's a mistake. 01:15:10.820 --> 01:15:13.640 align:middle line:90% This should be y, u, v. 01:15:13.640 --> 01:15:16.010 align:middle line:84% As usual, we just copy y to the left argument. 01:15:16.010 --> 01:15:16.970 align:middle line:90% We don't change it. 01:15:16.970 --> 01:15:23.290 align:middle line:90% And then x changes to x rem y. 01:15:23.290 --> 01:15:28.030 align:middle line:84% And then it turns out you can write, 01:15:28.030 --> 01:15:31.150 align:middle line:84% you can figure out how x rem y is 01:15:31.150 --> 01:15:35.290 align:middle line:84% relative to-- as a linear combination of a and b 01:15:35.290 --> 01:15:36.255 align:middle line:90% like this. 01:15:36.255 --> 01:15:39.048 align:middle line:90% 01:15:39.048 --> 01:15:40.215 align:middle line:90% It's kind of a little gross. 01:15:40.215 --> 01:15:47.890 align:middle line:90% 01:15:47.890 --> 01:15:54.140 align:middle line:84% So we started with representing x as s times a plus t times b. 01:15:54.140 --> 01:15:56.560 align:middle line:84% And we take s, and we subtract off 01:15:56.560 --> 01:16:00.910 align:middle line:84% a corrective term, which is x div y times either u or v, 01:16:00.910 --> 01:16:03.950 align:middle line:84% where u or v is what represented y. 01:16:03.950 --> 01:16:08.300 align:middle line:84% If you work it out, that's exactly what we were doing here. 01:16:08.300 --> 01:16:14.180 align:middle line:84% We started with one number, and we subtracted off the div times 01:16:14.180 --> 01:16:15.600 align:middle line:90% what we had before. 01:16:15.600 --> 01:16:18.890 align:middle line:84% So if you stare at this equation long enough, 01:16:18.890 --> 01:16:22.250 align:middle line:84% this is exactly what we're doing but in integer linear 01:16:22.250 --> 01:16:26.390 align:middle line:90% combination space of a and b. 01:16:26.390 --> 01:16:29.900 align:middle line:84% So I'm not going to justify this anymore. 01:16:29.900 --> 01:16:32.520 align:middle line:84% And I won't even show an example. 01:16:32.520 --> 01:16:34.430 align:middle line:90% But it gives you-- 01:16:34.430 --> 01:16:37.610 align:middle line:84% this is a particular way to compute it. 01:16:37.610 --> 01:16:40.110 align:middle line:84% And I will say at the beginning of an example, 01:16:40.110 --> 01:16:48.450 align:middle line:84% if you have 5, 1, 0 3, 0, 1, and you run this a few times, 01:16:48.450 --> 01:16:59.270 align:middle line:84% you get to 2, 1 minus 1, 1 minus 1, 2. 01:16:59.270 --> 01:17:04.840 align:middle line:84% And this part tells us that 1 equals minus 1 times 01:17:04.840 --> 01:17:08.420 align:middle line:90% 5 plus 2 times 3. 01:17:08.420 --> 01:17:09.910 align:middle line:90% I hope that is correct. 01:17:09.910 --> 01:17:14.020 align:middle line:90% Minus 5 plus 6 is 1, yep. 01:17:14.020 --> 01:17:17.080 align:middle line:84% OK, So this is how you represent 1 01:17:17.080 --> 01:17:19.610 align:middle line:84% as an integer linear combination of 5 and 3. 01:17:19.610 --> 01:17:23.710 align:middle line:84% 1 is interesting because it's the GCD of 5 and 3. 01:17:23.710 --> 01:17:25.930 align:middle line:84% This is connecting to the water pouring. 01:17:25.930 --> 01:17:28.090 align:middle line:84% Water pouring, we had a container 01:17:28.090 --> 01:17:29.900 align:middle line:90% of size 5, container of size 3. 01:17:29.900 --> 01:17:31.100 align:middle line:90% Their GCD is 1. 01:17:31.100 --> 01:17:35.050 align:middle line:84% They have basically no common divisors except 1 and minus 1. 01:17:35.050 --> 01:17:38.290 align:middle line:84% And we can represent the GCD as an integer linear combination 01:17:38.290 --> 01:17:41.150 align:middle line:90% of 5 and 3, namely this one. 01:17:41.150 --> 01:17:43.750 align:middle line:84% Therefore, we can represent 4 as an integer linear combination 01:17:43.750 --> 01:17:44.330 align:middle line:90% of 5 and 3. 01:17:44.330 --> 01:17:47.830 align:middle line:84% We just multiply both sides by 4, and we get minus 4 times 01:17:47.830 --> 01:17:51.980 align:middle line:84% 5 plus 8 times 3, is one way to do it. 01:17:51.980 --> 01:17:55.280 align:middle line:84% And that gives you at least a way to solve this problem. 01:17:55.280 --> 01:17:58.900 align:middle line:84% It turns out you can take eight copies of 3 01:17:58.900 --> 01:18:01.620 align:middle line:84% and subtract off four copies of 5. 01:18:01.620 --> 01:18:03.200 align:middle line:90% How do I do that? 01:18:03.200 --> 01:18:06.870 align:middle line:84% Well, I just fill this up 8 times in total. 01:18:06.870 --> 01:18:11.820 align:middle line:84% And each time I pour all that I can into the thing of size 5. 01:18:11.820 --> 01:18:15.890 align:middle line:84% And each time it overflows, I empty out the 5. 01:18:15.890 --> 01:18:18.320 align:middle line:84% And so that's exactly how I can build 01:18:18.320 --> 01:18:22.760 align:middle line:84% any integer linear combination of 5 and 3 01:18:22.760 --> 01:18:25.470 align:middle line:84% as long as one side is positive and the other is negative, 01:18:25.470 --> 01:18:27.800 align:middle line:84% which turns out to be necessary, then 01:18:27.800 --> 01:18:32.297 align:middle line:84% you can just take that many copies of filling up one bucket, 01:18:32.297 --> 01:18:34.130 align:middle line:84% pouring it into the other, removing whenever 01:18:34.130 --> 01:18:38.360 align:middle line:84% it overflows, and you will get exactly the number you wanted. 01:18:38.360 --> 01:18:44.180 align:middle line:84% For these numbers, pours of 3 into 5, will give you one. 01:18:44.180 --> 01:18:46.460 align:middle line:84% If you multiply this by 4, I get the number 4, 01:18:46.460 --> 01:18:50.210 align:middle line:90% and I save Manhattan, great. 01:18:50.210 --> 01:18:53.530 align:middle line:84% Good luck on your quiz on Thursday. 01:18:53.530 --> 01:18:59.000 align:middle line:90%