Dope Tech Episode 2

<<@GordonHugenay says : I'm assuming the points have to be distinct? Otherwise I have a counterexample: take 6 points and place 2 each at a corner of an equilateral triangle. Then there are 12 pairs of points at maximum distance.>> <<@eulerianorder6972 says : Hey Zach, Thanks for the cool problem and solution! I think there might be a slightly easier and more intuitive solution though. Think of this whole system as a graph in the following way: every point is a node and nodes have a vertex between them if they have the maximum distance( =1) between them. Now just consider the connected graphs, say G_1…,G_n with n_1,…,n_k number of nodes in total. Now, by induction each G_i has at most n_i nodes with max distance between them. So adding up, there are at most n_1 + …. + n_k = n nodes with the maximum distance between them. (You can check the base quickly)>> <<@greenmarble638 says : Thats one of the questions a mathematician thought of while high that eventually turns out to have a use in 4 different fields of science>> <<@Retro-bt2zc says : This is surreal lol Im expecting like a punchline because your voice sounds like its a joke but no, serious. LOL>> <<@agoofyahhchannel6870 says : sometimes i forget that this guy is really fucking smart and not some frat bro>> <<@DaJellyMoon says : What does it mean " 1 away" ? What is "1" in this case? Like 1 point away, 1 cm away? @Zach Star>> <<@Ridlay_ says : I never knew this channel existed until someone pointed it out on your other channel. Dang that’s impressive.>> <<@Ben.N says : pog>> <<@unknownentity5354 says : Could you do a video on the usages of abstract vector spaces from linear algebra?>> <<@tzimiscelord8483 says : I just want to swing by and say that I just discovered that you do both comedy and science and I'm super impressed. Keep grinding okay>> <<@mariogonzalez-jy6zr says : Maybe i'm getting something wrong, but is there any rule that doesn't allow for several points to be located in the same place? For instance, when it is reduced to x, a, c; if there where two points in location a wouldn't you have pairs a-x, a2-x, a-c, a2-c, and x,c. Those will be 5 max distance pairs with four points.>> <<@madmathematician4458 says : madguitargenius 🎸>> <<@LilCalebW says : Niice>> <<@jacopolattanzio8790 says : at least 3>> <<@TrayyaOnTwitch says : This is so different coming from the skit channel>> <<@JinaMukherjeeF says : Yooo had no clue u were a student.Most utubers r untalented>> <<@dominicbeaver1385 says : Question: If all of the points were in the exact same position, wouldn't that mean that they would all be the same distance away from each other? I mean, if you have 10 points all at the same x-y coordinates, wouldn't every pair distance be at the maximum? You said at 1:11 that if there are ten points, there are 45 possible pairs. As, in this situation, every pair would be the maximum, with 10 points wouldn't it possible to get 45 maximum distance pairs? Or maybe I just understood the question wrong, idk lol>> <<@bananaman-qj4nu says : What would happen if instead of being on a plane, the points were on a boat?>> <<@townofsalme says : Me watching this video knowing full well I suck at geometry: "Mmmm yes, very true.">> <<@th3m1st0cl3s says : Jigsaw frantically watching this in the background, trying to get a machine to work>> <<@ffc1a28c7 says : The statement at 3:32 is easily shown via the pigeonhole principle. You know that there will be >2n line segment ends, then if those ends are evenly distributed among the points, after 2n are distributed, each will have 2. Thus, at least 1 must have 3.>> <<@phillipotey9736 says : I'd like to see this for n dimentions>> <<@lukeskymaster says : I just wanted to see your main channel, and didn't know it was this, pretty neat>> <<@godfreypigott says : Odysee is a site that supports far-right propaganda. A leaked email from one of its executives states: _“a Nazi that makes videos about the superiority of the white race would not be grounds by itself for removal from the platform."_ Why would you support this garbage?>> <<@isws says : nice vid! Now do it in 3d _-_>> <<@azimuth4850 says : Isn't there a simpler proof? Start with 3 points. They must define an equilateral triangle, it's the only case. Label the points 1, 2, and 3. Imagine the 1st point is the center of a circle, the connect the arc of that circle between points 2 and 3. Do this same procedure for points 2 and 3, connecting circular arcs between the other two points with that point as the center of the circle. Now the only places you can add new "max distance pairs" is anywhere along these arcs. Except that each time you add a point, it ONLY pairs with the point at the center of the circle of the arc you placed it on, and does not form pairs with points on any other arc. Because geometrically that distance would be less than 1, even if only infinitesimally. Therefore you can never form more max distance pairs than points you add, because each point can, at maximum add only 1 new pair. Correct me if I'm wrong.>> <<@smg5120 says : Is it possible the universe doesn’t end or we find someone to cheat it and the death of the sun btw screw Conrad Murray worldwide, Michael Jackson the greatest>> <<@justinjustin7224 says : I feel this proof would have been simpler by drawing two points, a circle around each point that has a radius of the distance between the two points, then showing that the two points where the circles intersect have a greater distance than between the two original points. With just that, you'd show that 3 points can have 3 max distance lengths, but any more points added to the system cannot increase the number of max distance pairs by 2, which would be necessary for m to become greater than n. This simple construction was the entire basis of the response I left on the previous video.>> <<@tamirerez2547 says : For each n points there are m pairs with the largest distance. OK I got it. We must prove that m n So what options are left ???>> <<@johnchessant3012 says : Question: Can every proof by 'descent' (like this video's proof) be rephrased as a proof by induction? And is one more 'aesthetic' than the other?>> <<@johnchessant3012 says : 3:14 This part is easy. Add up all the numbers of max-distance pairs that each point is a part of. This should be twice the number of max-distance pairs since each pair is counted twice. So by assumption it's greater than 2n. But that would be impossible if each of the n points were part of at most 2 max-distance pairs. So at least one is part of 3 (or more) max-distance pairs.>> <<@kylecow1930 says : first two make shape from overlapping cirlces, whenever corners are saturated two maxes added but you can only make 2 corners twice, the rest of the time corners are formed by cutting off other corners so corners <= No. points, max is achived when all corners are saturated as adding a point to a corner adds 2 maxes and adds 1 point so as available corners is <= numberr of points available if you fill all the corners to make the maxes it still wont be enough as saturating each corner will only add 1>> <<@swerasnym says : A slight thing (at 2:00) you are stating that if we have 2 line segments of length 1 that are not crossing we can form a quadrilateral with intersecting diagonals. But that are missing not coverint e.g. the pairs {(0,0), (0,1)} and {(1, ½), (2, ½)} for which we can connect every point with every other one without any interactions. To resolve these we can note that if the convex hull of our 4 points form a triangle with a base of 1 with the corresponding height of at least 1. Thus one of the 2 remaining sides have to have a length > 1 and the statement that the segments must cross are still true!>> <<@sinecurve9999 says : Nice proof! What would happen if the points, instead of being on a plane, were on some surface with positive or negative curvature? How would the argument change?>> <<@Chalisque says : Re 3:20 is proved by the pigeonhole principle.>> <<@neobullseye1 says : What happens if you make a circle though? And I mean a full circle, not just a bunch of dots roughly in the shape of a circle. A circle has an infinite amount of points on it (which is relatively easy to prove, but not the point right now, so I'm going to skip over that). The longest distance between any given two points can be found between one point and the one point on the exact opposite side of the circle. Since all the points are on a circle, this logically goes for every single pair there is; no point has more than one partner, and no point has no partner at all. Thus, there would be exactly half as many pairs as there are points. But we said that there are an infinite amount of them, and half of infinity is still infinity. So... there would also be an infinite amount of pairs, correct?>> <<@gauravagrawal9265 says : Another way to prove: Just think about the maximum number of pairs possible. All points have to fall on the perimeter of the circle. Each pair will fall in its diameter (max distance between them). Any point outside will not be possible as it will exceed the diameter or our maximum distance. This will be true for every Even number of points. Number of pairs=n/2. Now, an odd number of points will also fall into a circle, but not in diameter. Just like 3 point maximum distance shown in video. In this case maximum 2 pairs are possible from 1 point. Number of pairs=n>> <<@yonatanbeer3475 says : Why can't theta > 60deg, and dist(a,c) > 1? That's ok, as long as all the other distances are 1 then (a,c) doesn't have to be 1, no? EDIT: I'm dumb, then (x,a) and (x,c) wouldn't be max distance. Stupid stupid.>> <<@justsomegoosewithinterneta4199 says : Cool as hecc>> <<@simonmaier9274 says : Dude that‘s actually cool>> <<@laneeardink9849 says : 97 points>> <<@nickseggie1047 says : 5 points>> <<@Inapainting says : id say 7 points>>