Each point is connected to its closest neighbor, how many connections can a single point have (max)? 

It would also be interesting to apply this to other metrics

My first guess is that it'd always be at least 1 less than the kissing number for that dimension

seems related to the kissing number problem, which is unsolved already in 5D

@@leo.simensen I think it is exactly the kissing number for that dimension, unless that number is tight, in which case it's 1 less than the kissing number.

What’s the Kiss9ng number?

The sketches have made it really hard for me to take his voice seriously. I keep expecting a joke.

@@alansmithee419 the entire time I was expecting to hear something like "PA < PB which also tells us that (PA + PB)*1000 < The amount of time you've spent with women"

5 cause you can't have 6, and 7 is impossible

My bad accidentally replied instead of writing a normal comment

@@briangeer1024 did you know you can delete comments?

@@o5-1-formerlycalvinlucien60 eh I don't think it really matters in the end

I did the same thing for 6 my way :D

i did the same. i thought about it before watching the video so i didn't know about the "unique distance" rule, so i thought of using equilateral triangles to see how many would fit in a circular arrangement around a point. and then i thought, okay, a circle is 360 degrees and the triangles are 60 degrees, so 360 divided by 60 is 7. got it. answer is 7. i felt so smart. it's always the stupid arithmetic errors that get you.

Yeah this is exactly how it was for me. Before I clicked on the video I was thinking "it's 6" then he said that the distances are not equal, so I knew the answer was 5.

same here

although if i were to be a bit more accurate, the first thing i did was think of an equilateral triangle, and then from that i got a hexagon

Its crazy how much high- quality content is present on youtube. Its such a great way to stay interested and learn overall

I didn't understand please can you explain? I thought of infinity when he said an infinite plain

​@@someone.-gi7zr Right, that is intentionally not very relevant. The infinite plane is often invoked to basically free up the possibility space, from the kinds of strange and unhelpful restrictions that can come with trying to assign an arbitrary bounding space. An unbounded space means we can try anything - It does NOT mean anything in particular will actually WORK. So, the proof - It's true that if we were just concerning ourselves with how many points a point could connect to generally, the answer would easily be infinity. However, we have a very specific constraint here - Each point must connect, specifically and exclusively, to the point CLOSEST to itself. We can't just make arbitrary connections. If that were the only constraint, a hexagonal arrangement, with a point representing the center of the hexagon, would allow us to get one point to have six connections. However, because that introduces some fuzzy arbitrary decision making, the "unique distance" constraint is added. This makes 6 impossible, but 5 is still achievable. I want you to think about a hexagon - If need be, draw this out on a piece of paper - And really THINK about where you would put a point to go from six points connected to one, to seven. Keep in mind, any new connection you make will potentially unmake other connections. That's the real trick here. If we put a point INSIDE the hexagon, we'll either connect it to the center and redirect one or more of our initial connections AWAY from the center, or we'll connect to one of the outer points, redirecting it away from the center. If we put the point OUTSIDE the hexagon, then it will always connect to an outer point, again robbing the center of a connection. If we put two or more points far enough away, they connect to each other, and have no affect on the original arrangement at all. I hope you see that even with infinite possibilities, by simply categorizing those possibilities and taking note of the properties of the CATEGORIES, we can do finite work to describe the infinite, and solve problems therein. There's an entire branch of mathematics all about that, but in this proof it's practically a tangential, hardly notable thing. At least, to a trained mathematician who's expecting this kind of infinite tomfoolery. I encourage you to play around with this problem with paper and pencil - Get a feel for various arrangements, see what you can make of it. Try the hexagon arrangement. Try interrupting it. Try a septagon and a pentagon. Try connecting some random points, by the rules described in the video. You may find the process more enlightening and invigorating than simply being told about it :)

sure you did buddy

@@uggupuggu try it, the method works

It's a poorly-posed puzzle because of that.

Well, that no two distances are the same was a given. So, you cannot have regular hexagon.

​@@AbiGail-ok7fc Probably answered from the title. It was a pretty obvious solution if you know what a regular hexagon is. The "distances must be unique" was only mentioned a minute or so into the video.

Holy shit, good video!

I agree and am doing the same. [Edited to add] Yay! :)

Same... 6 connection points would mean that all points are equally far from each other. But since the center one only connects to the closest is must be less than 6, thus 5

On a spherical surface (or more in general anywhere the curvature of the surface is positive) you get the same result as on a flat surface (max 5 lines) as long as your points are “close enough” to each other. On a hyperbolic surface (constant negative curvature), on the other hand, you can have as many lines you wish connecting to the central point, provided the points are sufficiently distant from each other.

To use this though you would need to prove that a hexagon is comprised of equilateral triangles. To prove that would probably look a bit like the proof in the video which uses more geometric primitives

The first assumption was that all the distances were unique

No two points are allowed to have the same distance as a different set of points

Watch the videos beginning again pls

you can't because the premise is that all the distances within the graph are different

We aren't quite able to do that, since at the start of the video he specifies that the distances are all unique. However, if we throw that constraint away, then we can sneak in a 6th point and make a regular hexagon by using similar reasoning to the video.

there can't be same length thus no

ah yeah, math mistake, sorry :D i dont know why i remember sqrt(3)/2 for hexagons

That's not graph theory, that's geometry.

He stated in the beginning of the video that all distances between points are unique. There are no equidistant points