you should do every chess gambit. There are 619 of them. I'm pretty sure that some of your videos are either outdated (or inaccurate in a way that seems outdated, like "actually this thing was proved to be unsolvable, therefore it's solved") or don't contain everything despite claiming to (like in the cases were less than ten examples are presented. comparing to other videos, you could do more). In the second case, just change the title to reflect that (don't say "every"). In the first case, just add an apostrophe to the title (no need to address it in the thumbnail, or at all).
What is the largest number n such that any n points on the plane can be covered by disjoint unit circles? Unsolved. For n=10 there's an elegant proof, so n is greater than 10. For n=60 a pattern of 60 points can be constructed in a triangular lattice that cannot be covered by disjoint unit circles, so n is less than 60.
It kinda makes sense that that's a hard problem. Basically, you have to keep track of the distances of lattice points from the origin. But calculating these distances for a certain "layer" of lattice points based off of the distances of the previous layers only, is quite complicated. The lattice simply isn't a radial configuration. So the Euclidean distance isn't quite the right tool for it. There are other notions of distance that work way better, but sadly circles in Euclidean geometry only care about Euclidean distances. It's like trying to screw in slotted-head screws when all you have is a cross-head screwdriver.
@@lonestarr1490 Is this really unsolved? If you have the equation for half a circle, like \sqrt{r^2 - x^2}, you could than proceed to count the height(y) for every integer(x) on the radius (remove the decimals) and sum them together. then double them. Then add the dots on the zero line. Isnt there a mathematical way to write this? Doesnt that count as solved? I know there is at least an mathematical way to write sum (like a for loop). In latex its written as \sum_{i=1}^{n} a_i
for circle packing, cant you use regular polygon meshes, like squares around triangles, and oth tessellations, to predict and pinpoint the centerpoints of non-regular circle packings?
Am i missing something about the binary disk packing problem? You can have the ratio approach 0 (1 disk has some size and the second one's size approaches 0, therefore the packing approaching perfection)
I mean, it's immediately pretty obvious that this isn't the optimal packing. Instead of having those smaller disks shrink down infinitesimally, there are some clear gaps between the larger disks that you could fill with the smaller ones, achieving a better packing. In fact, the video even explicitly points out at 6:58 that the uniform packing is worse than several of the binary packings.
kissing number in higher dimensions - if we say 4th dimension's kn is 24, then by that logic, the kn for 5th dimension should be 48...6thD kn - 96...and so and so...easy math.
@@kokiczdrzewny except you can't prove me wrong...mostly because common public didn't solve how to represent higher dimensions in graphs easily to understand...I did, therefore I say - it's just doubled.
@@kokiczdrzewny it can be 48 as same as it it 6 for 2D and 12 for 3D. the main issue is - mathematicians make things unnecessary complicated and then they have problem to work with their overcomplicated construct. remember mathematicians who came with easier ways to solve "unsolvable" construct.
@marian20012 the kn for the 1st dimension is 2 and last time i checked 6 isnt double of 2 so your function would just go 2, 6, 12, 24, 48, 96 and so on with that random 2 at the start do you really think mathematicians are that stupid to get so many possible ranges in many dimensions only for a simple "double the number" function to work for all of them?
The Paradox of the Circular Plane Contradictory: In Euclidean Plane Geometry, defining a circle as the set of points equidistant from a center point is paradoxically circular: C = {(x,y) : sqrt((x-a)^2 + (y-b)^2) = r} (Circle of radius r) This defines C using the algebraic distance function invoking C itself. Non-Contradictory: Infinitesimal Pluritopic Homotopy Theory C = {p : ∃q ∈ S1, p =r q} (Circle as monadic group quotient) Tπ = ⨀p ⨂q Γp,q(r) (Winding homotopy over relations) Defining circles C topologically as quotients of the monadic group S1 by pluralistic infinitesimal monadic relations Γp,q avoids circularity using homotopic methods.
No, there's nothing circular about defining a circle using Euclidean distance. The definition of this distance simply relies on the Pythagorean theorem, which doesn't include any circles.
@@knedl9796 Thank you. For I moment I thought I'm dumb. I got a point x in 2D space and I got a positive real number r. What prevents me from considering the set of all points of Euclidean distance r from x?
![I.N "HALLUCINATION" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/n5B5q1Hwt_U/mqdefault.jpg)
Let me know if there's a topic you'd like me to cover next. 😊
be less repeatative and add some personality
you should do every chess gambit. There are 619 of them.
I'm pretty sure that some of your videos are either outdated (or inaccurate in a way that seems outdated, like "actually this thing was proved to be unsolvable, therefore it's solved") or don't contain everything despite claiming to (like in the cases were less than ten examples are presented. comparing to other videos, you could do more). In the second case, just change the title to reflect that (don't say "every"). In the first case, just add an apostrophe to the title (no need to address it in the thumbnail, or at all).
Correction at 8:31 - a disc is only open if it does not contain any of its boundary. You can have sets that are neither open nor closed
?
indeed, you can also have sets that are both closed and open
Erdős-Oler Conjecture just means I can sneak a donut from every triangular donut box without anybody knowing
What is the largest number n such that any n points on the plane can be covered by disjoint unit circles? Unsolved. For n=10 there's an elegant proof, so n is greater than 10. For n=60 a pattern of 60 points can be constructed in a triangular lattice that cannot be covered by disjoint unit circles, so n is less than 60.
Flower from BFDI in the thumbnail 🔥🔥🔥‼‼
Looks like she's grown an extra petal.
What do you mean? I don’t even watch the show, and I still know that Flower has 5 petals, not 6.
@@FlamingSpiral20 whatever, close enough 😐
I was shocked that counting points on grid inside circle r is unsolved problem!
It kinda makes sense that that's a hard problem. Basically, you have to keep track of the distances of lattice points from the origin. But calculating these distances for a certain "layer" of lattice points based off of the distances of the previous layers only, is quite complicated.
The lattice simply isn't a radial configuration. So the Euclidean distance isn't quite the right tool for it. There are other notions of distance that work way better, but sadly circles in Euclidean geometry only care about Euclidean distances. It's like trying to screw in slotted-head screws when all you have is a cross-head screwdriver.
@@lonestarr1490 Is this really unsolved? If you have the equation for half a circle, like \sqrt{r^2 - x^2}, you could than proceed to count the height(y) for every integer(x) on the radius (remove the decimals) and sum them together. then double them. Then add the dots on the zero line. Isnt there a mathematical way to write this? Doesnt that count as solved? I know there is at least an mathematical way to write sum (like a for loop). In latex its written as \sum_{i=1}^{n} a_i
@@SubaruStaffanyou might have figured it out (I have no idea what you just said)
@@SubaruStaffani would assume its a closed formula which is unsolved for. i recently coded this up for my ICPC problem and it worked well till 1e18
@devd_rx whats the meaning of a closed formula, and why isnt what I wrote approved? :)
ill let you know that my kissing number is 0
Fascinating! Thanks!
for circle packing, cant you use regular polygon meshes, like squares around triangles, and oth tessellations, to predict and pinpoint the centerpoints of non-regular circle packings?
so you only need to find all regular polygon tessellations in order to also solve for all possible uniform packings with non regular disks/circles?
Awesome video, I am wondering how can the first conjecture not be solved yet though!
Am i missing something about the binary disk packing problem? You can have the ratio approach 0 (1 disk has some size and the second one's size approaches 0, therefore the packing approaching perfection)
I mean, it's immediately pretty obvious that this isn't the optimal packing. Instead of having those smaller disks shrink down infinitesimally, there are some clear gaps between the larger disks that you could fill with the smaller ones, achieving a better packing. In fact, the video even explicitly points out at 6:58 that the uniform packing is worse than several of the binary packings.
3:34 USA tst 2018 P6!
kissing number in higher dimensions - if we say 4th dimension's kn is 24, then by that logic, the kn for 5th dimension should be 48...6thD kn - 96...and so and so...easy math.
except it is not like that
@@kokiczdrzewny except you can't prove me wrong...mostly because common public didn't solve how to represent higher dimensions in graphs easily to understand...I did, therefore I say - it's just doubled.
@marian20012 except that the kissing number for the 5th dimension is in range of 40-44
so how can it be 48
@@kokiczdrzewny it can be 48 as same as it it 6 for 2D and 12 for 3D. the main issue is - mathematicians make things unnecessary complicated and then they have problem to work with their overcomplicated construct. remember mathematicians who came with easier ways to solve "unsolvable" construct.
@marian20012 the kn for the 1st dimension is 2
and last time i checked 6 isnt double of 2
so your function would just go
2, 6, 12, 24, 48, 96 and so on
with that random 2 at the start
do you really think mathematicians are that stupid to get so many possible ranges in many dimensions only for a simple "double the number" function to work for all of them?
1:14 Empty sum
Flower from bfdi!!!!
That last problem didn't sound easy at all
The Paradox of the Circular Plane
Contradictory:
In Euclidean Plane Geometry, defining a circle as the set of points equidistant from a center point is paradoxically circular:
C = {(x,y) : sqrt((x-a)^2 + (y-b)^2) = r} (Circle of radius r)
This defines C using the algebraic distance function invoking C itself.
Non-Contradictory:
Infinitesimal Pluritopic Homotopy Theory
C = {p : ∃q ∈ S1, p =r q} (Circle as monadic group quotient)
Tπ = ⨀p ⨂q Γp,q(r) (Winding homotopy over relations)
Defining circles C topologically as quotients of the monadic group S1 by pluralistic infinitesimal monadic relations Γp,q avoids circularity using homotopic methods.
No, there's nothing circular about defining a circle using Euclidean distance. The definition of this distance simply relies on the Pythagorean theorem, which doesn't include any circles.
@@knedl9796 Thank you. For I moment I thought I'm dumb.
I got a point x in 2D space and I got a positive real number r. What prevents me from considering the set of all points of Euclidean distance r from x?