@Rodhern
The problem with infinite chess boards is that it takes forever to promote.
@DanielDugovic
As a chess player, there are considerations relevant to chess, such as: - Given a pair of squares, is the king faster or the knight faster? (How does this change if we first randomly move the piece many times to determine its starting square?) - How many routes exist between two squares? - If optimal routes are blocked off, how fast is the second-shortest route?
@e-pluszak9419
One caveat, how exactly do we define "average" when the set is infinite? The paper you cited at 9:56 says "average ratio within a box, between king distance and knight distance", which is indeed well defined as the set is finite, I guess implicitely they assume "limit as the box sidelength approaches infinity", but more importantly: for a square box whose sides are parallel to axis and for example a square rotated at 45°, or an Euclidian circle etc. the answer will be massively different
@charlievane
so the king is just a couple of tiny donkeys in a robe, a (1,0) and a (1,1)
@choco_jack7016
i hate how you draw the knight paths vertical first for all 8 spaces.
@DarinBrownSJDCMath
2:33 "A knight on the rim is dim."
@GMPranav
Super knights are cool for math research, but they would be terrifying in a real chess game lol
@hotdogskid
If youre testing thumbnails out you gotta try using manim in the thumbnail its like clickbait but for recreational math people lol
@polyhistorphilomath
Workin' on our knight moves. Tryin' to make some front page drive-in news.
@duoasch
lets play on a toric board
@Macieks300
9:16 just a nitpick: gcd stands for greatest common divisor, not greatest common denominator
@NotSomeJustinWithoutAMoustache
This visualization at 1:06 with different "starting points" for the knight really helps illustrate why knights on the rim is so dim. In that initial middle placement there were only five 4-move away parts of the board, but in the actual starting placement or at position like 1:10 we see that nearly half the board takes four moves to reach.
@DrTrefor
Corrections (thanks to commenters!!) 1) In the video, I was arguing that if you knew a+b is odd AND you could get two squares to the right, then you could get to everything via symmetries. That's true, but the argument as stated only got us to HALF the darksquares (missing for instance (1,1) if you start at (0,0)). Here's the full argument: If you have an (a,b) knight, where, say, b is the odd one then using the "two steps to the left/right/up/down" argument we proved you can get to (0,1) or its symmetric equivalents and from there you can add to get anything like say (1,1)=(0,1)+(1,0) 2) When writing gcd(a,b)=1 I apparently said "denominator" not "divisor" - yikes! 3) At 11:40 I noted that if you know the estimates on the teal and blue triangles, by symmetry you get estimates everywhere. That's true, but my animation covering the full screen swapped the colors by mistake. It shouldn't be alternating, it should be two blue triangles then two teal triangles etc. This was my first video entirely created using Manim - the python library originally created by 3blue1brown. But I knew precisely zero python a month ago. I actually used this video's sponsor https://brilliant.org/TreforBazett to master the basics of python in about 2 weeks, and after that my foundation felt strong enough to tackle this video. Hope you enjoyed!
@anikethdesai
Please make videos on the Laplacian Operator and the Hessian Matrix and why it's used to find Maxima and Minima of a surface. Thank You, absolutely love your videos!!
@maxthexpfarmer3957
at 8:32 i would like to point out that being able to go 2 squares in any direction does not necessarily give the ability to go to any dark square, since going 1 square diagonally could still be out of the picture
@___d3p1
Amazing video and animation!
@mohammadhamdar9521
I had the honor to play chess with Tafula, he is unbeatable!
@PrimordialOracleOfManyWorlds
suppose you use 3D space, what will happen?
@mathematicskid
Sorry, but the proof around 9:00 is flawed. Being able to move by 2 doesn't mean you can reach all squares of the same color, only half. This is fixable by noting that if the knight moves by (2a, 2b+1) and its variants, and it can move by 2, it can move by (2a, 2b+1) - a * (2, 0) - b * (0, 2) = (0, 1) and thereby reach cell (c, d) by c * (1, 0) + d * (0,1), or by noting that any desired square of the opposite color has some square it can move to that's in the same half of the same color and some square that's in the opposite half (since the cells (-a, b) and (-b, a) away differ by (a-b, a-b), and a-b = a+b - 2b = odd - even = odd), and that as a result, any cell in the opposite color is reachable, and the opposite half of the same color is reachable through any cell of the opposite color.
@ChristianTáfula