Timeline for What is the limit of the "knight" distance on finer and finer chessboards?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
|
|
| Sep 29, 2011 at 3:56 | comment | added | Noam D. Elkies | Here's an exact formula for the Knight distance $d$ from $(0,0)$ to $(x,y)$. By symmetry we may assume $0\leq x\leq y$. Clearly $d$ satisfies $2d\geq y$, $3d\geq x+y$, and $d\equiv x+y\bmod2$. Then $d$ is the smallest integer satisfying these conditions, unless $(x,y)=(0,1)$ or $(2,2)$ when the distance is $3$ or $4$ respectively, exceeding the expected answer by $2$. These exceptions are well-known to chessplayers... On a finite chessboard the edges produce one more exception: the distance from a corner to its diagonal neighbor is $4$, not $2$ (the knight would have to leave the board). | |
| Sep 29, 2011 at 1:11 | history | edited | Qfwfq | CC BY-SA 3.0 |
edited body
|
| Sep 29, 2011 at 1:10 | vote | accept | Qfwfq | ||
| Sep 29, 2011 at 0:21 | comment | added | Michael Lugo | Similarly, one could define a "king distance", which has unit ball the square with corners at $(\pm 1, \pm 1)$ and a "rook distance", which has unit ball the square with corners at $(\pm 1, 0), (0, \pm 1)$. (The "rook distance" is really the distance which corresponds to a piece which can move $(\pm 1, 0)$ or $(0, \pm 1)$ on each move.) Somewhat surprisingly, the "bishop distance" defined in this way is the same as the king (or queen) distance... | |
| Sep 28, 2011 at 18:59 | answer | added | Will Sawin | timeline score: 13 | |
| Sep 28, 2011 at 18:35 | history | edited | Qfwfq | CC BY-SA 3.0 |
added 78 characters in body
|
| Sep 28, 2011 at 18:30 | comment | added | Simon Rose | Ah, that can't be right. This construction is constnt for points on the original lattice Z^2, so what I said makes no sense. | |
| Sep 28, 2011 at 18:28 | comment | added | Simon Rose | I admit that my intuition is that you should just get the normal Euclidean distance... It seems that for finer and finer chessboards, the L-shaped moves a night must make would disappear, and you would only see straight lines. | |
| Sep 28, 2011 at 18:18 | history | asked | Qfwfq | CC BY-SA 3.0 |