Timeline for Hexagonal rooks
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 13, 2017 at 23:33 | answer | added | Gerhard Paseman | timeline score: 2 | |
| Jul 31, 2012 at 19:35 | vote | accept | Jeremy Martin | ||
| Jul 31, 2012 at 13:57 | answer | added | Cristi Stoica | timeline score: 13 | |
| Jul 31, 2012 at 3:22 | answer | added | Noam D. Elkies | timeline score: 14 | |
| Jul 31, 2012 at 1:14 | comment | added | Patricia Hersh | Only loosely related: there's been an extensive study of simplicial complexes called chessboard complexes whose i-dimensional faces are the collections of i+1 nonattacking rooks on a chessboard. Now you make me curious about hexagonal chessboard complexes. The traditional chessboard complexes actually have come up in quite a few different contexts and have all sorts of torsion, etc. | |
| Jul 31, 2012 at 1:10 | answer | added | Will Sawin | timeline score: 6 | |
| Jul 31, 2012 at 1:00 | comment | added | Will Sawin | One of the coordinates must have an average value of no more than $n/3$ among all the rooks. The maximal number of distinct nonnegative integers whose avergae is $n/3$ is $2n/3+1$. This is a better bound. | |
| Jul 31, 2012 at 0:56 | comment | added | Will Sawin | Isn't $n+1$ an upper bound because no two of the rooks can agree on the first coordinate? | |
| Jul 31, 2012 at 0:25 | comment | added | David E Speyer | In particular, the rook placement $(2,1,0)$, $(1,0,2)$, $(0,2,1)$ for $n=3$ is better then the bound $(3+1)/2$ that Gerry proposes. I can't figure out how to generalize that example, though. | |
| Jul 31, 2012 at 0:15 | answer | added | Joseph O'Rourke | timeline score: 7 | |
| Jul 30, 2012 at 23:26 | comment | added | David E Speyer | I can prove the upper bound $3(n+2)(n+1)/(4n)$: Each rook threatens $2n$ "square"s, but each of the $(n+2)(n+1)/2$ squares may only be threatened $3$ times. | |
| Jul 30, 2012 at 23:23 | comment | added | David E Speyer | That's a nice idea, but I don't think it works. If you remove (2,1,0) from $G_3$, you get a path of length $3$, not a cycle. | |
| Jul 30, 2012 at 23:01 | comment | added | Gerry Myerson | The graph is regular, of degree $2n$. If you put a rook on $(n,0,0)$, the problem reduces to finding the maximum number of non-attacking rooks on the board of order $n-2$. I think the graph has enough symmetry that no matter where you put the first rook you reduce the problem to that of order $n-2$, but I'm not sure. If the graph does have that symmetry, then by induction you get more-or-less $n/2$ rooks ($(n+2)/2$ if $n$ is even, $(n+1)/2$ if $n$ is odd). | |
| Jul 30, 2012 at 22:27 | comment | added | user25199 | The bishop question is easy - it's 2n-2. Put bishops along two opposite edges and remove two at adjacent corners. This is maximal since there are only 2n-1 different North-East diagonals, and two of these are opposite corners so only 2n-2 can be filled. | |
| Jul 30, 2012 at 21:52 | comment | added | Zsbán Ambrus | Can you prove what the maximal number of mutually non-attacking bishops is on an ordinary n×n chessboard? | |
| Jul 30, 2012 at 19:59 | history | asked | Jeremy Martin | CC BY-SA 3.0 |