Timeline for Dissecting a square
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 29, 2011 at 11:31 | vote | accept | Colin D Wright | ||
| May 29, 2011 at 11:31 | |||||
| May 26, 2011 at 20:22 | comment | added | Colin D Wright | Or define two partitions to be equivalent if they agree on all but a nowhere dense set, which is probably the same, but I'd need to work hard to think about that clearly enough. Very helpful - thank you. | |
| May 26, 2011 at 20:06 | comment | added | Juris Steprans | Even connected is not enough though. For example let the square in question be centred at the origin and have width 2. For any subset $X\subseteq (0,1]$ consider the partition consisting of the open upper half of the square together with $X$ and $[-1,0) \setminus -X$. This again yields many connected partitions. So one should probably ask for partitions into open connected sets and define "partition" to mean you partition all but a closed nowhere dense subset of the square (so that you can ignore the boundaries). | |
| May 26, 2011 at 19:51 | comment | added | Juris Steprans | If you ask for all partitions into two pieces, the sets $X(f)$ and $Y(f)$ I describe do give you all solutions. If you ask for partitions into $n$ pieces then a similar argument, using $n^\text{th}$-roots of unity, also yields a description of all partitions. If you ask for connected pieces, that is a different question. | |
| May 26, 2011 at 18:48 | comment | added | Colin D Wright | Yes, but it's not all the solutions. Agreed that more solutions won't increase the cardinality, but not including all the solutions seems "sub-optimal." What if you additionally require the "pieces" to be connected (in some sense)? | |
| May 26, 2011 at 17:57 | history | answered | Juris Steprans | CC BY-SA 3.0 |