Timeline for Fill $N$ by $M$ grid with numbers in such a way that any given cells' neighbors are different
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 8, 2019 at 12:28 | comment | added | user33097 | @Hugh, no problem at all! I see from your history you like to make grammar edits whenever possible, which is great, so just wanted to make sure. Keep it up! | |
| Jan 8, 2019 at 6:20 | comment | added | user46002 | @tilper oh, I'm sorry. I was typing pretty carelessly and I must have missed it when I was proofreading. Sorry! | |
| Jan 8, 2019 at 5:41 | comment | added | Dr Xorile | Oh, I misunderstood the question. My bad | |
| Jan 7, 2019 at 21:52 | comment | added | Jaap Scherphuis | @DrXorile On a normal map in the 4-colour theorem, countries that touch only at a single point are not considered adjacent, because otherwise you could let 100 countries meet and you'd need 100 colours. Consider the four neighbours of a cell. These are supposed to be pairwise adjacent, because they all need to get different colours/numbers. On a normal map, you can make the top and bottom regions adjacent by putting a border line between them, but that cuts the left/right regions off from each other. But here the top/bottom adjacency crosses over the left/right - i.e. non-planar. | |
| Jan 7, 2019 at 19:47 | comment | added | user33097 | FYI @Hugh, the possessive form of "its" does not have an apostrophe. | |
| S Jan 7, 2019 at 19:43 | history | suggested | user33097 | CC BY-SA 4.0 |
Undoing a grammatical error introduced in the previous edit. The possessive form of "its" has no apostrophe.
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| Jan 7, 2019 at 19:41 | comment | added | user33097 | Took me a bit but I think this is more of a variant than a special case of the map coloring problem. | |
| Jan 7, 2019 at 19:16 | review | Suggested edits | |||
| S Jan 7, 2019 at 19:43 | |||||
| Jan 7, 2019 at 18:19 | answer | added | JMP | timeline score: 1 | |
| Jan 7, 2019 at 17:53 | vote | accept | iro otaku | ||
| Jan 7, 2019 at 17:47 | comment | added | hexomino | @DrXorile I'm with Jaap on this, I don't see the connection. For instance, you can change the problem from "square" cells to hexagonal ones where each hexagon has to border six, each containing different numbers. How would the analogue generalise? | |
| Jan 7, 2019 at 16:44 | comment | added | Dr Xorile | @Jaap, I cannot see how a 2d matrix could be anything but planar | |
| Jan 7, 2019 at 16:15 | comment | added | Jaap Scherphuis | @DrXorile I disagree. I think the graph corresponding to this problem is not planar when $N,M$ are large enough ($N=M=4$?), which is one of the prerequisites for the four-colour theorem to apply. The worst case may be $4$ nevertheless, but not because of that. | |
| S Jan 7, 2019 at 15:38 | history | suggested | user46002 | CC BY-SA 4.0 |
mathjax and grammar
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| Jan 7, 2019 at 15:19 | review | Suggested edits | |||
| S Jan 7, 2019 at 15:38 | |||||
| Jan 7, 2019 at 15:10 | comment | added | Dr Xorile | The worst case is 4 since this is just a special case of the map coloring problem | |
| Jan 7, 2019 at 15:02 | answer | added | Magic turtle | timeline score: 2 | |
| Jan 7, 2019 at 14:56 | answer | added | hexomino | timeline score: 4 | |
| Jan 7, 2019 at 11:40 | review | First posts | |||
| Jan 7, 2019 at 11:42 | |||||
| Jan 7, 2019 at 11:35 | history | asked | iro otaku | CC BY-SA 4.0 |