Timeline for Does one pieces of every kind of connected polyominoes P in $\mathbb{R}^2$ which has no hole cover a plane?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| May 2, 2017 at 0:57 | vote | accept | Takahiro Waki | ||
| May 1, 2017 at 15:11 | history | edited | Will Sawin | CC BY-SA 3.0 |
added 29 characters in body
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| May 1, 2017 at 15:07 | comment | added | Gerhard Paseman | Granted, if you are using only polyominoes without holes, then you want the unfilled region to have at most one hole. This can be done, but I still think "arbitrarily" is a word far from optimal in describing the process. Gerhard "It's How You Say It" Paseman, 2017.05.01. | |
| May 1, 2017 at 15:03 | comment | added | Gerhard Paseman | Because the unfilled region is connected and infinite, there are infinitely many polyominoes that will fit there. Some of them will be unused. Gerhard "So You Are Still Covered" Paseman, 2017.05.01. | |
| May 1, 2017 at 15:00 | comment | added | Will Sawin | @GerhardPaseman You should probably also require that the filled region be connected. If the unfilled region contains an shape like the letter P, where the line coming out is one square thick, there may be no way to fill it using polyominoes without holes that weren't used already. | |
| May 1, 2017 at 14:43 | comment | added | Gerhard Paseman | That is why with each placement, you require that the unfilled region be connected. Polyominoes with holes can be accommodated only if there are existing covered regions that fill those holes precisely, or combinations of unused polyominoes that (when placed together as a group) help maintain the invariant of the unfilled region being connected. Gerhard "Arbitrary Up To Some Constraints" Paseman, 2017.05.01. | |
| May 1, 2017 at 14:31 | comment | added | Gerhard Paseman | Note that you can modify the above to include some polyominoes with holes, so that a few more polyominoes are used. Gerhard "Not All Of Them, Unfortunately" Paseman, 2017.05.01. | |
| May 1, 2017 at 14:22 | comment | added | Will Sawin | @GerhardPaseman Sure, except you also need to specify that it is connected with no hole - otherwise you could produce a 1-square hole when the size 1 polyomino is already used, say. | |
| May 1, 2017 at 14:10 | comment | added | Gerhard Paseman | Of course it can be made to work. It may be more clear to say: On turn 2i, cover without overlap the smallest numbered uncovered cell with an unused polyomino so that the remaining unfilled region stays connected, and on turn 2i+1, place smallest numbered unused polyomino without overlap so that the remaining unfilled region stays connected, but otherwise arbitrarily. This then gives a tiling by induction. Gerhard "Not Much More Is Needed" Paseman, 2017.05.01. | |
| May 1, 2017 at 13:54 | comment | added | Will Sawin | @GerhardPaseman If I place it sufficiently far from the existing tiles, neither of these can happen. | |
| May 1, 2017 at 3:54 | comment | added | Gerhard Paseman | It can't be too arbitrary, as you don't want to make new holes, or cause a tile to be needed in more than one place. Gerhard "Patch Up Holes In Argument" Paseman, 2017.04.30. | |
| May 1, 2017 at 3:28 | history | answered | Will Sawin | CC BY-SA 3.0 |