Timeline for The Sudoku game: Solver-Spoiler variation
Current License: CC BY-SA 3.0
25 events
| when toggle format | what | by | license | comment | |
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| Aug 11, 2021 at 0:53 | history | edited | Tony Huynh |
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| May 6, 2018 at 12:51 | answer | added | yyw | timeline score: 1 | |
| May 6, 2018 at 0:06 | history | edited | Joel David Hamkins |
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| Apr 23, 2018 at 13:38 | answer | added | Tony Huynh | timeline score: 8 | |
| Apr 18, 2018 at 18:21 | comment | added | Joel David Hamkins | Ah, now I can do it using a group $G\oplus H$, where $G$ has size $\lambda$ and $H$ has size $\kappa$. So all the asymmetric infinite boards in fact have solutions. | |
| Apr 18, 2018 at 13:40 | comment | added | Joel David Hamkins | @GerhardPaseman Your question is very good. I can provide Sudoko solutions for $(\lambda\times\kappa)\times(\kappa\times\lambda)$ if both $\lambda$ and $\kappa$ are countable, which includes the infinite asymmetric case $(n\times\omega)\times(\omega\times n)$ for finite $n$. | |
| Apr 18, 2018 at 0:29 | comment | added | Joel David Hamkins | @GerhardPaseman Good question! Now that you mention it, I'm not actually sure. | |
| Apr 18, 2018 at 0:18 | comment | added | Gerhard Paseman | Can you show (for kappa,lambda) that Solver has a win if played unopposed? Gerhard "Solution Space May Be Empty" Paseman, 2018.04.17. | |
| Apr 17, 2018 at 23:12 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Updated question on asymmetric Sudoku
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| Apr 17, 2018 at 19:11 | comment | added | Joel David Hamkins | Ah, I see, thanks! That would affect my answer at your other question, since I didn't treat those cases. | |
| Apr 17, 2018 at 19:04 | comment | added | Christopher King | @JoelDavidHamkins Here are some examples of six by six sudoku: google.com/… In particular, any $nm \times nm$ sized board is possible. | |
| Apr 17, 2018 at 19:00 | comment | added | Joel David Hamkins | @PyRulez I don't quite follow you, since to my way of thinking, there is no $6\times 6$ Sudoku board, since it must be $n^2\times n^2$. Do you mean $36\times 36$? | |
| Apr 17, 2018 at 17:55 | comment | added | Christopher King | Related | |
| Apr 17, 2018 at 17:30 | comment | added | Christopher King | @JoelDavidHamkins Also, a nice follow up question is how close the solver-spoiler game is if the pie-rule is employed. | |
| Apr 17, 2018 at 17:29 | comment | added | Christopher King | @JoelDavidHamkins (The particular variant was one in which one player plays 2,4,6 and the other plays 1,3,5 on a 6x6 board. The first who can't move loses. Its a kind of symmetrical solver-spoiler game.) | |
| Apr 17, 2018 at 17:26 | comment | added | Joel David Hamkins | In that case, I think the question becomes extremely interesting! | |
| Apr 17, 2018 at 17:06 | comment | added | Christopher King | A friend and I where playing a variation on this, and the problem seems to be that solver can spoil the spoilers attempts to create an invalid position. Therefore, the spoiler's strategy will need to involve spoiling multiple things at once. | |
| Apr 17, 2018 at 15:40 | answer | added | Gerhard Paseman | timeline score: 2 | |
| Apr 17, 2018 at 15:27 | comment | added | Gerhard Paseman | If I get anything close to a strategy, I will post it. Gerhard "Will Try Planning My Mistakes" Paseman, 2018.04.17. | |
| Apr 17, 2018 at 15:25 | comment | added | Gerhard Paseman | The above suggests a strategy for Spoiler: develop but do not complete two traversal, and then spoil them either if Solver tries to complete a traversal, or if the constraints become narrow enough (but not too tight) to prevent both traversal from being spoiled. Alternatively, keep track of spoilable transversals. I conjecture the solutions space is such that Spoiler has a win before C/n of the n by n transversals have been filled where C is an absolute constant. Gerhard "It Doesn't Take Me Long" Paseman, 2018.04.17. | |
| Apr 17, 2018 at 15:20 | comment | added | Joel David Hamkins | Yes, I also expect the Spoiler to win on all the nontrivial finite boards. But can you make a general argument? We should not assume that the Solver is playing the symmetric strategy, since this is easy to defeat in this game variation. | |
| Apr 17, 2018 at 15:11 | comment | added | Gerhard Paseman | This is the variation I was considering when thinking about the symmetric strategy you proposed in another question. If the first player starts on an empty 4x4 board and the second player tries a symmetric strategy, the first player can spoil the solution by "breaking" a transversal. E.g. after both players play 1, then first player plays a three where a one would be expected. I imagine the Sudoku solution space is sparse enough that one can break any solution attempt. Gerhard "Goodness Knows I Have Frequently" Paseman, 2018.04.17. | |
| Apr 17, 2018 at 14:46 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 6 characters in body
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| Apr 17, 2018 at 14:40 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 91 characters in body
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| Apr 17, 2018 at 14:34 | history | asked | Joel David Hamkins | CC BY-SA 3.0 |