Timeline for Mathematical solution for a two-player single-suit trick taking game?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Feb 17, 2010 at 11:49 | comment | added | Ewan Delanoy | @ Qiaochu : I'm afraid there is no "normal condition" here, so the Sprague-Grundy theorem is not applicable. | |
| Feb 17, 2010 at 11:38 | vote | accept | Ewan Delanoy | ||
| Feb 17, 2010 at 1:49 | history | edited | Reid Barton | CC BY-SA 2.5 |
edited title
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| Feb 17, 2010 at 1:29 | answer | added | Reid Barton | timeline score: 21 | |
| Feb 17, 2010 at 1:00 | comment | added | Ori Gurel-Gurevich | Just a note: it is disadvantageous to play first, so I'm not sure whether the "natural" version should be that the winner of the previous trick plays first. | |
| Feb 17, 2010 at 0:15 | comment | added | Alon Amit | @Sam, I just corrected that. | |
| Feb 17, 2010 at 0:14 | history | edited | Alon Amit | CC BY-SA 2.5 |
added 1 characters in body
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| Feb 16, 2010 at 22:01 | comment | added | Sam Nead | Typo in second para - $I$ should range over subsets of $\{ 1, 2, \ldots, 2n \}$. | |
| Feb 16, 2010 at 21:24 | comment | added | Qiaochu Yuan | The Sprague-Grundy theorem might be applicable: en.wikipedia.org/wiki/Sprague%E2%80%93Grundy_theorem | |
| Feb 16, 2010 at 20:51 | history | asked | Ewan Delanoy | CC BY-SA 2.5 |