Timeline for A set-family game
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Nov 6, 2018 at 9:43 | vote | accept | Erel Segal-Halevi | ||
| Sep 10, 2018 at 19:29 | comment | added | Ilya Bogdanov | Sorry, I do not understand why this works for $G(2k,k+1)$. But, surely, this works for $G(2k+1,k+1)$. Moreover, in that case we conclude that $G(2k+1,k+1)=1/2$, since one may construct $F$ of two disjoint sets. | |
| Aug 7, 2018 at 11:01 | comment | added | JKreft | @ErelSegal-Halevi I believe so. As I noted above, I think it can be extended for any $G(2k,i)$ where $i\leq k+1$. | |
| Aug 7, 2018 at 11:00 | comment | added | JKreft | Rejected an edit that changed the proof to be incorrect. Note that in my version, A doesn't make an arbitrary first move, it just starts with B's optimal strategy as its first move. If B can force the win going second, A can force it going first. | |
| S Aug 7, 2018 at 11:00 | history | rollback | JKreft |
Rollback to Revision 1 - Edit approval overridden by post owner or moderator
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| Aug 7, 2018 at 9:11 | history | suggested | Erel Segal-Halevi | CC BY-SA 4.0 |
Suggesting an alternative wording of the same proof
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| Aug 7, 2018 at 7:32 | review | Suggested edits | |||
| S Aug 7, 2018 at 11:00 | |||||
| Aug 7, 2018 at 4:45 | comment | added | Erel Segal-Halevi | Cool proof. I wonder if this can be extended to other values, e.g, $G(4,2)$. | |
| Aug 6, 2018 at 23:25 | comment | added | JKreft | Yes, I was thinking player A player B. | |
| Aug 6, 2018 at 22:51 | comment | added | Gerry Myerson | I take it that A and B are your nicknames for Green and Red, respectively. | |
| Aug 6, 2018 at 15:46 | comment | added | JKreft | In fact, following this reasoning, I believe I can posit that $G(2k,i) \geq \frac{1}{2}$ if $i\leq k+1$. | |
| Aug 6, 2018 at 15:41 | history | answered | JKreft | CC BY-SA 4.0 |