Timeline for Yet another Erdős–Szekeres game
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 10, 2015 at 7:25 | vote | accept | Fedor Petrov | ||
| Mar 10, 2015 at 3:43 | answer | added | Michael Albert | timeline score: 8 | |
| Mar 9, 2015 at 23:32 | comment | added | Fedor Petrov | @GerryMyerson wow, they claim that in this game the first player wins for any $n\geq 4$ by some clever reason! researchgate.net/publication/2126205_Monotonic_Sequence_Games (Theorem 10) | |
| Mar 9, 2015 at 22:43 | comment | added | Gerry Myerson | You may be interested in Harary, Sagan, and West, Computer-aided analysis of monotonic sequence games, Atti Accad. Peloritana Pericoanti Cl. Sci. Fis. Mat. Natur. 61 (1983) 67-78, MR1006269 (90d:90100). | |
| Mar 9, 2015 at 16:17 | comment | added | Fedor Petrov | Well, let the exact question be "does number of turns grow as $\Omega(n^2)$ for large $n$"? | |
| Mar 9, 2015 at 16:11 | comment | added | Daniel Soltész | In a sense this is also a similar question: mathoverflow.net/questions/143547/a-ramsey-avoidance-game As there is a game, and a Ramsey-type result, that guarantees that the game ends in a finite number of steps, but it turned out that the game with optimal play "ends earlier". | |
| Mar 9, 2015 at 15:59 | comment | added | Daniel Soltész | For $n=3$, it is a first player win in four steps instead of five, so it is already better than Erdős-Szekeres. WLOG we can assume that the first palyer plays $1$, and the second player plays $3$. then the first player plays $2$ and wins. I am not sure what is the exact question. You would like a better bound than the Erdős-Szekeres, for all $n$ for optimal players? | |
| Mar 9, 2015 at 13:38 | history | asked | Fedor Petrov | CC BY-SA 3.0 |