Timeline for Who wins infinite Hex?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jul 6, 2023 at 19:50 | comment | added | Daniel Asimov | I am not sure at this point what my claim is, exactly, though we were looking at situations where a winner is determined after all hexagons in the tessellation are assigned one color or the other, and I think we excluded situations where a winner could be determined after only finitely many turns. | |
| Jul 6, 2023 at 16:55 | comment | added | Joel David Hamkins | @DanielAsimov Could you clarify exactly what is your claim? Why not the game where whoever plays in cell X wins? Or fix some finite subboard, and although the game takes place on the whole infinite board, declare the winner to be whoever wins that subboard. Aren't these games symmetrical and complete, etc.? If not, I'm not sure what your claim is exactly. Meanwhile, let me add that Davide and I do discuss various alternative winning conditions in our paper. For us, some of the important criteria were translation invariance, no both-players-win situation, and others. Take a look. | |
| Jul 4, 2023 at 22:33 | comment | added | Daniel Asimov | Oops, in the comment above I intended "any countable ordinal" (not "finite"). | |
| Jul 4, 2023 at 21:53 | comment | added | Joel David Hamkins | @DanielAsimov A big section in my paper with Davide is devoted to showing how various natural interpretations of the winning conditions admit various drawn outcomes. The no-infinite-game-values theorem we proved seems related to the claim you make. But meanwhile, we remain unsure of the complexity of the winning condition that we find most natural. Is it even a Borel game? If not, then we don't even know in general whether every position in infinite Hex has a winning strategy for one player or draw-or-better strategies for both. | |
| Jul 4, 2023 at 21:47 | comment | added | Daniel Asimov | Almost exactly 20 years ago I tried very hard, with algebraist George Bergman, to find a definition of Hex on the hexagonal tessellation of the plane that satisfied the standard Hex theorem (If all cells are filled, then exactly one player must win) with rules symmetrical with respect to the two players (except that one player must make the first move), and where any finite ordinal could index the moves. We concluded that there was no such game, to my great disappointment, with one sole exception: If the winner is determined by whoever plays on the very last unfilled cell. | |
| Jul 4, 2023 at 21:13 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
added 243 characters in body
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| Jul 4, 2023 at 20:31 | history | answered | Joel David Hamkins | CC BY-SA 4.0 |