Timeline for In the two-person Killing the Hydra game, what is the winning strategy?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| May 11, 2020 at 10:41 | history | became hot network question | |||
| May 11, 2020 at 8:50 | answer | added | Pedro Juan Soto | timeline score: 26 | |
| May 11, 2020 at 1:39 | comment | added | none | Is this necessarily a simple problem? Given an arbitrary non-drawn chess position, either White has a winning strategy from that position or Black does, but figuring out which one has it can be extremely difficult, and analysis must be specific to that position. These hydras can be much more complex than chess positions. You may have to just calculate it to the end. In fact maybe there could be a complexity result lurking in this question. I mean it takes stronger axioms than PA to prove that the game is guaranteed to end at all... | |
| May 10, 2020 at 22:44 | comment | added | Gro-Tsen | I suggest the following variant, which avoids making arbitrary assumptions about how the hydra grows heads, and which may be more amenable to an explicit description: first Alice plays as Hercules, then Bob plays as the Hydra (i.e., chooses the number of copies made), then Bob plays as Hercules, then Alice plays as the Hydra, and so on. | |
| May 10, 2020 at 21:13 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| May 10, 2020 at 21:01 | history | edited | Joel David Hamkins | CC BY-SA 4.0 |
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| May 10, 2020 at 21:01 | comment | added | omer | Maybe the analysis of Hackenbush game might be useful, and equivalently there a might be a natural way of analysing this game as version of Nim that grows more heaps? | |
| May 10, 2020 at 20:58 | comment | added | Joel David Hamkins | @NoahSchweber I'd love to hear more, if you can prove this. Meanwhile, I've described what I view as the standard Hydra rule set, and I don't know who wins. | |
| May 10, 2020 at 20:57 | comment | added | JoshuaZ | This is a wonderful question and I regret that I can only give it one upvote. | |
| May 10, 2020 at 20:53 | comment | added | Noah Schweber | I suspect that there's a fairly natural ruleset such that every clopen game of rank $<\epsilon_0$ can be represented as a two-person hydra game, possibly up to some fairly benign transformations; if so, for each ${\bf d}\le_T{\bf 0^{(\epsilon_0)}}$ there will be a game of that type all of whose winning strategies compute ${\bf d}$ (and since every clopen game of rank $<\epsilon_0$ has a winning strategy computable in ${\bf 0^{(\epsilon_0)}}$ that's optimal). | |
| May 10, 2020 at 20:40 | history | asked | Joel David Hamkins | CC BY-SA 4.0 |