Timeline for A winning move for the first player in $3 \times 3 \times \omega$ Ordinal Chomp
Current License: CC BY-SA 3.0
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| Jul 24, 2022 at 1:45 | comment | added | Thomas | I'm not sure I understand your question. 3x$(w+n)$ chomp is a first player win by the strategy stealing argument. I can even give you a computable algorithm to find the move. Things simplify dramatically when you go into infinite ordinals. 3 x $w^w$ is the position that is a second player win. | |
| Jun 20, 2022 at 14:26 | comment | added | Noah Schweber | @StevenStadnicki Is it even known (for example) that the set $$\{n: 3\times(\omega+n)-\mathsf{CHOMP}\mbox{ is a P1-win}\}$$ is computable? (I need the "$\omega+$" bit to avoid a trivial positive answer, since if the board is finite we can brute-force-check everything.) In general, is anything known about the computability theory of CHOMP determinacy on infinite boards? | |
| Dec 16, 2015 at 3:16 | comment | added | Thomas | I'm hoping that the complexity of the analysis will be lessened by the fact that I am only going up to $3 \times 3 \times \omega$, rather than trying to find which of $3 \times 3 \times \alpha$ is a P-position. | |
| Dec 15, 2015 at 16:10 | comment | added | Steven Stadnicki | I studied $3\times n$ myself for a while and I agree, it's an immensely complicated thing - even the question of whether the winning move is a height-1 or height-2 bite appears to be almost structureless (though see, e.g., emis.de/journals/INTEGERS/papers/fg7/fg7.pdf). Given the complexities in the analysis of $2\times 2\times\alpha$ chomp that showed that the $P$-position is (IIRC) $\alpha=\omega^3$, and that $3\times3\times\omega$ subsumes both of these cases, your problem is likely to be Hard. | |
| Dec 15, 2015 at 13:34 | comment | added | Thomas | Yes, the winning move is to $2 \times \omega$. $3 \times 3 \times \omega$ is the first unsolved infinite position. Also, if it happens to be a losing position, then (0, 3, 0) is an answer to (3, 0, 0) in $\omega \times \omega \times \omega$ chomp, which this is a special case of (and I am also trying to solve). | |
| Dec 15, 2015 at 12:51 | comment | added | Tony Huynh | Do you know what happens for $3 \times \omega$ Ordinal Chomp? | |
| Dec 15, 2015 at 9:29 | comment | added | Thomas | Unfortunately, the finite case also grows complicated very fast. The strategy stealing argument works for $3 \times 3 \times n$, so there is a winning move. Having already studied $3 \times n$ chomp, I can say that there probably isn't a nice general form for the winning moves in $3 \times 3 \times n$. One thing I was able to show in my analysis is that taking the "top" square in $3 \times n$ chomp is never a winning move, unless n = 2. I'll give the proof if anyone is interested. | |
| Dec 15, 2015 at 8:15 | comment | added | Steven Stadnicki | Have you studied $3\times3\times n$ boards for, say, $n\lt 10$ or so? That's where I'd start experimentally. | |
| S Dec 15, 2015 at 7:47 | history | suggested | user642796 | CC BY-SA 3.0 |
Improved title; including some information about the game; MathJaxified
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| Dec 15, 2015 at 7:40 | review | Suggested edits | |||
| S Dec 15, 2015 at 7:47 | |||||
| Dec 15, 2015 at 5:04 | history | asked | Thomas | CC BY-SA 3.0 |