Timeline for Checkmate in $\omega$ moves?
Current License: CC BY-SA 4.0
33 events
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| Jan 31 at 19:43 | comment | added | Akiva Weinberger | This has now been made into a video: youtu.be/b-Bb_TyhC1A | |
| Jan 30 at 20:29 | comment | added | Steven Stadnicki | This will (hopefully!) get buried in the much more directly pertinent comments, but I just wanted to add a tiny note that I really like the gender edits here. | |
| Jan 30 at 19:51 | history | edited | Johan Wästlund | CC BY-SA 4.0 |
added 6 characters in body
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| Jan 30 at 19:42 | history | edited | Johan Wästlund | CC BY-SA 4.0 |
added 6 characters in body
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| Jul 2, 2023 at 15:25 | answer | added | Andreas Tsevas | timeline score: 12 | |
| Jun 18, 2023 at 20:52 | comment | added | Wlod AA | What's the def. of the infinite chess board $\mathbb Z^2$ game?! | |
| Feb 5, 2023 at 2:45 | answer | added | Matthew Bolan | timeline score: 21 | |
| May 6, 2018 at 0:03 | history | edited | Joel David Hamkins |
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Nov 9, 2015 at 3:10 | comment | added | Toby Bartels | I'm claiming that it's an easy mistake to make, and I could understand the motivation of somebody who wanted to adopt this reversal deliberately (although in that case of course they should say so). As I said, the standard convention has good reasons, and I also would stick to it, but I can understand why somebody would switch them. | |
| Oct 13, 2015 at 22:13 | comment | added | Joel David Hamkins | @TobyBartels Are you excusing the notation of $2\omega$ for $\omega+\omega$? Although it may seem natural, and beginners with ordinals often want to do that, nevertheless there is a well-established notation here going back more than a century, by which $2\omega=\omega$ and $\omega\cdot 2=\omega+\omega$, and I find no sound reason not to use the established norms. Many of the posts on this thread, it seems to me, are using incorrect ordinal notation. | |
| Feb 16, 2014 at 18:22 | answer | added | Itai Bar-Natan | timeline score: 6 | |
| Feb 19, 2013 at 7:52 | comment | added | Toby Bartels | $2x$ in ordinary algebra is read aloud as ‘$2$ times $x$’, or equivalently twice $x$, meaning literally $x + x$. The reasons for writing ordinal multiplication the other way are good reasons, but it is still backwards from the usual convention, so it's quite natural to (accidentally or on purpose) write $2\omega$ for $\omega + \omega$. | |
| Feb 19, 2013 at 1:52 | answer | added | Joel David Hamkins | timeline score: 52 | |
| Jan 5, 2012 at 10:49 | answer | added | Philip Engel | timeline score: 6 | |
| May 4, 2011 at 12:57 | comment | added | Gerald Edgar | It is interesting that you write $2\omega+3$ and not $\omega 2 + 3$. | |
| May 3, 2011 at 20:32 | comment | added | Noam D. Elkies | Well I don't have $\omega^2$ yet. Doing that with the new family of mates-in-$N\omega$ would require a setup where Black can start by pulling the White Knight arbitrarily many moves away from the Black King. | |
| May 3, 2011 at 19:49 | answer | added | Noam D. Elkies | timeline score: 31 | |
| May 3, 2011 at 19:14 | comment | added | Johan Wästlund | $\omega^2$, wow! I'm looking forward to teaching ordinal arithmetic to the guys at the chess club! | |
| May 3, 2011 at 18:39 | comment | added | Noam D. Elkies | I can now get $N\omega$ on ${\bf Z}^2$ using only $O(1)$ pieces. Details coming below. So $\omega^2$ might be possible. | |
| May 3, 2011 at 14:53 | comment | added | Junkie | It seems to me that the current state of knowing sits: $N\omega$ is possible on ${1\over 4},{1\over 2}$-$\infty$ boards (Elkies), and $\omega$ is possible on the original all-infinite board ($Z^2$). But the $N\omega$ constructions start using $N$ pieces, and one suspicion is that $\omega^2$ is already not possible with finitely many pieces? | |
| May 3, 2011 at 3:25 | answer | added | Noam D. Elkies | timeline score: 10 | |
| May 1, 2011 at 21:49 | answer | added | Noam D. Elkies | timeline score: 48 | |
| May 1, 2011 at 20:57 | answer | added | Junkie | timeline score: 13 | |
| May 1, 2011 at 17:10 | vote | accept | Johan Wästlund | ||
| May 1, 2011 at 17:10 | vote | accept | Johan Wästlund | ||
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| Apr 30, 2011 at 11:33 | answer | added | Junkie | timeline score: 62 | |
| Apr 30, 2011 at 7:10 | answer | added | Jose Capco | timeline score: 0 | |
| Apr 29, 2011 at 18:36 | answer | added | Andreas Blass | timeline score: 12 | |
| Apr 29, 2011 at 18:34 | history | edited | Qiaochu Yuan | CC BY-SA 3.0 |
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| Apr 29, 2011 at 18:09 | answer | added | Richard Borcherds | timeline score: 3 | |
| Apr 29, 2011 at 17:54 | comment | added | Henry Towsner | If $\gamma$ could be large, this would lead to the possibility of positions where proving that White could force a mate had high consistency strength. I'd love to see that problem in the newspaper chess column: "Show that White can mate, using the existence of a measurable cardinal..." | |
| Apr 29, 2011 at 15:13 | history | asked | Johan Wästlund | CC BY-SA 3.0 |