Timeline for Can a group be a universal Turing machine?
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
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| Mar 15, 2015 at 17:02 | comment | added | Joel David Hamkins | @Ibrahim Given the input $a_p^i$, we can check if $p$ halts in $2i$ steps or fewer or not. | |
| Mar 15, 2015 at 16:39 | comment | added | Ibrahim Tencer | How can you compute the inverse of $a_p^{k/2}$? It should be equal to itself if you quotiented by $a_p^k$, but this requires knowing that it's finite order. | |
| Feb 14, 2012 at 5:58 | vote | accept | Terry Tao | ||
| Feb 14, 2012 at 0:51 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 1085 characters in body
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| Feb 13, 2012 at 22:37 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 248 characters in body
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| Feb 13, 2012 at 22:30 | comment | added | Joel David Hamkins | BS, that's precisely what I was hoping would be possible, but I think I'll have to rely on the professional group theorists for that... | |
| Feb 13, 2012 at 22:26 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Updated with better, direct construction; added 75 characters in body
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| Feb 13, 2012 at 22:25 | comment | added | Benjamin Steinberg | Looks good. Now can one not use some Higman-style embedding theorem to put your group into a fg or fp one? | |
| Feb 13, 2012 at 21:36 | comment | added | Joel David Hamkins | I modified the answer to address the two-dimensional version of the question. | |
| Feb 13, 2012 at 21:35 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
added 1209 characters in body; added 81 characters in body
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| Feb 13, 2012 at 21:20 | comment | added | Joel David Hamkins | BS, oh, you're right! It's only free abelian. If two generators are really desired, then perhaps the idea can be fixed by adding in another factor, or another factor on each coordinate? | |
| Feb 13, 2012 at 21:17 | history | edited | Joel David Hamkins | CC BY-SA 3.0 |
Fixed missing negation
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| Feb 13, 2012 at 21:16 | comment | added | Benjamin Steinberg | +1 anyway since this is nice. | |
| Feb 13, 2012 at 21:15 | comment | added | Benjamin Steinberg | Since G is abelian you never get a free group on 2 generators so Terry's specific problem is not 'officially solved' by this group. | |
| Feb 13, 2012 at 21:03 | history | answered | Joel David Hamkins | CC BY-SA 3.0 |