Timeline for Conjugate vertices and distinguishing properties
Current License: CC BY-SA 2.5
29 events
| when toggle format | what | by | license | comment | |
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| Apr 14, 2010 at 11:49 | history | edited | Hans-Peter Stricker | CC BY-SA 2.5 |
added 37 characters in body
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| Jan 6, 2010 at 22:42 | history | edited | Michael Lugo |
retag
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| Jan 6, 2010 at 15:00 | comment | added | Hans-Peter Stricker | I don't agree: the original question was how to find a distinguishing family of properties which by definition is in 1:1 correspondence with the orbits of $Aut(G)$. I only wanted to make this clearer. | |
| Jan 6, 2010 at 14:19 | comment | added | Mariano Suárez-Álvarez | It is not big deal to edit the question, but it becomes slightly inelegant rather fast to change what is actually being asked in question! The question has now become "how can one find the orbits of $Aut(G)$ efficiently?", which is quite unrelated to what was being ask originally... | |
| Jan 6, 2010 at 8:58 | history | edited | Hans-Peter Stricker | CC BY-SA 2.5 |
added motivation
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| Jan 5, 2010 at 23:38 | comment | added | Reid Barton | Nope. It's not really a big deal. | |
| Jan 5, 2010 at 22:00 | history | edited | Hans-Peter Stricker | CC BY-SA 2.5 |
further improved Definition 2
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| Jan 5, 2010 at 21:45 | comment | added | Hans-Peter Stricker | So it was my "fault". Can I roll this back? | |
| Jan 5, 2010 at 21:41 | comment | added | Reid Barton | Posts that are edited many times automatically become CW. (I don't know why.) | |
| Jan 5, 2010 at 21:33 | history | edited | Hans-Peter Stricker | CC BY-SA 2.5 |
improved definition 2
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| Jan 5, 2010 at 21:26 | comment | added | Hans-Peter Stricker | Now I found out: Yes, why is this CW? Was it me by accident? | |
| Jan 5, 2010 at 20:27 | comment | added | Hans-Peter Stricker | Sorry? (What is CW?) | |
| Jan 5, 2010 at 20:24 | history | edited | Hans-Peter Stricker | CC BY-SA 2.5 |
improved formatting
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| Jan 5, 2010 at 20:22 | comment | added | Mariano Suárez-Álvarez | I wonder why this is CW... | |
| Jan 5, 2010 at 20:21 | answer | added | Mariano Suárez-Álvarez | timeline score: 2 | |
| Jan 5, 2010 at 20:17 | history | edited | Hans-Peter Stricker |
added tag "names"
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| Jan 5, 2010 at 20:00 | history | edited | Hans-Peter Stricker | CC BY-SA 2.5 |
added 12 characters in body; Post Made Community Wiki
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| Jan 5, 2010 at 18:03 | history | edited | Hans-Peter Stricker | CC BY-SA 2.5 |
more examples
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| Jan 5, 2010 at 17:10 | history | edited | Hans-Peter Stricker | CC BY-SA 2.5 |
added 160 characters in body
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| Jan 5, 2010 at 17:05 | comment | added | Hans-Peter Stricker | I will provide some soon! | |
| Jan 5, 2010 at 16:55 | comment | added | Mariano Suárez-Álvarez | Is there a non trivial example? | |
| Jan 5, 2010 at 15:12 | comment | added | Hans-Peter Stricker | In the corrected version I hope to have said so. | |
| Jan 5, 2010 at 15:09 | history | edited | Hans-Peter Stricker | CC BY-SA 2.5 |
added 10 characters in body; added 2 characters in body
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| Jan 5, 2010 at 14:52 | history | edited | Hans-Peter Stricker | CC BY-SA 2.5 |
corrected definition 2; deleted 9 characters in body
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| Jan 5, 2010 at 14:27 | comment | added | Pete L. Clark | The property "x has exactly one neighbor" does not satisfy the defining condition for all graphs: just because two vertices have exactly one neighbor does not mean they are conjugate. If you want $\phi$ to have this property only with respect to a fixed graph $G$, you need to say so. | |
| Jan 5, 2010 at 14:22 | history | edited | Hans-Peter Stricker | CC BY-SA 2.5 |
rolled back to un-pleasant changes
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| Jan 5, 2010 at 14:10 | history | edited | Hans-Peter Stricker | CC BY-SA 2.5 |
added example
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| Jan 5, 2010 at 13:51 | history | edited | Hans-Peter Stricker | CC BY-SA 2.5 |
added 32 characters in body
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| Jan 5, 2010 at 13:42 | history | asked | Hans-Peter Stricker | CC BY-SA 2.5 |